The Practice of Statistics

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Using The Practice of Statistics, Fifth Edition, for Advanced Placement (AP®) Statistics (The percents in parentheses reflect coverage on the AP® exam.)

Topic Outline for AP® Statistics from the College Board’s AP ® Statistics Course Description

The Practice of Statistics, 5th ed. Chapter and Section references

I. Exploring data: describing patterns and departures from patterns (20%–30%) A. Constructing and interpreting graphical displays of distributions of univariate data (dotplot, stemplot, histogram, cumulative frequency plot) 1. Center and spread 2. Clusters and gaps 3. Outliers and unusual features 4. Shape B. Summarizing distributions of univariate data 1. Measuring center: median, mean 2. Measuring spread: range, interquartile range, standard deviation 3. Measuring position: quartiles, percentiles, standardized scores (z-scores) 4. Using boxplots 5. The effect of changing units on summary measures C. Comparing distributions of univariate data (dotplots, back-to-back stemplots, parallel boxplots) 1. Comparing center and spread 2. Comparing clusters and gaps 3. Comparing outliers and unusual features 4. Comparing shape D. Exploring bivariate data 1. Analyzing patterns in scatterplots 2. Correlation and linearity 3. Least-squares regression line 4. Residual plots, outliers, and influential points 5. Transformations to achieve linearity: logarithmic and power transformations E. Exploring categorical data 1. Frequency tables and bar charts 2. Marginal and joint frequencies for two-way tables 3. Conditional relative frequencies and association 4. Comparing distributions using bar charts

Dotplot, stemplot, histogram 1.2; Cumulative frequency plot 2.1 1.2 1.2 1.2 1.2 1.3 and 2.1 1.3 1.3 Quartiles 1.3; percentiles and z-scores 2.1 1.3 2.1 Dotplots and stemplots 1.2; boxplots 1.3 1.2 and 1.3 1.2 and 1.3 1.2 and 1.3 1.2 and 1.3 Chapter 3 and Section 12.2 3.1 3.1 3.2 3.2 12.2 Sections 1.1, 5.2, 5.3 1.1 (we call them bar graphs) Marginal 1.1; joint 5.2 1.1 and 5.3 1.1

II. Sampling and experimentation: planning and conducting a study (10%–15%) A. Overview of methods of data collection 1. Census 2. Sample survey 3. Experiment 4. Observational study B. Planning and conducting surveys 1. Characteristics of a well-designed and well-conducted survey 2. Populations, samples, and random selection 3. Sources of bias in sampling and surveys 4. Sampling methods, including simple random sampling, stratified random sampling, and cluster sampling C. Planning and conducting experiments 1. Characteristics of a well-designed and well-conducted experiment 2. Treatments, control groups, experimental units, random assignments, and replication 3. Sources of bias and confounding, including placebo effect and blinding 4. Completely randomized design 5. Randomized block design, including matched pairs design D. Generalizability of results and types of conclusions that can be drawn from ­ observational studies, experiments, and surveys

Sections 4.1 and 4.2 4.1 4.1 4.2 4.2 Section 4.1 4.1 4.1 4.1 4.1 Section 4.2 4.2 4.2 4.2 4.2 4.2 Section 4.3

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Using The Practice of Statistics, Fifth Edition, for Advanced Placement (AP®) Statistics (The percents in parentheses reflect coverage on the AP® exam.)

Topic Outline for AP® Statistics from the College Board’s AP ® Statistics Course Description

The Practice of Statistics, 5th ed. Chapter and Section references

III. Anticipating patterns: exploring random phenomena using probability and simulation (20%–30%) A. Probability 1. Interpreting probability, including long-run relative frequency interpretation 2. “Law of large numbers” concept 3. Addition rule, multiplication rule, conditional probability, and independence 4. Discrete random variables and their probability distributions, including binomial and geometric 5. Simulation of random behavior and probability distributions 6. Mean (expected value) and standard deviation of a random variable, and linear transformation of a random variable B. Combining independent random variables 1. Notion of independence versus dependence 2. Mean and standard deviation for sums and differences of independent random variables C. The Normal distribution 1. Properties of the Normal distribution 2. Using tables of the Normal distribution 3. The Normal distribution as a model for measurements D. Sampling distributions 1. Sampling distribution of a sample proportion 2. Sampling distribution of a sample mean 3. Central limit theorem 4. Sampling distribution of a difference between two independent sample proportions 5. Sampling distribution of a difference between two independent sample means 6. Simulation of sampling distributions 7. t distribution 8. Chi-square distribution

Chapters 5 and 6 5.1 5.1 Addition rule 5.2; other three topics 5.3 Discrete 6.1; Binomial and geometric 6.3 5.1 Mean and standard deviation 6.1; Linear transformation 6.2 Section 6.2 6.2 6.2 Section 2.2 2.2 2.2 2.2 Chapter 7; Sections 8.3, 10.1, 10.2, 11.1 7.2 7.3 7.3 10.1 10.2 7.1 8.3 11.1

IV. Statistical inference: estimating population parameters and testing hypotheses (30%–40%) A. Estimation (point estimators and confidence intervals) 1. Estimating population parameters and margins of error 2. Properties of point estimators, including unbiasedness and variability 3. Logic of confidence intervals, meaning of confidence level and confidence intervals, and properties of confidence intervals 4. Large-sample confidence interval for a proportion 5. Large-sample confidence interval for a difference between two proportions 6. Confidence interval for a mean 7. Confidence interval for a difference between two means (unpaired and paired) 8. Confidence interval for the slope of a least-squares regression line B. Tests of significance 1. Logic of significance testing, null and alternative hypotheses; P-values; one-and two-sided tests; concepts of Type I and Type II errors; concept of power 2. Large-sample test for a proportion 3. Large-sample test for a difference between two proportions 4. Test for a mean 5. Test for a difference between two means (unpaired and paired) 6. Chi-square test for goodness of fit, homogeneity of proportions, and independence (one- and two-way tables) 7. Test for the slope of a least-squares regression line

Chapter 8 plus parts of Sections 9.3, 10.1, 10.2, 12.1 8.1 8.1 8.1 8.2 10.1 8.3 Paired 9.3; unpaired 10.2 12.1 Chapters 9 and 11 plus parts of Sections 10.1, 10.2, 12.1 9.1; power in 9.2 9.2 10.1 9.3 Paired 9.3; unpaired 10.2 Chapter 11 12.1

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For the AP® Exam

The Practice of Statistics f i f t h E D ITI O N

AP® is a trademark registered by the College Board, which was not involved in the production of, and does not endorse, this product.

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Publisher: Ann Heath Assistant Editor: Enrico Bruno Editorial Assistant: Matt Belford Development Editor: Donald Gecewicz Executive Marketing Manager: Cindi Weiss Photo Editor: Cecilia Varas Photo Researcher: Julie Tesser Art Director: Diana Blume Cover Designers: Diana Blume, Rae Grant Text Designer: Patrice Sheridan Cover Image: Joseph Devenney/Getty Images Senior Project Editor: Vivien Weiss Illustrations: Network Graphics Production Manager: Susan Wein Composition: Preparé Printing and Binding: RR Donnelley

TI-83™, TI-84™, TI-89™, and TI-Nspire screen shots are used with permission of the publisher: © 1996, Texas Instruments Incorporated. TI-83™, TI-84™, TI-89™, and TI-Nspire Graphic Calculators are registered trademarks of Texas Instruments Incorporated. Minitab is a registered trademark of Minitab, Inc. Microsoft© and Windows© are registered trademarks of the Microsoft Corporation in the United States and other countries. Fathom Dynamic Statistics is a trademark of Key Curriculum, a McGraw-Hill Education Company.

M&M’S is a registered trademark of Mars, Incorporated and its affiliates. This trademark is used with permission. Mars, Incorporated is not associated with Macmillan Higher Education. Images printed with permission of Mars, Incorporated.

Library of Congress Control Number: 2013949503 ISBN-13: 978-1-4641-0873-0 ISBN-10: 1-4641-0873-0 © 2015, 2012, 2008, 2003, 1999 by W. H. Freeman and Company First printing 2014 All rights reserved Printed in the United States of America

W. H. Freeman and Company 41 Madison Avenue New York, NY 10010 Houndmills, Basingstoke RG21 6XS, England www.whfreeman.com

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For the AP® Exam

The Practice of Statistics

f i f t h E D ITI O N

Daren S. Starnes The Lawrenceville School Josh Tabor Canyon del Oro High School Daniel S. Yates Statistics Consultant David S. Moore Purdue University

W. H. Freeman and Company/BFW New York

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Contents About the Authors  vi To the Student  xii Overview: What Is Statistics?  xxi 1 Exploring Data

5 Probability: What Are the Chances?

xxxii

Introduction: Data Analysis: Making Sense of Data  2 1.1  Analyzing Categorical Data  7 1.2  Displaying Quantitative Data with Graphs  25 1.3  Describing Quantitative Data with Numbers  48 Free Response AP® Problem, Yay!  74 Chapter 1 Review  74 Chapter 1 Review Exercises  76 Chapter 1 AP® Statistics Practice Test  78 2 Modeling Distributions of Data

Introduction  84 2.1  Describing Location in a Distribution  85 2.2  Density Curves and Normal Distributions  103 Free Response AP® Problem, Yay!  134 Chapter 2 Review  134 Chapter 2 Review Exercises  136 Chapter 2 AP® Statistics Practice Test  137 3 Describing Relationships

140

344

Introduction  346 6.1  Discrete and Continuous Random Variables  347 6.2  Transforming and Combining Random Variables  363 6.3  Binomial and Geometric Random Variables  386 Free Response AP® Problem, Yay!  414 Chapter 6 Review  415 Chapter 6 Review Exercises  416 Chapter 6 AP® Statistics Practice Test  418 7 Sampling Distributions

Introduction  142 3.1  Scatterplots and Correlation  143 3.2 Least-Squares Regression 164 Free Response AP® Problem, Yay!  199 Chapter 3 Review  200 Chapter 3 Review Exercises  202 Chapter 3 AP® Statistics Practice Test  203 4 Designing Studies

Introduction  288 5.1  Randomness, Probability, and Simulation  289 5.2 Probability Rules 305 5.3  Conditional Probability and Independence  318 Free Response AP® Problem, Yay!  338 Chapter 5 Review  338 Chapter 5 Review Exercises  340 Chapter 5 AP® Statistics Practice Test  342 6 Random Variables

82

286

420

Introduction  422 7.1  What Is a Sampling Distribution?  424 7.2  Sample Proportions  440 7.3 Sample Means 450 Free Response AP® Problem, Yay!  464 Chapter 7 Review  465 Chapter 7 Review Exercises  466 Chapter 7 AP® Statistics Practice Test  468 Cumulative AP® Practice Test 2 470

206

Introduction  208 4.1  Sampling and Surveys  209 4.2 Experiments 234 4.3  Using Studies Wisely  266 Free Response AP® Problem, Yay!  275 Chapter 4 Review  276 Chapter 4 Review Exercises  278 Chapter 4 AP® Statistics Practice Test  279 Cumulative AP® Practice Test 1 282

8 Estimating with Confidence

474

Introduction  476 8.1  Confidence Intervals: The Basics  477 8.2  Estimating a Population Proportion  492 8.3  Estimating a Population Mean  507 Free Response AP® Problem, Yay!  530 Chapter 8 Review  531 Chapter 8 Review Exercises  532 Chapter 8 AP® Statistics Practice Test  534

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Section 2.1 Scatterplots and Correlations

9 Testing a Claim

536

Introduction  538 9.1  Significance Tests: The Basics  539 9.2  Tests about a Population Proportion  554 9.3  Tests about a Population Mean  574 Free Response AP® Problem, Yay!  601 Chapter 9 Review  602 Chapter 9 Review Exercises  604 Chapter 9 AP® Statistics Practice Test  605 10 Comparing Two Populations or Groups

13 Analysis of Variance 14 Multiple Linear Regression 15 Logistic Regression Photo Credits

608

11 Inference for Distributions

676

Introduction  678 11.1  Chi-Square Tests for Goodness of Fit  680 11.2  Inference for Two-Way Tables  697 Free Response AP® Problem, Yay!  730 Chapter 11 Review  731 Chapter 11 Review Exercises  732 Chapter 11 AP® Statistics Practice Test  734 12 More about Regression

PC-1

Notes and Data Sources

Introduction  610 10.1  Comparing Two Proportions  612 10.2  Comparing Two Means  634 Free Response AP® Problem, Yay!  662 Chapter 10 Review  662 Chapter 10 Review Exercises  664 Chapter 10 AP® Statistics Practice Test  666 Cumulative AP® Practice Test 3 669

of Categorical Data

Additional topics on CD-ROM and at http://www.whfreeman.com/tps5e

N/DS-1

Solutions

S-1

Appendices

A-1



Appendix A: About the AP® Exam  A-1 Appendix B: TI-Nspire Technology Corners  A-3

Formulas for AP® Statistics Exam

F-1

Tables

T-1



Table A: Standard Normal Probabilities  T-1 Table B: t Distribution Critical Values  T-3 Table C: Chi-Square Distribution Critical Values  T-4 Table D: Random Digits  T-5 Glossary/Glosario G-1 Index I-1

Technology Corners Reference 

Back Endpaper

Inference Summary 

Back Endpaper

736

Introduction  738 12.1  Inference for Linear Regression  739 12.2  Transforming to Achieve Linearity  765 Free Response AP® Problem, Yay!  793 Chapter 12 Review  794 Chapter 12 Review Exercises  795 Chapter 12 AP® Statistics Practice Test  797 Cumulative AP® Practice Test 4 800

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About the Authors

Daren S. Starnes is Mathematics Department Chair and holds the Robert S. and Christina Seix Dow Distinguished Master Teacher Chair in Mathematics at The Lawrenceville School near Princeton, New Jersey. He earned his MA in Mathematics from the University of Michigan and his BS in Mathematics from the University of North Carolina at Charlotte. Daren is also an alumnus of the North Carolina School of Science and Mathematics. Daren has led numerous one-day and weeklong AP® Statistics institutes for new and experienced AP® teachers, and he has been a Reader, Table Leader, and Question Leader for the AP® Statistics exam since 1998. Daren is a frequent speaker at local, state, regional, national, and international conferences. For two years, he served as coeditor of the Technology Tips column in the NCTM journal The Mathematics Teacher. From 2004 to 2009, Daren served on the ASA/NCTM Joint Committee on the Curriculum in Statistics and Probability (which he chaired in 2009). While on the committee, he edited the Guidelines for Assessment and Instruction in Statistics Education (GAISE) pre-K–12 report and coauthored (with Roxy Peck) Making Sense of Statistical Studies, a capstone module in statistical thinking for high school students. Daren is also coauthor of the popular text Statistics Through Applications, First and Second Editions.

Josh Tabor has enjoyed teaching general and AP® Statistics to high school students for more than 18 years, most recently at his alma mater, Canyon del Oro High School in Oro Valley, Arizona. He received a BS in Mathematics from Biola University, in La Mirada, California. In recognition of his outstanding work as an educator, Josh was named one of the five finalists for Arizona Teacher of the Year in 2011. He is a past member of the AP® Statistics Development Committee (2005–2009), as well as an experienced Table Leader and Question Leader at the AP® Statistics Reading. Each year, Josh leads one-week AP® Summer Institutes and one-day College Board workshops around the country and frequently speaks at local, national, and international conferences. In addition to teaching and speaking, Josh has authored articles in The Mathematics Teacher, STATS Magazine, and The Journal of Statistics Education. He is the author of the Annotated Teacher’s Edition and Teacher’s Resource Materials for The Practice of Statistics 4e and 5e, along with the Solutions Manual for The Practice of Statistics 5e. Combining his love of statistics and love of sports, Josh teamed with Christine Franklin to write Statistical Reasoning in Sports, an innovative textbook for on-level statistics courses.

Daniel S. Yates taught AP® Statistics in the Electronic Classroom (a distance-learning facility) affiliated with Henrico County Public Schools in Richmond, Virginia. Prior to high school teaching, he was on the mathematics faculty at Virginia Tech and Randolph-Macon College. He has a PhD in Mathematics Education from Florida State University. Dan received a College Board/ Siemens Foundation Advanced Placement Teaching Award in 2000.

David S. Moore is Shanti S. Gupta Distinguished Professor of Statistics (Emeritus) at Purdue University and was 1998 President of the American Statistical Association. David is an elected fellow of the American Statistical Association and of the Institute of Mathematical Statistics and an elected member of the International Statistical Institute. He has served as program director for statistics and probability at the National Science Foundation. He is the author of influential articles on statistics education and of several leading textbooks.

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Section 2.1 Scatterplots and Correlations

Content Advisory Board and Supplements Team Jason Molesky, Lakeville Area Public School, Lakeville, MN

Media Coordinator—Worked Examples, PPT lectures, Strive for a 5 Guide Jason has served as an AP® Statistics Reader and Table Leader since 2006. After teaching AP® Statistics for eight years and developing the FRAPPY system for AP® exam preparation, Jason moved into administration. He now serves as the Director of Program Evaluation and Accountability, overseeing the district’s research and evaluation, continuous improvement efforts, and assessment programs. Jason also provides professional development to statistics teachers across the United States and maintains the “Stats Monkey” Web site, a clearinghouse for AP® Statistics resources.

Tim Brown, The Lawrenceville School, Lawrenceville, NJ Content Advisor, Test Bank, TRM Tests and Quizzes Tim first piloted an AP® Statistics course the year before the first exam was administered. He has been an AP® Reader since 1997 and a Table Leader since 2004. He has taught math and statistics at The Lawrenceville School since 1982 and currently holds the Bruce McClellan Distinguished Teaching Chair. Doug Tyson, Central York High School, York, PA

Exercise Videos, Learning Curve Doug has taught mathematics and statistics to high school and undergraduate students for 22 years. He has taught AP® Statistics for 7 years and served as an AP® Reader for 4 years. Doug is the co-author of a curriculum module for the College Board, conducts student review sessions around the country, and gives workshops on teaching statistics.

Paul Buckley, Gonzaga College High School, Washington, DC

Exercise Videos Paul has taught high school math for 20 years and AP® Statistics for 12 years. He has been an AP® Statistics Reader for six years and helps to coordinate the integration of new Readers (Acorns) into the Reading process. Paul has presented at Conferences for AP®, NCTM, NCEA (National Catholic Education Association) and JSEA (Jesuit Secondary Education Association).

Leigh Nataro, Moravian Academy, Bethlehem, PA

Technology Corner Videos Leigh has taught AP® Statistics for nine years and has served as an AP® Statistics Reader for the past four years. She enjoys the challenge of writing multiple-choice questions for the College Board for use on the AP® Statistics exam. Leigh is a National Board Certified Teacher in Adolescence and Young Adulthood Mathematics and was previously named a finalist for the Presidential Award for Excellence in Mathematics and Science Teaching in New Jersey.

Ann Cannon, Cornell College, Mount Vernon, IA

Content Advisor, Accuracy Checker Ann has served as Reader, Table Leader, and Question Leader for the AP® Statistics exam for the past 13 years. She has taught introductory statistics at the college level for 20 years and is very active in the Statistics Education Section of the American Statistical Association, serving on the Executive Committee for two 3-year terms. She is co-author of STAT2: Building Models for a World of Data (W. H. Freeman and Company).

Michel Legacy, Greenhill School, Dallas, TX

Content Advisor, Strive for a 5 Guide Michael is a past member of the AP® Statistics Development Committee (2001–2005) and a former Table Leader at the Reading. He currently reads the Alternate Exam and is a lead teacher at many AP® Summer Institutes. Michael is the author of the 2007 College Board AP® Statistics Teacher’s Guide and was named the Texas 2009–2010 AP® Math/Science Teacher of the Year by the Siemens Corporation.

James Bush, Waynesburg University, Waynesburg, PA Learning Curve, Media Reviewer James has taught introductory and advanced courses in Statistics for over 25 years. He is currently a Professor of Mathematics at Waynesburg University and is the recipient of the Lucas Hathaway Teaching Excellence Award. James has served as an AP® Statistics Reader for the past seven years and conducts many AP® Statistics preparation workshops. Beth Benzing, Strath Haven High School, Wallingford/ Swarthmore School District, Wallingford, PA Activities Videos Beth has taught AP® Statistics for 14 years and has served as a Reader for the AP® Statistics exam for the past four years. She serves as Vice President on the board for the regional affiliate for NCTM in the Philadelphia area and is a moderator for an on-line course, Teaching Statistics with Fathom. Heather Overstreet, Franklin County High School,

Rocky Mount, VA TI-Nspire Technology Corners Heather has taught AP® Statistics for nine years and has served as an AP® Statistics Reader for the past six years. While working with Virginia Advanced Study Strategies, a program for promoting AP® math, science, and English courses in Virginia High Schools, she led many AP® Statistics Review Sessions and served as a Laying the Foundation trainer of teachers of pre-AP® math classes.

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Acknowledgments Teamwork: It has been the secret to the successful evolution of The Practice of Statistics (TPS) through its fourth and fifth editions. We are indebted to each and every member of the team for the time, energy, and passion that they have invested in making our collective vision for TPS a reality. To our team captain, Ann Heath, we offer our utmost gratitude. Managing a revision project of this scope is a Herculean task! Ann has a knack for troubleshooting thorny issues with an uncanny blend of forthrightness and finesse. She assembled an all-star cast to collaborate on the fifth edition of TPS and trusted each of us to deliver an excellent finished product. We hope you’ll agree that the results speak for themselves. Thank you, Ann, for your unwavering support, patience, and friendship throughout the production of these past two editions. Truth be told, we had some initial reservations about adding a Development Editor to our team starting with the fourth edition of TPS. Don Gecewicz quickly erased our doubts. His keen mind and sharp eye were evident in the many probing questions and insightful suggestions he offered at all stages of the project. Thanks, Don, for your willingness to push the boundaries of our thinking. Working behind the scenes, Enrico Bruno busily prepared the manuscript chapters for production. He did yeoman’s work in ensuring that the changes we intended were made as planned. For the countless hours that Enrico invested sweating the small stuff, we offer him our sincere appreciation. We are deeply grateful to Patrice Sheridan and to Diana Blume for their aesthetic contributions to the eye-catching design of TPS 5e. Our heartfelt thanks also go to Vivien Weiss and Susan Wein at W. H. Freeman for their skillful oversight of the production process. Patti Brecht did a superb job copyediting a very complex manuscript. A special thank you goes to our good friends on the high school sales and marketing staff at Bedford, Freeman, and Worth (BFW) Publishers. We feel blessed to have such enthusiastic professionals on our extended team. In particular, we want to thank our chief cheerleader, Cindi Weiss, for her willingness to promote The Practice of Statistics at every opportunity. We cannot say enough about the members of our Content Advisory Board and Supplements Team. This remarkable group is a veritable who’s who of the AP® Statistics community. We’ll start with Ann Cannon, who once again reviewed the statistical content of every chapter. Ann also checked the solutions to every exercise in the book. What a task! More than that, Ann offered us sage advice about virtually everything between the covers of TPS 5e. We are so grateful to Ann for all that she has done to enhance the quality of The Practice of Statistics over these past two editions. Jason Molesky, aka “Stats Monkey,” greatly expanded his involvement in the fifth edition by becoming our Media Coordinator in addition to his ongoing roles as author of the Strive for a 5 Guide and creator of the PowerPoint presentations for the book. Jason also graciously loaned us his Free Response AP® Problem, Yay! (FRAPPY!) concept for use in TPS 5e. We feel incredibly fortunate to have such a creative, energetic, and deeply thoughtful person at the helm as the media side of our project explodes in many new directions at once. Jason is surrounded by a talented media team. Doug Tyson and Paul Buckley have expertly recorded the worked exercise videos. Leigh Nataro has produced screencasts for the Technology Corners on the TI-83/84, TI-89, and TI-Nspire. Beth Benzing followed through on her creative suggestion to produce “how to” screencasts for teachers to accompany the Activities in the book. James Bush capably served as our expert reviewer for all of the media elements and is now partnering with Doug Tyson on a new Learning Curve media component. Heather Overstreet once

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Section 2.1 Scatterplots and Correlations

again compiled the TI-Nspire Technology Corners in Appendix B. We wish to thank the entire media team for their many contributions. Tim Brown, my Lawrenceville colleague, has been busy creating many new, high-quality assessment items for the fifth edition. The fruits of his labors are contained in the revised Test Bank and in the quizzes and tests that are part of the Teacher’s Resource Materials. We are especially thankful that Tim was willing to compose an additional cumulative test for Chapters 2 through 12. We offer our continued thanks to Michael Legacy for composing the superb questions in the Cumulative AP® Practice Tests and the Strive for a 5 Guide. Michael’s expertise as a twotime former member of the AP® Statistics Test Development Committee is invaluable to us. Although Dan Yates and David Moore both retired several years ago, their influence lives on in TPS 5e. They both had a dramatic impact on our thinking about how statistics is best taught and learned through their pioneering work as textbook authors. Without their early efforts, The Practice of Statistics would not exist! Thanks to all of you who reviewed chapters of the fourth and fifth editions of TPS and offered your constructive suggestions. The book is better as a result of your input. —Daren Starnes and Josh Tabor A final note from Daren: It has been a privilege for me to work so closely with Josh Tabor on TPS 5e. He is a gifted statistics educator; a successful author in his own right; and a caring parent, colleague, and friend. Josh’s influence is present on virtually every page of the book. He also took on the thankless task of revising all of the solutions for the fifth edition in addition to updating his exceptional Annotated Teacher’s Edition. I don’t know how he finds the time and energy to do all that he does! The most vital member of the TPS 5e team for me is my wonderful wife, Judy. She has read page proofs, typed in data sets, and endured countless statistical conversations in the car, in airports, on planes, in restaurants, and on our frequent strolls. If writing is a labor of love, then I am truly blessed to share my labor with the person I love more than anyone else in the world. Judy, thank you so much for making the seemingly impossible become reality time and time again. And to our three sons, Simon, Nick, and Ben—thanks for the inspiration, love, and support that you provide even in the toughest of times. A final note from Josh: When Daren asked me to join the TPS team for the fourth edition, I didn’t know what I was getting into. Having now completed another full cycle with TPS 5e, I couldn’t have imagined the challenge of producing a textbook—from the initial brainstorming sessions to the final edits on a wide array of supplementary materials. In all honesty, I still don’t know the full story. For taking on all sorts of additional tasks—managing the word-by-word revisions to the text, reviewing copyedits and page proofs, encouraging his co-author, and overseeing just about everything—I owe my sincere thanks to Daren. He has been a great colleague and friend throughout this process. I especially want to thank the two most important people in my life. To my wife Anne, your patience while I spent countless hours working on this project is greatly appreciated. I couldn’t have survived without your consistent support and encouragement. To my daughter Jordan, I look forward to being home more often and spending less time on my computer when I am there. I also look forward to when you get to use TPS 7e in about 10 years. For now, we have a lot of fun and games to catch up on. I love you both very much.

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Acknowledgments Fifth Edition Survey Participants and Reviewers Blake Abbott, Bishop Kelley High School, Tulsa, OK Maureen Bailey, Millcreek Township School District, Erie, PA Kevin Bandura, Lincoln County High School, Stanford, KY Elissa Belli, Highland High School, Highland, IN Jeffrey Betlan, Yough School District, Herminie, PA Nancy Cantrell, Macon County Schools, Franklin, NC Julie Coyne, Center Grove HS, Greenwood, IN Mary Cuba, Linden Hall, Lititz, PA Tina Fox, Porter-Gaud School, Charleston, SC Ann Hankinson, Pine View, Osprey, FL Bill Harrington, State College Area School District, State College, PA Ronald Hinton, Pendleton Heights High School, Pendleton, IN Kara Immonen, Norton High School, Norton, MA Linda Jayne, Kent Island High School, Stevensville, MD Earl Johnson, Chicago Public Schools, Chicago, IL Christine Kashiwabara, Mid-Pacific Institute, Honolulu, HI Melissa Kennedy, Holy Names Academy, Seattle, WA Casey Koopmans, Bridgman Public Schools, Bridgman, MI David Lee, SPHS, Sun Prairie, WI Carolyn Leggert, Hanford High School, Richland, WA Jeri Madrigal, Ontario High School, Ontario, CA Tom Marshall, Kents Hill School, Kents Hill, ME Allen Martin, Loyola High School, Los Angeles, CA Andre Mathurin, Bellarmine College Preparatory, San Jose, CA Brett Mertens, Crean Lutheran High School, Irvine, CA Sara Moneypenny, East High School, Denver, CO Mary Mortlock, The Harker School, San Jose, CA Mary Ann Moyer, Hollidaysburg Area School District, Hollidaysburg, PA Howie Nelson, Vista Murrieta HS, Murrieta, CA Shawnee Patry, Goddard High School, Wichita, KS Sue Pedrick, University High School, Hartford, CT Shannon Pridgeon, The Overlake School, Redmond, WA Sean Rivera, Folsom High, Folsom, CA Alyssa Rodriguez, Munster High School, Munster, IN Sheryl Rodwin, West Broward High School, Pembroke Pines, FL Sandra Rojas, Americas HS, El Paso, TX Christine Schneider, Columbia Independent School, Boonville, MO Amanda Schneider, Battle Creek Public Schools, Charlotte, MI Steve Schramm, West Linn High School, West Linn, OR Katie Sinnott, Revere High School, Revere, MA

Amanda Spina, Valor Christian High School, Highlands Ranch, CO Julie Venne, Pine Crest School, Fort Lauderdale, FL Dana Wells, Sarasota High School, Sarasota, FL Luke Wilcox, East Kentwood High School, Grand Rapids, MI Thomas Young, Woodstock Academy, Putnam, CT

Fourth Edition Focus Group Participants and Reviewers Gloria Barrett, Virginia Advanced Study Strategies, Richmond, VA David Bernklau, Long Island University, Brookville, NY Patricia Busso, Shrewsbury High School, Shrewsbury, MA Lynn Church, Caldwell Academy, Greensboro, NC Steven Dafilou, Springside High School, Philadelphia, PA Sandra Daire, Felix Varela High School, Miami, FL Roger Day, Pontiac High School, Pontiac, IL Jared Derksen, Rancho Cucamonga High School, Rancho Cucamonga, CA Michael Drozin, Munroe Falls High School, Stow, OH Therese Ferrell, I. H. Kempner High School, Sugar Land, TX Sharon Friedman, Newport High School, Bellevue, WA Jennifer Gregor, Central High School, Omaha, NE Julia Guggenheimer, Greenwich Academy, Greenwich, CT Dorinda Hewitt, Diamond Bar High School, Diamond Bar, CA Dorothy Klausner, Bronx High School of Science, Bronx, NY Robert Lochel, Hatboro-Horsham High School, Horsham, PA Lynn Luton, Duchesne Academy of the Sacred Heart, Houston, TX Jim Mariani, Woodland Hills High School, Greensburgh, PA Stephen Miller, Winchester Thurston High School, Pittsburgh, PA Jason Molesky, Lakeville Area Public Schools, Lakeville, MN Mary Mortlock, Harker School, San Jose, CA Heather Nichols, Oak Creek High School, Oak Creek, WI Jamis Perrett, Texas A&M University, College Station, TX Heather Pessy, Mount Lebanon High School, Pittsburgh, PA Kathleen Petko, Palatine High School, Palatine, IL Todd Phillips, Mills Godwin High School, Richmond, VA Paula Schute, Mount Notre Dame High School, Cincinnati, OH Susan Stauffer, Boise High School, Boise, ID Doug Tyson, Central York High School, York, PA Bill Van Leer, Flint High School, Oakton, VA Julie Verne, Pine Crest High School, Fort Lauderdale, FL

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Section 2.1 Scatterplots and Correlations

Steve Willot, Francis Howell North High School, St. Charles, MO Jay C. Windley, A. B. Miller High School, Fontana, CA

Reviewers of previous editions: Christopher E. Barat, Villa Julie College, Stevenson, MD Jason Bell, Canal Winchester High School, Canal Winchester, OH Zack Bigner, Elkins High School, Missouri City, TX Naomi Bjork, University High School, Irvine, CA Robert Blaschke, Lynbrook High School, San Jose, CA Alla Bogomolnaya, Orange High School, Pepper Pike, OH Andrew Bowen, Grand Island Central School District, Grand Island, NY Jacqueline Briant, Bishop Feehan High School, Attleboro, MA Marlys Jean Brimmer, Ridgeview High School, Bakersfield, CA Floyd E. Brown, Science Hill High School, Johnson City, TN James Cannestra, Germantown High School, Germantown, WI Joseph T. Champine, King High School, Corpus Christi, TX Jared Derksen, Rancho Cucamonga High School, Rancho Cucamonga, CA George J. DiMundo, Coast Union High School, Cambria, CA Jeffrey S. Dinkelmann, Novi High School, Novi, MI Ronald S. Dirkse, American School in Japan, Tokyo, Japan Cynthia L. Dishburger, Whitewater High School, Fayetteville, GA Michael Drake, Springfield High School, Erdenheim, PA Mark A. Fahling, Gaffney High School, Gaffney, SC David Ferris, Noblesville High School, Noblesville, IN David Fong, University High School, Irvine, CA Terry C. French, Lake Braddock Secondary School, Burke, VA Glenn Gabanski, Oak Park and River Forest High School, Oak Park, IL Jason Gould, Eaglecrest High School, Centennial, CO Dr. Gene Vernon Hair, West Orange High School, Winter Garden, FL Stephen Hansen, Napa High School, Napa, CA Katherine Hawks, Meadowcreek High School, Norcross, GA Gregory D. Henry, Hanford West High School, Hanford, CA Duane C. Hinders, Foothill College, Los Altos Hills, CA Beth Howard, Saint Edwards, Sebastian, FL Michael Irvin, Legacy High School, Broomfield, CO Beverly A. Johnson, Fort Worth Country Day School, Fort Worth, TX Matthew L. Knupp, Danville High School, Danville, KY Kenneth Kravetz, Westford Academy, Westford, MA Lee E. Kucera, Capistrano Valley High School, Mission Viejo, CA

Christina Lepi, Farmington High School, Farmington, CT Jean E. Lorenson, Stone Ridge School of the Sacred Heart, Bethesda, MD Thedora R. Lund, Millard North High School, Omaha, NE Philip Mallinson, Phillips Exeter Academy, Exeter, NH Dru Martin, Ramstein American High School, Ramstein, Germany Richard L. McClintock, Ticonderoga High School, Ticonderoga, NY Louise McGuire, Pattonville High School, Maryland Heights, MO Jennifer Michaelis, Collins Hill High School, Suwanee, GA Dr. Jackie Miller, Ohio State University Jason M. Molesky, Lakeville South High School, Lakeville, MN Wayne Nirode, Troy High School, Troy, OH Heather Pessy, Mount Lebanon High School, Pittsburgh, PA Sarah Peterson, University Preparatory Academy, Seattle, WA Kathleen Petko, Palatine High School, Palatine, IL German J. Pliego, University of St. Thomas Stoney Pryor, A&M Consolidated High School, College Station, TX Judy Quan, Alameda High School, Alameda, CA Stephanie Ragucci, Andover High School, Andover, MA James M. Reeder, University School, Hunting Valley, OH Joseph Reiken, Bishop Garcia Diego High School, Santa Barbara, CA Roger V. Rioux, Cape Elizabeth High School, Cape Elizabeth, ME Tom Robinson, Kentridge Senior High School, Kent, WA Albert Roos, Lexington High School, Lexington, MA Linda C. Schrader, Cuyahoga Heights High School, Cuyahoga Heights, OH Daniel R. Shuster, Royal High School, Simi Valley, CA David Stein, Paint Branch High School, Burtonsville, MD Vivian Annette Stephens, Dalton High School, Dalton, GA Charles E. Taylor, Flowing Wells High School, Tucson, AZ Reba Taylor, Blacksburg High School, Blacksburg, VA Shelli Temple, Jenks High School, Jenks, OK David Thiel, Math/Science Institute, Las Vegas, NV William Thill, Harvard-Westlake School, North Hollywood, CA Richard Van Gilst, Westminster Christian Academy, St. Louis, MO Joseph Robert Vignolini, Glen Cove High School, Glen Cove, NY Ira Wallin, Elmwood Park Memorial High School, Elmwood Park, NJ Linda L. Wohlever, Hathaway Brown School, Shaker Heights, OH

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To the Student Statistical Thinking and You The purpose of this book is to give you a working knowledge of the big ideas of statistics and of the methods used in solving statistical problems. Because data always come from a realworld context, doing statistics means more than just manipulating data. The Practice of Statistics (TPS), Fifth Edition, is full of data. Each set of data has some brief background to help you understand what the data say. We deliberately chose contexts and data sets in the examples and exercises to pique your interest. TPS 5e is designed to be easy to read and easy to use. This book is written by current high school AP® Statistics teachers, for high school students. We aimed for clear, concise explanations and a conversational approach that would encourage you to read the book. We also tried to enhance both the visual appeal and the book’s clear organization in the layout of the pages. Be sure to take advantage of all that TPS 5e has to offer. You can learn a lot by reading the text, but you will develop deeper understanding by doing Activities and Data Explorations and answering the Check Your Understanding questions along the way. The walkthrough guide on pages xiv–xx gives you an inside look at the important features of the text. You learn statistics best by doing statistical problems. This book offers many different types of problems for you to tackle. •

•

•

•

Section Exercises include paired odd- and even-numbered problems that test the same skill or concept from that section. There are also some multiple-choice questions to help prepare you for the AP® exam. Recycle and Review exercises at the end of each exercise set involve material you studied in previous sections. Chapter Review Exercises consist of free-response questions aligned to specific learning objectives from the chapter. Go through the list of learning objectives summarized in the Chapter Review and be sure you can say “I can do that” to each item. Then prove it by solving some problems. The AP® Statistics Practice Test at the end of each chapter will help you prepare for in-class exams. Each test has 10 to 12 multiple-choice questions and three freeresponse problems, very much in the style of the AP® exam. Finally, the Cumulative AP® Practice Tests after Chapters 4, 7, 10, and 12 provide challenging, cumulative multiple-choice and free-response questions like ones you might find on a midterm, final, or the AP® Statistics exam.

The main ideas of statistics, like the main ideas of any important subject, took a long time to discover and take some time to master. The basic principle of learning them is to be persistent. Once you put it all together, statistics will help you make informed decisions based on data in your daily life.

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Section 2.1 Scatterplots and Correlations

TPS and AP® Statistics The Practice of Statistics (TPS) was the first book written specifically for the Advanced Placement (AP®) Statistics course. Like the previous four editions, TPS 5e is organized to closely follow the AP® Statistics Course Description. Every item on the College Board’s “Topic Outline” is covered thoroughly in the text. Look inside the front cover for a detailed alignment guide. The few topics in the book that go beyond the AP® syllabus are marked with an asterisk (*). Most importantly, TPS 5e is designed to prepare you for the AP® Statistics exam. The entire author team has been involved in the AP® Statistics program since its early days. We have more than 80 years’ combined experience teaching introductory statistics and more than 30 years’ combined experience grading the AP® exam! Two of us (Starnes and Tabor) have served as Question Leaders for several years, helping to write scoring rubrics for free-response questions. Including our Content Advisory Board and Supplements Team (page vii), we have two former Test Development Committee members and 11 AP® exam Readers. TPS 5e will help you get ready for the AP® Statistics exam throughout the course by: •

•

•

•

Using terms, notation, formulas, and tables consistent with those found on the AP® exam. Key terms are shown in bold in the text, and they are defined in the Glossary. Key terms also are cross-referenced in the Index. See page F-1 to find “Formulas for the AP® Statistics Exam” as well as Tables A, B, and C in the back of the book for reference. Following accepted conventions from AP® exam rubrics when presenting model solutions. Over the years, the scoring guidelines for free-response questions have become fairly consistent. We kept these guidelines in mind when writing the solutions that appear throughout TPS 5e. For example, the four-step State-PlanDo-Conclude process that we use to complete inference problems in Chapters 8 through 12 closely matches the four-point AP® scoring rubrics. Including AP® Exam Tips in the margin where appropriate. We place exam tips in the margins and in some Technology Corners as “on-the-spot” reminders of common mistakes and how to avoid them. These tips are collected and summarized in Appendix A. Providing hundreds of AP® -style exercises throughout the book. We even added a new kind of problem just prior to each Chapter Review, called a FRAPPY (Free Response AP® Problem, Yay!). Each FRAPPY gives you the chance to solve an AP®-style free-response problem based on the material in the chapter. After you finish, you can view and critique two example solutions from the book’s Web site (www.whfreeman.com/tps5e). Then you can score your own response using a rubric provided by your teacher.

Turn the page for a tour of the text. See how to use the book to realize success in the course and on the AP® exam.

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143

section 3.1 scatterplots and correlation

increments starting with 135 cm. Refer to the sketch in the margin for comparison.

160

5. Plot each point from your class data table as accurately as you can on the graph. Compare your graph with those of your group members.

Height (cm)

READ THE TEXT and use the book’s features to help you grasp the big ideas. 155 150

6. As a group, discuss what the graph tells you about the relationship between hand span and height. Summarize your observations in a sentence or two.

145 140

7. Ask your teacher for a copy of the handprint found at the scene and the math department roster. Which math teacher does your group believe is the “prime suspect”? Justify your answer with appropriate statistical evidence.

135

15

15.5

16

16.5

17

Hand span (cm)

3.1

Read the LEARNING OBJECTIVES at the beginning of each section. Focus on mastering these skills and concepts as you work through the chapter.

17.5

Scatterplots and Correlation 145

section 3.1 scatterplots and correlation

WHAT You WILL LEARn

By the end of the section, you should be able to:

• Identify explanatory and response variables in situations Displaying Relationships: Scatterplots where one variable helps to explain or influences the other.

Mean Math score

• Interpret the correlation. • Understand the basic properties of correlation, 625 The most useful relationship between • Make a scatterplot to graph displayfor thedisplaying relationshipthe between including how the correlation is influenced by outliers. quantitative variables is a scatterplot. Figure 3.2 shows a 600 twotwo quantitative variables. • Use technology to calculate correlation. ofethe graduates in each state 168 C H• A PDescribe T scatterplot E R 3the direction, D s cpercent r i b iand nofghigh r eschool l at 575 form, strength of ai o n s h i p s • Explain why association does not imply causation. who took the SAT and the state’s mean SAT Math score in a relationship displayed in a scatterplot and identify outli550 recent year. We think that “percent taking” will help explain ers in a scatterplot. “mean score.” SoA“percent is the explanatory negativetaking” price doesn’t make muchvariable sense in this context. Look again at Figure 525 and “mean score” the response variable. want to see 3.8. Aistruck with 300,000 milesWe driven is far outside the set of x values for our data. 500 how mean score when percent taking changes, somiles driven and price remains linWe changes can’t say whether the relationship between 475 Most statistical studies examine data on more onemiles variable. we put percentear taking (theextreme explanatory variable) on the horiat such values. Predicting price for a truck with than 300,000 drivenFortunately, analysis of several-variable data builds on the the data toolsshow. we used to examine individual 450 zontal axis. Each point represents a single state. In Colorado, is an extrapolation of the relationship beyond what 0 90 10 20 30 40 50 60 70 80 variables. Thetheir principles that guide for example, 21% took the SAT, and mean SAT Math our work also remain the same: Percent taking SAT • the Plot the data, then score was 570. Find 21 on x (horizontal) axisadd andnumerical 570 on summaries. Often, using the regression DEFInItIon: Extrapolation theliney (vertical) axis. Colorado appears the point (21, and 570).departures from those patterns. FIGURE 3.2 Scatterplot of the • Look for as overall patterns to make a prediction for x = 0 is mean SAT Math score in each state Extrapolation is the use of a aregression line for pattern, predictionuse faraoutside the interval • When there’s regular overall simplified model to describe it. an extrapolation. That’s why the y against the percent of that state’s of values of the explanatory variable x used to obtain the line. Such predictions are intercept isn’t always statistically DEFInItIon: Scatterplot high school graduates who took the meaningful. often not accurate. SAT. The dotted lines intersect at A scatterplot shows the relationship between two quantitative variables measured the point (21, 570), the values for on the same individuals. The values of one variable horizontal Weappear think on thatthecar weight axis, helpsand explain accident deaths and that smoking influColorado. Few relationships are linear for all values of the explanatory variable. autio the values of the other variable appear on the vertical Each individual the data encesaxis. life expectancy. Ininthese relationships, the two variables play different roles. Don’t make predictions using values of x that are much larger or much appears as a point in the graph. Accident death rate and life expectancy are the response variables of interest. Car smaller than those that actually appear in your data. weight and number of cigarettes smoked are the explanatory variables.

Scan the margins for the purple notes, which represent the “voice of the teacher” giving helpful hints for being successful in the course.

Here’s a helpful way to remember: the eXplanatory variable goes on the x axis.

Look for the boxes with the blue bands. Some explain how to make graphs or set up calculations while others recap important concepts.

EXAMPLE

!

n

c

Explanatory and Response Variables

Take note of the green DEFINITION boxes that explain important vocabulary. Flip back to them to review key terms and their definitions.

Always plot the explanatory variable, if there is one, on the horizontal axis (the x axis) of a scatterplot. As a reminder, we usually call the explanatory DEFInItIon: Response variable variable,x explanatory variable Your understanding and the response variable y. If thereCheCk is no explanatory-response distinction, either Some data were collected on the weight of a male white laboratory rat for first 25 weeks A response variable measures an outcome of a study. An the explanatory variable variable can go on the horizontal axis. 155 section 3.1shows scatterplots and correlation after its birth.may A scatterplot of the weight grams)inand time since birth (in weeks) helpFor explain orproblems, predict(in changes a response variable. We used computer software to produce Figure 3.2. some you’ll a fairly strong, positive linear relationship. The linear regression equation weight = 100 + be expected to make scatterplots by40(time) hand. Here’s how to do it. models the data fairly well. 1. What is the slope of the regression line?Standardized Explain what itvalues have no units—in this example, they are no longer measured in points. How to MakE a ScattERplotmeans in context. – Somethe people like to writeExplain the To standardize 2. What’s y intercept? what it means in context. the number of wins, we use y = 8.08 and sy = 3.34. For correlation formula as 12 − 8.08 1. Decide which variable should3.goPredict on each theaxis. rat’s weight after 16 weeks. Show your work. = 1.17. Alabama’s number of wins (12) is 1.17 stanAlabama, zy = 1 zx zy r = you ∙ 3.34 2. Label and scale your axes. 4. Should Starnes-Yates5e_c03_140-205hr2.indd 143 9/30/13 4:44 PM n − 1 use this line to predict the rat’s weight at dard deviations above the mean number of wins for SEC teams. When we mulage 2 years? Use the equation to make the prediction and 3. Plot individual data values. to emphasize the product of thisare team’s think about the reasonableness of the result.tiply (There 454 two z-scores, we get a product of 1.2636. The correlation r is an standardized scores in the calculation. “average” of the products of the standardized scores for all the teams. Just grams in a pound.) The following example illustrates the process of constructing a scatterplot. as in the case of the standard deviation sx, the average here divides by one fewer than the number of individuals. Finishing the calculation reveals that r = 0.936 for the SEC teams.

Watch for CAUTION ICONS. They alert you to common mistakes that students make.

Residuals and the Least-Squares SEC Football What does correlation measure? The Fathom screen shots below proTHINK Make connections and Regression Line vide more detail. At the left is a scatterplot of the SEC football data with two lines Making a scatterplot

ABOUT ITcases, added—a vertical line at the group’s In most no line will pass exactly through allmean the points per game and a horizontal line deepen Atyour understanding at the mean number of wins of the group. Most of the points fall in the upper-right the end of the 2011 college football season, thepoints University Alabama in a of scatterplot. Because we use the line to predict or lower-left the graph. That is, teams with above-average points defeatedon Louisiana State deviation University for the national championship. In- errors“quadrants” y from x, the prediction we make areoferrors in y, the by reflecting the Vertical from the line peringame tend to have above-average numbers of wins, and teams with belowterestingly, both of these teams were from the Southeastern Conference vertical direction the scatterplot. A good regression line questions asked in the THINK average pointsofper numbers of wins that are below average. (SEC). Here are averageRegression number line of points scored per the game and nummakes vertical deviations the game points tend from to thehave line as 1 This confirms the positive association between the variables. ber of wins for each of the twelve teams in the SECsmall that season. as possible. yˆ = 38,257 2 0.1629x ABOUT IT passages. Below on the right is aFord scatterplot the standardized scores. To get this graph, Figure 3.9 shows a scatterplot of the F-150 of data 45,000

Price (in dollars)

40,000 35,000 30,000 25,000 20,000

Data point

15,000 10,000 5000

Starnes-Yates5e_c03_140-205hr2.indd 145

0

20,000 40,000

183 section 3.2 we least-squares regression transformed bothprediction the x- and errors the y-values with a regression line added. The are by subtracting their mean and dividing by their standard deviation. As we saw in Chapter 2, standardizing a data set converts the mean to 0 and the standard deviation to 1. That’s why the vertical and variables and their correlation. Exploring this method will highlight an important FIGURE 3.9 Scatterplot of the FordinF-150 data with a regression horizontal lines the right-hand graph are both at 0. relationship between the correlation and the slope oflinea should least-squares line added. A good regression make the regression prediction 60,000 80,000 100,000 120,000 140,000 160,000 line—and reveal why we include the word “regression” in theasexpression “leasterrors (shown as bold vertical4:44 segments) small as possible. 9/30/13 PM Miles driven squares regression line.”

How to calcUlatE tHE lEaSt-SqUaRES REGRESSIon lInE

Read the AP® EXAM TIPS. They give advice on how to be successful on the AP® exam.

ap® ExaM tIp The formula sheet for the AP® exam uses Starnes-Yates5e_c03_140-205hr2.indd 168 different notation for these sy equations: b1 = r and sx b0 = y– − b1 x– . That’s because the least-squares line is written as y^ = b0 + b1x . We prefer our simpler versions without the subscripts!

and y intercept

9/30/13 4:44 PM

sy b = r Notice that all the products of the standardized values will be positive—not sx surprising, considering the strong positive association between the variables. What if there was a negative association between two variables? Most of the points would a = y–be − in bx–the upper-left and lower-right “quadrants” and their z-score products would be negative, resulting in a negative correlation.

The formula for the y intercept comes from the fact that the least-squares regression line always passes through the point (x– , y– ). You discovered this in Step 4 of the Activity on page 170. Substituting (x– , y– ) into the equation y^ = a + bx produces the equation y– = a + bx– . Solving this equation for a gives the equation shown in How correlation behaves is more important than the details of the formula. Here’s the definition box, a = y– − bx– . what you in order to interpret correlation correctly. To see how these formulas work in practice, let’sneed looktoatknow an example.

Facts about Correlation

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We have data on an explanatory variable x and a response variable y for n individuals. From the data, calculate the means x– and y– and the standard deviations sx and sy of the two variables and their correlation r. The least-squares regression line is the line y^ = a + bx with slope

EXAMPLE

Using Feet to Predict Height Calculating the least-squares regression line

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538 426

CHAPTER 7

CHAPTER 9

TesTing a Claim

Sampling DiStributionS

Introduction

To make sense of sampling variability, we ask, “What would happen if we took many samples?” Here’s how to answer that question: • Take a large number of samples from the same population. • Calculate the statistic (like the sample mean x– or sample proportion p^ ) for each sample. • Make a graph of the values of the statistic. • Examine the distribution displayed in the graph for shape, center, and spread, as well as outliers or other unusual features. The following Activity gives you a chance to see sampling variability in action.

LEARN STATISTICS BY DOING STATISTICS

Confidence intervals are one of the two most common types of statistical inference. Use a confidence interval when your goal is to estimate a population parameter. The second common type of inference, called significance tests, has a different goal: to assess the evidence provided by data about some claim concerning a parameter. Here is an Activity that illustrates the reasoning of statistical tests.

Activity

Activity MATERIAlS: 200 colored chips, including 100 of the same color; large bag or other container

I’m a Great Free-Throw Shooter!

MATERIAlS:

Reaching for Chips

A basketball player claims to make 80% of the free throws that he attempts. We think he might be exaggerating. To test this claim, we’ll ask him to shoot some free throws—virtually—using The Reasoning of a Statistical Test applet at the book’s Web site. 1. Go to www.whfreeman.com/tps5e and launch the applet.

Computer with Internet access and projection capability

Before class, your teacher will prepare a population of 200 colored chips, with 100 PLET AP having the same color (say, red). The parameter is the actual proportion p of red chips in the population: p = 0.50. In this Activity, you will investigate sampling variability by taking repeated random samples of size 20 from the population. 1. After your teacher has mixed the chips thoroughly, each student in the class should take a sample of 20 chips and note the sample proportion p^ of red chips. When finished, the student should return all the chips to the bag, stir them up, and pass the bag to the next student. Note: If your class has fewer than 25 students, have some students take two 157 section 3.1 scatterplots and correlation samples. ^ andcourse, plot this 2. Each student should record the p-value in a chart on the board Of even giving means, standard deviations, and the correlation for “state value on a class dotplot. Label the graph scale from 0.10 to 0.90SAT withMath tick marks scores” and “percent taking” will not point out the clusters in Figure 3.2. spaced 0.05 units apart. Numerical summaries complement plots of data, but they do not replace them. 3. Describe what you see: shape, center, spread, and any outliers or other un144 CHAPTER 3 D e s c r i b i n g r e l at i o n s h i p s usual features.

EXAMPLE

You will often see explanatory variables

Scoring Skaters It is easiest toFigure identify explanatory and response variables when we actually

When Mr. Caldwell’s called class independent did the “Reaching for Chips” Activity, his 35 stuvariables and specify values of one variable see the howwhole it affectsstory another variable. For instance, Why correlation doesn’ttotell dents produced the graphresponse shown in Figurecalled 7.1. dependent Here’s what the class said about its variables 2. Set the to take 25 shots.researchers Click “Shoot.” How many of the 25 shots did to study the effect of alcohol on applet body temperature, gave several difdistribution of p^ -values. variables. Because the words Until a scandal at the 2002 Olympics brought change, figure skating was scored the player make?Then Do you have enough data tochange decide whether ferent amounts of from alcohol to mice. they measured theWe in eachthe player’s claim “independent” andis“dependent” have by judges on apeak scale 0.0 to 6.0. The scores were often controversial. have Shape: The graph roughly symmetric with a single is valid? minutes later. In this mouse’s temperature the scoresbody awarded by two judges,15 Pierre and Elena, for many skaters. How well do at 0.5.other meanings in statistics, we won’t 3. “Shoot” again for 25 more shots. Keep they amount agree? Weof calculate thatClick correlation between their scores is r = 0.9. doing But this until you are use them here. case, alcohol isthethe explanatory variable, player makes less than 80% of his shots or that the 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Center: The mean of our sample proportions is 0.499. This the change mean of Pierre’s scoresconvinced is 0.8 pointeither lower than Elena’s mean. and in body temperature isthat thethe response ^ player’s claim is true. How large a sample of shots did you need to make your p is the balance point of the distribution. variable. When we don’t specify the values of either decision? These facts don’t contradict each other. They simply give different kinds of inforfiGURe 7.1 Dotplot of sample proportions obtained by the Spread: The standard deviation of our sample proportions is variable both variables, there may mation. but The just meanobserve scores4.show that“Show Pierretrue awards lower scores than Elena. But Was your conclusion Click probability” to reveal the truth. 35 students in Mr. Caldwell’s class. 0.112. The values of p^ are typically about 0.112 from orbecause may away not be explanatory response Pierre gives everycorrect? skaterand a score about 0.8varipoint lower than Elena does, the mean. the correlation the same number to all values of either x ables. Whetherremains therehigh. are Adding depends on how If time permits, a newthe shooter repeat or plan y doesto notuse change the5.correlation. If both choose judges score same and skaters, the Steps 2 through 4. Is it you the data. Outliers: There are no obvious outliers or other unusual features. easier to because tell thatPierre the player is exaggerating whenperhis actual proportion of free competition is scored consistently and Elena agree on which throws made is closer to 0.8 or farther from 0.8? Of course, the class only took 35 different simple random samples of 20are chips. formances better than others. The high r shows their agreement. But if Pierre There are many, many possible SRSs of size 20 from a population scores of sizesome 200 (about skaters and Elena others, we should add 0.8 point to Pierre’s scores to 1.6 · 1027, actually). If we took every one of those possible samples, calculated arrive at athe fairvalue comparison. of p^ for each, and graphed all those p^ -values, then we’d have a sampling distribution.

Every chapter begins with a hands-on ACTIVITY that introduces the content of the chapter. Many of these activities involve collecting data and drawing conclusions from the data. In other activities, you’ll use dynamic applets to explore statistical concepts.

EXAMPLE

Linking SAT Math and Critical Reading Scores DATA EXPLORATION The SAT essay: Is longer better?

DATA EXPLORATIONS ask you to play the role of data detective. Your goal is to answer a puzzling, real-world question by examining data graphically and numerically.

Explanatory response? Following the debutor of the new SAT Writing test in March 2005, Dr. Les Perelman

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from the Massachusetts Institute of Technology stirred controversy by reporting,

Julie asks, “Can I predict a state’s mean SAT Math score if I know its mean SAT 10/10/13 4:59 PM of what a student wrote, the longer the essay, the “It appeared to me that regardless Critical Reading to know the mean SATpredictor Math and Critical higher the score.” score?” He went Jim on towants say, “I have neverhow found a quantifiable Reading scores this year 50 states are related to each in 25 years of grading that in wasthe anywhere as strong as this one. If youother. just graded

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them based on length without ever reading them, you’d be right over 90 percent

Problem: each student, variable and the response The table belowidentify shows the explanatory data that Dr. Perelman used to drawvariable his if possible. of the time.”3For conclusions.4 solution: Julie is treating the mean SAT Critical Reading score as the explanatory variable and

the mean SAT Math score asofthe response variable. Jim is simply interested in exploring the relationLength essay and score for a sample of SAT essays shipWords: between 460 the two there is variable. 422variables. 402 For 365him,357 278no clear 236 explanatory 201 168or response 156 133 Score:

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For 236 Practice 189 128Try Exercise

1

Score: 1 6 the6 goal 5is to 5show4 that 4changes 2 In many studies, in3 one or more explanatory variables actually cause changeswrite in aaresponse variable. However, other explanatoryDoes this mean that if students lot, they are guaranteed high scores? response relationships don’t involve direct causation. the alcohol and Carry out your own analysis of the data. How would youInrespond to each of mice study, Dr. Perelman’s claims?a change in body temperature. But there is no cause-and-effect alcohol actually causes relationship between SAT Math and Critical Reading scores. Because the scores are closely related, we can still use a state’s mean SAT Critical Reading score to predict its mean Math score. We will learn how to make such predictions in Section 3.2.

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CHECK YOUR UNDERSTANDING questions appear throughout the section. They help you to clarify definitions, concepts, and procedures. Be sure to check your answers in the back of the book.

CheCk Your understanding

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Identify the explanatory and response variables in each setting. 1. How does drinking beer affect the level of alcohol in people’s blood? The legal limit for driving in all states is 0.08%. In a study, adult volunteers drank different numbers of cans of beer. Thirty minutes later, a police officer measured their blood alcohol levels. 2. The National Student Loan Survey provides data on the amount of debt for recent college graduates, their current income, and how stressed they feel about college debt. A sociologist looks at the data with the goal of using amount of debt and income to explain the stress caused by college debt.

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EXAMPLES: Model statistical problems and how to solve them CHAPTER 3

D e s c r i b i n g r e l at i o n s h i p s

You will often see explanatory variables called independent variables and response variables called dependent variables. Because the words “independent” and “dependent” have other meanings in statistics, we won’t use them here.

EXAMPLE

It is easiest to identify explanatory and response variables when we actually specify values of one variable to see how it affects another variable. For instance, to study the effect of alcohol on body temperature, researchers gave several different amounts of alcohol to mice. Then they measured the change in each mouse’s body temperature 15 minutes later. In this case, amount of alcohol is the explanatory variable, and change in body temperature is the response variable. When we don’t specify the values of either variable but just observe both variables, there may or may not be explanatory and response variables. Whether there are depends on how you plan to use the data.

Exercises

section 3.1

The red number box next to the exercise directs you back to the page in the section where the model example appears.

Read through each example, and then try out the concept yourself by working the FOR PRACTICE exercise in the Section Exercises.

Need extra help? Examples and exercises marked with Linking SAT Math and Critical the PLAY ICON are Reading Scores supported by short video clips Explanatory or response? prepared by experienced AP® Julie asks, “Can I predict a state’s mean SAT Math score if I know its mean SAT teachers. The video guides Critical Reading score?” Jim wants to know how the mean SAT Math and Critical Reading scores this year in the 50 states are related to each other. 185 section 3.2 least-squares regression you through each step in the Problem: For each student, identify the explanatory variable and the response variable if possible. example and solution and solution: Julie is treating the mean SAT Critical Reading score as r. theIn explanatory variablethe andslope is equal to the correlation. The Fathom means b= other words, 2 the mean SAT Math score as the response variable. Jim isscreen simply interested in exploring the relationshot confirms these results. It shows that r = 0.49, so r =you !0.49 = 0.7, gives extra help when you ship between the two variables. For him, there is no clear explanatory or response variable. 159 section approximately the same value3.1 as thescatterplots slope of 0.697.and correlation need it. For Practice Try Exercise 1 In many studies, the goal is to show that changes in one or more explanatory variables actually cause changes in a response variable. However, other explanatoryresponse relationships don’t involve direct causation. In the alcohol and mice study, alcohol actually causes a change in body temperature. But there is no cause-and-effect 1. coralbetween reefs How changes in water how relationship SATsensitive Math andtoCritical Reading scores. Because the4.scores aremuch gas? Joan is concerned about the arestill coral findSAT out,Critical measure amount related, we can use reefs? a state’sTo mean Reading score to predict its of energy she uses to heat her home. The pg closely 144 temperature mean the Math score. of Wecorals will learn how to make suchthe predictions growth in aquariums where water in Section 3.2.graph below plots the mean number of cubic feet of temperature is controlled at different levels. Growth is gas per day that Joan used each month against the measured by weighing the coral before and after the average temperature that month (in degrees FahrenCheCk Your understanding experiment. What are the explanatory and response heit) for one heating season. Identifyvariables? the explanatory and response variables in each setting. Are they categorical or quantitative? 1000

Putting It All Together: Correlation and Regression It is now usual to remove only the tumor and nearby

1. How does drinking beer affect the level of alcohol in people’s blood? The legal limit 2.driving treating Early on, adult the most common for in all breast states iscancer 0.08%. In a study, volunteers drank different numbers of 800 forminutes breast cancer was removal of the breast. cans oftreatment beer. Thirty later, a police officer measured their blood alcohol levels. Gas consumed (cubic feet)

144

2. The National Student Loan Survey provides data on the amount of debt for recent 600 Inchange Chapter introduced a four-step process for organizing a statistics problem. nodes, followed by radiation. Thestressed in 1, collegelymph graduates, their current income, and how they feelwe about college debt. A Herethat is of another example four-step process in action. sociologist looks the to data with the goal ofexperiment using amount debt and income of to the explain policy wasatdue a large medical com400 the stress caused debt. Some breast cancer patients, pared the by twocollege treatments.

chosen at random, were given one or the other treatment. The patients were closely followed to see how long they lived following surgery. What are the explanatory and response variables? Are they categorical or quantitative?

EXAMPLE

3.

4-STEP EXAMPLES: By reading the 4-Step Examples and mastering the special “StatePlan-Do-Conclude” framework, you can develop good problemsolving skills and your ability to tackle more complex problems like those on the AP® exam.

Gesell Scores

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TEP iQ and grades Do studentsSwith higher IQ test scores Temperature (degrees Fahrenheit) Putting it all together tend to do better in school? The figure below shows Does the age at which a child begins to talk4:44 predict a later score on a test of mental 9/30/13 PM (a) Does the plot show a positive or negative association a scatterplot of IQ and school grade point average ability? A study of the development of young children recorded the age in months between the variables? Why does this make sense? (GPA) for all 78 seventh-grade studentsatin a rural which each of 21 children spoke their first word and their Gesell Adaptive Score, midwestern school. (GPA was recordedthe onresult a 12-point data appear inIsthe tablestrong? beof an aptitude(b) test What taken is much later.16 the form ofThe the relationship? it very scale with A+ = 12, A = 11, A− = 10,low, B+along = 9,with . . . , a scatterplot,Explain residual your plot, answers. and computer output. Should we use a 5 D− = 1, and F = 0.) linear model to predict a child’s Gesell score from his or her age at first word? If so, (c) Explain how accurate will our predictions be?what the point at the bottom right of the plot 12 represents.

Grade point average

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(c) At the bottom of the plot are several points that we might call outliers. One student in particular has a very low GPA despite an average IQ score. What are

Age5.(months) first word and Gesell score heavyatbackpacks Ninth-grade students at the Webb Schools aGE go on a backpacking trip each cHIlD ScoRE cHIlD aGEfall. Students ScoRE are divided into groups of15 size 8 by 95 8 11 hiking100 11selecting 102names from leaving, students and 71 9 a hat.8 Before 104 16 10 their backpacks 100 are one hiking 83 10 weighed. 20 The data 94 here are 17 from 12 105group in a recent year. Make a scatterplot by hand that shows 91 11 7 113 18 42 57 how backpack weight relates to body weight.

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bird colonies One of nature’s patterns connects the percent of adult birds in a colony that return from the previous year and the number of new adults that join the colony. Here are data for 13 colonies of sparrowhawks:6 9/30/13 4:45 PM

Percent return: 74 66 81 52 73 62 52 45 62 46 60 46 38 new adults:

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Make a scatterplot by hand that shows how the number

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EXERCISES: Practice makes perfect! 192

CHAPTER 3

section 3.2 Start by reading the SECTION SUMMARY to be sure that you understand the key concepts.

Most of the exercises are paired, meaning that odd- and even-numbered problems test the same skill or concept. If you answer an assigned problem incorrectly, try to figure out your mistake. Then see if you can solve the paired exercise.

Summary

A regression line is a straight line that describes how a response variable y changes as an explanatory variable x changes. You can use a regression line to predict the value of y for any value of x by substituting this x into the equation of the line. • The slope b of a regression line y^ = a + bx is the rate at which the predicted response y^ changes along the line as the explanatory variable x changes. Specifically, b is the predicted change in y when x increases by 1 unit. • The y intercept a of a regression line y^ = a + bx is the predicted response y^ when the explanatory variable x equals 0. This prediction is of no statistical use unless x can actually take values near 0. • Avoid extrapolation, the use of a regression line for prediction using values 193 section 3.2 least-squares regression of the explanatory variable outside the range of the data from which the line was calculated. • The most common method of fitting a line to a scatterplot is least squares. The least-squares regression line is the straight line y^ = a + bx that minimizes the sum of the squares of the vertical distances of the observed points from the line. • You can examine the fit of a regression line by studying the residuals, which arethe thesame differences between the observed predicted values of y. Befigure on the 35. What’s my line? You use bar of soap to in and Joan’s midwestern home. The below shows lookout for patterns in when the residual plot, thatwith a linear model line shower each morning. The bar weighs 80 grams thewhich originalindicate scatterplot the least-squares it is new. Its weight goesmay down bybe 6 grams per day on added. The equation of the least-squares line is not appropriate. average. What is the of the regression line of the residuals y^ =s1425 − 19.87x. • equation The standard deviation measures the typical size of the 197 section 3.2 least-squares regression for predicting weight from days of use? 900 prediction errors (residuals) when using the regression line. 36. What’s my line? An eccentric professor believes that 800 of the variation in the • The coefficient of determination r2 is the fraction a child with IQ 100 should have a reading test score 700 husbands and wives Refer toresponse Exercisevariable 61. If so, stumps shouldregression produce more beetle is accounted formore by least-squares on the ex-larvae. of 50 and predicts that reading score shouldthat increase 24 600 Here are the data: 2 planatory variable. Find rby and interpret this value in context. 1 point for every additional point of IQ. What 500 is thedata, equation of Interpret the regression line for • professor’s Thethis least-squares regression line of y on x is the line with slope b = r(sy/sx) For these s = 1.2. value. 2 4002 1 3 3 4 3 1 – 2– 5 1 3 predicting reading scoreand from IQ? a = y– − bx–. ThisStumps: line always passes through the point (x , y ). intercept the stock market Refer to Exercise 62. Beetle larvae: 10 300 30 12 24 36 40 43 11 27 56 18 40 37. gas mileage We expect a car’s highway gas mileagemust be interpreted with caution. Plot the data to be • Correlation and regression 200 2 Stumps: 2 1 2 2 35 1 401 45 4 150 2 55 1 604 Find rtoand interpret value in context. be related to this its city gas mileage. for all 1198 sure that theData relationship is roughly linear and to detect30 outliers. Also look for inlarvae: 25 8 21 14Temperature 16 6(degrees 54 Fahrenheit) 9 13 14 50 vehicles the8.3. government’s recent Fuel Economy fluential observations, individualBeetle points that substantially change the correlation For these data,ins = Interpret this value. Guide give the regression line: predicted highway (a) Identify the slope of the line and explain what Can we use a linearfor model to predict the of it or the regression line. Outliers in x are often influential the regression line.number Will impg bomb the final? We(city expect that students = 4.62 + 1.109 mpg). thisthe setting. beetlemeans larvae in from number of stumps? If so, how who do well on the midterm exam of in all, a course will not to conclude that there is a cause-and-effect rela• Most be careful accurate will our be?the Follow the four-step (a) What’s the well slopeon of this Interpret this value in context. Identify the predictions y intercept of line. Explain why it’s usually also do theline? final exam.between Gary Smith tionship two variables just(b) because they are strongly associated. •

section 3.2

63. (a) (b) 64. (a) (b) 65.

Exercises

Gas consumed (cubic feet)

Practice! Work the EXERCISES assigned by your teacher. Compare your answers to those in the Solutions Appendix at the back of the book. Short solutions to the exercises numbered in red are found in the appendix.

D e s c r i b i n g r e l at i o n s h i p s

process. risky to use this value as a prediction. What’sCollege the y intercept? why the value of(b) Pomona looked atExplain the exam scores of of the intercept not statistically meaningful. (c)and Usecalories the regression line to of predict theinamount all 346 studentsiswho took his statistics class over a 68. Fat The number calories a food of STEP 22 natural gas use in a monththe withamount an average 10-year period. Assumehighway that both the midterm item depends on Joan manywill factors, including (c) Find the predicted mileage for a carand that gets temperature of 30°F. final exam were of fat in the item. The data below show the amount 16 miles perscored gallonout in of the100 city.points. of fat (in grams) and the number of calories in 7 41. acid rain Refer to Exercise 39. Would it bebeef appropri(a) State equation of the least-squares 38. the iQ and reading scores Data onregression the IQ testline scores 25 sandwiches at McDonalds. ate to use the regression line to predict pH after 1000 if eachand student scored same the midterm and reading test the scores foron a group of fifth-grade 3.2 TECHNOLOGY months? Justify your answer. the final. children give the following regression line: predicted Sandwich Fat Calories CORNERs reading score = −33.4 + predicting 0.882(IQ score). ® how much gas? Refer to Exercise 40. Would it be 42. (b)TI-nspire The actual least-squares line for final29 550 Big Mac Instructions in Appendix B; HP Prime instructions on the book’s Web site appropriate to use the regression line to predict Joan’s ® exam score y from midterm-exam score x was (a) What’s the slope of this line? Interpret this value in 26 520 Quarter Pounder with Cheese natural-gas® consumption in a future month with an y^ = 46.6 + 0.41x. Predict the score of a student who context. with Cheeseof 65°F?page 42 171 Double Quarter Pounder 8.scored least-squares linesa student on the calculator average temperature Justify your 750 answer. 50 on theregression midterm and who scored (b) What’s the y intercept? Explain why the value of the Hamburger 9 175 250 9.100 residual on the calculator on the plots midterm. 43. least-squares idea The tablepage below gives a small intercept is not statistically meaningful. Cheeseburger 12 two lines 300 fits the set of data. Which of the following 199 section 3.2 least-squares regression (c) Explain how answers to partscore (b) illustrate regres(c) Find theyour predicted reading for a child with an Double Cheeseburger 440the leastdata better: y^ = 1 − x or y^ = 23 3 − 2x? Use sion toIQthe mean. score of 90. answer. (Note: McDouble squares criterion to justify your19 390 Neither it’s early expect that a baseball player 39.still acid rain Researchers studying acid rain who measured of these two lines is the least-squares regression line 0.93. (b)66.r2 will increase, s We will stay the same. battingofaverage in the in first month of wilderness the Can we use a linear model to predict the number of pg 166 a high 2 has the acidity precipitation a Colorado for these data.) 6.4. (c) r will increase, s will decrease. season willfor also have a high batting average rest of calories from the amount of fat? If so, how accurate will area 150 consecutive weeks. Aciditythe is measured Can’t tell without seeing the data. (d) r2 will stay theUsing same,66 s will stayLeague the same. −1 Follow1 the four-step x: 1 3 5 the season. Major playersThe our predictions be? process. by pH. Lower pH values show Baseball higher acidity. 23 Starnes-Yates5e_c03_140-205hr2.indd 192 9/30/13 4:45 PM 2 from the 2010 season, least-squares regression In addition to the regression line, the report on the (e) r will stay the same, s will adecrease. researchers observed a linear pattern over time. −1 measure −5 y: diabetes 2 People 0 with1 diabetes 69. Managing line was calculated to predict rest-of-season batting Mumbai measurements says that r 2 = 0.95. This They reported that the regression line pH = 5.43 − their fasting plasma glucose (FPG; measured in units Exercises 79 0.0053(weeks) and 80 refer to the following setting. 19 average y from first-month batting average x. Note: A 44. least-squares idea In Exercise 40, the line drawn suggests that fit the data well. of milligrams per milliliter) after fasting for at least player’s batting average is thethe proportion of times at on the scatterplot is the least-squares regression line. In its recent Fuel Economy Guide, Environmental although arm span and height are correlated, arm 8 hours. Another measurement, made at regular (a)that Identify the slope of the line and explain what it bat he gets a hit. A batting average over 0.300 is Explain the meaning of the phrase “least-squares” to Protection Agency gives data on 1152 vehicles. There are span does not predict height very accurately. medical checkups, is called HbA. This is roughly meansvery in this setting. considered good in Major with League Joan, who knows very little about statistics. a number of outliers, mainly vehicles veryBaseball. poor gas the percent of red blood cells that have a glucose height increases by !0.95 = 0.97 cm for each ad(b) Identify the intercepthowever, of the line and explain what it mileage. If we theyoutliers, the combined 45. acidattached. rain In the acid rainaverage study ofexposure Exercise to 39, the (a) State theignore equation of the least-squares regression line molecule It measures ditional centimeter of arm span. means in this setting. city andif highway gas had mileage of thebatting other 1120 or so pgglucose 169 actual pH measurement Week 50 was 5.08. Find each player the same average theverest of over a period of severalfor months. The table 95% of the relationship between height and arm span hicles isthe approximately Normal with mean 18.7 miles per (c) According to the regression line, what was the pH at interpret residual for FPG this week. season as he did in the first month of the season. belowand gives data onthe both HbA and for 18 diabetis accounted for by the regression line. gallon (mpg)the andend standard of this deviation study? 4.3 mpg. ics months had diabetes 46.fivehow muchafter gas?they Refer tocompleted Exercise 40.aDuring March, (b) The actual equation of the least-squares regression line 95% of the variation in height is accounted for by the 27 79. inismy chevrolet (2.2) The Chevrolet Malibu with education class. 40. how much gas? In Exercise 4 (page 159), we examthe average temperature was 46.4°F and Joan used 490 y^ = 0.245 + 0.109x. Predict the rest-of-season batting regression line. a four-cylinder has ahad combined ofmonthly ined relationship between themileage average cubic feet of gas per day. Find and interpret the residual average for the a engine player who a 0.200 gas batting average 95% of the height measurements are accounted for by 25the mpg. What percent of all vehicles worse gas temperature and the amount natural gas consumed forHbA this month. FPG HbA FPG first month of the season and for have aofplayer who had a the regression line. mileage than the Malibu? Subject (%) (mg/mL) Subject (%) (mg/mL) 0.400 batting average the first month of the season. One child in the Mumbai study had height 59 cm 80. the top 10% (2.2) How high must a vehicle’s gas 1 6.1 141 10 8.7 172 (c) Explain how your answers to part (b) illustrate regresand arm span 60 cm. This child’s residual is mileage be in order to fall in the top 10% of all 2 6.3 158 11 9.4 200 sion to the mean. vehicles? (The distribution omits a few high outliers, −3.2 cm. (c) −1.3 cm. (e) 62.2 cm. 3 6.4 112 12 10.4 271 67.mainly beavers andgas-electric beetles Dovehicles.) beavers benefit beetles? hybrid −2.2 cm. (d) 3.2 cm. STEP 4 6.8 153 13 10.6 103 9/30/13 4:45 PM Starnes-Yates5e_c03_140-205hr2.indd 193 Researchers out 23 circular(1.1) plots, Researchers each 4 meters 81. Marijuana andlaid traffic accidents Suppose that a tall child with arm span 120 cm and 5 7.0 134 14 10.7 172 diameter, ininterviewed an area where cutting in in New Zealand 907beavers driverswere at age 21. height 118 cm was added to the sample used in this down cottonwood trees. In each plot, they counted the 6 7.1 95 15 10.7 359 They had data on traffic accidents and they asked the study. What effect will adding this child have on the number of stumps fromuse. trees cut are by beavers drivers about marijuana Here data onand the the 7 7.5 96 16 11.2 145 correlation and the slope of the least-squares regresnumberofofaccidents clusters ofcaused beetleby larvae. Ecologists think numbers these drivers at age 8 7.7 78 17 13.7 147 sion line? pg 185 that the new sprouts from stumps are more tender 29 19, broken down by marijuana use at the same age: than 9 7.9 148 18 19.3 255 Correlation will increase, slope will increase. other cottonwood growth, so that beetles prefer them. Marijuana use per year Correlation will increase, slope will stay the same. never 1–10 times 11–50 times 51 ∙ times Correlation will increase, slope will decrease. Drivers 452 229 70 156 Correlation will stay the same, slope will stay the same. Accidents caused 59 36 15 50 Correlation will stay the same, slope will increase. 9/30/13 4:45 PM Starnes-Yates5e_c03_140-205hr2.indd 197 (a) Make a graph that displays the accident rate for each Suppose that the measurements of arm span and class. Is there evidence of an association between height were converted from centimeters to meters marijuana use and traffic accidents? by dividing each measurement by 100. How will this conversion affect the values of r2 and s? (b) Explain why we can’t conclude that marijuana use 2 causes accidents. r will increase, s will increase.

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Look for icons that appear next(c) to selected problems. (d) (e) will guide you to They 75. • an Example that models (a)the problem.

• (b)videos that provide stepby-step instructions for (c) solving the problem. (d)

• (e)earlier sections on which the problem draws (here, 76. Section 2.2). (a)

• (b)examples with the 77. 4-Step State-Plan-DoConclude way of solving problems. (a) (b) (c) (d) (e)

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Freshness (h)

220 210 200 190 180 170 160

(c) Calculate and interpret the residual for the flower that had 2 tablespoons of sugar and looked fresh for 204 hours. (d) Suppose that another group of students conducted a similar experiment using 12 flowers, but included different varieties in addition to carnations. Would you expect the value of r2 for the second group’s data to be greater than, less than, or about the same as the value of r2 for the first group’s data? Explain. After you finish, you can view two example solutions on the book’s Web site (www.whfreeman.com/tps5e). Determine whether you think each solution is “complete,” “substantial,” “developing,” or “minimal.” If the solution is not complete, what improvements would you suggest to the student who wrote it? Finally, your teacher will provide you with a scoring rubric. Score your response and note what, if anything, you would do differently to improve your own score.

REVIEW and PRACTICE for quizzes and tests 0

1

2

3

Sugar (tbsp)

(a) Briefly describe the association shown in the scatterplot. (b) The equation of the least-squares regression line for these data is y^ = 180.8 + 15.8x. Interpret the slope of the line in the context of the study.

Chapter Review Section 3.1: Scatterplots and Correlation In this section, you learned how to explore the relationship between two quantitative variables. As with distributions of a single variable, the first step is always to make a graph. A scatterplot is the appropriate type of graph to investigate associations between two quantitative variables. To describe a scatterplot, be sure to discuss four characteristics: direction, form, strength, and outliers. The direction of an association might be positive, negative, or neither. The form of an association can be linear or nonlinear. An association is strong if it closely follows a specific form. Finally, outliers are any points that clearly fall outside the pattern of the rest of the data. The correlation r is a numerical summary that describes the direction and strength of a linear association. When r > 0, the association is positive, and when r < 0, the association is negative. The correlation will always take values between −1 and 1, with r = −1 and r = 1 indicating a perfectly linear relationship. Strong linear associations have correlations near 1 or −1, while weak linear relationships have correlations near 0. However, it isn’t

Starnes-Yates5e_c03_140-205hr2.indd 200

Use the WHAT DID YOU LEARN? table to guide you to model examples and exercises to verify your mastery of each Learning Objective.

202

possible to determine the form of an association from only the correlation. Strong nonlinear relationships can have a correlation close to 1 or a correlation close to 0, depending on the association. You also learned that outliers can greatly affect value of theofcorrelation The difference between thethe observed value y and the and that value correlation does anot implyResiduals causation. predicted of y is called residual. are That the keyis, we can’t assumealmost that changes in one variable cause changes to understanding everything in this section. To find in the other justregression because line, theyfind have correlathe equation of the variable, least-squares thealine tion closethe to 1sum or –1. that minimizes of the squared residuals. To see if a linear model is appropriate, make a residual plot. If there is Sectionpattern 3.2: Least-Squares Regression no leftover in the residual plot, you know the model In this section, howfitstothe use least-squares is appropriate. To assessyou howlearned well a line data, calculate regressiondeviation lines as ofmodels for relationships the standard the residuals s to estimatebetween the size vari2 ables that have a linear association. It is of a typical prediction error. You can also calculate rimportant , which to understand the difference between the actual data and the model used to describe the data. For example, when you are interpreting the slope of a least-squares regression line, describe the predicted change in the y variable. To emphasize that the model only provides predicted values, least-squares regression lines are always expressed in terms of y^ instead of y.

Section

3.1

144

R3.4

3.1

145, 148

R3.4

Describe the direction, form, and strength of a relationship displayed in a scatterplot and recognize outliers in a scatterplot.

3.1

147, 148

R3.1

Interpret the correlation.

3.1

152

R3.3, R3.4

Understand the basic properties of correlation, including how the correlation is influenced by outliers.

3.1

152, 156, 157

R3.1, R3.2

Use technology to calculate correlation.

3.1

Activity on 152, 171

R3.4

Explain why association does not imply causation.

3.1

Discussion on 156, 190

R3.6

Interpret the slope and y intercept of a least-squares regression line.

3.2

166

R3.2, R3.4

Use the least-squares regression line to predict y for a given x. Explain the dangers of extrapolation.

3.2

167, Discussion on 168 (for extrapolation)

R3.2, R3.4, R3.5

Calculate and interpret residuals.

3.2

169

R3.3, R3.4

Explain the concept of least squares.

3.2

Discussion on 169

R3.5

Determine the equation of a least-squares regression line using technology or computer output.

3.2

Technology Corner on 171, 181

R3.3, R3.4

Construct and interpret residual plots to assess whether a linear model is appropriate.

3.2

Discussion on 175, 180

R3.3, R3.4

9/30/13 4:45 PM

3.2

180

R3.3, R3.5

Describe how the slope, y intercept, standard deviation of the residuals, and r 2 are influenced by outliers.

3.2

Discussion on 188

R3.1

Find the slope and y intercept of the least-squares regression line from the means and standard deviations of x and y and their correlation.

3.2

183

R3.5

r3.3 stats teachers’ cars A random sample of AP® Statistics teachers was asked to report the age (in years) and mileage of their primary vehicles. A scatterplot of the data, a least-squares regression printout, and a residual plot are provided below. Coef 3704 12188

S = 20870.5

A

SE Coef 8268 1492

R-Sq = 83.7%

T 0.45 8.17

P 0.662 0.000

140,000 120,000

30

100,000 80,000 60,000 40,000 20,000

10

0 0

0 100

200

300

400

500

600

(a) Describe the association shown in the scatterplot.

xviii

(b) Point A is the hippopotamus. What effect does this point have on the correlation, the equation of the least-squares regression line, and the standard deviation of the residuals? (c) Point B is the Asian elephant. What effect does this point have on the correlation, the equation of the least-squares regression line, and the standard deviation of the residuals?

r3.2 penguins diving A study of king penguins looked for a relationship between how deep the penguins Starnes-Yates5e_fm_i-xxiii_hr.indd 18 dive to seek food and how long they stay under 31

2

4

6

8

10

12

8

10

12

Tackle the CHAPTER REVIEW EXERCISES for practice in solving problems that test concepts from throughout the chapter.

Age

700

Gestation (days)

9/30/13 4:45 PM

60,000 50,000 40,000 30,000 20,000 10,000 0 10,000 20,000 30,000

Residual

0

201

R-Sq(adj) = 82.4%

160,000

Mileage

Life span (years)

B

20

Relevant Chapter Review Exercise(s)

Make a scatterplot to display the relationship between two quantitative variables.

r3.1 born to be old? Is there a relationship between the gestational period (time from conception to birth) of an animal and its average life span? The figure Predictor shows a scatterplot of the gestational period and avConstant201 Starnes-Yates5e_c03_140-205hr2.indd Age erage life span for 43 species of animals.30 40

Related Example on Page(s)

Identify explanatory and response variables in situations where one variable helps to explain or influences the other.

Chapter 3 Chapter Review Exercises These exercises are designed to help you review the important ideas and methods of the chapter.

summary statistics (the means and standard deviations of two variables and their correlation). As with the correlation, the equation of the least-squares regression line and the values of s and r2 can be greatly influenced by outliers, so be sure to plot the data and note any unusual values before making any calculations.

what Did You learn?

Learning Objective

Interpret the standard deviation of the residuals and r 2 and use these values to assess how well the least-squares regression line D e s c r i b i n g r e l at i o n s h i p s models the relationship between two variables.

CHAPTER 3

Review the CHAPTER SUMMARY to be sure that measures the fraction of the variation in the y variable that is you understand the key accounted for by its linear relationship with the x variable. You also learned how to obtain the equation of a leastinoutput eachandsection. squares regression line concepts from computer from

0

2

4

6

Age

(a) Give the equation of the least-squares regression line for these data. Identify any variables you use. (b) One teacher reported that her 6-year-old car had 65,000 miles on it. Find and interpret its residual.

11/20/13 7:44 PM

dicting when the cherry trees will bloom from the temperature. Which variable did you choose as the explanatory variable? Explain. (b) Use technology to calculate the correlation and the equation of the least-squares regression line. Interpret the correlation, slope, and y intercept of the line in this setting. (c) Suppose that the average March temperature this year was 8.2°C. Would you be willing to use the equation in part (b) to predict the date of first bloom? Explain.

(a) (b) (c) (d)

(d) Calculate and interpret the residual for the year when the average March temperature was 4.5°C. Show your work. (e) Use technology to help construct a residual plot. Describe what you see. r3.5 What’s my grade? In Professor Friedman’s economics course, the correlation between the students’ total scores prior to the final examination and their finalexamination scores is r = 0.6. The pre-exam totals for all students in the course have mean 280 and stanDesignin g deviation s t u D i e30. s The final-exam scores have mean dard 75 and standard deviation 8. Professor Friedman has

r3.6

®

the exam was 300. He decides to predict her finalexam score from her pre-exam total. Find the equation for the appropriate least-squares regression line for Professor Friedman’s prediction. Use the least-squares regression line to predict Julie’s final-exam score. Explain the meaning of the phrase “least squares” in the context of this question. Julie doesn’t think this method accurately predicts how well she did on the final exam. Determine r2. Use this result to argue that her actual score could have been much higher (or much lower) than the predicted value. calculating achievement The principal of a high school read a study that reported a high correlation between the number of calculators owned by high school students and their math achievement. Based on this study, he decides to buy each student at his school two calculators, hoping to improve their math achievement. Explain the flaw in the principal’s reasoning.

and the AP Exam

284

CHAPTER 4

AP1.11 You are planning an experiment to determine AP1.13 The frequency table below summarizes the times the effect of the brand of gasoline and the weight in the last month that patients at the emergency of a car on gas mileage measured in miles per room of a small-city hospital waited to receive gallon. You will use a single test car, adding medical attention. weights so that its total weight is 3000, 3500, or ® waiting time frequency 4000 pounds. The car will drive on a test track Less than 10 minutes 5 at each weight using each of Amoco, Marathon, At least 10 but less than 20 minutes 24 gasoline. is ithe way to C H A P T E Rand 4 Speedway Desig n i n gWhich s tSection uD e sI:best 282 Multiple choice Select the best answer for each question. At least 20 but less than 30 minutes 45 ® organize the study? 199 section 3.2 least-squares regression t3.1 Aand school guidance counselor examines theless number alcoholic (a) Start with 3000 pounds and Amoco run the At least 30 but than 40 minutes 38 beverages for each of 11 regions in Great of extracurricular activities that students do and their Britain was recorded. A scatterplot of spending on car on the test track. Then do 3500 and 4000 (b) Why was one adult chosen at random in each of caffeine normally ingested by that subject in one At least 40 but less than 50 minutes 19 grade point average. The guidance counselor says, alcohol versus spending on tobacco is shown below. pounds. to Marathon and go through theday. During the other study period, the subjects were household to respond to Change the survey? least 50 but less than 60 minutes (c) 0.93.that theAt correlation (b) r2 will increase, will stay the same. “The evidence indicates between Which 7of the following statements is strue? three weightscould in order. to Speedway given placebos. The order in which each subject (c) Explain how undercoverage leadThen to biaschange in At least 60a but less than 70 minutes 2 the more. number ofreceived extracurricular activities student par(d) 6.4. (c) r2 will increase, s will decrease. and do the three weights in order once the two types of capsules was randomized. 6.5 this sample survey. 2 ticipates in and his or her grade point average is close the following represents were of restricted during each of thepossible values Can’t diets tellWhich without seeing the data. (d) r will stay the same, s will stay the same. (b)start Start with and Amoco and run the The(e)subjects’ t4.14 Many people their day3000 with apounds jolt of caffeine 6.0 toMarathon zero.” A correct interpretation ofofthis the and mean times study Atfor theto end eachstatement 2-day study period,on car on theMost test track. Then 75.periods. In addition themedian regression line, thewaiting report the for the (e) r2 will stay the same, s will decrease. from coffee or a soft drink. experts agreechange that towould be thatsubjects were evaluated 2 in which emergency room last month? using a tapping task and then to Speedway without changing the Mumbai measurements says that r = 0.95. This 5.5 people who take in large amounts of caffeine each (a) active students tend to instructed be(a)students with poor grades, they were to press a button 200 times as fast weight. Then add weights to get 3500 pounds and Exercises 79 and 80 refer to the following setting. median = 27 minutes and mean = 24 minutes suggests that day may suffer from physical withdrawal symptoms 66 viceorder. versa. 5.0 as they could. go through the amounts three gasolines in theand same (b) median = 28 andcorrelated, mean = 30arm minutes if they stop ingesting their usual of caffeine. In its recent Fuel Economy Guide, the Environmental (a) although arm span andminutes height are (b) students with good grades tend to be students who Then change to 4000 pounds and do the three (a) How and why was blocking used in the design of = 35 minutes Researchers recruited 11 volunteers who were cafProtection Agency gives data on 1152 vehicles. There are (c) median = 31 minutes and mean span does not predict height very accurately. 4.5 are not involved in many extracurricular activities, gasolines in order again. this experiment? feine dependent and who were willing to take part in a number of outliers, mainly vehicles with very poor gas (d) increases median = minutes andcm mean 39 adminutes (b) height by 35 !0.95 = 0.97 for = each andthe vice versa. (c) Choose a gasolineThe at random, and run car(b) with a caffeine withdrawal experiment. experiment Why did researchers randomize the order in which mileage. If we ignore the outliers, however, the combined ditional centimeter ofactivities arm span. (e)extracurricular median = 45 minutes and mean = 46 minutes students in many are this gasoline at 3000, 3500, and (c) 4000 poundsinvolved in subjects was conducted on two 2-day periods that occurred received the two treatments? 3.0 city and 3.5 highway 4.0gas mileage 4.5 of the other 1120 or so ve(c) 95% of the relationship between height and arm span just as likely to get good grades as bad grades; the same is AP1.14 Boxplots of two data sets are shown. order. Choose one of the two remaining gasolines one week apart. During one of the 2-day periods, each hiclesTobacco is approximately Normal with mean 18.7 miles per (c) Could this experiment have been carried out in a is accounted for by the regression line. true forthen students involved in few extracurricular activities. and again run car at 3000, subject was givenatarandom capsule containing the the amount gallon (mpg) and standard deviation 4.3 mpg. double-blind manner? Explain. (a) The observation (4.5, 6.0) is an outlier. 3500, then 4000 pounds. Do the(d) same with thelinear(d) 95% of the variation in height is accounted there is no relationship between number of activPlotfor 1 by the 79. in of myachevrolet (2.2) The beChevrolet Malibu with (b) There is clear evidence negative association last gasoline. regression ities and grade point average for line. students at this school. a four-cylinder engine has a combined gas mileage of tween spending on alcohol and tobacco. (d) There are nine combinations of weight and gasoline. Plot 2 for by (e) involvement in (e) many extracurricular activities and are accounted 95% of the height measurements mpg. What line percent allplot vehicles have worse gas (c) The equation of the 25 least-squares for of this Run the car several times using each of these combigood grades go hand the in hand. regression line. mileage than the Malibu? would be approximately y^ = 10 − 2x. nations. Make all these runs in random order. 76. One child in the Mumbai study had height 59 cm (d) The correlation for r =(2.2) 0.99. How high must a vehicle’s gas 80.these thedata top is10% t3.2 The British government conducts regular surveys Randomly select amount of for weight and a brand and arm span 60 This child’s residual Section i: Multiple(e) Choice Choose thean best answer Questions AP1.1 to AP1.14. be incorner orderof tothe fallplot in the of household spending. TheBased average weekly house(e)is The below observation in themileage lower-right is top 10% of all oncm. the boxplots, which statement is of gasoline, and run the car on the test track. vehicles? line. (The distribution omits a few high outliers, hold spending (in pounds) on tobacco products and influential for the least-squares (a) −3.2 cm. (c) −1.3 cm. (e) 62.2 cm. true? AP1.1 You look at real Repeat estate ads housesa in Sarasota, thisfor process total of 30 times. AP1.4 For a certain experiment, the available experimenmainly hybrid gas-electric vehicles.) Therats, range of both plotsare is female about the same. cm. 3.2 four cm. Florida. Many houses range from $200,000 to tal (b) units−2.2 are(a) eight of(d) which AP1.12 A linear regression performed $400,000 in price. The few houses was on the water, using the five (F1, F3, andmeans are male (M1, M2, M3, (b)F4) The of both plots approximately 77.F2,Suppose that afour tall child with armare span 120 cm andequal. 81. Marijuana and traffic accidents (1.1) Researchers following data points: A(2, 22), B(10, 4), C(6, 14), in New Zealand interviewed 907 drivers at age 21. however, have prices up to $15 million. Which of M4). There are to cm be four treatment groups, A,used B, inPlot height 118 was added to the sample this1. (c) Plot 2 contains more data points than D(14, 2), best E(18, −4). The They had data on traffic accidents and they asked the the following statements describes theresidual distribu-for which of the C, and study. D. If aWhat randomized block design used,have on the effect will adding thisischild (d) The medians are approximately equal. Starnes-Yates5e_c03_140-205hr2.indd 203 9/30/13 PM on the five points has the largest absolute value? drivers about marijuana use. Here are4:45 data tion of home prices in Sarasota? with thecorrelation experimental unitsslope blocked gender, and the of thebyleast-squares regres(e) Plot 1 is more symmetric than Plot 2. numbers of accidents caused by these drivers at age (a) Ais most (b) likely B (c) C to(d) which of theline? following assignments of treatments sion (a) The distribution skewed the D left, (e) E ® age:29 19, broken down by marijuana use at the same is impossible? and the mean is greater than the median. (a) Correlation will increase, slope will increase. (a) A S (F1, M1), B S (F2, M2), (b) The distribution is most likely skewed to the left, Marijuana use per year (b) Correlation will increase, slope will stay the same. free Response Show all your work. Indicate clearly the methods CS (F3, M3), D S (F4,you M4)use, because you will be graded and theSection mean isii:less than the median. never 1–10 times 11–50 times 51 ∙ times (c) Correlation will increase, slope will decrease. on the correctness of your methods as well as on (b) the accuracy and completeness A S (F1, M2), B S (F2, M3), of your results and explanations. (c) The distribution is roughly symmetric with a few Drivers 452 229 70 156 (d)(F3, Correlation will stayM1) the same, slope will stay the same. CS M4), D S (F4, high outliers, and the mean is approximately equal 59 36 15 50 AP1.15 The manufacturer of exercise machines for fitness theStwo Noteslope that will higher scores indicate Accidents caused (e)(F1, Correlation will stay the same, increase. to the median. (c) A S M2), B (F3,machines. F2), designed newright, elliptical machinesC S gains in fitness. of arm span and (a) Make a graph that displays the accident rate for each 78.(F4, Suppose that measurements (d) The distributioncenters is mosthas likely skewedtwo to the M1), larger D Sthe (M3, M4) are meant to median. increase cardiovascular fitclass. Is there evidence of an association between height to meters and the mean isthat greater than the Machine A centimetersMachine B (d) A S (F4, M1),were B Sconverted (F2, M3),from ness. The two machines are being tested on 30 marijuana use and traffic accidents? by dividing each by (e) The distribution is most likely skewed to the right, M2), D S (F1,measurement M4) 0 100. 2How will this volunteers at a fitness center near the company’s C S (F3, 2 conversion affect the values of r and s? and the mean is less than the median. (b) Explain why we can’t conclude that marijuana use (e) A S (F4,2 M1), B S (F1, M4),5 4 1 0 headquarters. The volunteers are randomly ascauses accidents. (a)(F3, r will increase, s will increase. M2), D S (F2, M3) AP1.2 A child is 40 inches tall,towhich her at theand 90thuse it daily for C S signed one ofplaces the machines

Chapter 3 AP® Statistics Practice Test

Alcohol

Each chapter concludes with an AP STATISTICS PRACTICE TEST. This test includes about 10 AP -style multiple-choice questions and 3 free-response questions.

Cumulative AP® Practice Test 1

876320

2

Four CUMULATIVE AP TESTS simulate the real exam. They are placed after Chapters 4, 7, 10, and 12. The tests expand in length and content coverage from the first through the fourth.

159

percentile of all two children of similar age. The heights months. A measure of cardiovascular fitness For a biology project, you9measure 7 4 1 1 the weight 3 2489 AP1.5 in for children of this age form an approximately is administered at the start of the experiment and grams (g) and the tail length in millimeters (mm) of 61 4 257 Normal distribution with a mean of 38 inches. Based again at the end. The following table contains thea group of mice. The equation of the least-squares 5 359 on this information, what is in thethe standard deviation differences two scores (Afterof– Before) for line for predicting tail length from weight is the heights of all children of this age? predicted tail length = 20 + 3 × weight (a) 0.20 inches (c) 0.65 inches (e) 1.56 inches Which of the following is not correct? ® (b) 0.31 inches (d) 1.21 inches (a) The slope is 3, which indicates that a mouse’s AP1.3 A large set of test scores has mean 60 and standard weight should increase by about 3 grams for each ® The following problem is modeled after actual AP Statistics exam Two statistics students went to a flower shop and randeviation 18. If each score is doubled, and then 5 is additional millimeter of tail length. Starnes-Yates5e_c04_206-285hr.indd 284 9/17/13 4:44 PM free response questions. Your task is to generate a complete, condomly selected 12 carnations. When they got home, the subtracted from the result, the mean and standard (b) The predicted of a mouse that weighs cise response tail in 15length minutes. students prepared 12 identical vases with exactly the same deviation of the new scores are 38 grams is 134 millimeters. amount of water in each vase. They put one tablespoon of (a) mean 115; std. dev. 31. (d) mean 120; std. dev. 31. (c) By Directions: looking at the equation the least-squares line, Show all yourofwork. Indicate clearly the methods sugar in 3 vases, two tablespoons of sugar in 3 vases, and (b) mean 115; std. dev. 36. (e) mean 120; std. dev. 36. youyou canuse, seebecause that theyou correlation between weight will be scored on the correctness of your three tablespoons of sugar in 3 vases. In the remaining 200 C isH positive. AasP on T Ethe R accuracy 3 D e s c r i b i n g r e l at i o n s h i p s (c) mean 120; std. dev. 6. andmethods tail length as well and completeness of your 3 vases, they put no sugar. After the vases were prepared, results and explanations. the students randomly assigned 1 carnation to each vase

FRAPPY! Free Response AP Problem, Yay!

Learn how to answer free-response questions successfully by working the FRAPPY! The Free Response AP® Problem, Yay! that comes just before the Chapter Review in every chapter.

and observed how many hours each flower continued to look fresh. A scatterplot of the data is shown below.

Starnes-Yates5e_c04_206-285hr.indd 282

9/17/13 4:44 PM

240 230 Starnes-Yates5e_c03_140-205hr2.indd 199

Freshness (h)

220 210 200 190 180 170 160 0

1

2

3

Sugar (tbsp)

(a) Briefly describe the association shown in the scatterplot. (b) The equation of the least-squares regression line for these data is y^ = 180.8 + 15.8x. Interpret the slope of the line in the context of the study.

(c) Calculate and interpret the residual for the flower that had 2 tablespoons of sugar and looked fresh for 204 hours. (d) Suppose that another group of students conducted a similar experiment using 12 flowers, but in-9/30/13 cluded different varieties in addition to carnations. Would you expect the value of r2 for the second group’s data to be greater than, less than, or about the same as the value of r2 for the first group’s data? Explain.

4:45 PM

After you finish, you can view two example solutions on the book’s Web site (www.whfreeman.com/tps5e). Determine whether you think each solution is “complete,” “substantial,” “developing,” or “minimal.” If the solution is not complete, what improvements would you suggest to the student who wrote it? Finally, your teacher will provide you with a scoring rubric. Score your response and note what, if anything, you would do differently to improve your own score.

xix

Chapter Review Starnes-Yates5e_fm_i-xxiii_hr.indd 19

11/20/13 7:44 PM

Use Technology to discover and analyze Use technology as a tool for discovery and analysis. TECHNOLOGY CORNERS give step-by-step instructions for using the TI-83/84 and TI-89 calculator. 192 CHAPTER 3 D e s c r i b i n g r e l at i o n s h i p s Instructions for the TI-Nspire are in an end-of-book appendix. HP Prime instructions are on the book’s Web site and in the e-Book. 150

CHAPTER 3

D e s c r i b i n g r e l at i o n s h i p s

Summary

section 3.2

You can access video instructions for the Technology Corners through the e-Book or on the book’s Web site.

A regression line is a straight line that describes how a response variable y changes as an explanatory variable x changes. You can use a regression line to predict the value of y for any value of x by substituting this x into the equation of the line. TI-nspire instructions in Appendix B; HP Prime instructions on the book’s Web•site The slope b of a regression line y^ = a + bx is the rate at which the predicted response y^ changes along the line as the explanatory variable x changes. Specifically, Making scatterplots with technology is much easier than constructing them by hand. We’ll use the SEC football data b is the predicted change in y when x increases by 1 unit. from page 146 to show how to construct a scatterplot on a TI-83/84 or TI-89. • The y intercept a of a regression line y^ = a + bx is the predicted response y^ • Enter the data values into your lists. Put the points per game in L1/list1 and the number of wins in L2/list2. when the explanatory variable x equals 0. This prediction is of no statistical • Define a scatterplot in the statistics plot menu (press F2 on the TI-89). Specify the settings shown below. use unless x can actually take values near 0. • Avoid extrapolation, the use of a regression line for prediction using values of the explanatory variable outside the range of the data from which the line was calculated. • The most common method of fitting a line to a scatterplot is least squares. The least-squares regression line is the straight line y^ = a + bx that minimizes the sum of the squares of the vertical distances of the observed points from the line. • You can examine the fit of a regression line by studying the residuals, which are the differences between the observed and predicted values of y. Be on the • Use ZoomStat (ZoomData on the TI-89) to obtain a graph. The calculator will set the window dimensions automatically lookout for patterns in the residual plot, which indicate that a linear model by looking at the values in L1/list1 and L2/list2. may not be appropriate. • The standard deviation of the residuals s measures the typical size of the prediction errors (residuals) when using the regression line. • The coefficient of determination r2 is the fraction of the variation in the response variable that is accounted for by least-squares regression on the explanatory variable. The least-squares regression line of y on x is the line with slope b = r(sy/sx) 196 CHAPTER 3 D e s c r i b i n g r e l at i o n s h i p s• andonto intercept a = y– − bx–. This line always passes through the point (x–, y–). Notice that there are no scales on the axes and that the axes are not labeled. If you copy a scatterplot from your calculator your paper, make sure that you scale and label the axes. • Correlation and regression must be interpreted with caution. Plot the data to be 20 brain thatsure responds to physical pain goes up as distress that the relationship is roughly linear and to detect outliers. Also look for infrom social exclusion goes up. A scatterplot shows ap® ExaM10tIp If you are asked to make a scatterplot on a free-response question, be sure to fluential observations, individual points that substantially change the correlation a moderately strong, linear relationship. The figure label and scale both axes. Don’t just copy an unlabeled calculator graph directly onto yourorpaper. the regression line. Outliers in x are often influential for the regression line. below shows Minitab regression output for these data. 0 • Most of all, be careful not to conclude that there is a cause-and-effect rela10 tionship between two variables just because they are strongly associated. •

ScattERplotS on tHE calcUlatoR

Residual

7. TECHNOLOGY CORNER

Measuring Linear Association: Correlation 20

Some people refer to r as the “correlation coefficient.”

c

Find the Technology Corners easily by consulting the summary table at the end of each section or the complete table inside the back cover of the book.

59. Merlins breeding Exercise 13 (page 160) gives data on192 Starnes-Yates5e_c03_140-205hr2.indd the number of breeding pairs of merlins in an isolated area in each of seven years and the percent of males who returned the next year. The data show that the percent returning is lower after successful breeding seasons and that the relationship is roughly linear. The figure below shows Minitab regression output for these data. Regression Analysis: Percent return versus Breeding pairs Predictor Constant Breeding pairs S = 7.76227

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relationship with social distress score? Explain what each of these values means in this setting. (c) Use the information in the figure to find the correla58.DEFInItIon: Do heavier people burn more energy? Refer to Exercorrelation r tion r between social distress score and brain activity. cises 48 and 50. For the regression you performed earlier, do you know whether the sign of r is + or −? The correlation the direction and strength of the linear How relationship r2 = 0.768 and sr=measures 95.08. Explain what each of these between two quantitative variables. (d) Interpret the value of s in this setting. values means in this setting.

pg 181

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!

n

A scatterplot displays the direction, form, and strength of the relationship between two quantitative10variables. 11 12 Linear 13 14 relationships 15 16 17 are particularly important because a straight line is a simple pattern Agethat is quite common. A linear relationship is strong if the points lie close to a straight line and weak if they are widely scattered about a (a) Calculate and interpret the residual for the student 3.2 TECHNOLOGY line. Unfortunately, our eyes are not good judges of how strong a linear relationship is. who was 141 cm tall at age 10. (a) same Whatdata, is the equation of the least-squares regression CORNERs The two scatterplots in Figure 3.5 (on the facing page) show the autio (b) Is a linear model appropriate for these data? Explain. line for predicting brain activity from social distress but the graph on the right is drawn smaller inTI-nspire a large field. The right-hand Instructions in Appendix B; HP Prime instructions on the book’s Web site score? Use the equation to predict brain activity for (c) Interpret s. graph seemsthe to value show of a stronger linear relationship. distress score 2.0. 2 it’s value easy of to rbe scales or by the social amount of space (d) Because Interpret the . fooled by different 8. least-squares regression lines on the calculator What percent of the variation in brain activity among around the cloud of points in a 47 scatterplot, we need to use (b) a numerical measure 57. bird colonies Refer to Exercises and 49. For9. the residual plots on these the calculator subjects is accounted for by the straight-line to supplement graph. earlier, correlation is and the smeasure regression youthe performed r2 = 0.56 = 3.67. we use.

Coef 266.07 -6.650

SE Coef 52.15 1.736

R-Sq = 74.6%

T 5.10 -3.83

P 0.004 0.012

R-Sq(adj) = 69.5%

(a) What is the equation of the least-squares regression line for predicting the percent of males that return from the number of breeding pairs? Use the equation to predict the percent of returning males after a season with 30 breeding pairs. (b) What percent of the year-to-year variation in percent of returning males is accounted for by the straightline relationship with number of breeding pairs the previous year? (c) Use the information in the figure to find the correlation r between percent of males that return and number of breeding pairs. How do you know whether the sign of r is + or −? (d) Interpret the value of s in this setting. 60. Does social rejection hurt? Exercise 14 (page 161) gives data from a study that shows that social exclusion causes “real pain.” That is, activity in an area of the

61. husbands and wives The mean height of married American women in their early twenties is 64.5 inches and the standard deviation is 2.5 inches. The mean height of married men the same age is 68.5 inches, with standard deviation 2.7 inches. The correlation between 4:44 PMis about r = 0.5. the heights of husbands9/30/13 and wives

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(a) Find the equation of the least-squares regression line for predicting a husband’s height from his wife’s height for married couples in their early 20s. Show your work.

Other types of software displays, including Minitab, Fathom, and applet screen captures, appear throughout help youthink learn tobehavread 62. thethe stockbook marketto Some people that the ior of the stock market in January predicts its behavior many differentvariable kinds of for and the restinterpret of the year. Take the explanatory x to be the percent change in a stock market index in output. January and the response variable y to be the change (b) Suppose that the height of a randomly selected wife was 1 standard deviation below average. Predict the height of her husband without using the least-squares line. Show your work.

in the index for the entire year. We expect a positive correlation between x and y because the change during January contributes to the full year’s change. Calculation from data for an 18-year period gives x– = 1.75% sx = 5.36% y– = 9.07% sy = 15.35% r = 0.596

(a) Find the equation of the least-squares line for predicting full-year change from January change. Show your work. (b) Suppose that the percent change in a particular January was 2 standard deviations above average. Predict the percent change for the entire year, without using the least-squares line. Show your work.

12/2/13 11:15 AM

Overview

What Is Statistics?

Does listening to music while studying help or hinder learning? If an athlete fails a drug test, how sure can we be that she took a banned substance? Does having a pet help people live longer? How well do SAT scores predict college success? Do most people recycle? Which of two diets will help obese children lose more weight and keep it off? Should a poker player go “all in” with pocket aces? Can a new drug help people quit smoking? How strong is the evidence for global warming? These are just a few of the questions that statistics can help answer. But what is statistics? And why should you study it?

Statistics Is the Science of Learning from Data Data are usually numbers, but they are not “just numbers.” Data are numbers with a context. The number 10.5, for example, carries no information by itself. But if we hear that a family friend’s new baby weighed 10.5 pounds at birth, we congratulate her on the healthy size of the child. The context engages our knowledge about the world and allows us to make judgments. We know that a baby weighing 10.5 pounds is quite large, and that a human baby is unlikely to weigh 10.5 ounces or 10.5 kilograms. The context makes the number meaningful. In your lifetime, you will be bombarded with data and statistical information. Poll results, television ratings, music sales, gas prices, unemployment rates, medical study outcomes, and standardized test scores are discussed daily in the media. Using data effectively is a large and growing part of most professions. A solid understanding of statistics will enable you to make sound, data-based decisions in your career and everyday life.

Data Beat Personal Experiences It is tempting to base conclusions on your own experiences or the experiences of those you know. But our experiences may not be typical. In fact, the incidents that stick in our memory are often the unusual ones.

Do cell phones cause brain cancer? Italian businessman Innocente Marcolini developed a brain tumor at age 60. He also talked on a cellular phone up to 6 hours per day for 12 years as part of his job. Mr. Marcolini’s physician suggested that the brain tumor may have been caused by cell-phone use. So Mr. Marcolini decided to file suit in the Italian court system. A court ruled in his favor in October 2012. Several statistical studies have investigated the link between cell-phone use and brain cancer. One of the largest was conducted by the Danish Cancer Society. Over 350,000 residents of Denmark were included in the study. Researchers compared the brain-cancer rate for the cell-phone users with the rate in the general population. The result: no statistical difference in brain-cancer rates.1 In fact, most studies have produced similar conclusions. In spite of the evidence, many people (like Mr. Marcolini) are still convinced that cell phones can cause brain cancer. In the public’s mind, the compelling story wins every time. A statistically literate person knows better. Data are more reliable than personal experiences because they systematically describe an overall picture rather than focus on a few incidents.

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Where the Data Come from Matters Are you kidding me? The famous advice columnist Ann Landers once asked her readers, “If you had it to do over again, would you have children?” A few weeks later, her column was headlined “70% OF PARENTS SAY KIDS NOT WORTH IT.” Indeed, 70% of the nearly 10,000 parents who wrote in said they would not have children if they could make the choice again. Do you believe that 70% of all parents regret having children? You shouldn’t. The people who took the trouble to write Ann Landers are not representative of all parents. Their letters showed that many of them were angry with their children. All we know from these data is that there are some unhappy parents out there. A statistically designed poll, unlike Ann Landers’s appeal, targets specific people chosen in a way that gives all parents the same chance to be asked. Such a poll showed that 91% of parents would have children again. Where data come from matters a lot. If you are careless about how you get your data, you may ­announce 70% “No” when the truth is close to 90% “Yes.”

Who talks more—women or men? According to Louann Brizendine, author of The Female Brain, women say nearly three times as many words per day as men. Skeptical researchers devised a study to test this claim. They used electronic devices to record the talking patterns of 396 university students from Texas, Arizona, and Mexico. The device was programmed to record 30 seconds of sound every 12.5 minutes without the carrier’s knowledge. What were the results? According to a published report of the study in Scientific American, “Men showed a slightly wider variability in words uttered. . . . But in the end, the sexes came out just about even in the daily averages: women at 16,215 words and men at 15,669.”2  When asked where she got her figures, Brizendine admitted that she used unreliable sources.3 The most important information about any statistical study is how the data were produced. Only carefully designed studies produce results that can be trusted.

Always Plot Your Data Yogi Berra, a famous New York Yankees baseball player known for his unusual quotes, had this to say: “You can observe a lot just by watching.” That’s a motto for learning from data. A carefully chosen graph is often more instructive than a bunch of numbers.

Do people live longer in wealthier countries? The Gapminder Web site, www.gapminder.org, provides loads of data on the health and well-being of the world’s inhabitants. The graph on the next pages displays some data from Gapminder.4 The individual points represent all the world’s nations for which data are available. Each point shows the income per person and life expectancy in years for one country. We expect people in richer countries to live longer. The overall pattern of the graph does show this, but the relationship has an interesting shape. Life expectancy rises very quickly as personal income increases and then levels off. People in very rich countries like the United States live no longer than people in poorer but not extremely poor nations. In some less wealthy countries, people live longer than in the United States. Several other nations stand out in the graph. What’s special about each of these countries?

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80

Life expectancy

United States

Qatar

America

70

East Asia & Pacific Europe & Central Asia Gabon

Middle East & North Africa

60

South Asia

South Africa

Sub-Saharan Africa Equatorial Guinea 50

Graph of the life expectancy of people in many nations against each nation’s income per person in 2012.

00 ,0 90

00 ,0 80

00 ,0 70

00 ,0 60

00 ,0 50

00 ,0 40

00 ,0 30

00 ,0 20

10

,0

00

Botswana

Income per person in 2012

Variation Is Everywhere Individuals vary. Repeated measurements on the same individual vary. Chance outcomes—like spins of a roulette wheel or tosses of a coin—vary. Almost everything varies over time. Statistics provides tools for understanding variation.

Have most students cheated on a test? Researchers from the Josephson Institute were determined to find out. So they surveyed about 23,000 students from 100 randomly selected schools (both public and private) nationwide. The question they asked was “How many times have you cheated during a test at school in the past year?” Fifty-one percent said they had cheated at least once.5 If the researchers had asked the same question of all high school students, would exactly 51% have answered “Yes”? Probably not. If the Josephson Institute had selected a different sample of about 23,000 students to respond to the survey, they would probably have gotten a different estimate. Variation is everywhere! Fortunately, statistics provides a description of how the sample results will vary in relation to the actual population percent. Based on the sampling method that this study used, we can say that the estimate of 51% is very likely to be within 1% of the true population value. That is, we can be quite confident that between 50% and 52% of all high school students would say that they have cheated on a test. Because variation is everywhere, conclusions are uncertain. Statistics gives us a language for talking about uncertainty that is understood by statistically literate people everywhere.

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Chapter

1

Introduction 2 Data Analysis: Making Sense of Data Section 1.1 7 Analyzing Categorical Data Section 1.2 25 Displaying Quantitative Data with Graphs Section 1.3 48 Describing Quantitative Data with Numbers Free Response AP® Problem, YAY! 74 Chapter 1 Review 74 Chapter 1 Review Exercises 76 Chapter 1 AP® Statistics Practice Test 78

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Exploring Data case study

Do Pets or Friends Help Reduce Stress? If you are a dog lover, having your dog with you may reduce your stress level. Does having a friend with you reduce stress? To examine the effect of pets and friends in stressful situations, researchers recruited 45 women who said they were dog lovers. Fifteen women were assigned at random to each of three groups: to do a stressful task alone, with a good friend present, or with their dogs present. The stressful task was to count backward by 13s or 17s. The woman’s average heart rate during the task was one measure of the effect of stress. The table below shows the data.1 Average heart rates during stress with a pet (P), with a friend (F), and for the control group (C) Group

Rate

Group

P

 69.169

P

F

 99.692

Rate

Group

Rate

Group

Rate

68.862

C

84.738

C

75.477

C

87.231

C

84.877

C

62.646

P

 70.169

P

64.169

P

58.692

P

70.077

C

 80.369

C

91.754

P

79.662

F

88.015

C

 87.446

C

87.785

P

69.231

F

81.600

P

 75.985

F

91.354

C

73.277

F

86.985

F

 83.400

F

100.877

C

84.523

F

92.492

F

102.154

C

77.800

C

70.877

P

72.262

P

86.446

P

97.538

F

89.815

P

65.446

F

80.277

P

85.000

F

98.200

C

90.015

F

101.062

F

76.908

C

99.046

F

97.046

P

69.538

Based on the data, does it appear that the presence of a pet or friend reduces heart rate during a stressful task? In this chapter, you’ll develop the tools to help answer this question.

1

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2



CHAPTER 1

E x p l o r i n g Data

Introduction

What You Will Learn

Data Analysis: Making Sense of Data By the end of the section, you should be able to:

• Identify the individuals and variables in a set of data.

• Classify variables as categorical or quantitative.

Statistics is the science of data. The volume of data available to us is overwhelming. For example, the Census Bureau’s American Community Survey collects data from 3,000,000 housing units each year. Astronomers work with data on tens of millions of galaxies. The checkout scanners at Walmart’s 10,000 stores in 27 countries record hundreds of millions of transactions every week. In all these cases, the data are trying to tell us a story—about U.S. households, objects in space, or Walmart shoppers. To hear what the data are saying, we need to help them speak by organizing, displaying, summarizing, and asking questions. That’s data analysis.

Individuals and Variables Any set of data contains information about some group of individuals. The characteristics we measure on each individual are called variables.

Definition: Individuals and variables Individuals are the objects described by a set of data. Individuals may be people, animals, or things. A variable is any characteristic of an individual. A variable can take different values for different individuals. A high school’s student data base, for example, includes data about every currently enrolled student. The students are the individuals described by the data set. For each individual, the data contain the values of variables such as age, gender, grade point average, homeroom, and grade level. In practice, any set of data is accompanied by background information that helps us understand the data. When you first meet a new data set, ask yourself the following questions: 1. Who are the individuals described by the data? How many individuals are there? 2. What are the variables? In what units are the variables recorded? Weights, for example, might be recorded in grams, pounds, thousands of pounds, or kilograms. We could follow a newspaper reporter’s lead and extend our list of questions to include Why, When, Where, and How were the data produced? For now, we’ll focus on the first two questions. Some variables, like gender and grade level, assign labels to individuals that place them into categories. Others, like age and grade point average (GPA), take numerical values for which we can do arithmetic. It makes sense to give an average GPA for a group of students, but it doesn’t make sense to give an “average” gender.

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3

Introduction Data Analysis: Making Sense of Data

Definition: Categorical variable and quantitative variable A categorical variable places an individual into one of several groups or categories. A quantitative variable takes numerical values for which it makes sense to find an average.

EXAMPLE

c

Not every variable that takes number values is quantitative. Zip code is autio one ­example. Although zip codes are numbers, it doesn’t make sense to talk about the average zip code. In fact, zip codes place individuals (people or dwellings) into categories based on location. Some variables—such as gender, race, and o­ ccupation—are categorical by nature. Other categorical variables are created by grouping values of a quantitative variable into classes. For instance, we could classify people in a data set by age: 0–9, 10–19, 20–29, and so on. The proper method of analysis for a variable depends on whether it is categorical or quantitative. As a result, it is important to be able to distinguish these two types of variables. The type of data determines what kinds of graphs and which numerical summaries are appropriate.

!

n

AP® EXAM TIP ​If you learn to distinguish categorical from quantitative variables now, it will pay big rewards later. You will be expected to analyze categorical and quantitative variables correctly on the AP® exam.

Census at School Data, individuals, and variables CensusAtSchool is an international project that collects data about primary and secondary school students using surveys. Hundreds of thousands of students from Australia, Canada, New Zealand, South Africa, and the United Kingdom have taken part in the project since 2000. Data from the surveys are available at the project’s Web site (www.censusatschool.com). We used the site’s “Random Data Selector” to choose 10 Canadian students who completed the survey in a recent year. The table below displays the data.

There is at least one suspicious value in the data table. We doubt that the girl who is 166  cm tall really has a wrist circumference of 65 mm (about 2.6 inches). Always look to be sure the values make sense!

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Language spoken

Handed

Height (cm)

Male

1

Right

175

180

In person

Ontario

Female

1

Right

162.5

160

In person

Alberta

Male

1

Right

178

174

Facebook

Ontario

Male

2

Right

169

160

Cell phone

Ontario

Female

2

Right

166

 65

In person

Nunavut

Male

1

Right

168.5

160

Text messaging

Ontario

Female

1

Right

166

165

Cell phone

Ontario

Male

4

Left

157.5

147

Text Messaging

Ontario

Female

2

Right

150.5

187

Text Messaging

Ontario

Female

1

Right

171

180

Text Messaging

Province

Gender

Saskatchewan

Wrist Preferred circum. (mm) communication

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4



CHAPTER 1

E x p l o r i n g Data

Problem: (a) Who are the individuals in this data set? (b) What variables were measured? Identify each as categorical or quantitative. (c) Describe the individual in the highlighted row. We’ll see in Chapter 4 why choosing at random, as we did in this example, is a good idea.

To make life simpler, we sometimes refer to “categorical data” or “quantitative data” instead of identifying the variable as categorical or quantitative.

Solution: (a) The individuals are the 10 randomly selected Canadian students who participated in the

CensusAtSchool survey. (b) The seven variables measured are the province where the student lives (categorical), gender (categorical), number of languages spoken (quantitative), dominant hand (categorical), height (quantitative), wrist circumference (quantitative), and preferred communication method (categorical). (c) ​This student lives in Ontario, is male, speaks four languages, is left-handed, is 157.5 cm tall (about 62 inches), has a wrist circumference of 147 mm (about 5.8 inches), and prefers to communicate via text messaging. For Practice Try Exercise 3 Most data tables follow the format shown in the example—each row is an individual, and each column is a variable. Sometimes the individuals are called cases. A variable generally takes values that vary (hence the name “variable”!). Categorical variables sometimes have similar counts in each category and sometimes don’t. For instance, we might have expected similar numbers of males and ­females in the CensusAtSchool data set. But we aren’t surprised to see that most students are right-handed. Quantitative variables may take values that are very close ­together or values that are quite spread out. We call the pattern of variation of a variable its distribution.

Definition: Distribution The distribution of a variable tells us what values the variable takes and how often it takes these values. Section 1.1 begins by looking at how to describe the distribution of a single categorical variable and then examines relationships between categorical variables. Sections 1.2 and 1.3 and all of Chapter 2 focus on describing the distribution of a quantitative variable. Chapter 3 investigates relationships between two quantitative variables. In each case, we begin with graphical displays, then add numerical summaries for a more complete description.

How to Explore Data • Begin by examining each variable by itself. Then move on to study relationships among the variables. • Start with a graph or graphs. Then add numerical summaries.

Check Your Understanding

Jake is a car buff who wants to find out more about the vehicles that students at his school drive. He gets permission to go to the student parking lot and record some data. Later, he does some research about each model of car on the Internet. Finally, Jake

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Introduction Data Analysis: Making Sense of Data

5

makes a spreadsheet that includes each car’s model, year, color, number of cylinders, gas mileage, weight, and whether it has a navigation system. 1. Who are the individuals in Jake’s study? 2. ​What variables did Jake measure? Identify each as categorical or quantitative.

From Data Analysis to Inference Sometimes, we’re interested in drawing conclusions that go beyond the data at hand. That’s the idea of inference. In the CensusAtSchool example, 9 of the 10 randomly selected Canadian students are right-handed. That’s 90% of the sample. Can we conclude that 90% of the population of Canadian students who participated in CensusAtSchool are right-handed? No. If another random sample of 10 students was selected, the percent who are right-handed might not be exactly 90%. Can we at least say that the actual population value is “close” to 90%? That depends on what we mean by “close.” The following Activity gives you an idea of how statistical inference works.

ACTIVITY MATERIALS: Bag with 25 beads (15 of one color and 10 of another) or 25 identical slips of paper (15 labeled “M” and 10 labeled “F”) for each student or pair of students

Hiring discrimination—it just won’t fly! An airline has just finished training 25 pilots—15 male and 10 female—to ­become captains. Unfortunately, only eight captain positions are available right now. Airline managers announce that they will use a lottery to determine which pilots will fill the available positions. The names of all 25 pilots will be written on identical slips of paper. The slips will be placed in a hat, mixed thoroughly, and drawn out one at a time until all eight captains have been identified. A day later, managers announce the results of the lottery. Of the 8 captains chosen, 5 are female and 3 are male. Some of the male pilots who weren’t selected suspect that the lottery was not carried out fairly. One of these pilots asks your statistics class for advice about whether to file a grievance with the pilots’ union. The key question in this possible discrimination case seems to be: Is it plausible (believable) that these results happened just by chance? To find out, you and your classmates will simulate the lottery process that airline managers said they used. 1.  Mix the beads/slips thoroughly. Without looking, remove 8 beads/slips from the bag. Count the number of female pilots selected. Then return the beads/slips to the bag. 2.  Your teacher will draw and label a number line for a class dotplot. On the graph, plot the number of females you got in Step 1. 3.  Repeat Steps 1 and 2 if needed to get a total of at least 40 simulated lottery results for your class. 4.  Discuss the results with your classmates. Does it seem believable that airline managers carried out a fair lottery? What advice would you give the male pilot who contacted you? 5.  Would your advice change if the lottery had chosen 6 female (and 2 male) pilots? What about 7 female pilots? Explain.

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6



CHAPTER 1

E x p l o r i n g Data

Our ability to do inference is determined by how the data are produced. Chapter 4 discusses the two main methods of data production—sampling and experiments—and the types of conclusions that can be drawn from each. As the Activity illustrates, the logic of inference rests on asking, “What are the chances?” Probability, the study of chance behavior, is the topic of Chapters 5 through 7. We’ll introduce the most common inference techniques in Chapters 8 through 12.

Introduction

Summary •

•

•

Introduction

A data set contains information about a number of individuals. Individuals may be people, animals, or things. For each individual, the data give values for one or more variables. A variable describes some characteristic of an individual, such as a person’s height, gender, or salary. Some variables are categorical and others are quantitative. A categorical variable assigns a label that places each individual into one of several groups, such as male or female. A quantitative variable has numerical values that measure some characteristic of each individual, such as height in centimeters or salary in dollars. The distribution of a variable describes what values the variable takes and how often it takes them.

Exercises

1. Protecting wood  How can we help wood surfaces resist weathering, especially when restoring historic wooden buildings? In a study of this question, researchers prepared wooden panels and then exposed them to the weather. Here are some of the variables recorded: type of wood (yellow poplar, pine, cedar); type of water repellent (solvent-based, water-based); paint thickness (millimeters); paint color (white, gray, light blue); weathering time (months). Identify each variable as categorical or quantitative. 2. Medical study variables  Data from a medical study contain values of many variables for each of the people who were the subjects of the study. Here are some of the variables recorded: gender (female or male); age (years); race (Asian, black, white, or other); smoker (yes or no); systolic blood pressure (millimeters of mercury); level of calcium in the blood (micrograms per milliliter). Identify each as categorical or quantitative. 3. A class survey  Here is a small part of the data set that describes the students in an AP® Statistics class. The pg 3 data come from anonymous responses to a questionnaire filled out on the first day of class.

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The solutions to all exercises numbered in red are found in the Solutions Appendix, starting on page S-1.

Homework time (min)

Favorite music

Pocket change (cents)

Gender

Hand

Height (in.)

F

L

65

200

Hip-hop

 50

M

L

72

 30

Country

 35

M

R

62

 95

Rock

 35

F

L

64

120

Alternative

  0

M

R

63

220

Hip-hop

  0

F

R

58

 60

Alternative

 76

F

R

67

150

Rock

215

(a) What individuals does this data set describe? (b) What variables were measured? Identify each as categorical or quantitative. (c) Describe the individual in the highlighted row. 4. Coaster craze  Many people like to ride roller coasters. Amusement parks try to increase attendance by building exciting new coasters. The following table displays data on several roller coasters that were opened in a recent year.2

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7

Section 1.1 Analyzing Categorical Data

Roller coaster

Type

Height (ft)

Design

Speed (mph)

Duration (s)  70

Weight (lb)

Age (yr)

Travel to work (min)

Wild Mouse

Steel

49.3

Sit down

28

Terminator

Wood

95

Sit down

Manta

Steel

140

Flying

Gender

Income last year ($)

50.1

180

187

66

 0

Ninth grade

1

24,000

56

155

158

66 54

n/a

High school grad

2

0

10

Assoc. degree

2

11,900

School

Prowler

Wood

102.3

Sit down

51.2

150

176

Diamondback

Steel

230

Sit down

80

180

339

37

10

Assoc. degree

1

6000

 91

27

10

Some college

2

30,000

(a) ​What individuals does this data set describe?

155

18

n/a

High school grad

2

0

(b) What variables were measured? Identify each as categorical or quantitative.

213

38

15

Master’s degree

2

125,000

(c) ​Describe the individual in the highlighted row.

194

40

 0

High school grad

1

800

221

18

20

High school grad

1

2500

193

11

n/a

Fifth grade

1

0

5. Ranking colleges  Popular magazines rank colleges and universities on their “academic quality” in serving undergraduate students. Describe two categorical variables and two quantitative variables that you might record for each institution. 6. Students and TV  You are preparing to study the television-viewing habits of high school students. Describe two categorical variables and two quantitative variables that you might record for each student. Multiple choice: Select the best answer. Exercises 7 and 8 refer to the following setting. At the Census Bureau Web site www.census.gov, you can view detailed data collected by the American Community Survey. The following table includes data for 10 people chosen at random from the more than 1 million people in households contacted by the survey. “School” gives the highest level of education completed.

1.1 What You Will Learn

7. The individuals in this data set are (a) ​households. (b) ​people. (c) ​adults. (d) ​120 variables. (e) ​columns. 8. This data set contains (a) ​7 variables, 2 of which are categorical. (b) ​7 variables, 1 of which is categorical. (c) 6 variables, 2 of which are categorical. (d) 6 variables, 1 of which is categorical. (e) None of these.

Analyzing Categorical Data By the end of the section, you should be able to:

• Display categorical data with a bar graph. Decide if it would be appropriate to make a pie chart. • Identify what makes some graphs of categorical data deceptive. • Calculate and display the marginal distribution of a categorical variable from a two-way table.

• Calculate and display the conditional distribution of a categorical variable for a particular value of the other categorical variable in a two-way table. • Describe the association between two categorical variables by comparing appropriate conditional distributions.

The values of a categorical variable are labels for the categories, such as “male” and “female.” The distribution of a categorical variable lists the categories and gives either the count or the percent of individuals who fall within each category. Here’s an example.

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CHAPTER 1

EXAMPLE

E x p l o r i n g Data

Radio Station Formats Distribution of a categorical variable The radio audience rating service Arbitron places U.S radio stations into categories that describe the kinds of programs they broadcast. Here are two different tables showing the distribution of station formats in a recent year:3 Frequency table Format

Relative frequency table

Count of stations

Format

Percent of stations

Adult contemporary

1556

Adult contemporary

Adult standards

1196

Adult standards

8.6

Contemporary hit

4.1

Contemporary hit

569

11.2

Country

2066

Country

14.9

News/Talk/Information

2179

News/Talk/Information

15.7

Oldies

1060

Oldies

Religious

2014

Religious

7.7 14.6

Rock

869

Rock

6.3

Spanish language

750

Spanish language

5.4

Other formats Total

1579 13,838

Other formats

11.4

Total

99.9

In this case, the individuals are the radio stations and the variable being measured is the kind of programming that each station broadcasts. The table on the left, which we call a frequency table, displays the counts (frequencies) of stations in each format category. On the right, we see a relative frequency table of the data that shows the percents (relative frequencies) of stations in each format category. It’s a good idea to check data for consistency. The counts should add to 13,838, the total number of stations. They do. The percents should add to 100%. In fact, they add to 99.9%. What happened? Each percent is rounded to the nearest tenth. The exact percents would add to 100, but the rounded percents only come close. This is roundoff error. Roundoff errors don’t point to mistakes in our work, just to the effect of rounding off results.

Bar Graphs and Pie Charts Columns of numbers take time to read. You can use a pie chart or a bar graph to display the distribution of a categorical variable more vividly. Figure 1.1 illustrates both displays for the distribution of radio stations by format. Pie charts show the distribution of a categorical variable as a “pie” whose slices are sized by the counts or percents for the categories. A pie chart must include all the categories that make up a whole. In the radio station example, we needed the “Other formats” category to complete the whole (all radio stations) and allow us to make a pie chart. Use a pie chart only when you want to emphasize each

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9

Section 1.1 Analyzing Categorical Data This bar has height 14.9% because 14.9% of the radio stations have a “Country” format.

This slice occupies 14.9% of the pie because 14.9% of the radio stations have a “Country” format. Contemporary hit

14

Percent of stations

Country

16

Adult standards

Adult contemporary

News/Talk

Other Oldies

12 10 8 6 4 2

(a)

Spanish

(b)

r th e

ish

O

an

oc k R

Rock

Sp

on t A dS ta n C on H it C ou nt N r ew y s/T al k O ld ie R s el ig io us

0

A dC

Religious

Radio station format

FIGURE 1.1  ​(a) Pie chart and (b) bar graph of U.S. radio stations by format.

category’s relation to the whole. Pie charts are awkward to make by hand, but technology will do the job for you.

Bar graphs are also called bar charts.

EXAMPLE

Bar graphs represent each category as a bar. The bar heights show the category counts or percents. Bar graphs are easier to make than pie charts and are also easier to read. To convince yourself, try to use the pie chart in Figure 1.1 to estimate the percent of radio stations that have an “Oldies” format. Now look at the bar graph—it’s easy to see that the answer is about 8%. Bar graphs are also more flexible than pie charts. Both graphs can display the distribution of a categorical variable, but a bar graph can also compare any set of quantities that are measured in the same units.

Who Owns an MP3 Player? Choosing the best graph to display the data Portable MP3 music players, such as the Apple iPod, are popular—but not equally popular with people of all ages. Here are the percents of people in various age groups who own a portable MP3 player, according to an Arbitron survey of 1112 randomly selected people.4

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CHAPTER 1

E x p l o r i n g Data

% who own mp3 player

Age group (years)

Percent owning an MP3 player

12 to 17

54

18 to 24

30

25 to 34

30

35 to 54

13

55 and older

5

60

Problem: (a) Make a well-labeled bar graph to display the data. Describe what you see.

50

(b) ​Would it be appropriate to make a pie chart for these data? Explain.

40

Solution:

30

(a) We start by labeling the axes: age group goes on the horizontal axis, and percent who own an MP3 player goes on the vertical axis. For the vertical scale, which is measured in percents, we’ll start at 0 and go up to 60, with tick marks for every 10. Then for each age category, we draw a bar with height corresponding to the percent of survey respondents who said they have an MP3 player. Figure 1.2 shows the completed bar graph. It appears that MP3 players are more popular among young people and that their popularity generally decreases as the age category increases. (b) ​Making a pie chart to display these data is not appropriate because each percent in the table refers to a different age group, not to parts of a single whole.

20 10

0

12–17 18–24 25–34 35–54

55+

Age group (years)

FIGURE 1.2  ​Bar graph comparing the percents of several age groups who own portable MP3 players.

For Practice Try Exercise

15

Graphs: Good and Bad Bar graphs compare several quantities by comparing the heights of bars that represent the quantities. Our eyes, however, react to the area of the bars as well as to their height. When all bars have the same width, the area (width × height) varies in proportion to the height, and our eyes receive the right impression. When you draw a bar graph, make the bars equally wide. Artistically speaking, bar graphs are a bit dull. It is tempting to replace the bars with pictures for greater eye appeal. Don’t do it! The following example shows why.

EXAMPLE

Who Buys iMacs? Beware the pictograph! When Apple, Inc., introduced the iMac, the company wanted to know whether this new computer was expanding Apple’s market share. Was the iMac mainly being bought by previous Macintosh owners, or was it being purchased by first-time computer buyers and by previous PC users who were switching over? To find out, Apple hired a firm to conduct a survey of 500 iMac customers. Each customer was categorized as a new computer purchaser, a previous PC owner, or a previous Macintosh owner. The table summarizes the survey results.5

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Section 1.1 Analyzing Categorical Data

Previous ownership

300

200

Count

Percent (%)

None

85

17.0

PC

60

12.0

Macintosh

355

71.0

Total

500

100.0

Problem: (a) Here’s a clever graph of the data that uses pictures instead of the more

100

None

Windows

traditional bars. How is this graph misleading? (b) Two possible bar graphs of the data are shown below. Which one could be considered deceptive? Why?

Macintosh

Percent

Previous computer

80

80

70

70

60

60

50

Percent

Number of buyers

400

40 30

40 30

20

20

10 0

50

None

PC

Macintosh

Previous computer

10

None

PC

Macintosh

Previous computer

Solution: (a) Although the heights of the pictures are accurate, our eyes respond to the area of the pictures. The

pictograph makes it seem like the percent of iMac buyers who are former Mac owners is at least ten times higher than either of the other two categories, which isn’t the case. (b) The bar graph on the right is misleading. By starting the vertical scale at 10 instead of 0, it looks like the percent of iMac buyers who previously owned a PC is less than half the percent who are first-time computer buyers. We get a distorted impression of the relative percents in the three categories.

17

autio

!

n

There are two important lessons to be learned from this example: (1) beware the pictograph, and (2) watch those scales.

c

For Practice Try Exercise

Two-Way Tables and Marginal Distributions We have learned some techniques for analyzing the distribution of a single categorical variable. What do we do when a data set involves two categorical variables? We begin by examining the counts or percents in various categories for one of the variables. Here’s an example to show what we mean.

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CHAPTER 1

EXAMPLE

E x p l o r i n g Data

I’m Gonna Be Rich! Two-way tables A survey of 4826 randomly selected young adults (aged 19 to 25) asked, “What do you think the chances are you will have much more than a middle-class income at age 30?” The table below shows the responses.6 Young adults by gender and chance of getting rich Gender Male

Total

96

98

194

Some chance but probably not

426

286

712

A 50-50 chance

696

720

1416

A good chance

663

758

1421

Almost certain

486

597

1083

2367

2459

4826

Opinion Almost no chance

Total

Female

This is a two-way table because it describes two categorical variables, gender and opinion about becoming rich. Opinion is the row variable because each row in the table describes young adults who held one of the five opinions about their chances. Because the opinions have a natural order from “Almost no chance” to “Almost certain,” the rows are also in this order. Gender is the column variable. The entries in the table are the counts of individuals in each opinion-by-gender class. How can we best grasp the information contained in the two-way table above? First, look at the distribution of each variable separately. The distribution of a categorical variable says how often each outcome occurred. The “Total” column at the right of the table contains the totals for each of the rows. These row totals give the distribution of opinions about becoming rich in the entire group of 4826 young adults: 194 thought that they had almost no chance, 712 thought they had just some chance, and so on. (If the row and column totals are missing, the first thing to do in studying a two-way table is to calculate them.) The distributions of opinion alone and gender alone are called marginal distributions because they appear at the right and bottom margins of the two-way table.

Definition: Marginal distribution The marginal distribution of one of the categorical variables in a two-way table of counts is the distribution of values of that variable among all individuals described by the table.

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Section 1.1 Analyzing Categorical Data

13

Percents are often more informative than counts, especially when we are comparing groups of different sizes. We can display the marginal distribution of opinions in percents by dividing each row total by the table total and converting to a percent. For instance, the percent of these young adults who think they are almost certain to be rich by age 30 is almost certain total 1083 = = 0.224 = 22.4% table total 4826

EXAMPLE

I’m Gonna Be Rich! Examining a marginal distribution Problem: (a) ​Use the data in the two-way table to calculate the marginal distribution (in percents) of opinions. (b) ​Make a graph to display the marginal distribution. Describe what you see.

Solution: (a) ​We can do four more calculations like the one shown above to obtain the marginal distribution of

opinions in percents. Here is the complete distribution. Response

Percent

Almost no chance

194 = 4.0% 4826

30

Some chance

712 = 14.8% 4826

20

A 50–50 chance

1416 = 29.3% 4826

10

A good chance

1421 = 29.4% 4826

Almost certain

1083 = 22.4% 4826

Percent

Chance of being wealthy by age 30

0

Almost Some 50–50 Good Almost none chance chance chance certain

Survey response

FIGURE 1.3  ​Bar graph showing the marginal distribution of opinion about chance of being rich by age 30.

(b) ​Figure 1.3 is a bar graph of the distribution of opinion among these young adults. It seems that many young adults are optimistic about their future income. Over 50% of those who responded to the survey felt that they had “a good chance” or were “almost certain” to be rich by age 30. For Practice Try Exercise

19

Each marginal distribution from a two-way table is a distribution for a single categorical variable. As we saw earlier, we can use a bar graph or a pie chart to display such a distribution.

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CHAPTER 1

E x p l o r i n g Data

Check Your Understanding Country Superpower

U.K.

U.S.

Fly

54

45

Freeze time

52

44

Invisibility

30

37

Superstrength

20

23

Telepathy

44

66

A random sample of 415 children aged 9 to 17 from the United Kingdom and the United States who completed a CensusAtSchool survey in a recent year was selected. Each student’s country of origin was recorded along with which superpower they would most like to have: the ability to fly, ability to freeze time, invisibility, superstrength, or telepathy (ability to read minds). The data are summarized in the table.7 1. Use the two-way table to calculate the marginal distribution (in percents) of superpower preferences. 2. Make a graph to display the marginal distribution. Describe what you see.

Relationships between Categorical Variables: Conditional Distributions The two-way table contains much more information than the two marginal distributions of opinion alone and gender alone. Marginal distributions tell us nothing about the relationship between two variables. To describe a relationship between two categorical variables, we must calculate some well-chosen percents from the counts given in the body of the table. Young adults by gender and chance of getting rich Gender Female

Male

Total

96

98

194

Some chance but probably not

426

286

712

A 50-50 chance

696

720

1416

A good chance

663

758

1421

Almost certain

486

597

1083

2367

2459

4826

Opinion Almost no chance

Total Conditional distribution of opinion among women Response

Percent

Almost no chance

96 = 4.1% 2367

Some chance

426 = 18.0% 2367

A 50-50 chance

696 = 29.4% 2367

A good chance

663 = 28.0% 2367

Almost certain

486 = 20.5% 2367

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We can study the opinions of women alone by looking only at the “Female” column in the two-way table. To find the percent of young women who think they are almost certain to be rich by age 30, divide the count of such women by the total number of women, the column total: women who are almost certain 486 = = 0.205 = 20.5% column total 2367 Doing this for all five entries in the “Female” column gives the conditional distribution of opinion among women. See the table in the margin. We use the term “conditional” because this distribution describes only young adults who satisfy the condition that they are female.

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Section 1.1 Analyzing Categorical Data

15

Definition: Conditional distribution A conditional distribution of a variable describes the values of that variable among individuals who have a specific value of another variable. There is a separate conditional distribution for each value of the other variable.

Now let’s examine the men’s opinions.

EXAMPLE

I’m Gonna Be Rich! Calculating a conditional distribution Problem:  Calculate the conditional distribution of opinion among the young men. Solution:  To find the percent of young men who think they are almost certain to be rich by age

30, divide the count of such men by the total number of men, the column total: men who are almost certain 597 = = 24.3% column total 2459

If we do this for all five entries in the “Male” column, we get the conditional distribution shown in the table. Conditional distribution of opinion among men Response

Percent

Almost no chance

98 = 4.0% 2459

Some chance

286 = 11.6% 2459

A 50-50 chance

720 = 29.3% 2459

A good chance

758 = 30.8% 2459

Almost certain

597 = 24.3% 2459

For Practice Try Exercise

21

There are two sets of conditional distributions for any two-way table: one for the column variable and one for the row variable. So far, we have looked at the conditional distributions of opinion for the two genders. We could also examine the five conditional distributions of gender, one for each of the five opinions, by looking separately at the rows in the original two-way table. For instance, the conditional distribution of gender among those who responded “Almost certain” is Female 486 = 44.9% 1083

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Male 597 = 55.1% 1083

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16

CHAPTER 1

E x p l o r i n g Data

That is, of the young adults who said they were almost certain to be rich by age 30, 44.9% were female and 55.1% were male. Because the variable “gender” has only two categories, comparing the five conditional distributions amounts to comparing the percents of women among young adults who hold each opinion. Figure 1.4 makes this comparison in a bar graph. The bar heights do not add to 100%, because each bar represents a different group of people.

70

Percent of women in the opinion group

60 50 40 30 20 10 0

Almost Some A 50–50 A good Almost none chance chance chance certain

Opinion

FIGURE 1.4  ​Bar graph comparing the percents of females among those who hold each opinion about their chance of being rich by age  30.

THINK ABOUT IT

Which conditional distributions should we compare?  Our goal all along has been to analyze the relationship between gender and opinion about chances of becoming rich for these young adults. We started by examining the conditional distributions of opinion for males and females. Then we looked at the conditional distributions of gender for each of the five opinion categories. Which of these two gives us the information we want? Here’s a hint: think about whether changes in one variable might help explain changes in the other. In this case, it seems reasonable to think that gender might influence young adults’ opinions about their chances of getting rich. To see whether the data support this idea, we should compare the conditional distributions of opinion for women and men. Software will calculate conditional distributions for you. Most programs allow you to choose which conditional distributions you want to compute.

1. Technology Corner

Analyzing two-way tables

Figure 1.5 presents the two conditional distributions of opinion, for women and for men, and also the marginal distribution of opinion for all of the young adults. The distributions agree (up to rounding) with the results in the last two examples.

FIGURE 1.5  ​Minitab output for the two-way table of young adults by gender and chance of being rich, along with each entry as a percent of its column total. The “Female” and “Male” columns give the conditional distributions of opinion for women and men, and the “All” column shows the marginal distribution of opinion for all these young adults.

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Section 1.1 Analyzing Categorical Data

17

Putting It All Together: Relationships Between Categorical Variables Now it’s time to complete our analysis of the relationship between gender and opinion about chances of becoming rich later in life.

Example

Women’s and Men’s Opinions Conditional distributions and relationships Problem:  Based on the survey data, can we conclude that young men and women differ in their opinions about the likelihood of future wealth? Give appropriate evidence to support your answer. solution:  We suspect that gender might influence a young adult’s opinion about the chance of getting rich. So we’ll compare the conditional distributions of response for men alone and for women alone. Response

Chance of being wealthy by age 30 40

Female Male

Percent

30 20 10 0

Almost Some 50–50 Good Almost none chance chance chance certain

Opinion

FIGURE 1.6  ​Side-by-side bar graph comparing the opinions of males and females.

100

Almost certain

Percent

90 80

A good chance

70

A 50–50 chance

60

Some chance

50

Almost none

40 30 20

Percent of Females

Percent of Males

Almost no chance

96 = 4.1% 2367

98 = 4.0% 2459

Some chance

426 = 18.0% 2367

286 = 11.6% 2459

A 50-50 chance

696 = 29.4% 2367

720 = 29.3% 2459

A good chance

663 = 28.0% 2367

758 = 30.8% 2459

Almost certain

486 = 20.5% 2367

597 = 24.3% 2459

We’ll make a side-by-side bar graph to compare the opinions of males and females. Figure 1.6 displays the completed graph. Based on the sample data, men seem somewhat more optimistic about their future income than women. Men were less likely to say that they have “some chance but probably not” than women (11.6% vs. 18.0%). Men were more likely to say that they have “a good chance” (30.8% vs. 28.0%) or are “almost certain” (24.3% vs. 20.5%) to have much more than a middle-class income by age 30 than women were. For Practice Try Exercise

25

We could have used a segmented bar graph to compare the distributions of male and female responses in the previous example. Figure 1.7 shows the completed graph. Each bar has five segments—one for each of the opinion categories. It’s fairly difficult to compare the percents of males and females in each category because the “middle” segments in the two bars start at different locations on the vertical axis. The side-by-side bar graph in Figure 1.6 makes comparison easier.

10 0

Female

Opinion

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Male

FIGURE 1.7  ​Segmented bar graph comparing the opinions of males and females.

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CHAPTER 1

E x p l o r i n g Data

Both graphs provide evidence of an association between gender and opinion about future wealth in this sample of young adults. Men more often rated their chances of becoming rich in the two highest categories; women said “some chance but probably not” much more frequently.

Definition: Association We say that there is an association between two variables if knowing the value of one variable helps predict the value of the other. If knowing the value of one variable does not help you predict the value of the other, then there is no association between the variables.

Can we say that there is an association between gender and opinion in the population of young adults? Making this determination requires formal inference, which will have to wait a few chapters.

THINK ABOUT IT

What does “no association” mean?  Figure 1.6 (page 17) suggests

an association between gender and opinion about future wealth for young adults. Knowing that a young adult is male helps us predict his opinion: he is more likely than a female to say “a good chance” or “almost certain.” What would the graph look like if there was no association between the two variables? In that case, knowing that a young adult is male would not help us predict his opinion. He would be no more or less likely than a female to say “a good chance” or “almost certain” or any of the other possible responses. That is, the conditional distributions of opinion about becoming rich would be the same for males and females. The segmented bar graphs for the two genders would look the same, too.

Check Your Understanding

U.K.

U.S.

Fly

54

45

Freeze time

52

44

Invisibility

30

37

Superstrength

20

23

Telepathy

44

66

1. Find the conditional distributions of superpower preference among students from the United Kingdom and the United States. 2. Make an appropriate graph to compare the conditional distributions. 3. Is there an association between country of origin and superpower preference? Give appropriate evidence to support your answer.

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autio

!

n

There’s one caution that we need to offer: even a strong association between two categorical variables can be influenced by other variables lurking in the background. The Data Exploration that follows gives you a chance to explore this idea using a famous (or infamous) data set.

c

Country Superpower

Let’s complete our analysis of the data on superpower preferences from the previous Check Your Understanding (page 14). Here is the two-way table of counts once again.

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19

Section 1.1 Analyzing Categorical Data

DATA EXPLORATION A Titanic disaster In 1912 the luxury liner Titanic, on its first voyage across the Atlantic, struck an iceberg and sank. Some passengers got off the ship in lifeboats, but many died. The two-way table below gives information about adult passengers who lived and who died, by class of travel. Class of Travel Survival status

First class

Second class

Third class

197 122

 94 167

151 476

Lived Died

Here’s another table that displays data on survival status by gender and class of travel. Class of Travel First class

Second class

Third class

Female

Male

Female

Male

Female

Male

Lived

140

 57

80

 14

76

 75

Died

  4

118

13

154

89

387

Survival status

The movie Titanic, starring Leonardo DiCaprio and Kate Winslet, suggested the following: • First-class passengers received special treatment in boarding the lifeboats, while some other passengers were prevented from doing so (especially thirdclass passengers). • Women and children boarded the lifeboats first, followed by the men. 1. What do the data tell us about these two suggestions? Give appropriate graphical and numerical evidence to support your answer. 2. How does gender affect the relationship between class of travel and survival status? Explain.

Section 1.1

Summary •

•

•

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The distribution of a categorical variable lists the categories and gives the count (frequency) or percent (relative frequency) of individuals that fall within each category. Pie charts and bar graphs display the distribution of a categorical variable. Bar graphs can also compare any set of quantities measured in the same units. When examining any graph, ask yourself, “What do I see?” A two-way table of counts organizes data about two categorical variables measured for the same set of individuals. Two-way tables are often used to summarize large amounts of information by grouping outcomes into categories.

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CHAPTER 1

E x p l o r i n g Data

•

•

•

The row totals and column totals in a two-way table give the marginal distributions of the two individual variables. It is clearer to present these distributions as percents of the table total. Marginal distributions tell us nothing about the relationship between the variables. There are two sets of conditional distributions for a two-way table: the distributions of the row variable for each value of the column variable, and the distributions of the column variable for each value of the row variable. You may want to use a side-by-side bar graph (or possibly a segmented bar graph) to display conditional distributions. There is an association between two variables if knowing the value of one variable helps predict the value of the other. To see whether there is an association between two categorical variables, compare an appropriate set of conditional distributions. Remember that even a strong association between two categorical variables can be influenced by other variables.

1.1 T echnology Corner

TI-Nspire instructions in ApTI-

1. Analyzing two-way tables

Section 1.1

Exercises

9. Cool car colors  The most popular colors for cars and light trucks change over time. Silver passed green in 2000 to become the most popular color worldwide, then gave way to shades of white in 2007. Here is the distribution of colors for vehicles sold in North America in 2011.8 Color

Percent of vehicles

White

23

Black

18

Silver

16

Gray

13

Red

10

Blue

 9

Brown/beige

 5

Yellow/gold

 3

Green

 2

(a) What percent of vehicles had colors other than those listed? (b) Display these data in a bar graph. Be sure to label your axes.

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page 16

(c) Would it be appropriate to make a pie chart of these data? Explain. 10. Spam  Email spam is the curse of the Internet. Here is a compilation of the most common types of spam:9 Type of spam

Percent

Adult

19

Financial

20

Health

 7

Internet

 7

Leisure

 6

Products

25

Scams

 9

Other

??

(a) ​What percent of spam would fall in the “Other” category? (b) Display these data in a bar graph. Be sure to label your axes. (c) Would it be appropriate to make a pie chart of these data? Explain.

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21

Section 1.1 Analyzing Categorical Data 11. Birth days  Births are not evenly distributed across the days of the week. Here are the average numbers of babies born on each day of the week in the United States in a recent year:10 Day

Births

Sunday

7374

Monday

11,704

Tuesday

13,169

Wednesday

13,038

Thursday

13,013

Friday

12,664

Saturday

14. Which major?  About 1.6 million first-year students enroll in colleges and universities each year. What do they plan to study? The pie chart displays data on the percents of first-year students who plan to major in several discipline areas.13 About what percent of first-year students plan to major in business? In social science?

Technical

Other

Arts/ humanities Biological sciences

Professional

8459

Business

(a) Present these data in a well-labeled bar graph. Would it also be correct to make a pie chart? (b) Suggest some possible reasons why there are fewer births on weekends. 12. Deaths among young people  Among persons aged 15 to 24 years in the United States, the leading causes of death and number of deaths in a recent year were as follows: accidents, 12,015; homicide, 4651; suicide, 4559; cancer, 1594; heart disease, 984; congenital defects, 401.11

Social science Education Physical sciences Engineering

15. Buying music online  Young people are more likely than older folk to buy music online. Here are the percents of people in several age groups who bought music online in a recent year:14

pg 9

(a) Make a bar graph to display these data.

Age group

(b) To make a pie chart, you need one additional piece of information. What is it?

12 to 17 years

24%

18 to 24 years

21%

25 to 34 years

20%

35 to 44 years

16%

13. Hispanic origins  Below is a pie chart prepared by the Census Bureau to show the origin of the more than 50 million Hispanics in the United States in 2010.12 About what percent of Hispanics are Mexican? Puerto Rican? Percent Distribution of Hispanics by Type: 2010

Bought music online

45 to 54 years

10%

55 to 64 years

 3%

65 years and over

 1%

(a) Explain why it is not correct to use a pie chart to display these data. (b) Make a bar graph of the data. Be sure to label your axes. Puerto Rican Mexican

Cuban Central American South American

Other Hispanic

Comment: You see that it is hard to determine numbers from a pie chart. Bar graphs are much easier to use. (The Census Bureau did include the percents in its pie chart.)

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16. The audience for movies  Here are data on the percent of people in several age groups who attended a movie in the past 12 months:15 Age group

Movie attendance

18 to 24 years

83%

25 to 34 years

73%

35 to 44 years

68%

45 to 54 years

60%

55 to 64 years

47%

65 to 74 years

32%

75 years and over

20%

(a) ​Display these data in a bar graph. Describe what you see.

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E x p l o r i n g Data

(b) W ​ ould it be correct to make a pie chart of these data? Why or why not? (c) ​A movie studio wants to know what percent of the total audience for movies is 18 to 24 years old. Explain why these data do not answer this question. 17. Going to school  Students in a high school statistics class were given data about the main method of transportation to school for a group of 30 students. They produced the pictograph shown.

pg 10

20. Smoking by students and parents  Here are data from a survey conducted at eight high schools on smoking among students and their parents:17 Neither parent smokes

Car Key: = 1 Cycler = 2 Cars = 7 Bus takers = 2 Walkers

(a) ​How is this graph misleading? (b) ​Make a new graph that isn’t misleading. 18. Oatmeal and cholesterol  Does eating oatmeal reduce cholesterol? An advertisement included the following graph as evidence that the answer is “Yes.” Representative cholesterol point drop 210

206 204

1168

1823

1380

 188

 416

 400

(a) H ​ ow many students are described in the two-way table? What percent of these students smoke? (b) ​Give the marginal distribution (in percents) of parents’ smoking behavior, both in counts and in percents. 21. Attitudes toward recycled products  Exercise 19 gives data on the opinions of people who have and have not bought coffee filters made from recycled paper. To see the relationship between opinion and experience with the product, find the conditional distributions of opinion (the response variable) for buyers and nonbuyers. What do you conclude?

pg 15

22. Smoking by students and parents  Refer to Exercise 20. Calculate three conditional distributions of students’ smoking behavior: one for each of the three parental smoking categories. Describe the relationship between the smoking behaviors of students and their parents in a few sentences.

202 200 198 Week 1 Week 2 Week 3 Week 4

(a) ​How is this graph misleading? (b) M ​ ake a new graph that isn’t misleading. What do you conclude about the relationship between eating oatmeal and cholesterol reduction? 19. Attitudes toward recycled products  Recycling is supposed to save resources. Some people think recycled products are lower in quality than other products, a fact that makes recycling less practical. People who use a recycled product may have different opinions from those who don’t use it. Here are data on attitudes toward coffee filters made of recycled paper from a sample of people who do and don’t buy these filters:16

23. Popular colors—here and there  Favorite vehicle colors may differ among countries. The side-by-side bar graph shows data on the most popular colors of cars in a recent year for the United States and Europe. Write a few sentences comparing the two distributions.

pg 13

30

Percent

20

Buy recycled filters?

ol d

n

en

/g w

lo

w

re Ye l

ro /b

ge

G

y

43

ed

 9

R

Lower

ei

25

B

 7

e

The same

0

rl

29

5

pe a

No

20

10

te /

Yes

Higher

15

W hi

Think quality is

U.S. Europe

25

ra

196

Both parents smoke

Student smokes

lu

Cholesterol

208

One parent smokes

Student does not smoke

G

STATE COLLEGE BUS SERVICE

Walk

er

STATE COLLEGE BUS SERVICE

lv

STATE COLLEGE BUS SERVICE

Si

Bus

ck

Mode of transport

Cycle

(a) ​How many people does this table describe? How many of these were buyers of coffee filters made of recycled paper? (b) Give the marginal distribution (in percents) of opinion about the quality of recycled filters. What percent of the people in the sample think the quality of the recycled product is the same or higher than the quality of other filters?

B

CHAPTER 1

la



B

22

Color

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23

Section 1.1 Analyzing Categorical Data 24. Comparing car colors  Favorite vehicle colors may differ among types of vehicle. Here are data on the most popular colors in a recent year for luxury cars and for SUVs, trucks, and vans. Color

Luxury cars (%)

SUVs, trucks, vans (%)

Black

22

13

Silver

16

16

White pearl

14

1

Gray

12

13

White

11

25

Blue

7

10

Red

7

11

Yellow/gold

6

1

Green

3

4

Beige/brown

2

6

25. Snowmobiles in the park  Yellowstone National Park surveyed a random sample of 1526 winter visitors to the park. They asked each person whether they owned, rented, or had never used a snowmobile. Respondents were also asked whether they belonged to an environmental organization (like the Sierra Club). The two-way table summarizes the survey responses.

pg 17

Environmental Club



Yes

Total

Never used

 445

212

 657

Snowmobile renter

 497

 77

 574

Snowmobile owner

 279

 16

 295

Total

1221

305

1526

Do these data suggest that there is an association between environmental club membership and snowmobile use among visitors to Yellowstone National Park? Give appropriate evidence to support your answer.

26. Angry people and heart disease  People who get angry easily tend to have more heart disease. That’s the conclusion of a study that followed a random sample of 12,986 people from three locations for about four years. All subjects were free of heart disease at the beginning of the study. The subjects took the Spielberger Trait Anger Scale test, which measures how prone a person is to sudden anger. Here are data for the 8474 people in the sample who had normal blood pressure. CHD stands for “coronary heart disease.” This includes people who had heart attacks and those who needed medical treatment for heart disease.

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Moderate anger

High anger

Total

CHD

  53

 110

 27

 190

No CHD

3057

4621

606

8284

Total

3110

4731

633

8474



Do these data support the study’s conclusion about the relationship between anger and heart disease? Give appropriate evidence to support your answer.

Multiple choice: Select the best answer for Exercises 27 to 34. Exercises 27 to 30 refer to the following setting. The National Survey of Adolescent Health interviewed several thousand teens (grades 7 to 12). One question asked was “What do you think are the chances you will be married in the next ten years?” Here is a two-way table of the responses by gender:18

(a) Make a graph to compare colors by vehicle type. (b) Write a few sentences describing what you see.

No

Low anger

Female

Male

Almost no chance

119

103

Some chance, but probably not

150

171

A 50-50 chance

447

512

A good chance

735

710

Almost certain

1174

756

27. The percent of females among the respondents was (a) 2625. (c) ​about 46%. (e) ​None of these. (b) ​4877. (d) ​about 54%. 28. Your percent from the previous exercise is part of (a) ​the marginal distribution of females. (b) the marginal distribution of gender. (c) the marginal distribution of opinion about marriage. (d) the conditional distribution of gender among adolescents with a given opinion. (e) ​the conditional distribution of opinion among adolescents of a given gender. 29. What percent of females thought that they were almost certain to be married in the next ten years? (a) About 16% (c)  About 40% (e)  About 61% (b) About 24% (d)  About 45% 30. Your percent from the previous exercise is part of (a) the marginal distribution of gender. (b) the marginal distribution of opinion about marriage. (c) the conditional distribution of gender among adolescents with a given opinion. (d) the conditional distribution of opinion among adolescents of a given gender. (e) the conditional distribution of “Almost certain” among females.

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24

CHAPTER 1

E x p l o r i n g Data

31. For which of the following would it be inappropriate to display the data with a single pie chart? (a) The distribution of car colors for vehicles purchased in the last month. (b) The distribution of unemployment percentages for each of the 50 states. (c) The distribution of favorite sport for a sample of 30 middle school students. (d) The distribution of shoe type worn by shoppers at a local mall. (e) The distribution of presidential candidate preference for voters in a state.

Favorite subject

(a) (b) (c) (d) (e)

The subjects are not listed in the correct order. This distribution should be displayed with a pie chart. The vertical axis should show the percent of students. The vertical axis should start at 0 rather than 100. The foreign language bar should be broken up by language.

33. In the 2010–2011 season, the Dallas Mavericks won the NBA championship. The two-way table below displays the relationship between the outcome of each game in the regular season and whether the Mavericks scored at least 100 points. 100 or more points

Fewer than 100 points

Total

Win

43

14

57

Loss

 4

21

25

Total

47

35

82

Which of the following is the best evidence that there is an association between the outcome of a game and whether or not the Mavericks scored at least 100 points? (a) The Mavericks won 57 games and lost only 25 games. (b) The Mavericks scored at least 100 points in 47 games and fewer than 100 points in only 35 games. (c) The Mavericks won 43 games when scoring at least 100 points and only 14 games when scoring fewer than 100 points.

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34. The following partially complete two-way table shows the marginal distributions of gender and handedness for a sample of 100 high school students. Male Right

Total

Female

 90

x

 10

Left Total

40

60

100



If there is no association between gender and handedness for the members of the sample, which of the following is the correct value of x? (a) 20. (b) 30. (c) 36. (d) 45. (e) Impossible to determine without more information.

ts ar e Fi n

F la ore ng ig ua n ge

S st ocia ud l ie s

ng

ie

nc

lis h

e

h

at

M

E

280 260 240 220 200 180 160 140 120 100

Sc

Number of students

32. The following bar graph shows the distribution of favorite subject for a sample of 1000 students. What is the most serious problem with the graph?

(d) The Mavericks won a higher proportion of games when scoring at least 100 points (43/47) than when they scored fewer than 100 points (14/35). (e) The combination of scoring 100 or more points and winning the game occurred more often (43 times) than any other combination of outcomes.

35. Marginal distributions aren’t the whole story Here are the row and column totals for a two-way table with two rows and two columns: a

b

 50

c

d

 50

60

40

100

Find two different sets of counts a, b, c, and d for the body of the table that give these same totals. This shows that the relationship between two variables cannot be obtained from the two individual distributions of the variables. 36. Fuel economy (Introduction)  Here is a small part of a data set that describes the fuel economy (in miles per gallon) of model year 2012 motor vehicles: Make and model

Vehicle type

Transmission Number of City Highway type cylinders mpg mpg

Aston Martin Two-seater Vantage

Manual

8

14

20

Honda Civic Hybrid

Subcompact

Automatic

4

44

44

Toyota Prius

Midsize

Automatic

4

51

48

Chevrolet Impala

Large

Automatic

6

18

30

(a) ​What are the individuals in this data set? (b) ​What variables were measured? Identify each as categorical or quantitative.

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Section 1.2 Displaying Quantitative Data with Graphs

Displaying Quantitative Data with Graphs

1.2 What You Will Learn

By the end of the section, you should be able to:

• Make and interpret dotplots and stemplots of quantitative data.

• Identify the shape of a distribution from a graph as roughly symmetric or skewed. • Make and interpret histograms of quantitative data. • Compare distributions of quantitative data using dotplots, stemplots, or histograms.

• Describe the overall pattern (shape, center, and spread) of a distribution and identify any major departures from the pattern (outliers).

To display the distribution of a categorical variable, use a bar graph or a pie chart. How can we picture the distribution of a quantitative variable? In this section, we present several types of graphs that can be used to display quantitative data.

Dotplots One of the simplest graphs to construct and interpret is a dotplot. Each data value is shown as a dot above its location on a number line. We’ll show how to make a dotplot using some sports data.

EXAMPLE

Gooooaaaaallllll! How to make a dotplot How good was the 2012 U.S. women’s soccer team? With players like Abby Wambach, Megan Rapinoe, and Hope Solo, the team put on an impressive showing en route to winning the gold medal at the 2012 Olympics in London. Here are data on the number of goals scored by the team in the 12 months prior to the 2012 Olympics.19 1

3

1

14

13

4

3

4

2

5

2

0

1

3

4

 3

 4

2

4

3

1

2

4

2

4

Here are the steps in making a dotplot: •  Draw a horizontal axis (a number line) and label it with the variable name. In this case, the variable is number of goals scored.

0

2

4

6

•  Scale the axis. Start by looking at the minimum and maximum values of the variable. For these data, the minimum number of goals scored was 0, and the maximum was 14. So we mark our scale from 0 to 14, with tick marks at every whole-number value. 8 10 12 14

Number of goals scored

FIGURE 1.8  ​A dotplot of goals scored by the U.S. women’s soccer team in 2012.

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•  Mark a dot above the location on the horizontal axis corresponding to each data value. Figure 1.8 displays a completed dotplot for the soccer data.

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CHAPTER 1

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Making a graph is not an end in itself. The purpose of a graph is to help us ­understand the data. After you make a graph, always ask, “What do I see?” Here is a general strategy for interpreting graphs of quantitative data.

How to Examine the Distribution of a Quantitative Variable In any graph, look for the overall pattern and for striking departures from that pattern. • You can describe the overall pattern of a distribution by its shape, center, and spread. • An important kind of departure is an outlier, an individual value that falls outside the overall pattern. You’ll learn more formal ways of describing shape, center, and spread and identifying outliers soon. For now, let’s use our informal understanding of these ideas to examine the graph of the U.S. women’s soccer team data. Shape:  The dotplot has a peak at 4, a single main cluster of dots between 0 and 5, and a large gap between 5 and 13. The main cluster has a longer tail to the left of the peak than to the right. What does the shape tell us? The U.S. women’s soccer team scored between 0 and 5 goals in most of its games, with 4 being the most common value (known as the mode). Center:  The “midpoint” of the 25 values shown in the graph is the 13th value if we count in from either end. You can confirm that the midpoint is at 3. What does this number tell us? In a typical game during the 2012 season, the U.S. women’s soccer team scored about 3 goals. When describing a distribution of quantitative data, don’t forget your SOCS (shape, outliers, center, spread)!

Spread:  The data vary from 0 goals scored to 14 goals scored.

EXAMPLE

Are You Driving a Gas Guzzler?

Outliers:  The games in which the women’s team scored 13 goals and 14 goals clearly stand out from the overall pattern of the distribution. So we label them as possible outliers. (In Section 1.3, we’ll establish a procedure for determining whether a particular value is an outlier.)

Interpreting a dotplot The Environmental Protection Agency (EPA) is in charge of determining and reporting fuel economy ratings for cars (think of those large window stickers on a new car). For years, consumers complained that their actual gas mileages were noticeably lower than the values reported by the EPA. It seems that the EPA’s tests—all of which are done on computerized devices to ensure consistency—did not consider things like outdoor temperature, use of the air conditioner, or realistic acceleration and braking by drivers. In 2008 the EPA changed the method for measuring a vehicle’s fuel economy to try to give more accurate e­stimates. The following table displays the EPA estimates of highway gas mileage in miles per gallon (mpg) for a sample of 24 model year 2012 midsize cars.20

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Section 1.2 Displaying Quantitative Data with Graphs

Model

mpg

Model

mpg

Model

mpg

Acura RL

24

Dodge Avenger

30

Mercedes-Benz E350

30

Audi A8

28

Ford Fusion

25

Mitsubishi Galant

30

Bentley Mulsanne

18

Hyundai Elantra

40

Nissan Maxima

26

BMW 550I

23

Jaguar XF

23

Saab 9-5 Sedan

28

Buick Lacrosse

27

Kia Optima

34

Subaru Legacy

31

Cadillac CTS

27

Lexus ES 350

28

Toyota Prius

48

Chevrolet Malibu

33

Lincoln MKZ

27

Volkswagen Passat

31

Chrysler 200

30

Mazda 6

31

Volvo S80

26

Figure 1.9 shows a dotplot of the data: FIGURE 1.9  Dotplot displaying EPA estimates of highway gas mileage for model year 2012 midsize cars.

20

25

30

35

HwyMPG

40

45

Problem:  Describe the shape, center, and spread of the distribution. Are there any outliers? Solution: Shape:  The dotplot has a peak at 30 mpg and a main cluster of values from 23 to 34 mpg. There are

The 2012 Nissan Leaf, an electric car, got an EPA estimated 92 miles per gallon on the highway. With the U.S. government’s plan to raise the fuel economy standard to an average of 54.5 mpg by 2025, even more alternative-fuel vehicles like the Leaf will have to be developed.

large gaps between 18 and 23, 34 and 40, 40 and 48 mpg. Center:  The midpoint of the 24 values shown in the graph is 28. So a typical model year 2012 midsize car in the sample got about 28 miles per gallon on the highway. Spread:  The data vary from 18 mpg to 48 mpg. Outliers:  We see two midsize cars with unusually high gas mileage ratings: the Hyundai Elantra (40 mpg) and the Toyota Prius (48 mpg). The Bentley Mulsanne stands out for its low gas mileage rating (18 mpg). All three of these values seem like clear outliers. For Practice Try Exercise

39

Describing Shape When you describe a distribution’s shape, concentrate on the main features. Look for major peaks, not for minor ups and downs in the graph. Look for clusters of values and obvious gaps. Look for potential outliers, not just for the smallest and largest observations. Look for rough symmetry or clear skewness.

Skewed to the left!

For his own safety, which way should Mr. Starnes go “skewing”?

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Definition: Symmetric and skewed distributions A distribution is roughly symmetric if the right and left sides of the graph are approximately mirror images of each other. A distribution is skewed to the right if the right side of the graph (containing the half of the observations with larger values) is much longer than the left side. It is skewed to the left if the left side of the graph is much longer than the right side.

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CHAPTER 1

E x p l o r i n g Data

c

EXAMPLE

!

n

For brevity, we sometimes say “left-skewed” instead of “skewed to the left” and “right-skewed” instead of “skewed to the right.” We could also describe a distribution with a long tail to the left as “skewed toward negative values” or “negatively skewed” and a distribution with a long right tail as “positively skewed.” The direction of skewness is the direction of the long tail, not the direc- autio tion where most observations are clustered. See the drawing in the margin on page 27 for a cute but corny way to help you keep this straight.

Die Rolls and Quiz Scores Describing shape Figure 1.10 displays dotplots for two different sets of quantitative data. Let’s practice describing the shapes of these distributions. Figure 1.10(a) shows the results of rolling a pair of fair, six-sided dice and finding the sum of the up-faces 100 times. This distribution is roughly symmetric. The dotplot in Figure 1.10(b) shows the scores on an AP® Statistics class’s first quiz. This distribution is skewed to the left.

(a)

(b)

2

4

6

8

10

12

Dice rolls

70

75

80

85

90

95

100

Score

FIGURE 1.10  ​Dotplots displaying different shapes: (a) roughly symmetric; (b) skewed to the left.

Although the dotplots in the previous example have different shapes, they do have something in common. Both are unimodal, that is, they have a single peak: the graph of dice rolls at 7 and the graph of quiz scores at 90. (We don’t count minor ups and downs in a graph, like the “bumps” at 9 and 11 in the dice rolls dotplot, as “peaks.”) Figure 1.11 is a dotplot of the duration (in minutes) of 220

FIGURE 1.11  ​Dotplot displaying duration (in minutes) of Old Faithful eruptions. This graph has a bimodal shape.

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Section 1.2 Displaying Quantitative Data with Graphs

29

eruptions of the Old Faithful geyser. We would describe this distribution’s shape as roughly symmetric and bimodal because it has two clear peaks: one near 2 minutes and the other near 4.5 minutes. (Although we could continue the pattern with “trimodal” for three peaks and so on, it’s more common to refer to distributions with more than two clear peaks as multimodal.)

THINK ABOUT IT

What shape will the graph have?  Some variables have distributions

with predictable shapes. Many biological measurements on individuals from the same species and gender—lengths of bird bills, heights of young women—have roughly symmetric distributions. Salaries and home prices, on the other hand, usually have right-skewed distributions. There are many moderately priced houses, for example, but the few very expensive mansions give the distribution of house prices a strong right skew. Many distributions have irregular shapes that are neither symmetric nor skewed. Some data show other patterns, such as the two peaks in Figure 1.11. Use your eyes, describe the pattern you see, and then try to explain the pattern.

Check Your Understanding

The Fathom dotplot displays data on the number of siblings reported by each student in a statistics class.

1. Describe the shape of the distribution. 2. Describe the center of the distribution. 3. Describe the spread of the distribution. 4. Identify any potential outliers.

Comparing Distributions Some of the most interesting statistics questions involve comparing two or more groups. Which of two popular diets leads to greater long-term weight loss? Who texts more—males or females? Does the number of people living in a household differ among countries? As the following example suggests, you should always discuss shape, center, spread, and possible outliers whenever you compare distributions of a quantitative variable.

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CHAPTER 1

EXAMPLE

E x p l o r i n g Data

Household Size: U.K. versus South Africa Comparing distributions How do the numbers of people living in households in the United Kingdom (U.K.) and South Africa compare? To help answer this question, we used CensusAtSchool’s “Random Data Selector” to choose 50 students from each country. ­Figure 1.12 is a dotplot of the household sizes reported by the survey respondents.

Problem:  Compare the distributions of household size for these two countries. Solution: 

South Africa

Shape:  The distribution of household size for the U.K. sample is roughly symmetric and unimodal, while the distribution for the South Africa sample is skewed to the right and unimodal. Center:  Household sizes for the South African students tended to be larger than for the U.K. students. The midpoints of the household sizes for the two groups are 6 people and 4 people, respectively. Spread:  The household sizes for the South African students vary more (from 3 to 26 people) than for the U.K. students (from 2 to 6 people). Outliers:  There don’t appear to be any outliers in the U.K. distribution. The South African distribution seems to have two outliers in the right tail of the distribution—students who reported living in households with 15 and 26 people. Place

0

5

10

15

20

25

30

25

30

Household size

U.K.

AP® EXAM TIP ​When comparing distributions of quantitative data, it’s not enough just to list values for the center and spread of each distribution. You have to explicitly compare these values, using words like “greater than,” “less than,” or “about the same as.”

FIGURE 1.12  ​Dotplots of household size for random samples of 50 students from the United Kingdom and South Africa.

0

5

10

15

20

Household size

For Practice Try Exercise

43

Notice that we discussed the distributions of household size only for the two samples of 50 students in the previous example. We might be interested in w ­ hether the sample data give us convincing evidence of a difference in the population distributions of household size for South Africa and the United Kingdom. We’ll have to wait a few chapters to decide whether we can reach such a conclusion, but our

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Section 1.2 Displaying Quantitative Data with Graphs

31

ability to make such an inference later will be helped by the fact that the students in our samples were chosen at random.

Stemplots Another simple graphical display for fairly small data sets is a stemplot (also called a stem-and-leaf plot). Stemplots give us a quick picture of the shape of a distribution while including the actual numerical values in the graph. Here’s an example that shows how to make a stemplot.

EXAMPLE

How Many Shoes? Making a stemplot How many pairs of shoes does a typical teenager have? To find out, a group of AP® Statistics students conducted a survey. They selected a random sample of 20 female students from their school. Then they recorded the number of pairs of shoes that each respondent reported having. Here are the data: 50 26 26 31 57 19 24 22 23 38 13 50 13 34 23 30 49 13 15 51

Here are the steps in making a stemplot. Figure 1.13 displays the process. •  Separate each observation into a stem, consisting of all but the final digit, and a leaf, the final digit. Write the stems in a vertical column with the smallest at the top, and draw a vertical line at the right of this column. Do not skip any stems, even if there is no data value for a particular stem. For these data, the tens digits are the stems, and the ones digits are the 1 1 93335 1 33359 leaves. The stems run from 1 to 5. Key: 4|9 represents 2 2 664233 2 233466 a female student •  Write each leaf in the row to the right of its stem. 3 3 1840 3 0148 who reported having 4 4 9 4 9 For example, the female student with 50 pairs of shoes 49 pairs of shoes. 5 5 0701 5 0017 would have stem 5 and leaf 0, while the student with Stems Add leaves Order leaves Add a key 31 pairs of shoes would have stem 3 and leaf 1. •  Arrange the leaves in increasing order out from the FIGURE 1.13  ​Making a stemplot of the shoe data. (1) Write stem. the stems. (2) Go through the data and write each leaf on the •  Provide a key that explains in context what the stems proper stem. (3) Arrange the leaves on each stem in order out and leaves represent. from the stem. (4) Add a key.

The AP® Statistics students in the previous example also collected data from a random sample of 20 male students at their school. Here are the numbers of pairs of shoes reported by each male in the sample: 14  7 6 5 12 38 8  7 10 10 10 11 4 5 22   7 5 10 35  7

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What would happen if we tried the same approach as before: using the first digits as stems and the last digits as leaves? The completed stemplot is shown in Figure 1.14(a). What shape does this distribution have? It is difficult to tell with so few stems. We can get a better picture of male shoe ownership by splitting stems. In Figure 1.14(a), the values from 0 to 9 are placed on the “0” stem. Figure 1.14(b) shows another stemplot of the same data. This time, values having leaves 0 through 4 are placed on one stem, while values ending in 5 through 9 are placed on another stem. Now we can see the single peak, the cluster of values between 4 and 14, and the large gap between 22 and 35 more clearly. What if we want to compare the number of pairs of shoes that males and females have? That calls for a back-to-back stemplot with common stems. The leaves on each side are ordered out from the common stem. Figure 1.15 is a back-to-back stemplot for the male and female shoe data. Note that we have used the split stems from Figure 1.14(b) as the common stems. The values on the right are the male data from Figure 1.14(b). The values on the left are the female data, ordered out from the stem from right to left. We’ll ask you to compare these two distributions shortly. 0 1 2 3

4555677778 0000124 2 58

(a)

0 0 1 1 2 2 3 3

4 555677778 0000124 2

58

(b) Key: 2|2 represents a male student with 22 pairs of shoes.

Notice that we include this stem even though it contains no data.

FIGURE 1.14  ​Two stemplots showing the male shoe data. Figure 1.14(b) improves on the stemplot of Figure 1.14(a) by splitting stems.

• • •

Instead of rounding, you can also truncate (remove one or more digits) when data have too many digits. The teacher’s salary of $42,549 would truncate to $42,000.

•

Females

333 95 4332 66 410 8 9 100 7

0 0 1 1 2 2 3 3 4 4 5 5

Males 4 555677778 0000124 2

Key: 2|2 represents a male student with 22 pairs of shoes.

58

FIGURE 1.15  ​Back-to-back stemplot comparing numbers of pairs of shoes for male and female students at a school.

Here are a few tips to consider before making a stemplot: Stemplots do not work well for large data sets, where each stem must hold a large number of leaves. There is no magic number of stems to use, but five is a good minimum. Too few or too many stems will make it difficult to see the distribution’s shape. If you split stems, be sure that each stem is assigned an equal number of possible leaf digits (two stems, each with five possible leaves; or five stems, each with two possible leaves). You can get more flexibility by rounding the data so that the final digit after rounding is suitable as a leaf. Do this when the data have too many digits. For example, in reporting teachers’ salaries, using all five digits (for example, $42,549) would be unreasonable. It would be better to round to the nearest thousand and use 4 as a stem and 3 as a leaf.

Check Your Understanding 1. Use the back-to-back stemplot in Figure 1.15 to write a few sentences comparing the number of pairs of shoes owned by males and females. Be sure to address shape, center, spread, and outliers.

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Section 1.2 Displaying Quantitative Data with Graphs 6 7 8 9 10 11 12 13 14 15 16

8 8 79 08 15566 012223444457888999 01233333444899 02666 23 8

Key: 8|8 represents a state in which 8.8% of residents are 65 and older.

33

Multiple choice: Select the best answer for Questions 2 through 4. Here is a stemplot of the percents of residents aged 65 and older in the 50 states and the District of Columbia. The stems are whole percents and the leaves are tenths of a percent. 2. The low outlier is Alaska. What percent of Alaska residents are 65 or older? (a) 0.68   (b) ​6.8  

(c) ​8.8  

(d) ​16.8  

(e) ​68

3. ​Ignoring the outlier, the shape of the distribution is (a) skewed to the right. (b) ​​skewed to the left.

(c)  skewed to the middle. (d) ​​bimodal.

(e) ​roughly symmetric.

4. The center of the distribution is close to (a) 11.6%.   (b) 12.0%.   (c) ​12.8%.   (d) ​13.3%.   (e) ​6.8% to 16.8%.

Histograms Quantitative variables often take many values. A graph of the distribution is clearer if nearby values are grouped together. One very common graph of the distribution of a quantitative variable is a histogram. Let’s look at how to make a histogram using data on foreign-born residents in the United States.

Example

Foreign-Born Residents Making a histogram What percent of your home state’s residents were born outside the U ­ nited States? A few years ago, the country as a whole had 12.5% foreign-born residents, but the states varied from 1.2% in West Virginia to 27.2% in California. The following table presents the data for all 50 states.21 The individuals in this data set are the states. The variable is the percent of a state’s residents who are foreign-born. It’s much easier to see from a graph than from the table how your state compared with other states.

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State Percent

State Percent

State Percent

Alabama 2.8 Alaska 7.0 Arizona 15.1 Arkansas 3.8 California 27.2 Colorado 10.3 Connecticut 12.9 Delaware 8.1 Florida 18.9 Georgia 9.2 Hawaii 16.3 Idaho 5.6 Illinois 13.8 Indiana 4.2 Iowa 3.8 Kansas 6.3 Kentucky 2.7

Louisiana 2.9 Maine 3.2 Maryland 12.2 Massachusetts 14.1 Michigan 5.9 Minnesota 6.6 Mississippi 1.8 Missouri 3.3 Montana 1.9 Nebraska 5.6 Nevada 19.1 New Hampshire 5.4 New Jersey 20.1 New Mexico 10.1 New York 21.6 North Carolina 6.9 North Dakota 2.1

Ohio 3.6 Oklahoma 4.9 Oregon 9.7 Pennsylvania 5.1 Rhode Island 12.6 South Carolina 4.1 South Dakota 2.2 Tennessee 3.9 Texas 15.9 Utah 8.3 Vermont 3.9 Virginia 10.1 Washington 12.4 West Virginia 1.2 Wisconsin 4.4 Wyoming 2.7

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Here are the steps in making a histogram: •  Divide the data into classes of equal width. The data in the table vary from 1.2 to 27.2, so we might choose to use classes of width 5, beginning at 0: 0–5 5–10 10–15 15–20 20–25 25–30 But we need to specify the classes so that each individual falls into exactly one class. For instance, what if exactly 5.0% of the residents in a state were born outside the United States? Because a value of 0.0% would go in the 0–5 class, we’ll agree to place a value of 5.0% in the 5–10 class, a value of 10.0% in the 10–15 class, and so on. In reality, then, our classes for the percent of foreign-born residents in the states are 0 to 0. • mY = a + bmX. • sY = |b|sX (because b could be a negative number).

The bottom two rules in the summary box don’t just apply to means and standard deviations. Linear transformations have similar effects on other measures of center or location (median, quartiles, percentiles) and spread (range, IQR). Whether we’re dealing with data or random variables, the effects of a linear transformation are the same. Note that these results apply to both discrete and continuous random variables.

EXAMPLE

The Baby and the Bathwater Linear transformations Problem:  One brand of bathtub comes with a dial to set the water temperature. When the “babysafe” setting is selected and the tub is filled, the temperature X of the water follows a Normal distribution with a mean of 34°C and a standard deviation of 2°C. (a) Define the random variable Y to be the water temperature in degrees Fahrenheit 9 (recall that F = C + 32) when the dial is set on “babysafe.” Find the mean and standard 5 deviation of Y. (b) According to Babies R Us, the temperature of a baby’s bathwater should be between 90°F and 100°F. Find the probability that the water temperature on a randomly selected day when the “babysafe” setting is used meets the Babies R Us recommendation. Show your work.

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Solution: 9 (a) According to the formula for converting Celsius to Fahrenheit, Y = X + 32. We could also 5 9 write this in the form Y = 32 + X . The mean of Y  is 5 9 9 mY = 32 + mX = 32 + (34) = 93.2°F 5 5 The standard deviation of Y  is

N(93.2,3.6)

90 93.2 Temperature (°F)

FIGURE 6.8  T​ he Normal probability distribution of the random variable Y = the temperature (in °F) of the bathwater when the dial is set on “babysafe.” The shaded area is the probability that the water temperature is between 90°F and 100°F.

9 9 sY = sX = (2) = 3.6°F 5 5

(b) Step 1: State the distribution and the values of interest. The linear transformation doesn’t change the shape of the probability distribution, so the random variable Y is Normally distributed with a mean of 93.2 and a standard deviation of 3.6. We want to find 380P(90 ≤ Y ≤ 100). The shaded area in Figure 6.8 shows the desired probability. Step 2: Perform calculations—show your work! To find this area, we can standardize the boundary values and use Table A:

100

90 − 93.2 100 − 93.2 = −0.89  and  z = = 1.89 3.6 3.6 Then P (−0.89 ≤ Z ≤ 1.89) = 0.9706 − 0.1867 = 0.7839. Using technology: The command normalcdf(lower:90, upper:100, m:93.2, s:3.6) gives an area of 0.7835. Step 3: Answer the question. There’s about a 78% chance that the water temperature meets the recommendation on a randomly selected day. z=

For Practice Try Exercise

45

Combining Random Variables So far, we have looked at settings that involved a single random variable. Many interesting statistics problems require us to combine two or more random variables.

EXAMPLE

Pete’s Jeeps and Erin’s Adventures When one random variable isn’t enough Earlier, we examined the probability distribution for the random variable X = the number of passengers on a randomly selected half-day trip with Pete’s Jeep Tours. Here’s a brief recap: 3 4 5 6 No. of passengers xi: 2 0.15 0.25 0.35 0.20 0.05 Probability pi: Mean: mX = 3.75  Standard deviation: sX = 1.0897

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0.35

Probability

0.30 0.25 0.20 0.15 0.10 0.05 0.00 0

1

2

3

4

5

6

Number of passengers (X)

Pete’s sister Erin,Starnes/Yates/Moore: who lives near The a tourist in another part of the country, is Practicearea of Statistics, 4E Fig.: 06_06 Perm. Fig.: 6023She decides to join the business, runimpressed by theNew success of Pete’s business. Pass: 2010-04-14 ning tours on theFirst same days as Pete in her slightly smaller vehicle, under the name 2nd Pass: 2010-05-04 Erin’s Adventures. After a year of steady bookings, Erin discovers that the number 3rd Pass: 2010-05-07 of passengers Y on her half-day tours has the following probability distribution. Figure 6.9 displays this distribution as a h ­ istogram. No. of passengers yi: 2 3 4 5 0.3 0.4 0.2 0.1 Probability pi:

0.5

Probability

0.4

Mean: mY = 3.10   Standard deviation: sY = 0.943

0.3

How many total passengers T will Pete and Erin have on their tours on a randomly selected day? To answer this question, we need to know about the distribution of the random variable T = X + Y.

0.2 0.1 0 0

1

2

3

4

5

Number of passengers (Y )

FIGURE 6.9  The probability distribution of the random variable Y = the number of passengers on Erin’s trip on a randomly chosen day.

How many more or fewer passengers D will Pete have than Erin on a randomly selected day? To answer this question, we need to know about the distribution of the random variable D = X – Y.

Starnes/Yates/Moore: The Practice of Statistics, 4E New Fig.: 06-10 Perm. Fig.: 6030 First Pass: 2010-04-16 2nd Pass: 2010-05-04 3rd Pass: 2010-05-07

As the example suggests, we want to investigate what happens when we add or subtract random variables.

Sums of random variables  How many total passengers T can Pete and

Erin expect to have on their tours on a randomly selected day? Because Pete averages mX = 3.75 passengers per trip and Erin averages mY = 3.10 passengers per trip, they will average a total of mT = 3.75 + 3.10 = 6.85 passengers per day. We can generalize this result for any two random variables as follows: if T = X + Y, then mT = mX + mY. In other words, the expected value (mean) of the sum of two random variables is equal to the sum of their expected values (means).

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Mean of the Sum of Random Variables For any two random variables X and Y, if T = X + Y, then the expected value of T is E(T) = mT = mX + mY In general, the mean of the sum of several random variables is the sum of their means.

How much variability is there in the total number of passengers who go on Pete’s and Erin’s tours on a randomly chosen day? Let’s think about the possible values of T = X + Y. The number of passengers X on Pete’s tour is between 2 and 6, and the number of passengers Y on Erin’s tour is between 2 and 5. So the total number of passengers T is between 4 and 11. Thus, the range of T is 11 − 4 = 7. How is this value related to the ranges of X and Y? The range of X is 4 and the range of Y is 3, so range of T = range of X + range of Y That is, there’s more variability in the values of T than in the values of X or Y alone. This makes sense, because the variation in X and the variation in Y both contribute to the variation in T. What about the standard deviation sT? If we had the probability distribution of the random variable T, then we could calculate s T. Let’s try to construct this probability distribution starting with the smallest possible value, T = 4. The only way to get a total of 4 passengers is if Pete has X = 2 passengers and Erin has Y = 2 passengers. We know that P(X = 2) = 0.15 and that P(Y = 2) = 0.3. If the two events X = 2 and Y = 2 are independent, then we can multiply these two probabilities. Otherwise, we’re stuck. In fact, we can’t calculate the probability for any value of T unless X and Y are independent random variables.

Definition: Independent random variables If knowing whether any event involving X alone has occurred tells us nothing about the occurrence of any event involving Y alone, and vice versa, then X and Y are ­independent random variables.

Probability models often assume independence when the random variables describe outcomes that appear unrelated to each other. You should always ask whether the assumption of independence seems reasonable. For instance, it’s reasonable to treat the random variables X = number of passengers on Pete’s trip and Y = number of passengers on Erin’s trip on a randomly chosen day as independent, because the siblings operate their trips in different parts of the country. Now we can calculate the probability distribution of the total number of passengers that day.

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EXAMPLE

Pete’s Jeep Tours and Erin’s ­Adventures Sum of two random variables Let T = X + Y, as before. Because X and Y are independent ­random variables, P(T = 4) = P(X = 2 and Y = 2) = P(X = 2) × P(Y = 2) = (0.15)(0.3) = 0.045. There are two ways to get a total of T = 5 passengers on a randomly selected day: X = 3, Y = 2 or X = 2, Y = 3. So P(T = 5) = P(X = 2 and Y = 3) + P(X = 3 and Y = 2) = (0.15)(0.4) + (0.25)(0.3) = 0.06 + 0.075 = 0.135. We can construct the probability distribution by listing all combinations of X and Y that yield each possible value of T and adding the corresponding probabilities. Here is the result. Value ti: 4 5 6 7 8 9 10 11 Probability pi: 0.045 0.135 0.235 0.265 0.190 0.095 0.030 0.005 You can check that the probabilities add to 1. A histogram of the probability distribution is shown in Figure 6.10. 0.300

Probability

0.250

FIGURE 6.10  ​The probability distribution of the random variable T = the total number of passengers on Pete’s and Erin’s trips on a randomly chosen day.

0.200 0.150 0.100 0.050 0.000 0

1

2

3

4

5

6

7

8

9

10

11

Total number of passengers (T)

∙

The mean of T is m T = tipi = (4)(0.045) + (5)(0.135) + . . . + (11)(0.005) = 6.85. Recall that mX = 3.75 and mY = 3.10. Our calculation confirms that What about

Starnes/Yates/Moore: The Practice of Statistics, 4E m06-11 + mYFig.: = 3.75 T = mX Perm. New Fig.: 6032 + 3.10 = 6.85 Pass: 2010-04-16 the First variance of T? It’s 2nd Pass: 2010-05-04 2 3rd Pass: 2010-05-07 2

∙

sT = (ti − mT) pi = (4 − 6.85)2(0.045) + (5 − 6.85)2(0.135) + . . . + (11 − 6.85)2(0.005) = 2.0775

Recalling that s2X = 1.1875 and s2Y = 0.89, we see that 1.1875 + 0.89 = 2.0775. That is, s2T = s2X + s2Y To find the standard deviation of T, take the square root of the variance sT = !2.0775 = 1.441

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Section 6.2 Transforming and Combining Random Variables

As the preceding example illustrates, when we add two independent random variables, their variances add. Standard deviations do not add. For Pete’s and Erin’s passenger totals, sX + sY = 1.0897 + 0.943 = 2.0327 This is very different from sT = 1.441.

Variance of the Sum of Independent Random Variables For any two independent random variables X and Y, if T = X + Y, then the variance of T is s2T = s2X + s2Y In general, the variance of the sum of several independent random variables is the sum of their variances.

c

EXAMPLE

!

n

You might be wondering whether there’s a formula for computing the variance of the sum of two random variables that are not independent. There is, autio but it’s beyond the scope of this course. Just remember that you can add variances only if the two random variables are independent and that you can never add standard deviations.

SAT Scores The role of independence A college uses SAT scores as one criterion for admission. Experience has shown that the distribution of SAT scores among its entire population of applicants is such that SAT Math score X: SAT Critical Reading score Y:

mX = 519 mY = 507

sX = 115 sY = 111

Problem:  What are the mean and standard deviation of the total score X + Y  for a randomly selected applicant to this college? Solution:  The mean total score is

mX+Y = mX + mY = 519 + 507 = 1026 The variance and standard deviation of the total cannot be computed from the information given. SAT Math and Critical Reading scores are not independent, because students who score high on one exam tend to score high on the other also. For Practice Try Exercise

47

The next example involves two independent random variables and some transformations.

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R a n d o m Va r i a b l e s

Pete’s and Erin’s Tours Rules for adding random variables Earlier, we defined X = the number of passengers that Pete has and Y = the number of passengers that Erin has on a randomly selected day. Recall that mX = 3.75, sX = 1.0897    mY = 3.10, sY = 0.943 Pete charges $150 per passenger and Erin charges $175 per passenger. Problem:  Calculate the mean and the standard deviation of the total amount that Pete and Erin collect on a randomly chosen day. Solution:  Let W = the total amount collected. Then W = 150X + 175Y. If we let C = 150X and G = 175Y, then we can write W as the sum of two random variables: W = C + G. We can use what we learned earlier about the effect of multiplying by a constant to find the mean and standard deviation of C and G. For C = 150X, mC = 150mX = 150(3.75) = $562.50 and sC = 150(1.0897) = $163.46 For G = 175Y, mG = 175mY = 175(3.10) = $542.50 and sG = 175(0.943) = $165.03 We know that the mean of the sum of two random variables equals the sum of their means: mW = mC + mG = 562.50 + 542.50 = 1105 On average, Pete and Erin expect to collect a total of $1105 per day. Because the number of passengers X and Y are independent random variables, so are the amounts of money collected C and G. Therefore, the variance of W is the sum of the variances of C and G. s2W = s2C + s2G = (163.46)2 + (165.03)2 = 53,954.07 To get the standard deviation, we take the square root of the variance: sW = !53,954.07 = 232.28

The standard deviation of the total amount they collect is $232.28. For Practice Try Exercise

51

We can extend our rules for adding random variables to situations involving repeated observations from the same chance process. For instance, suppose a gambler plays two games of roulette, each time placing a $1 bet on either red or black. What can we say about his total gain (or loss) from playing two games? Earlier, we showed that if X = the amount gained on a single $1 bet on red or black, then mX = −$0.05 and sX = $1.00. Because we’re interested in the player’s total gain over two games, we’ll define X1 as the amount he gains from the first game and X2 as the amount he gains from the second game. Then, his total gain T = X1 + X2. Both X1 and X2 have the

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same probability distribution as X and, therefore, the same mean (−$0.05) and standard deviation ($1.00). The player’s expected gain in two games is mT = mX1 + mX2 = (−$0.05) + (−$0.05) = −$0.10 Because knowing the result of one game tells the player nothing about the result of the other game, X1 and X2 are independent random variables. As a result, s2T = s2X1 + s2X2 = (1.00)2 + (1.00)2 = 2.00 and the standard deviation of the player’s total gain is sT = !2.00 = $1.41

THINK ABOUT IT

X1 + X2 is not the same as 2X  ​At the beginning of the section, we asked

whether a roulette player would be better off placing two separate $1 bets on red or a single $2 bet on red. The player’s total gain from two $1 bets on red is T = X1 + X2. This sum of random variables has mean m T = −$0.10 and standard deviation sT = $1.41. Now think about what happens if the gambler places a $2 bet on red in a single game of roulette. Because the random variable X represents a player’s gain from a $1 bet, the random variable Y = 2X represents his gain from a $2 bet. What’s the player’s expected gain from a single $2 bet on red? It’s mY = 2mX = 2(−$0.05) = −$0.10 That’s the same as his expected gain from playing two games of roulette with a $1 bet each time. But the standard deviation of the player’s gain from a single $2 bet is sY = 2sX = 2($1.00) = $2.00 Compare this result to sT = $1.41. There’s more variability in the gain from a single $2 bet than in the total gain from two $1 bets. Let’s take this one step further. Would it be better for the player to place a single $100 bet on red or to play 100 games and bet $1 each time on red? For the single $100 bet, the mean and standard deviation of the amount gained would be mean = 100mX = 100(−$0.05) = −$5.00 standard deviation = 100sX = 100($1.00) = $100.00 For 100 games with a $1 bet, the mean and standard deviation of the amount gained would be mean = mX1 + mX2 + c+ mX100 = (−$0.05) + (−$0.05) + c+ (−$0.05) = −$5.00 variance = s2X1 + s2X2 + c+ s2X100 = (1)2 + (1)2 + c+ (1)2 = 100 standard deviation = !100 = $10.00

The player has a much better chance of winning (or losing) big with a single $100 bet than with 100 separate $1 bets. Of course, the casino accepts thousands of bets each day, so it can count on being fairly close to its expected return of 5 cents per dollar bet.

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CHECK YOUR UNDERSTANDING

A large auto dealership keeps track of sales and lease agreements made during each hour of the day. Let X = the number of cars sold and Y = the number of cars leased during the first hour of business on a randomly selected Friday. Based on previous records, the probability distributions of X and Y are as follows: Cars sold xi: 0 1 2 3 Probability pi: 0.3 0.4 0.2 0.1 Mean: mX = 1.1   Standard deviation: sX = 0.943 Cars leased yi: 0 1 2 Probability pi: 0.4 0.5 0.1 Mean: mY = 0.7   Standard deviation: sY = 0.64 Define T = X + Y. Assume that X and Y are independent. 1. Find and interpret mT. 2. Compute sT. Show your work. 3. The dealership’s manager receives a $500 bonus for each car sold and a $300 bonus for each car leased. Find the mean and standard deviation of the manager’s total bonus B. Show your work.

Differences of random variables  Now that we’ve examined sums of

random variables, it’s time to investigate the difference of two random variables. Let’s start by looking at the difference in the number of passengers that Pete and Erin have on their tours on a randomly selected day, D = X − Y. Because Pete averages mX = 3.75 passengers per trip and Erin averages mY = 3.10 passengers per trip, the average difference is mD = 3.75 − 3.10 = 0.65 passengers. That is, Pete averages 0.65 more passengers per day than Erin does. We can generalize this result for any two random variables as follows: if D = X − Y, then mD = mX − mY. In other words, the mean (expected value) of the difference of two random variables is equal to the difference of their means (expected values).

Mean of the Difference of Random Variables For any two random variables X and Y, if D = X − Y, then the mean of D is mD = E(D) = mX − mY

c

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!

n

The order of subtraction is important. If we had defined D = Y − X, then autio mD = mY − mX = 3.10 − 3.75 = −0.65. In other words, Erin averages 0.65 fewer passengers than Pete does on a randomly chosen day. Earlier, we saw that the variance of the sum of two independent random ­variables is the sum of their variances. Can you guess what the variance of the difference of two independent random variables will be? (If you were thinking something like “the difference of their variances,” think again!) Let’s return to the jeep tours scenario. On a randomly selected day, the number of passengers X on Pete’s tour is between 2 and 6, and the number of passengers Y on Erin’s tour is between

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377

2 and 5. So the difference in the number of passengers D = X − Y is between −3 and 4. Thus, the range of D is 4 − (−3) = 7. How is this value related to the ranges of X and Y? The range of X is 4 and the range of Y is 3, so range of D = range of X + range of Y As with sums of random variables, there’s more variability in the values of the difference D than in the values of X or Y alone. This should make sense, because the variation in X and the variation in Y both contribute to the variation in D. If you follow the process we used earlier with the random variable T = X + Y, you can build the probability distribution of D = X – Y. Here it is. –3 –2 –1 0 1 2 3 4 Value di: Probability pi: 0.015 0.055 0.145 0.235 0.260 0.195 0.080 0.015 You can use the probability distribution to confirm that: 1. mD = mX − mY = 3.75 − 3.10 = 0.65 2. s2D = s2X + s2Y = 1.1875 + 0.89 = 2.0775 3. sD = !2.0775 = 1.441

Result 2 shows that, just like with addition, when we subtract two independent random variables, variances add.

Variance of the Difference of Random Variables For any two independent random variables X and Y, if D = X − Y, then the variance of D is s2D = s2X + s2Y

Let’s put our new rules for subtracting random variables to use in a familiar setting.

EXAMPLE

Pete’s Jeep Tours and Erin’s ­Adventures Difference of random variables We have defined several random variables related to Pete’s and Erin’s tour businesses. For a randomly selected day, C = amount of money that Pete collects G = amount of money that Erin collects Here are the means and standard deviations of these random variables:

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mC = 562.50 sC = 163.46

mG = 542.50 sG = 165.03

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Problem:  Calculate the mean and the standard deviation of the difference D = C − G in the amounts that Pete and Erin collect on a randomly chosen day. Interpret each value in context. Solution:  We know that the mean of the difference of two random variables is the difference of their means. That is, mD = mC − mG = 562.50 − 542.50 = 20.00 On average, Pete collects $20 more per day than Erin does. Some days the difference will be more than $20, other days it will be less, but the average difference after lots of days will be about $20. Because the number of passengers X and Y are independent random variables, so are the amounts of money collected C and G. Therefore, the variance of D is the sum of the variances of C and G: s2D = s2C + s2G = (163.46)2 + (165.03)2 = 53,954.07 The value sD = $232.28 in the example should look familiar. It’s the same value we got earlier when we calculated the standard deviation of the total amount that Pete and Erin collect on a randomly chosen day, sT = $232.28.

sD = !53,954.07 = 232.28

The standard deviation of the difference in the amounts collected by Pete and Erin is $232.28. Even though the average difference in the amounts collected is $20, the difference on individual days will typically vary from the mean by about $232. For Practice Try Exercise

55

CHECK YOUR UNDERSTANDING

A large auto dealership keeps track of sales and lease agreements made during each hour of the day. Let X = the number of cars sold and Y = the number of cars leased during the first hour of business on a randomly selected Friday. Based on previous records, the probability distributions of X and Y are as follows: Cars sold xi: 0 1 2 3 Probability pi: 0.3 0.4 0.2 0.1 Mean: mX = 1.1   Standard deviation: sX = 0.943 Cars leased yi: 0 1 2 Probability pi: 0.4 0.5 0.1 Mean: mY = 0.7   Standard deviation: sY = 0.64 Define D = X − Y. Assume that X and Y are independent. 1. Find and interpret mD. 2. Compute sD. Show your work. 3. The dealership’s manager receives a $500 bonus for each car sold and a $300 bonus for each car leased. Find the mean and standard deviation of the difference in the manager’s bonus for cars sold and leased. Show your work.

Combining Normal Random Variables So far, we have concentrated on finding rules for means and variances of random variables. If a random variable is Normally distributed, we can use its mean

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and standard deviation to compute probabilities. The earlier example on young women’s heights (page 357) shows the method. What if we combine two Normal random variables?

EXAMPLE

A Computer Simulation Sums and differences of Normal random variables We used Fathom software to simulate taking independent SRSs of 1000 observations from each of two Normally distributed random variables, X and Y. ­Figure 6.11(a) shows the results. The random variable X is N(3, 0.9) and the random variable Y is N(1, 1.2). What do we know about the sum and difference of these two random variables? The histograms in Figure 6.11(b) came from adding and subtracting the values of X and Y for the 1000 randomly generated observations from each distribution.

​(a)

​(b) FIGURE 6.11  ​(a) Histograms showing the results of randomly selecting 1000 values from two different Normal random variables X and Y. (b) Histograms of the sum and difference of the 1000 randomly selected values of X and Y.

Let’s summarize what we see: Sum X + Y

Difference X − Y

Shape: Looks approximately Normal

Looks approximately Normal

Center: About 4, which makes sense because

About 2, which makes sense because

mX + Y = mX + mY = 3+1 = 4

mX − Y = mX − mY = 3 − 1 = 2

Spread: The spreads of the two distributions are about the same. That makes sense because s2X+Y = s2X−Y = s2X + s2Y

As the previous example illustrates, any sum or difference of independent Normal random variables is also Normally distributed. The mean and standard deviation of the resulting Normal distribution can be found using the appropriate rules for means and variances.

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Give Me Some Sugar! Sums of Normal random variables Mr. Starnes likes sugar in his hot tea. From experience, he needs between 8.5 and 9 grams of sugar in a cup of tea for the drink to taste right. While making his tea one morning, Mr. Starnes adds four randomly selected packets of sugar. Suppose the amount of sugar in these packets follows a Normal distribution with mean 2.17 grams and standard deviation 0.08 grams.

PROBLEM:  What’s the probability that Mr. Starnes’s tea tastes right? SOLUTION: Step 1:  State the distribution and the values of interest. Let X = the amount of sugar in a randomly selected packet. Then X1 = amount of sugar in Packet 1, X2 = amount of sugar in Packet 2, X3 = amount of sugar in Packet 3, and X4 = amount of sugar in Packet 4. Each of these random variables has a Normal distribution with mean 2.17 grams and standard deviation 0.08 grams. We’re interested in the total amount of sugar that Mr. Starnes puts in his tea, which is given by T = X1 + X2 + X3 + X4. The random variable T is a sum of four independent Normal random variables. So T follows a Normal distribution with mean mT = mX1 + mX2 + mX3 + mX4 = 2.17 + 2.17 + 2.17 + 2.17 = 8.68 grams and variance s2T = s2X1 + s2X2 + s2X3 + s2X4 = (0.08)2 + (0.08)2 + (0.08)2 + (0.08)2 = 0.0256 The standard deviation of T is sT = !0.0256 = 0.16 grams

We want to find the probability that the total amount of sugar in Mr. Starnes’s tea is between 8.5 and 9 grams. Figure 6.12 shows this probability as the area under a Normal curve. N(8.68,0.16)

Step 2:  Perform calculations—show your work! To find this area, we can standardize the boundary values and use Table A: z=

8.50 8.68 9.00 Total amount of sugar (grams)

FIGURE 6.12  ​Normal distribution of the total amount of sugar in Mr. Starnes’s tea.

8.5 − 8.68 9 − 8.68 = −1.13  and  z = = 2.00 0.16 0.16

Then P(−1.13 ≤ Z ≤ 2.00) = 0.9772 − 0.1292 = 0.8480. Using Technology : The command normalcdf(lower:8.5, upper:9, m:8.68, s:0.16) gives an area of 0.8470. Step 3:  Answer the question. There’s about an 85% chance that Mr. Starnes’s tea will taste right. For Practice Try Exercise

61

Here’s an example that involves subtracting two Normal random variables.

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EXAMPLE

381

Put a Lid on It! Differences of Normal random variables The diameter C of a randomly selected large drink cup at a fast-food restaurant follows a Normal distribution with a mean of 3.96 inches and a standard deviation of 0.01 inches. The diameter L of a randomly selected large lid at this restaurant follows a Normal distribution with mean 3.98 inches and standard deviation 0.02 inches. For a lid to fit on a cup, the value of L has to be bigger than the value of C, but not by more than 0.06 inches.

PROBLEM:  What’s the probability that a randomly selected large lid will fit on a randomly chosen

large drink cup?

SOLUTION: Step 1:  State the distribution and the values of interest. We’ll define the random variable D = L − C to represent the difference between the lid’s diameter and the cup’s diameter. The random variable D is the difference of two independent Normal random variables. So D follows a Normal distribution with mean mD  = mL − mC  = 3.98 − 3.96 = 0.02 and variance s2D = s2L + s2C = (0.02)2 + (0.01)2 = 0.0005 The standard deviation of D is sD = !0.0005 = 0.0224

We want to find the probability that the difference D is between 0 and 0.06 inches. Figure 6.13 shows this probability as the area under a Normal curve. Step 2:  Perform calculations—show your work! To find this area, we can standardize the boundary values and use Table A: N(0.02,0.0224)

0 0.02 0.06 Difference in diameter (lid - cup), in inches

FIGURE 6.13  Normal distribution of the difference in lid diameter and cup diameter at a fast-food restaurant.

z=

0 − 0.02 0.06 − 0.02 = −0.89  and  z = = 1.79 0.0224 0.0224

Then P(−0.89 ≤ Z ≤ 1.79) = 0.9633 − 0.1867 = 0.7766. Using Technology :  The command normalcdf(lower:0, upper:0.06, m:0.02, s:0.0224) gives an area of 0.7770. Step 3:  Answer the question. There’s about a 78% chance that a randomly selected large lid will fit on a randomly chosen large drink cup at this fast-food restaurant. Roughly 22% of the time, the lid won’t fit. This seems like an unreasonably high chance of getting a lid that doesn’t fit. Maybe the restaurant should find a new supplier! For Practice Try Exercise

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Summary •

Adding a positive constant a to (subtracting a from) a random variable increases (decreases) the mean of the random variable by a but does not affect its standard deviation or the shape of its probability distribution. • Multiplying (dividing) a random variable by a positive constant b multiplies (divides) the mean of the random variable by b and the standard deviation by b but does not change the shape of its probability distribution. • A linear transformation of a random variable involves adding or subtracting a constant a, multiplying or dividing by a constant b, or both. We can write a linear transformation of the random variable X in the form Y = a + bX. The shape, center, and spread of the probability distribution of Y are as follows: Shape: Same as the probability distribution of X if b > 0. Center: mY = a + bmX Spread: sY = |b|sX • If X and Y are any two random variables,  mX + Y = mX + mY:  The mean of the sum of two random variables is the sum of their means.  mX − Y = mX − mY:  The mean of the difference of two random variables is the difference of their means. • If X and Y are independent random variables, then knowing the value of one variable tells you nothing about the value of the other. In that case, variances add:  s2X+Y = s2X + s2Y :  The variance of the sum of two independent random variables is the sum of their variances.  s2X−Y = s2X + s2Y :  The variance of the difference of two independent random variables is the sum of their variances. • The sum or difference of independent Normal random variables follows a Normal distribution.

Section 6.2

Exercises

35. Crickets  The length in inches of a cricket chosen at random from a field is a random variable X with mean 1.2 inches and standard deviation 0.25 inches. Find the mean and standard deviation of the length Y of a randomly chosen cricket from the field in centimeters. There are 2.54 centimeters in an inch. 36. Men’s heights  A report of the National Center for Health Statistics says that the height of a 20-year-old man chosen at random is a random variable H with mean 5.8 feet and standard deviation 0.24 feet. Find the mean and standard deviation of the height J of a

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randomly selected 20-year-old man in inches. There are 12 inches in a foot. 37. Get on the boat!  A small ferry runs every half hour from one side of a large river to the other. The number of cars X on a randomly chosen ferry trip has the probability distribution shown below. You can check that mX = 3.87 and sX = 1.29. Cars: 0 1 2 3 4 5 Probability: 0.02 0.05 0.08 0.16 0.27 0.42

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(a) The cost for the ferry trip is $5. Make a graph of the probability distribution for the random variable M = money collected on a randomly selected ferry trip. Describe its shape.

40. Easy quiz 

(b) Find and interpret mM.

(c) What shape would the probability distribution of G have? Justify your answer.

(c) Find and interpret sM. 38. Skee Ball  Ana is a dedicated Skee Ball player (see photo) who always rolls for the 50-point slot. The probability distribution of Ana’s score X on a single roll of the ball is shown below. You can check that mX = 23.8 and sX = 12.63. Score: 10 20 30 40 50 Probability: 0.32 0.27 0.19 0.15 0.07

(a) A player receives one ticket from the game for every 10 points scored. Make a graph of the probability distribution for the random variable T = number of tickets Ana gets on a randomly selected throw. Describe its shape. (b) Find and interpret mT. (c) Find and interpret sT. Exercises 39 and 40 refer to the following setting. Ms. Hall gave her class a 10-question multiple-choice quiz. Let X = the number of questions that a randomly selected student in the class answered correctly. The computer output below gives information about the probability distribution of X. To determine each student’s grade on the quiz (out of 100), Ms. Hall will multiply his or her number of correct answers by 5 and then add 50. Let G = the grade of a randomly chosen student in the class. N

Mean

Median

30

7.6

8.5

StDev Min Max 1.32

4

10

Q1

Q3

8

9

(a) Find the median of G. Show your method. (b) Find the IQR of G. Show your method.

41. Get on the boat!  Refer to Exercise 37. The ferry company’s expenses are $20 per trip. Define the random variable Y to be the amount of profit (money collected minus expenses) made by the ferry company on a randomly selected trip. That is, Y = M − 20. (a) Find and interpret the mean of Y. (b) Find and interpret the standard deviation of Y. 42. The Tri-State Pick 3  Most states and Canadian provinces have government-sponsored lotteries. Here is a simple lottery wager, from the Tri-State Pick 3 game that New Hampshire shares with Maine and Vermont. You choose a number with 3 digits from 0 to 9; the state chooses a three-digit winning number at random and pays you $500 if your number is chosen. Because there are 1000 numbers with three digits, you have probability 1/1000 of winning. Taking X to be the amount your ticket pays you, the probability distribution of X is Payoff: $0 $500 Probability: 0.999 0.001 (a) Show that the mean and standard deviation of X are mX = $0.50 and sX = $15.80. (b) If you buy a Pick 3 ticket, your winnings are W = X − 1, because it costs $1 to play. Find the mean and standard deviation of W. Interpret each of these values in context. 43. Get on the boat!  Based on the analysis in Exercise 41, the ferry company decides to increase the cost of a trip to $6. We can calculate the company’s profit Y on a randomly selected trip from the number of cars X. Find the mean and standard deviation of Y. Show your work. 44. Making a profit  Rotter Partners is planning a major investment. From experience, the amount of profit X (in millions of dollars) on a randomly selected investment of this type is uncertain, but an estimate gives the following probability distribution: Profit: 1 1.5 2 4 10 Probability: 0.1 0.2 0.4 0.2 0.1

39. Easy quiz  (a) Find the mean of G. Show your method. (b) Find the standard deviation of G. Show your method. (c) How do the variance of X and the variance of G compare? Justify your answer.

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Based on this estimate, mX = 3 and sX = 2.52. Rotter Partners owes its lender a fee of $200,000 plus 10% of the profits X. So the firm actually retains Y = 0.9X − 0.2 from the investment. Find the mean and standard deviation of Y. Show your work.

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45. Too cool at the cabin?  During the winter months, pg 368 the temperatures at the Starneses’ Colorado cabin can stay well below freezing (32°F or 0°C) for weeks at a time. To prevent the pipes from freezing, Mrs. Starnes sets the thermostat at 50°F. She also buys a digital thermometer that records the indoor temperature each night at midnight. Unfortunately, the thermometer is programmed to measure the temperature in degrees Celsius. Based on several years’ worth of data, the temperature T in the cabin at midnight on a randomly selected night follows a Normal distribution with mean 8.5°C and standard deviation 2.25°C. (a) Let Y = the temperature in the cabin at midnight on a randomly selected night in degrees Fahrenheit (recall that F = (9/5)C + 32). Find the mean and standard deviation of Y.

49. Get on the boat!  Refer to Exercise 41. Find the expected value and standard deviation of the total amount of profit made on two randomly selected days. Show your work. 50. The Tri-State Pick 3  Refer to Exercise 42. Suppose you buy one Pick 3 ticket on each of two consecutive days. Find the expected value and standard deviation of your total winnings. Show your work. 51. Essay errors  Typographical and spelling errors can be pg 374 either “nonword errors” or “word errors.” A nonword error is not a real word, as when “the” is typed as “teh.” A word error is a real word, but not the right word, as when “lose” is typed as “loose.” When students are asked to write a 250-word essay (without spell-checking), the number of nonword errors X in a randomly selected essay has the following probability distribution: Value: 0 1 2 3 4 Probability: 0.1 0.2 0.3 0.3 0.1

(b) Find the probability that the midnight temperature in the cabin is below 40°F. Show your work.

mX = 2.1  sX = 1.136

46. Cereal  A company’s single-serving cereal boxes advertise 9.63 ounces of cereal. In fact, the amount of cereal X in a randomly selected box follows a Normal distribution with a mean of 9.70 ounces and a standard deviation of 0.03 ounces. (a) Let Y = the excess amount of cereal beyond what’s advertised in a randomly selected box, measured in grams (1 ounce = 28.35 grams). Find the mean and standard deviation of Y.

The number of word errors Y has this probability distribution: Value: 0 1 2 3 Probability: 0.4 0.3 0.2 0.1 mY = 1.0  sY = 1.0

(b) Find the probability of getting at least 3 grams more cereal than advertised. Show your work. 47. His and her earnings  Researchers randomly select a pg 373 married couple in which both spouses are employed. Let X be the income of the husband and Y be the income of the wife. Suppose that you know the means mX and mY and the variances s2X and s2Y of both variables.

52.

(a) Is it reasonable to take the mean of the total income X + Y to be mX + mY? Explain your answer. (b) Is it reasonable to take the variance of the total income to be s2X + s2Y ? Explain your answer. 48. Rainy days  Imagine that we randomly select a day from the past 10 years. Let X be the recorded rainfall on this date at the airport in Orlando, Florida, and Y be the recorded rainfall on this date at Disney World just outside Orlando. Suppose that you know the means mX and mY and the variances s2X and s2Y of both variables. (a) Is it reasonable to take the mean of the total rainfall X + Y to be mX + mY? Explain your answer. (b) Is it reasonable to take the variance of the total rainfall to be s2X + s2Y ? Explain your answer.

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53. (a)

(b)

54.

Assume that X and Y are independent. An English professor deducts 3 points from a student’s essay score for each nonword error and 2 points for each word error. Find the mean and standard deviation of the total score deductions for a randomly selected essay. Show your work. The Tri-State Pick 3  Refer to Exercise 42. You and a friend decide to play Pick 3, but with two different strategies. Your friend buys a $1 Pick 3 ticket on each of five consecutive days. You bet $5 on a single number on your Pick 3 ticket. Find the mean and standard deviation of the total winnings for you and your friend. Show your work. Essay errors  Refer to Exercise 51. Find the mean and standard deviation of the difference Y − X in the number of errors made by a randomly selected student. Interpret each value in context. From the information given, can you find the probability that a randomly selected student makes more word errors than nonword errors? If so, find this probability. If not, explain why not. Study habits  The Survey of Study Habits and Attitudes (SSHA) is a psychological test that measures academic motivation and study habits. The distribution of SSHA scores among the women at a college has mean 120 and standard deviation 28, and the distribution of scores among male students has mean

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Section 6.2 Transforming and Combining Random Variables 105 and standard deviation 35. You select a single male student and a single female student at random and give them the SSHA test. (a) Find the mean and standard deviation of the difference (female minus male) between their scores. Interpret each value in context. (b) From the information given, can you find the probability that the woman chosen scores higher than the man? If so, find this probability. If not, explain why you cannot. 55. Essay scores  Refer to Exercise 51. Find the mean and standard deviation of the difference in score deductions (nonword – word) for a randomly selected essay. Show your work.

pg 377

56. The Tri-State Pick 3  Refer to Exercise 52. Find the mean and standard deviation of the difference (you – your friend) in winnings. Show your work. Exercises 57 and 58 refer to the following setting. In Exercises 14 and 18 of Section 6.1, we examined the probability distribution of the random variable X = the amount a life insurance company earns on a randomly chosen 5-year term life policy. Calculations reveal that mX = $303.35 and sX = $9707.57. 57. Life insurance  The risk of insuring one person’s life is reduced if we insure many people. Suppose that we insure two 21-year-old males, and that their ages at death are independent. If X1 and X2 are the insurer’s income from the two insurance policies, the insurer’s average income W on the two policies is X1 + X2 2 Find the mean and standard deviation of W. (You see that the mean income is the same as for a single policy but the standard deviation is less.) W=



58. Life insurance  If four 21-year-old men are insured, the insurer’s average income is X1 + X2 + X3 + X4 4 where Xi is the income from insuring one man. Assuming that the amount of income earned on individual policies is independent, find the mean and standard deviation of V. (If you compare with the results of Exercise 57, you should see that averaging over more insured individuals reduces risk.) V=

59. Time and motion  A time-and-motion study measures the time required for an assembly-line worker to perform a repetitive task. The data show that the time required to bring a part from a bin to its position on an automobile chassis varies from car to car according to a Normal distribution with mean 11 seconds and standard deviation 2 seconds. The time required to attach the part to the chassis follows a Normal distri-

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bution with mean 20 seconds and standard deviation 4 seconds. The study finds that the times required for the two steps are independent. A part that takes a long time to position, for example, does not take more or less time to attach than other parts. (a) What is the distribution of the time required for the entire operation of positioning and attaching a randomly selected part? (b) Management’s goal is for the entire process to take less than 30 seconds. Find the probability that this goal will be met for a randomly selected part. 60. Electronic circuit  The design of an electronic circuit for a toaster calls for a 100-ohm resistor and a 250-ohm resistor connected in series so that their resistances add. The components used are not perfectly uniform, so that the actual resistances vary independently according to Normal distributions. The resistance of 100-ohm resistors has mean 100 ohms and standard deviation 2.5 ohms, while that of 250-ohm resistors has mean 250 ohms and standard deviation 2.8 ohms. (a) What is the distribution of the total resistance of the two components in series for a randomly selected toaster? (b) Find the probability that the total resistance for a randomly selected toaster lies between 345 and 355 ohms. 61. Swim team  Hanover High School has the best pg 380 women’s swimming team in the region. The 400-meter freestyle relay team is undefeated this year. In the 400-meter freestyle relay, each swimmer swims 100 meters. The times, in seconds, for the four swimmers this season are approximately Normally distributed with means and standard deviations as shown. Assume that the swimmer’s individual times are independent. Find the probability that the total team time in the 400-meter freestyle relay for a randomly selected race is less than 220 seconds. Swimmer

Mean

Std. dev.

Wendy

55.2

2.8

Jill

58.0

3.0

Carmen

56.3

2.6

Latrice

54.7

2.7

62. Toothpaste  Ken is traveling for his business. He has a new 0.85-ounce tube of toothpaste that’s supposed to last him the whole trip. The amount of toothpaste Ken squeezes out of the tube each time he brushes varies according to a Normal distribution with mean 0.13 ounces and standard deviation 0.02 ounces. If Ken brushes his teeth six times on a randomly selected trip, what’s the probability that he’ll use all the toothpaste in the tube?

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63. Auto emissions  The amount of nitrogen oxides pg 381 (NOX) present in the exhaust of a particular type of car varies from car to car according to a Normal distribution with mean 1.4 grams per mile (g/mi) and standard deviation 0.3 g/mi. Two randomly selected cars of this type are tested. One has 1.1 g/mi of NOX; the other has 1.9 g/mi. The test station attendant finds this difference in emissions between two similar cars surprising. If the NOX levels for two randomly chosen cars of this type are independent, find the probability that the difference is at least as large as the value the attendant observed. 64. Loser buys the pizza  Leona and Fred are friendly competitors in high school. Both are about to take the ACT college entrance examination. They agree that if one of them scores 5 or more points better than the other, the loser will buy the winner a pizza. Suppose that in fact Fred and Leona have equal ability, so that each score varies Normally with mean 24 and standard deviation 2. (The variation is due to luck in guessing and the accident of the specific questions being familiar to the student.) The two scores are independent. What is the probability that the scores differ by 5 or more points in either direction? Multiple choice: Select the best answer for Exercises 65 and 66, which refer to the following setting. The number of calories in a 1-ounce serving of a certain breakfast cereal is a random variable with mean 110 and standard deviation 10. The number of calories in a cup of whole milk is a random variable with mean 140 and standard deviation 12. For breakfast, you eat 1 ounce of the cereal with 1/2 cup of whole milk. Let T be the random variable that represents the total number of calories in this breakfast.

6.3 What You Will Learn

65. The mean of T is (a) 110. (b) 140. (c) 180. (d) 195. (e) 250. 66. The standard deviation of T is (a) 22.

(b) 16. (c) 15.62. (d) 11.66. (e) 4.

67. Statistics for investing (3.1)  Joe’s retirement plan invests in stocks through an “index fund” that follows the behavior of the stock market as a whole, as measured by the Standard & Poor’s (S&P) 500 stock index. Joe wants to buy a mutual fund that does not track the index closely. He reads that monthly returns from Fidelity Technology Fund have correlation r = 0.77 with the S&P 500 index and that Fidelity Real Estate Fund has correlation r = 0.37 with the index. (a) Which of these funds has the closer relationship to returns from the stock market as a whole? How do you know? (b) Does the information given tell Joe anything about which fund has had higher returns? 68. Buying stock (5.3, 6.1)  You purchase a hot stock for $1000. The stock either gains 30% or loses 25% each day, each with probability 0.5. Its returns on consecutive days are independent of each other. You plan to sell the stock after two days. (a) What are the possible values of the stock after two days, and what is the probability for each value? What is the probability that the stock is worth more after two days than the $1000 you paid for it? (b) What is the mean value of the stock after two days? (Comment: You see that these two criteria give different answers to the question “Should I invest?”)

Binomial and Geometric Random Variables By the end of the section, you should be able to:

• Determine whether the conditions for using a binomial random variable are met. • Compute and interpret probabilities involving binomial distributions. • Calculate the mean and standard deviation of a binomial random variable. Interpret these values in context.

• Find probabilities involving geometric random ­variables. • When appropriate, use the Normal approximation to the binomial distribution to calculate probabilities.*

*This topic is not required for the AP® Statistics exam. Some teachers prefer to discuss this topic when presenting the sampling distribution of pˆ (Chapter 7).

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When the same chance process is repeated several times, we are often interested in whether a particular outcome does or doesn’t happen on each repetition. Here are some examples: • To test whether someone has extrasensory perception (ESP), choose one of four cards at random—a star, wave, cross, or circle. Ask the person to identify the card without seeing it. Do this a total of 50 times and see how many cards the person identifies correctly. Chance process: choose a card at random. Outcome of interest: person identifies card correctly. Random variable: X = number of correct identifications. • A shipping company claims that 90% of its shipments arrive on time. To test this claim, take a random sample of 100 shipments made by the company last month and see how many arrived on time. Chance process: Randomly select a shipment and check when it arrived. Outcome of interest: arrived on time. Random variable: Y = number of on-time shipments. • In the game of Pass the Pigs, a player rolls a pair of pig-shaped dice. On each roll, the player earns points according to how the pigs land. If the player gets a “pig out,” in which the two pigs land on opposite sides, she loses all points earned in that round and must pass the pigs to the next player. A player can choose to stop rolling at any point during her turn and to keep the points that she has earned before passing the pigs. Chance process: roll the pig dice. Outcome of interest: pig out. Random variable: T = number of rolls until the player pigs out. Some random variables, like X and Y in the first two examples above, count the number of times the outcome of interest occurs in a fixed number of repetitions. They are called binomial random variables. Other random variables, like T in the Pass the Pigs setting, count the number of repetitions of the chance process it takes for the outcome of interest to occur. They are known as geometric random variables. These two special types of discrete random variables are the focus of this section.

Binomial Settings and Binomial Random Variables What do the following scenarios have in common? • Toss a coin 5 times. Count the number of heads. • Spin a roulette wheel 8 times. Record how many times the ball lands in a red slot. • Take a random sample of 100 babies born in U.S. hospitals today. Count the number of females. In each case, we’re performing repeated trials of the same chance process. The number of trials is fixed in advance. Also, knowing the outcome of one trial tells us nothing about the outcome of any other trial. That is, the trials are independent. We’re interested in the number of times that a specific event (we’ll call it a “success”) occurs. Our chances of getting a “success” are the same on each trial. When these conditions are met, we have a binomial setting.

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Definition: Binomial setting A binomial setting arises when we perform several independent trials of the same chance process and record the number of times that a particular outcome occurs. The four conditions for a binomial setting are

The boldface letters in the box give you a helpful way to remember the conditions for a binomial setting: just check the BINS!

• Binary? The possible outcomes of each trial can be classified as “success” or “failure.” • Independent? Trials must be independent; that is, knowing the result of one trial must not tell us anything about the result of any other trial. • Number? The number of trials n of the chance process must be fixed in advance. • Success? There is the same probability p of success on each trial. When checking the Binary condition, note that there can be more than two possible outcomes per trial—a roulette wheel has numbered slots of three colors: red, black, and green. If we define “success” as having the ball land in a red slot, then “failure” occurs when the ball lands in a black or a green slot. Think of tossing a coin n times as an example of the binomial setting. Each toss gives either heads or tails. Knowing the outcome of one toss doesn’t change the probability of a head on any other toss, so the tosses are independent. If we call heads a success, then p is the probability of a head and remains the same as long as we toss the same coin. For tossing a fair coin, p is 0.5. The number of heads we count is a binomial random variable X. The probability distribution of X is called a binomial distribution.

Definition: Binomial random variable and binomial distribution The count X of successes in a binomial setting is a binomial random variable. The probability distribution of X is a binomial distribution with parameters n and p, where n is the number of trials of the chance process and p is the probability of a success on any one trial. The possible values of X are the whole numbers from 0 to n. Later in the section, we’ll learn how to assign probabilities to outcomes and how to find the mean and standard deviation of a binomial random variable. For now, it’s important to be able to distinguish situations in which a binomial distribution does and doesn’t apply.

EXAMPLE

From Blood Types to Aces Binomial settings and random variables Problem:  Here are three scenarios involving chance behavior. In each case, determine whether or not the given random variable has a binomial distribution. Justify your answer. (a) Genetics says that children receive genes from each of their parents independently. Each child of a particular set of parents has probability 0.25 of having type O blood. Suppose these parents have 5 children. Let X = the number of children with type O blood.

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(b) Shuffle a deck of cards. Turn over the first 10 cards, one at a time. Let Y = the number of aces you observe. (c) Shuffle a deck of cards. Turn over the top card. Put the card back in the deck, and shuffle again. Repeat this process until you get an ace. Let W = the number of cards required.

Solution: 

The Independent condition involves conditional probabilities. P(2nd card ace | 1st card ace) = 3/51 ≠ P(2nd card ace) = 4/52, so the trials are not independent. The Success condition is about unconditional probabilities. P(k th card in a shuffled deck is an ace) = 4/52, so this condition is met.

(a) To see if this is a binomial setting, we’ll check the BINS: • Binary? “Success” = has type O blood. “Failure” = doesn’t have type O blood. • Independent? Knowing one child’s blood type tells you nothing about another’s because children inherit genes determining blood type independently from each of their parents. • Number? There are n = 5 trials of this chance process. • Success? The probability of a success is p = 0.25 on each trial. This is a binomial setting. Because X counts the number of successes, it is a binomial random variable with parameters n = 5 and p = 0.25. (b) Let’s check the BINS: • Binary? “Success” = get an ace. “Failure” = don’t get an ace. • Independent? No. If the first card you turn over is an ace, then the next card is less likely to be an ace because you’re not replacing the top card in the deck. Similarly, if the first card isn’t an ace, the second card is more likely to be an ace. • Number? There are n = 10 trials of this chance process. • Success? The probability that any particular card in a shuffled deck is an ace is p = 4/52. Because the trials are not independent, this is not a binomial setting. (c) Let’s check the BINS: • Binary? “Success” = get an ace. “Failure” = don’t get an ace. • Independent? Because you are replacing the card in the deck and shuffling each time, the result of one trial does not tell you anything about the outcome of any other trial. • Number? The number of trials is not set in advance. You could get an ace on the first card you turn over, or it may take many cards to get an ace. • Success? The probability of getting an ace is p = 4/52 on each trial. Because there is no fixed number of trials, this is not a binomial setting. For Practice Try Exercises

69

and

71

Part (c) of the example raises an important point about binomial random variables. Besides checking the BINS, make sure that you’re being asked to count the number of successes in a certain number of trials. In part (c), you’re asked to count the number of trials until you get a success. That can’t be a binomial random variable. (As you’ll see later, W is a geometric random variable.)

CHECK YOUR UNDERSTANDING

For each of the following situations, determine whether the given random variable has a binomial distribution or not. Justify your answer. 1. Shuffle a deck of cards. Turn over the top card. Put the card back in the deck, and shuffle again. Repeat this process 10 times. Let X = the number of aces you observe.

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R a n d o m Va r i a b l e s 2. Choose students at random from your class. Let Y = the number who are over 6 feet tall. 3. Roll a fair die 100 times. Sometime during the 100 rolls, one corner of the die chips off. Let W = the number of 5s you roll.

Binomial Probabilities In a binomial setting, we can define a random variable (say, X) as the number of successes in n independent trials. What’s the probability distribution of X? Let’s see if an example can help shed some light on this question.

EXAMPLE

Inheriting Blood Type Calculating binomial probabilities Each child of a particular set of parents has probability 0.25 of having type O blood. Genetics says that children receive genes from each of their parents independently. If these parents have 5 children, the count X of children with type O blood is a binomial random variable with n = 5 trials and probability p = 0.25 of success on each trial. In this setting, a child with type O blood is a “success” (S) and a child with another blood type is a “failure” (F). What’s P(X = 0)? That is, what’s the probability that none of the 5 children has type O blood? It’s the chance that all 5 children don’t have type O blood. The probability that any one of this couple’s children doesn’t have type O blood is 1 − 0.25 = 0.75 (complement rule). By the multiplication rule for independent events (Chapter 5), P(X = 0) = P(FFFFF) = (0.75)(0.75)(0.75)(0.75)(0.75) = (0.75)5 = 0.2373 How about P(X = 1)? There are several different ways in which exactly 1 of the 5 children could have type O blood. For instance, the first child born might have type O blood, while the remaining 4 children don’t have type O blood. The probability that this happens is P(SFFFF) = (0.25)(0.75)(0.75)(0.75)(0.75) = (0.25)(0.75)4 Alternatively, Child 2 could be the one that has type O blood. The corresponding probability is P(FSFFF) = (0.75)(0.25)(0.75)(0.75)(0.75) = (0.25)(0.75)4 There are three more possibilities to consider: P(FFSFF) = (0.75)(0.75)(0.25)(0.75)(0.75) = (0.25)(0.75)4 P(FFFSF) = (0.75)(0.75)(0.75)(0.25)(0.75) = (0.25)(0.75)4 P(FFFFS) = (0.75)(0.75)(0.75)(0.75)(0.25) = (0.25)(0.75)4

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In all, there are five different ways in which exactly 1 child would have type O blood, each with the same probability of occurring. As a result, P(X = 1) = P(exactly 1 child with type O blood) = P(SFFFF) + P(FSFFF) + P(FFSFF) + P(FFFSF) + P(FFFFS) = 5(0.25)(0.75)4 = 0.39551 There’s about a 40% chance that exactly 1 of the couple’s 5 children will have type O blood.

Let’s continue with the scenario from the previous example. What if we wanted to find P(X = 2), the probability that exactly 2 of the couple’s children have type O blood? Because the method doesn’t depend on the specific setting, we use “S” for success and “F” for failure for short. Do the work in two stages, as shown in the example. • Find the probability that a specific 2 of the 5 tries—say, the first and the third—give successes. This is the outcome SFSFF. Because tries are independent, the multiplication rule for independent events applies. The probability we want is P(SFSFF) = P(S)P(F)P(S)P(F)P(F) = (0.25)(0.75)(0.25)(0.75)(0.75) = (0.25)2(0.75)3 • Observe that any one arrangement of 2 S’s and 3 F’s has this same probability. This is true because we multiply together 0.25 twice and 0.75 three times whenever we have 2 S’s and 3 F’s. The probability that X = 2 is the probability of getting 2 S’s and 3 F’s in any arrangement whatsoever. Here are all the possible arrangements: SSFFF  SFSFF  SFFSF  SFFFS  FSSFF FSFSF  FSFFS  FFSSF  FFSFS  FFFSS There are 10 of them, all with the same probability. The overall probability of 2 successes is therefore P(X = 2) = 10(0.25)2(0.75)3 = 0.26367 The pattern of this calculation works for any binomial probability. That is, P(X = k) = P(exactly k successes in n trials) = number of arrangements # pk(1 − p)n−k To use this formula, we must count the number of arrangements of k successes in n observations. This number is called the binomial coefficient. We use the following fact to do the counting without actually listing all the arrangements.

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Definition: Binomial coefficient The number of ways of arranging k successes among n observations is given by the binomial coefficient n n! a b= k k!(n − k)!

for k = 0, 1, 2, . . . , n where

n! = n(n−1)(n−2) · . . . · (3)(2)(1) and 0! = 1. The formula for binomial coefficients uses the factorial notation. For any positive whole number n, its factorial n! is n! = n(n − 1)(n − 2) · … · (3)(2)(1) We also define 0! = 1. The larger of the two factorials in the denominator of a binomial coefficient will cancel much of the n! in the numerator. For example, the binomial coefficient we need to find the probability that exactly 2 of the couple’s 5 children inherit type O blood is

12. TECHNOLOGY CORNER

5

c

The binomial coefficient is (2) not related to the fraction 52 . A helpful autio way to remember its meaning is to read it as “5 choose 2.” Binomial coefficients have many uses, but we are interested in them only as an aid to finding binomial probabilities. If you need to compute a binomial coefficient, use your calculator.

!

n

Some people prefer the notation 5 5C2 instead of ( 2) for the binomial coefficient.

(5)(4)(3)(2)(1) (5)(4) 5 5! = = = 10 a b= 2 2!3! (2)(1)(3)(2)(1) (2)(1)

Binomial coefficients on the calculator TI-Nspire instructions in Appendix B; HP Prime instructions on the book’s Web site. 5

To calculate a binomial coefficient like (2) on the TI-83/84 or TI-89, proceed as ­follows:

TI-83/84

TI-89

• Type 5, press MATH , arrow over to PRB, choose nCr, and press ENTER . Then type 2 and press ENTER again to execute the command 5 nCr 2.

• From the home screen, press 2nd 5 (MATH), choose Probability, nCr(, and press ENTER . Complete the command nCr(5,2) and press ENTER .

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The binomial coefficient ( k) counts the number of different ways in which k successes can be arranged among n trials. The binomial probability P(X = k) is this count multiplied by the probability of any one specific arrangement of the k successes. Here is the result we seek.

Binomial Probability FORMULA If X has the binomial distribution with n trials and probability p of success on each trial, the possible values of X are 0, 1, 2, . . . , n. If k is any one of these values, n P(X = k) = a bpk(1 − p)n−k k With our formula in hand, we can now calculate any binomial probability.

EXAMPLE

Inheriting Blood Type Using the binomial probability formula Problem:  Each child of a particular set of parents has probability 0.25 of having type O blood. Suppose the parents have 5 children. (a) Find the probability that exactly 3 of the children have type O blood. (b) Should the parents be surprised if more than 3 of their children have type O blood? Justify your answer. Solution:  Let X = the number of children with type O blood. We know that X has a binomial distribution with n = 5 and p = 0.25. (a) We want to find P(X = 3).

(5)

P(X = 3) = 3 (0.25)3(0.75)2 = 10(0.25)3(0.75)2 = 0.08789 There is about a 9% chance that exactly 3 of the 5 children have type O blood. (b) To answer this question, we need to find P(X > 3).

(5)

(5)

P(X > 3) = P(X = 4) + P(X = 5) = 4 (0.25)4(0.75)1 + 5 (0.25)5(0.75)0 = 5(0.25)4(0.75)1 + 1(0.25)5(0.75)0 = 0.01465 + 0.00098 = 0.01563 Because there’s only about a 1.5% chance of having more than 3 children with type O blood, the parents should definitely be surprised if this happens. For Practice Try Exercises

75

and

77

We could also use the calculator’s binompdf and binomcdf commands to perform the calculations in the previous example. The following Technology Corner shows how to do it.

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13. Technology Corner

​Binomial probability on the calculator TI-Nspire instructions in Appendix B; HP Prime instructions on the book’s Web site.

There are two handy commands on the TI-83/84 and TI-89 for finding binomial probabilities: binompdf and binomcdf. The inputs for both commands are the number of trials n, the success probability p, and the values of interest for the binomial random variable X. binompdf(n,p,k) computes P(X = k) binomcdf(n,p,k) computes P(X ≤ k) Let’s use these commands to confirm our answers in the previous example. (a)  Find the probability that exactly 3 of the children have type O blood.

TI-83/84

TI-89

• Press 2nd VARS (DISTR) and choose binompdf(.

• In the Stats/List Editor, Press F5 (Distr)and choose Binomial Pdf.

• OS 2.55 or later: In the dialog box, enter these values: trials:5, p:0.25, x value:3, choose Paste, and then press ENTER . Older OS: Complete the command binompdf(5,0.25,3) and press ENTER .



• In the dialog box, enter these values: Num Trials, n:5, Prob Success,p:0.25, X value:3, and then choose ENTER .



These results agree with our previous answer using the binomial probability formula: 0.08789. (b)  Should the parents be surprised if more than 3 of their children have type O blood? To find P(X > 3), use the complement rule: P(X > 3) = 1 – P(X ≤ 3) = 1 – binomcdf(5,0.25,3) • Press 2nd VARS (DISTR) and choose binomcdf(.

• In the Stats/List Editor, Press F5 (Distr) and choose Binomial Cdf….

• OS 2.55 or later: In the dialog box, enter these values: trials:5, p:0.25, x value:3, choose Paste, and then press ENTER . Subtract this result from 1 to get the answer. Older OS: Complete the command binomcdf(5,0.25,3)and press ENTER . Subtract this result from 1 to get the answer.

• In the dialog box, enter these values: Num Trials, n:5, Prob Success,p:0.25, lower value:0, upper value:3 and then choose ENTER . Subtract this result from 1 to get the answer.

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We could also have done the calculation for part (b) as P(X > 3) = P(X = 4) + P(X = 5) = binompdf(5,0.25,4) + binompdf(5,0.25,5) = 0.01465 + 0.00098 = 0.01563.

    Now we subtract from 1 to get the desired answer: 1 – 0.984375 = 0.015625. This result agrees with our previous answer using the binomial probability formula: 0.01563. AP® EXAM TIP ​Don’t rely on “calculator speak” when showing your work on free-response questions. Writing binompdf(5,0.25,3) = 0.08789 will not earn you full credit for a binomial probability calculation. At the very least, you must indicate what each of those calculator inputs represents. For example, “I used binompdf(trials:5,p:0.25,x value:3).”

Note the use of the complement rule to find P(X > 3) in the Technology Corner: P(X > 3) = 1 – P(X ≤ 3). This is necessary because the calculator’s binomcdf(n,p,k)command only computes the probability of getting k or fewer successes in n trials. Students often have trouble identifying the correct third input for the binomcdf command when a question asks them to find the probability of getting less than, more than, or at least so many successes. Here’s a helpful tip to avoid making such a mistake: write out the possible values of the variable, circle the ones you want to find the probability of, and cross out the rest. In the previous example, X can take values from 0 to 5 and we want to find P(X > 3): 0  1  2  3  4  5 Crossing out the values from 0 to 3 shows why the correct calculation is 1 – P(X ≤ 3). Take another look at the solution in the blood-type example. The structure is much like the one we used when doing Normal calculations. Here is a revised summary box that describes the process.

How to Find Binomial Probabilities Step 1: State the distribution and the values of interest. Specify a binomial distribution with the number of trials n, success probability p, and the values of the variable clearly identified. Step 2: Perform calculations—show your work! Do one of the following: (i) Use the binomial probability formula to find the desired probability; or (ii) use the binompdf or binomcdf command and label each of the inputs. Step 3: Answer the question. Here’s an example that shows the method at work.

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Example

R a n d o m Va r i a b l e s

Free Lunch? Binomial calculations A local fast-food restaurant is running a “Draw a three, get it free” lunch promotion. After each customer orders, a touch-screen display shows the message “Press here to win a free lunch.” A computer program then simulates one card being drawn from a standard deck. If the chosen card is a 3, the customer’s order is free. Otherwise, the customer must pay the bill.

PROBLEM:

(a) All 12 players on a school’s basketball team place individual orders at the restaurant. What is the probability that exactly 2 of them win a free lunch? (b) If 250 customers place lunch orders on the first day of the promotion, what’s the probability that fewer than 10 win a free lunch?

SOLUTION:

(a) Step 1: State the distribution and the values of interest. Let X = the number of players who win a free lunch. There are 12 independent trials of the chance process, each with success probability 4/52 (because there are 4 threes in a standard deck of 52 cards). So X has a binomial distribution with n = 12 and p = 4/52. We want to find P(X = 2). Step 2: Perform calculations—show your work! The binomial probability formula gives P(X = 2) = a

12 4 2 48 10 b a b a b = 0.1754 2 52 52

Using technology: The command binompdf(trials:12,p:4/52,x value:2)also gives 0.1754. Step 3: Answer the question. There is about a 17.5% probability that exactly 2 players will win a free lunch. (b) Step 1: State the distribution and the values of interest. Let Y = the number of customers who win a free lunch. There are 250 independent trials of the chance process, each with success probability 4/52. So Y  has a binomial distribution with n = 250 and p = 4/52. We want to find P(Y < 10). Step 2: Perform calculations—show your work! The values of Y that interest us are 0 1 2 3 4 5 6 7 8 9 10 11 12 . . . 250 To use the binomial formula, we would have to add up the probabilities for Y = 0, Y = 1, . . . , Y = 9. That’s too much work! The better option is to use technology: P(Y < 10) = P(Y ≤ 9) = binomcdf(trials:250,p:4/52,x value:9) = 0.00613 Step 3: Answer the question. There is almost no chance that fewer than 10 customers will win a free lunch. If this actually happened, the customers should be suspicious about the restaurant’s claim. For Practice Try Exercise

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397

Check Your Understanding

To introduce her class to binomial distributions, Mrs. Desai gives a 10-item, multiplechoice quiz. The catch is, students must simply guess an answer (A through E) for each question. Mrs. Desai uses her computer’s random number generator to produce the answer key, so that each possible answer has an equal chance to be chosen. Patti is one of the students in this class. Let X = the number of Patti’s correct guesses. 1. Show that X is a binomial random variable. 2. Find P(X = 3). Explain what this result means. 3. To get a passing score on the quiz, a student must guess correctly at least 6 times. Would you be surprised if Patti earned a passing score? Compute an appropriate probability to support your answer.

Mean and Standard Deviation of a Binomial Distribution What does the probability distribution of a binomial random variable look like? The table below shows the possible values and corresponding probabilities for X  = the number of children with type O blood. This is a binomial random variable with n = 5 and p = 0.25. Figure 6.14 shows a histogram of the probability distribution. Value xi: 0 1 2 3 4 5 Probability pi: 0.23730 0.39551 0.26367 0.08789 0.01465 0.00098 Let’s describe what we see. Shape: The probability distribution of X is skewed to the right. Because the chance that any one of the couple’s children inherits type O blood is 0.25, it’s quite likely that 0, 1, or 2 of the children will have type O blood. Larger values of X are much less likely. Center: The median number of children with type O blood is 1 because that’s where the 50th percentile of the distribution falls. How about the mean? It’s mX = xipi = (0)(0.23730) + (1)(0.39551) + c+

0.45 0.40

Probability

0.35 0.30 0.25 0.20 0.15 0.10

∙

0.05 0.00 0

1

2

3

4

5

(5)(0.00098) = 1.25

Number of children with type O blood (X)

So the expected number of children with type O blood in

FIGURE 6.14  ​Histogram showing families like this one with 5 children is 1.25. the probability distribution of the Spread: The variance of X is binomial random variable X = Starnes/Yates/Moore: The Practice of Statistics, 4E number of children with type O s2X = (xi − mX)2pi New Fig.: 06-15 Perm. Fig.: 6049 blood in a family with 5 children. First Pass: 2010-04-16 2

∙

2nd Pass: 2010-05-04

= (0 − 1.25) (0.23730) + (1 − 1.25)2(0.39551) + c+ (5 − 1.25)2(0.00098) = 0.9375

So the standard deviation of X is sX = "0.9375 = 0.968

The number of children with type O blood will typically differ from the mean by about 0.968 in families like this one with 5 children.

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Did you think about why the mean is mX = 1.25? Because each child has a 0.25 chance of inheriting type O blood, we’d expect one-fourth of the 5 children to have this blood type. In other words, mX = 5(0.25) = 1.25. This method can be used to find the mean of any binomial random variable with parameters n and p: mX = np There are fairly simple formulas for the variance and standard deviation, too, but they aren’t as easy to explain: s2X = np(1 − p)  and  sX = "np(1 − p)

For our family with 5 children,

s2X = 5(0.25)(0.75) = 0.9375  and  sX = "0.9375 = 0.968

Mean and Standard Deviation of a Binomial Random Variable If a count X of successes has the binomial distribution with number of trials n and probability of success p, the mean and standard deviation of X are

THINK ABOUT IT

autio

!

n

Remember that these formulas work only for binomial distributions. They can’t be used for other distributions.

c

mX = np sX = "np(1 − p)

Where do the binomial mean and variance formulas come from?  ​We can derive the formulas for the mean and variance of a binomi-

al distribution using what we learned about combining random variables in ­Section 6.2. Let’s start with the random variable B that’s described by the following probability distribution: Value bi: Probability pi:

0

1

1−p

p

You can think of B as representing the result of a single trial of some chance process. If a success occurs (probability p), then B = 1. If a failure occurs, then B = 0. Notice that the mean of B is mB =

∙b p = (0)(1 − p) + (1)(p) = p i i

and that the variance of B is s2B =

∙(b − m ) p = (0 − p) (1 − p) + (1 − p) p i

B

2

i

2



= p(1 − p)[p + (1 − p)]



= p(1 − p)

2

Now consider the random variable X = B1 + B2 + . . . + Bn. We can think of X as counting the number of successes in n independent trials of this chance ­process, with each trial having success probability p. In other words, X is a bino-

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mial random variable with parameters n and p. By the rules from Section 6.2, the mean of X is mX = mB + mB + c+ mB = p + p + c+ p = np 1

2

n

and the variance of X is

s2X = s2B1 + s2B2 + c+ s2Bn = p(1 − p) + p(1 − p) + c+ p(1 − p) = np(1 − p)

The standard deviation of X is therefore sX = "np(1 − p)

Example

Bottled Water versus Tap Water Binomial distribution in action Mr. Bullard’s AP® Statistics class did the Activity on page 346. There were 21 students in the class. If we assume that the students in his class could not tell tap water from bottled water, then each one was basically guessing, with a 1/3 chance of being correct. Let X = the number of students who correctly identified the cup containing the different type of water. Problem: (a) Explain why X is a binomial random variable. (b) Find the mean and standard deviation of X. Interpret each value in context. (c) Of the 21 students in the class, 13 made correct identifications. Are you convinced that Mr. Bullard’s students could tell bottled water from tap water? Justify your answer.

Solution: (a) Assuming that students were just guessing, the Activity consisted of 21 repetitions of this chance process. Let’s check the BINS: • Binary? On each trial, “success” = correct identification; “failure” = incorrect identification. • Independent? One student’s result should tell us nothing about any other student’s result. • Number? There were n = 21 trials. • Success? Each student had a p = 1/3 chance of guessing correctly. Because X is counting the number of successes in this binomial setting, it is a binomial random variable. (b) The mean of X is mX = np = 21(1/3) = 7 If the Activity were repeated many times with groups of 21 students who were just guessing, the average number of students who guess correctly would be about 7. The standard deviation of X is sX = !np(1 − p) = !21(1∙3)(2∙3) = 2.16

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If the Activity were repeated many times with groups of 21 students who were just guessing, the number of correct identifications would typically differ from 7 by about 2.16. (c) The class’s result corresponds to X = 13, a value that’s nearly 3 standard deviations above the mean. How likely is it that 13 or more of Mr. Bullard’s students would guess correctly? It’s P(X ≥ 13) = 1 - P(X ≤ 12) Using the calculator’s binomcdf (trials:21,p:1/3,x value:12) command: P(X ≥ 13) = 1 - 0.9932 = 0.0068 The students had less than a 1% chance of getting so many right if they were all just guessing. This is strong evidence that some of the students in the class could tell bottled water from tap water. For Practice Try Exercise

85

Figure 6.15 shows the probability distribution for the number of correct guesses in Mr. Bullard’s class if no one can tell bottled water from tap water. As you can see from the graph, the chance of 13 or more guessing correctly is quite small.

Probability

0.2

Although the histogram is slightly rightskewed (there’s a long tail that extends out to X = 21), it looks like a Normal density curve might fit the bulk of the distribution fairly well.

0.1

0.0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

Number of correct guesses

FIGURE 6.15   ​Histogram of the probability distribution for the binomial random variable X = number of correct guesses in Mr. Bullard’s class. Starnes/Yates/Moore: The Practice of Statistics, 4E New Fig.: 06-16 Perm. Fig.: 6051 First Pass: 2010-04-16 2nd Pass:Check 2010-05-04 previous Your Understanding (page 397)

Check Your Understanding

Refer to the about Mrs. Desai’s special multiple-choice quiz on binomial distributions. We defined X = the number of Patti’s correct guesses. 1. Find mX. Interpret this value in context. 2. Find sX. Interpret this value in context. 3. What’s the probability that the number of Patti’s correct guesses is more than 2 standard deviations above the mean? Show your method.

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Binomial Distributions in Statistical Sampling The binomial distributions are important in statistics when we wish to make inferences about the proportion p of successes in a population. Here is an example involving a familiar product.

Example

Bad Flash Drives Binomial distributions and sampling A supplier inspects an SRS of 10 flash drives from a shipment of 10,000 flash drives. Suppose that (unknown to the supplier) 2% of the flash drives in the shipment are defective. Count the number X of bad flash drives in the sample. This is not quite a binomial setting. Removing 1 flash drive changes the proportion of bad flash drives remaining in the shipment. The conditional probability that the second flash drive chosen is bad changes when we know whether the first is good or bad. But removing 1 flash drive from a shipment of 10,000 changes the makeup of the remaining 9999 flash drives very little. The distribution of X is very close to the binomial distribution with n = 10 and p = 0.02. To illustrate this, let’s compute the probability that none of the 10 flash drives is defective. Using the binomial distribution, it’s P(X = 0) = a

10 b (0.02)0(0.98)10 = 0.8171 0

The actual probability of getting no defective flash drives is P(no defectives) =

9800 9799 9798 9791 × × × c× = 0.8170 10,000 9999 9998 9991

Those two probabilities are quite close! Almost all real-world sampling, such as taking an SRS from a population of interest, is done without replacement. As the previous example illustrates, sampling without replacement leads to a violation of the independence condition. The flash drives example shows how we can use binomial distributions in the statistical setting of selecting an SRS. When the population is much larger than the sample, a count of successes in an SRS of size n has approximately the binomial distribution with n equal to the sample size and p equal to the proportion of successes in the population. What counts as “much larger”? In practice, the binomial distribution gives a good approximation as long as we don’t sample more than 10% of the population. We refer to this as the 10% condition.

10% Condition When taking an SRS of size n from a population of size N, we can use a binomial distribution to model the count of successes in the sample as long 1 as n ≤ N. 10

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Here’s an example that shows why it’s important to check the 10% condition before calculating a binomial probability. You might recognize the setting from the first activity in the book (page 5).

Example

Hiring Discrimination—It Just Won’t Fly! Sampling without replacement An airline has just finished training 25 first officers—15 male and 10 female— to become captains. Unfortunately, only eight captain positions are available right now. Airline managers decide to use a lottery to determine which pilots will fill the available positions. Of the 8 captains chosen, 5 are female and 3 are male. Problem:  Explain why the probability that 5 female pilots are chosen in a fair lottery is not 8 P(X = 5) = a b(0.40)5(0.60)3 = 0.124 5

(The correct probability is 0.106.) Solution:  The managers are sampling without replacement when they do the lottery. There’s a 0.40 chance that the first pilot selected for a captain position is female. Once that person is chosen, the probability that the next pilot selected will be female is no longer 0.40. The binomial formula assumes that the conditional probability of success stays constant at 0.40 throughout the eight trials of this chance process. This calculation will be approximately correct if the success probability doesn’t change too much—as long as we don’t sample more than 10% of the population. In this case, managers are sampling 8 out of 25 pilots—almost 1/3 of the population. That explains why the binomial probability is off by about 17% (0.018/0.106) from the correct answer. For Practice Try Exercise

87

The Normal approximation to binomial distributions*  As n gets

larger, something interesting happens to the shape of the binomial distribution. Figure 6.16 shows histograms of binomial distributions for different values of n and p. As the number of observations n becomes larger, the binomial distribution gets close to a Normal distribution. You can investigate the relationship between n and p yourself using the Normal Approximation to Binomial Distributions applet at the book’s Web site.

PLET AP

0.165

0.000

0.117

0.000 0

(a)

0.148

Probability

0.234

Probability

Probability

0.330

8

n = 10, p = 0.8

10

0.000 0

(b)

0.074

16

n = 20, p = 0.8

20

0

(c)

40

50

n = 50, p = 0.8

FIGURE 6.16  ​Histograms of binomial distributions with (a) n = 10 and p = 0.8, (b) n = 20 and p = 0.8, and (c) n = 50 and p = 0.8. As n increases, the shape of the probability distribution gets more and more Normal. ® Starnes/Yates/Moore: The Practice of Statistics, Starnes/Yates/Moore: Practice of Statistics, 4E Starnes/Yates/Moore: The Practice Statistics, 4E *This 4E topic is not required for The the AP Statistics exam. Some teachers prefer to discuss this topicofwhen New Fig.: 06-17a Perm. Fig.: 6054 New Fig.: 06-17b Perm. Fig.: New Fig.: 06-17c Perm. Fig.: 6056 presenting the sampling distribution of pˆ 6055 (Chapter 7). First Pass: 2010-04-16 First Pass: 2010-04-16 First Pass: 2010-04-16 2nd Pass: 2010-05-04 2nd Pass: 2010-05-04 2nd Pass: 2010-05-04

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When n is large, we can use Normal probability calculations to approximate binomial probabilities. Here are the facts.

Normal Approximation for Binomial Distributions: The Large Counts Condition Suppose that a count X of successes has the binomial distribution with n trials and success probability p. When n is large, the distribution of X is approximately Normal with mean: mX = np and standard deviation: sX = !np(1 − p)

As a rule of thumb, we will use the Normal approximation when n is so large that np ≥ 10  and  n(1 − p) ≥ 10 That is, the expected number of successes and failures are both at least 10. We refer to this as the Large Counts condition. The Normal approximation is easy to remember because it says to act as if X is Normal with exactly the same mean and standard deviation as the binomial. The accuracy of the Normal approximation improves as the sample size n increases. It is most accurate for any fixed n when p is close to 1/2 and least accurate when p is near 0 or 1. This is why the rule of thumb in the box depends on p as well as n.

Example

Attitudes toward Shopping Normal approximation to a binomial Sample surveys show that fewer people enjoy shopping than in the past. A survey asked a nationwide random sample of 2500 adults if they agreed or disagreed that “I like buying new clothes, but shopping is often frustrating and time-consuming.”11 The population that the poll wants to draw conclusions about is all U.S. residents aged 18 and over. Problem:  Suppose that exactly 60% of all adult U.S. residents would say “Agree” if asked the same question. Let X = the number in the sample who agree. (a) Show that X is approximately a binomial random variable. (b) Check the conditions for using a Normal approximation in this setting. (c) Use a Normal distribution to estimate the probability that 1520 or more of the sample agree.

Solution: (a) Let’s check the BINS. • Binary? Success = agree that shopping is frustrating, failure = don’t agree. • Independent? The trials are not independent: the conditional probability of a success changes due to the sampling without replacement. But the 10% condition is met because 2500 people is much less than 10% of all U.S. adult residents. • Number? There are n = 2500 trials of this chance process. • Success? There is the same probability of selecting an adult who agrees on each trial: p = 0.6.

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So the number in our sample who agree that shopping is frustrating is a random variable X having roughly the binomial distribution with n = 2500 and p = 0.6. (b) We need to check the Large Counts condition. Because np = 2500(0.6) = 1500 and n(1 − p) = 2500(0.4) = 1000 are both at least 10, we should be safe using the Normal approximation. (c) Step 1: State the distribution and the values of interest. Act as though the count X has the Normal distribution with the same mean and standard deviation as the binomial distribution: m = np = 2500(0.6) = 1500  and  s = !np(1 − p) = !2500(0.6)(0.4) = 24.49

We want to find P(X ≥ 1520). Figure 6.17 shows this probability as the area under a Normal curve. Step 2: Perform calculations—show your work! Standardizing the boundary value gives N(1500,24.49) 1520 − 1500 z= = 0.82 24.49 Using Table A, the probability we want is P(Z ≥ 0.82) = 1 - 0.7939 = 0.2061 1500 1520

Using technology: The command normalcdf(lower:1520, upper: 10000, m:1500, s:24.49) gives an area of 0.2071. FIGURE 6.17  Normal distribution Step 3: Answer the question. There is about a 21% chance of getting a sample in which 1520 to approximate the binomial or more agree with the statement. probability of getting 1520 or more successes when n = 2500 and p = 0.6.

For Practice Try Exercise

91

We can also find the probability that 1520 or more of the sample agree that shopping is often frustrating and time-consuming using the command 1binomcdf(2500,0.6,1519), which yields 0.2131. The Normal approximation, 0.2061, misses the more accurate binomial probability by about 0.007.

Geometric Random Variables In a binomial setting, the number of trials n is fixed in advance, and the binomial random variable X counts the number of successes. The possible values of X are 0, 1, 2, . . . , n. In other situations, the goal is to repeat a chance process until a success occurs: • Roll a pair of dice until you get doubles. • In basketball, attempt a three-point shot until you make one. • Keep placing a $1 bet on the number 15 in roulette until you win. These are all examples of a geometric setting. Although the number of trials isn’t fixed in advance, the trials are independent and the probability of success remains constant.

Definition: Geometric setting A geometric setting arises when we perform independent trials of the same chance process and record the number of trials it takes to get one success. On each trial, the probability p of success must be the same.

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Here’s an Activity your class can try that involves a geometric setting.

Activity MATERIALS: Calculator or computer random number generator to select student names and days of the week

​Is This Your Lucky Day? Your teacher is planning to give you 10 problems for homework. As an alternative, you can agree to play the Lucky Day Game. Here’s how it works. A student will be selected at random from your class and asked to pick a day of the week (for instance, Thursday). Then your teacher will use technology to randomly choose a day of the week as the “lucky day.” If the student picks the correct day, the class will have only one homework problem. If the student picks the wrong day, your teacher will select another student from the class at random. The chosen student will pick a day of the week and your teacher will use technology to choose a “lucky day.” If this student gets it right, the class will have two homework problems. The game continues until a student correctly guesses the lucky day. Your teacher will assign a number of homework problems that is equal to the total number of guesses made by members of your class. Are you ready to play the Lucky Day Game? 1.  Decide as a class about whether to “gamble” on the number of homework problems you will receive. You have 30 seconds. 2.  Play the Lucky Day Game and see what happens! In a geometric setting, if we define the random variable Y to be the number of trials needed to get the first success, then Y is called a geometric random variable. The probability distribution of Y is a geometric distribution.

Definition: Geometric random variable and geometric distribution The number of trials Y that it takes to get a success in a geometric setting is a ­geometric random variable. The probability distribution of Y is a geometric ­distribution with parameter p, the probability of a success on any trial. The possible values of Y are 1, 2, 3, . . . . As with binomial random variables, it’s important to be able to distinguish situations in which a geometric distribution does and doesn’t apply.

Example

The Lucky Day Game Geometric settings and random variables The random variable of interest in this game is Y = the number of picks it takes to correctly match the lucky day. Each pick is one trial of the chance process. Knowing the result of one student’s pick tells us nothing about the result of any other pick. On each trial, the probability of a correct pick is 1/7. This is a geometric setting. Because Y counts the number of trials to get the first success, it is a geometric random variable with parameter p = 1/7.

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What is the probability that the first student picks correctly and wins the Lucky Day Game? It’s P(Y = 1) = 1/7. That’s also the class’s chance of getting only one homework problem. For the class to have two homework problems, the first student selected must pick an incorrect day of the week and the second student must pick the lucky day correctly. The probability that this happens is P(Y = 2) = (6/7)(1/7) = 0.1224. Likewise, P(Y = 3) = (6/7)(6/7)(1/7) = 0.1050. In general, the probability that the first correct pick comes on the kth trial is P(Y = k) = (6/7)k − 1(1/7). Let’s summarize what we’ve learned about calculating a geometric probability.

Geometric Probability FORMULA If Y has the geometric distribution with probability p of success on each trial, the possible values of Y are 1, 2, 3, . . . . If k is any one of these values, P(Y = k) = (1 − p)k − 1p With our formula in hand, we can now compute any geometric probability.

Example

The Lucky Day Game Calculating geometric probabilities Problem:  Let the random variable Y  be defined as in the previous example. (a) Find the probability that the class receives exactly 10 homework problems as a result of playing the Lucky Day Game. (b) Find P(Y < 10) and interpret this value in context. Solution:  Y  = the number of attempts it takes to get a correct pick = the number of homework problems. (a) P(Y = 10) = (6/7)9(1/7) = 0.0357. (b) P(Y < 10) = P(Y = 1) + P(Y = 2) + P(Y = 3) + . . . + P(Y = 9) = 1/7 + (6/7)(1/7) + (6/7)2(1/7) + . . . + (6/7)8(1/7) = 0.7503. There’s about a 75% chance that the class will get less homework by playing the Lucky Day Game. For Practice Try Exercise 97 As you probably guessed, we used the calculator’s geometpdf and geometcdf commands for the computations in the previous example. The following Technology Corner shows you how we did it.

14. T echnology Corner

​Geometric probability on the calculator TI-Nspire instructions in Appendix B; HP Prime instructions on the book’s Web site.

There are two handy commands on the TI-83/84 and TI-89 for finding geometric probabilities: geometpdf and geometcdf. The inputs for both commands are the success probability p and the value(s) of interest for the geometric random variable Y. geometpdf(p,k) computes P(Y = k) geometcdf(p,k) computes P(Y ≤ k)

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Let’s use these commands to confirm our answers in the previous example. (a)  Find the probability that the class receives exactly 10 homework problems as a result of playing the Lucky Day Game.

TI-83/84

TI-89

• Press 2nd VARS (DISTR) and choose geometpdf(.

• In the Stats/List Editor, Press F5 (Distr) and choose Geometric Pdf....

• OS 2.55 or later: In the dialog box, enter these values: p:1/7, x value:10, choose Paste, and then press ENTER . Older OS: Complete the command geometpdf(1/7,10)and press ENTER .



• In the dialog box, enter these values: Prob Success, p:1/7, X value:10, and then choose ENTER .



These results agree with our previous answer using the geometric probability formula: 0.0357. (b) Find P(Y < 10) and interpret this value in context. To find P(Y < 10), use the geometcdf command: P(Y < 10) = P(Y ≤ 9) = geometcdf(1/7,9) • Press 2nd VARS (DISTR) and choose geometcdf(. • OS 2.55 or later: In the dialog box, enter these values: p:1/7, x value:9, choose Paste, and then press ENTER . Older OS: Complete the command geomet cdf(1/7,9)and press ENTER .



• In the Stats/List Editor, Press F5 (Distr) and choose Geometric Cdf.... • In the dialog box, enter these values: Prob Success, p:1/7, Lower value:0, Upper value:9, and then choose ENTER .



These results agree with our previous answer using the geometric probability formula: 0.7503.

The table below shows part of the probability distribution of Y. We can’t show the entire distribution, because the number of trials it takes to get the first success could be a very large number. .  .  . Value yi: 1 2 3 4 5 6 7 8 9 Probability pi: 0.143 0.122 0.105 0.090 0.077 0.066 0.057 0.049 0.042 Figure 6.18 is a histogram of the probability distribution for values of Y from 1 to 26. Let’s describe what we see.

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Shape: The heavily right-skewed shape is characteristic of any geometric distribution. That’s because the most likely value of a geometric random variable is 1. The probability of each successive value decreases by a factor of (1 − p).

0.16 0.14

Probability

0.12 0.10 0.08 0.06 0.04 0.02 0.00 0

5

10

15

20

Number of picks required (Y)

FIGURE 6.18  ​Histogram showing the probability distribution of the geometric random variable Y = number of trials needed for students to pick correctly in the Lucky Day Game.

25

Center: The mean of Y is mY = 7. (Due to the infinite number of possible values of Y, the calculation of the mean is beyond the scope of this text.) If the class played the Lucky Day Game many times, they would receive an average of 7 homework problems. It’s no coincidence that p = 1/7 and mY = 7. With probability of success 1/7 on each trial, we’d expect it to take an average of 7 trials to get the first success.

Spread: The standard deviation of Y is sY = 6.48. (Due to the infinite number of possible values of Y, the calculation of the standard deviation is beyond the scope of this text.) If the class played the Lucky Day game many times, the number of homework problems they receive would typically differ from 7 by about 6.5 problems. That could mean a lot of homework! We can generalize the result for the mean of a geometric random variable.

Mean (Expected Value) of a Geometric Random Variable If Y is a geometric random variable with probability of success p on each 1 trial, then its mean (expected value) is mY = E(Y) = . That is, the expected p number of trials required to get the first success is 1/p.

Check Your Understanding

Suppose you roll a pair of fair, six-sided dice until you get doubles. Let T = the number of rolls it takes. 1. Show that T is a geometric random variable. 2. Find P(T = 3). Interpret this result in context. 3. ​In the game of Monopoly, a player can get out of jail free by rolling doubles within 3 turns. Find the probability that this happens.

case closed

A Jury of Your Peers? In the chapter-opening Case Study on page 345, a defense attorney challenged the jury-pool selection process in his accused client’s trial. Here are the facts: • About 7.28% of the citizens in the court’s jurisdiction were black. • The jury pool had between 60 and 100 members, 3 of whom were black. Use what you have learned in this chapter to help answer the following questions.

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For now, assume that the court carried out a proper random-selection process to obtain a jury pool with 100 members. 1. Let X = the number of black citizens in the jury pool. What distribution does the random variable X have? Justify your answer. 2. Find the mean and standard deviation of X. Interpret these values in context. 3. If a jury pool has 3 or fewer blacks, should we be suspicious that the court did not carry out the random selection process correctly? Compute P(X ≤ 3) and use this result to support your answer. What if the jury pool had 60 members? Assume once again that the court carried out a proper random-selection process. Let Y = the number of black citizens in the jury pool. 4. Without doing any calculations, decide if P(Y ≤ 3) is greater than, equal to, or less than P(X ≤ 3). Justify your answer. 5. Using the logic of Question 4, explain why you do not have to consider jury pools with 61, 62, . . . , 99 members to render a verdict about whether or not the jury-selection process was carried out properly. What is your verdict?

Section 6.3

Summary •

•

A binomial setting consists of n independent trials of the same chance process, each resulting in a success or a failure, with probability of success p on each trial. Remember to check the BINS! The count X of successes is a binomial random variable. Its probability distribution is a binomial distribution. The binomial coefficient n! n a b= k k!(n − k)!

counts the number of ways k successes can be arranged among n trials. The factorial of n is n! = n(n − 1)(n − 2) · … · (3)(2)(1) •

•

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for positive whole numbers n, and 0! = 1. If X has the binomial distribution with parameters n and p, the possible values of X are the whole numbers 0, 1, 2, . . . , n. The binomial probability of observing k successes in n trials is n P(X = k) = a bpk(1 − p)n−k k Binomial probabilities are best found using technology.

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• •

•

•

•

The mean and standard deviation of a binomial random variable X are mX = np   sX = !np(1 − p)

The binomial distribution with n trials and probability p of success gives a good approximation to the count of successes in an SRS of size n from a large population containing proportion p of successes. This is true as long as the sample size n is no more than 10% of the population size N (the 10% condition). The Normal approximation to the binomial distribution* says that if X is a count of successes having the binomial distribution with parameters n and p, then when n is large, X is approximately Normally distributed with mean np and standard deviation !np(1 − p). We will use this approximation when np ≥ 10 and n(1 − p) ≥ 10 (the Large Counts condition). A geometric setting consists of repeated trials of the same chance process in which the probability p of success is the same on each trial, and the goal is to count the number of trials it takes to get one success. If Y = the number of trials required to obtain the first success, then Y is a geometric random variable. Its probability distribution is called a geometric distribution. If Y has the geometric distribution with probability of success p, the possible values of Y are the positive integers 1, 2, 3, . . .  . The geometric probability that Y takes any value is P(Y = k) = (1 − p)k − 1p

•

The mean (expected value) of a geometric random variable Y is mY = 1/p.

*This topic is not required for the AP® Statistics exam. Some teachers prefer to discuss this topic when presenting the sampling distribution of pˆ (Chapter 7).

6.3 T echnology Corners TI-Nspire Instructions in Appendix B; HP Prime instructions on the book’s Web site. 12. Binomial coefficients on the calculator 13. Binomial probability on the calculator 14.  Geometric probability on the calculator

Section 6.3

page 392 page 394 page 406

Exercises

In Exercises 69 to 72, determine whether the given random variable has a binomial distribution. Justify your answer.

flower seeds from Seed Depot and plants them in her garden. Let X = the number of seeds that germinate.

69. Sowing seeds  Seed Depot advertises that its new flower seeds have an 85% chance of germinating (growing). Suppose that the company’s claim is true. Judy gets a packet with 20 randomly selected new

70. Long or short?  Put the names of all the students in your class in a hat. Mix them up, and draw four names without looking. Let Y = the number whose last names have more than six letters.

pg 388

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Section 6.3 Binomial and Geometric Random Variables 71. Lefties  Exactly 10% of the students in a school are left-handed. Select students at random from the school, one at a time, until you find one who is left-handed. Let V = the number of students chosen.

pg 388

72. Taking the train  According to New Jersey Transit, the 8:00 a.m. weekday train from Princeton to New York City has a 90% chance of arriving on time on a randomly selected day. Suppose this claim is true. Choose 6 days at random. Let W = the number of days on which the train arrives late. 73. Binomial setting?  A binomial distribution will be approximately correct as a model for one of these two settings and not for the other. Explain why by briefly discussing both settings. (a) When an opinion poll calls residential telephone numbers at random, only 20% of the calls reach a person. You watch the random digit dialing machine make 15 calls. X is the number that reach a person. (b) When an opinion poll calls residential telephone numbers at random, only 20% of the calls reach a live person. You watch the random digit dialing machine make calls. Y is the number of calls needed to reach a live person. 74. Binomial setting?  A binomial distribution will be approximately correct as a model for one of these two sports settings and not for the other. Explain why by briefly discussing both settings. (a) A National Football League kicker has made 80% of his field goal attempts in the past. This season he attempts 20 field goals. The attempts differ widely in distance, angle, wind, and so on. (b) A National Basketball Association player has made 80% of his free-throw attempts in the past. This season he takes 150 free throws. Basketball free throws are always attempted from 15 feet away with no interference from other players. 75. Elk  Biologists estimate that a baby elk has a 44% chance of surviving to adulthood. Assume this estimate is correct. Suppose researchers choose 7 baby elk at random to monitor. Let X = the number who survive to adulthood. Use the binomial probability formula to find P(X = 4). Interpret this result in context.

pg 393

76. Rhubarb  Suppose you purchase a bundle of 10 bareroot rhubarb plants. The sales clerk tells you that 5% of these plants will die before producing any rhubarb. Assume that the bundle is a random sample of plants and that the sales clerk’s statement is accurate. Let Y = the number of plants that die before producing any rhubarb. Use the binomial probability formula to find P(Y = 1). Interpret this result in context.

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77. Elk  Refer to Exercise 75. How surprising would it be for more than 4 elk in the sample to survive to adulthood? Calculate an appropriate probability to support your answer.

pg 393

78. Rhubarb  Refer to Exercise 76. Would you be surprised if 3 or more of the plants in the bundle die before producing any rhubarb? Calculate an appropriate probability to support your answer. 79. Sowing seeds  Refer to Exercise 69. pg 396

(a) Find the probability that exactly 17 seeds germinate. Show your work. (b) If only 12 seeds actually germinate, should Judy be suspicious that the company’s claim is not true? Compute P(X ≤ 12) and use this result to support your answer. 80. Taking the train  Refer to Exercise 72. (a) ​ Find the probability that the train arrives late on exactly 2 days. Show your work. (b) ​ Would you be surprised if the train arrived late on 2 or more days? Compute P(W ≥ 2) and use this result to support your answer. 81. Random digit dialing  When an opinion poll calls a residential telephone number at random, there is only a 20% chance that the call reaches a live person. You watch the random digit dialing machine make 15 calls. Let X = the number of calls that reach a live person. (a) Find and interpret mX. (b) Find and interpret sX. 82. Lie detectors  A federal report finds that lie detector tests given to truthful persons have probability about 0.2 of suggesting that the person is deceptive.12 A company asks 12 job applicants about thefts from previous employers, using a lie detector to assess their truthfulness. Suppose that all 12 answer truthfully. Let X = the number of people who the lie detector says are being deceptive. (a) Find and interpret mX. (b) Find and interpret sX. 83. Random digit dialing  Refer to Exercise 81. Let Y = the number of calls that don’t reach a live person. (a) Find the mean of Y. How is it related to the mean of X? Explain why this makes sense. (b) Find the standard deviation of Y. How is it related to the standard deviation of X? Explain why this makes sense.

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84. Lie detectors  Refer to Exercise 82. Let Y = the number of people who the lie detector says are telling the truth.

88. Scrabble  In the game of Scrabble, each player begins by drawing 7 tiles from a bag containing 100 tiles. There are 42 vowels, 56 consonants, and 2 blank tiles in the bag. Cait chooses her 7 tiles and is surprised to discover that all of them are vowels. Can we use a binomial distribution to approximate this probability? Justify your answer.

(a) Find P(Y ≥ 10). How is this related to P(X ≤ 2)? Explain. (b) Calculate mY and sY. How do they compare with mX and sX? Explain why this makes sense. 85. 1 in 6 wins  As a special promotion for its 20-ounce bottles of soda, a soft drink company printed a message on the inside of each cap. Some of the caps said, “Please try again,” while others said, “You’re a winner!” The company advertised the promotion with the slogan “1 in 6 wins a prize.” Suppose the company is telling the truth and that every 20-ounce bottle of soda it fills has a 1-in-6 chance of being a winner. Seven friends each buy one 20-ounce bottle of the soda at a local convenience store. Let X = the number who win a prize.

89. 10% condition  To use a binomial distribution to approximate the count of successes in an SRS, why do we require that the sample size n be no more than 10% of the population size N?

pg 399

90. *Large Counts condition  To use a Normal distribution to approximate binomial probabilities, why do we require that both np and n(1 − p) be at least 10? 91. *On the Web  What kinds of Web sites do males aged 18 to 34 visit most often? Half of male Internet users in this age group visit an auction site such as eBay at least once a month.13 A study of Internet use interviews a random sample of 500 men aged 18 to 34. Let X = the number in the sample who visit an auction site at least once a month.

pg 403

(a) Explain why X is a binomial random variable. (b) Find the mean and standard deviation of X. Interpret each value in context. (c) The store clerk is surprised when three of the friends win a prize. Is this group of friends just lucky, or is the company’s 1-in-6 claim inaccurate? Compute P(X ≥ 3) and use the result to justify your answer. 86. Aircraft engines  Engineers define reliability as the probability that an item will perform its function under specific conditions for a specific period of time. A certain model of aircraft engine is designed so that each engine has probability 0.999 of performing properly for an hour of flight. Company engineers test an SRS of 350 engines of this model. Let X = the number that operate for an hour without failure. (a) Explain why X is a binomial random variable. (b) Find the mean and standard deviation of X. Interpret each value in context. (c) Two engines failed the test. Are you convinced that this model of engine is less reliable than it’s supposed to be? Compute P(X ≤ 348) and use the result to justify your answer. 87. Airport security  The Transportation Security Administration (TSA) is responsible for airport safety. On some pg 402 flights, TSA officers randomly select passengers for an extra security check before boarding. One such flight had 76 passengers—12 in first class and 64 in coach class. Some passengers were surprised when none of the 10 passengers chosen for screening were seated in first class. Can we use a binomial distribution to approximate this probability? Justify your answer.

(a) Show that X is approximately a binomial random variable. (b) Check the conditions for using a Normal approximation in this setting. (c) Use a Normal distribution to estimate the probability that at least 235 of the men in the sample visit an online auction site at least once a month. 92. *Checking for survey errors  One way of checking the effect of undercoverage, nonresponse, and other sources of error in a sample survey is to compare the sample with known facts about the population. About 12% of American adults identify themselves as black. Suppose we take an SRS of 1500 American adults and let X be the number of blacks in the sample. (a) Show that X is approximately a binomial random variable. (b) Check the conditions for using a Normal approximation in this setting. (c) Use a Normal distribution to estimate the probability that the sample will contain between 165 and 195 blacks. 93. Using Benford’s law  According to Benford’s law (Exercise 5, page 359), the probability that the first digit of the amount on a randomly chosen invoice is a 1 or a 2 is 0.477. Suppose you examine an SRS of 90 invoices from a vendor and find 29 that have first digits 1 or 2. Do you suspect that the invoice

*This topic is not required for the AP® Statistics exam. Some teachers prefer to discuss this topic when presenting the sampling distribution of pˆ (Chapter 7).

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Section 6.3 Binomial and Geometric Random Variables amounts are not genuine? Compute an appropriate probability to support your answer. 94. A .300 hitter  In baseball, a 0.300 hitter gets a hit in 30% of times at bat. When a baseball player hits 0.300, fans tend to be impressed. Typical Major Leaguers bat about 500 times a season and hit about 0.260. A hitter’s successive tries seem to be independent. Could a typical Major Leaguer hit 0.300 just by chance? Compute an appropriate probability to support your answer. 95. Geometric or not?  Determine whether each of the following scenarios describes a geometric setting. If so, define an appropriate geometric random variable. (a) A popular brand of cereal puts a card with 1 of 5 famous NASCAR drivers in each box. There is a 1/5 chance that any particular driver’s card ends up in any box of cereal. Buy boxes of the cereal until you have all 5 drivers’ cards. (b) In a game of 4-Spot Keno, Lola picks 4 numbers from 1 to 80. The casino randomly selects 20 winning numbers from 1 to 80. Lola wins money if she picks 2 or more of the winning numbers. The probability that this happens is 0.259. Lola decides to keep playing games of 4-Spot Keno until she wins some money. 96. Geometric or not?  Determine whether each of the following scenarios describes a geometric setting. If so, define an appropriate geometric random variable. (a) Shuffle a standard deck of playing cards well. Then turn over one card at a time from the top of the deck until you get an ace. (b) Lawrence likes to shoot a bow and arrow in his free time. On any shot, he has about a 10% chance of hitting the bull’s-eye. As a challenge one day, Lawrence decides to keep shooting until he gets a bull’s-eye.

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99. Using Benford’s law  According to Benford’s law (Exercise 5, page 359), the probability that the first digit of the amount of a randomly chosen invoice is an 8 or a 9 is 0.097. Suppose you examine randomly selected invoices from a vendor until you find one whose amount begins with an 8 or a 9. (a) How many invoices do you expect to examine until you get one that begins with an 8 or 9? Justify your answer. (b) In fact, you don’t get an amount starting with an 8 or 9 until the 40th invoice. Do you suspect that the invoice amounts are not genuine? Compute an appropriate probability to support your answer. 100. Roulette  Marti decides to keep placing a $1 bet on number 15 in consecutive spins of a roulette wheel until she wins. On any spin, there’s a 1-in-38 chance that the ball will land in the 15 slot. (a) How many spins do you expect it to take until Marti wins? Justify your answer. (b) Would you be surprised if Marti won in 3 or fewer spins? Compute an appropriate probability to support your answer. Multiple choice: Select the best answer for Exercises 101 to 105. 101. Joe reads that 1 out of 4 eggs contains salmonella bacteria. So he never uses more than 3 eggs in cooking. If eggs do or don’t contain salmonella independently of each other, the number of contaminated eggs when Joe uses 3 chosen at random has the following distribution: (a) binomial; n = 4 and p = 1/4 (b) binomial; n = 3 and p = 1/4 (c) binomial; n = 3 and p = 1/3

97. 1-in-6 wins  Alan decides to use a different strategy for the 1-in-6 wins game of Exercise 85. He keeps buying one 20-ounce bottle of the soda at a time until he gets a winner.

(d) geometric; p = 1/4

(a) Find the probability that he buys exactly 5 bottles. Show your work.

Exercises 102 and 103 refer to the following setting. A fastfood restaurant runs a promotion in which certain food items come with game pieces. According to the restaurant, 1 in 4 game pieces is a winner.

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(b) Find the probability that he buys no more than 8 bottles. Show your work. 98. Cranky mower  To start her old lawn mower, Rita has to pull a cord and hope for some luck. On any particular pull, the mower has a 20% chance of starting. (a) Find the probability that it takes her exactly 3 pulls to start the mower. Show your work. (b) Find the probability that it takes her more than 10 pulls to start the mower. Show your work.

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(e) geometric; p = 1/3

102. If Jeff gets 4 game pieces, what is the probability that he wins exactly 1 prize? 4 (a) 0.25 (d)  a b(0.25)3(0.75)1 1 (b) 1.00 (c)

(e)  (0.75)3(0.25)1

4 a b(0.25)1(0.75)3 1

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103. If Jeff keeps playing until he wins a prize, what is the probability that he has to play the game exactly 5 times? (a) (0.25)5 (b) (0.75)4 (c) (0.75)5

(d) (0.75)4(0.25) 5 (e) a b(0.75)4(0.25) 1

104. Each entry in a table of random digits like Table D has probability 0.1 of being a 0, and the digits are independent of one another. If many lines of 40 random digits are selected, the mean and standard deviation of the number of 0s will be approximately (a) mean = 0.1, standard deviation = 0.05. (b) mean = 0.1, standard deviation = 0.1. (c) mean = 4, standard deviation = 0.05. (d) mean = 4, standard deviation = 1.90. (e) mean = 4, standard deviation = 3.60. 105. *In which of the following situations would it be appropriate to use a Normal distribution to approximate probabilities for a binomial distribution with the given values of n and p? (a) n = 10, p = 0.5 (b) n = 40, p = 0.88 (c) n = 100, p = 0.2 (d) n = 100, p = 0.99 (e) n = 1000, p = 0.003

106. Spoofing (4.2)  To collect information such as passwords, online criminals use “spoofing” to direct Internet users to fraudulent Web sites. In one study of Internet fraud, students were warned about spoofing and then asked to log in to their university account starting from the university’s home page. In some cases, the login link led to the genuine dialog box. In others, the box looked genuine but in fact was linked to a different site that recorded the ID and password the student entered. The box that appeared for each student was determined at random. An alert student could detect the fraud by looking at the true Internet address displayed in the browser status bar, but most just entered their ID and password. Is this study an experiment? Why? What are the explanatory and response variables? 107. Smoking and social class (5.3)  As the dangers of smoking have become more widely known, clear class differences in smoking have emerged. British government statistics classify adult men by occupation as “managerial and professional” (43% of the population), “intermediate” (34%), or “routine and manual” (23%). A survey finds that 20% of men in managerial and professional occupations smoke, 29% of the intermediate group smoke, and 38% in routine and manual occupations smoke.14 (a) Use a tree diagram to find the percent of all adult British men who smoke. (b) Find the percent of male smokers who have routine and manual occupations.

*This topic is not required for the AP® Statistics exam. Some teachers prefer to discuss this topic when presenting the sampling distribution of pˆ (Chapter 7).

FRAPPY!  Free Response AP® Problem, Yay! The following problem is modeled after actual AP® Statistics exam free response questions. Your task is to generate a complete, concise response in 15 minutes.

Directions: Show all your work. Indicate clearly the methods you use, because you will be scored on the correctness of your methods as well as on the accuracy and completeness of your results and explanations. Buckley Farms produces homemade potato chips that it sells in bags labeled “16 ounces.” The total weight of each bag follows an approximately Normal distribution with a mean of 16.15 ounces and a standard deviation of 0.12 ounces. (a) If you randomly selected 1 bag of these chips, what is the probability that the total weight is less than 16 ounces? (b) If you randomly selected 10 bags of these chips, what is the probability that exactly 2 of the bags will have a total weight less than 16 ounces?

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(c) Buckley Farms ships its chips in boxes that contain 6 bags. The empty boxes have a mean weight of 10 ounces and a standard deviation of 0.05 ounces. Calculate the mean and standard deviation of the total weight of a box containing 6 bags of chips. (d) Buckley Farms decides to increase the mean weight of each bag of chips so that only 5% of the bags have weights that are less than 16 ounces. Assuming that the standard deviation remains 0.12 ounces, what mean weight should Buckley Farms use? After you finish, you can view two example solutions on the book’s Web site (www.whfreeman.com/tps5e). Determine whether you think each solution is “complete,” “substantial,” “developing,” or “minimal.” If the solution is not complete, what improvements would you suggest to the student who wrote it? Finally, your teacher will provide you with a scoring rubric. Score your response and note what, if anything, you would do differently to improve your own score.

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Chapter Review Section 6.1: Discrete and Continuous Random Variables A random variable assigns numerical values to the outcomes of a chance process. The probability distribution of a random variable describes its possible values and their probabilities. There are two types of random variables: discrete and continuous. Discrete random variables take on a fixed set of values with gaps in between. Continuous random variables take on all values in an interval of numbers. As in Chapter 1, we are often interested in the shape, center, and spread of a probability distribution. The shape of a discrete probability distribution can be identified by graphing a probability histogram, with the height of each bar representing the probability of a single value. The center is usually identified by the mean (expected value) of the random variable. The expected value is the average value of the random variable if the chance process is repeated many times. The spread of a probability distribution is usually identified by the standard deviation, which describes how much the values of a random variable typically differ from the mean value, in many repetitions of the chance process. Continuous probability distributions, such as the Normal distribution, describe the distribution of continuous random variables. A density curve is used to display a continuous probability distribution. Probabilities for continuous random variables are determined by finding the area under the density curve and above the values of interest. Section 6.2: Transforming and Combining Random Variables In this section, you learned how linear transformations of a random variable affect the shape, center, and spread of its probability distribution. As you learned in Chapter 2, a linear transformation does not change the shape (unless you multiply by a negative number) but can change the center and spread depending on the type of transformation. Multiplying (or dividing) each value of the random variable by a positive constant b multiplies (divides) the mean and standard deviation by b. Adding a constant a to (subtracting a from) each value of the random variable adds a to (subtracts a from) the mean but doesn’t change the standard deviation. You also learned how to calculate the mean and standard deviation for a combination of two or more random variables. If you are adding two random variables, X and Y, the mean and standard deviation of X + Y are mX+Y = mX + mY and sX+Y = "s2X + s2Y

Likewise, if you are subtracting two random variables, X and Y, the mean and standard deviation of X – Y are mX−Y = mX − mY and sX−Y = "s2X + s2Y

The formulas for the standard deviation of X + Y and X – Y are only correct if X and Y are independent, that is, if knowing the value of one variable doesn’t provide any additional information about the other variable. Also, if X and

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Y are both Normally distributed, then X + Y and X – Y are both Normally distributed as well. To determine which formulas to use for a particular problem, it is important to be able to distinguish linear transformations and combinations of random variables. Linear transformations take the values of one random variable and add, subtract, multiply, or divide them by a constant. Combinations of random variables take two or more random variables and add or subtract them. When a problem involves both linear transformations and a combination of random variables, remember to do the linear transformations first. Section 6.3: Binomial and Geometric Random Variables In this section, you learned about two common types of discrete random variables, binomial random variables and geometric random variables. Binomial random variables count the number of successes in a fixed number of trials (n), whereas geometric random variables count the number of trials needed to get one success. Otherwise, the binomial and geometric settings have the same conditions: there must be two possible outcomes for each trial (success or failure), the trials must be independent, and the probability of success p must stay the same throughout all trials. To calculate probabilities for a binomial distribution with n trials and probability of success p on each trial, use technology or the binomial probability formula n P(X = k) = a bpk(1 − p)n−k k The mean and standard deviation of a binomial random variable X are mX = np and sX = !np(1 − p)

The shape of a binomial distribution depends on both the number of trials n and the probability of success p. When the number of trials is large enough that both np and n(1 – p) are at least 10, the distribution of the binomial random variable X has an approximately Normal distribution. Be sure to check the large counts condition before using a Normal approximation to a binomial distribution. A common application of the binomial distribution is when we count the number of times a particular outcome occurs in a random sample from some population. Because sampling is almost always done without replacement, the independence condition is violated. However, if the sample size is a small fraction of the population size (less than 10%), the lack of independence isn’t a concern. Be sure to check the 10% condition when sampling is done without replacement before using a binomial distribution. Finally, to calculate probabilities for a geometric distribution with probability of success p on each trial, use technology or the geometric probability formula P(Y = k) = (1 − p)k−1p

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What Did You Learn? Learning Objective

Section

Related Example on Page(s)

Relevant Chapter Review Exercise(s)

Compute probabilities using the probability distribution of a discrete random variable.

6.1

349

R6.1

Calculate and interpret the mean (expected value) of a discrete random variable.

6.1

350, 352

R6.1, R6.3

Calculate and interpret the standard deviation of a discrete random variable.

6.1

353

R6.1, R6.3

Compute probabilities using the probability distribution of certain c­ ontinuous random variables.

6.1

355, 357

Describe the effects of transforming a random variable by adding or subtracting a constant and multiplying or dividing by a constant.

6.2

365, 366, 368

R6.2, R6.3

Find the mean and standard deviation of the sum or difference of independent random variables.

6.2

372, 373, 374, 377

R6.3, R6.4

Find probabilities involving the sum or difference of independent Normal random variables.

6.2

380, 381

R6.4

Determine whether the conditions for using a binomial random variable are met.

6.3

388

R6.5

R6.4

Compute and interpret probabilities involving binomial distributions.

6.3

390, 393, 396

R6.6

Calculate the mean and standard deviation of a binomial random variable. Interpret these values in context.

6.3

399

R6.5

Find probabilities involving geometric random variables.

6.3

406

R6.7

*When appropriate, use the Normal approximation to the binomial distribution to calculate probabilities.

6.3

403

R6.8

*This topic is not required for the AP® Statistics exam.

Chapter 6 Chapter Review Exercises These exercises are designed to help you review the important ideas and methods of the chapter. R6.1 ​ Knees  Patients receiving artificial knees often experience pain after surgery. The pain is measured on a subjective scale with possible values of 1 (low) to 5 (high). Let X be the pain score for a randomly selected patient. The following table gives part of the probability distribution for X. Value: 1 2 3 4 5 Probability: 0.1 0.2 0.3 0.3 ??

(a) Find P(X = 5). (b) Is pain score a discrete or continuous random variable? Explain. (c) Find P(X ≤ 2). Is this the same as P(X < 2)? Explain. (d) Compute the expected pain score and the standard deviation of the pain scores. R6.2 ​ A glass act  In a process for manufacturing glassware, glass stems are sealed by heating them in a flame. Let X be the temperature (in degrees Celsius) for a randomly chosen glass. The mean and standard deviation of X are mX = 550°C and sX = 5.7°C.

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Chapter Review Exercises (a) Is temperature a discrete or continuous random variable? Explain. (b) How is P(X < 540) related to P(X ≤ 540)? Explain. (c) The target temperature is 550°C. What are the mean and standard deviation of the number of degrees off target, D = X − 550? (d) A manager asks for results in degrees Fahrenheit. The conversion of X into degrees Fahrenheit is given by 9 Y = X + 32. What are the mean mY and the standard 5 deviation sY of the temperature of the flame in the Fahrenheit scale? R6.3 ​ Keno  In a game of 4-Spot Keno, the player picks 4 numbers from 1 to 80. The casino randomly selects 20 winning numbers from 1 to 80. The table below shows the possible outcomes of the game and their probabilities, along with the amount of money (Payout) that the player wins for a $1 bet. If X = the payout for a single $1 bet, you can check that mX = $0.70 and sX = $6.58. Matches: 0 1 2 3 4 Payout xi : $0 $0 $1 $3 $120 Probability pi : 0.308 0.433 0.213 0.043 0.003

(a) Interpret the values of mX and sX in context. (b) Jerry places a single $5 bet on 4-Spot Keno. Find the expected value and the standard deviation of his winnings. (c) Marla plays five games of 4-Spot Keno, betting $1 each time. Find the expected value and the standard deviation of her total winnings. (d) Based on your answers to (b) and (c), which player would the casino prefer? Justify your answer. R6.4 ​Applying torque  A machine fastens plastic screw-on caps onto containers of motor oil. If the machine applies more torque than the cap can withstand, the cap will break. Both the torque applied and the strength of the caps vary. The capping-machine torque T follows a Normal distribution with mean 7 inch-pounds and standard deviation 0.9 inch-pounds. The cap strength C (the torque that would break the cap) follows a Normal distribution with mean 10 inch-pounds and standard deviation 1.2 inch-pounds. (a) What is the probability that a randomly selected cap has a strength greater than 11 inch-pounds? (b) Explain why it is reasonable to assume that the cap strength and the torque applied by the machine are independent.

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(c) Let the random variable D = C − T. Find its mean and standard deviation. (d) What is the probability that a randomly selected cap will break while being fastened by the machine? Show your work. Exercises R6.5 and R6.6 refer to the following setting. According to Mars, Incorporated, 20% of its plain M&M’S candies are orange. Assume that the company’s claim is true. Suppose that you reach into a large bag of plain M&M’S (without looking) and pull out 8 candies. Let X = the number of orange candies you get. R6.5 ​ Orange M&M’S (a) Explain why it is reasonable to use the binomial distribution for probability calculations involving X. (b) Find and interpret the expected value of X. (c) Find and interpret the standard deviation of X. R6.6 ​ Orange M&M’S (a) Would you be surprised if none of the candies were orange? Compute an appropriate probability to support your answer. (b) How surprising would it be to get 5 or more orange candies? Compute an appropriate probability to support your answer. R6.7 ​ Sushi Roulette  In the Japanese game show Sushi Roulette, the contestant spins a large wheel that’s divided into 12 equal sections. Nine of the sections have a sushi roll, and three have a “wasabi bomb.” When the wheel stops, the contestant must eat whatever food is on that section. To win the game, the contestant must eat one wasabi bomb. Find the probability that it takes 3 or fewer spins for the contestant to get a wasabi bomb. Show your method clearly. R6.8* ​Is this coin balanced?  While he was a prisoner of war during World War II, John Kerrich tossed a coin 10,000 times. He got 5067 heads. If the coin is perfectly balanced, the probability of a head is 0.5. (a) Find the mean and the standard deviation of the number of heads in 10,000 tosses, assuming the coin is perfectly balanced. (b) Explain why the Normal approximation is appropriate for calculating probabilities involving the number of heads in 10,000 tosses. (c) Is there reason to think that Kerrich’s coin was not balanced? To answer this question, use a Normal distribution to estimate the probability that tossing a balanced coin 10,000 times would give a count of heads at least this far from 5000 (that is, at least 5067 heads or at most 4933 heads).

*This topic is not required for the AP® Statistics exam. Some teachers prefer to discuss this topic when presenting the sampling distribution of pˆ (Chapter 7).

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Chapter 6 AP® Statistics Practice Test Section I: Multiple Choice  ​Select the best answer for each question. Questions T6.1 to T6.3 refer to the following setting. A psychologist studied the number of puzzles that subjects were able to solve in a five-minute period while listening to soothing music. Let X be the number of puzzles completed successfully by a randomly chosen subject. The psychologist found that X had the following probability distribution: Value: 1 2 3 4 Probability: 0.2 0.4 0.3 0.1

T6.1 What is the probability that a randomly chosen subject completes more than the expected number of puzzles in the five-minute period while listening to soothing music? (a) ​0.1   (b) ​0.4   (c) 0.8   (d) ​1   (e) ​Cannot be determined T6.2 The standard deviation of X is 0.9. Which of the following is the best interpretation of this value? (a) About 90% of subjects solved 3 or fewer puzzles. (b) About 68% of subjects solved between 0.9 puzzles less and 0.9 puzzles more than the mean. (c) The typical subject solved an average of 0.9 puzzles. (d) The number of puzzles solved by subjects typically differed from the mean by about 0.9 puzzles. (e) The number of puzzles solved by subjects typically differed from one another by about 0.9 puzzles. T6.3 ​ Let D be the difference in the number of puzzles solved by two randomly selected subjects in a fiveminute period. What is the standard deviation of D? (a) ​ 0  (b) ​0.81   (c) ​0.9   (d) ​1.27   (e) ​1.8 T6.4 ​ Suppose a student is randomly selected from your school. Which of the following pairs of random variables are most likely independent? (a) ​X = student’s height; Y = student’s weight (b) X = student’s IQ; Y = student’s GPA (c) ​X = student’s PSAT Math score; Y = student’s PSAT Verbal score (d) X = average amount of homework the student does per night; Y = student’s GPA (e) X = average amount of homework the student does per night; Y = student’s height

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T6.5 ​ A certain vending machine offers 20-ounce bottles of soda for $1.50. The number of bottles X bought from the machine on any day is a random variable with mean 50 and standard deviation 15. Let the random variable Y equal the total revenue from this machine on a given day. Assume that the machine works properly and that no sodas are stolen from the machine. What are the mean and standard deviation of Y? (a) mY = $1.50, sY = $22.50 (b) ​mY = $1.50, sY = $33.75 (c) ​mY = $75, sY = $18.37 (d) ​mY = $75, sY = $22.50 (e) ​mY = $75, sY = $33.75 T6.6 The weight of tomatoes chosen at random from a bin at the farmer’s market follows a Normal distribution with mean m = 10 ounces and standard deviation s = 1 ounce. Suppose we pick four tomatoes at random from the bin and find their total weight T. The random variable T is (a) Normal, with mean 10 ounces and standard deviation 1 ounce. (b) Normal, with mean 40 ounces and standard deviation 2 ounces. (c) Normal, with mean 40 ounces and standard deviation 4 ounces. (d) binomial, with mean 40 ounces and standard deviation 2 ounces. (e) binomial, with mean 40 ounces and standard deviation 4 ounces. T6.7 ​ Which of the following random variables is geometric? (a) ​The number of times I have to roll a die to get two 6s. (b) ​The number of cards I deal from a well-shuffled deck of 52 cards until I get a heart. (c) ​The number of digits I read in a randomly selected row of the random digits table until I find a 7. (d) ​The number of 7s in a row of 40 random digits. (e) ​The number of 6s I get if I roll a die 10 times. T6.8 ​ Seventeen people have been exposed to a particular disease. Each one independently has a 40% chance of contracting the disease. A hospital has the capacity to handle 10 cases of the disease. What is the probability that the hospital’s capacity will be exceeded? (a) ​0.011 (b) 0.035 (c) ​0.092

(d) ​0.965 (e) ​0.989

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AP® Statistics Practice Test T6.9 ​ The figure shows the probability distribution of a discrete random variable X. Note that the cursor is on the histogram bar representing a value of 6. Which of the following best describes this random variable?

(a) ​Binomial with n = 8, p = 0.1 (b) ​Binomial with n = 8, p = 0.3 (c) ​Binomial with n = 8, p = 0.8 (d) ​Geometric with p = 0.1 (e) ​Geometric with p = 0.2

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T6.10 ​A test for extrasensory perception (ESP) involves asking a person to tell which of 5 shapes—a circle, star, triangle, diamond, or heart—appears on a hidden computer screen. On each trial, the computer is equally likely to select any of the 5 shapes. Suppose researchers are testing a person who does not have ESP and so is just guessing on each trial. What is the probability that the person guesses the first 4 shapes incorrectly but gets the fifth correct? (a) ​1/5 4 4 (b) a b 5

4 4 1 (c) ​a b # a b 5 5

5 4 4 1 (d) a b# a b #a b 1 5 5 (e) ​4/5

Section II: Free Response  ​Show all your work. Indicate clearly the methods you use, because you will be graded on the correctness of your methods as well as on the accuracy and completeness of your results and explanations. T6.11 Let Y denote the number of broken eggs in a randomly selected carton of one dozen “store brand” eggs at a local supermarket. Suppose that the probability distribution of Y is as follows. Value yi : 0 1 2 3 4 Probability pi : 0.78 0.11 0.07 0.03 0.01

(a) ​What is the probability that at least 10 eggs in a randomly selected carton are unbroken? (b) ​Calculate and interpret mY. (c) ​Calculate and interpret sY. Show your work. (d) A quality control inspector at the store keeps looking at randomly selected cartons of eggs until he finds one with at least 2 broken eggs. Find the probability that this happens in one of the first three cartons he inspects. T6.12 ​ Ladies Home Journal magazine reported that 66% of all dog owners greet their dog before greeting their spouse or children when they return home at the end of the workday. Assume that this claim is true. Suppose 12 dog owners are selected at random. Let X = the number of owners who greet their dogs first. (a) ​ Explain why it is reasonable to use the binomial distribution for probability calculations involving X. (b) ​ Only 4 of the owners in the random sample greeted their dogs first. Does this give convincing evidence against the Ladies Home Journal claim? Calculate an appropriate probability to support your answer.

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T6.13 ​Ed and Adelaide attend the same high school, but are in different math classes. The time E that it takes Ed to do his math homework follows a Normal distribution with mean 25 minutes and standard deviation 5 minutes. Adelaide’s math homework time A follows a Normal distribution with mean 50 minutes and standard deviation 10 minutes. Assume that E and A are independent random variables. (a) Randomly select one math assignment of Ed’s and one math assignment of Adelaide’s. Let the random variable D be the difference in the amount of time each student spent on their assignments: D = A − E. Find the mean and the standard deviation of D. Show your work. (b) ​Find the probability that Ed spent longer on his assignment than Adelaide did on hers. Show your work. T6.14 ​ According to the Census Bureau, 13% of American adults (aged 18 and over) are Hispanic. An opinion poll plans to contact an SRS of 1200 adults. (a) ​What is the mean number of Hispanics in such samples? What is the standard deviation? (b) Should we be suspicious if the sample selected for the opinion poll contains 15% Hispanic people? Compute an appropriate probability to support your answer.

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Chapter

7

Introduction 422 Section 7.1 424 What Is a Sampling Distribution? Section 7.2 Sample Proportions

440

Section 7.3 Sample Means

450

Free Response AP® Problem, Yay! 464 Chapter 7 Review 465 Chapter 7 Review Exercises 466 Chapter 7 AP® Statistics Practice Test 468 Cumulative AP® Practice Test 2 470

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Sampling Distributions case study

Building Better Batteries Everyone wants to have the latest technological gadget. That’s why iPods, digital cameras, smartphones, Game Boys, and the Wii have sold millions of units. These devices require lots of power and can drain batteries quickly. Battery manufacturers are constantly searching for ways to build longer-lasting b ­ atteries. A particular manufacturer produces AA batteries that are designed to last an average of 17 hours with a standard deviation of 0.8 hours. Quality control ­inspectors select a random sample of 50 batteries during each hour of production, and they then drain them under conditions that mimic normal use. Here are the lifetimes (in hours) of the batteries from one such sample: 16.73 15.60 16.31 17.57 16.14 17.28 16.67 17.28 17.27 17.50 15.46 16.50 16.19 15.59 17.54 16.46 15.63 16.82 17.16 16.62 16.71 16.69 17.98 16.36 17.80 16.61 15.99 15.64 17.20 17.24 16.68 16.55 17.48 15.58 17.61 15.98 16.99 16.93 16.01 17.54 17.41 16.91 16.60 16.78 15.75 17.31 16.50 16.72 17.55 16.46 Do these data suggest that the production process is working properly? Is it safe for plant managers to send out all the batteries produced in this hour for sale? In this chapter, you will develop the tools you need to help answer questions like this.

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Introduction The battery manufacturer in the Case Study could find the true mean lifetime m of all the batteries produced in an hour. Quality control inspectors would simply measure the lifetime of each battery (by draining it) and then calculate the average. With this method, the company would know the truth about the population mean m, but it would have no batteries left to sell! Instead of taking a census, the manufacturer collects data from a random sample of 50 batteries produced that hour. The company’s goal is to use the sample mean lifetime x– to estimate the unknown population mean m. This is an example of statistical inference: we use information from a sample to draw conclusions about a larger population. To make such an inference, we need to know how close the sample mean x– is likely to be to the population mean m. After all, different random samples of 50 batteries from the same hour of production would yield different values of x– . How can we describe this sampling distribution of possible x– -values? We can think of x– as a random variable because it takes numerical values that describe the outcomes of the random sampling process. As a result, we can examine its probability distribution using what we learned in Chapter 6. This same reasoning applies to other types of inference settings. Here are a few examples. • Each month, the Current Population Survey (CPS) interviews a random sample of individuals in about 60,000 U.S. households. The CPS uses the proportion of unemployed people in the sample p^ to estimate the national unemployment rate p. • Tom is cooking a large turkey breast for a holiday meal. He wants to be sure that the turkey is safe to eat, which requires a minimum internal temperature of 165°F. Tom uses a thermometer to measure the temperature of the turkey meat at four randomly chosen points. If the minimum reading in the sample is 170°F, can Tom safely serve the turkey? • How much do gasoline prices vary in a large city? To find out, a reporter records the price per gallon of regular unleaded gasoline at a random sample of 10 gas stations in the city on the same day. The range (maximum – minimum) of the prices in the sample is 25 cents. What can the reporter say about the range of gas prices at all the city’s stations? The following Activity gives you a chance to estimate an unknown population value based on data from a random sample.

Activity Materials: Tags or pieces of cardstock numbered 1 to N; small brown paper bag; index card and prelabeled graph grid for each team; prizes for the winners

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​The German Tank Problem During World War II, the Allies captured several German tanks. Each tank had a serial number on it. Allied commanders wanted to know how many tanks the ­Germans had so that they could allocate their forces appropriately. They sent the serial numbers of the captured tanks to a group of mathematicians in Washington, D.C., and asked for an estimate of the total number of German tanks N. In this Activity, you and your teammates will play the role of the mathematicians.

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Introduction

More recently, people used the mathematicians’ method from the German tank problem to estimate the number of iPhones produced.

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1. ​Your teacher will create tags numbered 1, 2, 3, . . . , N to represent the German tanks and place them in a bag. The class will be divided into teams of three or four students. 2. ​The teacher will mix the tags well and ask four students to draw one tag each from the bag. Each selected tag represents the serial number of a captured German tank. All four numbers should be written on the board for everyone to see. Return the tags to the bag. 3. ​Each team will have 15 minutes to come up with a statistical formula for estimating the total number of tanks N in the bag. You should have time to try several ideas. When you are satisfied with your method, calculate your estimate of N. Write your team members’ names, your formula, and your estimate on the index card provided. 4. ​When time is called, each team must give its index card to your teacher. The teacher will make a chart on the board showing the formulas and estimates. Each team will have one minute to explain why it chose the formula it did. 5. ​The teacher will reveal the actual number of tanks. Which team came closest to the correct answer? 6. ​What if the Allies had captured four other German tanks? Which team’s formula would consistently produce the best estimate? Students should help choose nine more simple random samples of four tanks from the bag. After each sample is taken, the four serial numbers chosen should be written on the board, and the tags should be returned to the bag and mixed thoroughly. 7. ​Each team should use its formula to estimate the total number of tanks N for each of the nine new samples. The team should then make a dotplot of its 10 estimates. 8. ​Compare the teams’ dotplots. As a class, decide which team used the best method for estimating the number of tanks.

Sampling distributions are the key to inference when data are produced by random sampling. Because the results of random samples include an element of chance, we can’t guarantee that our inferences are correct. What we can guarantee is that our methods usually give correct answers. The reasoning of statistical inference rests on asking, “How often would this method give a correct answer if I used it very many times?” If our data come from random sampling, the laws of probability answer the question “What would happen if we did this many times?” Section 7.1 presents the basic ideas of sampling distributions. The most ­common applications of statistical inference involve proportions and means. Section 7.2 focuses on sampling distributions of sample proportions. Section 7.3 investigates sampling distributions of sample means.

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7.1 What You Will Learn

What Is a Sampling Distribution? By the end of the section, you should be able to:

• Distinguish between a parameter and a statistic. • Use the sampling distribution of a statistic to evaluate a claim about a parameter. • Distinguish among the distribution of a population, the distribution of a sample, and the sampling distribution of a statistic.

• Determine whether or not a statistic is an unbiased estimator of a population parameter. • Describe the relationship between sample size and the variability of a statistic.

What is the average income of American households? Each March, the government’s Current Population Survey (CPS) asks detailed questions about income. The random sample of about 60,000 households contacted in March 2012 had a mean “total money income” of $69,677 in 2011.1 (The median income was lower, of course, at $50,054.) That $69,677 describes the sample, but we use it to estimate the mean income of all households.

Parameters and Statistics As we begin to use sample data to draw conclusions about a wider population, we must be clear about whether a number describes a sample or a population. For the sample of households contacted by the CPS, the mean income was x– = $69,677. The number $69,677 is a statistic because it describes this one CPS sample. The population that the poll wants to draw conclusions about is all 121 million U.S. households. In this case, the parameter of interest is the mean income m of all these households. We don’t know the value of this parameter.

Definition: Parameter, statistic A parameter is a number that describes some characteristic of the population. A statistic is a number that describes some characteristic of a sample.

It is common practice to use Greek letters for parameters and Roman letters for statistics. In that case, the population proportion would be p (pi, the Greek letter for “p”) and the sample proportion would be p. We’ll stick with the notation that’s used on the AP® exam, however.

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The value of a parameter is usually not known because we cannot examine the entire population. The value of a statistic can be computed directly from the sample data. We often use a statistic to estimate an unknown parameter. Remember s and p: statistics come from samples, and parameters come from populations. As long as we were doing data analysis, the distinction between population and sample rarely came up. Now, however, it is essential. The notation we use should reflect this distinction. For instance, we write m (the Greek letter mu) for the population mean and x– for the sample mean. We use p to represent a population proportion. The sample proportion p^ is used to estimate the unknown parameter p.

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Section 7.1 What Is a Sampling Distribution?

Example

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From Ghosts to Cold Cabins Parameters and statistics Problem:  Identify the population, the parameter, the sample, and the statistic in each of the ­following settings. (a) The Gallup Poll asked a random sample of 515 U.S. adults whether or not they believe in ghosts. Of the respondents, 160 said “Yes.”2 (b) During the winter months, the temperatures outside the Starneses’ cabin in Colorado can stay well below freezing (32°F, or 0°C) for weeks at a time. To prevent the pipes from freezing, Mrs. Starnes sets the thermostat at 50°F. She wants to know how low the temperature actually gets in the cabin. A digital thermometer records the indoor temperature at 20 randomly chosen times during a given day. The minimum reading is 38°F. Solution: (a) The population is all U.S. adults, and the parameter of interest is p, the proportion of all U.S. adults who believe in ghosts. The sample is the 515 people who were interviewed in this Gallup Poll. 160 = 0.31, the proportion of the sample who say they believe in ghosts. The statistic is p^ = 515 (b) The population is all times during the day in question; the parameter of interest is the true minimum temperature in the cabin that day. The sample consists of the 20 temperature readings at randomly selected times. The statistic is the sample minimum, 38°F. For Practice Try Exercise

1

Check Your Understanding

Each boldface number in Questions 1 and 2 is the value of either a parameter or a statistic. In each case, state which it is and use appropriate notation to describe the number. 1. On Tuesday, the bottles of Arizona Iced Tea filled in a plant were supposed to contain an average of 20 ounces of iced tea. Quality control inspectors sampled 50 bottles at random from the day’s production. These bottles contained an average of 19.6 ounces of iced tea. 2. On a New York–to–Denver flight, 8% of the 125 passengers were selected for random security screening before boarding. According to the Transportation Security Administration, 10% of passengers at this airport are chosen for random screening.

Sampling Variability How can x– , based on a sample of only a few thousand of the 121 million A ­ merican households, be an accurate estimate of m? After all, a second random sample taken at the same time would choose different households and likely produce a different value of x– . This basic fact is called sampling variability: the value of a statistic varies in repeated random sampling.

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To make sense of sampling variability, we ask, “What would happen if we took many samples?” Here’s how to answer that question: • Take a large number of samples from the same population. • Calculate the statistic (like the sample mean x– or sample proportion p^ ) for each sample. • Make a graph of the values of the statistic. • Examine the distribution displayed in the graph for shape, center, and spread, as well as outliers or other unusual features. The following Activity gives you a chance to see sampling variability in action.

Activity

Reaching for Chips

Materials: 200 colored chips, including 100 of the same color; large bag or other container

Before class, your teacher will prepare a population of 200 colored chips, with 100 having the same color (say, red). The parameter is the actual proportion p of red chips in the population: p = 0.50. In this Activity, you will investigate sampling variability by taking repeated random samples of size 20 from the population. 1. ​After your teacher has mixed the chips thoroughly, each student in the class should take a sample of 20 chips and note the sample proportion p^ of red chips. When finished, the student should return all the chips to the bag, stir them up, and pass the bag to the next student. Note:  If your class has fewer than 25 students, have some students take two ­samples. 2. ​Each student should record the p^ -value in a chart on the board and plot this value on a class dotplot. Label the graph scale from 0.10 to 0.90 with tick marks spaced 0.05 units apart. 3. ​Describe what you see: shape, center, spread, and any outliers or other unusual features.

When Mr. Caldwell’s class did the “Reaching for Chips” Activity, his 35 students produced the graph shown in Figure 7.1. Here’s what the class said about its ­distribution of p^ -values. Shape:  The graph is roughly symmetric with a single peak at 0.5. 0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

^ p

FIGURE 7.1  ​Dotplot of sample proportions obtained by the 35 students in Mr. Caldwell’s class.

Center:  The mean of our sample proportions is 0.499. This is the balance point of the distribution. Spread:  The standard deviation of our sample proportions is 0.112. The values of p^ are typically about 0.112 away from the mean.

Outliers:  There are no obvious outliers or other unusual ­features. Of course, the class only took 35 different simple random samples of 20 chips. There are many, many possible SRSs of size 20 from a population of size 200 (about 1.6 · 1027, actually). If we took every one of those possible samples, calculated the value of p^ for each, and graphed all those p^ -values, then we’d have a sampling distribution.

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Section 7.1 What Is a Sampling Distribution?

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Definition: Sampling distribution The sampling distribution of a statistic is the distribution of values taken by the ­statistic in all possible samples of the same size from the same population. It’s usually too difficult to take all possible samples of size n to obtain the sampling distribution of a statistic. Instead, we can use simulation to imitate the process of taking many, many samples and create an approximate sampling distribution.

EXAMPLE

Reaching for Chips Simulating a sampling distribution We used Fathom software to simulate choosing 500 SRSs of size n = 20 from a population of 200 chips, 100 red and 100 blue. Figure 7.2 is a dotplot of the values of p^ , the sample proportion of red chips, from these 500 samples.

Problem:

p^

FIGURE 7.2  ​Dotplot of the sample proportion p^ of red chips in 500 simulated SRSs, ­created by Fathom software.

(a) There is one dot on the graph at 0.15. Explain what this value represents. (b) Describe the distribution. Are there any obvious outliers? (c) Would it be surprising to get a sample proportion of 0.85 or higher in an SRS of size 20 when p = 0.5? Justify your answer. (d) Suppose your teacher prepares a bag with 200 chips and claims that half of them are red. A classmate takes an SRS of 20 chips; 17 of them are red. What would you conclude about your teacher’s claim? Explain.

Solution: (a) In one SRS of 20 chips, there were 3 red chips. So p^ = 3/20 = 0.15 for this sample. (b) Shape: Symmetric, unimodal, and somewhat bell-shaped. Center: Around 0.5. Spread: The values of p^ fall mostly between 0.25 and 0.75. Outliers: One sample with p^ = 0.15 stands out. (c) It is very unlikely to obtain an SRS of 20 chips in which p^ = 0.85 from a population in which p = 0.5. A value of p^ this large or larger never occurred in 500 simulated samples. (d) This student’s result gives strong evidence against the teacher’s claim. As noted in part (c), it is very unlikely to get a sample proportion of 0.85 or higher when p = 0.5. For Practice Try Exercise

9

Strictly speaking, the sampling distribution is the ideal pattern that would emerge if we looked at all possible samples of size 20 from our population of chips. A distribution obtained from simulating a smaller number of random samples, like the 500 values of p^ in Figure 7.2, is only an approximation to the sampling distribution. One of the uses of probability theory in statistics is to obtain sampling distributions without simulation. We’ll get to the theory later. The interpretation of a sampling distribution is the same, however, whether we obtain it by simulation or by the mathematics of probability.

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Figure 7.3 illustrates the process of choosing many random samples of 20 chips and finding the sample proportion of red chips p^ for each one. Follow the flow of the figure from the population at the left, to choosing an SRS and finding the p^ for this sample, to collecting together the p^ ’s from many samples. The first sample has p^ = 0.40. The second sample is a different group of chips, with p^ = 0.55, and so on. The dotplot at the right of the figure shows the distribution of the values of p^ from 500 separate SRSs of size 20. This dotplot displays the approximate sampling distribution of the statistic p^ . Distributions of sample data 14

Frequency

12

Population distribution Parameter: p = 0.50

^ = 8 = 0.40 p 20

4 0

Approximate sampling distribution

Blue Red Color

12 10

SRS n = 20

Frequency

Frequency

100 80 60 40 20 0

8 6 2

SRS n = 20

120

10

8 4 2 0

Blue Red Color

^ = 11 = 0.55 p 20

6

Blue Red Color

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 p^

FIGURE 7.3  ​The idea of a sampling distribution: take many samples from the same population, collect the p^ ’s from all the samples, and display the distribution of the p^ ’s. The dotplot shows the results of 500 samples.

c

!

n

AP® EXAM TIP ​Terminology matters. Don’t say “sample distribution” when you mean sampling distribution. You will lose credit on free response questions for misusing statistical terms.

As Figure 7.3 shows, there are three distinct distributions involved when we sample repeatedly and measure a variable of interest. The population distribution gives the values of the variable for all individuals in the population. In this case, the individuals are the 200 chips and the variable we’re recording is color. Our ­parameter of interest is the proportion of red chips in the population, p  = 0.50. Each random sample that we take consists of 20 chips. The distribution of sample data shows the values of the variable “color” for the individuals in the sample. For each sample, we record a value for the statistic p^ , the sample proportion of red chips. Finally, we collect the values of p^ from all possible samples of the same size and display them in the sampling distribution. Be careful: The population distribution and the distribution of sample autio data describe individuals. A sampling distribution describes how a statistic varies in many samples from the population.

Check Your Understanding

Mars, Incorporated, says that the mix of colors in its M&M’S® Milk Chocolate Candies is 24% blue, 20% orange, 16% green, 14% yellow, 13% red, and 13% brown. Assume that the company’s claim is true. We want to examine the proportion of orange M&M’S in repeated random samples of 50 candies. 1. ​Graph the population distribution. Identify the individuals, the variable, and the ­parameter of interest.

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Section 7.1 What Is a Sampling Distribution?

2. ​Imagine taking an SRS of 50 M&M’S. Make a graph showing a possible distribution of the sample data. Give the value of the appropriate statistic for this sample.

Color

R ed Ye llo w

n

ge

O

ra n

n

re e G

B

B

lu e

20 18 16 14 12 10 8 6 4 2 0

ro w

Frequency

3. ​Which of the graphs that follow could be the approximate sampling distribution of the statistic? Explain your choice.

 

0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 ^ p

 

0.20

0.30

0.40

0.50

0.60

^ p

Describing Sampling Distributions The fact that statistics from random samples have definite sampling distributions allows us to answer the question “How trustworthy is a statistic as an estimate of a parameter?” To get a complete answer, we consider the shape, center, and spread of the sampling distribution. For reasons that will be clear later, we’ll save shape for last.

Center: Biased and unbiased estimators  ​Let’s return to the familiar

chips example. How well does the sample proportion of red chips estimate the population proportion of red chips, p = 0.5? The dotplot in the margin shows the approximate sampling distribution of p^ once again. We noted earlier that the center of this distribution is very close to 0.5, the parameter value. In fact, if we took all possible samples of 20 chips from the population, calculated p^ for each sample, and then found the mean of all those p^ -values, we’d get exactly 0.5. For this reason, we say that p^ is an unbiased estimator of p. 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 ^ p

Definition: Unbiased estimator A statistic used to estimate a parameter is an unbiased estimator if the mean of its sampling distribution is equal to the value of the parameter being estimated. If we take many samples, the value of an unbiased estimator will sometimes exceed the value of the parameter and sometimes be less. However, because the sampling distribution of the statistic is centered at the true value, we will not consistently overestimate or underestimate the parameter. This is consistent with our definition of bias from Chapter 4. We will confirm in Section 7.2 that the sample proportion p^ is an unbiased estimator of the population proportion p. This is a very helpful result if we’re dealing with a categorical variable (like color). With quantitative variables, we might be ­interested in estimating the population mean, m ­ edian, minimum, maximum, Q1, Q3, variance, standard deviation, IQR, or range. Which (if any) of these are unbiased estimators? The following Activity should shed some light on this question.

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Activity

Sampling heights

Materials:

In this Activity, you will use a population of quantitative data to investigate whether a given statistic is an unbiased estimator of its corresponding population parameter. 1. ​Each student should write his or her height (in inches) neatly on a small piece of cardstock and then place it in the bag. 2. ​After your teacher has mixed the cards thoroughly, each student in the class should take a sample of four cards and record the heights of the four chosen students. When finished, the student should return the cards to the bag, mix them up, and pass the bag to the next student. Note:  If your class has fewer than 25 students, have some students take two ­samples. 3. ​For your SRS of four students from the class, calculate the sample mean x– and the sample range (maximum – minimum) of the heights. Then go to the board and record the heights of the four students in your sample, the sample mean x– , and the sample range in a chart like the one below.

Small piece of cardstock for each student; bag

Height (in.)

Sample mean (x– )

Sample range (max - min)

62, 75, 68, 63

67

75 - 62 = 13

4. ​Plot the values of your sample mean and sample range on the two class ­dotplots drawn by your teacher. 5. ​Once everyone has finished, find the population mean m and the population range. 6. ​Based on your approximate sampling distributions of x– and the sample range, which statistic appears to be an unbiased estimator? Which appears to be a biased estimator?

When Mrs. Washington’s class did the “Sampling Heights” Activity, they produced the graphs shown in Figure 7.4. Her students concluded that the sample mean x– is probably an unbiased estimator of the population mean m. Their reason: the center of the approximate sampling distribution of x– , 65.67 inches, is close to the population mean of 66.07 inches. On the other hand, Mrs. Washington’s Approximate sampling distribution of sample range (n = 4)

Approximate sampling distribution of x (n = 4)

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Center: mean = 65.67 inches Population mean m = 66.07 inches

Center: mean = 10.12 inches Population range = 21 inches

20

22

FIGURE 7.4  ​Results from Mrs. Washington’s class. The sample mean appears to be an unbiased estimator. The sample range appears to be a biased estimator.

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students decided that the sample range is a biased estimator of the population range. Why? Because the center of the sampling distribution for this statistic was 10.12 inches, much less than the corresponding parameter value of 21 inches. To confirm the class’s conclusions, we used Fathom software to simulate taking 250 SRSs of n = 4 students. For each sample, we plotted the mean height x– and the range of the heights. Figure 7.5 shows the approximate sampling distributions for these two statistics. It looks like the class was right: x– is an unbiased estimator of m, but the sample range is clearly a biased estimator. The range of the sample heights tends to be much lower, on average, than the population range. Population mean µ  66.07 inches

Population range  21 inches

Center: mean  66.15 inches

Center: mean  10.21 inches

FIGURE 7.5  ​Results from a Fathom simulation of 250 SRSs of size n = 4 from the students in Mrs. Washington’s class. The sample mean is an unbiased estimator. The sample range is a biased estimator.

THINK ABOUT IT

Why do we divide by n − 1 when calculating the sample ­variance?  ​In Chapter 1, we introduced the sample variance s2x as a measure of

spread for a set of quantitative data. The idea of s2x is simple: it’s a number that describes the “average” squared deviation of the values in the sample from their mean x–. It probably surprised you when we computed this average by dividing by n − 1 1 (xi − x–)2. instead of n. Now we’re ready to tell you why we defined s2x = n−1 In an inference setting involving a quantitative variable, we might be interested in estimating the variance s2 of the population distribution. The most logical choice for our estimator is the sample variance s2x . We used Fathom software to take 500 SRSs of size n = 4 from the population distribution of heights in Mrs. Washington’s class. Note that the population variance is s2 = 22.19. For each sample, we recorded the value of two statistics:

∙

FIGURE 7.6  ​Results from a Fathom simulation of 500 SRSs of size n = 4 from the population distribution of heights in Mrs. Washington’s class. The sample variance sx2 (labeled “var” in the figure) is an unbiased estimator. The “varn” statistic (dividing by n instead of n − 1) is a biased estimator.

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var = s2x = varn =

∙

1 (xi − x–)2 (the sample variance) n−1

∙

1 (xi − x–)2 n

Figure 7.6 shows the approximate sampling distributions of these two statistics. We used histograms to show the overall pattern more clearly. The vertical lines mark the means of these two ­distributions.

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We can see that “varn” is a biased estimator of the population variance. The mean of its sampling distribution (marked with a blue line segment) is clearly less than the value of the population parameter, 22.19. However, the statistic “var” (otherwise known as the sample variance s2x ) is an unbiased estimator. Its values are centered at 22.19. That’s why we divide by n − 1 and not n when calculating the sample variance: to get an unbiased estimator of the population variance.

Spread: Low variability is better!  ​To get a trustworthy estimate of an unknown population parameter, start by ­using a statistic that’s an unbiased estimator. This ensures that you won’t consistently overestimate or underestimate the parameter. ­Unfortunately, using an unbiased estimator doesn’t guarantee that the value of your statistic will be close to the actual parameter value. The following example illustrates what we mean.

EXAMPLE

Who Watches Survivor? Why sample size matters Television executives and companies who advertise on TV are interested in how many viewers watch particular shows. According to Nielsen ratings, ­Survivor was one of the most-watched television shows in the United States during every week that it aired. Suppose that the true proportion of U.S. adults who have watched Survivor is p = 0.37. The top dotplot in Figure 7.7 shows the results of drawing 400 SRSs of size n = 100 from a population with p = 0.37. We see that a sample of 100 people ­ arameter. That is why a Gallup often gave a p^ quite far from the population p Poll asked not 100, but 1000 people whether they had watched Survivor. Let’s repeat our simulation, this time taking 400 SRSs of size n = 1000 from a population with proportion p = 0.37 who have watched Survivor. The bottom dotplot in Figure 7.7 displays the distribution of the 400 values of p^ from these new samples. Both graphs are drawn on the same horizontal scale to make comparison easy.

n = 100

n = 1000

FIGURE 7.7  ​The approximate sampling distribution of the sample proportion p^ from SRSs of size n = 100 and n = 1000 drawn from a population with proportion p = 0.37 who have watched Survivor. Both dotplots show the results of 400 SRSs.

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0.25

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Section 7.1 What Is a Sampling Distribution?

We can see that the spread of the top dotplot in Figure 7.7 is much greater than the spread of the bottom dotplot. With samples of size 100, the values of p^ vary from 0.25 to 0.54. The standard deviation of these p^ -values is about 0.05. Using SRSs of size 1000, the values of p^ only vary from 0.328 to 0.412. The standard deviation of these p^ -values is about 0.015, so most random samples of 1000 people give a p^ that is within 0.03 of the actual population parameter, p = 0.37.

c

!

n

The sample proportion p^ from a random sample of any size is an unbiased ­estimator of the parameter p. As we can see from the previous example, though, larger random samples have a clear advantage. They are much more likely to produce an estimate close to the true value of the parameter. Said another way, larger random samples give us more precise estimates than smaller random samples. That’s because a large random sample gives us more information about the underlying population than a smaller sample does. autio Taking a larger sample doesn’t fix bias. Remember that even a very large voluntary response sample or convenience sample is worthless because of bias. There are general rules for describing how the spread of the sampling distribution of a statistic decreases as the sample size increases. In Sections 7.2 and 7.3, we’ll reveal these rules for the sampling distributions of p^ and x– . One important and surprising fact is that the variability of a statistic in repeated sampling does not depend very much on the size of the population.

Variability of a Statistic The variability of a statistic is described by the spread of its sampling distribution. This spread is determined mainly by the size of the random sample. Larger samples give smaller spreads. The spread of the sampling distribution does not depend much on the size of the population, as long as the population is at least 10 times larger than the sample. Why does the size of the population have little influence on the behavior of statistics from random samples? Imagine sampling harvested corn by thrusting a scoop into a large sack of corn kernels. The scoop doesn’t know whether it is surrounded by a bag of corn or by an entire truckload. As long as the corn is well mixed (so that the scoop selects a random sample), the variability of the result depends only on the size of the scoop. The fact that the variability of a statistic is controlled by the size of the sample has important consequences for designing samples. Suppose a researcher wants to estimate the proportion of all U.S. adults who use Twitter regularly. A random sample of 1000 or 1500 people will give a fairly precise estimate of the parameter because the sample size is large. Now consider another researcher who wants to estimate the proportion of all Ohio State University students who use Twitter regularly. It can take just as large an SRS to estimate the proportion of Ohio State University students who use Twitter regularly as to estimate with equal precision the proportion of all U.S. adults who use Twitter regularly. We can’t expect to need a smaller SRS at Ohio State just because there are about 60,000 Ohio State students and about 235 million adults in the United States.

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Check Your Understanding

   The histogram above left shows the intervals (in minutes) between eruptions of Old ­Faithful geyser for all 222 recorded eruptions during a particular month. For this population, the median is 75 minutes. We used Fathom software to take 500 SRSs of size 10 from the population. The 500 values of the sample median are displayed in the histogram above right. The mean of these 500 values is 73.5. 1. ​Is the sample median an unbiased estimator of the population median? Justify your answer. 2. ​Suppose we had taken samples of size 20 instead of size 10. Would the spread of the sampling distribution be larger, smaller, or about the same? Justify your answer. 3. ​Describe the shape of the sampling distribution.

Bias, variability, and shape  ​We can think of the true value of the popu-

High bias, low variability

(a)

lation parameter as the bull’s-eye on a target and of the sample statistic as an arrow fired at the target. Both bias and variability describe what happens when we take many shots at the target. Bias means that our aim is off and we consistently miss the bull’s-eye in the same direction. Our sample values do not center on the population value. High variability means that repeated shots are widely scattered on the target. Repeated samples do not give very similar results. Figure 7.8 shows this target illustration of the two types of error. Notice that low variability (shots are close together) can accompany high bias (shots are consistently away from the bull’s-eye in one direction). And low or no bias Low bias, high variability (shots center on the bull’s-eye) can accompany high variability (shots are widely scattered). Ideally, we’d like our (b) estimates to be accurate (unbiased) and precise (have low variability). See Figure 7.8(d). The following example attempts to tie these ideas together in a familiar setting.

High bias, high variability

The ideal: no bias, low variability

(c)

(d)

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FIGURE 7.8  ​Bias and variability. (a) High bias, low variability. (b) Low bias, high variability. (c) High bias, high variability. (d) The ideal: no bias, low variability.

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Section 7.1 What Is a Sampling Distribution?

Example

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The German Tank Problem Evaluating estimators: Shape, center, spread Refer to the Activity on page 422. Mrs. Friedman’s student teams came up with four different methods for estimating the number of tanks in the bag: (1) “maxmin” = maximum + minimum, (2) “meanpl2sd” = x– + 2sx, (3) “twicemean” = 2x– , and (4) “twomedian” = 2(median). She added one more method, called “partition.” Figure 7.9 shows the results of taking 250 SRSs of 4 tanks and recording the value of the five statistics for each sample. The vertical line marks the actual value of the population parameter N: there were 342 tanks in the bag.

Problem:  Use the information in Figure 7.9 to help answer these questions. (a) Which of the four statistics proposed by the student teams is the best estimator? Justify your answer. (b) Why was the partition method, which uses the statistic (5/4) · maximum, recommended by the mathematicians in Washington, D.C.? Solution: (a) ​Meanpl2sd is a biased estimator: the center of its sampling distribution is too high. This statistic produces consistent over­ estimates of the number of tanks. The other three statistics proposed by the students appear to be unbiased estimators. All three FIGURE 7.9  ​Results from a Fathom simulation of 250 SRSs of sampling distributions have roughly symmetric shapes, so these 4 tanks. The approximate sampling distributions of five ­different statistics are about equally likely to underestimate or overestimate statistics are shown. the number of tanks. Because maxmin has the smallest variability among the three, it would generally produce estimates that are closer to the actual number of tanks. Among the students’ proposed statistics, maxmin would be the best estimator. (b) ​The partition method uses a statistic (5/4 · maximum) that is an unbiased estimator and that has much less variability than any of the student teams’ statistics. Its sampling distribution is leftskewed, so the mean of the distribution is less than its median. Because more than half of the dots in the graph are to the right of the mean, the statistic is more likely to overestimate than underestimate the number of tanks. The mathematicians believed that it would be better to err on the side of caution and give the military commanders an estimate that is slightly too high. For Practice Try Exercise

19

The lesson about center and spread is clear: given a choice of statistics to e­ stimate an unknown parameter, choose one with no or low bias and minimum variability. Shape is a more complicated issue. We have seen sampling distributions that are left-skewed, right-skewed, roughly symmetric, and even approximately Normal. The same statistic can have sampling distributions with different shapes depending on the population distribution and the sample size. Our advice: be sure to consider the shape of the sampling distribution before doing inference.

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Sampling Distributions

Summary • •

•

• •

Section 7.1

A parameter is a number that describes a population. To estimate an unknown parameter, use a statistic calculated from a sample. The population distribution of a variable describes the values of the variable for all individuals in a population. The sampling distribution of a statistic describes the values of the statistic in all possible samples of the same size from the same population. Don’t confuse the sampling distribution with a distribution of sample data, which gives the values of the variable for all individuals in a particular sample. A statistic can be an unbiased estimator or a biased estimator of a parameter. A statistic is a biased estimator if the center (mean) of its sampling distribution is not equal to the true value of the parameter. The variability of a statistic is described by the spread of its sampling distribution. Larger samples give smaller spread. When trying to estimate a parameter, choose a statistic with low or no bias and minimum variability.

Exercises

For Exercises 1 and 2, identify the population, the parameter, the sample, and the statistic in each setting. pg 425

1. Healthy living (a) A random sample of 1000 people who signed a card saying they intended to quit smoking were contacted 9 months later. It turned out that 210 (21%) of the sampled individuals had not smoked over the past 6 months.

For each boldface number in Exercises 3 to 6, (1) state whether it is a parameter or a statistic and (2) use appropriate notation to describe each number; for example, p = 0.65. 3.

(b) Tom is cooking a large turkey breast for a holiday meal. He wants to be sure that the turkey is safe to eat, which requires a minimum internal temperature of 165°F. Tom uses a thermometer to measure the temperature of the turkey meat at four randomly chosen points. The minimum reading in the sample is 170°F.

Get your bearings  A large container is full of ball bearings with mean diameter 2.5003 centimeters (cm). This is within the specifications for acceptance of the container by the purchaser. By chance, an inspector chooses 100 bearings from the container that have mean diameter 2.5009 cm. Because this is outside the specified limits, the container is mistakenly rejected.

4.

2. The economy (a) Each month, the Current Population Survey interviews a random sample of individuals in about 60,000 U.S. households. One of their goals is to estimate the national unemployment rate. In October 2012, 7.9% of those interviewed were unemployed.

Voters  Voter registration records show that 41% of voters in a state are registered as Democrats. To test a random digit dialing device, you use it to call 250 randomly chosen residential telephones in the state. Of the registered voters contacted, 33% are registered Democrats.

5.

Unlisted numbers  A telemarketing firm in a large city uses a device that dials residential telephone numbers in that city at random. Of the first 100 numbers dialed, 48% are unlisted. This is not surprising because 52% of all residential phones in the city are unlisted.

6.

How tall?  A random sample of female college students has a mean height of 64.5 inches, which is greater than the 63-inch mean height of all adult American women.

(b) How much do gasoline prices vary in a large city? To find out, a reporter records the price per gallon of regular unleaded gasoline at a random sample of 10 gas stations in the city on the same day. The range (maximum – minimum) of the prices in the sample is 25 cents.

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Section 7.1 What Is a Sampling Distribution? Exercises 7 and 8 refer to the small population {2, 6, 8, 10, 10, 12} with mean m = 8 and range 10. 7. Sampling distribution (a) List all 15 possible SRSs of size n = 2 from the population. Find the value of x– for each sample. (b) Make a graph of the sampling distribution of x– . Describe what you see. 8. Sampling distribution (a) List all 15 possible SRSs of size n = 2 from the population. Find the value of the range for each sample.

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distribution applies at their high school, an AP® Statistics class takes an SRS of 20 of the 300 16-year-old females at the school and measures their heights. What values of the sample mean x– would be consistent with the population distribution being N(64, 2.5)? To find out, we used Fathom software to simulate choosing 250 SRSs of size n = 20 students from a population that is N(64, 2.5). The figure below is a dotplot of the sample mean height x– of the students in each sample.

(b) Make a graph of the sampling distribution of the sample range. Describe what you see. 9. pg 427

Doing homework  A school newspaper article claims that 60% of the students at a large high school did all their assigned homework last week. Some skeptical AP® Statistics students want to investigate whether this claim is true, so they choose an SRS of 100 students from the school to interview. What values of the sample proportion p^ would be consistent with the claim that the population proportion of students who completed all their homework is p = 0.60? To find out, we used Fathom software to simulate choosing 250 SRSs of size n = 100 students from a population in which p = 0.60. The figure below is a dotplot of the sample proportion p^ of students who did all their homework.

(a) There is one dot on the graph at 62.4. Explain what this value represents. (b) Describe the distribution. Are there any obvious outliers? (c) Would it be surprising to get a sample mean of 64.7 or more in an SRS of size 20 when m = 64? Justify your answer. (d) Suppose that the average height of the 20 girls in the class’s actual sample is x– = 64.7. What would you conclude about the population mean height m for the 16-year-old females at the school? Explain. 11. Doing homework  Refer to Exercise 9. (a) Make a bar graph of the population distribution given that the newspaper’s claim is correct. (b) Sketch a possible graph of the distribution of sample data for the SRS of size 100 taken by the AP® Statistics students.

(a) There is one dot on the graph at 0.73. Explain what this value represents. (b) Describe the distribution. Are there any obvious outliers? (c) Would it be surprising to get a sample proportion of 0.45 or lower in an SRS of size 100 when p = 0.6? Justify your answer. (d) Suppose that 45 of the 100 students in the actual sample say that they did all their homework last week. What would you conclude about the newspaper article’s claim? Explain. 10. Tall girls  According to the National Center for Health Statistics, the distribution of heights for 16-year-old females is modeled well by a Normal density curve with mean m = 64 inches and ­standard deviation s = 2.5 inches. To see if this

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12. Tall girls  Refer to Exercise 10. (a) Make a graph of the population distribution. (b) Sketch a possible dotplot of the distribution of sample data for the SRS of size 20 taken by the AP® Statistics class. Exercises 13 and 14 refer to the following setting. During the winter months, outside temperatures at the Starneses’ cabin in Colorado can stay well below freezing (32°F, or 0°C) for weeks at a time. To prevent the pipes from freezing, Mrs. Starnes sets the thermostat at 50°F. The manufacturer claims that the thermostat allows variation in home temperature that follows a Normal distribution with s = 3°F. To test this claim, Mrs. Starnes programs her digital thermometer to take an SRS of n = 10 r­ eadings during a 24-hour period. Suppose the thermostat is

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­ orking properly and that the actual temperatures in the w cabin vary according to a Normal distribution with mean m = 50°F and standard deviation s = 3°F. 13. Cold cabin?  The Fathom screen shot below shows the results of taking 500 SRSs of 10 temperature readings from a population distribution that is N(50, 3) and recording the sample variance s2x each time.

17. A sample of teens  Refer to Exercise 15. The sample mean x– is an unbiased estimator of the population mean m no matter what size SRS the study chooses. Explain to someone who knows nothing about statistics why a large random sample will give more trustworthy results than a small random sample. 18. Predict the election  Refer to Exercise 16. The sample proportion p^ is an unbiased estimator of the population proportion p no matter what size random sample the polling organization chooses. Explain to someone who knows nothing about statistics why a large random sample will give more trustworthy results than a small random sample. 19. Bias and variability  The figure below shows histograms of four sampling distributions of different statistics intended to estimate the same parameter.

pg 435 Sample variance

(a) Describe the approximate sampling distribution. (b) Suppose that the variance from an actual sample is s2x = 25. What would you conclude about the thermostat manufacturer’s claim? Explain. 14. Cold cabin?  The Fathom screen shot below shows the results of taking 500 SRSs of 10 temperature readings from a population distribution that is N(50, 3) and recording the sample minimum each time.

Population parameter (i)

Population parameter (ii)

Sample minimum

(a) Describe the approximate sampling distribution. (b) Suppose that the minimum of an actual sample is 40°F. What would you conclude about the thermostat manufacturer’s claim? Explain. 15. A sample of teens  A study of the health of teenagers plans to measure the blood cholesterol levels of an SRS of 13- to 16-year-olds. The researchers will report the mean x– from their sample as an estimate of the mean cholesterol level m in this population. Explain to someone who knows little about statistics what it means to say that x– is an unbiased estimator of m. 16. Predict the election  A polling organization plans to ask a random sample of likely voters who they plan to vote for in an upcoming election. The researchers will report the sample proportion p^ that favors the incumbent as an estimate of the population proportion p that favors the incumbent. Explain to someone who knows little about statistics what it means to say that p^ is an unbiased estimator of p.

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Population parameter (iii)

Population parameter (iv)

(a) Which statistics are unbiased estimators? Justify your answer. (b) Which statistic does the best job of estimating the parameter? Explain. 20. IRS audits  The Internal Revenue Service plans to examine an SRS of individual federal income tax returns. The parameter of interest is the proportion of all returns claiming itemized deductions. Which would be better for estimating this parameter: an SRS of 20,000 returns or an SRS of 2000 returns? Justify your answer.

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Section 7.1 What Is a Sampling Distribution? Multiple choice: Select the best answer for Exercises 21 to 24. 21. At a particular college, 78% of all students are receiving some kind of financial aid. The school newspaper selects a random sample of 100 students and 72% of the respondents say they are receiving some sort of financial aid. Which of the following is true? (a) 78% is a population and 72% is a sample. (b) 72% is a population and 78% is a sample. (c) 78% is a parameter and 72% is a statistic. (d) 72% is a parameter and 78% is a statistic. (e) 78% is a parameter and 100 is a statistic. 22. A statistic is an unbiased estimator of a parameter when (a) the statistic is calculated from a random sample. (b) in a single sample, the value of the statistic is equal to the value of the parameter. (c) in many samples, the values of the statistic are very close to the value of the parameter. (d) in many samples, the values of the statistic are centered at the value of the parameter. (e) in many samples, the distribution of the statistic has a shape that is approximately Normal. 23. In a residential neighborhood, the median value of a house is $200,000. For which of the following sample sizes is the sample median most likely to be above $250,000? (a) n = 10

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apparatus measures bone mineral density (BMD). BMD is usually reported in standardized form. The standardization is based on a population of healthy young adults. The World Health Organization (WHO) criterion for osteoporosis is a BMD score that is 2.5 standard deviations below the mean for young adults. BMD measurements in a population of people similar in age and gender roughly follow a Normal distribution. (a) What percent of healthy young adults have osteoporosis by the WHO criterion? (b) Women aged 70 to 79 are, of course, not young adults. The mean BMD in this age group is about −2 on the standard scale for young adults. Suppose that the standard deviation is the same as for young adults. What percent of this older population has osteoporosis? 26. Squirrels and their food supply (3.2) Animal species produce more offspring when their supply of food goes up. Some animals appear able to anticipate unusual food abundance. Red squirrels eat seeds from pinecones, a food source that sometimes has very large crops. Researchers collected data on an index of the abundance of pinecones and the average number of offspring per female over 16 years.3 Computer output from a leastsquares regression on these data and a residual plot are shown below. Predictor Coef SE Coef T P Constant 1.4146 0.2517 5.62 0.000 Cone index 0.4399 0.1016 4.33 0.001 S = 0.600309 R-Sq = 57.2% R-Sq(adj) = 54.2%

(b) n = 50 (c) n = 100 (d) n = 1000 (e) Impossible to determine without more information. 24. Increasing the sample size of an opinion poll will reduce the (a) bias of the estimates made from the data collected in the poll. (b) variability of the estimates made from the data collected in the poll. (c) effect of nonresponse on the poll. (d) variability of opinions in the sample. (e) variability of opinions in the population. 25. Dem bones (2.2)  Osteoporosis is a condition in which the bones become brittle due to loss of minerals. To diagnose osteoporosis, an elaborate

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(a) Give the equation for the least-squares regression line. Define any variables you use. (b) Is a linear model appropriate for these data? Explain. (c) Interpret the values of r2 and s in context.

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7.2 What You Will Learn

Sample Proportions By the end of the section, you should be able to:

• Find the mean and standard deviation of the sampling distribution of a sample proportion p^ . Check the 10% condition before calculating sp^ . • Determine if the sampling distribution of p^ is approximately Normal.

• If appropriate, use a Normal distribution to calculate probabilities involving p^ .

What proportion of U.S. teens know that 1492 was the year in which Columbus “discovered” America? A Gallup Poll found that 210 out of a random sample of 501 American teens aged 13 to 17 knew this historically important date.4 The sample proportion 210 = 0.42 501 is the statistic that we use to gain information about the unknown population proportion p. Because another random sample of 501 teens would likely result in a different estimate, we can only say that “about” 42% of U.S. teenagers know that Columbus discovered America in 1492. In this section, we’ll use sampling distributions to clarify what “about” means. p^ =

The Sampling Distribution of p^ How good is the statistic p^ as an estimate of the parameter p? To find out, we ask, “What would happen if we took many samples?” The sampling distribution of p^ answers this question. How do we determine the shape, center, and spread of the sampling distribution of p^ ? Let’s start with a simulation.

Activity Materials: Computer with Internet ­access—one for the class or one per pair of students

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The Candy Machine Imagine a very large candy machine filled with orange, brown, and yellow candies. When you insert money, the machine dispenses a sample of candies. In this Activity, you will use an applet to investigate the sample-to-sample variability in the proportion of orange candies dispensed by the machine.

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Section 7.2 Sample Proportions

FIGURE 7.10  ​The result of ­taking one SRS of 25 candies from a large candy machine in which 45% of the candies are orange.

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1. ​Launch the Reese’s Pieces® applet at www.rossmanchance. com. Change the population proportion of orange candies to p = 0.45 (the applet calls this value p instead of p). 2. ​Click on the “Draw Samples” button. An animated simple ­random sample of n = 25 candies should be dispensed. Figure 7.10 shows the results of one such sample. Was your sample proportion of orange candies close to the actual population proportion, p = 0.45? Look at the value of p^ in the applet window. 3. ​Click “Draw Samples” 9 more times, so that you have a total of 10 sample results. Look at the dotplot of your p^ -values. What is the mean of your 10 sample proportions? What is their standard deviation? 4. ​To take many more samples quickly, enter 390 in the “number of samples” box. Click on the Animate box to turn the animation off. Then click “Draw Samples.” You have now taken a total of 400 samples of 25 candies from the machine. Describe the shape, center, and spread of the approximate sampling distribution of p^ shown in the dotplot. 5. ​ How would the sampling distribution of the sample proportion p^ change if the machine dispensed n = 50 candies each time instead of 25? “Reset” the ­applet. Take 400 samples of 50 candies. Describe the shape, center, and spread of the approximate sampling distribution. 6. ​How would the sampling distribution of p^ change if the proportion of orange candies in the machine was p = 0.15 instead of p = 0.45? Does your answer depend on whether n = 25 or n = 50? Use the applet to investigate these ­questions. Then write a brief summary of what you learned. 7.  For what combinations of n and p is the sampling distribution of p^ approximately Normal? Use the applet to investigate. Figure 7.11 shows one set of possible results from Step 4 of “The Candy Machine” Activity. Let’s describe what we see. Shape:  Roughly symmetric and somewhat bell-shaped. It looks as though a Normal curve would approximate this distribution fairly well. Center:  The mean of the 400 sample proportions is 0.449. This is quite close to the actual population proportion, p = 0.45. Spread:  The standard deviation of the 400 values of p^ from these samples is 0.105. The dotplot in Figure 7.11 is the approximate sampling distribution of p^ . If we took all possible SRSs of n = 25 candies from the machine and graphed the value of p^ for each sample, then we’d have the sampling distribution of p^ . We can get an idea of its shape, center, and spread from Figure 7.11. FIGURE 7.11  ​The result of taking 400 SRSs of 25 candies from a large candy machine in which 45% of the candies are orange. The dotplot shows the ­approximate sampling distribution of p^ .

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Sampling Candies Effect of n and p on shape, center, and spread In a similar way, we can explore the sampling distribution of p^ when n = 50 (Step 5 of the Activity). As Figure 7.12 shows, the dotplot is once again roughly symmetric and somewhat bell-shaped. This graph is also centered at about 0.45. With samples of size 50, however, there is less spread in the values of p^ . The standard deviation in Figure 7.12 is 0.070. For the samples of size 25 in Figure 7.11, it is 0.105. To repeat what we said earlier, larger samples give the sampling distribution a smaller spread.

FIGURE 7.12  ​The approximate sampling distribution of p^ for 400 SRSs of 50 candies from a population in which p = 0.45 of the candies are orange.

What if the actual proportion of orange candies in the ­machine were p = 0.15? Figure 7.13(a) shows the a­ pproximate sampling ­ otice that the dotplot is slightly distribution of p^ when n = 25. N right-skewed. The graph is centered close to the population parameter, p = 0.15. As for the spread, it’s similar to the standard deviation in Figure 7.12, where n = 50 and p = 0.45. If we increase the sample size to n = 50, the sampling distribution of p^ should show less variability. The standard deviation in Figure 7.13(b) confirms this. Note that we can’t just visually compare the graphs because the horizontal scales are different. The dotplot is more symmetrical than the graph in Figure 7.13(a) and is once again centered at a value that is close to p = 0.15.

FIGURE 7.13  ​The result of taking 400 SRSs of (a) size n = 25 and (b) size n = 50 candies from a large candy machine in which 15% of the candies are orange. The dotplots show the approximate sampling distribution of p^ in each case.

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What have we learned so far about the sampling distribution of p^ ? Shape:  In some cases, the sampling distribution of p^ can be approximated by a Normal curve. This seems to depend on both the sample size n and the population proportion p. Center:  The mean of the distribution is mp^ = p. This makes sense because the sample proportion p^ is an unbiased estimator of p. Spread:  For a specific value of p, the standard deviation sp^ gets smaller as n gets larger. The value of sp^ depends on both n and p. To sort out the details of shape and spread, we need to make an important connection between the sample proportion p^ and the number of “successes” X in the sample. In the candy machine example, we started by taking repeated SRSs of n = 25 candies from a population with proportion p = 0.45 of orange candies. For any such sample, we can think of each candy that comes out of the machine as a trial of this chance process. A “success” occurs when we get an orange candy. Let X = the number of orange candies obtained. As long as the number of candies in the machine is very large, X will have close to a binomial distribution with n = 25 and p = 0.45. (Refer to the 10% condition on page 401.) The sample proportion of successes is closely related to X: p^ =

THINK ABOUT IT

count of successes in sample X = n size of sample

How is the sampling distribution of p^ related to the binomial count X ?  ​From Chapter 6, we know that the mean and standard deviation of a binomial random variable X are

mX = np  and  sX = "np(1 − p)

Because p^ = X/n = (1/n)X, we’re just multiplying the random variable X by a constant (1/n) to get the random variable p^ . Recall from Chapter 6 that multiplying by a constant multiplies both the mean and the standard deviation of the new random variable by that constant. We have 1 mp^ = (np) = p (confirming that p^ is an unbiased estimator of p) n



np(1 − p) p(1 − p) 1 sp^ = !np(1 − p) = = 2 n n Å Å n

(as sample size increases, spread decreases)

That takes care of center and spread. What about shape? Multiplying a random variable by a constant doesn’t change the shape of the probability distribution. So the sampling distribution of p^ will have the same shape as the distribution of the binomial random variable X. If you studied the optional material in Chapter 6 about the Normal approximation to a binomial distribution, then you already know the punch line. Whenever np and n(1 - p) are at least 10, a Normal distribution can be used to approximate the sampling distribution of p^ .

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Here’s a summary of the important facts about the sampling distribution of p^ .

Sampling Distribution of a Sample Proportion Choose an SRS of size n from a population of size N with proportion p of successes. Let p^ be the sample proportion of successes. Then: • The mean of the sampling distribution of p^ is mp^ = p. • The standard deviation of the sampling distribution of p^ is sp^ =

p(1 − p) n Å

1 N. 10 • As n increases, the sampling distribution of p^ becomes approximately ­Normal. Before you perform Normal calculations, check that the L ­ arge Counts condition is satisfied: np ≥ 10 and n(1 − p) ≥ 10. as long as the 10% condition is satisfied: n ≤

Figure 7.14 displays the facts in a form that helps you recall the big idea of a sampling distribution. The mean of the sampling distribution of p^ is the true value of the population proportion p. The standard deviation of p^ gets smaller as the sample size n increases. In fact, because the sample size n is under the square root sign, we’d have to take a sample four times as large to cut the standard deviation in half.

SRS size n SRS size n SRS size n

p^ Standard deviation p(1– p) n

p^ p^

Mean p

Population proportion p of successes

Values of p^

FIGURE 7.14  ​Select a large SRS from a population in which proportion p are successes. The sampling distribution of the proportion p^ of successes in the sample is approximately Normal. The mean is p and the standard deviation is !p(1 − p)∙n.

The two conditions in the preceding box are very important. (1) Large Counts condition: If we assume that the sampling distribution of p^ is approximately Normal when it isn’t, any calculations we make using a Normal distribution will be flawed. (2) 10% condition: When we’re sampling without replacement from a (finite) population, the observations are not independent, because knowing the outcome of one trial helps us predict the outcome of future trials. But the standard deviation formula assumes that the observations are independent. If we sample too large a fraction of the population, our calculated value of sp^ will be inaccurate.

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Because larger random samples give better information, it sometimes makes sense to sample more than 10% of a population. In such a case, there’s a more accurate formula for calculating the standard deviation sp^ . It uses something called a finite population correction (FPC). We’ll avoid situations that require the FPC in this text.

Check Your Understanding

About 75% of young adult Internet users (ages 18 to 29) watch online videos. Suppose that a sample survey contacts an SRS of 1000 young adult Internet users and calculates the proportion p^ in this sample who watch online videos. 1. What is the mean of the sampling distribution of p^ ? Explain. 2. Find the standard deviation of the sampling distribution of p^ . Check that the 10% condition is met. 3. Is the sampling distribution of p^ approximately Normal? Check that the Large Counts condition is met. 4. If the sample size were 9000 rather than 1000, how would this change the sampling distribution of p^ ?

Using the Normal Approximation for p^ Inference about a population proportion p is based on the sampling distribution of p^ . When the sample size is large enough for np and n(1 − p) to both be at least 10 (the Large Counts condition), the sampling distribution of p^ is approximately Normal. In that case, we can use a Normal distribution to calculate the probability of obtaining an SRS in which p^ lies in a specified interval of values. Here is an example.

Example

Going to College Normal calculations involving p^ A polling organization asks an SRS of 1500 first-year college students how far away their home is. Suppose that 35% of all first-year students attend college within 50 miles of home.

PROBLEM:  Find the probability that the random sample of 1500 students will give a result

within 2 percentage points of this true value. Show your work.

SOLUTION: Step 1: State the distribution and the values of interest.  We want to find the probability that p^ falls between 0.33 and 0.37 (within 2 percentage points, or 0.02, of 0.35). In symbols, that’s P(0.33 ≤ p^ ≤ 0.37). We have an SRS of size n = 1500 drawn from a population in which the proportion p = 0.35 attend college within 50 miles of home. What do we know about the sampling ­distribution of p^? •  Its mean is mp^ = p = 0.35.

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•  What about the standard deviation? We need to check the 10% condition. To use the standard deviation formula we derived, the population must contain at least 10(1500) = 15,000 people. There are over 1.7 million first-year college students, so

N(0.35,0.0123) pˆ = 0.0123

sp^ =

0.3

0.34

0.32 0.33

0.38 pˆ = 0.35 0.36 Value of pˆ 0.37

0.40

FIGURE 7.15  ​The Normal approximation to the sampling distribution of p^ .

•  Can we use a Normal distribution to approximate the sampling distribution of p^? Check the Large Counts condition: np = 1500(0.35) = 525 and n(1 − p) = 1500(0.65) = 975. Both are much larger than 10, so the Normal approximation will be quite accurate. Figure 7.15 shows the Normal distribution that we’ll use with the area of interest shaded and the mean, standard deviation, and boundary values labeled. Step 2:  Perform calculations—show your work!  The standardized scores for the two boundary values are 0.33 − 0.35 0.37 − 0.35 = −1.63 and z = = 1.63 0.0123 0.0123 Figure 7.16 shows the area under the standard Normal curve corresponding to these standardized values. Using Table A, the desired probability is z=

Standard Normal curve Probability = 0.0516

P(0.33 ≤ p^ ≤ 0.37) = P(−1.63 ≤ Z ≤ 1.63)

Probability = 0.8968

z = –1.63

p(1 − p) (0.35)(0.65) = = 0.0123 n Å Å 1500

z = 1.63

Probability = 0.9484

FIGURE 7.16  ​Probabilities as areas under the standard Normal curve.

= 0.9484 − 0.0516 = 0.8968

Using technology: The command normalcdf(lower:0.33, upper:0.37, m:0.35, s:0.0123) gives an area of 0.8961. Step 3:  Answer the question.  About 90% of all SRSs of size 1500 will give a result within 2 percentage points of the truth about the population. For Practice Try Exercise

Section 7.2

Summary •

•

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39

When we want information about the population proportion p of ­successes, we often take an SRS and use the sample proportion p^ to estimate the unknown parameter p. The sampling distribution of p^ describes how the sample proportion varies in all possible samples from the population. The mean of the sampling distribution of p^ is equal to the population proportion p. That is, p^ is an unbiased estimator of p.

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Section 7.2 Sample Proportions •

•

Section 7.2

The standard deviation of the sampling distribution of p^ is !p(1 − p)∙n for an SRS of size n. This formula can be used if the population is at least 10 times as large as the sample (the 10% condition). The standard deviation of p^ gets smaller as the sample size n gets larger. Because of the square root, a sample four times larger is needed to cut the standard deviation in half. When the sample size n is large, the sampling distribution of p^ is close to a Normal distribution with mean p and standard deviation !p(1 − p)∙n. In practice, use this Normal approximation when both np ≥ 10 and n(1 − p) ≥ 10 (the Large Counts condition).

Exercises

27. The candy machine  Suppose a large candy machine has 45% orange candies. Use Figures 7.11 and 7.12 (pages 441 and 442) to help answer the following questions. (a) Would you be surprised if a sample of 25 candies from the machine contained 8 orange candies (that’s 32% orange)? How about 5 orange candies (20% orange)? Explain. (b) Which is more surprising: getting a sample of 25 candies in which 32% are orange or getting a sample of 50 candies in which 32% are orange? Explain. 28. The candy machine  Suppose a large candy machine has 15% orange candies. Use Figure 7.13 (page 442) to help answer the following questions. (a) Would you be surprised if a sample of 25 candies from the machine contained 8 orange candies (that’s 32% orange)? How about 5 orange candies (20% orange)? Explain. (b) Which is more surprising: getting a sample of 25 candies in which 32% are orange or getting a sample of 50 candies in which 32% are orange? Explain. 29. The candy machine  Suppose a large candy machine has 45% orange candies. Imagine taking an SRS of 25 candies from the machine and observing the sample proportion p^ of orange candies. (a) What is the mean of the sampling distribution of p^ ? Why? (b) Find the standard deviation of the sampling distribution of p^ . Check to see if the 10% condition is met.

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(c) Is the sampling distribution of p^ approximately Normal? Check to see if the Large Counts ­condition is met. (d) If the sample size were 100 rather than 25, how would this change the sampling distribution of p^ ? 30. The candy machine  Suppose a large candy ­machine has 15% orange candies. Imagine taking an SRS of 25 candies from the machine and observing the sample proportion p^ of orange candies. (a) What is the mean of the sampling distribution of p^ ? Why? (b) Find the standard deviation of the sampling distribution of p^ . Check to see if the 10% condition is met. (c) Is the sampling distribution of p^ approximately ­Normal? Check to see if the Large Counts condition is met. (d) If the sample size were 225 rather than 25, how would this change the sampling distribution of p^ ? 31. Airport security  The Transportation Security Administration (TSA) is responsible for airport safety. On some flights, TSA officers randomly select passengers for an extra security check before boarding. One such flight had 76 passengers—12 in first class and 64 in coach class. TSA officers selected an SRS of 10 passengers for screening. Let p^ be the proportion of first-class passengers in the sample. (a) Is the 10% condition met in this case? Justify your answer. (b) Is the Large Counts condition met in this case? Justify your answer.

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32. Scrabble  In the game of Scrabble, each player begins by drawing 7 tiles from a bag containing 100 tiles. There are 42 vowels, 56 consonants, and 2 blank tiles in the bag. Cait chooses an SRS of 7 tiles. Let p^ be the proportion of vowels in her sample. (a) Is the 10% condition met in this case? Justify your answer. (b) Is the Large Counts condition met in this case? Justify your answer.

­attended church during the past week. Let p^ be the proportion of people in the sample who attended church. A newspaper report claims that 40% of all U.S. adults went to church last week. Suppose this claim is true. (a) What is the mean of the sampling distribution of p^ ? Why? (b) Find the standard deviation of the sampling ­distribution of p^ . Check to see if the 10% condition is met.

In Exercises 33 and 34, explain why you cannot use the methods of this section to find the desired probability.

(c) Is the sampling distribution of p^ approximately Normal? Check to see if the Large Counts ­condition is met.

33. Hispanic workers  A factory employs 3000 unionized workers, of whom 30% are Hispanic. The 15-member union executive committee contains 3 Hispanics. What would be the probability of 3 or fewer Hispanics if the executive committee were chosen at random from all the workers?

(d) Of the poll respondents, 44% said they did attend church last week. Find the probability of obtaining a sample of 1785 adults in which 44% or more say they attended church last week if the newspaper report’s claim is true. Does this poll give convincing evidence against the claim? Explain.

34. Studious athletes  A university is concerned about the academic standing of its intercollegiate athletes. A study committee chooses an SRS of 50 of the 316 athletes to interview in detail. Suppose that 40% of the athletes have been told by coaches to neglect their studies on at least one occasion. What is the probability that at least 15 in the sample are among this group?

37. Do you drink the cereal milk?  What sample size would be required to reduce the standard deviation of the sampling distribution to one-half the value you found in Exercise 35(b)? Justify your answer.

35. Do you drink the cereal milk? A USA Today Poll asked a random sample of 1012 U.S. adults what they do with the milk in the bowl after they have eaten the cereal. Let p^ be the proportion of people in the sample who drink the cereal milk. A spokesman for the dairy industry claims that 70% of all U.S. adults drink the cereal milk. Suppose this claim is true. (a) What is the mean of the sampling distribution of p^ ? Why? (b) Find the standard deviation of the sampling ­distribution of p^ . Check to see if the 10% ­condition is met. (c) Is the sampling distribution of p^ approximately Normal? Check to see if the Large Counts condition is met. (d) Of the poll respondents, 67% said that they drink the cereal milk. Find the probability of obtaining a sample of 1012 adults in which 67% or fewer say they drink the cereal milk if the milk industry ­spokesman’s claim is true. Does this poll give convincing evidence against the claim? Explain. 36. Do you go to church?  The Gallup Poll asked a random sample of 1785 adults whether they

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38. Do you go to church?  What sample size would be required to reduce the standard deviation of the sampling distribution to one-third the value you found in Exercise 36(b)? Justify your answer. 39. Students on diets  A sample survey interviews an SRS of 267 college women. Suppose that 70% of college women have been on a diet within the past 12 months. What is the probability that 75% or more of the women in the sample have been on a diet? Show your work.

pg 445

40. Who owns a Harley?  Harley-Davidson motorcycles make up 14% of all the motorcycles registered in the United States. You plan to interview an SRS of 500 motorcycle owners. How likely is your sample to contain 20% or more who own Harleys? Show your work. 41. On-time shipping  A mail-order company advertises that it ships 90% of its orders within three working days. You select an SRS of 100 of the 5000 orders received in the past week for an audit. The audit reveals that 86 of these orders were shipped on time. (a) If the company really ships 90% of its orders on time, what is the probability that the proportion in an SRS of 100 orders is 0.86 or less? Show your work. (b) A critic says, “Aha! You claim 90%, but in your sample the on-time percentage is lower than that.

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So the 90% claim is wrong.” Explain in simple language why your probability calculation in (a) shows that the result of the sample does not refute the 90% claim.

(b) np = 225 and n(1 − p) = 525 are both at least 10.

42. Underage drinking  The Harvard College Alcohol Study finds that 67% of college students support efforts to “crack down on underage drinking.” Does this result hold at a large local college? To find out, college administrators survey an SRS of 100 students and find that 62 support a crackdown on underage drinking.

(e) the sampling distribution of p^ always has this shape.

(a) Suppose that the proportion of all students attending this college who support a crackdown is 67%, the same as the national proportion. What is the probability that the proportion in an SRS of 100 students is 0.62 or less? Show your work. (b) A writer in the college’s student paper says that “support for a crackdown is lower at our school than nationally.” Write a short letter to the editor explaining why the survey does not support this conclusion.

(c) a random sample was chosen. (d) the athletes’ responses are quantitative. 46. In a congressional district, 55% of the registered voters are Democrats. Which of the following is equivalent to the probability of getting less than 50% Democrats in a random sample of size 100? (a) PaZ < (b) P Z < £

(c) P Z < £

(d) PaZ < Multiple choice: Select the best answer for Exercises 43 to 46. Exercises 43 to 45 refer to the following setting. The magazine Sports Illustrated asked a random sample of 750 Division I college athletes, “Do you believe performanceenhancing drugs are a problem in college sports?” Suppose that 30% of all Division I athletes think that these drugs are a problem. Let p^ be the sample proportion who say that these drugs are a problem. 43. Which of the following are the mean and standard deviation of the sampling distribution of the sample proportion p^ ? (a) Mean = 0.30, SD = 0.017 (b) Mean = 0.30, SD = 0.55 (c) Mean = 0.30, SD = 0.0003 (d) Mean = 225, SD = 12.5 (e) Mean = 225, SD = 157.5 44. Decreasing the sample size from 750 to 375 would multiply the standard deviation by (a) 2.

(c) 1/2.

(b) !2.

(d) 1∙!2.

(e) ​none of these.

45. The sampling distribution of p^ is approximately ­Normal because (a) there are at least 7500 Division I college athletes.

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(e) PaZ <

0.50 − 0.55 b 100 0.50 − 0.55

0.55(0.45) ≥ Å 100 0.55 − 0.50

0.55(0.45) ≥ Å 100

0.50 − 0.55 b !100(0.55)(0.45) 0.55 − 0.50 b !100(0.55)(0.45)

47. Sharing music online (5.2)  A sample survey reports that 29% of Internet users download music files ­online, 21% share music files from their computers, and 12% both download and share music.5 Make a Venn diagram that displays this information. What percent of Internet users neither download nor share music files? 48. California’s endangered animals (4.1)  The California Department of Fish and Game publishes a list of the state’s endangered animals. The reptiles on the list are given below. Desert tortoise Olive Ridley sea turtle Island night lizard Flat-tailed horned lizard Green sea turtle Leatherback sea turtle Alameda whip snake

Southern rubber boa Loggerhead sea turtle Barefoot banded gecko Coachella Valley fringe-toed lizard Blunt-nosed leopard lizard Giant garter snake San Francisco garter snake

(a) Describe how you would use Table D at line 111 to choose an SRS of 3 of these reptiles to study. (b) Use your method from part (a) to select your sample. Identify the reptiles you chose.

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7.3 What You Will Learn

Sample Means By the end of the section, you should be able to:

• Find the mean and standard deviation of the sampling distribution of a sample mean x– . Check the 10% condition before calculating sx–. • Explain how the shape of the sampling distribution of x– is affected by the shape of the population distribution and the sample size.

• If appropriate, use a Normal distribution to calculate probabilities involving x– .

Sample proportions arise most often when we are interested in categorical variables. We then ask questions like “What proportion of U.S. adults have watched Survivor?” or “What percent of the adult population attended church last week?” But when we record quantitative variables—household i­ncome, lifetime of car brake pads, blood pressure—we are interested in other statistics, such as the median or mean or standard deviation of the variable. The sample mean x– is the most common statistic computed from quantitative data. This section describes the sampling distribution of the sample mean. The following Activity and the subsequent example give you a sense of what lies ahead.

Activity Materials: Large container with several hundred pennies

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Penny for Your Thoughts Your teacher will assemble a large population of pennies of various ages.6 In this Activity, your class will investigate the sampling distribution of the mean year x– in a sample of pennies for SRSs of several different sizes. Then, you will compare these distributions of the mean year with the population distribution. 1.  Your teacher will provide a dotplot of the population distribution of penny years. 2. ​ Have each member of the class take an SRS of 5 pennies from the population and record the year on each penny. Be sure to replace these coins in the container before the next student takes a sample. If your class has fewer than 25 students, have each person take two samples. 3. ​ Calculate the mean year x– of the 5 pennies in your sample. 4. ​Make a class dotplot of the sample mean years for SRSs of size 5 ­using the same scale as you did for the population distribution. Use x– ’s instead of dots when making the graph. 5. ​Repeat the process in Steps 2 to 4 for samples of size 25. Use the same scale for your dotplot and place it beside the graph for samples of size 5. 6. ​Compare the population distribution with the two approximate sampling distributions of x– . What do you notice about shape, center, and spread as the sample size increases?

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Section 7.3 Sample Means

Example

451

Making Money

A first look at the sampling distribution of x– Figure 7.17(a) is a histogram of the earnings of a population of 61,742 households that had earned income greater than zero in a recent year.7 As we expect, the distribution of earned incomes is strongly skewed to the right and very spread out. The right tail of the distribution is even longer than the histogram shows because there are too few high incomes for their bars to be visible on this scale. We cut off the earnings scale at $400,000 to save space. The mean earnings for these 61,742 households was m = $69,750.

30

30

25

25

Percent of samples

Percent of households

Take an SRS of 100 households. The mean earnings in this sample is x– = $66,807. That’s less than the mean of the population. Take another SRS of size 100. The mean for this sample is x– = $70,820. That’s higher than the mean of the population. What would happen if we did this many times? Figure 7.17(b) is a histogram of the mean earnings for 500 samples, each of size n = 100. The scales in Figures 7.17(a) and 7.17(b) are the same, for easy comparison. Although the distribution of individual earnings is skewed and very spread out, the distribution of sample means is roughly symmetric and much less spread out. Both distributions are centered at m = $69,750.

20 15 10 5

Because both histograms use the same scales, you can directly compare this graph with the one to the left.

20 15 10 5

0

0 0

(a)

200 300 400 µ 100 Household earnings (thousands of dollars)

0

(b)

200 300 400 µ 100 Mean household earnings for samples of size 100 (thousands of dollars)

FIGURE 7.17  ​(a) The distribution of earned income in a population of 61,472 households. (b) The distribution of the mean earnings x– for 500 SRSs of n = 100 households from this population.

This example illustrates an important fact that we will make precise in this ­section: averages are less variable than individual observations.

The Sampling Distribution of x–: Mean and Standard Deviation Figure 7.17 suggests that when we choose many SRSs from a population, the sampling distribution of the sample mean is centered at the population mean m and is less spread out than the population distribution. Here are the facts.

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Mean and Standard Deviation of the Sampling Distribution of x– Suppose that x– is the mean of an SRS of size n drawn from a large population with mean m and standard deviation s. Then: • The mean of the sampling distribution of x– is mx– = m. • The standard deviation of the sampling distribution of x– is s sx– = !n as long as the 10% condition is satisfied: n ≤ 101 N.

AP® EXAM TIP ​Notation matters. The symbols p^ , x– , p, m, s, mp^ , sp^ , m x– , and s x– all have specific and different meanings. Either use notation correctly—or don’t use it at all. You can expect to lose credit if you use incorrect notation.

Example

The behavior of x– in repeated samples is much like that of the sample proportion p^ : • The sample mean x– is an unbiased estimator of the population mean m. • The values of x– are less spread out for larger samples. Their standard deviation decreases at the rate !n, so you must take a sample four times as large to cut the standard deviation of the distribution of x– in half.

• You should use the formula s∙!n for the standard deviation of x– only when the population is at least 10 times as large as the sample (the 10% condition). Notice that these facts about the mean and standard deviation of x– are true no matter what shape the population distribution has.

This Wine Stinks

Mean and standard deviation of x– Sulfur compounds such as dimethyl sulfide (DMS) are sometimes present in wine. DMS causes “off-odors” in wine, so winemakers want to know the odor threshold, the lowest concentration of DMS that the human nose can detect. Extensive studies have found that the DMS odor threshold of adults follows a distribution with mean m = 25 micrograms per liter and standard deviation s = 7 micrograms per liter. Suppose we take an SRS of 10 adults and determine the mean odor threshold x– for the individuals in the sample.

Problem:

(a) What is the mean of the sampling distribution of x– ? Explain. (b) What is the standard deviation of the sampling distribution of x– ? Check that the 10% condition is met.

Solution:

(a) Because x– is an unbiased estimator of m, mx– = m = 25 micrograms per liter. s 7 (b) The standard deviation is sx– = = = 2.214 because there are at least 10(10) = 100 !n !10 adults in the population. For Practice Try Exercise

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THINK ABOUT IT

453

Can we confirm the formulas for the mean and standard ­deviation of x– ?  ​Choose an SRS of size n from a population, and measure a

variable X on each individual in the sample. Call the individual measurements X1, X2, . . . , Xn. If the population is large relative to the sample, we can think of these Xi’s as independent random variables, each with mean m and standard deviation s. Because 1 x– = (X1 + X2 + c+ Xn) n we can use the rules for random variables from Chapter 6 to find the mean and 1 standard deviation of x– . If we let T = X1 + X2 + . . . + Xn, then x– = T. n Using the addition rules for means and variances, we get mT = mX + mX + c+ mX = m + m + c+ m = nm 1

2

n

s2T = s2X1 + s2X2 + c+ s2Xn = s2 + s2 + c+ s2 = ns2 1 sT = "ns2 = s!n

Because x– is just a constant multiple of the random variable T, 1 1 mx– = mT = (nm) = m n n 1 1 1 s n 1 sx– = sT = (s!n) = s =s =s = n n Å n2 Ån !n !n

Sampling from a Normal Population We have described the mean and standard deviation of the sampling distribution of a sample mean x– but not its shape. That’s because the shape of the distribution of x– depends on the shape of the population distribution. In one important case, there is a simple relationship between the two distributions. The following Activity shows what we mean.

Activity Materials: Computer with Internet ­access—one for the class or one per pair of students

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Exploring the Sampling Distribution of x– for a Normal Population Professor David Lane of Rice University has developed a wonderful applet for ­investigating the sampling distribution of x– . It’s dynamic, and it’s fun to play with. In this Activity, you’ll use Professor Lane’s applet to explore the shape of the sampling distribution when the population is Normally distributed. 1. ​ Search for “online statbook sampling distributions applet” and go to the Web site. When the BEGIN button appears on the left side of the screen, click on it. You will then see a yellow page entitled “Sampling Distributions” like the one in the following figure.

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2. ​There are choices for the population distribution: Normal, uniform, skewed, and custom. The default is Normal. Click the “Animated” button. What happens? Click the button several more times. What do the black boxes represent? What is the blue square that drops down onto the plot below? What does the red horizontal band under the population histogram tell us? Look at the left panel. Important numbers are displayed there. Did you notice that the colors of the numbers match up with the objects to the right? As you make things happen, the numbers change accordingly, like an automatic scorekeeper. 3. ​Click on “Clear lower 3” to start clean. Then click on the “1,000” button under “Sample:” repeatedly until you have simulated taking 10,000 SRSs of size n = 5 from the population (look for “Reps = 10000” on the left panel in black letters). Answer these questions: •  Does the approximate sampling distribution (blue bars) have a recognizable shape? Click the box next to “Fit normal.” •  Compare the mean of the approximate sampling distribution with the mean of the population. •  How is the standard deviation of the approximate sampling distribution related to the standard deviation of the population? 4. ​Click “Clear lower 3.” Use the drop-down menus to set up the bottom graph to display the mean for samples of size n = 20. Then sample 10,000 times. How do the two distributions of x– compare: shape, center, and spread? 5. ​What have you learned about the shape of the sampling distribution of x– when the population has a Normal shape?

As the previous Activity demonstrates, if the population distribution is Normal, then so is the sampling distribution of x– . This is true no matter what the sample size is.

Sampling Distribution of a Sample Mean from a Normal Population Suppose that a population is Normally distributed with mean m and standard deviation s. Then the sampling distribution of x– has the Normal distribution with mean m and standard deviation (provided the 10% condition is met) s∙!n.

We already knew the mean and standard deviation of the sampling distribution. All we have added is the Normal shape. Now we have enough information to calculate probabilities involving x– when the population distribution is Normal.

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Example

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Young Women’s Heights Finding probabilities involving the sample mean Problem:  The height of young women follows a Normal distribution with mean m = 64.5 inches

and standard deviation s = 2.5 inches. (a) Find the probability that a randomly selected young woman is taller than 66.5 inches. Show your work. (b) Find the probability that the mean height of an SRS of 10 young women exceeds 66.5 inches. Show your work.

Solution: (a) Step 1: State the distribution and the values of interest. Let X be the height of a randomly selected young woman. The random variable X follows a Normal distribution with m = 64.5 inches and s = 2.5 inches. We want to find P(X > 66.5). Figure 7.18 shows the distribution (purple curve) with the area of interest shaded and the mean, standard deviation, and boundary value labeled. Step 2: Perform calculations—show your work! The standardized score for the boundary 66.5 − 64.5 = 0.80. Using Table A, P(X > 66.5) = P(Z > 0.80) = 1 − 0.7881 = value is z = 2.5 0.2119. Using technology: The command normalcdf(lower:66.5, upper:10000, m:64.5, s:2.5) gives an area of 0.2119. Step 3: Answer the question. The probability of choosing Sampling a young woman at random whose height exceeds 66.5 inches is distribution N(64.5,0.79) about 0.21. of x (b) Step 1: State the distribution and the values of Population interest. For an SRS of 10 young women, the sampling distribudistribution tion of their sample mean height x– will have mean m x– = m = 64.5 N(64.5,2.5) inches. The 10% condition is met because there are at least 10(10) = 100 young women in the population. So the standard deviation is s 2.5 s x– = = = 0.79. Because the population distribution µ = 64.5 in. 66.5 in. !n !10 is Normal, the values of x– will follow an N(64.5, 0.79) distribution. We FIGURE 7.18  ​The sampling distribution of the mean height x– want to find P ( x– > 66.5) inches. Figure 7.18 shows the distribution for SRSs of 10 young women compared with the population distribution of young women’s heights. (blue curve) with the area of interest shaded and the mean, standard deviation, and boundary value labeled. Figure 7.18 compares the population Step 2: Perform calculations—show your work! The standardized score for the boundary distribution and the sampling – value is distribution of x . It also shows 66.5 − 64.5 = 2.53. z= the areas corresponding to the 0.79 probabilities that we computed. You can see that it is much less likely for the average height of 10 randomly selected young women to exceed 66.5 inches than it is for the height of one randomly selected young woman to exceed 66.5 inches.

Using Table A. P ( x– > 66.5) = P(Z > 2.53) = 1 − 0.9943 = 0.0057. Using technology: The command normalcdf(lower:66.5, upper:10000, m:64.5, s:0.79) gives an area of 0.0057. Step 3: Answer the question. It is very unlikely (less than a 1% chance) that we would choose an SRS of 10 young women whose average height exceeds 66.5 inches. For Practice Try Exercise

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The fact that averages of several observations are less variable than individual observations is important in many settings. For example, it is common practice to repeat a measurement several times and report the average of the results. Think of the results of n repeated measurements as an SRS from the population of outcomes we would get if we repeated the measurement forever. The average of the n results (the sample mean x– ) is less variable than a single measurement.

Check Your Understanding

The length of human pregnancies from conception to birth varies according to a distribution that is approximately Normal with mean 266 days and standard deviation 16 days. 1. Find the probability that a randomly chosen pregnant woman has a pregnancy that lasts for more than 270 days. Show your work. Suppose we choose an SRS of 6 pregnant women. Let x– = the mean pregnancy length for the sample. 2. What is the mean of the sampling distribution of x– ? Explain.

3. Compute the standard deviation of the sampling distribution of x– . Check that the 10% condition is met. 4. Find the probability that the mean pregnancy length for the women in the sample exceeds 270 days. Show your work.

The Central Limit Theorem Most population distributions are not Normal. The household incomes in ­Figure 7.17(a) on page 451, for example, are strongly skewed. Yet Figure 7.17(b) suggests that the distribution of means for samples of size 100 is approximately Normal. What is the shape of the sampling distribution of x– when the population distribution isn’t Normal? The following Activity sheds some light on this question.

Activity Materials: Computer with Internet ­access—one for the class or one per pair of students

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Exploring the Sampling Distribution of x– for a Non-Normal Population Let’s use the sampling distributions applet from the previous A ­ ctivity (page 453) to investigate what happens when we start with a non-Normal population distribution. 1. ​Go to the Web site and launch the applet. Select “Skewed” population. Set the bottom two graphs to display the mean—one for samples of size 2 and the other for samples of size 5. Click the Animated button a few times to be sure you see what’s happening. Then “Clear lower 3” and take 10,000 SRSs. Describe what you see. 2. ​Change the sample sizes to n = 10 and n =16 and repeat Step 1. What do you notice?

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3. ​Now change the sample sizes to n = 20 and n = 25 and take 10,000 more samples. Did this confirm what you saw in Step 2? 4. ​Clear the page, and select “Custom” distribution. Click on a point on the population graph to insert a bar of that height. Or click on a point on the horizontal axis, and drag up to define a bar. Make a distribution that looks as strange as you can. (Note: You can shorten a bar or get rid of it completely by clicking on the top of the bar and dragging down to the axis.) Then repeat Steps 1 to 3 for your custom distribution. Cool, huh? 5. ​Summarize what you learned about the shape of the sampling distribution of x– .

It is a remarkable fact that as the sample size increases, the sampling distribution of x– changes shape: it looks less like that of the population and more like a Normal distribution. When the sample size is large enough, the sampling distribution of x– is very close to Normal. This is true no matter what shape the population distribution has, as long as the population has a finite standard deviation s. This famous fact of probability theory is called the central limit theorem (sometimes abbreviated as CLT).

Definition: Central limit theorem (CLT) Draw an SRS of size n from any population with mean m and finite standard deviation s. The central limit theorem (CLT) says that when n is large, the sampling distribution of the sample mean x– is approximately Normal.

c

Example

!

n

How large a sample size n is needed for the sampling distribution of x– to be close to Normal depends on the population distribution. More observations are required if the shape of the population distribution is far from Normal. In that case, the sampling distribution of x– will also be very non-Normal if the utio sample size is small. Be sure you understand what the CLT does—and a doesn’t—say.

A Strange Population Distribution The CLT in action We used the sampling distribution applet to create a population distribution with a very strange shape. See the graph at the top of the next page.

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Figure 7.19 below shows the approximate sampling distribution of the sample mean x– for SRSs of size (a) n = 2, (b) n = 5, (c) n = 10, and (d) n = 25. As n increases, the shape becomes more Normal. For SRSs of size 2, the sampling distribution is very non-Normal. The distribution of x– for 10 observations is slightly skewed to the right but already resembles a Normal curve. By n = 25, the sampling distribution is even more Normal. The contrast between the shapes of the population distribution and the distribution of the mean when n = 10 or 25 is striking.

(a)

(b)

(c)

(d) FIGURE 7.19  ​The central limit theorem in action: the distribution of sample means x– from a strongly non-Normal population becomes more Normal as the sample size increases. (a) The ­distribution of x– for samples of size 2. (b) The distribution of x– for samples of size 5. (c) The distribution of x– for samples of size 10. (d) The distribution of x– for samples of size 25.

As the previous example illustrates, even when the population distribution is very non-Normal, the sampling distribution of x– often looks approximately Normal with sample sizes as small as n = 25. To be safe, we’ll require that n be at least 30 to invoke the CLT. With that issue settled, we can now state the Normal/ Large Sample condition for sample means.

Normal/large sample Condition for Sample Means • If the population distribution is Normal, then so is the sampling distribution of x– . This is true no matter what the sample size n is. • If the population distribution is not Normal, the central limit theorem tells us that the sampling distribution of x– will be approximately Normal in most cases if n ≥ 30. The central limit theorem allows us to use Normal probability calculations to answer questions about sample means from many observations even when the population distribution is not Normal.

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Example

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Servicing Air Conditioners Calculations using the CLT Your company has a contract to perform preventive maintenance on thousands of air-conditioning units in a large city. Based on service records from the past year, the time (in hours) that a technician requires to complete the work follows a strongly right-skewed distribution with m = 1 hour and s = 1 hour. In the coming week, your company will service an SRS of 70 air-conditioning units in the city. You plan to budget an average of 1.1 hours per unit for a technician to complete the work. Will this be enough?

PROBLEM:  What is the probability that the average maintenance time x– for 70 units exceeds

1.1 hours? Show your work.

SOLUTION: Step 1:  State the distribution and the values of interest. The sampling distribution of the sample mean time x– spent working on 70 units has •  mean m x– = m = 1 hour 1 s N(1,0.12) •  standard deviation s x– = = 0.12 because the = !70 !70 10% condition is met (there are more than 10(70) = 700 airconditioning units in the population) •  an approximately Normal shape because the Normal/Large Sample condition is met: n = 70 ≥ 30 The distribution of x– is therefore approximately N(1, 0.12). We want to find P( x– > 1.1). Figure 7.20 shows the Normal curve with the area of 1 1.1 interest shaded and the mean, standard deviation, and boundary value Average maintenance time (hours) labeled. FIGURE 7.20  ​The Normal approximation from the central limit theorem for the average time needed to maintain an air conditioner.

Step 2:  Perform calculations—show your work! The standardized score for the boundary value is 1.1 − 1 z= = 0.83 0.12 Using Table A, P( x– > 1.1) = P(Z > 0.83) = 1 − 0.7967 = 0.2033. Using technology: The command normalcdf(lower:1.1, upper:10000, m:1, s:0.12) gives an area of 0.2023. Step 3:  Answer the question. If you budget 1.1 hours per unit, there is about a 20% chance that the technicians will not complete the work within the budgeted time. You will have to decide if this risk is worth taking or if you should schedule more time for the work. For Practice Try Exercise

63

Figure 7.21 on the next page summarizes the facts about the sampling distribution of x– . It reminds us of the big idea of a sampling distribution. Keep taking random samples of size n from a population with mean m. Find the sample mean x– for each sample. ­Collect all the x– ’s and display their distribution: the sampling distribution of x– . Sampling distributions are the key to understanding statistical inference. Keep this figure in mind for future reference.

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SRS size n SRS size n SRS size n

σ n

x x x

µ

Population Mean µ Std. dev. σ

Values of x

FIGURE 7.21  ​The sampling distribution of a sample mean x– has mean m and standard deviation s∙!n. It has a Normal shape if the population distribution is Normal. If the population distribution isn’t Normal, the sampling distribution of x– is approximately Normal if the sample size is large enough.

case closed

Building Better Batteries Refer to the chapter-opening Case Study on page 421. Assuming the process is working properly, the population distribution of battery lifetimes has mean m = 17 hours and standard deviation s = 0.8. We don’t know the shape of the population distribution. 1. M  ake an appropriate graph to display the sample data. ­Describe what you see. 2. Assume that the battery production process is working properly. Describe the shape, center, and spread of the sampling distribution of x– for random samples of 50 batteries. Justify your answers. For the random sample of 50 batteries, the average lifetime was x– = 16.718 hours. 3. Find the probability of obtaining a random sample of 50 batteries with a mean lifetime of 16.718 hours or less if the production process is working properly. Show your work. Based on your answer, do you believe that the process is working properly? Why or why not? The plant manager also wants to know what proportion p of all the batteries produced that day lasted less than 16.5 hours, which he has declared “unsuitable.” From past experience, about 27% of batteries made at the plant are unsuitable. If the manager does not find convincing evidence that the proportion of unsuitable batteries p produced that day is greater than 0.27, the whole batch of batteries will be shipped to customers.

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4. Assume that the actual proportion of unsuitable batteries produced that day is p = 0.27. Describe the shape, center, and spread of the sampling distribution of p^ for random samples of 50 batteries. Justify your answers. For the random sample of 50 batteries, the sample proportion with lifetimes less than 16.5 hours was p^ = 0.32. 5. Find the probability of obtaining a random sample of 50 batteries in which 32% or more of the batteries are unsuitable if p = 0.27. Show your work. Based on your answer, should the entire batch of batteries be shipped to customers? Why or why not?

Section 7.3

Summary •

• •

•

•

Section 7.3

When we want information about the population mean m for some variable, we often take an SRS and use the sample mean x– to estimate the unknown parameter m. The sampling distribution of x– describes how the statistic x– varies in all possible samples of the same size from the population. The mean of the sampling distribution is m, so x– is an unbiased estimator of m. The standard deviation of the sampling distribution of x– is s∙!n for an SRS of size n if the population has standard deviation s. That is, averages are less variable than individual observations. This formula can be used if the population is at least 10 times as large as the sample (10% condition). Choose an SRS of size n from a population with mean m and standard deviation s. If the population distribution is Normal, then so is the sampling distribution of the sample mean x– . If the population distribution is not Normal, the central limit theorem (CLT) states that when n is large, the sampling distribution of x– is approximately Normal. We can use a Normal distribution to calculate approximate probabilities for events involving x– whenever the Normal/Large Sample condition is met: • If the population distribution is Normal, so is the sampling distribution of x– . • If n ≥ 30, the CLT tells us that the sampling distribution of x– will be approximately Normal in most cases.

Exercises

49. Songs on an iPod  David’s iPod has about 10,000 songs. The distribution of the play times for these songs is heavily skewed to the right with a mean of 225 seconds and a standard deviation of 60 seconds.

pg 452

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Suppose we choose an SRS of 10 songs from this population and calculate the mean play time x– of these songs. What are the mean and the standard deviation of the sampling distribution of x– ? Explain.

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50. Making auto parts  A grinding machine in an auto parts plant prepares axles with a target diameter m = 40.125 millimeters (mm). The machine has some variability, so the standard deviation of the diameters is s = 0.002 mm. The machine operator inspects a random sample of 4 axles each hour for quality control purposes and records the sample mean diameter x– . Assuming that the process is working properly, what are the mean and standard deviation of the sampling distribution of x– ? Explain. 51. Songs on an iPod  Refer to Exercise 49. How many songs would you need to sample if you wanted the standard deviation of the sampling distribution of x– to be 30 seconds? Justify your answer. 52. Making auto parts  Refer to Exercise 50. How many axles would you need to sample if you wanted the standard deviation of the sampling distribution of x– to be 0.0005 mm? Justify your answer. 53. Larger sample  Suppose that the blood cholesterol level of all men aged 20 to 34 follows the Normal distribution with mean m = 188 milligrams per deciliter (mg/dl) and standard deviation s = 41 mg/dl. (a) Choose an SRS of 100 men from this population. Describe the sampling distribution of x– . (b) Find the probability that x– estimates m within ±3 mg/dl. (This is the probability that x– takes a value between 185 and 191 mg/dl.) Show your work. (c) Choose an SRS of 1000 men from this population. Now what is the probability that x– falls within ±3 mg/dl of m? Show your work. In what sense is the larger sample “better”? 54. Dead battery?  A car company has found that the lifetime of its batteries varies from car to car according to a Normal distribution with mean m = 48 months and standard deviation s = 8.2 months. The company installs a new brand of battery on an SRS of 8 cars. (a) If the new brand has the same lifetime distribution as the previous type of battery, describe the sampling distribution of the mean lifetime x– . (b) The average life of the batteries on these 8 cars turns out to be x– = 42.2 months. Find the probability that the sample mean lifetime is 42.2 months or less if the lifetime distribution is unchanged. What conclusion would you draw? 55. Bottling cola  A bottling company uses a filling machine to fill plastic bottles with cola. The bottles are supposed to contain 300 milliliters (ml). In fact, the contents vary according to a Normal distribution with mean m = 298 ml and standard deviation s = 3 ml.

pg 455

(a) What is the probability that a randomly selected bottle contains less than 295 ml? Show your work.

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(b) What is the probability that the mean contents of six randomly selected bottles are less than 295 ml? Show your work. 56. Cereal  A company’s cereal boxes advertise 9.65 ounces of cereal. In fact, the amount of cereal in a randomly selected box follows a Normal distribution with mean m = 9.70 ounces and standard deviation s = 0.03 ounces. (a) What is the probability that a randomly selected box of the cereal contains less than 9.65 ounces of cereal? Show your work. (b) Now take an SRS of 5 boxes. What is the probability that the mean amount of cereal x– in these boxes is 9.65 ounces or less? Show your work. 57. What does the CLT say?  Asked what the central limit theorem says, a student replies, “As you take larger and larger samples from a population, the histogram of the sample values looks more and more Normal.” Is the student right? Explain your answer. 58. What does the CLT say?  Asked what the central limit theorem says, a student replies, “As you take larger and larger samples from a population, the spread of the sampling distribution of the sample mean decreases.” Is the student right? Explain your answer. 59. Songs on an iPod  Refer to Exercise 49. (a) Explain why you cannot safely calculate the probability that the mean play time x– is more than 4 minutes (240 seconds) for an SRS of 10 songs. (b) Suppose we take an SRS of 36 songs instead. ­Explain how the central limit theorem allows us to find the probability that the mean play time is more than 240 seconds. Then calculate this probability. Show your work. 60. Lightning strikes  The number of lightning strikes on a square kilometer of open ground in a year has mean 6 and standard deviation 2.4. The ­National Lightning Detection Network (NLDN) uses automatic sensors to watch for lightning in a random sample of 10 one-square-kilometer plots of land. (a) What are the mean and standard deviation of the sampling distribution of x– , the sample mean number of strikes per square ­kilometer? (b) Explain why you cannot safely calculate the probability that x– < 5 based on a sample of size 10. (c) Suppose the NLDN takes a random sample of n = 50 square kilometers instead. Explain how the central limit theorem allows us to find the probability that the mean number of lightning strikes per square kilometer is less than 5. Then calculate this probability. Show your work.

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Section 7.3 Sample Means 61. Airline passengers get heavier  In response to the increasing weight of airline passengers, the Federal Aviation Administration (FAA) told airlines to assume that passengers average 190 pounds in the summer, including clothes and carry-on baggage. But passengers vary, and the FAA did not specify a standard deviation. A reasonable standard deviation is 35 pounds. Weights are not Normally distributed, especially when the population includes both men and women, but they are not very non-Normal. A commuter plane carries 30 passengers.

Multiple choice: Select the best answer for Exercises 65 to 68. 65. Scores on the mathematics part of the SAT exam in a recent year were roughly Normal with mean 515 and standard deviation 114. You choose an SRS of 100 students and average their SAT Math scores. Suppose that you do this many, many times. Which of the following are the mean and standard deviation of the sampling distribution of x– ? (a) Mean = 515, SD = 114

(a) Explain why you cannot calculate the probability that a randomly selected passenger weighs more than 200 pounds.

(b) Mean = 515, SD = 114∙!100

(b) Find the probability that the total weight of 30 randomly selected passengers exceeds 6000 pounds. Show your work. (Hint: To apply the central limit theorem, restate the problem in terms of the mean weight.)

(d) Mean = 515/100, SD = 114∙!100

62. How many people in a car?  A study of rush-hour traffic in San Francisco counts the number of people in each car entering a freeway at a suburban interchange. Suppose that this count has mean 1.5 and standard deviation 0.75 in the population of all cars that enter at this interchange during rush hours.

66. Why is it important to check the 10% condition before calculating probabilities involving x– ?

(a) Could the exact distribution of the count be ­Normal? Why or why not? (b) Traffic engineers estimate that the capacity of the interchange is 700 cars per hour. Find the probability that 700 randomly selected cars at this freeway entrance will carry more than 1075 people. Show your work. (Hint: Restate this event in terms of the mean number of people x– per car.) 63. More on insurance  An insurance company claims pg 459 that in the entire population of homeowners, the mean annual loss from fire is m = $250 and the standard deviation of the loss is s = $1000. The distribution of losses is strongly right-skewed: many policies have $0 loss, but a few have large losses. An auditor examines a random sample of 10,000 of the company’s policies. If the company’s claim is correct, what’s the probability that the average loss from fire in the sample is no greater than $275? Show your work. 64. Bad carpet  The number of flaws per square yard in a type of carpet material varies with mean 1.6 flaws per square yard and standard deviation 1.2 flaws per square yard. The population distribution cannot be Normal, because a count takes only whole-number values. An inspector studies a random sample of 200 square yards of the material, records the number of flaws found in each square yard, and calculates x– , the mean number of flaws per square yard inspected. Find the probability that the mean number of flaws exceeds 1.8 per square yard. Show your work.

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(c) Mean = 515/100, SD = 114/100 (e) Cannot be determined without knowing the 100 scores.

(a) To reduce the variability of the sampling distribution of x– . (b) To ensure that the distribution of x– is approximately Normal. (c) To ensure that we can generalize the results to a larger population. (d) To ensure that x– will be an unbiased estimator of m. (e) To ensure that the observations in the sample are close to independent. 67. A newborn baby has extremely low birth weight (ELBW) if it weighs less than 1000 grams. A study of the health of such children in later years examined a random sample of 219 children. Their mean weight at birth was x– = 810 grams. This sample mean is an unbiased estimator of the mean weight m in the population of all ELBW babies, which means that (a) in all possible samples of size 219 from this ­population, the mean of the values of x– will equal 810. (b) in all possible samples of size 219 from this ­population, the mean of the values of x– will equal m. (c) as we take larger and larger samples from this population, x– will get closer and closer to m. (d) in all possible samples of size 219 from this ­population, the values of x– will have a distribution that is close to Normal. (e) the person measuring the children’s weights does so without any error.

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68. The number of hours a lightbulb burns before failing varies from bulb to bulb. The population distribution of burnout times is strongly skewed to the right. The central limit theorem says that

for the civilian population aged 25 years and over in a recent year. The table entries are counts in thousands of people.

(a) as we look at more and more bulbs, their average burnout time gets ever closer to the mean m for all bulbs of this type. (b) the average burnout time of a large number of bulbs has a sampling distribution with the same shape (strongly skewed) as the population distribution. (c) the average burnout time of a large number of bulbs has a sampling distribution with similar shape but not as extreme (skewed, but not as strongly) as the population distribution. (d) the average burnout time of a large number of bulbs has a sampling distribution that is close to Normal. (e) the average burnout time of a large number of bulbs has a sampling distribution that is exactly Normal. Exercises 69 to 72 refer to the following setting. In the language of government statistics, you are “in the labor force” if you are available for work and either working or actively seeking work. The unemployment rate is the proportion of the labor force (not of the entire population) who are unemployed. Here are data from the Current Population Survey

Highest education

Total population

In labor force

Employed

Didn’t finish high school

27,669

12,470

11,408

High school but no college

59,860

37,834

35,857

Less than bachelor’s degree

47,556

34,439

32,977

College graduate

51,582

40,390

39,293

69. Unemployment (1.1)  Find the unemployment rate for people with each level of education. How does the unemployment rate change with education? 70. Unemployment (5.1)  What is the probability that a randomly chosen person 25 years of age or older is in the labor force? Show your work. 71. Unemployment (5.3)  If you know that a randomly chosen person 25 years of age or older is a college graduate, what is the probability that he or she is in the labor force? Show your work. 72. Unemployment (5.3)  Are the events “in the labor force” and “college graduate” independent? Justify your answer.

FRAPPY!  Free Response AP® Problem, Yay! The following problem is modeled after actual AP® Statistics exam free response questions. Your task is to generate a complete, concise response in 15 minutes.

Directions: Show all your work. Indicate clearly the methods you use, because you will be scored on the correctness of your methods as well as on the accuracy and completeness of your results and explanations. The principal of a large high school is concerned about the number of absences for students at his school. To investigate, he prints a list showing the number of absences during the last month for each of the 2500 students at the school. For this population of students, the distribution of absences last month is skewed to the right with a mean of m = 1.1 and a standard deviation of s = 1.4. Suppose that a random sample of 50 students is selected from the list printed by the principal and the sample mean number of absences is calculated. (a) What is the shape of the sampling distribution of the sample mean? Explain.

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(b) What are the mean and standard deviation of the sampling distribution of the sample mean? (c) What is the probability that the mean number of absences in a random sample of 50 students is less than 1? (d) Because the population distribution is skewed, the principal is considering using the median number of absences last month instead of the mean number of absences to summarize the distribution. Describe how the principal could use a simulation to estimate the standard deviation of the sampling distribution of the sample median for samples of size 50. After you finish, you can view two example solutions on the book’s Web site (www.whfreeman.com/tps5e). Determine whether you think each solution is “complete,” “substantial,” “developing,” or “minimal.” If the solution is not complete, what improvements would you suggest to the student who wrote it? Finally, your teacher will provide you with a scoring rubric. Score your response and note what, if anything, you would do differently to improve your own score.

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Chapter Review Section 7.1: What Is a Sampling Distribution? In this section, you learned the “big ideas” of sampling distributions. The first big idea is the difference between a statistic and a parameter. A parameter is a number that describes some characteristic of a population. A statistic estimates the value of a parameter using a sample from the population. Making the distinction between a statistic and a parameter will be crucial throughout the rest of the course. The second big idea is that statistics vary. For example, the mean weight in a sample of high school students is a variable that will change from sample to sample. This means that statistics have distributions, but parameters do not. The distribution of a statistic in all possible samples of the same size is called the sampling distribution of the statistic. The third big idea is the distinction between the distribution of the population, the distribution of the sample, and the sampling distribution of a sample statistic. Reviewing the illustration on page 428 will help you understand the difference between these three distributions. When you are writing your answers, be sure to indicate which distribution you are referring to. Don’t make ambiguous statements like “the distribution will become less variable.” The fourth big idea is how to describe a sampling distribution. To adequately describe a sampling distribution, you need to address shape, center, and spread. If the center (mean) of the sampling distribution is the same as the value of the parameter being estimated, then the statistic is called an unbiased estimator. An estimator is unbiased if it doesn’t consistently under- or overestimate the parameter in many samples. Ideally, the spread of a sampling distribution will be very small, meaning that the statistic provides precise estimates of the parameter. Larger sample sizes result in sampling distributions with smaller spreads. Section 7.2: Sample Proportions In this section, you learned about the shape, center, and spread of the sampling distribution of a sample proportion. When the Large Counts condition (np ≥ 10 and n(1 – p) ≥ 10) is met, the sampling distribution of p^ will

be approximately Normal. The mean of the sampling distribution of p^ is mp^ = p, the population proportion. As a result, the sample proportion p^ is an unbiased estimator of the population proportion p. When the 10% 1 condition an ≤ Nb is met, the standard deviation 10 of the sampling distribution of the sample proportion is p(1 − p) . This formula tells us that the variability sp^ = n Å of the distribution of p^ is smaller when the sample size is larger. Section 7.3: Sample Means In this section, you learned about the shape, center, and spread of the sampling distribution of a sample mean. When the population is Normal, the sampling distribution of x– will also be Normal for any sample size. When the population is not Normal and the sample size is small, the sampling distribution of x– will resemble the population shape. However, the central limit theorem says that the sampling distribution of x– will become approximately Normal for larger sample sizes (typically when n ≥ 30), no matter what the population shape. When you are using a Normal distribution to calculate probabilities involving the sampling distribution of x– , make sure that the Normal/ Large Sample condition is met. The mean of the sampling distribution of x– is mx– = m, the population mean. As a result, the sample mean x– is an unbiased estimator of the population mean m. When 1 the 10% condition an ≤ Nb is met, the standard de10 viation of the sampling distribution of the sample mean is s sx– = . This formula tells us that the variability of the !n distribution of x– is smaller when the sample size is larger. Finally, when you are using a Normal distribution to calculate probabilities involving the sampling distribution of p^ or x– , make sure that you (1) state the distribution and values of interest, (2) perform calculations—show your work, and (3) answer the question.

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What Did You Learn? Learning Objective

Section

Related Example on Page(s)

Relevant Chapter Review Exercise(s)

Distinguish between a parameter and a statistic.

7.1

425

R7.1

Use the sampling distribution of a statistic to evaluate a claim about a parameter.

7.1

427

R7.5, R7.7

Distinguish among the distribution of a population, the distribution of a sample, and the sampling distribution of a statistic.

7.1

Discussion on 428

R7.2

Determine whether or not a statistic is an unbiased estimator of a population parameter.

7.1

Discussion on 430–431; 435

R7.3

Describe the relationship between sample size and the variability of a statistic.

7.1

432

R7.3

Find the mean and standard deviation of the sampling distribution of a sample proportion p^ . Check the 10% condition before calculating sp^ .

7.2

445

R7.4

Determine if the sampling distribution of p^ is approximately Normal.

7.2

445

R7.4

If appropriate, use a Normal distribution to calculate probabilities involving p^ .

7.2

445

Find the mean and standard deviation of the sampling distribution of a sample mean x– . Check the 10% condition before calculating s x– .

R7.4, R7.5

7.3

452

Explain how the shape of the sampling distribution of x– is affected by the shape of the population distribution and the sample size.

R7.6

7.3

457

R7.6, R7.7

If appropriate, use a Normal distribution to calculate probabilities involving x– .

7.3

455, 459

R7.6, R7.7

Chapter 7  Chapter Review Exercises These exercises are designed to help you review the important ideas and methods of the chapter. R7.1 Bad eggs  Sale of eggs that are contaminated with salmonella can cause food poisoning in consumers. A large egg producer takes an SRS of 200 eggs from all the eggs shipped in one day. The laboratory reports that 9 of these eggs had salmonella contami-

nation. Unknown to the producer, 3% of all eggs shipped had salmonella. Identify the population, the parameter, the sample, and the statistic. Exercises R7.2 and R7.3 refer to the following setting. ­ esearchers in Norway analyzed data on the birth R weights of 400,000 newborns over a 6-year period. The distribution of birth weights is approximately Normal

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Chapter Review Exercises with a mean of 3668 grams and a standard deviation of 511 grams.8 In this population, the range (maximum – minimum) of birth weights is 3417 grams. We used Fathom software to take 500 SRSs of size n = 5 and calculate the range (maximum – minimum) for each sample. The dotplot below shows the results.

R7.2 Birth weights (a) Sketch a graph that displays the distribution of birth weights for this population. (b) Sketch a possible graph of the distribution of birth weights for an SRS of size 5. (c) In the graph above, there is a dot at approximately 2750. Explain what this value represents. R7.3 Birth weights (a) Is the sample range an unbiased estimator of the population range? Give evidence from the graph above to support your answer. (b) Explain how we could decrease the variability of the sampling distribution of the sample range. R7.4. Do you jog?  The Gallup Poll once asked a random sample of 1540 adults, “Do you happen to jog?” Suppose that the true proportion of all adults who jog is p = 0.15. (a) What is the mean of the sampling distribution of p^ ? Justify your answer. (b) Find the standard deviation of the sampling distribution of p^ . Check that the 10% condition is met. (c) Is the sampling distribution of p^ approximately Normal? Justify your answer. (d) Find the probability that between 13% and 17% of a random sample of 1540 adults are joggers. R7.5 Bag check  Thousands of travelers pass through the airport in Guadalajara, Mexico, each day. Before leaving the airport, each passenger must pass through the Customs inspection area. Customs agents want to be sure that passengers do not bring illegal items into the country. But they do not have time to search every traveler’s luggage. Instead, they require each person to press a button. Either a red

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or a green bulb lights up. If the red light shows, the passenger will be searched by Customs agents. A green light means “go ahead.” Customs agents claim that the proportion of all travelers who will be stopped (red light) is 0.30, because the light has probability 0.30 of showing red on any push of the button. To test this claim, a concerned citizen watches a random sample of 100 travelers push the button. Only 20 get a red light. (a) Assume that the Customs agents’ claim is true. Find the probability that the proportion of travelers who get a red light is as small as or smaller than the result in this sample. Show your work. (b) Based on your results in (a), do you believe the Customs agents’ claim? Explain. R7.6 IQ tests  The Wechsler Adult Intelligence Scale (WAIS) is a common “IQ test” for adults. The distribution of WAIS scores for persons over 16 years of age is approximately Normal with mean 100 and standard deviation 15. (a) What is the probability that a randomly chosen ­individual has a WAIS score of 105 or higher? Show your work. (b) Find the mean and standard deviation of the sampling distribution of the average WAIS score x– for an SRS of 60 people. (c) What is the probability that the average WAIS score of an SRS of 60 people is 105 or higher? Show your work. (d) Would your answers to any of parts (a), (b), or (c) be affected if the distribution of WAIS scores in the adult population were distinctly non-Normal? Explain. R7.7 Detecting gypsy moths  The gypsy moth is a serious threat to oak and aspen trees. A state agriculture department places traps throughout the state to detect the moths. Each month, an SRS of 50 traps is inspected, the number of moths in each trap is recorded, and the mean number of moths is calculated. Based on years of data, the distribution of moth counts is discrete and strongly skewed, with a mean of 0.5 and a standard deviation of 0.7. (a) Explain why it is reasonable to use a Normal distribution to approximate the sampling distribution of x– for SRSs of size 50. (b) Estimate the probability that the mean number of moths in a sample of size 50 is greater than or equal to 0.6. (c) In a recent month, the mean number of moths in an SRS of size 50 was 0.6. Based on this result, should the state agricultural department be worried that the moth population is getting larger in their state? Explain.

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Chapter 7  AP® Statistics Practice Test Section I: Multiple Choice  ​Select the best answer for each question. T7.1 A study of voting chose 663 registered voters at random shortly after an election. Of these, 72% said they had voted in the election. Election records show that only 56% of registered voters voted in the election. Which of the following statements is true about the boldface numbers? (a) 72% is a sample; 56% is a population. (b) 72% and 56% are both statistics. (c) 72% is a statistic and 56% is a parameter. (d) 72% is a parameter and 56% is a statistic. (e) 72% and 56% are both parameters. T7.2 The Gallup Poll has decided to increase the size of its random sample of voters from about 1500 people to about 4000 people right before an election. The poll is d ­ esigned to estimate the proportion of voters who favor a new law banning smoking in public buildings. The effect of this increase is to (a) reduce the bias of the estimate. (b) increase the bias of the estimate. (c) reduce the variability of the estimate. (d) increase the variability of the estimate. (e) reduce the bias and variability of the estimate. T7.3 Suppose we select an SRS of size n = 100 from a large population having proportion p of successes. Let p^ be the proportion of successes in the sample. For which value of p would it be safe to use the Normal approximation to the sampling distribution of p^ ? (a) 0.01  (b) 0.09  (c) 0.85  (d) 0.975  (e) 0.999 T7.4 The central limit theorem is important in statistics because it allows us to use the Normal distribution to find probabilities involving the sample mean (a) if the sample size is reasonably large (for any ­population). (b) if the population is Normally distributed and the ­sample size is reasonably large. (c) if the population is Normally distributed (for any sample size). (d) if the population is Normally distributed and the population standard deviation is known (for any sample size). (e) if the population size is reasonably large (whether the population distribution is known or not).

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T7.5 The number of undergraduates at Johns Hopkins University is approximately 2000, while the number at Ohio State University is approximately 60,000. At both schools, a simple random sample of about 3% of the undergraduates is taken. Each sample is used to estimate the proportion p of all students at that university who own an iPod. Suppose that, in fact, p = 0.80 at both schools. Which of the following is the best conclusion? (a) The estimate from Johns Hopkins has less sampling variability than that from Ohio State. (b) The estimate from Johns Hopkins has more sampling variability than that from Ohio State. (c) The two estimates have about the same amount of ­sampling variability. (d) It is impossible to make any statement about the ­sampling variability of the two estimates because the students surveyed were different. (e) None of the above. T7.6 A researcher initially plans to take an SRS of size n from a population that has mean 80 and standard deviation 20. If he were to double his sample size (to 2n), the standard deviation of the sampling distribution of the sample mean would be multiplied by (a) !2.  (b)  1∙!2.  (c) ​2.  (d) ​1/2.  (e)  1∙!2n.

T7.7 The student newspaper at a large university asks an SRS of 250 undergraduates, “Do you favor eliminating the carnival from the term-end celebration?” All in all, 150 of the 250 are in favor. Suppose that (unknown to you) 55% of all undergraduates favor eliminating the carnival. If you took a very large number of SRSs of size n = 250 from this population, the sampling distribution of the sample proportion p^ would be (a) exactly Normal with mean 0.55 and standard deviation 0.03. (b) approximately Normal with mean 0.55 and standard deviation 0.03. (c) exactly Normal with mean 0.60 and standard deviation 0.03. (d) approximately Normal with mean 0.60 and standard deviation 0.03. (e) heavily skewed with mean 0.55 and standard deviation 0.03.

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AP® Statistics Practice Test T7.8 Which of the following statements about the sampling distribution of the sample mean is incorrect? (a) The standard deviation of the sampling distribution will decrease as the sample size increases. (b) The standard deviation of the sampling distribution is a measure of the variability of the sample mean among r­ epeated samples. (c) The sample mean is an unbiased estimator of the population mean. (d) The sampling distribution shows how the sample mean will vary in repeated samples. (e) The sampling distribution shows how the sample was distributed around the sample mean. T7.9 A machine is designed to fill 16-ounce bottles of shampoo. When the machine is working properly, the amount poured into the bottles follows a Normal distribution with mean 16.05 ounces and standard deviation 0.1 ounce. Assume that the machine is working properly. If four bottles are randomly selected and the number of ounces in each bottle is measured, then there is about

469

a 95% chance that the sample mean will fall in which of the following intervals? (a) 16.05 to 16.15 ounces (d) ​15.90 to 16.20 ounces (b) 16.00 to 16.10 ounces (e) ​15.85 to 16.25 ounces (c) 15.95 to 16.15 ounces T7.10 Suppose that you are a student aide in the library and agree to be paid according to the “random pay” system. Each week, the librarian flips a coin. If the coin comes up heads, your pay for the week is $80. If it comes up tails, your pay for the week is $40. You work for the library for 100 weeks. Suppose we choose an SRS of 2 weeks and calculate your average earnings x– . The shape of the sampling distribution of x– will be (a) Normal. (b) approximately Normal. (c) right-skewed. (d) left-skewed. (e) symmetric but not Normal.

Section II: Free Response  ​Show all your work. Indicate clearly the methods you use, because you will be graded on the correctness of your methods as well as on the accuracy and completeness of your results and explanations. T7.11 Below are histograms of the values taken by three sample statistics in several hundred samples from the same population. The true value of the population parameter is marked with an arrow on each histogram.

A

B

C

Which statistic would provide the best estimate of the ­parameter? Justify your answer. Starnes/Yates/Moore: The Practice of Statistics, 4E

T7.12 households NewThe Fig.: amount 07_UN20 thatPerm. Fig.: 7065pay service providers access to the Internet varies quite a bit, but the Firstfor Pass: 2010-06-21

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mean monthly fee is $38 and the standard deviation is $10. The distribution is not Normal: many households pay a base rate for low-speed access, but some pay much more for faster connections. A sample survey asks an SRS of 500 households with Internet access how much they pay. Let x– be the mean amount paid. (a) Explain why you can’t determine the probability that the amount a randomly selected household pays for access to the Internet exceeds $39. (b) What are the mean and standard deviation of the ­sampling distribution of x– ? (c) What is the shape of the sampling distribution of x– ? Justify your answer. (d) Find the probability that the average fee paid by the sample of households exceeds $39. Show your work. T7.13 According to government data, 22% of American children under the age of six live in households with incomes less than the official poverty level. A study of learning in early childhood chooses an SRS of 300 children. Find the probability that more than 20% of the sample are from poverty-level households. Be sure to check that you can use the Normal approximation.

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Cumulative AP® Practice Test 2 Section I: Multiple Choice  Choose the best answer for each question. AP2.1 The five-number summary for a data set is given by min = 5, Q1 = 18, median = 20, Q3 = 40, max = 75. If you wanted to construct a boxplot for the data set (that is, one that would show outliers, if any existed), what would be the maximum possible length of the right-side “whisker”? (a) 33  (b) 35   (c) ​45   (d) ​53   (e) ​55 AP2.2 The probability distribution for the number of heads in four tosses of a coin is given by Number of heads: 0 1 2 3 4 Probability: 0.0625 0.2500 0.3750 0.2500 0.0625

The probability of getting at least one tail in four tosses of a coin is (a) 0.2500. (c) ​0.6875. (e) ​0.0625. (b) 0.3125. (d) 0.9375. AP2.3 In a certain large population of adults, the distribution of IQ scores is strongly left-skewed with a mean of 122 and a standard deviation of 5. Suppose 200 adults are randomly selected from this population for a market research study. The distribution of the sample mean of IQ scores is (a) left-skewed with mean 122 and standard deviation 0.35. (b) exactly Normal with mean 122 and standard deviation 5. (c) exactly Normal with mean 122 and standard deviation 0.35. (d) approximately Normal with mean 122 and standard deviation 5. (e) approximately Normal with mean 122 and standard deviation 0.35. AP2.4 A 10-question multiple-choice exam offers 5 choices for each question. Jason just guesses the answers, so he has probability 1/5 of getting any one answer correct. You want to perform a simulation to determine the number of correct answers that Jason gets. One correct way to use a table of random digits to do this is the following: (a) One digit from the random digit table simulates one answer, with 5 = right and all other digits = wrong. Ten digits from the table simulate 10 answers. (b) One digit from the random digit table simulates one answer, with 0 or 1 = right and all other digits = wrong. Ten digits from the table simulate 10 answers. (c) One digit from the random digit table simulates one answer, with odd = right and even = wrong. Ten digits from the table simulate 10 answers.

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(d) One digit from the random digit table simulates one answer, with 0 or 1 = right and all other digits = wrong, ignoring repeats. Ten digits from the table simulate 10 answers. (e) Two digits from the random digit table simulate one answer, with 00 to 20 = right and 21 to 99 = wrong. Ten pairs of digits from the table simulate 10 answers. AP2.5 Suppose we roll a fair die four times. The probability that a 6 occurs on exactly one of the rolls is 1 3 5 1 1 1 5 3 (a) 4a b a b (c)  4a b a b 6 6 6 6 1 3 5 1 (b) a b a b 6 6

1 1 5 3 (d)  a b a b 6 6

1 1 5 3 (e)  6a b a b 6 6

AP2.6 You want to take an SRS of 50 of the 816 students who live in a dormitory on a college campus. You label the students 001 to 816 in alphabetical order. In the table of random digits, you read the entries 95592  94007  69769  33547  72450  16632  81194  14873

The first three students in your sample have labels (a) 955, 929, 400. (d) ​929, 400, 769. (b) 400, 769, 769. (e) ​400, 769, 335. (c) 559, 294, 007. AP2.7 The number of unbroken charcoal briquets in a 20-pound bag filled at the factory follows a Normal distribution with a mean of 450 briquets and a standard ­deviation of 20 briquets. The company expects that a certain number of the bags will be underfilled, so the company will replace for free the 5% of bags that have too few briquets. What is the minimum number of unbroken briquets the bag would have to contain for the company to avoid having to replace the bag for free? (a) 404   (b) ​411   (c) ​418   (d) ​425   (e) ​448 AP2.8 You work for an advertising agency that is preparing a new television commercial to appeal to women. You have been asked to design an experiment to compare the effectiveness of three versions of the commercial. Each subject will be shown one of the three versions and then asked about her attitude toward the product. You think there may be large differences between women who are employed and those who are not. Because of these differences, you should use (a) ​ a block design, but not a matched pairs design. (b) a completely randomized design. (c) a matched pairs design.

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Cumulative AP® Practice Test 2 (d) a simple random sample. (e) a stratified random sample. AP2.9 Suppose that you have torn a tendon and are facing surgery to repair it. The orthopedic surgeon explains the risks to you. Infection occurs in 3% of such operations, the repair fails in 14%, and both infection and failure occur together 1% of the time. What is the probability that the operation is successful for someone who has an operation that is free from infection? (a) 0.8342 (c)  0.8600​ (e) ​0.9900 (b) 0.8400 (d) ​0.8660 AP2.10 Social scientists are interested in the association between high school graduation rate (HSGR, measured as a percent) and the percent of U.S. families living in poverty (POV). Data were collected from all 50 states and the District of Columbia, and a regression analysis was conducted.

ñ

The resulting least-squares regression line is given by POV = 59.2 − 0.620(HSGR) with r2 = 0.802. Based on the information, which of the following is the best interpretation for the slope of the leastsquares regression line? (a) For each 1% increase in the graduation rate, the percent of families living in poverty is predicted to decrease by approximately 0.896. (b) For each 1% increase in the graduation rate, the percent of families living in poverty is predicted to decrease by approximately 0.802. (c) For each 1% increase in the graduation rate, the percent of families living in poverty is predicted to decrease by approximately 0.620. (d) For each 1% increase in the percent of families living in poverty, the graduation rate is predicted to increase by approximately 0.802. (e) For each 1% increase in the percent of families living in poverty, the graduation rate is predicted to decrease by approximately 0.620. Here is a dotplot of the adult literacy rates in 177 countries in a recent year, according to the United Nations. For example, the lowest literacy rate was 23.6%, in the African country of Burkina Faso. Mali had the next lowest literacy rate at 24.0%. Use the graph to answer Questions AP2.11 to AP2.13.

20

40

60

Literacy rate

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80

100

471

AP2.11 The overall shape of this distribution is (a) clearly skewed to the right. (b) clearly skewed to the left. (c) roughly symmetric. (d) uniform. (e) There is no clear shape. AP2.12 The mean of this distribution (don’t try to find it) will be (a) very close to the median. (b) greater than the median. (c) less than the median. (d) You can’t say, because distribution isn’t symmetric. (e) You can’t say, because the distribution isn’t Normal. AP2.13 Based on the shape of this distribution, what ­measures of center and spread would be most appropriate to report? (a) The mean and standard deviation (b) The mean and the interquartile range (c) The median and the standard deviation (d) The median and the interquartile range (e) The mean and the range AP2.14 The correlation between the age and height of children under the age of 12 is found to be r = 0.60. Suppose we use the age x of a child to predict the height y of the child. What can we conclude? (a) The height is generally 60% of a child’s weight. (b) About 60% of the time, age will accurately predict height. (c) Thirty-six percent of the variation in height is accounted for by the linear model relating height to age. (d) For every 1 year older a child is, the regression line predicts an increase of 0.6 feet in height. (e) Thirty-six percent of the time, the least-squares regression line accurately predicts height from age. AP2.15 An agronomist wants to test three different types of fertilizer (A, B, and C) on the yield of a new variety of wheat. The yield will be measured in bushels per acre. Six 1-acre plots of land were randomly assigned to each of the three fertilizers. The treatment, experimental unit, and response variable are, respectively, (a) a specific fertilizer, bushels per acre, a plot of land. (b) a plot of land, bushels per acre, a specific fertilizer. (c) random assignment, a plot of land, wheat yield. (d) a specific fertilizer, a plot of land, wheat yield. (e) a specific fertilizer, the agronomist, wheat yield. AP2.16 According to the U.S. Census, the proportion of adults in a certain county who owned their own home was 0.71. An SRS of 100 adults in a certain

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section of the county found that 65 owned their home. Which one of the following represents the approximate probability of ­obtaining a sample of 100 adults in which fewer than 65 own their home, assuming that this section of the county has the same overall proportion of adults who own their home as does the entire county?

Assuming that the weights are independent, what is the standard deviation of the total weight of the boxes that are shipped from this source? (a) 1.84 (b) 2.60

(c) ​3.02 (d) ​3.40

(e) ​9.10

AP2.20 A grocery chain runs a prize game by giving each customer a ticket that may win a prize when the 0.65 − 0.71 100 65 35 box is scratched off. Printed on the ticket is a (a)  a b (0.71) (0.29) (d)  P Z < ° 65 dollar value ($500, $100, $25) or the statement (0.65)(0.35) ¢ “This ticket is not a winner.” Monetary prizes can Å 100 be redeemed for groceries at the store. Here is the 100 0.65 − 0.71 probability distribution of the amount won on a (b)  a P Z< b (0.29)65(0.71)35 (e)  (0.71)(0.29) ¢ randomly selected ticket: 65 ° !100 Amount won: $500 $100 $25 $0 0.65 − 0.71 (c)  P Z < Probability: 0.01 0.05 0.20 0.74 ° (0.71)(0.29) ¢ Å

100



AP2.17 Which one of the following would be a correct interpretation if you have a z-score of +2.0 on an exam? (a) It means that you missed two questions on the exam. (b) It means that you got twice as many questions correct as the average student. (c) It means that your grade was 2 points higher than the mean grade on this exam. (d) It means that your grade was in the upper 2% of all grades on this exam. (e) It means that your grade is 2 standard deviations above the mean for this exam. AP2.18 Records from a random sample of dairy farms yielded the information below on the number of male and female calves born at various times of the day. Day

Evening

Night

Total

Males

129

15

117

261

Females

118

18

116

252

Total

247

33

233

513

Which of the following are the mean and standard deviation, respectively, of the winnings?

(a) $15.00, $2900.00 (b) $15.00, $53.85 (c) $15.00, $26.93 (d) $156.25, $53.85 (e) $156.25, $26.93 AP2.21 A large company is interested in improving the ­efficiency of its customer service and decides to examine the length of the business phone calls made to clients by its sales staff. A cumulative relative frequency graph is shown below from data collected over the past year. According to the graph, the shortest 80% of calls will take how long to complete?



What is the probability that a randomly selected calf was born in the night or was a female? 369 485 116 116 116 (a)   (b)    (c)    (d)    (e)  513 513 513 252 233 ​AP2.19 When people order books from a popular online source, they are shipped in standard-sized boxes. Suppose that the mean weight of the boxes is 1.5 pounds with a standard deviation of 0.3 pounds, the mean weight of the packing material is 0.5 pounds with a standard deviation of 0.1 pounds, and the mean weight of the books shipped is 12 pounds with a standard deviation of 3 pounds.

(a) Less than 10 minutes (b) At least 10 minutes (c) Exactly 10 minutes (d) At least 5.5 minutes (e) Less than 5.5 minutes

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Section II: Free Response  Show all your work. Indicate clearly the methods you use, because you will be graded on the correctness of your methods as well as on the accuracy and completeness of your results and explanations. AP2.22 A health worker is interested in determining if omega-3 fish oil can help reduce cholesterol in adults. She obtains permission to examine the health records of 200 people in a large medical clinic and classifies them according to whether or not they take omega-3 fish oil. She also obtains their latest cholesterol readings and finds that the mean cholesterol reading for those who are taking ­omega-3 fish oil is 18 points lower than the mean for the group not taking omega-3 fish oil. (a) Is this an observational study or an experiment? Justify your answer. (b) Explain the concept of confounding in the context of this study and give one example of a variable that could be confounded with whether or not people take omega-3 fish oil. (c) Researchers find that the 18-point difference in the mean cholesterol readings of the two groups is statistically significant. Can they conclude that omega-3 fish oil is the cause? Why or why not?

least one of the specimens contains type A blood from the White ethnic group. AP2.24 Every 17 years, swarms of cicadas emerge from the ground in the eastern United States, live for about six weeks, and then die. (There are several different “broods,” so we experience cicada eruptions more often than every 17 years.) There are so many cicadas that their dead bodies can serve as fertilizer and increase plant growth. In a study, a researcher added 10 dead cicadas under 39 randomly selected plants in a natural plot of American bellflowers on the forest floor, leaving other plants undisturbed. One of the response variables measured was the size of seeds produced by the plants. Here are the boxplots and summary statistics of seed mass (in milligrams) for 39 cicada plants and 33 undisturbed (control) plants:

AP2.23 There are four major blood types in humans: O, A, B, and AB. In a study conducted using blood specimens from the Blood Bank of Hawaii, individuals were classified according to blood type and ethnic group. The ethnic groups were Hawaiian, Hawaiian-White, Hawaiian-Chinese, and White. Suppose that a blood bank specimen is selected at random. Ethnic Group Blood type

Hawaiians

Hawaiian- HawaiianWhite Chinese

White

Total

O

1903

4469

2206

53,759

62,337

A

2490

4671

2368

50,008

59,537

B

178

 606

 568

16,252

17,604

AB

99

 236

 243

  5001

5579

4670

9982

5385

125,020

145,057

Total

(a) Find the probability that the specimen contains type O blood or comes from the Hawaiian-Chinese ethnic group. Show your work. (b) What is the probability that the specimen contains type AB blood, given that it comes from the Hawaiian ethnic group? Show your work. (c) Are the events “type B blood” and “Hawaiian ethnic group” independent? Give appropriate statistical evidence to support your answer. (d) Now suppose that two blood bank specimens are selected at random. Find the probability that at

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Variable: n Minimum Q1 Median Q3 Maximum Cicada plants: 39 0.17 0.22 0.25 0.28 0.35 Control plants: 33 0.14 0.19 0.25 0.26 0.29

(a) Write a few sentences comparing the distributions of seed mass for the two groups of plants. (b) Based on the graphical displays, which distribution has the larger mean? Justify your answer. (c) Explain the purpose of the random assignment in this study. (d) Name one benefit and one drawback of only using American bellflowers in the study. AP2.25 In a city library, the mean number of pages in a novel is 525 with a standard deviation of 200. Approximately 30% of the novels have fewer than 400 pages. Suppose that you randomly select 50 novels from the library. (a) What is the probability that the total number of pages is fewer than 25,000? Show your work. (b) What is the probability that at least 20 of the novels have fewer than 400 pages? Show your work.

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Chapter

8

Introduction 476 Section 8.1 477 Confidence Intervals: The Basics Section 8.2 492 Estimating a Population Proportion Section 8.3 507 Estimating a Population Mean Free Response AP® Problem, YAY! 530 Chapter 8 Review 531 Chapter 8 Review Exercises 532 Chapter 8 AP® Statistics Practice Test 534

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Estimating with Confidence case study

Need Help? Give Us a Call!

Frequency

If your cable television goes out, you phone the cable company to get it fixed. Does a real person answer your call? These days, probably not. It is far more likely that you will get an automated response. You will probably be offered several options, such as: to order cable service, press 1; for questions about your bill, press 2; to add new channels, press 3; (and finally) to speak with a customer ser35 vice agent, press 4. Customers will get frustrated 30 if they have to wait too long before speaking to a 25 live person. So companies try hard to minimize 20 the time required to connect to a customer ser15 vice representative. 10 A large bank decided to study the call response times in its customer service department. The 5 bank’s goal was to have a representative answer an 0 incoming call in less than 30 seconds. Figure 8.1 is a 0.0 7.5 15.0 22.5 30.0 37.5 45.0 Call response time (seconds) histogram of the response times in a random sample of 241 calls to the bank’s ­customer service center in FIGURE 8.1  ​Histogram showing the response time (in seconds) a given month. What does the graph suggest about at a bank’s customer service center for a random sample of 241 calls. how well the bank is meeting its goal? Starnes/Yates/Moore: The Practice of Statistics, 4E New Fig.: 08_01 Perm. Fig.: 8002 the First sample Pass: 2010-06-28

By the end of this chapter, you will be able to use data to make inferences about the proportion p of all calls to the customer service department that are answered within 30 seconds and the mean call response time m.

475

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Introduction How long does a new model of laptop battery last? What proportion of college undergraduates have engaged in binge drinking? How much does the weight of a quarter-pound hamburger at a fast-food restaurant vary after cooking? These are the types of questions we would like to be able to answer. It wouldn’t be practical to determine the lifetime of every laptop battery, to ask all college undergraduates about their drinking habits, or to weigh every burger after cooking. Instead, we choose a sample of individuals (batteries, college students, burgers) to represent the population and collect data from those individuals. Our goal in each case is to use a sample statistic to estimate an unknown population parameter. From what we learned in Chapter 4, if we randomly select the sample, we should be able to generalize our results to the population of interest. We cannot be certain that our conclusions are correct—a different sample would probably yield a different estimate. Statistical inference uses the language of probability to express the strength of our conclusions. Probability allows us to take chance variation due to random selection or random assignment into account. The following Activity gives you an idea of what lies ahead.

Activity Materials: TI-83/84 or TI-89 with display capability

The Mystery Mean In this Activity, each team of three to four students will try to estimate the mystery value of the population mean m that your teacher entered before class.1 1. ​Before class, your teacher will store a value of m (represented by M) in the display calculator. The teacher will then clear the home screen so you can’t see the value of M. 2. ​With the class watching, the teacher will execute the following command: mean(randNorm(M,20,16)). This tells the calculator to choose an SRS of 16 observations from a Normal population with mean M and standard deviation 20 and then compute the mean x– of those 16 sample values. Is the sample mean shown likely to be equal to the mystery mean M? Why or why not? 3. ​Now for the challenge! Your group must determine an interval of reasonable values for the population mean m. Use the result from Step 2 and what you learned about sampling distributions in the previous chapter. 4. ​Share your team’s results with the class.

In this chapter and the next, we will meet the two most common types of formal statistical inference. Chapter 8 concerns confidence intervals for estimating the value of a parameter. Chapter 9 presents significance tests, which assess the evidence for a claim about a parameter. Both types of inference are based on the sampling distributions of statistics. That is, both report probabilities that state what would happen if we used the inference method many times.

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Section 8.1 examines the idea of a confidence interval. We start by presenting the reasoning of confidence intervals in a general way that applies to estimating any unknown parameter. In Section 8.2, we show how to estimate a population proportion. Section 8.3 focuses on confidence intervals for a population mean.

8.1 What You Will Learn

Confidence Intervals: The Basics By the end of the section, you should be able to:

• Determine the point estimate and margin of error from a confidence interval. • Interpret a confidence interval in context. • Interpret a confidence level in context.

• Describe how the sample size and confidence level affect the length of a confidence interval. • Explain how practical issues like nonresponse, undercoverage, and response bias can affect the interpretation of a confidence interval.

Mr. Schiel’s class did the mystery mean Activity from the Introduction. The TI screen shot displays the information that the students received about the unknown population mean m. Here is a summary of what the class said about the calculator output: • The population distribution is Normal and its standard deviation is s = 20. • A simple random sample of n = 16 observations was taken from this p ­ opulation. – • The sample mean is x = 240.80. If we had to give a single number to estimate the value of M that Mr. Schiel chose, what would it be? Such a value is known as a point estimate. How about 240.80? That makes sense, because the sample mean x– is an unbiased estimator of the population mean m. We are using the statistic x– as a point estimator of the parameter m.

Definition: Point estimator and point estimate A point estimator is a statistic that provides an estimate of a population parameter. The value of that statistic from a sample is called a point estimate.

As we saw in Chapter 7, the ideal point estimator will have no bias and low variability. Here’s an example involving some of the more common point estimators.

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E s t i m at i n g w i t h C o n f i d e n ce

From Batteries to Smoking Point estimators Problem:  In each of the following settings, determine the point estimator you would use and calculate the value of the point estimate. (a) Quality control inspectors want to estimate the mean lifetime m of the AA batteries produced in an hour at a factory. They select a random sample of 50 batteries during each hour of production and then drain them under conditions that mimic normal use. Here are the lifetimes (in hours) of the batteries from one such sample: 16.73 15.60 16.31 17.57 16.14 17.28 16.67 17.28 17.27 17.50 15.46 16.50 16.19 15.59 17.54 16.46 15.63 16.82 17.16 16.62 16.71 16.69 17.98 16.36 17.80 16.61 15.99 15.64 17.20 17.24 16.68 16.55 17.48 15.58 17.61 15.98 16.99 16.93 16.01 17.54 17.41 16.91 16.60 16.78 15.75 17.31 16.50 16.72 17.55 16.46

(b) What proportion p of U.S. high school students smoke? The 2011 Youth Risk Behavioral Survey questioned a random sample of 15,425 students in grades 9 to 12. Of these, 2792 said they had smoked cigarettes at least one day in the past month. (c) The quality control inspectors in part (a) want to investigate the variability in battery lifetimes by estimating the population variance s2.

Solution:

(a) Use the sample mean x– as a point estimator for the population mean m. For these data, our point estimate is x– = 16.718 hours. (b) Use the sample proportion p^ as a point estimator for the population proportion p. For this 2792 = 0.181. survey, our point estimate is p^ = 15,425 (c) Use the sample variance s2x as a point estimator for the population variance s2. For the battery life data, our point estimate is s2x = 0.441 hours2. For Practice Try Exercises

1

and

3

The Idea of a Confidence Interval Is the value of the population mean m that Mr. Schiel entered in his calculator exactly 240.80? Probably not. Because x– = 240.80, we guess that m is “somewhere around 240.80.” How close to 240.80 is m likely to be? To answer this question, we ask another: How would the sample mean x– vary if we took many SRSs of size 16 from this same population? The sampling distribution of x– describes how the values of x– vary in repeated samples. Recall the facts about this sampling distribution from Chapter 7: Shape: Because the population distribution is Normal, so is the sampling distribution of x– . Thus, the Normal/Large Sample condition is met. Center: The mean of the sampling distribution of x– is the same as the unknown mean m of the entire population. That is, mx– = m. Spread: The standard deviation of the sampling distribution of x– for samples of size n = 16 is 20 s =5 = sx– = !n !16

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because the 10% condition is met—we are sampling from an infinite population in this case. Figure 8.2 summarizes these facts. Sampling distribution of x σ = 20

6 n=1 SRS 16 SRS n =

SRS n = 16

x = 240.80

µ (unknown)

x = 246.05 x = 248.85

σ =5 − n

µ

Values of x

Population

FIGURE 8.2  ​The sampling distribution of the mean score x– for SRSs of 16 observations from a Normally distributed population with unknown mean m and standard deviation s = 20.

The next example gives the reasoning of statistical estimation in a nutshell.

EXAMPLE

The Mystery Mean Moving beyond a point estimate When Mr. Schiel’s class discussed the results of the mystery mean Activity, students used the following logic to come up with an “interval estimate” for the unknown population mean m.

1. The sample mean x– = 240.80 is our point estimate for m. We don’t expect x– to be exactly equal to m, so we want to say how precise this estimate is. 2. In repeated samples, the values of x– follow a Normal distribution with mean m and standard deviation 5, as in Figure 8.2. 3. The 95 part of the 68–95–99.7 rule for Normal distributions says that x– is within 2(5) = 10 (that’s 2 standard deviations) of the population mean m in about 95% of all samples of size n = 16. See Figure 8.3. Sampling distribution 4. Whenever x– is within 10 points of m, m is within 10 of x points of x– . This happens in about 95% of all possible samples. So the interval from x– − 10 to x– + 10 “captures” the population mean m in about 95% of all samples of size 16. Probability ≈ 0.95 5. If we estimate that m lies somewhere in the interval from x– − 10 = 240.80 − 10 = 230.80 to µ − 10

µ

µ + 10

FIGURE 8.3  ​In about 95% of all samples, x– lies within ±10 of the unknown population mean m. So m also lies within ±10 of x– in those samples.

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x– + 10 = 240.80 + 10 = 250.80 we’d be calculating this interval using a method that captures the true m in about 95% of all possible samples of this size.

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The big idea is that the sampling distribution of x– tells us how close to m the sample mean x– is likely to be. Statistical estimation just turns that information around to say how close to x– the unknown population mean m is likely to be. In the mystery mean example, the value of m is usually within 2(5) = 10 of x– for SRSs of size 16. Because the class’s sample mean was x– = 240.80, the interval 240.80 ± 10 gives an approximate 95% confidence interval for m. There are several ways to write a confidence interval. We can give the interval for Mr. Schiel’s mystery mean as 240.80 ± 10, as 230.80 to 250.80, or as (230.80, 250.80). All the confidence intervals we will meet have a form similar to this: point estimate ± margin of error The point estimate (x– = 240.80 in our example) is our best guess for the value of the unknown parameter. The margin of error, 10, shows how close we believe our guess is, based on the variability of the estimate in repeated SRSs of size 16. We say that our confidence level is about 95% because the interval x– ± 10 catches the unknown parameter in about 95% of all possible samples. A confidence interval is sometimes referred to as an interval estimate. This is consistent with our earlier use of the term point estimate.

Definition: Confidence interval, margin of error, confidence level A C% confidence interval gives an interval of plausible values for a parameter. The interval is calculated from the data and has the form point estimate ± margin of error The difference between the point estimate and the true parameter value will be less than the margin of error in C% of all samples. The confidence level C gives the overall success rate of the method for calculating the confidence interval. That is, in C% of all possible samples, the method would yield an interval that captures the true parameter value.

c

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!

n

The interval from 230.80 to 250.80 gives the set of plausible values for Mr. Schiel’s mystery mean m with 95% confidence. We wouldn’t be surprised if any of the values in this interval turned out to be the actual value of m. Plausible does not mean the same thing as possible. You could argue autio that just about any value of a parameter is possible. A plausible value of a parameter is a reasonable or believable value based on the data. There is a trade-off between the confidence level and the amount of information provided by a confidence interval, as the cartoon below illustrates. We usually choose a confidence level of 90% or higher because we want to be quite sure of our conclusions. The most common confidence level is 95%.

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Interpreting Confidence Intervals and Confidence Levels Our 95% confidence interval for Mr. Schiel’s mystery mean was (230.80, 250.80). How do we interpret this interval? We say, “We are 95% confident that the interval from 230.80 to 250.80 captures the mystery mean chosen by Mr. Schiel.”

Interpreting Confidence Intervals To interpret a C% confidence interval for an unknown parameter, say, “We are C% confident that the interval from _____ to _____ captures the [parameter in context].” Here’s an example that involves interpreting a confidence interval for a proportion.

Example

Who Will Win the Election? Interpreting a confidence interval Two weeks before a presidential election, a polling organization asked a random sample of registered voters the following question: “If the presidential election were held today, would you vote for candidate A or candidate B?” Based on this poll, the 95% confidence interval for the population proportion who favor candidate A is (0.48, 0.54).

PROBLEM: (a) Interpret the confidence interval. (b) What is the point estimate that was used to create the interval? What is the margin of error? (c) Based on this poll, a political reporter claims that the majority of registered voters favor candidate A. Use the confidence interval to evaluate this claim.

SOLUTION: (a) We are 95% confident that the interval from 0.48 to 0.54 captures the true proportion of all registered voters who favor candidate A in the election. (b) A confidence interval has the form point estimate ± margin of error The point estimate is at the midpoint of the interval. Here, the point estimate is p^ = 0.51. The margin of error gives the distance from the point estimate to either end of the interval. So the margin of error for this interval is 0.03. (c) Any value from 0.48 to 0.54 is a plausible value for the population proportion p that favors candidate A. Because there are plausible values of p less than 0.50, the confidence interval does not give convincing evidence to support the reporter’s claim that a majority (more than 50%) of registered voters favor candidate A. For Practice Try Exercise

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The following Activity gives you a chance to explore the meaning of the confidence level.

Activity Materials: Computer with Internet connection and display capability PLET AP

The Confidence Intervals Applet The Confidence Intervals applet at the book’s Web site will quickly generate many confidence intervals. In this Activity, you will use the applet to investigate the idea of a confidence level. 1. ​Go to www.whfreeman.com/tps5e and launch the applet. Use the default settings: confidence level 95% and sample size n = 20. 2. ​Click “Sample” to choose an SRS and display the resulting confidence interval. Did the interval capture the population mean m (what the applet calls a “hit”)? Do this a total of 10 times. How many of the intervals captured the population mean m? Note: So far, you have used the applet to take 10 SRSs, each of size n = 20. Be sure you understand the difference between sample size and the number of samples taken. 3. ​Reset the applet. Click “Sample 25” twice to choose 50 SRSs and display the confidence intervals based on those samples. How many captured the parameter m? Keep clicking “Sample 25” and observe the value of “Percent hit.” What do you notice? 4.  Repeat Step 3 using a 90% confidence level. 5.  Repeat Step 3 using an 80% confidence level. 6.  Summarize what you have learned about the relationship between confidence level and “Percent hit” after taking many samples. We will investigate the effect of changing the sample size later.

As the Activity confirms, the confidence level is the overall capture rate if the method is used many times. Figure 8.4 illustrates the behavior of the confidence interval x– ± 10 for Mr. Schiel’s mystery mean. Starting with the population, imagine taking many SRSs of 16 observations. The first sample has x– = 240.80, the second has x– = 246.05, the third has x– = 248.85, and so on. The sample mean varies from sample to sample, but when we use the formula x– ± 10 to get an interval based on each sample, about 95% of these intervals capture the unknown population mean m.

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σ = 20

6 n=1 SRS 16 SRS n =

SRS n = 16

µ

Population

Many SRSs

x ± 10 = 240.80 ± 10 x ± 10 = 246.05 ± 10 x ± 10 = 248.85 ± 10

About 95% of these intervals capture the unknown mean µ of the population.

Many confidence intervals

FIGURE 8.4  ​To say that x– ± 10 is a 95% confidence interval for the population mean m is to say that, in repeated samples, about 95% of these intervals capture m.

Sampling distribution of x

µ

Values of x

This interval misses the true µ. The others all capture µ .

FIGURE 8.5  ​Twenty-five samples of the same size from the same population gave these 95% confidence intervals. In the long run, about 95% of samples give an interval that captures the population mean m.

Starnes/Yates/Moore: The Practice of Statistics, 4E New Fig.: 08_05 Perm. Fig.: 8014 First Pass: 2010-06-28 Interpreting 2nd Pass: 2010-07-12

Figure 8.5 illustrates the idea of a confidence interval in a different form. It shows the result of drawing many SRSs from the same population and calculating a 95% confidence interval from each sample. The center of each interval is at x– and therefore varies from sample to sample. The sampling distribution of x– appears at the top of the figure to show the long-term pattern of this variation. The 95% confidence intervals from 25 SRSs appear below. Here’s what you should notice: • The center x– of each interval is marked by a dot. • The distance from the dot to either endpoint of the interval is the margin of error. • 24 of these 25 intervals (that’s 96%) contain the true value of m. If we took many samples, about 95% of the resulting confidence intervals would capture m. Figures 8.4 and 8.5 give us the insight we need to interpret a confidence level.

Confidence Levels

To say that we are 95% confident is shorthand for “If we take many samples of the same size from this population, about 95% of them will result in an interval that captures the actual parameter value.”

c

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!

n

The confidence level tells us how likely it is that the method we are using will produce an interval that captures the population parameter if we use it many times. However, in practice we tend to calculate only a single confidence interval for a given situation. The confidence level does not tell us the chance autio that a particular confidence interval captures the population parameter. Instead, the confidence interval gives us a set of plausible values for the parameter.

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E s t i m at i n g w i t h C o n f i d e n ce

The Mystery Mean Interpreting a confidence level The confidence level in the mystery mean ­example—roughly 95%—tells us that if we take many SRSs of size 16 from Mr. Schiel’s mystery population, the interval x– ± 10 will capture the population mean m for about 95% of those samples.

Be sure you understand the basis for our confidence. There are only two possibilities: 1. The interval from 230.80 to 250.80 contains the population mean m. 2. The interval from 230.80 to 250.80 does not contain the population mean m. Our SRS was one of the few samples for which x– is not within 10 points of the true m. Only about 5% of all samples result in a confidence interval that fails to capture m. We cannot know whether our sample is one of the 95% for which the interval x– ± 10 catches m or whether it is one of the unlucky 5%. The statement that we are “95% confident” that the unknown m lies between 230.80 and 250.80 is shorthand for saying, “We got these numbers by a method that gives correct results 95% of the time.”

THINK ABOUT IT

What’s the probability that our 95% confidence interval captures the para­ meter? ​It’s not 95%! Before we execute the command

mean(randNorm​(M,20,16)), we have a 95% chance of getting a sample mean that’s within 2sx– of the mystery m, which would lead to a confidence interval that captures m. Once we have chosen a random sample, the sample mean x– either is or isn’t within 2sx– of m. And the resulting confidence interval either does or doesn’t contain m. After we construct a confidence interval, the probability that it captures the population parameter is either 1 (it does) or 0 (it doesn’t). We interpret confidence intervals and confidence levels in much the same way whether we are estimating a population mean, proportion, or some other parameter.

Example

Do You Use Twitter? Interpreting a confidence interval and a confidence level The Pew Internet and American Life Project asked a random sample of 2253 U.S. adults, “Do you ever . . . use Twitter or another service to share updates about yourself or to see updates about others?” Of the sample, 19% said “Yes.” ­A ccording to Pew, the resulting 95% confidence interval is (0.167, 0.213). 2

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Problem:  Interpret the confidence interval and the confidence level. Solution: Confidence interval:  We are 95% confident that the interval from 0.167 to 0.213 captures the true proportion p of all U.S. adults who use T­ witter or another service for updates.

Confidence level:  If many samples of 2253 U.S. adults were taken, the resulting confidence intervals

would capture the true proportion of all U.S. adults who use Twitter or another service for updates for about 95% of those samples. For Practice Try Exercise

c

Confidence intervals are statements about parameters. In the previous example, it would be wrong to say, “We are 95% confident that the interval from 0.167 to 0.213 contains the sample proportion of U.S. adults who use Twitter.” Why? Because we know that the sample proportion, p^ = 0.19, is in the interval. Likewise, in the mystery mean example, it would be wrong to say that “95% of the values are between 230.80 and 250.80,” whether we are referring to the sample or the populaautio tion. All we can say is, “Based on Mr. Schiel’s sample, we believe that the population mean is somewhere between 230.80 and 250.80.” When interpreting a confidence interval, make it clear that you are describing a parameter and not a statistic. And be sure to include context.

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AP® EXAM TIP ​On a given problem, you may be asked to interpret the confidence interval, the confidence level, or both. Be sure you understand the difference: the confidence interval gives a set of plausible values for the parameter and the confidence level describes the long-run capture rate of the method.

15

Check Your Understanding

How much does the fat content of Brand X hot dogs vary? To find out, researchers measured the fat content (in grams) of a random sample of 10 Brand X hot dogs. A 95% confidence interval for the population standard deviation s is 2.84 to 7.55. 1. Interpret the confidence interval. 2. Interpret the confidence level. 3. True or false: The interval from 2.84 to 7.55 has a 95% chance of containing the actual population standard deviation s. Justify your answer.

Constructing a Confidence Interval Why settle for 95% confidence when estimating an unknown parameter? Do ­larger random samples yield “better” intervals? The Confidence Intervals applet might shed some light on these questions. PLET AP

Activity Materials: Computer with Internet connection and display capability

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The Confidence Intervals Applet In this Activity, you will use the applet to explore the relationship between the confidence level, the sample size, and the confidence interval. 1. ​Go to www.whfreeman.com/tps5e and launch the Confidence Intervals applet. Use the default settings: confidence level 95% and sample size n = 20. Click “Sample 25.”

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2. ​Change the confidence level to 99%. What happens to the length of the confidence intervals? 3. ​Now change the confidence level to 90%. What happens to the length of the confidence intervals? 4. ​Finally, change the confidence level to 80%. What happens to the length of the confidence intervals? 5.  Summarize what you learned about the relationship between the confidence level and the length of the confidence intervals for a fixed sample size. 6.  Reset the applet and change the confidence level to 95%. What happens to the length of the confidence intervals as you increase the sample size? 7.  Does increasing the sample size increase the capture rate (percent hit)? Use the applet to investigate.

As the Activity illustrates, the price we pay for greater confidence is a wider interval. If we’re satisfied with 80% confidence, then our interval of plausible values for the parameter will be much narrower than if we insist on 90%, 95%, or 99% confidence. But we’ll also be much less confident in our estimate. Taking the idea of confidence to an extreme, what if we want to estimate with 100% confidence the proportion p of all U.S. adults who use Twitter? That’s easy: we’re 100% confident that the interval from 0 to 1 captures the true population proportion! The activity also shows that we can get a more precise estimate of a parameter by increasing the sample size. Larger samples yield narrower confidence intervals. This result holds for any confidence level. Let’s look a bit more closely at the method we used earlier to calculate an approximate 95% confidence interval for Mr. Schiel’s mystery mean. We started with point estimate ± margin of error Our point estimate came from the sample statistic x– = 240.80. What about the margin of error? Because the population distribution is Normal, so is the sampling distribution of x– . About 95% of the values of x– will lie within 2 standard deviations (2sx– ) of the mystery mean m. See the figure below. We could rewrite our interval as Sampling distribution of x

σ =5 − n

µ (unknown)

Values of x

Starnes/Yates/Moore: The Practice of Statistics, 4E New Fig.: 08_UN08 Perm. Fig.: 8019

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240.80 ± 2 # 5 = x– ± 2 # s x– This leads to the more general formula for a confidence ­interval: statistic ± (critical value) · (standard deviation of statistic) The critical value is a multiplier that makes the interval wide enough to have the stated capture rate. The critical value depends on both the confidence level C and the sampling distribution of the statistic.

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Calculating a Confidence Interval The confidence interval for estimating a population parameter has the form statistic ± (critical value) · (standard deviation of statistic) where the statistic we use is the point estimator for the parameter. The confidence interval for the mystery mean m of Mr. Schiel’s population illustrates several important properties that are shared by all confidence intervals in common use. The user chooses the confidence level, and the margin of error follows from this choice. We would like high confidence and also a small margin of error. High confidence says that our method almost always gives correct answers. A small margin of error says that we have pinned down the parameter quite precisely. Our general formula for a confidence interval is Margin of error

Statistic ± (critical value) · (standard deviation of statistic)

s and !n p(1 − p) . So as the sample sp^ = Å n size n increases, the standard deviation of the statistic decreases. Recall that sx– =

We can see that the margin of error depends on the critical value and the standard deviation of the statistic. The critical value is tied directly to the confidence level: greater confidence requires a larger critical value. The standard deviation of the statistic depends on the sample size n: larger samples give more precise estimates, which means less variability in the statistic. So the margin of error gets smaller when: • The confidence level decreases. There is a trade-off between the confidence level and the margin of error. To obtain a 80% confidence smaller margin of error from the same data, you must be 95% confidence willing to accept lower confidence. Earlier, we found that a 95% confidence interval for Mr. Schiel’s mystery mean m is 230.80 to 250.80. The 80% confidence interval for m is 230 232 234 236 238 240 242 244 246 248 250 252 254 234.39 to 247.21. Figure 8.6 compares these two intervals. FIGURE 8.6  ​80% and 95% confidence intervals for • The sample size n increases. Increasing the sample size n Mr. Schiel’s mystery mean. Higher confidence requires a longer interval. reduces the margin of error for any fixed confidence level. Starnes/Yates/Moore: The Practice of Statistics, 4E New Fig.: 08_06 Perm. Fig.: 8021 First Pass: 2010-06-28 2nd Pass: 2010-07-12

Using Confidence Intervals Wisely Our goal in this section has been to introduce you to the big ideas of confidence intervals without getting bogged down in details. You may have noticed that we only calculated intervals in a contrived setting: estimating an unknown population mean m when we somehow knew the population standard deviation s. In practice, when we don’t know m, we don’t know s either. We’ll learn to construct confidence intervals for a population mean in this more realistic setting in Section 8.3. First, we will study confidence intervals for a population proportion p in Section 8.2. Although it is possible to estimate other parameters, confidence intervals for means and proportions are the most common tools in everyday use.

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c c

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Section 8.1

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Here are two important cautions to keep in mind when constructing and interpreting confidence intervals. • Our method of calculation assumes that the data come from an SRS of autio size n from the population of interest. Other types of random samples (stratified or cluster, say) might be preferable to an SRS in a given setting, but they require more complex calculations than the ones we’ll use. • The margin of error in a confidence interval covers only chance variation due to random sampling or random assignment. The margin of e­ rror is autio obtained from the sampling distribution. It indicates how close our estimate is likely to be to the unknown parameter if we repeat the random sampling or random assignment process many times. Practical difficulties, such as undercoverage and nonresponse in a sample survey, can lead to additional errors that may be larger than this chance variation. Remember this unpleasant fact when reading the results of an opinion poll or other sample survey. The way in which a survey or experiment is conducted influences the trustworthiness of its results in ways that are not included in the announced margin of error.

Summary •

•

•

To estimate an unknown population parameter, start with a statistic that provides a reasonable guess. The chosen statistic is a point estimator for the parameter. The specific value of the point estimator that we use gives a point estimate for the parameter. A C% confidence interval uses sample data to estimate an unknown population parameter with an indication of how precise the estimate is and of how confident we are that the result is correct. A confidence interval gives an interval of plausible values for the parameter. The interval is computed from the data and has the form point estimate ± margin of error When calculating a confidence interval, it is common to use the form statistic ± (critical value) · (standard deviation of statistic)

•

•

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To interpret a C% confidence interval, say, “We are C% confident that the interval from ____ to ____ captures the [parameter in context].” Be sure that your interpretation describes a parameter and not a statistic. The confidence level C is the success rate of the method that produces the interval. If you use 95% confidence intervals often, in the long run about 95% of your intervals will contain the true parameter value. You don’t know whether a 95% confidence interval calculated from a particular set of data actually captures the true parameter value.

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Section 8.1 Confidence Intervals: The Basics •

•

Section 8.1

Other things being equal, the margin of error of a confidence interval gets smaller as • the confidence level C decreases. • the sample size n increases. Remember that the margin of error for a confidence interval includes only chance variation, not other sources of error like nonresponse and undercoverage.

Exercises mainly basic arithmetic and the ability to apply it to realistic problems. Scores on the test range from 0 to 500. For example, a person who scores 233 can add the amounts of two checks appearing on a bank deposit slip; someone scoring 325 can determine the price of a meal from a menu; a person scoring 375 can transform a price in cents per ounce into dollars per pound.4 Suppose that you give the NAEP test to an SRS of 840 people from a large population in which the scores have mean 280 and standard deviation s = 60. The mean x– of the 840 scores will vary if you take repeated samples.

In Exercises 1 to 4, determine the point estimator you would use and calculate the value of the point estimate. 1. pg 478

Got shoes?  How many pairs of shoes, on average, do female teens have? To find out, an AP® Statistics class conducted a survey. They selected an SRS of 20 female students from their school. Then they recorded the number of pairs of shoes that each student reported having. Here are the data:

50

26

26

31

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19

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3. pg 478

4.

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Got shoes?  The class in Exercise 1 wants to estimate the variability in the number of pairs of shoes that female students have by estimating the population variance s2. Going to the prom  Tonya wants to estimate what proportion of the seniors in her school plan to attend the prom. She interviews an SRS of 50 of the 750 seniors in her school and finds that 36 plan to go to the prom. Reporting cheating  What proportion of students are willing to report cheating by other students? A student project put this question to an SRS of 172 undergraduates at a large university: “You witness two students cheating on a quiz. Do you go to the professor?” Only 19 answered “Yes.”3 NAEP scores  Young people have a better chance of full-time employment and good wages if they are good with numbers. How strong are the quantitative skills of young Americans of working age? One source of data is the National Assessment of Educational Progress (NAEP) Young Adult Literacy Assessment Survey, which is based on a nationwide probability sample of households. The NAEP survey includes a short test of quantitative skills, covering

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(a) Describe the shape, center, and spread of the sampling distribution of x– . (b) Sketch the sampling distribution of x– . Mark its mean and the values 1, 2, and 3 standard deviations on either side of the mean. (c) According to the 68–95–99.7 rule, about 95% of all values of x– lie within a distance m of the mean of the sampling distribution. What is m? Shade the region on the axis of your sketch that is within m of the mean. (d) Whenever x– falls in the region you shaded, the population mean m lies in the confidence interval x– ± m. For what percent of all possible samples does the interval capture m? 6.

Auto emissions  Oxides of nitrogen (called NOX for short) emitted by cars and trucks are important contributors to air pollution. The amount of NOX emitted by a particular model varies from vehicle to vehicle. For one light-truck model, NOX emissions vary with mean m = 1.8 grams per mile and standard deviation s = 0.4 gram per mile. You test an SRS of 50 of these trucks. The sample mean NOX level x– will vary if you take repeated samples.

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(a) Describe the shape, center, and spread of the sampling distribution of x– . (b) Sketch the sampling distribution of x– . Mark its mean and the values 1, 2, and 3 standard deviations on either side of the mean. (c) According to the 68–95–99.7 rule, about 95% of all values of x– lie within a distance m of the mean of the sampling distribution. What is m? Shade the region on the axis of your sketch that is within m of the mean.

(c) Based on this poll, Gallup claims that more than half of U.S. adults want to lose weight. Use the confidence interval to evaluate this claim. 11. How confident?  The figure below shows the result of taking 25 SRSs from a Normal population and constructing a confidence interval for each sample. Which confidence level—80%, 90%, 95%, or 99%— do you think was used? Explain.

(d) Whenever x– falls in the region you shaded, the unknown population mean m lies in the confidence interval x– ± m. For what percent of all possible samples does the interval capture m? 7.

NAEP scores  Refer to Exercise 5. Below your sketch, choose one value of x– inside the shaded region and draw its corresponding confidence interval. Do the same for one value of x– outside the shaded region. What is the most important difference between these intervals? (Use Figure 8.5, on page 483, as a model for your drawing.)

8.

Auto emissions  Refer to Exercise 6. Below your sketch, choose one value of x– inside the shaded region and draw its corresponding confidence interval. Do the same for one value of x– outside the shaded region. What is the most important difference between these intervals? (Use Figure 8.5, on page 483, as a model for your drawing.)

9.

Prayer in school A New York Times/CBS News Poll asked a random sample of U.S. adults the question, “Do you favor an amendment to the Constitution that would permit organized prayer in public schools?” Based on this poll, the 95% confidence interval for the population proportion who favor such an amendment is (0.63, 0.69).

pg 481

(a) Interpret the confidence interval. (b) What is the point estimate that was used to create the interval? What is the margin of error? (c) Based on this poll, a reporter claims that more than two-thirds of U.S. adults favor such an amendment. Use the confidence interval to evaluate this claim. 10. Losing weight  A Gallup Poll asked a random sample of U.S. adults, “Would you like to lose weight?” Based on this poll, the 95% confidence interval for the population proportion who want to lose weight is (0.56, 0.62).5 (a) Interpret the confidence interval. (b) What is the point estimate that was used to create the interval? What is the margin of error?

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12. How confident?  The figure below shows the result of taking 25 SRSs from a Normal population and constructing a confidence interval for each sample. Which confidence level—80%, 90%, 95%, or 99%— do you think was used? Explain.

13. Prayer in school  Refer to Exercise 9. The news article goes on to say: “The theoretical errors do not take into account c additional error resulting from the various practical difficulties in taking any survey of public opinion.” List some of the “practical difficulties” that may cause errors which are not included in the ±3 percentage point margin of error. 14. Losing weight  Refer to Exercise 10. As Gallup indicates, the 3 percentage point margin of error for this poll includes only sampling variability (what they call “sampling error”). What other potential sources of error (Gallup calls these “nonsampling errors”) could affect the accuracy of the 95% confidence interval? 15. Shoes  The AP® Statistics class in Exercise 1 also pg 484 asked an SRS of 20 boys at their school how many

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Section 8.1 Confidence Intervals: The Basics pairs of shoes they have. A 95% confidence interval for the difference in the population means (girls – boys) is 10.9 to 26.5. Interpret the confidence interval and the confidence level. 16. Lying online  Many teens have posted profiles on sites such as Facebook. A sample survey asked random samples of teens with online profiles if they included false information in their profiles. Of 170 younger teens (ages 12 to 14) polled, 117 said “Yes.” Of 317 older teens (ages 15 to 17) polled, 152 said “Yes.”6 A 95% confidence interval for the difference in the population proportions (younger teens – older teens) is 0.120 to 0.297. Interpret the confidence interval and the confidence level. 17. Shoes  Refer to Exercise 15. Does the confidence interval give convincing evidence of a difference in the population mean number of pairs of shoes for boys and girls at the school? Justify your answer. 18. Lying online  Refer to Exercise 16. Does the confidence interval give convincing evidence of a difference in the population proportions of younger and older teens who include false information in their profiles? Justify your answer. 19. Explaining confidence  A 95% confidence interval for the mean body mass index (BMI) of young American women is 26.8 ± 0.6. Discuss whether each of the following explanations is correct. (a) We are confident that 95% of all young women have BMI between 26.2 and 27.4. (b) We are 95% confident that future samples of young women will have mean BMI between 26.2 and 27.4. (c) Any value from 26.2 to 27.4 is believable as the true mean BMI of young American women. (d) If we take many samples, the population mean BMI will be between 26.2 and 27.4 in about 95% of those samples. (e) The mean BMI of young American women cannot be 28.

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(d) If we take many samples, about 95% of them will contain the interval (107.8, 116.2). (e) The probability that the interval (107.8, 116.2) captures m is either 0 or 1, but we don’t know which. Multiple choice: Select the best answer for Exercises 21 to 24. Exercises 21 and 22 refer to the following setting. A researcher plans to use a random sample of families to estimate the mean monthly family income for a large population. 21. The researcher is deciding between a 95% confidence level and a 99% confidence level. Compared to a 95% confidence interval, a 99% confidence interval will be (a) narrower and would involve a larger risk of being incorrect. (b) wider and would involve a smaller risk of being incorrect. (c) narrower and would involve a smaller risk of ­being incorrect. (d) wider and would involve a larger risk of being incorrect. (e) wider and would have the same risk of being incorrect. 22. The researcher is deciding between a sample of size n = 500 and a sample of size n = 1000. Compared to using a sample size of n = 500, a 95% confidence interval based on a sample size of n = 1000 will be (a) narrower and would involve a larger risk of being incorrect. (b) wider and would involve a smaller risk of being incorrect. (c) narrower and would involve a smaller risk of being incorrect. (d) wider and would involve a larger risk of being incorrect. (e) narrower and would have the same risk of being incorrect. 23. In a poll, I. ​ Some people refused to answer questions.

20. Explaining confidence  The admissions director from Big City University found that (107.8, 116.2) is a 95% confidence interval for the mean IQ score of all freshmen. Discuss whether each of the following explanations is correct.

II. ​People without telephones could not be in the sample.

(a) There is a 95% probability that the interval from 107.8 to 116.2 contains m.

(a) ​I only

(c) ​III only  

(b) ​II only​

(d) ​I, II, and III

(b) There is a 95% chance that the interval (107.8, 116.2) contains x– . (c) This interval was constructed using a method that produces intervals that capture the true mean in 95% of all possible samples.

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III. ​Some people never answered the phone in several calls. Which of these possible sources of bias is included in the ±2% margin of error announced for the poll? (e) ​None of these

24. You have measured the systolic blood pressure of an SRS of 25 company employees. A 95% confidence interval for the mean systolic blood pressure for the employees of this company is (122, 138). Which of the following statements is true?

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(a) 95% of the sample of employees have a systolic blood pressure between 122 and 138. (b) 95% of the population of employees have a ­systolic blood pressure between 122 and 138. (c) If the procedure were repeated many times, 95% of the resulting confidence intervals would contain the population mean systolic blood pressure.

­ regnant. Result: no connection between leukemia p and exposure to magnetic fields of the kind produced by power lines was found.7 (a) Was this an observational study or an experiment? Justify your answer. (b) Does this study show that living near power lines doesn’t cause cancer? Explain.

(d) If the procedure were repeated many times, 95% of the time the population mean systolic blood pressure would be between 122 and 138.

26. Sisters and brothers (3.1, 3.2)  How strongly do physical characteristics of sisters and brothers correlate? Here are data on the heights (in inches) of 11 adult pairs:8

(e) If the procedure were repeated many times, 95% of the time the sample mean systolic blood pressure would be between 122 and 138.

Brother:

71 68 66 67 70 71 70 73 72 65 66

Sister:

69 64 65 63 65 62 65 64 66 59 62

25. Power lines and cancer (4.2, 4.3)  Does living near power lines cause leukemia in children? The National Cancer Institute spent 5 years and $5 million gathering data on this question. The researchers compared 638 children who had leukemia with 620 who did not. They went into the homes and measured the magnetic fields in children’s bedrooms, in other rooms, and at the front door. They recorded facts about power lines near the family home and also near the mother’s residence when she was

8.2 What You Will Learn

(a) Construct a scatterplot using brother’s height as the explanatory variable. Describe what you see. (b) Use your calculator to compute the least-squares regression line for predicting sister’s height from brother’s height. Interpret the slope in context. (c) Damien is 70 inches tall. Predict the height of his sister Tonya. (d) Do you expect your prediction in (c) to be very accurate? Give appropriate evidence to support your answer.

Estimating a Population ­Proportion By the end of the section, you should be able to:

• State and check the Random, 10%, and Large Counts conditions for constructing a confidence interval for a population proportion. • Determine critical values for calculating a C % confidence interval for a population proportion using a table or technology.

• Construct and interpret a confidence interval for a population proportion. • Determine the sample size required to obtain a C % confidence interval for a population proportion with a specified margin of error.

In Section 8.1, we saw that a confidence interval can be used to estimate an unknown population parameter. We are often interested in estimating the proportion p of some outcome in the population. Here are some examples: • What proportion of U.S. adults are unemployed right now? • What proportion of high school students have cheated on a test?

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• What proportion of pine trees in a national park are infested with beetles? • What proportion of college students pray daily? • What proportion of a company’s laptop batteries last as long as the company claims? This section shows you how to construct and interpret a confidence interval for a population proportion. The following Activity gives you a taste of what lies ahead.

Activity Materials: Several thousand small plastic beads of at least two colors, container to hold all the beads, small cup for sampling, several small bowls

The Beads Before class, your teacher will prepare a large population of different-colored beads and put them into a container that you cannot see inside. Your goal is to estimate the actual proportion of beads in the population that have a particular color (say, red). 1. ​As a class, discuss how to use the cup provided to get a simple random sample of beads from the container. Think this through carefully, because you will get to take only one sample. 2. ​Have one student take an SRS of beads. Separate the beads into two groups: those that are red and those that aren’t. Count the number of beads in each group. 3. ​Determine a point estimate for the unknown population proportion p of red beads in the container. 4. ​Now for the challenge: each team of three to four students will be given about 10 minutes to find a 95% confidence interval for the parameter p. Be sure to consider any conditions that are required for the methods you use. 5. ​Compare the results with those of the other teams in the class. Discuss any problems you encountered and how you dealt with them.

Conditions for Estimating p Before constructing a confidence interval for p, you should check some important conditions: • Random: The data should come from a well-designed random sample or randomized experiment. Otherwise, there’s no scope for inference to a population (sampling) or inference about cause and effect (experiment). If we can’t draw conclusions beyond the data at hand, there’s not much point in constructing a confidence interval! Another important reason for random selection or random assignment is to introduce chance into the data-production process. We can model chance behavior with a probability distribution, like the sampling distributions of Chapter 7. The probability distribution helps us calculate a confidence interval. ~~ 10%: The procedure for calculating confidence intervals assumes that individual observations are independent. Well-conducted studies that use random sampling or random assignment can help ensure independent

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observations. However, our formula for the standard deviation of the p(1 − p) , acts as if we are sampling sampling distribution of p^ , sp^ = n Å with replacement from a population. That’s rarely the case. When sampling without replacement from a finite population, be sure to check that 1 n ≤ N. Sampling more than 10% of the population would give a more 10 precise estimate of the parameter p but would require us to use a different formula to calculate the standard deviation of the sampling distribution. • Large Counts: The method that we use to construct a confidence interval for p depends on the fact that the sampling distribution of p^ is approximately Normal. From what we learned in Chapter 7, we can use the Normal approximation to the sampling distribution of p^ as long as np ≥ 10 and n(1 – p) ≥ 10. In practice, of course, we don’t know the value of p. If we did, we wouldn’t need to construct a confidence interval for it! So we cannot check if np and n(1 – p) are at least 10. In large random samples, p^ will tend to be close to p. So we replace p by p^ in checking the Large Counts condition: np^ ≥ 10 and n(1 − p^ ) ≥ 10.

Conditions for Constructing a Confidence Interval About a Proportion • Random: The data come from a well-designed random sample or randomized experiment. 1 ~~ 10%: When sampling without replacement, check that n ≤ N. 10 • Large Counts: Both np^ and n(1 − p^ ) are at least 10. When Mr. Vignolini’s class did the beads Activity, they got 107 red beads and 144 white beads. Their point estimate for the unknown proportion p of red beads in the population is p^ =

107 = 0.426 251

Let’s see how the conditions play out for Mr. Vignolini’s class.

EXAMPLE

The Beads Checking conditions Mr. Vignolini’s class wants to construct a confidence interval for the true proportion p of red beads in the container. Recall that the class’s sample had 107 red beads and 144 white beads.

Problem:  Check that the conditions for constructing a confidence interval for p are met. Solution:  There are three conditions to check:

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•  Random:  The class took an SRS of 251 beads from the container. ºº 10%:  Because the class sampled without replacement, they need to check that there are at least 10(251) = 2510 beads in the population. Mr. Vignolini reveals that there are 3000 beads in the container. •  Large Counts:  To use a Normal approximation for the sampling distribution of p^, we need both np and n (1 − p) to be at least 10. Because we don’t know p, we check np^ = 251a

107 144 107 b = 107 ≥ 10 and n(1 − p^ ) = 251a1 − b = 251a b = 144 ≥ 10 251 251 251

That is, the counts of successes (red beads) and failures (non-red beads) are both at least 10. All conditions are met, so it should be safe for the class to construct a confidence i­nterval.

For Practice Try Exercise

27

Notice that np^ and n(1 − p^ ) should be whole numbers. You don’t really need to calculate these values since they are just the number of successes and failures in the sample. In the previous example, we could address the Large Counts condition simply by saying, “The numbers of successes (107) and failures (144) in the sample are both at least 10.”

What happens if one of the conditions is violated?  If the data

PLET AP

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come from a convenience sample or a poorly designed experiment, there’s no point constructing a confidence interval for p. Violation of the Random condition severely limits our ability to make any inference beyond the data at hand. The figure below shows a screen shot from the Confidence Intervals for Proportions applet at the book’s Web site, www.whfreeman.com/tps5e. We set n = 20 and p = 0.25. The Large Counts condition is not met because np = 20(0.25) = 5. We used the applet to generate 1000 95% confidence intervals for p. Only 902 of those 1000 intervals contained p = 0.25, a capture rate of 90.2%. When the Large Counts condition is violated, the capture rate will be lower than the one advertised by the confidence level if the method is used many times.

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That leaves just the 10% condition when sampling without replacement from a finite population. Large random samples give more precise estimates than small random samples. So randomly selecting more than 10% of a population should be a good thing! Unfortunately, the formula for the standard deviation of p^ that we p(1 − p) developed in Chapter 7, sp^ = , is not correct when the 10% condition n Å is violated. The formula gives a value that is too large. Confidence intervals based on this formula are longer than they need to be. If many 95% confidence intervals for a population proportion are constructed in this way, more than 95% of them will capture p. The actual capture rate is greater than the reported confidence level when the 10% condition is violated.

Check Your Understanding

In each of the following settings, check whether the conditions for calculating a confidence interval for the population proportion p are met. 1. An AP® Statistics class at a large high school conducts a survey. They ask the first 100 students to arrive at school one morning whether or not they slept at least 8 hours the night before. Only 17 students say “Yes.” 2. A quality control inspector takes a random sample of 25 bags of potato chips from the thousands of bags filled in an hour. Of the bags selected, 3 had too much salt.

Constructing a Confidence Interval for p When the conditions are met, the sampling distribution of p^ will be approximately p(1 − p) Normal with mean mp^ = p and standard deviation sp^ = . Figure 8.7 n Å displays this distribution. Inference about a population proportion p is based on the sampling distribution of p^ . We can use the general formula from Section 8.1 to construct Standard deviation a confidence interval for an unknown population proportion p:

Sampling distribution ˆ of p

p(1 − p)/n

statistic ± (critical value) · (standard deviation of statistic) The sample proportion p^ is the statistic we use to estimate p. ­ oing so makes sense if the data came from a well-designed ranD dom sample or randomized experiment (the Random condition). The standard deviation of the sampling distribution of p^ is

Mean p

FIGURE 8.7  Suppose that a population has proportion p of successes. When the conditions for inference are met, the sampling distribution of the proportion of successes p^ in a sample is approximately Normal with mean p and standard p (1 − p) deviation . Å n Some books refer to sp^ as the “standard error” of p^ and to what we call the standard ­error as the “estimated standard error.”

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sp^ =

p(1 − p) n Å

if the 10% condition is met. Because we don’t know the value of p, we replace it with the sample proportion p^ . The resulting quantity is called the standard error (SE) of the sample proportion p^ .

p^ (1 − p^) n Å ^ It describes how close the sample proportion p will typically be to the population proportion p in repeated SRSs of size n. SEp^ =

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Definition: Standard error When the standard deviation of a statistic is estimated from data, the result is called the standard error of the statistic.

Standard Normal curve

Probability = 0.95

Area = 0.025

−z∗ = −1.96

Area = 0.025

z∗ = 1.96

0

FIGURE 8.8  ​Finding the critical value for a 95% ­confidence interval. The correct value is z * = 1.96, which is more precise than the value of 2 we had been using for 95% confidence.

How do we get the critical value for our confidence interval? If the Large Counts condition is met, we can use a Normal curve. For the approximate 95% confidence intervals of Section 8.1, we used a critical value of 2 based on the 68–95–99.7 rule for Normal distributions. We can get a more precise critical value from Table A or a calculator. As Figure 8.8 shows, the central 95% of the standard Normal distribution is marked off by 2 points, z* = 1.96 and −z* = −1.96. We use the * to remind you that this is a critical value, not a standardized score that has been calculated from data. To find a level C confidence interval, we need to catch the central C% under the standard Normal curve. Here’s an example that shows how to get the critical value z* for a different confidence level.

Starnes/Yates/Moore: The Practice of Statistics, 4E New Fig.: 08_08 Perm. Fig.: 8026 First Pass: 2010-06-28

EXAMPLE

80% Confidence Finding a critical value

Probability = 0.1

Problem:  Use Table A or technology to find the critical value z* for an 80% confidence interval. Assume that the Large Counts condition is met. Solution:  For an 80% confidence level, we need to capture the central 80% of the standard Normal distribution. In capturing the Standard Normal curve central 80%, we leave out 20%, or 10% in each tail. So the desired ­critical value z* is the point with area 0.1 to its right under the standard Normal curve. Figure 8.9 shows the details in picture form. Search the body of Table A to find the point −z* with area 0.1 to its Probability = 0.1 left. The closest entry is z = −1.28. (See the excerpt from Table A Probability = 0.8 below.) So the critical value we want is z* = 1.28.

-z* = -1.28

0

z* = 1.28

FIGURE 8.9  ​Finding the critical value for an 80% confidence interval.

z

.07

.08

.09

−1.3

.0853

.0838

.0823

−1.2

.1020

.1003

.0985

−1.1

.1210

.1190

.1170

Using technology: The command invNorm(area:0.1,m:0, s:1) gives z = −1.28. The critical value is z* = 1.28, which matches what we got from Table A. For Practice Try Exercise

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Technically, the correct formula for a confidence interval is statistic ± (critical value) · (standard error of statistic). We are following the convention used on the AP® Statistics exam formula sheet.

Once we find the critical value z*, our confidence interval for the population proportion p is statistic ± (critical value) · (standard deviation of statistic) p^ (1 − p^) = p^ ± z* n Å Notice that we replaced the standard deviation of p^ with the formula for its standard error. The resulting interval is sometimes called a one-sample z interval for a population proportion.

One-Sample z Interval for a Population Proportion When the conditions are met, a C% confidence interval for the unknown proportion p is p^ ± z* Å

p^ (1 − p^) n

where z* is the critical value for the standard Normal curve with C% of its area between −z* and z*. Now we can get the desired confidence interval for Mr. Vignolini’s class.

EXAMPLE

The Beads Calculating a confidence interval for p Problem:  Mr. Vignolini’s class took an SRS of beads from the container and got 107 red beads and 144 white beads. (a) Calculate and interpret a 90% confidence interval for p. (b) Mr. Vignolini claims that exactly half of the beads in the container are red. Use your result from part (a) to comment on this claim. Solution:  We checked conditions for calculating the interval earlier. (a) Our confidence interval has the form statistic ± (critical value) · (standard deviation of statistic) p^ (1 − p^) = p^ ± z* n Å

Standard Normal curve

Area = 0.90

Area = 0.05

-z*= -1.645

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0

Area = 0.05

z*= 1.645

The sample statistic is p^ = 107∙251 = 0.426. Now let’s find the critical value. From Table A, we look for the point with area 0.05 to its left. As the excerpt from Table A shows, this point is between z = −1.64 and z = −1.65. The calculator’s invNorm(area:0.05,m:0, s:1) gives z = −1.645. So we use z* = 1.645 as our critical value. z

.03

.04

.05

−1.7

.0418

.0409

.0401

−1.6

.0516

.0505

.0495

−1.5

.0630

.0618

.0606

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The resulting 90% confidence interval is = 0.426±1.645 = 0.426±0.051 = (0.375, 0.477)

(0.426)(1 − 0.426) Å 251

We are 90% confident that the interval from 0.375 to 0.477 captures the true proportion of red beads in Mr. Vignolini’s container. (b) The confidence interval in part (a) gives a set of plausible values for the population proportion of red beads. Because 0.5 is not contained in the interval, it is not a plausible value for p. We have good reason to doubt Mr. Vignolini’s claim. For Practice Try Exercise

35

Check Your Understanding

Alcohol abuse has been described by college presidents as the number one problem on campus, and it is an important cause of death in young adults. How common is it? A survey of 10,904 randomly selected U.S. college students collected information on drinking behavior and alcohol-related problems.9 The researchers defined “frequent binge drinking” as having five or more drinks in a row three or more times in the past two weeks. According to this definition, 2486 students were classified as frequent binge drinkers. 1. Identify the parameter of interest. 2. Check conditions for constructing a confidence interval for the parameter. 3. Find the critical value for a 99% confidence interval. Show your method. Then calculate the interval. 4. Interpret the interval in context.

Putting It All Together: The Four-Step Process Taken together, the examples about Mr. Vignolini’s class and the beads Activity show you how to get a confidence interval for an unknown population proportion p. We can use the familiar four-step process whenever a problem asks us to construct and interpret a confidence interval.

Confidence Intervals: A Four-Step Process STEP

4

State: What parameter do you want to estimate, and at what confidence level? Plan:  Identify the appropriate inference method. Check conditions. Do:  If the conditions are met, perform calculations. Conclude:  Interpret your interval in the context of the problem.

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The next example illustrates the four-step process in action.

EXAMPLE

Teens Say Sex Can Wait

STEP

Confidence interval for p

4

The Gallup Youth Survey asked a random sample of 439 U.S. teens aged 13 to 17 whether they thought young people should wait to have sex until marriage.10 Of the sample, 246 said “Yes.” Construct and interpret a 95% confidence interval for the proportion of all teens who would say “Yes” if asked this question.

State:  We want to estimate the true proportion p of all 13- to 17-year-olds in the United States who would say that young people should wait to have sex until they get married with 95% confidence. Plan:  We should use a one-sample z interval for p if the conditions are met. •  Random:  Gallup surveyed a random sample of U.S. teens. ºº 10%:  Because Gallup is sampling without replacement, we need to check the 10% condition: there are at least 10(439) = 4390 U.S. teens aged 13 to 17. •  Large Counts:  We check the counts of “successes” and “failures”: np^ = 246 ≥ 10  and  n (1 − p^ ) = 193 ≥ 10 ®

AP EXAM TIP ​If a freeresponse question asks you to construct and interpret a confidence interval, you are expected to do the entire four-step process. That includes clearly defining the parameter, identifying the procedure, and checking conditions.

Do:  The sample statistic is p^ = 246/439 = 0.56. A 95% confidence interval for p is given by p^ ± z* Å

p^ (1 − p^) n

(0.56)(0.44) Å 439 = 0.56±0.046 = 0.56±1.96

= (0.514, 0.606)

Conclude:  We are 95% confident that the interval from 0.514 to 0.606 captures the true proportion of 13- to 17-year-olds in the United States who would say that teens should wait until marriage to have sex. For Practice Try Exercise

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c

Remember that the margin of error in a confidence interval includes only autio sampling variability! There are other sources of error that are not taken into account. As is the case with many surveys, we are forced to assume that the teens answered truthfully. If they didn’t, then our estimate may be biased. Other problems like nonresponse and question wording can also affect the results of this or any other poll. Lesson: Sampling beads is much easier than sampling people! Your calculator will handle the “Do” part of the four-step process, as the following Technology Corner illustrates.

!

n

Simulation studies have shown that a variation of our method for calculating a 95% confidence interval for p can result in closer to a 95% capture rate in the long run, especially for small sample sizes. This simple adjustment, first suggested by Edwin Bidwell Wilson in 1927, is sometimes called the “plus four” estimate. Just pretend we have four additional observations, two of which are successes and two of which are failures. Then calculate the “plus four interval” using the plus four estimate in place of p^ in our usual formula.

39

AP® EXAM TIP ​You may use your calculator to compute a confidence interval on the AP® exam. But there’s a risk involved. If you just give the calculator answer with no work, you’ll get either full credit for the “Do” step (if the interval is correct) or no credit (if it’s wrong). If you opt for the calculator-only method, be sure to name the procedure (e.g., one-proportion z interval) and to give the interval (e.g., 0.514 to 0.607).

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15. TECHNOLOGY CORNER

501

Confidence interval for a population proportion TI-Nspire instructions in Appendix B; HP Prime instructions on the book’s Web site.

The TI-83/84 and TI-89 can be used to construct a confidence interval for an unknown population proportion. We’ll demonstrate using the previous ­example. Of n = 439 teens surveyed, X = 246 said they thought that young people should wait to have sex until after marriage. To construct a confidence interval: TI-83/84 TI-89 • In the Statistics/List Editor, press 2nd F2 ([F7]) and choose 1-PropZInt.

• Press STAT , then choose TESTS and 1-PropZInt.

• When the 1-PropZInt screen appears, enter x = 246, n = 439, and confidence level 0.95.





• Highlight “Calculate” and press ENTER . The 95% confidence interval for p is reported, along with the sample proportion p^ and the sample size, as shown here.





Choosing the Sample Size In planning a study, we may want to choose a sample size that allows us to estimate a population proportion within a given margin of error. National survey organizations like the Gallup Poll typically sample between 1000 and 1500 American adults, who are interviewed by telephone. Why do they choose such sample sizes? The margin of error (ME) in the confidence interval for p is ME = z*

p^ (1 − p^) n Å

Here, z* is the standard Normal critical value for the confidence level we want. Because the margin of error involves the sample proportion of successes p^ , we have to guess the value of p^ when choosing n. Here are two ways to do this: 1. Use a guess for p^ based on a pilot study or on past experience with similar studies. You should do several calculations that cover the range of p^ -values you might get. 2. Use p^ = 0.5 as the guess. The margin of error ME is largest when p^ = 0.5, so this guess is conservative in the sense that if we get any other p^ when we do our study, we will get a margin of error smaller than planned.

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Once you have a guess for p^ , the formula for the margin of error can be solved to give the sample size n needed.

Sample Size for Desired Margin of Error To determine the sample size n that will yield a C% confidence interval for a population proportion p with a maximum margin of error ME, solve the following inequality for n: z*

p^ (1 − p^) ≤ ME n Å

where p^ is a guessed value for the sample proportion. The margin of error will always be less than or equal to ME if you use p^ = 0.5.

Here’s an example that shows you how to determine the sample size.

EXAMPLE

Customer Satisfaction Determining sample size A company has received complaints about its customer service. The managers intend to hire a consultant to carry out a survey of customers. Before contacting the consultant, the company president wants some idea of the sample size that she will be required to pay for. One critical question is the degree of satisfaction with the company’s customer service, measured on a 5-point scale. The president wants to estimate the proportion p of customers who are satisfied (that is, who choose either “somewhat satisfied” or “very satisfied,” the two highest levels on the 5-point scale). She decides that she wants the estimate to be within 3% (0.03) at a 95% confidence level. How large a sample is needed?

Problem:  Determine the sample size needed to estimate p within 0.03 with 95% confidence. Solution:  The critical value for 95% confidence is z* = 1.96. We have no idea about the true

proportion p of satisfied customers, so we decide to use p^ = 0.5 as our guess. Because the company president wants a margin of error of no more than 0.03, we need to solve the inequality 0.5(1 − 0.5) ≤ 0.03 n Å for n. Multiplying both sides by !n and then dividing both sides by 0.03 yields 1.96

Squaring both sides gives

1.96 !0.5(1 − 0.5) ≤ !n 0.03

1.96 2 b (0.5)(1 − 0.5) ≤ n 0.03 1067.111 ≤ n We round up to 1068 respondents to ensure that the margin of error is no more than 3%. a

For Practice Try Exercise

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503

Why not round to the nearest whole number—in this case, 1067? Because a smaller sample size will result in a larger margin of error, possibly more than the desired 3% for the poll. If you want a 2.5% margin of error rather than 3%, then n≥a

1.96 2 b (0.5)(1 − 0.5) = 1536.64 1 n = 1537 0.025

For a 2% margin of error, the sample size you need is n≥a

1.96 2 b (0.5)(1 − 0.5) = 2401 0.02

As usual, smaller margins of error call for larger samples. News reports frequently describe the results of surveys with sample sizes between 1000 and 1500 and a margin of error of about 3%. These surveys generally use sampling procedures more complicated than a simple random sample, so the calculation of confidence intervals is more involved than what we have studied in this section. The calculations of the previous example still give you a rough idea of how such surveys are planned.

CHECK YOUR UNDERSTANDING

Refer to the previous example about the company’s customer satisfaction survey. 1. In the company’s prior-year survey, 80% of customers surveyed said they were “somewhat satisfied” or “very satisfied.” Using this value as a guess for p^ , find the sample size needed for a margin of error of 3% at a 95% confidence level. 2. What if the company president demands 99% confidence instead? Determine how this would affect your answer to Question 1.

Section 8.2

Summary •

The conditions for constructing a confidence interval about a population proportion are • Random:  The data were produced by a well-designed random sample or randomized experiment. ~~ 10%: When sampling without replacement, we check that the population is at least 10 times as large as the sample. Large Counts:  The sample is large enough that np^ and n(1 − p^ ), the counts of successes and failures in the sample, are both at least 10. Confidence intervals for a population proportion p are based on the sampling distribution of the sample proportion p^ . When the conditions for inference are met, the sampling distribution of p^ is approximately Normal with mean p and standard deviation !p(1 − p)∙n. •

•

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•

In practice, we use the sample proportion p^ to estimate the unknown parameter p. We therefore replace the standard deviation of p^ with its standard error when constructing a confidence interval. The C% confidence interval for p is p^ ± z* Å

• STEP

4 •

p^ (1 − p^) n

where z* is the standard Normal critical value with C% of its area between −z* and z*. When constructing a confidence interval, follow the four-step process: STATE: What parameter do you want to estimate, and at what confidence level? PLAN:  Identify the appropriate inference method. Check conditions. DO:  If the conditions are met, perform calculations. CONCLUDE:  Interpret your interval in the context of the problem. The sample size needed to obtain a confidence interval with approximate margin of error ME for a population proportion involves solving z*

p^ (1 − p^) ≤ ME n Å

for n, where p^ is a guessed value for the sample proportion, and z* is the critical value for the confidence level you want. Use p^ = 0.5 if you don’t have a good idea about the value of p^ .

8.2 T echnology Corner TI-Nspire Instructions in Appendix B; HP Prime instructions on the book’s Web site. 15.  Confidence interval for a population proportion

Section 8.2

Exercises

For Exercises 27 to 30, check whether each of the conditions is met for calculating a confidence interval for the population proportion p. 27. Rating school food  Latoya wants to estimate what pg 494 proportion of the seniors at her boarding high school like the cafeteria food. She interviews an SRS of 50 of the 175 seniors living in the dormitory. She finds that 14 think the cafeteria food is good. 28. High tuition costs  Glenn wonders what proportion of the students at his school believe that tuition is too high. He interviews an SRS of 50 of the 2400 students

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at his college. Thirty-eight of those interviewed think tuition is too high. 29. AIDS and risk factors  In the National AIDS Behavioral Surveys sample of 2673 adult heterosexuals, 0.2% had both received a blood transfusion and had a sexual partner from a group at high risk of AIDS. We want to estimate the proportion p in the population who share these two risk factors. 30. Whelks and mussels  The small round holes you often see in sea shells were drilled by other sea creatures, who ate the former dwellers of the shells.

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Section 8.2 Estimating a Population ­Proportion Whelks often drill into mussels, but this behavior appears to be more or less common in different locations. Researchers collected whelk eggs from the coast of Oregon, raised the whelks in the laboratory, then put each whelk in a container with some delicious mussels. Only 9 of 98 whelks drilled into a mussel.11 The researchers want to estimate the proportion p of Oregon whelks that will spontaneously drill into mussels.

STEP

(a) Calculate and interpret a 95% confidence ­interval for the population proportion p that would report texting with their friends every day. (b) Is it plausible that the true proportion of American teens who text with their friends every day is 0.45? Use your result from part (a) to support your answer.

31. 98% confidence Find z* for a 98% confidence interval using Table A or your calculator. Show your method.

37. Binge drinking  Describe a possible source of error that is not included in the margin of error for the 99% confidence interval in Exercise 35.

32. 93% confidence Find z* for a 93% confidence interval using Table A or your calculator. Show your method.

(a) Identify the population and parameter of interest.

38. Teens’ texting  Describe a possible source of error that is not included in the margin of error for the 95% confidence interval in Exercise 36.

(d) Interpret the interval in context. 34. Reporting cheating  What proportion of students are willing to report cheating by other students? A student project put this question to an SRS of 172 undergraduates at a large university: “You witness two students cheating on a quiz. Do you go to the professor?” Only 19 answered “Yes.”12 (a) Identify the population and parameter of ­interest. (b) Check conditions for constructing a confidence interval for the parameter. (c) Construct a 99% confidence interval for p. Show your method. (d) Interpret the interval in context. 35. Binge drinking  In a recent National Survey of Drug Use and Health, 2312 of 5914 randomly selected full-time U.S. college students were classified as binge drinkers.13

pg 498 STEP

4

39. How common is SAT coaching?  A random sample of students who took the SAT college entrance examination twice found that 427 of the respondents had paid for coaching courses and that the remaining 2733 had not.14 Construct and interpret a 99% confidence interval for the proportion of coaching among students who retake the SAT.

pg 500

STEP

(b) Check conditions for constructing a confidence interval for the parameter. (c) Construct a 90% confidence interval for p. Show your method.

36. Teens’ texting  A Pew Internet and American Life Project survey found that 392 of 799 randomly ­selected teens reported texting with their friends every day.

4

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33. Going to the prom  Tonya wants to estimate what proportion of her school’s seniors plan to attend the prom. She interviews an SRS of 50 of the 750 seniors in her school and finds that 36 plan to go to the prom.

505

STEP

4

40. 2010 begins  In January 2010 a Gallup Poll asked a random sample of adults, “In general, are you satisfied or dissatisfied with the way things are going in the United States at this time?” In all, 256 said that they were satisfied and the remaining 769 said they were not. Construct and interpret a 90% confidence interval for the proportion of adults who are satisfied with how things are going.

4

41. Equality for women?  Have efforts to promote equality for women gone far enough in the United States? A poll on this issue by the cable network MSNBC contacted 1019 adults. A newspaper article about the poll said, “Results have a margin of sampling error of plus or minus 3 percentage points.”15 (a) The news article said that 65% of men, but only 43% of women, think that efforts to promote equality have gone far enough. Explain why we do not have enough information to give confidence intervals for men and women separately.

(a) Calculate and interpret a 99% confidence ­interval for the population proportion p that are binge drinkers.

(b) Would a 95% confidence interval for women alone have a margin of error less than 0.03, about equal to 0.03, or greater than 0.03? Why? (You see that the news article’s statement about the margin of error for poll results is a bit misleading.)

(b) A newspaper article claims that 45% of full-time U.S. ­college students are binge drinkers. Use your result from part (a) to comment on this claim.

42. A TV poll  A television news program conducts a call-in poll about a proposed city ban on handgun ownership. Of the 2372 calls, 1921 oppose the ban.

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The station, following recommended practice, makes a confidence statement: “81% of the Channel 13 Pulse Poll sample opposed the ban. We can be 95% confident that the true proportion of citizens opposing a handgun ban is within 1.6% of the sample result.” Is the station’s conclusion justified? Explain. 43. Can you taste PTC?  PTC is a substance that has a strong bitter taste for some people and is tasteless for others. The ability to taste PTC is inherited. About 75% of Italians can taste PTC, for example. You want to estimate the proportion of Americans who have at least one Italian grandparent and who can taste PTC.

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(a) How large a sample must you test to estimate the proportion of PTC tasters within 0.04 with 90% confidence? Answer this question using the 75% estimate as the guessed value for p^ . (b) Answer the question in part (a) again, but this time use the conservative guess p^ = 0.5. By how much do the two sample sizes differ? 44. School vouchers  A national opinion poll found that 44% of all American adults agree that parents should be given vouchers that are good for education at any public or private school of their choice. The result was based on a small sample. (a) How large an SRS is required to obtain a margin of error of 0.03 (that is, ±3%) in a 99% confidence ­interval? Answer this question using the previous poll’s result as the guessed value for p^ . (b) Answer the question in part (a) again, but this time use the conservative guess p^ = 0.5. By how much do the two sample sizes differ? 45. Election polling  Gloria Chavez and Ronald Flynn are the candidates for mayor in a large city. We want to estimate the proportion p of all registered voters in the city who plan to vote for Chavez with 95% confidence and a margin of error no greater than 0.03. How large a random sample do we need? Show your work. 46. Starting a nightclub  A college student organization wants to start a nightclub for students under the age of 21. To assess support for this proposal, they will select an SRS of students and ask each respondent if he or she would patronize this type of establishment. What sample size is required to obtain a 90% confidence interval with an approximate margin of error of 0.04? Show your work. 47. Teens and their TV sets  According to a Gallup Poll report, 64% of teens aged 13 to 17 have TVs in their rooms. Here is part of the footnote to this report:

These results are based on telephone interviews with a randomly selected national sample of 1028 teenagers

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in the Gallup Poll Panel of households, aged 13 to 17. For results based on this sample, one can say . . . that the maximum error attributable to sampling and other random effects is ±3 percentage points. In addition to sampling error, question wording and practical difficulties in conducting surveys can introduce error or bias into the findings of public opinion polls.16 (a) We omitted the confidence level from the ­footnote. Use what you have learned to determine the confidence level, assuming that Gallup took an SRS. (b) Give an example of a “practical difficulty” that could lead to biased results for this survey. 48. Gambling and the NCAA  Gambling is an issue of great concern to those involved in college athletics. Because of this concern, the National Collegiate Athletic Association (NCAA) surveyed randomly selected student athletes concerning their gamblingrelated behaviors.17 Of the 5594 Division I male athletes in the survey, 3547 reported participation in some gambling behavior. This includes playing cards, betting on games of skill, buying lottery tickets, betting on sports, and similar activities. A report of this study cited a 1% margin of error. (a) The confidence level was not stated in the report. Use what you have learned to find the confidence level, assuming that the NCAA took an SRS. (b) The study was designed to protect the anonymity of the student athletes who responded. As a result, it was not possible to calculate the number of students who were asked to respond but did not. How does this fact affect the way that you interpret the results? Multiple choice: Select the best answer for Exercises 49 to 52. 49. A Gallup Poll found that only 28% of American adults expect to inherit money or valuable possessions from a relative. The poll’s margin of error was ±3 percentage points at a 95% confidence level. This means that (a) the poll used a method that gets an answer within 3% of the truth about the population 95% of the time. (b) the percent of all adults who expect an inheritance is between 25% and 31%. (c) if Gallup takes another poll on this issue, the results of the second poll will lie between 25% and 31%. (d) there’s a 95% chance that the percent of all adults who expect an inheritance is between 25% and 31%. (e) Gallup can be 95% confident that between 25% and 31% of the sample expect an inheritance.

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Section 8.3 Estimating a Population Mean 50. Most people can roll their tongues, but many can’t. The ability to roll the tongue is genetically determined. Suppose we are interested in determining what proportion of students can roll their tongues. We test a simple random sample of 400 students and find that 317 can roll their tongues. The margin of error for a 95% confidence interval for the true proportion of tongue rollers among students is closest to (a) 0.0008.

(c) 0.03.

(b) 0.02.

(d) 0.04.

(e) 0.05.

51. You want to design a study to estimate the proportion of students at your school who agree with the statement, “The student government is an effective organization for expressing the needs of students to the administration.” You will use a 95% confidence interval, and you would like the margin of error to be 0.05 or less. The minimum sample size required is (a) 22. (b) 271. (c) 385. (d) 769. (e) 1795. 52. A newspaper reporter asked an SRS of 100 residents in a large city for their opinion about the mayor’s job performance. Using the results from the sample, the C% confidence interval for the proportion of all residents in the city who approve of the mayor’s job performance is 0.565 to 0.695. What is the value of C? (a) 82

(b) 86 (c) 90 (d) 95 (e) 99

Exercises 53 and 54 refer to the following setting. The following table displays the number of accidents at a factory during each hour of a 24-hour shift (1 = 1:00 a.m.).

8.3 What You Will Learn

1 2 3 4 5 6 7 8 9 10 11 12

Number of accidents

Hour

Number of accidents

5 8 17 31 24 18 12 7 1 0 2 14

13 14 15 16 17 18 19 20 21 22 23 24

21 12 10 1 0 1 3 21 23 18 11 2

53. Accidents happen (1.2, 3.1) (a) Construct a plot that displays the distribution of the number of accidents effectively. (b) Construct a plot that shows the relationship between the number of accidents and the time when they ­occurred. (c) Describe something that the plot in part (a) tells you about the data that the plot in part (b) does not. (d) Describe something that the plot in part (b) tells you about the data that the plot in part (a) does not. 54. Accidents happen (1.3)  Plant managers are concerned that the number of accidents may be significantly higher during the midnight to 8:00 a.m. shift than during the 4:00 p.m. to midnight shift. What would you tell them? Give appropriate statistical evidence to support your conclusion.

Estimating a Population Mean By the end of the section, you should be able to:

• State and check the Random, 10%, and Normal/ Large Sample conditions for constructing a confidence ­interval for a population mean. • Explain how the t distributions are different from the standard Normal distribution and why it is necessary to use a t distribution when calculating a confidence interval for a population mean.

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Hour

• Determine critical values for calculating a C% confidence interval for a population mean using a table or technology. • Construct and interpret a confidence interval for a population mean. • Determine the sample size required to obtain a C% confidence interval for a population mean with a ­specified margin of error.

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Inference about a population proportion usually arises when we study categorical variables. We learned how to construct and interpret confidence intervals for a population proportion p in Section 8.2. To estimate a population mean, we have to record values of a quantitative variable for a sample of individuals. It makes sense to try to estimate the mean amount of sleep that students at a large high school got last night but not their mean eye color! In this section, we’ll examine confidence intervals for a population mean m.

The Problem of Unknown s Mr. Schiel’s class did the mystery mean Activity (page 476) and got a value of x– = 240.80 from an SRS of size 16, as shown. Their task was to estimate the unknown population mean m. They knew that the population distribution was Normal and that its standard deviation was s = 20. Their estimate was based on the sampling distribution of x– . Figure 8.10 shows this Normal sampling d ­ istribution once again. Sampling distribution of x

σ =5 − n

FIGURE 8.10  ​The Normal sampling distribution of x– for the mystery mean Activity.

µ (unknown)

Values of x

To calculate a 95% confidence interval for m, we use our familiar formula: Starnes/Yates/Moore: The Practice of Statistics, 4E New Fig.: 08_10 Fig.: 8037 statistic ± (criticalPerm. value) · (standard deviation First Pass: 2010-06-28 2nd Pass: 2010-07-12

of statistic)

The critical value, z* = 1.96, tells us how many standardized units we need to go out to catch the middle 95% of the sampling distribution. Our interval is s 20 = 240.80 ± 1.96 # = 240.80 ± 9.80 = (231.00, 250.60) x– ± z* # !n !16

We call such an interval a one-sample z interval for a population mean. This method isn’t very useful in practice, however. In most real-world settings, if we don’t know the population mean m, then we don’t know the population standard deviation s either. How do we estimate m when the population standard deviation s is unknown? Our best guess for the value of s is the sample standard deviation sx. Maybe we could use the one-sample z interval for a population mean with sx in place of s: x– ± z* Let’s try it.

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sx !n

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ACTIVITY Materials: TI-83/84 for each student

509

Calculator BINGO! A farmer wants to estimate the mean weight (in grams) of all tomatoes grown on his farm. To do so, he will select a random sample of 4 tomatoes, calculate the mean weight (in grams), and use the sample mean x– to create a 99% confidence interval for the population mean m. Suppose that the weights of tomatoes on his farm are approximately Normally distributed with a mean of 100 grams and a standard deviation of 40 grams. 1.  Use your calculator to simulate taking an SRS of size 4 from this s = population and creating a one-sample z interval for m: x– ± z* !n 40 . Enter the command shown below and press ENTER. x– ± 2.576 !4 randNorm(100,40,4)→L1:ZInterval 40,mean(L1),4,99 Check to see whether the resulting interval captures m = 100. If it does not, shout “BINGO!”

To get the randNorm command, press MATH and arrow to PRB. The ZInterval command is in the Catalog. To get the mean and stdDev commands, press 2nd STAT (LIST) and arrow to the MATH menu.

Keep pressing ENTER to generate more 99% confidence intervals. Check each interval to see whether it captures m = 100. If it does not, shout “BINGO!” If this method of constructing confidence intervals is working properly, about what percent of the time should you get a BINGO? Does the method seem to be working? The method in Step 1 works well if we know the population standard deviation s. That’s rarely the case in real life. What happens if we use the sample standard deviation sx in place of s when calculating a confidence interval for the population mean? 2.  Use your calculator to simulate taking an SRS of size 4 from this population and sx sx = x– ± 2.576 . creating a “modified” one-sample z interval for m: x– ± z* !n !4 Enter the command shown below and press ENTER. randNorm(100,40,4)→L1:ZInterval stdDev(L1),mean(L1),4,99 Check to see whether the resulting interval captures m = 100. If it does not, shout “BINGO!” Keep pressing ENTER to generate more 99% confidence intervals. Check each interval to see whether it captures m = 100. If it does not, shout “BINGO!” If this method of constructing confidence intervals is working properly, about what percent of the time should you get a BINGO? Does the method seem to be working?

The figure on the next page shows the results of using an applet from www. rossmanchance.com to repeatedly construct confidence intervals as described in Step 2 of the Activity. Of the 1000 intervals constructed, only 923 (that’s 92.3%) captured the population mean. That’s far below our desired 99% confidence level. What went wrong? The intervals that missed (those in red) came from samples with small standard deviations sx and from samples in which x– was far from the population mean m. In those cases, multiplying sx∙!4 by z* = 2.576 didn’t produce long enough intervals to reach m = 100. We need to multiply by a larger critical value to achieve a 99% capture rate. But what critical value should we use?

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When s Is Unknown: The t Distributions

When the sampling distribution of x– is close to Normal, we can find probabilities involving x– by standardizing: Recall that a statistic is a number computed from sample data. We know that the sample mean x– is a statistic. So is the standardized value x– − m z= . The sampling distribution s∙!n of z shows the values this statistic takes in all possible SRSs of size n from the population.

z=

x– − m s∙!n

Recall that the sampling distribution of x– has mean m and standard deviation s∙!n, as shown in Figure 8.11(a). What are the shape, center, and spread of the sampling distribution of the new statistic z? From what we learned in Chapter 6, subtracting the constant m from the values of the random variable x– shifts the distribution left by m units, making the mean 0. This transformation doesn’t affect the shape or spread of the distribution. Dividing Sampling distribution of x

σ − n

Standard Normal curve z=

µ (unknown)

x−µ

σ − n

σ=1

Values of x

µ=0

(a)

(b)

FIGURE 8.11  ​(a) Sampling distribution of x– when the Normal/Large Sample condition is met. (b) Standardized values of x– lead to the statistic z, which follows the standard Normal distribution. Starnes/Yates/Moore: The Practice of Statistics, 4E New Fig.: 08_11 Perm. Fig.: 8039 First Pass: 2010-06-28 2nd Pass: 2010-07-12 Starnes-Yates5e_c08_474-535hr2.indd 510

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511

by the constant s∙!n keeps the mean at 0, makes the standard deviation 1, and leaves the shape unchanged. As shown in Figure 8.11(b), z has the standard Normal distribution N(0, 1). Therefore, we can use Table A or a calculator to find the related probability involving z. That’s how we have gotten the critical values for our confidence intervals so far. When we don’t know s, we estimate it using the sample standard deviation sx. What happens now when we standardize? ?? =

x– − m sx∙!n

To find out, let’s start with a Normal population having mean m = 100 and standard deviation s = 5. We’ll simulate choosing an SRS of size n = 4 and calculating the sample mean x– . Then we will standardize the result in two ways: z=

x– − 100 x– − 100   and  ?? = 5∙!4 sx∙!4

Figure 8.12 shows the results of taking 500 SRSs of size n = 4 and standardizing the value of the sample mean x– in both ways. The values of z follow a standard Normal distribution, as expected. The standardized values we get, using the sample standard deviation sx in place of the population standard deviation s, show much greater spread. In fact, in a few samples, the statistic ?? =

x– − m sx∙!n

took values below −6 or above 6. This statistic has a distribution that is new to us, called a t distribution. It has a different shape than the standard Normal curve: still symmetric with a single peak at 0, but with much more area in the tails.

FIGURE 8.12  ​Fathom simulation showing standardized values of the sample mean x– in 500 SRSs. The statistic z follows a standard Normal distribution. Replacing s with sx yields a statistic with much greater variability that doesn’t follow the standard Normal curve. The Normal probability plot for the t statistic shows the departure from Normality in the tails of the t distribution.

The statistic t has the same interpretation as any standardized statistic: it says how far x– is from its mean m in standard deviation units. There is a different t distribution for each sample size. We specify a particular t distribution by giving its degrees of freedom (df). When we perform inference about a population mean m using a t distribution, the appropriate degrees of freedom are found by subtracting 1 from the sample size n, making df = n − 1. We will write the t distribution with n − 1 degrees of freedom as tn−1 for short.

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The t distribution and the t inference procedures were invented by William S. Gosset (1876–1937). Gosset worked for the Guinness brewery, and his goal in life was to make better beer. He used his new t procedures to find the best varieties of barley and hops. Gosset’s statistical work helped him become head brewer. ­Because Gosset published under the pen name “Student,” you will often see the t distribution called “Student’s t ” in his honor.

The t Distributions; Degrees of Freedom Draw an SRS of size n from a large population that has a Normal distribution with mean m and standard deviation s. The statistic t=

x– − m sx∙!n

has the t distribution with degrees of freedom df = n − 1. When the population distribution isn’t Normal, this statistic will have approximately a tn−1 distribution if the sample size is large enough. Figure 8.13 compares the density curves of the standard Normal distribution and the t distributions with 2 and 9 degrees of freedom. The figure illustrates these facts about the t distributions: t distributions have more area in the tails than the standard Normal distribution.

t, 2 degrees of freedom t, 9 degrees of freedom Standard Normal

FIGURE 8.13  ​Density curves for the t distributions with 2 and 9 degrees of freedom and the standard Normal distribution. All are symmetric with center 0. The t ­distributions are somewhat more spread out.

0

• The density curves of the t distributions are similar in shape to the standard Starnes/Yates/Moore: The Practice of Statistics, 4E Normal curve. They are symmetric about 0, single-peaked, and bell-shaped. New Fig.: 08_13 Perm. Fig.: 8046 • The spread ofFirst thePass: t distributions is a bit greater than that of the standard Nor2010-06-28 mal distribution. The t distributions in Figure 8.13 have more probability in the tails and less in the center than does the standard Normal. This is true because substituting the estimate sx for the fixed parameter s introduces more variation into the statistic. • As the degrees of freedom increase, the t density curve approaches the standard Normal density curve ever more closely. This happens because sx estimates s more accurately as the sample size increases. So using sx in place of s causes little extra variation when the sample is large. Table B in the back of the book gives critical values t* for the t distributions. Each row in the table contains critical values for the t distribution whose degrees of freedom appear at the left of the row. For convenience, several of the more common confidence levels C are given at the bottom of the table. By looking down any column, you can check that the t critical values approach the Normal critical values z* as the degrees of freedom increase. When you use Table B to determine the correct value of t* for a given confidence interval, all you need to know are the confidence level C and the degrees

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of freedom (df). Unfortunately, Table B does not include every possible sample size. When the actual df does not appear in the table, use the greatest df available that is less than your desired df. This guarantees a wider confidence interval than you need to justify a given confidence level. Better yet, use technology to find an accurate value of t* for any df.

EXAMPLE

Finding t* Critical Values Using Table B Problem:  What critical value t* from Table B should be used in constructing a confidence i­ nterval for the population mean in each of the following settings? (a) A 95% confidence interval based on an SRS of size n = 12. (b) A 90% confidence interval from a random sample of 48 observations. Solution: (a) In Table B, we consult the row corresponding to df = 12 − 1 = 11. We move across that row to the entry that is directly above 95% confidence level on the bottom of the chart. The desired critical value is t * = 2.201. (b) With 48 observations, we want to find the t* critical value for df = 48 − 1 = 47 and 90% ­confidence. There is no df = 47 row in Table B, so we use the more conservative df = 40. The ­corresponding critical value is t* = 1.684.

The bottom row of Table B gives z * critical values and is labeled as df = `. That’s because the t distributions approach the standard Normal distribution as the degrees of freedom approach infinity.

Upper-tail probability p

Upper-tail probability p df

.05

.025

.02

.01

df

.10

.05

.025

.02

10

1.812

2.228

2.359

2.764

30

1.310

1.697

2.042

2.147

11

1.796

2.201

2.328

2.718

40

1.303

1.684

2.021

2.123

12

1.782

2.179

2.303

2.681

50

1.299

1.676

2.009

2.109

∞

1.645

1.960

2.054

2.326

∞

1.282

1.645

1.960

2.054

90%

95%

96%

98%

80%

90%

95%

96%

Confidence level C

Confidence level C

For Practice Try Exercise

55

For part (a) of the example, the corresponding standard Normal critical value for 95% confidence is z* = 1.96. We have to go out farther than 1.96 standard deviations to capture the central 95% of the t distribution with 11 degrees of freedom. Technology will quickly produce t* critical values for any sample size.

16. TECHNOLOGY CORNER

Inverse t on the calculator TI-Nspire instructions in Appendix B; HP Prime instructions on the book’s Web site.

Most newer TI-84 and TI-89 calculators allow you to find critical values t* using the inverse t command. As with the calculator’s inverse Normal command, you have to enter the area to the left of the desired critical value. Let’s use the inverse t command to find the critical values in parts (a) and (b) of the example.

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TI-83/84 • Press 2nd VARS (DISTR) and choose invT(. • For part (a), OS 2.55 or later: In the dialog box, enter these values: area:.025, df:11, choose Paste, and then press ENTER . Older OS: Complete the command invT(.025,11)and press ENTER . • For part (b), use the command invT(.05,47).

•

TI-89 In the Statistics/List Editor, press F5 , choose Inverse and Inverse t....

•

For part (a), enter Area: .025 and Deg of Freedom, df: 11, and then press ENTER .

•

For part (b), use Area:.05 and df: 47.

  Note that the t critical values are t* = 2.201 and t* = 1.678, respectively.

CHECK YOUR UNDERSTANDING

Use Table B to find the critical value t* that you would use for a confidence interval for a population mean m in each of the following settings. If possible, check your answer with technology. 1. A 96% confidence interval based on a random sample of 22 observations. 2. A 99% confidence interval from an SRS of 71 observations.

Conditions for Estimating m As with proportions, you should check some important conditions before constructing a confidence interval for a population mean. Two of the conditions should be familiar by now.The Random condition is crucial for doing inference. If the data don’t come from a well-designed random sample or randomized experiment, you can’t draw conclusions about a larger population or about cause and effect. When sampling without replacement, the 10% condition ensures that our formula for the standard deviation of the statistic x– sx– =

s !n

is approximately correct. The method we use to construct a confidence interval for μ depends on the fact that the sampling distribution of x– is approximately Normal. From Chapter 7, we know that the sampling distribution of x– is Normal if the population distribution is Normal. When the population distribution is not Normal, the central limit theorem tells us that the sampling distribution of x– will be approximately Normal if the sample size is large enough (n ≥ 30). Be sure to check this Normal/Large Sample condition before calculating a confidence interval.

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Conditions for Constructing a Confidence Interval About a Mean • Random: The data come from a well-designed random sample or ­randomized experiment. 1 ~~ 10%:  When sampling without replacement, check that n ≤ N. 10 • Normal/Large Sample: The population has a Normal distribution or the sample size is large (n ≥ 30). If the population distribution has unknown shape and n < 30, use a graph of the sample data to assess the Normality of the population. Do not use t procedures if the graph shows strong skewness or outliers.

Larger samples improve the accuracy of critical values from the t distributions when the population is not Normal. This is true for two reasons: 1. The sampling distribution of x– for large sample sizes is close to Normal. 2. As the sample size n grows, the sample standard deviation sx will give a more sx s accurate estimate of s. This is important because we use in place of !n !n when doing calculations.

The Normal/Large Sample condition is obviously met if we know that the population distribution is Normal or that the sample size is at least 30. What if we don’t know the shape of the population distribution and n < 30? In that case, we have to graph the sample data. Our goal is to answer the question, “Is it reasonable to believe that these data came from a Normal population?” How should graphs of data from small samples look if the population has a Normal distribution? The following Activity sheds some light on this question.

ACTIVITY Materials: TI-83/84 or TI-89 for each student

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Sampling from a Normal Population Let’s use the calculator to simulate choosing random samples of size n = 20 from a Normal distribution with m = 100 and s = 15 and then to plot the data. 1.  Choose an SRS of 20 observations from this Normal population. TI-83/84: Press MATH , arrow to PRB and choose randNorm(. Complete the command randNorm(100,15,20)nL1 and press ENTER . TI-89: Press CATALOG F3 (Flash Apps), press alpha 2 (R) to jump to the r’s, and choose randNorm(… Complete the command tistat.randNorm(100,15,20)nlist1 and press ENTER . 2.  Make a histogram, a boxplot, and a Normal probability plot of the data in L1/list1. Do you see any obvious departures from Normality in the graphs of the sample data?

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    3.  Repeat Steps 1 and 2 several times. Do the graphs of the sample data always look approximately Normal when the population distribution is Normal? 4.  Compare the results with those of your classmates. How easy do you think it will be to use a graph of sample data to determine whether or not a population has a Normal distribution?

Did you expect that a random sample from a Normal population would yield a graph that looked Normal? Unfortunately, that’s usually not the case. The figure below shows boxplots from three different SRSs of size 20 chosen in Step 3 of the Activity. The left-hand graph is skewed to the right. The right-hand graph shows three outliers in the sample. Only the middle graph looks symmetric and has no outliers.

 

 

c

EXAMPLE

!

n

As the Activity shows, it is very difficult to use a graph of sample data to assess the Normality of a population distribution. If the graph has a skewed shape or if there are outliers present, it could be because the population distribution isn’t Normal. Skewness or outliers could also occur naturally in a ranautio dom sample from a Normal population. To be safe, you should only use a t distribution for small samples with no outliers or strong skewness. What constitutes strong skewness in a distribution? The following example gives you some idea.

GPAs, Wood, and SATs Can we use t? PROBLEM:  Determine if we can safely use a t* critical value to calculate a confidence interval for the population mean in each of the following settings. (a) To estimate the average GPA of students at your school, you randomly select 50 students from classes you take. Figure 8.14(a) is a histogram of their GPAs. (b) How much force does it take to pull wood apart? Figure 8.14(b) shows a stemplot of the force (in pounds) required to pull apart a random sample of 20 pieces of Douglas fir.

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517

(c) Suppose you want to estimate the mean SAT Math score at a large high school. Figure 8.14(c) is a boxplot of the SAT Math scores for a random sample of 20 students at the school.

(a)

(b)

23 24 25 26 27 28 29 30 31 32 33

0 0 5 Key: 31/3 = 313 pounds 7 259 399 033677 0236

(c)

FIGURE 8.14  Can we use a t distribution for these data? (a) GPAs of 50 randomly selected students in your classes at school. (b) Force required to pull apart a random sample of 20 pieces of Douglas fir. (c) SAT Math scores for a random sample of 20 students at a large high school.

SOLUTION: (a) No. Although the histogram is roughly symmetric with no outliers, the random sample of 50 students was only from your classes and not from all students at your school. So we should not use these data to calculate a confidence interval for the mean GPA of all students at the school. (b) No. The graph is strongly skewed to the left with possible low outliers. We cannot trust a critical value from a t distribution with df = 19 in this case. (c) Yes. The distribution is only moderately skewed to the right and there are no outliers present. AP® EXAM TIP  If a question on the AP® exam asks you to calculate a confidence interval, all the conditions should be met. However, you are still required to state the conditions and show evidence that they are met.

For Practice Try Exercise

THINK ABOUT IT

59

What’s the difference between “strongly skewed” and “moderately skewed”?  Look at the stemplot in Figure 8.14(b) and the boxplot

in Figure 8.14(c). Compare the distance from the maximum to the median and from the median to the minimum in both graphs. In Figure 8.14(b), maximum – median = 336 – 319.5 = 16.5 and median – minimum = 319.5 – 230 = 89.5. The half of the stemplot with smaller values is more than five times as long as the half of the stemplot with larger values. In Figure 8.14(c), maximum – median ≈ 775 – 525 = 250 and median – minimum ≈ 525 – 375 = 150. The right half of the boxplot is not quite twice as long as the left half.

There is no accepted rule of thumb for identifying strong skewness. For that reason, we have chosen the data sets in examples and exercises to avoid borderline cases. You should be able to tell easily if strong skewness is present in a graph of data from a small sample.

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Constructing a Confidence Interval for m When the conditions are met, the sampling distribution of x– has roughly a Normal distribution with mean m and standard deviation s∙!n. Because we don’t know s, we estimate it by the sample standard deviation sx. We then estimate the sx . This value is standard deviation of the sampling distribution with SEx– = !n called the standard error of the sample mean x– , or just the standard error of the mean.

As with proportions, some books refer to the standard deviation of the sampling distribution of x– as the “standard error” and what we call the standard error of the mean as the “estimated standard error.” The standard error of the mean is often abbreviated SEM.

Definition:  Standard error of the sample mean sx The standard error of the sample mean x– is SE x– = , where sx is the sample !n – standard deviation. It describes how far x will typically be from m in repeated SRSs of size n.

To construct a confidence interval for m, replace the standard deviation s∙!n of x– by its standard error sx∙!n in the formula for the one-sample z interval for a population mean. Use critical values from the t distribution with n – 1 degrees of freedom in place of the z critical values. That is, statistic ± (critical value) # (standard deviation of statistic) sx = x– ± t* !n

This one-sample t interval for a population mean is similar in both reasoning and computational detail to the one-sample z interval for a population proportion of Section 8.2. So we will now pay more attention to questions about using these methods in practice.

The One-Sample t Interval for a Population Mean When the conditions are met, a C% confidence interval for the unknown mean m is sx x– ± t* !n

where t* is the critical value for the tn –1 distribution, with C% of the area between –t* and t*.

The following example shows you how to construct a confidence interval for a population mean when s is unknown. By now, you should recognize the four-step process.

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EXAMPLE

Auto Pollution

STEP

A one-sample t interval for m

4

Environmentalists, government officials, and vehicle manufacturers are all interested in studying the auto exhaust emissions produced by motor vehicles. The major pollutants in auto exhaust from gasoline engines are hydrocarbons, carbon monoxide, and nitrogen oxides (NOX). Researchers collected data on the NOX levels (in grams/mile) for a random sample of 40 light-duty engines of the same type. The mean NOX reading was 1.2675 and the standard deviation was 0.3332.18

Problem: (a) Construct and interpret a 95% confidence interval for the mean amount of NOX emitted by lightduty engines of this type. (b) The Environmental Protection Agency (EPA) sets a limit of 1.0 gram/mile for average NOX emissions. Are you convinced that this type of engine violates the EPA limit? Use your interval from (a) to support your answer.

SOLUTION: (a)  STATE: We want to estimate the true mean amount m of NOX emitted by all light-duty engines of this type at a 95% confidence level. PLAN:  We should construct a one-sample t interval for m if the conditions are met.

•  Random:  The data come from a random sample of 40 light-duty engines of this type. ºº 10%:  We are sampling without replacement, so we need to assume that there are at least 10(40) = 400 light-duty engines of this type. •  Normal/Large Sample:  We don’t know whether the population distribution of NOX emissions is Normal. Because the sample size is large (n = 40 ≥ 30), we should be safe using a t distribution.

Do:  The formula for the one-sample t interval is

sx !n From the information given, x– = 1.2675 g/mi and sx = 0.3332 g/mi. To find the critical value t*, we use the t distribution with df = 40 – 1 = 39. Unfortunately, there is no row corresponding to 39 degrees of freedom in Table B. We can’t pretend we have a larger sample size than we actually do, so we use the more conservative df = 30. At a 95% confidence level, the critical value is t* = 2.042. So the 95% confidence interval for m is x– ± t*

Upper-tail probability p df

.05

.025

.02

29

1.699

2.045

2.150

30

1.697

2.042

2.147

40

1.684

2.021

2.123

90%

95%

96%

Confidence level C

x– ± t*

sx 0.3332 = 1.2675 ± 2.042 = 1.2675 ± 0.1076 = (1.1599, 1.3751) !n !40

Using technology: The command invT(.025,39) gives t = –2.023. Using the critical value t* = 2.023 for the 95% confidence interval gives x– ± t*

sx 0.3332 = 1.2675 ± 2.023 = 1.2675 ± 0.1066 = (1.1609, 1.3741) !n !40

This interval is slightly narrower than the one found using Table B. CONCLUDE:  We are 95% confident that the interval from 1.1609 to 1.3741 grams/mile captures the true mean level of nitrogen oxides emitted by this type of light-duty engine.

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(b) The confidence interval from (a) tells us that any value from 1.1609 to 1.3741 g/mi is a plausible value of the mean NOX level m for this type of engine. Because the entire interval exceeds 1.0, it appears that this type of engine violates EPA limits. For Practice Try Exercise

65

Now that we’ve calculated our first confidence interval for a population mean m, it’s time to make a simple observation. Inference for proportions uses z; inference for means uses t. That’s one reason why distinguishing categorical from quantitative variables is so important. Here is another example, this time with a smaller sample size.

EXAMPLE

Video Screen Tension

STEP

Constructing a confidence interval for m

4

A manufacturer of high-resolution video terminals must control the tension on the mesh of fine wires that lies behind the surface of the viewing screen. Too much tension will tear the mesh, and too little will allow wrinkles. The tension is measured by an electrical device with output readings in millivolts (mV). Some variation is inherent in the production process. Here are the tension readings from a random sample of 20 screens from a single day’s production: 269.5 297.0 269.6 283.3 304.8 280.4 233.5 257.4 317.5 327.4 264.7 307.7 310.0 343.3 328.1 342.6 338.8 340.1 374.6 336.1

Construct and interpret a 90% confidence interval for the mean tension m of all the screens produced on this day.

State:  We want to estimate the true mean tension m of all the video terminals produced this day with 90% confidence. Plan:  If the conditions are met, we should use a one-sample t interval to estimate m.

220

260

300

340

Screen tension

(a)

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380

250

250 300

300 350 350

ScreenScreen tension tension

Expected z-score

Expected z-score

•  Random:  We are told that the data come from a random sample of 20 screens produced that day. ºº 10%:  Because we are sampling without replacement, we must assume that at least 10(20) = 200 video terminals were produced this day. •  Normal/Large Sample:  Because the sample size is small (n = 20), we must check whether it’s reasonable to believe that the population distribution is Normal. So we examine the sample data. Figure 8.15 shows (a) a dotplot, (b) a boxplot, and (c) a Normal probability plot of the tension 2 1 0 –1 –2

2 1 0 –1 –2 250

250 300

300 350 350

ScreenScreen tension tension

(b) (c) FIGURE 8.15  ​(a) A dotplot, (b) boxplot, and (c) Normal probability plot of the video screen tension readings.

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When the sample size is small (n < 30), as in this example, the Normal condition is about the shape of the population distribution. We look at a graph of the sample data to see if it’s believable that the data came from a Normal population.

Upper-tail probability p df

.10

.05

.025

18

1.330

1.734

2.101

19

1.328

1.729

2.093

20

1.325

1.725

2.086

80%

90%

95%

521

readings in the sample. Neither the dotplot nor the boxplot shows strong skewness or any outliers. The Normal probability plot looks roughly linear. These graphs give us no reason to doubt the Normality of the population. DO:  We used our calculator to find the mean and standard deviation of the tension readings for the 20 screens in the sample: x– = 306.32 mV and sx = 36.21 mV. We use the t distribution with df = 19 to find the critical value. For a 90% confidence level, the critical value is t* = 1.729. So the 90% confidence interval for m is sx 36.21 = 306.32 ± 14.00 = (292.32, 320.32) = 306.32 ± 1.729 x– ± t* !20 !n Using technology:  The calculator’s invT(.05,19) gives t = –1.729, which matches the t* = 1.729 critical value we got from the table. CONCLUDE:  We are 90% confident that the interval from 292.32 to 320.32 mV captures the true mean tension in the entire batch of video terminals produced that day. For Practice Try Exercise

Confidence level C

69

AP® Exam Tip  It is not enough just to make a graph of the data on your calculator when assessing Normality. You must sketch the graph on your paper to receive credit. You don’t have to draw multiple graphs—any appropriate graph will do.

As you probably guessed, your calculator will compute a one-sample t interval for a population mean from sample data or summary statistics.

17. TECHNOLOGY CORNER

One-sample t intervals for m on the calculator TI-Nspire instructions in Appendix B; HP Prime instructions on the book’s Web site.

Confidence intervals for a population mean using t distributions can be constructed on the TI-83/84 and TI-89, thus avoiding the use of Table B. Here is a brief summary of the techniques when you have the actual data values and when you have only numerical summaries. TI-83/84 TI-89 1. Using summary statistics (see auto pollution example, page 519) From the home screen, From inside the Statistics/List ­Editor, • Press STAT , arrow over to TESTS, and choose TInterval ... .

• Press 2nd F2 ([F7]) to go into the ­intervals (Ints) menu, then choose TInterval ... .

• On the TInterval screen, ­adjust your settings as shown and choose Calculate.

• Choose “Stats” as the Data Input Method.

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• On the TInterval screen, adjust your settings as shown and press ENTER .

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2. ​Using raw data (see video screen tension example, page 520) Enter the 20 video screen tension readings data in L1/list1. Proceed to the TInterval screen as in Step 1, but choose Data as the input method. Then adjust your settings as shown and calculate the interval.

CHECK YOUR UNDERSTANDING

Biologists studying the healing of skin wounds measured the rate at which new cells closed a cut made in the skin of an anesthetized newt. Here are data from a random sample of 18 newts, measured in micrometers (millionths of a meter) per hour:19 29 27 34 40 22 28 14 35 26 35 12 30 23 18 11 22 23 33 Calculate and interpret a 95% confidence interval for the mean healing rate m.

Choosing the Sample Size A wise user of statistics never plans data collection without planning the inference at the same time. You can arrange to have both high confidence and a small margin of error by taking enough observations. When the population standard deviation s is unknown and conditions are met, the C % confidence interval for m is x– ± t*

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sx !n

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523

where t* is the critical value for confidence level C and degrees of freedom df = n – 1. The margin of error (ME) of the confidence interval is ME = t*

sx !n

To determine the sample size for a desired margin of error, it makes sense to set the expression for ME less than or equal to the specified value and solve the inequality for n. There are two problems with this approach: 1. We don’t know the sample standard deviation sx because we haven’t produced the data yet. 2. The critical value t* depends on the sample size n that we choose. The second problem is more serious. To get the correct value of t*, we need to know the sample size. But that’s what we’re trying to find! There is no easy solution to this problem. One alternative (the one we recommend!) is to come up with a reasonable estimate for the population standard deviation s from a similar study that was done in the past or from a small-scale pilot study. By pretending that s is known, we can use the one-sample z interval for m: x– ± z*

s !n

Using the appropriate standard Normal critical value z* for confidence level C, we can solve z*

s ≤ ME !n

for n. Here is a summary of this strategy.

Choosing Sample Size for a Desired Margin of Error When Estimating m There are other methods of determining sample size that do not require us to use a known value of the population standard deviation s. These methods are beyond the scope of this text. Our advice: consult with a statistician when planning your study!

To determine the sample size n that will yield a C% confidence interval for a population mean with a specified margin of error ME: • Get a reasonable value for the population standard deviation s from an earlier or pilot study. • Find the critical value z* from a standard Normal curve for confidence level C. • Set the expression for the margin of error to be less than or equal to ME and solve for n: z*

s ≤ ME !n

The procedure is best illustrated with an example.

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EXAMPLE

How Many Monkeys? Determining sample size from margin of error Researchers would like to estimate the mean cholesterol level m of a particular v­ ariety of monkey that is often used in laboratory experiments. They would like their estimate to be within 1 milligram per deciliter (mg/dl) of the true value of m at a 95% confidence level. A previous study involving this variety of monkey suggests that the standard deviation of cholesterol level is about 5 mg/dl.

Problem:  Obtaining monkeys for research is time-consuming, expensive, and controversial. What is the minimum number of monkeys the researchers will need to get a satisfactory estimate? Solution:  For 95% confidence, z* = 1.96. We will use s = 5 as our best guess for the standard deviation of the monkeys’ cholesterol level. Set the expression for the margin of error to be at most 1 and solve for n : 1.96

5 ≤1 !n

(1.96)(5) ≤ !n 1 96.04 ≤ n Remember: always round up to the next whole number when finding n.

Because 96 monkeys would give a slightly larger margin of error than desired, the researchers would need 97 monkeys to estimate the cholesterol levels to their satisfaction. (On learning the cost of getting this many monkeys, the researchers might want to consider studying rats instead!) For Practice Try Exercise

73

c

!

n

Taking observations costs time and money. The required sample size autio may be impossibly expensive. Notice that it is the size of the sample that determines the margin of error. The size of the population does not influence the sample size we need. This is true as long as the population is much larger than the sample.

CHECK YOUR UNDERSTANDING

Administrators at your school want to estimate how much time students spend on homework, on average, during a typical week. They want to estimate m at the 90% confidence level with a margin of error of at most 30 minutes. A pilot study indicated that the standard deviation of time spent on homework per week is about 154 minutes. How many students need to be surveyed to meet the administrators’ goal? Show your work.

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case closed

Need Help? Give Us a Call! Refer to the chapter-opening Case Study on page 475. The bank manager wants to know whether or not the bank’s ­customer service agents generally met the goal of answering incoming calls in less than 30 seconds. We can approach this question in two ways: by estimating the proportion p of all calls that were answered within 30 seconds or by estimating the mean response time m. Some graphs and numerical summaries of the data are provided below. 35

Frequency

30 25 20 15 10 5 0 0.0

7.5

15.0

22.5

30.0

37.5

Call response time (seconds)

45.0

0

10

20

30

40

50

Call response time (seconds)

Descriptive Statistics: Call response time (sec) Variable N Mean SE Mean StDev Minimum Q1 Median Q3 Maximum Call response time (sec) 241 18.353 0.758 11.761 1.000 9.000 16.000 25.000 49.000

1. Describe the distribution of call response times for the random sample of 241 calls. Starnes/Yates/Moore: The Practice of Statistics, 4E what proportion of the call response times in the sample New Fig.: 08_UN33 2. Perm.About Fig.: 8067 First Pass: 2010-06-28 were less than 30 seconds? Explain how you got your answer. 3. The bank’s manager would like to estimate the true proportion p of calls to the bank’s customer service center that are answered in less than 30 seconds. (a) What conditions must be met to calculate a 95% confidence interval for p? Show that the conditions are met in this case. (b) Explain the meaning of 95% confidence in this setting. (c) A 95% confidence interval for p is (0.783, 0.877). Give the margin of error and show how it was calculated. (d) Interpret the interval from part (c) in context. 4. Construct and interpret a 95% confidence interval for the true mean response time of calls to the bank’s customer service center. 5. Is the customer service center meeting its goal of answering calls in less than 30 seconds? Give appropriate evidence to support your answer.

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Section 8.3

Summary •

Confidence intervals for the mean m of a Normal population are based on the sample mean x– of an SRS. If we somehow know s, we use the z critical value and the standard Normal distribution to help calculate confidence intervals.

•

In practice, we usually don’t know s. Replace the standard deviation s∙!n of the sampling distribution of x– by the standard error SEx– = sx∙!n and use the t distribution with n − 1 degrees of freedom (df). There is a t distribution for every positive degrees of freedom. All t distributions are unimodal, symmetric, and centered at 0. The t distributions approach the standard Normal distribution as the number of degrees of freedom increases. The conditions for constructing a confidence interval about a population mean are

•

•

•

Random:  The data were produced by a well-designed random sample or randomized experiment. ~~ 10%:  When sampling without replacement, check that the population is at least 10 times as large as the sample.

Normal/Large Sample:  The population distribution is Normal or the sample size is large (n ≥ 30). When the sample size is small (n < 30), examine a graph of the sample data for any possible departures from Normality in the population. You should be safe using a t distribution as long as there is no strong skewness and no outliers are present. When conditions are met, a C% confidence interval for the mean m is given by the one-sample t interval sx x– ± t* !n •

•

STEP

4

•

•

The critical value t* is chosen so that the t curve with n − 1 degrees of freedom has C% of the area between −t* and t*. Follow the four-step process—State, Plan, Do, Conclude—whenever you are asked to construct and interpret a confidence interval for a population mean Remember: inference for proportions uses z; inference for means uses t. The sample size needed to obtain a confidence interval with approximate margin of error ME for a population mean involves solving z*

s ≤ ME !n

for n, where the standard deviation s is a reasonable value from a previous or pilot study, and z* is the critical value for the level of confidence we want.

8.3 T echnology Corners TI-Nspire Instructions in Appendix B; HP Prime instructions on the book’s Web site. 16. Inverse t on the calculator 17. One-sample t intervals for m on the calculator

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Section 8.3

527

Exercises

55. Critical values  What critical value t* from Table B would you use for a confidence interval for the population mean in each of the following situations? (a) A 95% confidence interval based on n = 10 ­randomly selected observations (b) A 99% confidence interval from an SRS of 20 observations (c) A 90% confidence interval based on a random sample of 77 individuals 56. Critical values  What critical value t* from Table B should be used for a confidence interval for the population mean in each of the following situations? (a) A 90% confidence interval based on n = 12 randomly selected ­observations (b) A 95% confidence interval from an SRS of 30 observations (c) A 99% confidence interval based on a random sample of size 58 57. Pulling wood apart  How heavy a load (pounds) is needed to pull apart pieces of Douglas fir 4 inches long and 1.5 inches square? A random sample of 20 similar pieces of Douglas fir from a large batch was selected for a science class. The Fathom boxplot below shows the class’s data. Explain why it would not be wise to use a t critical value to construct a confidence interval for the population mean m.

pg 513

58. Weeds among the corn  Velvetleaf is a particularly annoying weed in cornfields. It produces lots of seeds, and the seeds wait in the soil for years until conditions are right for sprouting. How many seeds do velvetleaf plants produce? The Fathom histogram below shows the counts from a random sample of 28 plants that came up in a cornfield when no herbicide was used.20 Explain why it would not be wise to use a t critical value to construct a confidence interval for the mean number of seeds m produced by velvetleaf plants.

59. Should we use t?  Determine whether we can safely use a t* critical value to calculate a confidence interval for the population mean in each of the following settings. (a) We collect data from a random sample of adult residents in a state. Our goal is to estimate the overall percent of adults in the state who are college graduates. (b) The coach of a college men’s basketball team ­records the resting heart rates of the 15 team members. We use these data to construct a confidence interval for the mean resting heart rate of all male students at this college. (c) Do teens text more than they call? To find out, an AP® Statistics class at a large high school collected data on the number of text messages and calls sent or received by each of 25 randomly selected students. The Fathom boxplot below displays the difference (texts − calls) for each student.

pg 516

60. Should we use t?  Determine whether we can safely use a t* critical value to calculate a confidence interval for the population mean in each of the following settings. (a) We want to estimate the average age at which U.S. presidents have died. So we obtain a list of all U.S. presidents who have died and their ages at death. (b) How much time do students spend on the ­Internet? We collect data from the 32 members of our AP® Statistics class and calculate the mean amount of time that each student spent on the ­Internet ­yesterday. (c) Judy is interested in the reading level of a medical journal. She records the length of a random sample of 100 words. The Minitab histogram below displays the data.

61. Blood pressure  A medical study finds that x– = 114.9 and sx = 9.3 for the seated systolic blood pressure of the 27 members of one treatment group. What is the standard error of the mean? Interpret this value in context.

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62. Travel time to work  A study of commuting times reports the travel times to work of a random sample of 20 employed adults in New York State. The mean is x– = 31.25 minutes, and the standard deviation is sx = 21.88 minutes. What is the standard error of the mean? Interpret this value in context. 63. Willows in Yellowstone  Writers in some fields summarize data by giving x– and its standard error rather than x– and sx. Biologists studying willow plants in Yellowstone National Park reported their results in a table with columns labeled x– ± SE. The table entry for the heights of willow plants (in centimeters) in one region of the park was 61.55 ± 19.03.21 The researchers measured a total of 23 plants. (a) Find the sample standard deviation sx for these measurements. Show your work. (b) A hasty reader believes that the interval given in the table is a 95% confidence interval for the mean height of willow plants in this region of the park. Find the actual confidence level for the given interval. 64. Blink  When two lights close together blink alternately, we “see” one light moving back and forth if the time between blinks is short. What is the longest interval of time between blinks that preserves the illusion of motion? Ask subjects to turn a knob that slows the blinking until they “see” two lights rather than one light moving. A report gives the results in the form “mean plus or minus the standard error of the mean.”22 Data for 12 subjects are summarized as 251 ± 45 (in milliseconds). (a) Find the sample standard deviation sx for these measurements. Show your work. (b) A hasty reader believes that the interval given in the report is a 95% confidence interval for the population mean. Find the actual confidence level for the given interval. 65. Bone loss by nursing mothers  Breast-feeding mothpg 519 ers secrete calcium into their milk. Some of the calcium may come from their bones, so mothers may lose bone mineral. Researchers measured the percent STEP change in bone mineral content (BMC) of the spines of 47 randomly selected mothers during three months of breast-feeding.23 The mean change in BMC was −3.587% and the standard deviation was 2.506%. (a) Construct and interpret a 99% confidence interval to estimate the mean percent change in BMC in the population. (b) Based on your interval from part (a), do these data give good evidence that on the average nursing ­mothers lose bone mineral? Explain. 66. Reading scores in Atlanta  The Trial Urban District STEP Assessment (TUDA) is a government-sponsored study of student achievement in large urban school districts. TUDA gives a reading test scored from 0 to 500. A score of 243 is a “basic” reading level and a score of 281 is “proficient.” Scores for a random sample of

(a) (b)

STEP

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4

(a)

(b)

STEP

68.

4

(a)

(b)

4

4

67.

69. pg 520

STEP

4

1470 eighth-graders in Atlanta had x– = 240 with standard deviation 42.17.24 Calculate and interpret a 99% confidence interval for the mean score of all Atlanta eighth-graders. Based on your interval from part (a), is there good evidence that the mean for all Atlanta eighth-graders is less than the basic level? Explain. Men and muscle  Ask young men to estimate their own degree of body muscle by choosing from a set of 100 photos. Then ask them to choose what they believe women prefer. The researchers know the actual degree of muscle, measured as kilograms per square meter of fat-free mass, for each of the photos. They can therefore measure the difference between what a subject thinks women prefer and the subject’s own self-image. Call this difference the “muscle gap.” Here are summary statistics for the muscle gap from a random sample of 200 American and European young men: x– = 2.35 and sx =2.5.25 Calculate and interpret a 95% confidence interval for the mean size of the muscle gap for the population of American and European young men. A graph of the sample data is strongly skewed to the right. Explain why this information does not invalidate the interval you calculated in part (a). A big-toe problem  A bunion on the big toe is fairly uncommon in youth and often requires surgery. Doctors used X-rays to measure the angle (in degrees) of deformity on the big toe in a random sample of 37 patients under the age of 21 who came to a medical center for surgery to correct a bunion. The angle is a measure of the seriousness of the deformity. For these 37 patients, the mean angle of deformity was 24.76 degrees and the standard deviation was 6.34 degrees. A dotplot of the data revealed no outliers or strong skewness.26 Construct and interpret a 90% confidence interval for the mean angle of deformity in the population of all such patients. Researchers omitted one patient with a deformity angle of 50 degrees from the analysis due to a measurement issue. What effect would including this outlier have on the confidence interval in part (a)? Justify your answer without doing any calculations. Give it some gas!  Computers in some vehicles calculate various quantities related to performance. One of these is fuel efficiency, or gas mileage, usually expressed as miles per gallon (mpg). For one vehicle equipped in this way, the miles per gallon were recorded each time the gas tank was filled and the computer was then reset.27 Here are the mpg values for a random sample of 20 of these records:

15.8 13.6 15.6 19.1 22.4 15.6 22.5 17.2 19.4 22.6 19.4 18.0 14.6 18.7 21.0 14.8 22.6 21.5 14.3 20.9

Construct and interpret a 95% confidence interval for the mean fuel efficiency m for this vehicle.

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Section 8.3 Estimating a Population Mean

STEP

70. Vitamin C content  Several years ago, the U.S. Agency for International Development provided 238,300 metric tons of corn-soy blend (CSB) for emergency relief in countries throughout the world. CSB is a highly nutritious, low-cost fortified food. As part of a study to evaluate appropriate vitamin C levels in this food, measurements were taken on samples of CSB produced in a factory.28 The following data are the amounts of vitamin C, measured in milligrams per 100 grams (mg/100 g) of blend, for a random sample of size 8 from one production run:

4

26  31  23  22  11  22  14  31



Construct and interpret a 95% confidence interval for the mean amount of vitamin C m in the CSB from this production run.

71. Paired tires  Researchers were interested in comparing two methods for estimating tire wear. The first method used the amount of weight lost by a tire. The second method used the amount of wear in the grooves of the tire. A random sample of 16 tires was obtained. Both methods were used to estimate the total distance traveled by each tire. The table below provides the two estimates (in thousands of miles) for each tire.29 (a) Construct and interpret a 95% confidence interval for the mean difference m in the estimates from these two methods in the population of tires. (b) Does your interval in part (a) give convincing evidence of a difference in the two methods of ­estimating tire wear? Justify your answer. Tire Weight Groove 1 2 3 4 5 6 7 8

45.9 41.9 37.5 33.4 31.0 30.5 30.9 31.9

35.7 39.2 31.1 28.1 24.0 28.7 25.9 23.3

Diff. 10.2  2.7  6.4  5.3  7.0  1.8  5.0  8.6

Tire 9 10 11 12 13 14 15 16

Weight Groove 30.4 27.3 20.4 24.5 20.9 18.9 13.7 11.4

23.1 23.7 20.9 16.1 19.9 15.2 11.5 11.2

Diff.   7.3   3.6 −0.5   8.4   1.0   3.7   2.2   0.2

72. Water  Trace metals found in wells affect the taste of drinking water, and high concentrations can pose a health risk. Researchers measured the concentration of zinc (in milligrams/liter) near the top and the bottom of 10 randomly selected wells in a large region. The data are provided in the table below.30 (a) Construct and interpret a 95% confidence interval for the mean difference m in the zinc concentrations from these two locations in the wells. (b) Does your interval in part (a) give convincing evidence of a difference in zinc concentrations at the top and bottom of wells in the region? Justify your answer. Well: Bottom:

1 2 3 4 5 6 7 8 9 10 0.430 0.266 0.567 0.531 0.707 0.716 0.651 0.589 0.469 0.723

Top:

0.415 0.238 0.390 0.410 0.605 0.609 0.632 0.523 0.411 0.612

Difference: 0.015 0.028 0.177 0.121 0.102 0.107 0.019 0.066 0.058 0.111

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73. Estimating BMI  The body mass index (BMI) of all American young women is believed to follow a Normal distribution with a standard deviation of about 7.5. How large a sample would be needed to estimate the mean BMI m in this population to within ±1 with 99% confidence? Show your work. 74. The SAT again  High school students who take the SAT Math exam a second time generally score higher than on their first try. Past data suggest that the score increase has a standard deviation of about 50 points. How large a sample of high school students would be needed to estimate the mean change in SAT score to within 2 points with 95% confidence? Show your work.

pg 524

Multiple choice: Select the best answer for Exercises 75 to 78. 75. One reason for using a t distribution instead of the standard Normal curve to find critical values when calculating a level C confidence interval for a population mean is that (a) z can be used only for large samples. (b) z requires that you know the population standard deviation s. (c) z requires that you can regard your data as an SRS from the population. (d) z requires that the sample size is at most 10% of the population size. (e) a z critical value will lead to a wider interval than a t critical value. 76. You have an SRS of 23 observations from a large population. The distribution of sample values is roughly symmetric with no outliers. What critical value would you use to obtain a 98% confidence interval for the mean of the population? (a)  2.177  (b) 2.183  (c) 2.326  (d) 2.500  (e) 2.508 77. A quality control inspector will measure the salt content (in milligrams) in a random sample of bags of potato chips from an hour of production. Which of the following would result in the smallest margin of error in estimating the mean salt content m? (a) 90% confidence; n = 25 (b) 90% confidence; n = 50 (c) 95% confidence; n = 25 (d) 95% confidence; n = 50 (e) n = 100 at any confidence level 78. Scientists collect data on the blood cholesterol levels (milligrams per deciliter of blood) of a random sample of 24 laboratory rats. A 95% confidence interval for the mean blood cholesterol level m is 80.2 to 89.8. Which of the following would cause the most worry about the validity of this interval? (a) There is a clear outlier in the data. (b) A stemplot of the data shows a mild right skew.

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(c) You do not know the population standard ­deviation s. (d) The population distribution is not exactly N ­ ormal. (e) None of these are a problem when using a t interval. 79. Watching TV (6.1, 7.3)  Choose a young person (aged 19 to 25) at random and ask, “In the past seven days, how many days did you watch television?” Call the response X for short. Here is the probability distribution for X:31 Days: Probability:

0 1 2 3 4 5 6 0.04 0.03 0.06 0.08 0.09 0.08 0.05

7 ???

(a) What is the probability that X = 7? Justify your answer. (b) Calculate the mean of the random variable X. Interpret this value in context. (c) Suppose that you asked 100 randomly selected young people (aged 19 to 25) to respond to the question and found that the mean x– of their responses was 4.96. Would this result surprise you? Justify your answer. 80. Price cuts (4.2)  Stores advertise price reductions to attract customers. What type of price cut is most attractive? Experiments with more than one factor allow insight into interactions between the factors. A study of the attractiveness of advertised price discounts had two factors: percent of all foods on sale (25%, 50%, 75%, or 100%) and whether the discount was stated precisely (as in, for example, “60% off”) or as a range (as in “40% to 70% off”). Subjects rated the attractiveness of the sale on a scale of 1 to 7.

(a) Describe a completely randomized design using 200 student subjects. (b) Explain how you would use the partial table of random digits below to assign subjects to treatment groups. Then use your method to select the first 3 subjects for one of the treatment groups. Show your work clearly on your paper. 45740 41807 65561 33302 07051 93623 18132 09547 12975 13258 13048 45144 72321 81940 00360 02428

(c) The figure below shows the mean ratings for the eight treatments formed from the two factors.32 Based on these results, write a careful description of how percent on sale and precise discount versus range of discounts influence the attractiveness of a sale. 5

Precise Range Mean attractiveness score

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4

3

2

1 25

50

75

100

Percent of goods on sale

FRAPPY!  Free Response AP® Problem, Yay! The following problem is modeled after actual AP® Statistics exam free response questions. Your task is to generate a complete, concise response in 15 minutes.

Directions: Show all your work. Indicate clearly the methods you use, because you will be scored on the correctness of your methods as well as on the accuracy and completeness of your results and explanations. Members at a popular fitness club currently pay a $40 per month membership fee. The owner of the club wants to raise the fee to $50 but is concerned that some members will leave the gym if the fee increases. To investigate, the owner plans to survey a random sample of the club members and construct a 95% confidence interval for the proportion of all members who would quit if the fee was raised to $50. (a) Explain the meaning of “95% confidence” in the context of the study. (b) After the owner conducted the survey, he calculated the confidence interval to be 0.18 ± 0.075.

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Starnes/Yates/Moore: The Practice of Statistics, 4E Interpret interval in the context New Fig.: 08_UN39 thisPerm. Fig.: 8074 First Pass: 2010-06-28 study.

of the

(c) According to the club’s accountant, the fee increase will be worthwhile if fewer than 20% of the members quit. According to the interval from part (b), can the owner be confident that the fee increase will be worthwhile? Explain. (d) One of the conditions for calculating the confidence interval in part (b) is that np^ ≥ 10 and n(1 − p^) ≥ 10. Explain why it is necessary to check this condition.

After you finish, you can view two example solutions on the book’s Web site (www.whfreeman.com/tps5e). Determine whether you think each solution is “complete,” “substantial,” “developing,” or “minimal.” If the solution is not complete, what improvements would you suggest to the student who wrote it? Finally, your teacher will provide you with a scoring rubric. Score your response and note what, if anything, you would do differently to improve your own score.

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Chapter Review Section 8.1: Confidence Intervals: The Basics In this section, you learned that a point estimate is the single best guess for the value of a population parameter. You also learned that a confidence interval provides an interval of plausible values for a parameter. To interpret a confidence interval, say, “We are C% confident that the interval from ___ to ___ captures the [parameter in context],” where C is the confidence level of the interval. The confidence level C describes the percentage of confidence intervals that we expect to capture the value of the parameter. To interpret a C% confidence level, we say, “If we took many samples of the same size and used them to construct C% confidence intervals, about C% of those intervals would capture the [parameter in context].” Confidence intervals are formed by including a margin of error on either side of the point estimate. The size of the margin of error is determined by several factors, including the confidence level C and the sample size n. Increasing the sample size n makes the standard deviation of our ­estimate smaller, decreasing the margin of error. Increasing the confidence level C makes the margin of error larger, to ensure that the capture rate of the interval increases to C%. Remember that the margin of error only accounts for sampling variability—it does not account for any bias in the data collection process. Section 8.2: Estimating a Population Proportion In this section, you learned how to construct and interpret confidence intervals for a population proportion. Several important conditions must be met for this type of confidence interval to be valid. First, the data used to calculate the interval must come from a well-designed random sample or randomized experiment (the Random condition). When the sample is taken without replacement from the population, the sample size should be no more than 10% of the population size (the 10% condition). Finally, the observed number of successes np^ and observed number of failures n(1 − p^) must both be at least 10 (the Large Counts condition). The formula for calculating a confidence interval for a population proportion is p^ (1 − p^) n Å

p^ ± z*

where p^ is the sample proportion, z* is the critical value, and n is the sample size. The value of z* is based on the confidence level C. To find z*, use Table A or technology to determine the values of z* and –z* that capture the middle C% of the standard Normal distribution. The four-step process (State, Plan, Do, Conclude) is perfectly suited for problems that ask you to construct and interpret a confidence interval. You should state the parameter you are estimating and at what confidence level, plan your work

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by naming the type of interval you will use and checking the appropriate conditions, do the calculations, and make a conclusion in the context of the problem. You can use technology for the Do step, but make sure that you identify the procedure you are using and type in the values correctly. Finally, an important part of planning a study is determining the size of the sample to be selected. The necessary sample size is based on the confidence level, the proportion of successes, and the desired margin of error. To calculate the minimum sample size, solve the following inequality for n, where p^ is a guessed value for the sample proportion: p^ (1 − p^) z* ≤ ME n Å If you do not have an approximate value of p^ from a previous study or a pilot study, use p^ = 0.5 to determine the sample size that will yield a value less than or equal to the desired margin of error. Section 8.3: Estimating a Population Mean In this section, you learned how to construct and interpret confidence intervals for a population mean. Remember that you have to check conditions before doing calculations. The Random and 10% conditions are the same as those for proportions. There’s one new condition for means: the population must be Normally distributed or the sample size must be at least 30 (the Normal/Large Sample condition). If the population shape is unknown and the sample size is less than 30, graph the sample data and check for strong skewness or outliers. If there is no strong skewness or outliers, it is reasonable to assume that the population distribution is approximately Normal. The formula for calculating a confidence interval for a population mean is sx x– ± t* !n where x– is the sample mean, t* is the critical value, sx is the sample standard deviation, and n is the sample size. We use a t critical value instead of a z critical value when the population standard deviation is unknown—which is almost always the case. The value of t* is based on the confidence level C and the degrees of freedom (df = n – 1). To find t*, use Table B or technology to determine the values of t* and –t* that capture the middle C% of the appropriate t distribution. The t distributions are bell-shaped, symmetric, and centered at 0. However, they are more variable and have a shape slightly different from that of the standard Normal distribution. You also learned how to estimate the sample size when planning a study, as in Section 8.2. To calculate the minimum sample size, solve the following inequality for n, where s is a guessed value for the population standard deviation: s z* ≤ ME !n 531 11/13/13 3:13 PM

What Did You Learn? Learning Objective Determine the point estimate and margin of error from a confidence interval.

Section

Related Example on Page(s)

Relevant Chapter Review Exercise(s)

8.1

481

R8.2

Interpret a confidence interval in context.

8.1

481, 484

R8.3, R8.4, R8.6, R8.7

Interpret a confidence level in context.

8.1

484

R8.2

Describe how the sample size and confidence level affect the length of a confidence interval.

8.1

Discussion on 487

R8.9

Explain how practical issues like nonresponse, undercoverage, and response bias can affect the interpretation of a confidence interval.

8.1

Discussion on 488

R8.6

State and check the Random, 10%, and Large Counts conditions for constructing a confidence interval for a population proportion.

8.2

494

R8.3

Determine critical values for calculating a C% confidence interval for a population proportion using a table or technology.

8.2

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R8.1

Construct and interpret a confidence interval for a population proportion.

8.2

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R8.3, R8.6

Determine the sample size required to obtain a C% confidence interval for a population proportion with a specified margin of error.

8.2

502

R8.5

State and check the Random, 10%, and Normal/Large Sample conditions for constructing a confidence interval for a population mean.

8.3

516

R8.4

Explain how the t distributions are different from the standard Normal distribution and why it is necessary to use a t distribution when calculating a confidence interval for a population mean.

8.3

Discussion on 511–512

R8.10

Determine critical values for calculating a C% confidence interval for a population mean using a table or technology.

8.3

513

R8.1

Construct and interpret a confidence interval for a population mean.

8.3

519–520

R8.4, R8.7

Determine the sample size required to obtain a C% confidence interval for a population mean with a specified margin of error.

8.3

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R8.8

Chapter 8 Chapter Review Exercises These exercises are designed to help you review the important ideas and methods of the chapter. R8.1 It’s critical  Find the appropriate critical value for constructing a confidence interval in each of the following settings. (a) Estimating a population proportion p at a 94% confidence level based on an SRS of size 125.

(b) Estimating a population mean m at a 99% confidence level based on an SRS of size 58. R8.2 Batteries  A company that produces AA batteries tests the lifetime of a random sample of 30 batteries using a special device designed to imitate real-world use. Based on the testing, the company makes the following statement: “Our AA batteries last an average of 430 to 470 minutes, and our confidence in that interval is 95%.”33

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Chapter Review Exercises (a) Determine the point estimate, margin of error, standard error, and sample standard ­deviation. (b) A reporter translates the statistical announcement into “plain English” as follows: “If you buy one of this company’s AA batteries, there is a 95% chance that it will last between 430 and 470 minutes.” Comment on this interpretation. (c) Your friend, who has just started studying statistics, claims that if you select 40 more AA batteries at random from those manufactured by this company, there is a 95% probability that the mean lifetime will fall between 430 and 470 minutes. Do you agree? Explain. (d) Give a statistically correct interpretation of the confidence level that could be published in a newspaper report. R8.3 We love football!  A recent Gallup Poll conducted telephone interviews with a random sample of adults aged 18 and older. Data were obtained for 1000 people. Of these, 37% said that football is their favorite sport to watch on television. (a) Define the parameter p in this setting. Explain to someone who knows nothing about statistics why we can’t just say that 37% of all adults would say that football is their favorite sport to watch on television. (b) Check the conditions for constructing a confidence interval for p. (c) Construct a 95% confidence interval for p. (d) Interpret the interval in context. R8.4 Smart kids  A school counselor wants to know how smart the students in her school are. She gets funding from the principal to give an IQ test to an SRS of 60 of the over 1000 students in the school. The mean IQ score was 114.98 and the standard deviation was 14.80.34 (a) Define the parameter m in this setting. (b) Check the conditions for constructing a confidence interval for m. (c) Construct a 90% confidence interval for the mean IQ score of students at the school. (d) Interpret your result from part (c) in context. R8.5 Do you go to church?  The Gallup Poll plans to ask a random sample of adults whether they attended a ­religious service in the last 7 days. How large a sample would be ­required to obtain a margin of error of at most 0.01 in a 99% confidence interval for the population proportion who would say that they attended a religious service? Show your work. R8.6 Running red lights  A random digit dialing telephone survey of 880 drivers asked, “Recalling the last ten traffic lights you drove through, how many of them were red when you entered the intersections?” Of the 880 respondents, 171 admitted that at least one light had been red.35

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(a) Construct and interpret a 95% confidence interval for the population proportion. (b) Nonresponse is a practical problem for this ­survey—only 21.6% of calls that reached a live person were completed. Another practical problem is that people may not give truthful answers. What is the likely direction of the bias: Do you think more or fewer than 171 of the 880 respondents really ran a red light? Why? Are these sources of bias included in the margin of error? R8.7 Engine parts  Here are measurements (in millimeters) of a critical dimension on an SRS of 16 of the more than 200 auto engine crankshafts produced in one day: 224.120 224.001 224.017 223.982 223.989 223.961 223.960 224.089 223.987 223.976 223.902 223.980 224.098 224.057 223.913 223.999

(a) Construct and interpret a 95% confidence interval for the process mean at the time these crankshafts were produced. (b) The process mean is supposed to be m = 224 mm but can drift away from this target during production. Does your interval from part (a) suggest that the process mean has drifted? Explain. R8.8 Good wood?  A lab supply company sells pieces of Douglas fir 4 inches long and 1.5 inches square for force experiments in science classes. From experience, the strength of these pieces of wood follows a Normal distribution with standard deviation 3000 pounds. You want to estimate the mean load needed to pull apart these pieces of wood to within 1000 pounds with 95% confidence. How large a sample is needed? Show your work. R8.9 It’s about ME  Explain how each of the following would affect the margin of error of a confidence interval, if all other things remained the same. (a) Increasing the confidence level (b) Quadrupling the sample size R8.10 t time  When constructing confidence intervals for a population mean, we almost always use critical values from a t distribution rather than the standard Normal distribution. (a) When is it necessary to use a t critical value rather than a z critical value when constructing a confidence interval for a population mean? (b) Describe two ways that the t distributions are different from the standard Normal distribution. (c) Explain what happens to the t distributions as the degrees of freedom increase.

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CHAPTER 8

E s t i m at i n g w i t h C o n f i d e n ce

Chapter 8 AP® Statistics Practice Test Section I: Multiple Choice  ​Select the best answer for each question. T8.1 The Gallup Poll interviews 1600 people. Of these, 18% say that they jog regularly. The news report adds: “The poll had a margin of error of plus or minus three percentage points at a 95% confidence level.” You can safely conclude that (a) 95% of all Gallup Poll samples like this one give ­answers within ±3% of the true population value. (b) the percent of the population who jog is certain to be between 15% and 21%. (c) 95% of the population jog between 15% and 21% of the time. (d) we can be 95% confident that the sample proportion is captured by the confidence interval. (e) if Gallup took many samples, 95% of them would find that 18% of the people in the sample jog. T8.2 The weights (in pounds) of three adult males are 160, 215, and 195. The standard error of the mean of these three weights is (a) 190.  (b) ​27.84.  (c) ​ 22.73.  (d) ​ 16.07.  (e) ​13.13. T8.3 In preparing to construct a one-sample t interval for a population mean, suppose we are not sure if the population distribution is Normal. In which of the following circumstances would we not be safe constructing the interval based on an SRS of size 24 from the population? (a) A stemplot of the data is roughly bell-shaped. (b) A histogram of the data shows slight skewness. (c) A boxplot of the data has a large outlier. (d) The sample standard deviation is large. (e) A Normal probability plot of the data is fairly linear. T8.4 Many television viewers express doubts about the validity of certain commercials. In an attempt to answer their critics, Timex Group USA wishes to estimate the proportion of consumers who believe what is shown in Timex television commercials. Let p represent the true proportion of consumers who believe what is shown in Timex television commercials. What is the

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smallest number of consumers that Timex can survey to guarantee a margin of error of 0.05 or less at a 99% confidence level? (a) 550  (b) ​600   (c) ​650   (d) ​700   (e) ​750 T8.5 You want to compute a 90% confidence interval for the mean of a population with unknown population standard deviation. The sample size is 30. The value of t* you would use for this interval is (a) 1.645.  (b) ​ 1.699.  (c) ​1.697.  (d) ​1.96.  (e) ​2.045. T8.6 A radio talk show host with a large audience is interested in the proportion p of adults in his listening area who think the drinking age should be lowered to eighteen. To find this out, he poses the following question to his listeners: “Do you think that the drinking age should be reduced to eighteen in light of the fact that eighteen-year-olds are eligible for military service?” He asks listeners to phone in and vote “Yes” if they agree the drinking age should be lowered and “No” if not. Of the 100 people who phoned in, 70 answered “Yes.” Which of the following conditions for inference about a proportion using a confidence interval are violated?

I. The data are a random sample from the population of interest. II. The population is at least 10 times as large as the ­sample. III. n is so large that both np^ and n(1 − p^) are at least 10. (a) I only (b) II only

(c) ​III only (d) ​I and II only

(e) ​I, II, and III

T8.7 A 90% confidence interval for the mean m of a population is computed from a random sample and is found to be 9 ± 3. Which of the following could be the 95% confidence interval based on the same data? (a) 9 ± 2   (b) ​9 ± 3   (c) ​9 ± 4   (d) 9 ± 8 (e) Without knowing the sample size, any of the above answers could be the 95% confidence interval.

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AP® Statistics Practice Test T8.8 Suppose we want a 90% confidence interval for the average amount spent on books by freshmen in their first year at a major university. The interval is to have a margin of error of $2. Based on last year’s book sales, we estimate that the standard deviation of the amount spent will be close to $30. The number of observations required is closest to (a) 25.  (b) ​30.   (c) ​608.   (d) ​609.   (e) ​865. T8.9 A telephone poll of an SRS of 1234 adults found that 62% are generally satisfied with their lives. The announced margin of error for the poll was 3%. Does the margin of error account for the fact that some adults do not have telephones? (a) Yes. The margin of error includes all sources of error in the poll. (b) Yes. Taking an SRS eliminates any possible bias in ­estimating the population proportion. (c) Yes. The margin of error includes undercoverage but not nonresponse. (d) No. The margin of error includes nonresponse but not undercoverage.

535

(e) No. The margin of error only includes sampling ­variability. T8.10 A Census Bureau report on the income of Americans says that with 90% confidence the median income of all U.S. households in a recent year was $57,005 with a margin of error of ±$742. This means that (a) 90% of all households had incomes in the range $57,005 ± $742. (b) we can be sure that the median income for all households in the country lies in the interval $57,005 ± $742. (c) 90% of the households in the sample interviewed by the Census Bureau had incomes in the interval $57,005 ± $742. (d) the Census Bureau got the result $57,005 ± $742 ­using a method that will capture the true median income 90% of the time when used repeatedly. (e) 90% of all possible samples of this same size would result in a sample median that falls within $742 of $57,005.

Section II: Free Response  ​Show all your work. Indicate clearly the methods you use, because you will be graded on the correctness of your methods as well as on the accuracy and completeness of your results and explanations. T8.11 The U.S. Forest Service is considering additional restrictions on the number of vehicles allowed to enter ­Yellowstone National Park. To assess public reaction, the service asks a random sample of 150 visitors if they favor the proposal. Of these, 89 say “Yes.” (a) Construct and interpret a 99% confidence interval for the proportion of all visitors to Yellowstone who favor the restrictions. (b) Based on your work in part (a), can the U.S. Forest ­Service conclude that more than half of visitors to Yellowstone National Park favor the proposal? Justify your answer. T8.12 How many people live in South African households? To find out, we collected data from an SRS of 48 out of the over 700,000 South African students who took part in the CensusAtSchool survey project. The mean number of people living in a household was 6.208; the standard deviation was 2.576.

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(a) Is the Normal/Large Sample condition met in this case? Justify your answer. (b) Maurice claims that a 95% confidence interval for 2.576 . Explain !47 why this interval is wrong. Then give the correct interval.

the population mean is 6.208±1.96

T8.13 A milk processor monitors the number of bacteria per milliliter in raw milk received at the factory. A random sample of 10 one-milliliter specimens of milk supplied by one producer gives the following data: 5370 4890 5100 4500 5260 5150 4900 4760 4700 4870



Construct and interpret a 90% confidence interval for the population mean m.

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Chapter

9

Introduction 538 Section 9.1 539 Significance Tests: The Basics Section 9.2 554 Tests about a Population Proportion Section 9.3 574 Tests about a Population Mean Free Response AP® Problem, YAY! 601 Chapter 9 Review 602 Chapter 9 Review Exercises 604 Chapter 9 AP® Statistics Practice Test 605

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Testing a Claim case study

Do You Have a Fever? Sometimes when you’re sick, your forehead feels really warm. You might have a fever. How can you find out whether you do? By taking your temperature, of course. But what temperature should the ­thermometer show if you’re healthy? Is this temperature the same for everyone? Several years ago, researchers conducted a study to determine whether the “accepted” value for n ­ ormal body temperature, 98.6°F, is accurate. They used an oral thermometer to measure the temperatures of a random sample of healthy men and women aged 18 to 40. As is often the case, the researchers did not provide their original data. Allen Shoemaker, from Calvin College, produced a data set with the same properties as the original temperature readings. His data set consists of one oral temperature reading for each of 130 randomly chosen, healthy 18- to 40-year-olds.1 A dotplot of Shoemaker’s temperature data is shown below. We have added a vertical line at 98.6°F for reference.

96

97

98

99

Temperature (°F)

100

101

Exploratory data analysis revealed several interesting facts about this data set: • The mean temperature was x– = 98.25°F. • The standard deviation of the temperature readings was sx = 0.73°F. • 62.3% of the temperature readings were less than 98.6°F.

Based on the results of this study, can we conclude that “normal” body temperature in the population of healthy 18- to 40-year-olds is not 98.6°F? By the end of this chapter, you will have developed the necessary tools for answering this question.

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CHAPTER 9

Testing a Claim

Introduction Confidence intervals are one of the two most common types of statistical ­inference. Use a confidence interval when your goal is to estimate a population para­meter. The second common type of inference, called significance tests, has a ­different goal: to assess the evidence provided by data about some claim concerning a ­parameter. Here is an Activity that illustrates the reasoning of statistical tests.

Activity Materials: Computer with Internet access and projection ­capability PLET AP

I’m a Great Free-Throw Shooter! A basketball player claims to make 80% of the free throws that he attempts. We think he might be exaggerating. To test this claim, we’ll ask him to shoot some free throws—virtually—using The Reasoning of a Statistical Test applet at the book’s Web site. 1.  Go to www.whfreeman.com/tps5e and launch the applet.

2. ​Set the applet to take 25 shots. Click “Shoot.” How many of the 25 shots did the player make? Do you have enough data to decide whether the player’s claim is valid? 3. ​Click “Shoot” again for 25 more shots. Keep doing this until you are ­convinced either that the player makes less than 80% of his shots or that the player’s claim is true. How large a sample of shots did you need to make your decision? 4. ​Click “Show true probability” to reveal the truth. Was your conclusion ­correct? 5. ​If time permits, choose a new shooter and repeat Steps 2 through 4. Is it easier to tell that the player is exaggerating when his actual proportion of free throws made is closer to 0.8 or farther from 0.8?

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Section 9.1 Significance Tests: The Basics

539

In the free-throw shooter Activity, the parameter of interest is the proportion p of free throws that the player will make if he shoots forever. Our player claims that p = 0.80. To test his claim, we let the applet simulate 25 shots. If the player makes only 40% of his free throws (10 of 25 shots made), we have fairly strong evidence that he doesn’t shoot 80%. But what if he makes 76% of his free throws (19 of 25 shots made)? This provides some evidence that his true long-term percent may be less than 80%, but it’s not nearly as convincing as p^ = 0.40. Statistical tests weigh the evidence against a claim like p = 0.8 and in favor of a counter-claim like p < 0.80. Section 9.1 focuses on the underlying logic of statistical tests. Once the foundation is laid, we consider the implications of using these tests to make decisions— about everything from free-throw shooting to the effectiveness of a new drug. In Section 9.2, we present the details of performing a test about a population proportion. Section 9.3 shows how to test a claim about a population mean. Along the way, we examine the connection between confidence intervals and tests.

9.1 What You Will Learn

Significance Tests: The Basics By the end of the section, you should be able to:

• State the null and alternative hypotheses for a ­significance test about a population parameter. • Interpret a P-value in context.

• Determine whether the results of a study are ­statistically significant and make an appropriate ­conclusion using a significance level. • Interpret a Type I and a Type II error in context and give a consequence of each.

A significance test is a formal procedure for using observed data to decide between two competing claims (also called hypotheses). The claims are often statements about a parameter, like the population proportion p or the population mean m. Let’s start by taking a closer look at how to state hypotheses.

Stating Hypotheses

Remember: the null hypothesis is the dull hypothesis!

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A significance test starts with a careful statement of the claims we want to compare. In our free-throw shooter example, the virtual player claims that his long-run proportion of made free throws is p = 0.80. This is the claim we seek evidence against. We call it the null hypothesis, abbreviated H0. Usually, the null hypothesis is a statement of “no difference.” For the free-throw shooter, no difference from what he claimed gives H0 : p = 0.80. The claim we hope or suspect to be true instead of the null hypothesis is called the alternative hypothesis. We abbreviate the alternative hypothesis as Ha. In this case, we believe the player might be exaggerating, so our alternative hypothesis is Ha : p < 0.80.

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CHAPTER 9

Testing a Claim

Definition: Null hypothesis H0, alternative hypothesis Ha The claim we weigh evidence against in a statistical test is called the null ­hypothesis (H0). Often the null hypothesis is a statement of “no difference.” The claim about the population that we are trying to find evidence for is the alternative hypothesis (Ha). Some people insist that all three possibilities—greater than, less than, and equal to—be accounted for in the hypotheses. For the free-throw shooter example, since the alternative hypothesis is Ha : p < 0.80, they would write the null hypothesis as H0 : p ≥ 0.80. In spite of the mathematical appeal of covering all three cases, we use only the value p = 0.80 in our calculations. So we’ll stick with H0 : p = 0.80.

Example

In the free-throw shooter example, our hypotheses are H0 : p = 0.80 Ha : p < 0.80 where p is the long-run proportion of made free throws. The alternative hypothesis is one-sided because we are interested only in whether the player is overstating his free-throw shooting ability. Because Ha expresses the effect that we hope to find evidence for, it is sometimes easier to begin by stating Ha and then set up H0 as the statement that the hoped-for effect is not present. Here is an example in which the alternative hypothesis is two-sided.

Juicy Pineapples Stating hypotheses At the Hawaii Pineapple Company, managers are interested in the size of the pineapples grown in the company’s fields. Last year, the mean weight of the pineapples harvested from one large field was 31 ounces. A ­different irrigation system was installed in this field after the growing season. ­Managers wonder how this change will affect the mean weight of p ­ ineapples grown in the field this year.

Problem:  State appropriate hypotheses for performing a significance test. Be sure to define the parameter of interest. Solution:  The parameter of interest is the mean weight m of all pineapples grown in the field this year. Because managers wonder whether the mean weight of this year’s pineapples will differ from last year’s mean weight of 31 ounces, the alternative hypothesis is two-sided; that is, either m < 31 or m > 31. For simplicity, we write this as m ≠ 31. The null hypothesis says that there is no difference in the mean weight of the pineapples after the irrigation system was changed. That is, H0 : m = 31 Ha : m ≠ 31

1

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autio

!

n

The hypotheses should express the hopes or suspicions we have ­before we see the data. It is cheating to look at the data first and then frame hypotheses to fit what the data show. For example, the data for the pineapple study

c

For Practice Try Exercise

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Section 9.1 Significance Tests: The Basics

showed that x– = 31.935 ounces for a random sample of 50 pineapples grown in the field this year. You should not change the alternative hypothesis to Ha : m > 31 after looking at the data. If you do not have a specific direction firmly in mind in advance, use a two-sided alternative hypothesis.

Definition: One-sided alternative hypothesis and two-sided alternative hypothesis The alternative hypothesis is one-sided if it states that a parameter is larger than the null hypothesis value or if it states that the parameter is smaller than the null ­value. It is two-sided if it states that the parameter is different from the null ­hypothesis value (it could be either larger or smaller).

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The null hypothesis has the form H0 : parameter = value. The alternative ­hypothesis has one of the forms Ha : parameter < value, Ha : parameter > value, or Ha : parameter ≠ value. To determine the correct form of Ha, read the problem carefully. Hypotheses always refer to a population, not to a sample. Be sure to autio state H0 and Ha in terms of population parameters. It is never correct to write a hypothesis about a sample statistic, such as p^ = 0.64 or x– > 85.

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It is common to refer to a significance test with a one-sided alternative hypothesis as a one-sided test or one-tailed test and to a test with a two-sided alternative hypothesis as a two-sided test or two-tailed test.

Check Your Understanding

For each of the following settings, (a) describe the parameter of interest, and (b) state ­appropriate hypotheses for a significance test. 1. According to the Web site sleepdeprivation.com, 85% of teens are getting less than eight hours of sleep a night. Jannie wonders whether this result holds in her large high school. She asks an SRS of 100 students at the school how much sleep they get on a ­typical night. In all, 75 of the responders said less than 8 hours. 2. As part of its 2010 census marketing campaign, the U.S. Census Bureau advertised “10 questions, 10 minutes—that’s all it takes.” On the census form itself, we read, “The U.S. Census Bureau estimates that, for the average household, this form will take about 10 minutes to complete, including the time for reviewing the instructions and answers.” We suspect that the actual time it takes to complete the form may be longer than advertised.

The Reasoning of Significance Tests A significance test is sometimes referred to as a test of significance, a hypothesis test, or a test of hypotheses.

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Significance tests ask if sample data give convincing evidence against the null hypothesis and in favor of the alternative hypothesis. A test answers the question, “How likely is it to get a result like this just by chance when the null hypothesis is true?” The answer comes in the form of a probability. Here is an activity that introduces the underlying logic of significance tests.

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ACTIVITY MATERIALS: Copy of pie chart with 80% shaded and paper clip for each student

I’m a Great Free-Throw Shooter! Our virtual basketball player in the previous Activity claimed to be an 80% ­free-throw shooter. Suppose that he shoots 50 free throws and makes 32 of them. 32 His sample proportion of made shots is p^ = = 0.64. This result suggests that he 50 may really make less than 80% of his free throws in the long run. But do we have convincing evidence that p < 0.80? In this activity, you and your classmates will perform a simulation to find out. 1.  Make a spinner that gives the shooter an 80% chance of making a free throw. Using the pie chart provided by your teacher, label the 80% region “made shot” and the 20% region “missed shot.” Straighten out one of the ends of a paper clip so that there is a loop on one side and a pointer on the other. On a flat surface, place a pencil through the loop, and put the tip of the pencil on the center of the pie chart. Then flick the paper clip and see where the pointed end lands. 2.  Simulate a random sample of 50 shots. Flick the paper clip 50 times, and count the number of times that the pointed end lands in the “made shot” region. 3.  Compute the sample proportion p^ of made shots in your simulation from Step 2. Plot this value on the class dotplot drawn by your teacher. 4.  Repeat Steps 2 and 3 as needed to get at least 40 trials of the simulation for your class. 5.  Based on the results of your simulation, how likely is it for an 80% shooter to make 64% or less when he shoots 50 free throws? Does the observed p^ = 0.64 result give convincing evidence that the player is exaggerating?

Our reasoning in the Activity is based on asking what would happen if the player’s claim (p = 0.80) were true and we observed many samples of 50 free throws. We used Fathom software to simulate 400 sets of 50 shots assuming that the player is really an 80% shooter. Figure 9.1 shows a dotplot of the results. Each dot on the graph represents the proportion of made shots in one set of 50 attempts. For example, if the player makes 43/50 shots in one trial, the dot would be placed at p^ = 0.86. You can say how strong the evidence against the player’s claim is by giving the probability that he would make as few In 400 sets of 50 shots, as 32 out of 50 free throws if he really makes 80% in the long there were only 3 sets run. Based on the simulation, our estimate of this probabilwhen our shooter made ity is 3/400 = 0.0075. The observed statistic, p^ = 0.64, is so as few as or fewer than the observed p^ = 0.64 unlikely if the actual parameter value is p = 0.80 that it gives convincing evidence that the player’s claim is not true. Be sure that you understand why this evidence is convincing. There are two possible explanations of the fact that our virtual player made only p^ = 32/50 = 0.64 of his free throws: 0.5 0.6 0.7 0.8 0.9 ^ 1. The null hypothesis is correct (p = 0.8), and just by p chance, a very unlikely outcome occurred. ^ = 0.64 p 2. The alternative hypothesis is correct—the population FIGURE 9.1  Fathom dotplot of the sample proportion p^ ­proportion is less than 0.8, so the sample result is not an of free throws made by an 80% shooter in 400 sets of unlikely outcome. 50 shots. Figure 9.1 shows what sample proportions are likely to occur by chance alone, assuming that p = 0.80.

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Explanation 1 might be correct—the result of our random sample of 50 shots could be due to chance alone. But the probability that such a result would o­ ccur by chance is so small (less than 1 in a 100) that we are quite confident that ­Explanation 2 is right. Statistical tests use an elaborate vocabulary, but the basic idea is: an outcome that would rarely happen if the null hypothesis were true is good evidence that the null hypothesis is not true.

Interpreting P -Values The idea of stating a null hypothesis that we want to find evidence against seems odd at first. It may help to think of a criminal trial. The defendant is “innocent until proven guilty.” That is, the null hypothesis is innocence and the prosecution must try to provide convincing evidence against this hypothesis and in favor of the alternative hypothesis: guilt. That’s exactly how statistical tests work, although in statistics we deal with evidence provided by data and use a probability to say how strong the evidence is. The null hypothesis H0 states the claim that we are seeking evidence against. The probability that measures the strength of the evidence against H0 and in favor of Ha is called a P-value.

Definition: P -value The probability, computed assuming H0 is true, that the statistic (such as p^ or x–  ) would take a value as extreme as or more extreme than the one actually observed, in the direction specified by Ha, is called the P -value of the test. Small P-values are evidence against H0 because they say that the observed result is unlikely to occur when H0 is true. Large P-values fail to give convincing evidence against H0 and in favor of Ha because they say that the observed result is likely to occur by chance alone when H0 is true. We’ll show you how to calculate P-values later. For now, let’s focus on interpreting them.

Example

I’m a Great Free-Throw Shooter! Interpreting a P-value The P-value is the probability of getting a sample result at least as extreme as the one we did if H0 were true. Because the alternative hypothesis is Ha : p < 0.80, the sample results that count as “at least as extreme” are those with p^ ≤ 0.64. In other words, the P-value is the conditional probability P(p^ ≤ 0.64 | p = 0.80). Earlier, we used a simulation to estimate this probability as 3/400 = 0.0075. So if H0 is true and the player makes 80% of his free throws in the long run, there’s about a 0.0075 probability that the player would make 32 or fewer of 50 shots by chance alone. The small probability gives strong evidence against H0 and in favor of the alternative Ha : p < 0.80 because it would be so unlikely for this result to occur just by chance if H0 were true.

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The alternative hypothesis sets the direction that counts as evidence against H0. In the previous example, only values of p^ that are much less than 0.80 count as evidence against the null hypothesis because the alternative is one-sided on the low side. If the alternative is two-sided, both directions count.

Example

Healthy Bones Interpreting a P-value Calcium is a vital nutrient for healthy bones and teeth. The National Institutes of Health (NIH) recommends a calcium intake of 1300 mg per day for teenagers. The NIH is concerned that teenagers aren’t getting enough calcium. Is this true? Researchers want to perform a test of H0 : m = 1300 Ha : m < 1300 where m is the true mean daily calcium intake in the population of teenagers. They ask a random sample of 20 teens to record their food and drink consumption for 1 day. The researchers then compute the calcium intake for each student. Data analysis reveals that x– = 1198 mg and sx = 411 mg. After checking that conditions were met, researchers performed a significance test and obtained a P-value of 0.1404.

Problem:  (a) Explain what it would mean for the null hypothesis to be true in this setting. (b) Interpret the P-value in context.

SOLUTION: (a) In this setting, H0 : m = 1300 says that the mean daily calcium intake in the population of teenagers is 1300 mg. If H0 is true, then teenagers are getting enough calcium, on average. (b) Assuming that the mean daily calcium intake in the teen population is 1300 mg, there is a 0.1404 probability of getting a sample mean of 1198 mg or less just by chance in a random sample of 20 teens. For Practice Try Exercise

11

Statistical Significance The final step in performing a significance test is to draw a conclusion about the competing claims you were testing. We will make one of two decisions based on the strength of the evidence against the null hypothesis (and in favor of the alternative hypothesis)—reject H0 or fail to reject H0. If our sample result is too unlikely to have happened by chance assuming H0 is true, then we’ll reject H0 and say that there is convincing evidence for Ha. Otherwise, we will fail to reject H0 and say that there is not convincing evidence for Ha. This wording may seem unusual at first, but it’s consistent with what happens in a criminal trial. Once the jury has weighed the evidence against the null

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Example

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hypothesis of innocence, they return one of two verdicts: “guilty” (reject H0) or “not guilty” (fail to reject H0). A not-guilty verdict doesn’t guarantee that the defendant is innocent, just that there’s not convincing evidence of guilt. Likewise, a fail-to-reject H0 decision in a significance test doesn’t mean that H0 is autio true. For that reason, you should never “accept H0” or use language implying that you believe H0 is true.

Free Throws and Healthy Bones Drawing conclusions In the free-throw shooter example, because the estimated P-value of 0.0075 is so small, there is strong evidence against the null hypothesis H0 : p = 0.80. For that reason, we would reject H0 in favor of the alternative Ha : p < 0.80. We have convincing evidence that the virtual player makes fewer than 80% of his free throws. For the teen calcium study, however, the large P-value of 0.1404 gives weak evidence against H0 : m = 1300 and in favor of Ha : m < 1300. We therefore fail to reject H0. Researchers do not have convincing evidence that teens are getting less than 1300 mg of calcium per day, on average.

In a nutshell, our conclusion in a significance test comes down to P-value small ® reject H0 ® convincing evidence for Ha (in context) P-value large ® fail to reject H0 ® not convincing evidence for Ha (in context) There is no rule for how small a P-value is required to reject H0—it’s a matter of judgment and depends on the specific circumstances. But we can compare the P-value with a fixed value that we regard as decisive, called the significance level. We write it as a, the Greek letter alpha. If we choose a = 0.05, we are requiring that the data give evidence against H0 so strong that it would happen less than 5% of the time just by chance when H0 is true. If we choose a = 0.01, we are insisting on stronger evidence against the null hypothesis, a result that would occur less often than 1 in every 100 times by chance alone when H0 is true. In Chapter 4, we said that an observed result is “statistically significant” if it would rarely occur by chance alone. When our P-value is less than the chosen a in a significance test, we say that the result is statistically significant at level a.

Definition: Statistically significant at level a

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autio

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“Significant” in the statistical sense does not necessarily mean “important.” It means simply “not likely to happen just by chance.” The significance level a makes “not likely” more exact.

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If the P-value is smaller than alpha, we say that the results of a study are statistically significant at level a. In that case, we reject the null hypothesis H0 and conclude that there is convincing evidence in favor of the alternative hypothesis Ha.

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Significance at level 0.01 is often expressed by the statement “The results were significant (P < 0.01).” Here, P stands for the P-value. The actual P-value is more informative than a statement of significance because it allows us to assess significance at any level we choose. For example, a result with P = 0.03 is significant at the a = 0.05 level but is not significant at the a = 0.01 level. When we use a fixed significance level to draw a conclusion in a statistical test, P-value < a ® reject H0 ® convincing evidence for Ha (in context) P-value ≥ a ® fail to reject H0 ® not convincing evidence for Ha (in context)

Example

Better Batteries Statistical significance A company has developed a new deluxe AAA battery that is supposed to last longer than its regular AAA battery.2 However, these new batteries are more expensive to produce, so the company would like to be convinced that they really do last longer. Based on years of experience, the company knows that its regular AAA ­batteries last for 30 hours of continuous use, on average. The company selects an SRS of 15 new batteries and uses them continuously until they are c­ ompletely drained. The sample mean lifetime is x– = 33.9 hours. A significance test is ­performed ­using the hypotheses H0 : m = 30 hours Ha : m > 30 hours

AP® EXAM TIP The conclusion to a significance test should always include three components: (1) an explicit comparison of the P-value to a stated significance level, (2) a decision about the null hypothesis: reject or fail to reject H0, and (3) a statement in the context of the problem about whether or not there is convincing evidence for Ha.

where m is the true mean lifetime of the new deluxe AAA batteries. The resulting P-value is 0.0729.

Problem:  What conclusion would you make for each of the following significance levels? Justify

your answer. (a) a = 0.10

(b) a = 0.05

SOLUTION: (a) Because the P-value, 0.0729, is less than a = 0.10, we reject H0. We have convincing evidence that the company’s deluxe AAA batteries last longer than 30 hours, on average. (b) Because the P-value, 0.0729, is greater than a = 0.05, we fail to reject H0. We do not have convincing evidence that the company’s deluxe AAA batteries last longer than 30 hours, on average. For Practice Try Exercise

13

In practice, the most commonly used significance level is a = 0.05. This is mainly due to Sir Ronald A. Fisher, a famous statistician who worked on agricultural experiments in England during the early twentieth century. Fisher was the first to suggest deliberately using random assignment in an experiment. In a paper published in 1926, Fisher wrote that it is convenient to draw the line at

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about the level at which we can say: “Either there is something in the treatment, or a coincidence has occurred such as does not occur more than once in twenty trials.”3 Sometimes it may be preferable to choose a = 0.01 or a = 0.10, for reasons we will discuss shortly. Warning: if you are going to draw a conclusion based on a significance level a, then a should be stated before the data are produced. Otherwise, a deceptive user of statistics might set an a level after the data have been analyzed in an attempt to manipulate the conclusion. This is just as inappropriate as choosing an alternative hypothesis to be one-sided in a particular direction after looking at the data.

THINK ABOUT IT

How do you choose a significance level?  The purpose of a significance test is to give a clear statement of the strength of evidence provided by the data against the null hypothesis and in favor of the alternative hypothesis. The P-value does this. But how small a P-value is convincing evidence against the null hypothesis? This depends mainly on two circumstances: • How plausible is H0? If H0 represents an assumption that the people you must convince have believed for years, strong evidence (a very small P-value) will be needed to persuade them. • What are the consequences of rejecting H0? If rejecting H0 in favor of Ha means making an expensive change of some kind, you need strong evidence that the change will be beneficial. These criteria are a bit subjective. Different people will insist on different levels of significance. Giving the P-value allows each of us to decide individually if the evidence is sufficiently strong. Users of statistics have often emphasized standard significance levels such as 10%, 5%, and 1%. The 5% level, a = 0.05, is very common. For example, courts have tended to accept 5% as a standard in discrimination cases.4 Beginning users of statistical tests generally find it easier to compare a P-value to a significance level than to interpret the P-value correctly in context. For that reason, we will include stating a significance level as a required part of every significance test. We’ll also ask you to explain what a P-value means in a variety of settings.

Type I and Type II Errors When we draw a conclusion from a significance test, we hope our conclusion will be correct. But sometimes it will be wrong. There are two types of mistakes we can make. We can reject the null hypothesis when it’s actually true, known as a Type I error, or we can fail to reject H0 when the alternative hypothesis is true, which is a Type II error.

Definition: Type I error and Type II error If we reject H0 when H0 is true, we have committed a Type I error. If we fail to reject H0 when Ha is true, we have committed a Type II error.

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The possibilities are summarized in Figure 9.2. If H0 is true, our conclusion is correct if we fail to reject H0, but it is a Type I error if we reject H0. If Ha is true, our conclusion is correct if we H0 true Ha true reject H0, but is a Type II error if we fail to reject H0. Only one Correct Conclusion Reject H0 Type I error error is possible at a time. conclusion based on It is important to be able to describe Type I and Type II errors sample Correct Type II error Fail to reject H0 in the context of a problem. Considering the consequences of conclusion each of these types of error is also important, as the following FIGURE 9.2  The two types of errors in significance tests. example shows. Truth about the population

Example

Perfect Potatoes Type I and Type II errors A potato chip producer and its main supplier agree that each shipment of potatoes must meet certain quality standards. If the producer determines that more than 8% of the potatoes in the shipment have “blemishes,” the truck will be sent away to get another load of potatoes from the supplier. Otherwise, the entire truckload will be used to make potato chips. To make the decision, a supervisor will inspect a random sample of potatoes from the shipment. The producer will then perform a significance test using the hypotheses H0 : p = 0.08 Ha : p > 0.08 where p is the actual proportion of potatoes with blemishes in a given truckload.

Problem:  Describe a Type I and a Type II error in this setting, and explain the consequences of

each.

SOLUTION:  A Type I error occurs if we reject H0 when H0 is true. That would happen if the

Here’s a helpful reminder to keep the two types of errors straight. “Fail to” goes with Type II.

producer finds convincing evidence that the proportion of potatoes with blemishes is greater than 0.08 when the actual proportion is 0.08 (or less). Consequence: The potato-chip producer sends the truckload of acceptable potatoes away, which may result in lost revenue for the supplier. Furthermore, the producer will have to wait for another shipment of potatoes before producing the next batch of potato chips. A Type II error occurs if we fail to reject H0 when Ha is true. That would happen if the producer does not find convincing evidence that more than 8% of the potatoes in the shipment have blemishes when that is actually the case. Consequence: The producer uses the truckload of potatoes to make potato chips. More chips will be made with blemished potatoes, which may upset customers. For Practice Try Exercise

21

Which is more serious: a Type I error or a Type II error? That depends on the situation. For the potato-chip producer, a Type II error could result in upset customers, leading to decreased sales. A Type I error, turning away a shipment even though 8% or less of the potatoes have blemishes, may not have much impact if additional shipments of potatoes can be obtained fairly easily. However, the supplier won’t be too happy with a Type I error.

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Check Your Understanding Refer to the “Better Batteries” example on page 546. 1. ​Describe a Type I error in this setting. 2. Describe a Type II error in this setting. 3. Which type of error is more serious in this case? Justify your answer.

Error Probabilities  ​We can assess the performance of a significance test by looking at the probabilities of the two types of error. That’s because statistical inference is based on asking, “What would happen if I repeated the data-production process many times?” We cannot (without inspecting the whole truckload) guarantee that good shipments of potatoes will never be sent away and bad shipments will never be used to make chips. But we can think about our chances of making each of these mistakes.

Example

Perfect Potatoes Type I error probability For the truckload of potatoes in the previous example, we were testing H0 : p = 0.08 Ha : p > 0.08 where p is the proportion of all potatoes with blemishes in the shipment. Suppose that the potato-chip producer decides to carry out this test based on a random sample of 500 potatoes using a 5% significance level (a = 0.05). A Type I error is to reject H0 when H0 is actually true. If our sample results in a value of p^ that is much larger than 0.08, we will reject H0. How large would p^ need to be? The 5% significance level tells us to count results that could happen less than 5% of the time by chance if H0 is true as evidence that H0 is false. Assuming H0 : p = 0.08 is true, the sampling distribution of p^ will have Shape: Approximately Normal because np = 500(0.08) = 40 and n(1 − p)= 500(0.92) = 460 are both at least 10. Center: mp^ = p = 0.08 p(1 − p) 0.08(0.92) = = 0.01213, assuming that there are at n Å Å 500 least 10(500) = 5000 potatoes in the shipment.

Spread: sp^ =

Figure 9.3 on the next page shows the Normal curve that approximates this sampling distribution.

The shaded area in the right tail of Figure 9.3 is 5%. We used the calculator command invNorm(area: .95, m: .08, s: .01213) to get the boundary value p^ = 0.10. ­Values of p^ to the right of the green line at p^ = 0.10 will cause us to reject H0 even though H0 is true. This will happen in 5% of all possible samples. That is, the probability of making a Type I error is 0.05.

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Sampling distribution of ^ if H : p = 0.08 is true p 0

Fail to reject H0

Reject H0 If H0 is true, a decision to reject H0 based on the data is a Type I error. P(Type I error) = α 0.05

0.0437

0.0558

0.0679

0.0800

0.0921

0.1042

0.1163

Values of p^ ^ = 0.10 p

FIGURE 9.3  The probability of a Type I error (shaded area) is the probability of r­ ejecting H0 : p = 0.08 when H0 is actually true.

The probability of a Type I error is the probability of rejecting H0 when it is true. As the previous example showed, this is exactly the significance level of the test.

Significance and Type I Error The significance level a of any fixed-level test is the probability of a Type I ­error. That is, a is the probability that the test will reject the null hypothesis H0 when H0 is actually true. Consider the consequences of a Type I error before choosing a significance level.

What about Type II errors? We’ll discuss them at the end of Section 9.2, after you have learned how to carry out a significance test.

Section 9.1

Summary • •

•

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A significance test assesses the evidence provided by data against a null ­hypothesis H0 and in favor of an alternative hypothesis Ha. The hypotheses are usually stated in terms of population parameters. Often, H0 is a statement of no change or no difference. The alternative hypothesis states what we hope or suspect is true. A one-sided alternative Ha says that a parameter differs from the null hypothesis value in a specific direction. A two-sided alternative Ha says that a parameter differs from the null value in either direction.

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•

• •

•

•

Section 9.1

The reasoning of a significance test is as follows. Suppose that the null hypothesis is true. If we repeated our data production many times, would we often get data as inconsistent with H0, in the direction specified by Ha, as the data we actually have? If the data are unlikely when H0 is true, they provide evidence against H0 and in favor of Ha. The P-value of a test is the probability, computed supposing H0 to be true, that the statistic will take a value at least as extreme as the observed result in the direction specified by Ha. Small P-values indicate strong evidence against H0. To calculate a P-value, we must know the sampling distribution of the test statistic when H0 is true. If the P-value is smaller than a specified value a (called the significance level), the data are statistically significant at level a. In that case, we can reject H0 and say that we have convincing evidence for Ha. If the P-value is greater than or equal to a, we fail to reject H0 and say that we do not have convincing evidence for Ha. A Type I error occurs if we reject H0 when it is in fact true. In other words, the data give convincing evidence for Ha when the null hypothesis is correct. A Type II error occurs if we fail to reject H0 when Ha is true. In other words, the data don’t give convincing evidence for Ha, even though the alternative hypothesis is correct. In a fixed level a significance test, the probability of a Type I error is the significance level a.

Exercises handed. He wonders if the proportion of lefties at his large community college is really 12%. Simon chooses an SRS of 100 students and records whether each student is right- or left-handed.

In Exercises 1 to 6, each situation calls for a significance test. State the appropriate null hypothesis H0 and ­alternative hypothesis Ha in each case. Be sure to define your parameter each time. 1. pg 540

Attitudes  The Survey of Study Habits and Attitudes (SSHA) is a psychological test that measures students’ attitudes toward school and study habits. Scores range from 0 to 200. The mean score for U.S. college students is about 115. A teacher suspects that older students have better attitudes toward school. She gives the SSHA to an SRS of 45 of the over 1000 students at her college who are at least 30 years of age.

2.

Anemia  Hemoglobin is a protein in red blood cells that carries oxygen from the lungs to body tissues. People with less than 12 grams of hemoglobin per deciliter of blood (g/dl) are anemic. A public health official in Jordan suspects that Jordanian children are at risk of anemia. He measures a random sample of 50 children.

3.

Lefties  Simon reads a newspaper report claiming that 12% of all adults in the United States are left-

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4.

Don’t argue!  A Gallup Poll report revealed that 72% of teens said they seldom or never argue with their friends.5 Yvonne wonders whether this result holds true in her large high school. So she surveys a random sample of 150 students at her school.

5.

Cold cabin?  During the winter months, the temperatures at the Colorado cabin owned by the Starnes family can stay well below freezing (32°F or 0°C) for weeks at a time. To prevent the pipes from freezing, Mrs. Starnes sets the thermostat at 50°F. The manufacturer claims that the thermostat allows variation in home temperature of s = 3°F. Mrs. Starnes suspects that the manufacturer is overstating how well the thermostat works.

6.

Ski jump  When ski jumpers take off, the distance they fly varies considerably depending on their speed, skill, and wind conditions. Event organizers

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must position the landing area to allow for differences in the distances that the athletes fly. For a particular competition, the organizers estimate that the variation in distance flown by the athletes will be s = 10 meters. An experienced jumper thinks that the organizers are underestimating the variation.

15. Attitudes  Refer to Exercise 11. What conclusion would you make if a = 0.05? If a = 0.01? Justify your answers.

In Exercises 7 to 10, explain what’s wrong with the stated hypotheses. Then give correct hypotheses. 7.

8.

9.

Better parking  A change is made that should ­improve student satisfaction with the parking situation at a local high school. Right now, 37% of students approve of the parking that’s provided. The null hypothesis H0 : p > 0.37 is tested against the alternative Ha : p = 0.37.

16. Anemia  Refer to Exercise 12. What conclusion would you make if a = 0.05? If a = 0.01? Justify your answers. 17. Interpreting a P-value  When asked to explain the meaning of the P-value in Exercise 13, a student says, “This means there is about a 22% chance that the null hypothesis is true.” Explain why the student’s explanation is wrong.

Better parking  A change is made that should improve student satisfaction with the parking situation at your school. Right now, 37% of students approve of the parking that’s provided. The null hypothesis H0 : p^ = 0.37 is tested against the alternative Ha : p^ ≠ 0.37.

18. Interpreting a P-value  When asked to explain the meaning of the P-value in Exercise 14, a student says, “There is a 0.0291 probability of getting a sample proportion of p^ = 96/150 = 0.64 by chance alone.” Explain why the student’s explanation is wrong.

Birth weights  In planning a study of the birth weights of babies whose mothers did not see a doctor before delivery, a researcher states the hypotheses as

19. Drawing conclusions  A student performs a test of H0 : p = 0.75 versus Ha : p > 0.75 and gets a P-value of 0.99. The student writes: “Because the P-value is greater than 0.75, we reject H0. The data prove that Ha is true.” Explain what is wrong with this conclusion.

H0 : x– = 1000 grams Ha : x– < 1000 grams 10. Birth weights  In planning a study of the birth weights of babies whose mothers did not see a doctor before delivery, a researcher states the hypotheses as H0 : m < 1000 grams

20. Drawing conclusions  A student performs a test of H0: p = 0.5 versus Ha: p ≠ 0.5 and gets a P-value of 0.63. The student writes: “Because the P-value is greater than a = 0.05, we accept H0. The data provide convincing evidence that the null hypothesis is true.” Explain what is wrong with this conclusion.

Ha : m = 900 grams 11. Attitudes  In the study of older students’ attitudes pg 544 from Exercise 1, the sample mean SSHA score was 125.7 and the sample standard deviation was 29.8. A significance test yields a P-value of 0.0101.

Exercises 21 and 22 refer to the following setting. Slow response times by paramedics, firefighters, and policemen can have serious consequences for accident victims. In the case of life-threatening injuries, victims generally need medical attention within 8 minutes of the accident. Several cities have begun to monitor emergency response times. In one such city, the mean response time to all ­accidents involving life-threatening injuries last year was m = 6.7 minutes. Emergency personnel arrived within 8 minutes on 78% of all calls involving life-threatening injuries last year. The city manager shares this information and encourages these first responders to “do better.” At the end of the year, the city manager selects an SRS of 400 calls involving life-threatening injuries and examines the response times.

(a) Explain what it would mean for the null hypothesis to be true in this setting. (b) Interpret the P-value in context. 12. Anemia  For the study of Jordanian children in ­Exercise 2, the sample mean hemoglobin level was 11.3 g/dl and the sample standard deviation was 1.6 g/dl. A significance test yields a P-value of 0.0016. (a) Explain what it would mean for the null hypothesis to be true in this setting. (b) Interpret the P-value in context. 13. Lefties  Refer to Exercise 3. In Simon’s SRS, 16 of the pg 546 students were left-handed. A significance test yields a P-value of 0.2184. What conclusion would you make if a = 0.10? If a = 0.05? Justify your answers.

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21. Awful accidents pg 548

(a) State hypotheses for a significance test to determine whether the average response time has decreased. Be sure to define the parameter of interest.

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Section 9.1 Significance Tests: The Basics (b) Describe a Type I error and a Type II error in this setting, and explain the consequences of each.

(a) H0 : m = 18; Ha : m ≠ 18.

(c) Which is more serious in this setting: a Type I error or a Type II error? Justify your answer.

(c) H0 : m < 18; Ha : m = 18.

(b) H0 : m = 18; Ha : m > 18.

22. Awful accidents

(d) H0 : m = 18; Ha : m < 18.

(a) State hypotheses for a significance test to determine whether first responders are arriving within 8 minutes of the call more often. Be sure to define the parameter of interest.

(e) H0 : x– = 18; Ha : x– < 18.

(b) Describe a Type I error and a Type II error in this setting and explain the consequences of each. (c) Which is more serious in this setting: a Type I error or a Type II error? Justify your answer. 23. Opening a restaurant  You are thinking about o­ pening a restaurant and are searching for a good l­ocation. From research you have done, you know that the mean income of those living near the restaurant must be over $85,000 to support the type of upscale restaurant you wish to open. You decide to take a simple random sample of 50 people living near one potential location. Based on the mean income of this sample, you will decide whether to open a restaurant there.6 (a) State appropriate null and alternative hypotheses. Be sure to define your parameter. (b) Describe a Type I and a Type II error, and explain the consequences of each. 24. Blood pressure screening  Your company markets a computerized device for detecting high blood pressure. The device measures an individual’s blood pressure once per hour at a randomly selected time throughout a 12-hour period. Then it calculates the mean systolic (top number) pressure for the sample of measurements. Based on the sample results, the device determines whether there is convincing evidence that the individual’s actual mean systolic pressure is greater than 130. If so, it recommends that the person seek medical attention. (a) State appropriate null and alternative hypotheses in this setting. Be sure to define your parameter. (b) Describe a Type I and a Type II error, and explain the consequences of each. Multiple choice: Select the best answer for Exercises 25 to 28. 25. Experiments on learning in animals sometimes measure how long it takes mice to find their way through a maze. The mean time is 18 seconds for one particular maze. A researcher thinks that a loud noise will cause the mice to complete the maze faster. She measures how long each of 10 mice takes with a noise as stimulus. The appropriate hypotheses for the significance test are

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553

Exercises 26–28 refer to the following setting. Members of the city council want to know if a majority of city ­residents supports a 1% increase in the sales tax to fund road repairs. To investigate, they survey a random sample of 300 city residents and use the results to test the following hypotheses: H0 : p = 0.50 Ha : p > 0.50 where p is the proportion of all city residents who support a 1% increase in the sales tax to fund road repairs. 26. A Type I error in the context of this study occurs if the city council (a) finds convincing evidence that a majority of residents supports the tax increase, when in reality there isn’t convincing evidence that a majority supports the increase. (b) finds convincing evidence that a majority of residents supports the tax increase, when in reality at most 50% of city residents support the increase. (c) finds convincing evidence that a majority of residents supports the tax increase, when in reality more than 50% of city residents do support the increase. (d) does not find convincing evidence that a majority of residents supports the tax increase, when in reality more than 50% of city residents do support the increase. (e) does not find convincing evidence that a majority of residents supports the tax increase, when in reality at most 50% of city residents do support the increase. 27. In the sample, p^ = 158/300 = 0.527. The resulting P-value is 0.18. What is the correct interpretation of this P-value? (a) Only 18% of the city residents support the tax increase. (b) There is an 18% chance that the majority of residents supports the tax increase. (c) Assuming that 50% of residents support the tax increase, there is an 18% probability that the sample proportion would be 0.527 or higher by chance alone. (d) Assuming that more than 50% of residents support the tax increase, there is an 18% probability that the sample proportion would be 0.527 or higher by chance alone. (e) Assuming that 50% of residents support the tax increase, there is an 18% chance that the null ­hypothesis is true by chance alone.

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28. Based on the P-value in Exercise 27, which of the following would be the most appropriate conclusion? (a) Because the P-value is large, we reject H0. We have convincing evidence that more than 50% of city residents support the tax increase. (b) Because the P-value is large, we fail to reject H0. We have convincing evidence that more than 50% of city residents support the tax increase. (c) Because the P-value is large, we reject H0. We have convincing evidence that at most 50% of city residents support the tax increase. (d) Because the P-value is large, we fail to reject H0. We have convincing evidence that at most 50% of city residents support the tax increase. (e) Because the P-value is large, we fail to reject H0. We do not have convincing evidence that more than 50% of city residents support the tax increase. 29. Women in math (5.3)  Of the 24,611 degrees in mathematics given by U.S. colleges and universities in a recent year, 70% were bachelor’s degrees, 24% were master’s degrees, and the rest were doctorates. M ­ oreover,

9.2 What You Will Learn

women earned 43% of the bachelor’s degrees, 41% of the master’s degrees, and 29% of the doctorates.7 (a) How many of the mathematics degrees given in this year were earned by women? Justify your answer. (b) Are the events “degree earned by a woman” and “degree was a bachelor’s degree” independent? Justify your answer using appropriate probabilities. (c) If you choose 2 of the 24,611 mathematics degrees at random, what is the probability that at least 1 of the 2 degrees was earned by a woman? Show your work. 30. Explaining confidence (8.2)  Here is an explanation from a newspaper concerning one of its opinion polls. Explain what is wrong with the following ­statement. For a poll of 1,600 adults, the variation due to sampling error is no more than three percentage points either way. The error margin is said to be valid at the 95 percent confidence level. This means that, if the same questions were repeated in 20 polls, the results of at least 19 surveys would be within three percentage points of the results of this survey.

Tests about a Population Proportion By the end of the section, you should be able to:

• State and check the Random, 10%, and Large Counts conditions for performing a significance test about a population proportion. • Perform a significance test about a population ­proportion.

• Interpret the power of a test and describe what factors affect the power of a test. • Describe the relationship among the probability of a Type I error (significance level), the probability of a Type II error, and the power of a test.

Confidence intervals and significance tests are based on the sampling distributions of statistics. That is, both use probability to say what would happen if we used the inference method many times. Section 9.1 presented the reasoning of significance tests, including the idea of a P-value. In this section, we focus on the details of testing a claim about a population proportion.

Carrying Out a Significance Test In Section 9.1, we met a virtual basketball player who claimed to make 80% of his free throws. We thought that he might be exaggerating. In an SRS of 50 shots, the player made only 32. His sample proportion of made free throws was therefore p^ =

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32 = 0.64 50

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Section 9.2 Tests about a Population Proportion

555

This result is much lower than what he claimed. Does it provide convincing evidence against the player’s claim? To find out, we need to perform a significance test of H0 : p = 0.80 Ha : p < 0.80 where p = the actual proportion of free throws that the shooter makes in the long run.

Conditions  ​In Chapter 8, we introduced conditions that should be met be-

fore we construct a confidence interval for an unknown population proportion: ­Random, 10% when sampling without replacement, and Large Counts. These same conditions must be verified before carrying out a significance test. The Large Counts condition for proportions requires that both np and n(1 − p) be at least 10. Because we assume H0 is true when performing a significance test, we use the parameter value specified by the null hypothesis (sometimes called p0) when checking the Large Counts condition. In this case, the Large Counts condition says that the expected count of successes np0 and failures n(1 − p0) are both at least 10.

Conditions for Performing a Significance Test about a Proportion • Random: The data come from a well-designed random sample or randomized experiment. 1 ~~ 10%: When sampling without replacement, check that n ≤ N. 10 • Large Counts: Both np0 and n(1 − p0) are at least 10. Here’s an example that shows how to check the conditions.

Example

I’m a Great Free-Throw Shooter! Checking conditions Problem:  Check the conditions for performing a significance test of the virtual basketball

player’s claim.

SOLUTION:  The required conditions are •  Random:  We can view this set of 50 computer-generated shots as a simple random sample from the population of all possible shots that the virtual shooter takes. ºº 10%:  We’re not sampling without replacement from a finite population (because the applet can keep on shooting), so we don’t need to check the 10% condition. (Note that the outcomes of individual shots are independent because they are determined by the computer’s random number generator.) •  Large Counts: Assuming H0 is true, p = 0.80. Then np0 = (50)(0.80) = 40 and n (1 − p0) = (50)(0.20) = 10 are both at least 10, so this condition is met. For Practice Try Exercise

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31

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N(0.80,0.0566)

If the null hypothesis H0 : p = 0.80 is true, then the player’s sample proportion p^ of made free throws in an SRS of 50 shots would vary according to an approximately Normal sampling distribution with mean mp^ = p = 0.80 and standard deviation p(1 − p) (0.80)(0.20) = = 0.0566. n Å Å 50

sp^ =

Figure 9.4 displays this distribution. We have added the play32 = 0.64. er’s sample result, p^ = 50 0.6302

0.6868

0.7434

0.8000

0.8566

0.9132

0.9698

Calculations: Test Statistic and P-Value  A sig-

nificance test uses sample data to measure the strength of evidence against H0 and in favor of Ha. Here are some principles that apply to most tests: FIGURE 9.4  Normal approximation to the sampling distribution of the proportion p^ of made shots in random • The test compares a statistic calculated from sample data samples of 50 free throws by an 80% shooter. with the value of the parameter stated by the null hypothesis. • Values of the statistic far from the null parameter value in the direction specified by the alternative hypothesis give strong evidence against H0. • To assess how far the statistic is from the parameter, standardize the statistic. This standardized value is called the test statistic: ^ = 0.64 p

On the AP® exam formula sheet, this value is referred to as the “standardized test statistic.”

test statistic =

statistic − parameter standard deviation of statistic

Definition: Test statistic A test statistic measures how far a sample statistic diverges from what we would expect if the null hypothesis H0 were true, in standardized units. That is, test statistic =

statistic − parameter standard deviation of statistic

The test statistic says how far the sample result is from the null parameter value, and in what direction, on a standardized scale. You can use the test statistic to find the P-value of the test, as the following example shows.

Example

I’m a Great Free-Throw Shooter! Computing the test statistic Problem:  In an SRS of 50 free throws, the virtual player made 32. (a) Calculate the test statistic. (b) Find the P-value using Table A or technology. Show this result as an area under a standard Normal curve.

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Section 9.2 Tests about a Population Proportion

SOLUTION: (a) His sample proportion of made shots is p^ = 0.64. Standardizing, we get test statistic =

statistic − parameter

standard deviation of statistic 0.64 − 0.80 0.64 − 0.80 z= = = −2.83 0.0566 (0.80)(0.20)

Å 50 (b) The shaded area under the curve in Figure 9.5(a) shows the P-value. Figure 9.5(b) shows the corresponding area on the standard Normal curve, which displays the distribution of the z test statistic. From Table A, we find that the P-value is P (Z ≤ −2.83) = 0.0023. Using technology: The command normalcdf (lower: −10000, upper: −2.83, m : 0, s : 1) also gives a P-value of 0.0023. (a)

(b)

N(0.80,0.0566)

P-value = 0.0023

P-value = 0.0023

0.6302

p^ = 0.64

0.6868

N(0,1)

0.7434

0.8000

0.8566

0.9132

0.9698

–3

–2

–1

0

1

2

3

z = – 2.83

FIGURE 9.5  The shaded area shows the P-value for the player’s sample proportion of made shots (a) on the Normal approximation to the sampling distribution of p^ from Figure 9.4 and (b) on the standard Normal curve.

If H0 is true and the player makes 80% of his free throws in the long run, there’s only about a 0.0023 probability that he would make 32 or fewer of 50 shots by chance alone. For Practice Try Exercise

35

Earlier, we used simulation to estimate the P-value as 0.0075. As the example shows, the P-value is even smaller, 0.0023. So if H0 is true, and the player makes 80% of his free throws in the long run, there’s only about a 0.0023 probability that the player would make 32 or fewer of 50 shots by chance alone. This result confirms our earlier ­decision to reject H0 and gives convincing evidence that the player is exaggerating.

The One-Sample z  Test for a Proportion To perform a significance test, we state hypotheses, check conditions, calculate a test statistic and P-value, and draw a conclusion in the context of the problem. The four-step process is ideal for organizing our work.

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Significance Tests: A Four-Step Process STEP

4

State: What hypotheses do you want to test, and at what significance level? Define any parameters you use. Plan:  Choose the appropriate inference method. Check conditions. Do:  If the conditions are met, perform calculations. • Compute the test statistic. • Find the P-value. Conclude:  Make a decision about the hypotheses in the context of the problem. When the conditions are met—Random, 10%, and Large Counts—the sampling distribution of p^ is approximately Normal with p(1 − p) mean mp^ = p  and  standard deviation sp^ = n Å

For confidence intervals, we substitute p^ for p in the standard deviation formula to obtain the standard error. When performing a significance test, however, the null hypothesis specifies a value for p, which we call p0. We assume that this value is correct when performing our calculations. If we standardize the statistic p^ by subtracting its mean and dividing by its standard deviation, we get the test statistic: test statistic = z=

statistic − parameter standard deviation of statistic p^ − p0 p0(1 − p0) n Å

This z statistic has approximately the standard Normal distribution when H0 is true and the conditions are met. P-values therefore come from the standard ­Normal distribution. Here is a summary of the details for a one-sample z test for a proportion. The AP® Statistics course outline calls this test a large-sample test for a proportion because it is based on a Normal approximation to the sampling distribution of p^ that becomes more accurate as the sample size increases.

One-Sample z Test for a Proportion Suppose the conditions are met. To test the hypothesis H0 : p = p0, compute the z statistic p^ − p0 z= p0(1 − p0) n Å Find the P-value by calculating the probability of getting a z statistic this large or larger in the direction specified by the alternative hypothesis Ha: Ha : p > p 0

z

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Ha : p = p0

Ha : p < p 0

  

z

  

–z

z

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559

Section 9.2 Tests about a Population Proportion Here is an example of the test in action.

Example

One Potato, Two Potato

STEP

Performing a significance test about p

4

The potato-chip producer of Section 9.1 has just received a truckload of potatoes from its main supplier. Recall that if the producer finds convincing evidence that more than 8% of the potatoes in the shipment have blemishes, the truck will be sent away to get another load from the supplier. A supervisor selects a random sample of 500 potatoes from the truck. An inspection reveals that 47 of the potatoes have blemishes. Carry out a significance test at the a = 0.05 significance level. What should the producer conclude?

STATE:  We want to perform a test of H0 : p  =  0.08 Ha : p  >  0.08 where p is the actual proportion of potatoes in this shipment with blemishes. We’ll use an a = 0.05 significance level. N(0,1)

Area = 0.1251

–3

–2

–1

0

1

2

3

z = 1.15

FIGURE 9.6  ​The P-value for the one-sided test. AP® EXAM TIP ​When a significance test leads to a fail to reject H0 decision, as in this example, be sure to interpret the results as “we don’t have convincing evidence to conclude Ha.” Saying anything that sounds like you believe H0 is (or might be) true will lead to a loss of credit. And don’t write text-message-type responses, like “FTR the H0.”

•  Test statistic  z =

PLAN:  If conditions are met, we should do a one-sample z test for the population proportion p. •  Random:  The supervisor took a random sample of 500 potatoes from the shipment. ºº 10%: It seems reasonable to assume that there are at least 10(500) = 5000 potatoes in the shipment. •  Large Counts: Assuming H0 : p = 0.08 is true, the expected counts of blemished and unblemished potatoes are np0 = 500(0.08) = 40 and n(1 − p0) = 500(0.92) = 460, respectively. Because both of these values are at least 10, we should be safe doing Normal calculations. DO:  The sample proportion of blemished potatoes is p^ = 47/500 = 0.094 p^ − p0 p0(1 − p0) n Å

=

0.094 − 0.08 0.08(0.92) Å 500

= 1.15

•  P-value  Figure 9.6 displays the P-value as an area under the standard Normal curve for this one-sided test. Table A gives the P-value as P(Z ≥ 1.15) = 1 − 0.8749 = 0.1251. •  Using technology:  The command normalcdf(lower:1.15, upper:10000, m:0, s:1) also gives a P-value of 0.1251.

CONCLUDE:  Because our P-value, 0.1251, is greater than a = 0.05, we fail to reject H0. There is not convincing evidence that the shipment contains more than 8% blemished potatoes. As a result, the producer will use this truckload of potatoes to make potato chips. For Practice Try Exercise

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39

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The preceding example reminds us why significance tests are important. The sample proportion of blemished potatoes was p^ = 47/500 = 0.094. This result gave evidence against H0 in favor of Ha. To see whether such an outcome is unlikely to occur just by chance when H0 is true, we had to carry out a significance test. The P-value told us that a sample proportion this large or larger would occur in about 12.5% of all random samples of 500 potatoes when H0 is true. So we can’t rule out sampling variability as a plausible explanation for getting a sample proportion of p^ = 0.094.

What happens when the data don’t support Ha?  ​Suppose the supervisor had inspected a random sample of 500 potatoes from the shipment and found 33 with blemishes. This yields a sample proportion of p^ = 33/500 = 0.066. Such a sample doesn’t give any evidence to support the alternative hypothesis Ha : p > 0.08! There’s no need to continue with the significance test. The conclusion is clear: we should fail to reject H0 : p = 0.08. This truckload of potatoes will be used by the potato-chip producer. If you weren’t paying attention, you might end up carrying out the test. Let’s see what would happen. The corresponding test statistic is

THINK ABOUT IT

z=

Area = 0.8749

N(0,1)

–3

–2

–1

0

1

2

z = –1.15

FIGURE 9.7  The P-value for the one-sided test.

p^ − p0 p0(1 − p0) n Å

=

0.066 − 0.08 0.08(0.92) Å 500

= −1.15

What’s the P-value? It’s the probability of getting a z statistic this large or larger in the direction specified by Ha, P(Z ≥ −1.15). ­Figure 9.7 shows this P-value as an area under the standard Normal curve. Using Table A or technology, the P-value is 1 − 0.1251 = 0.8749. There’s about an 87.5% chance of getting a sample proportion as large as or larger than p^ = 0.066 if p = 0.08. As a result, we would fail to reject H0. Same conclusion, but with lots of unnecessary work!

3

Always check to see whether the data give evidence against H0 in the direction specified by Ha before you do calculations.

Check Your Understanding

According to the National Campaign to Prevent Teen and Unplanned Pregnancy, 20% of teens aged 13 to 19 say that they have electronically sent or posted sexually suggestive images of themselves.8 The counselor at a large high school worries that the actual figure might be higher at her school. To find out, she administers an anonymous survey to a random sample of 250 of the school’s 2800 students. All 250 respond, and 63 admit to sending or posting sexual images. Carry out a significance test at the a = 0.05 significance level. What conclusion should the counselor draw?

Your calculator will handle the “Do” part of the four-step process, as the following Technology Corner illustrates. However, be sure to read the AP® Exam Tip on the next page.

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Section 9.2 Tests about a Population Proportion

18. Technology Corner

561

One-proportion z  test on the calculator TI-Nspire instructions in Appendix B; HP Prime instructions on the book’s Web site.

The TI-83/84 and TI-89 can be used to test a claim about a population proportion. We’ll demonstrate using the ­previous example. In a random sample of size n = 500, the supervisor found X = 47 potatoes with blemishes. To ­perform a significance test:

TI-83/84 TI-89

• Press STAT , then choose TESTS and 1-PropZTest.

• I n the Statistics/List Editor, press 2nd F1 ([F6]) and choose 1-PropZTest.

• On the 1-PropZTest screen, enter the values shown : p0 = 0.08, x = 47, and n = 500. Specify the alternative hypothesis as “prop > p0.” Note: x is the number of successes and n is the number of trials. Both must be whole numbers!





• If you select “Calculate” and press ENTER , you will see that the test statistic is z = 1.15 and the P-value is 0.1243.





• If you select the “Draw” option, you will see the screen shown here. Compare these results with those in the example on page 559.



AP® EXAM TIP  You can use your calculator to carry out the mechanics of a significance test on the AP® exam. But there’s a risk involved. If you just give the calculator answer with no work, and one or more of your values is incorrect, you will probably get no credit for the “Do” step. If you opt for the calculator-only method, be sure to name the procedure (oneproportion z test) and to report the test statistic (z = 1.15) and P-value (0.1243).

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Two-Sided Tests Both the free-throw shooter and blemished-potato examples involved one-sided tests. The P-value in a one-sided test is the area in one tail of a standard Normal ­distribution—the tail specified by Ha. In a two-sided test, the alternative hypothesis has the form Ha : p ≠ p0. The P-value in such a test is the probability of getting a sample proportion as far as or farther from p0 in either direction than the observed value of p^ . As a result, you have to find the area in both tails of a standard Normal distribution to get the P-value. The following example shows how this process works.

Example

Nonsmokers

STEP

A two-sided test

4

According to the Centers for Disease Control and Prevention (CDC) Web site, 50% of high school students have never smoked a cigarette. Taeyeon wonders whether this national result holds true in his large, urban high school. For his AP® Statistics class project, Taeyeon surveys an SRS of 150 students from his school. He gets responses from all 150 students, and 90 say that they have never smoked a cigarette. What should Taeyeon conclude? Give appropriate evidence to support your answer.

STATE:  We want to perform a significance test using the hypotheses H0 : p = 0.50 Ha : p ≠ 0.50 where p = the proportion of all students in Taeyeon’s school who would say they have never smoked cigarettes. Because no significance level was stated, we will use a = 0.05. PLAN:  If conditions are met, we’ll do a one-sample z test for the population proportion p. •  Random:  Taeyeon surveyed an SRS of 150 students from his school. ºº 10%:  It seems reasonable to assume that there are at least 10(150) = 1500 students in a large high school. •  Large Counts: Assuming H0  : p = 0.50 is true, the expected counts of smokers and nonsmokers in the sample are np0 = 150(0.50) = 75 and n (1 − p0) = 150(0.50) = 75. Because N(0,1) both of these values are at least 10, we should be safe doing Normal calculations. DO:  The sample proportion is p^ = 90/150 = 0.60. •  Test statistic

P-value = 2(0.0071) = 0.0142

z=

–3

–2

–1

0

1

z = -2.45

FIGURE 9.8  The P-value for the two-sided test.

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2

3 z = 2.45

p^ − p0 p0(1 − p0) n Å

=

0.60 − 0.50 0.50(0.50) Å 150

= 2.45

•  P-value  Figure 9.8 displays the P-value as an area under the standard Normal curve for this two-sided test. To compute this P-value, we find the area in one tail and double it. Table A gives P (z ≥ 2.45) = 0.0071 (the right-tail area). So the desired P-value is 2(0.0071) = 0.0142.

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563

Using technology: The calculator's 1-PropZTest gives z = 2.449 and P-value = 0.0143. CONCLUDE:  Because our P-value, 0.0143, is less than a = 0.05, we reject H0. We have convincing evidence that the proportion of all students at Taeyeon’s school who would say they have never smoked differs from the national result of 0.50. For Practice Try Exercise

45

Check Your Understanding

According to the National Institute for Occupational Safety and Health, job stress poses a major threat to the health of workers. A news report claims that 75% of restaurant employees feel that work stress has a negative impact on their personal lives.9 Managers of a large restaurant chain wonder whether this claim is valid for their employees. A random sample of 100 employees finds that 68 answer “Yes” when asked, “Does work stress have a negative impact on your personal life?” Is this good reason to think that the proportion of all employees in this chain who would say “Yes” differs from 0.75? Support your answer with a significance test.

Why Confidence Intervals Give More Information The result of a significance test begins with a decision to reject H0 or fail to reject H0. In Taeyeon’s smoking study, for instance, the data led us to reject H0 : p = 0.50 because we found convincing evidence that the proportion of students at his school who would say they have never smoked cigarettes differs from the national value. We’re left wondering what the actual proportion p might be. A confidence interval might shed some light on this issue.

Example

Nonsmokers A confidence interval gives more info Taeyeon found that 90 of an SRS of 150 students said that they had never smoked a cigarette. We checked the conditions for performing the significance test earlier. Before we construct a confidence interval for the population proportion p, we should check that both np^ and n(1 − p^ ) are at least 10. Because the number of successes and the number of failures in the sample are 90 and 60, respectively, we can proceed with calculations. Our 95% confidence interval is p^ ± z*

p^ (1 − p^) 0.60(0.40) = 0.60 ± 1.96 = 0.60 ± 0.078 = (0.522, 0.678) n Å Å 150

We are 95% confident that the interval from 0.522 to 0.678 captures the true proportion of students at Taeyeon’s high school who would say that they have never smoked a cigarette.

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The confidence interval in this example is much more informative than the significance test we performed earlier. The interval gives the values of p that are plausible based on the sample data. We would not be surprised if the true proportion of students at Taeyeon’s school who would say they have never smoked cigarettes was as low as 0.522 or as high as 0.678. However, we would be surprised if the true proportion was 0.50 because this value is not contained in the confidence interval. Figure 9.9 gives computer output from Minitab software that includes both the results of the significance test and the confidence interval.

FIGURE 9.9  Minitab output for the two-sided significance test at a = 0.05 and a 95% confidence interval for Taeyeon’s smoking study.

There is a link between confidence intervals and two-sided tests. The 95% confidence interval (0.522, 0.678) gives an approximate set of p0’s that would not be rejected by a two-sided test at the a = 0.05 significance level. With proportions, the link isn’t perfect because the standard error used for the confidence interval is based on the sample proportion p^ , while the denominator of the test statistic is based on the value p0 from the null hypothesis. Test statistic: z =

p^ − p0 p0(1 − p0) n Å

    Confidence interval: p^ ± z*

p^ (1 − p^) n Å

The big idea is still worth considering: a two-sided test at significance level a and a 100(1 − a)% confidence interval (a 95% confidence interval if a = 0.05) give similar information about the population parameter.

Check Your Understanding

The figure below shows Minitab output from a significance test and confidence interval for the restaurant worker data in the previous Check Your Understanding (page 563). Explain how the confidence interval is consistent with, but gives more information than, the test.

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Type II Error and the Power of a Test A significance test makes a Type II error when it fails to reject a null hypothesis H0 that really is false. There are many values of the parameter that make the alternative hypothesis Ha true, so we concentrate on one value. Consider the potato-chip example on page 559 that involved a test of H0 : p = 0.08 versus Ha : p > 0.08. If the true proportion of blemished potatoes in the shipment was p = 0.09, we made a Type II error by failing to reject H0 based on the sample data. Of course, we also made a Type II error if p = 0.11 or p = 0.15. The probability of making a Type II error depends on several factors, including the actual value of the parameter. In the potato-chip example, our test will be more likely to reject H0 : p = 0.08 in favor of Ha : p > 0.08 if the true proportion of blemished potatoes in the shipment is p = 0.11 than if it is p = 0.09. Why? because p = 0.11 is farther away from the null value than is p = 0.09. So we will be less likely to make a Type II error if 11% of potatoes in the shipment are blemished than if only 9% are blemished. A high probability of Type II error for a specific alternative parameter value means that the test is not sensitive enough to usually detect that alternative. It is more common to report the probability that a significance test does reject H0 when an alternative parameter value is true. This probability is called the power of the test against that specific alternative.

Definition: Power The power of a test against a specific alternative is the probability that the test will reject H0 at a chosen significance level a when the specified alternative value of the parameter is true. As the following example illustrates, Type II error and power are closely linked.

Example

Perfect Potatoes Type II error and power The potato-chip producer wonders whether the significance test of H0 : p = 0.08 versus Ha : p > 0.08 based on a random sample of 500 potatoes has enough power to detect a shipment with, say, 11% blemished potatoes. In this case, a particular Type II error is to fail to reject H0 : p = 0.08 when p = 0.11. Figure 9.10 on the next page shows two sampling distributions of p^ , one when p = 0.08 and the other when p = 0.11. Earlier, we decided to reject H0 if our sample yielded a value of p^ to the right of the green line at p^ = 0.10. That decision was based on using a significance level (Type I error probability) of a = 0.05. Now look at the sampling distribution for p = 0.11. The shaded area to the right of the green line represents the probability of correctly rejecting H0 when p = 0.11. That is, the power of this test to detect p = 0.11 is about 0.76. In other words, the potato-chip producer has roughly a 3-in-4 chance of rejecting a truckload with 11% blemished potatoes based on a random sample of 500 potatoes from the shipment. We would fail to reject H0 if the sample proportion p^ falls to the left of the green line. The white area under the bottom Normal distribution shows the probability

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of failing to reject H0 when H0 is false. This is the probability of a Type II error. The potato-chip producer has about a 1-in-4 chance of failing to send away a ­shipment with 11% blemished potatoes. Normal approximation to the sampling distribution of p^ if H0 : p = 0.08 is true Fail to reject H0

Reject H0 If H0 is true, a decision to reject H0 based on the data is a Type I error. P(Type I error) = α 0.05

0.0437

0.0558

0.0679

0.0800

0.0921

0.1042

0.1163

Values of p^ ^ = 0.10 p

Normal approximation to the sampling distribution of p^ if H0 is false and p = 0.11 is true

If H0 is false, a decision to fail to reject H0 based on the data is a Type II error. P(Type II error) = 1 – 0.7626 = 0.2374

The power of the test to detect that p = 0.11

0.7626

0.0680

0.0820

0.0960

0.1100

0.1240

0.1380

0.1520

Values of p^ ^ = 0.10 p

FIGURE 9.10  In the bottom graph, the power of the test (shaded area) is the probability that it correctly rejects H0 : p = 0.08 when the truth is p = 0.11. In this case, power = 0.7626. The probability of making a Type II error (white area) is 1 − 0.7626 = 0.2374.

After reading the example, you might be wondering whether 0.76 is a high power or a low power. That depends on how certain the potato-chip producer wants to be to detect a shipment with 11% blemished potatoes. The power of a test against a specific alternative value of the parameter (like p = 0.11) is a number between 0 and 1. A power close to 0 means the test has almost no chance of correctly detecting that H0 is false. A power near 1 means the test is very likely to reject H0 in favor of Ha when H0 is false. The significance level of a test is the probability of reaching the wrong conclusion when the null hypothesis is true. The power of a test to detect a specific alternative is the probability of reaching the right conclusion when that a­ lternative is true. We can just as easily describe the test by giving the probability of making a Type II error (sometimes called b).

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Power and Type II Error The power of a test against any alternative is 1 minus the probability of a Type II error for that alternative; that is, power = 1 − b. Calculating a Type II error probability or power by hand is possible but ­unpleasant. It’s better to let technology do the work for you.

ACTIVITY MATERIALS: Computer with Internet access and projection ­capability

What Affects the Power of a Test? A virtual basketball player claims to make 80% of his free throws. Suppose that the player is exaggerating—he really makes less than 80% in the long run. We have the computer player shoot 50 shots and record the sample proportion p^ of made free throws. We then use the sample result to perform a test of H0 : p = 0.80 Ha : p < 0.80 at the a = 0.05 significance level. How can we increase the power of the test to detect that the player is exaggerating? In this Activity, we will use an applet to investigate. 1.  Go to www.amstat.org/publications/jse/v11n3/java/Power and select the ­Proportions applet at the bottom of the page. 2.  Adjust the applet settings as follows: choose “One Sample” for the Test, “Less Than” for the Alternative, enter 0.8 for p1, 50 for n1, and 0.05 for a. The Null distribution should appear in the applet window.

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3.  Let’s assume for now that the virtual player really makes 65% of his free throws (p = 0.65). Drag your mouse to the left in the applet screen and watch as an Alternative distribution appears. Keep dragging until the Delta value in the top panel shows –0.1500. This sets the alternative parameter value to be 0.15 less than the null parameter value of 0.80. Click your mouse to set the Alternative distribution. The Power of the test to detect p = 0.65 is shown in the top panel: 0.8009. 4.  Sample size  Change the sample size from n = 50 shots to n = 100 shots. What happens in the bottom panel of the applet? Does the power increase or decrease? Explain why this makes sense. 5.  Significance level  Reset the sample size to n = 50. (a)  Change the significance level to a = 0.01. What happens in the bottom panel of the applet? Does the power increase or decrease? (b) Make a guess about what will happen if you change the significance level to a = 0.10. Use the applet to test your conjecture. (c) Explain what the results in parts (a) and (b) tell you about the relationship between Type I error probability and Type II error probability. 6.  Difference between null parameter value and alternative parameter value  Reset the sample size to n = 50 and the significance level to a = 0.05. Will we be more likely to detect that the player is really a 65% shooter or that he is really a 70% shooter? Use your mouse to adjust the location of the Alternative ­distribution. How does the power change? Explain why this makes sense.

Step 5 of the Activity reveals an important link between Type I and Type II e­ rror probabilities. Because P(Type I error) = a, increasing the significance level increases the chance of making a Type I error. As the applet shows, this change also increases the power of the test. Because P(Type II error) = 1 – Power, higher power means a smaller chance of making a Type II error. So increasing the Type I error probability a decreases the Type II error probability b. By the same logic, decreasing the chance of a Type I error results in a higher chance of a Type II error. Let’s summarize what the Activity reveals about how to increase the power of a significance test to detect when H0 is false and Ha is true.

Many researchers who design statistical studies refer to the difference that’s important to detect as the effect size.

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• Increase the sample size. As Step 4 of the Activity confirms, we get better information about the virtual player’s free-throw shooting from a random sample of 100 shots than from a random sample of 50 shots. Increasing the sample size decreases the spread of both the Null and Alternative distributions. This change decreases the amount of overlap between the two distributions, making it easier to detect a difference between the null and alternative parameter values. • Increase the significance level a. Using a larger value of a increases the area of the green and blue “reject H0” regions in both the Null and Alternative distributions. This change makes it more likely to get a sample proportion that leads us to correctly reject the null hypothesis when the shooter is ­exaggerating. • Increase the difference between the null and alternative parameter values that is important to detect. Step 6 of the Activity shows that it is easier to ­detect large differences between the null and alternative parameter values than smaller differences. The size of difference that is important to detect is usually determined by experts in the field, so the statistician usually gets little or no input on this factor.

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In addition to these three factors, we can also gain power by making wise choices when collecting data. For example, using blocking in an experiment or stratified random sampling can greatly increase the power of a test in some circumstances. Our best advice for maximizing the power of a test is to choose as high an a level (Type I error probability) as you are willing to risk and as large a sample size as you can afford.

CHECK YOUR UNDERSTANDING

Refer to the Perfect Potatoes example on page 565. 1. Which is more serious for the potato-chip producer in this setting: a Type I error or a Type II error? Based on your answer, would you choose a significance level of a = 0.01, 0.05, or 0.10? 2. Tell if each of the following would increase or decrease the power of the test. Justify your answers. (a) Change the significance level to a = 0.10. (b) Take a random sample of 250 potatoes instead of 500 potatoes. (c) Insist on being able to detect that p = 0.10 instead of p = 0.11.

Section 9.2

Summary •

•

The conditions for performing a significance test of H0 : p = p0 are: • Random: The data were produced by a well-designed random sample or randomized experiment. ~~ 10%: When sampling without replacement, check that the population is at least 10 times as large as the sample. • Large Counts: The sample is large enough to satisfy np0 ≥ 10 and n(1 – p0) ≥ 10 (that is, the expected counts of successes and failures are both at least 10). The one-sample z test for a population proportion is based on the test statistic z=

STEP

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4

p^ − p0

p0(1 − p0) n Å with P-values calculated from the standard Normal distribution. • Follow the four-step process when you are asked to carry out a significance test:  State: What hypotheses do you want to test, and at what significance level? Define any parameters you use. Plan:  Choose the appropriate inference method. Check conditions. Do:  If the conditions are met, perform calculations. • Compute the test statistic. • Find the P-value. Conclude:  Make a decision about the hypotheses in the context of the problem.

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•

•

• •

Confidence intervals provide additional information that significance tests do not—namely, a set of plausible values for the true population parameter p. A two-sided test of H0 : p = p0 at significance level a gives roughly the same conclusion as a 100(1 – a)% confidence interval. The power of a significance test against a specific alternative is the probability that the test will reject H0 when the alternative is true. Power measures the ability of the test to detect an alternative value of the parameter. For a specific alternative, P(Type II error) = 1 − power. There is an important link between the probabilities of Type I and Type II error in a significance test: as one increases, the other decreases. We can increase the power of a significance test by increasing the sample size, increasing the significance level, or increasing the difference that is important to detect between the null and alternative parameter values.

9.2 T echnology Corner TI-Nspire Instructions in Appendix B; HP Prime instructions on the book’s Web site. 18.  One-proportion z test on the calculator

Section 9.2

page 561

Exercises SRS of 200 of the college’s 15,000 living alumni to perform a test of H0 : p = 0.99 versus Ha : p < 0.99.

In Exercises 31 and 32, check that the conditions for ­carrying out a one-sample z test for the population ­proportion p are met. 31. Home computers  Jason reads a report that says pg 555 80% of U.S. high school students have a computer at home. He believes the proportion is smaller than 0.80 at his large rural high school. Jason chooses an SRS of 60 students and records whether they have a computer at home. 32. Walking to school  A recent report claimed that 13% of students typically walk to school.10 DeAnna thinks that the proportion is higher than 0.13 at her large elementary school, so she surveys a random sample of 100 students to find out. In Exercises 33 and 34, explain why we aren’t safe carrying out a one-sample z test for the population proportion p. 33. No test  You toss a coin 10 times to perform a test of H0 : p = 0.5 that the coin is balanced against Ha : p ≠ 0.5. 34. No test  A college president says, “99% of the alumni support my firing of Coach Boggs.” You contact an

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pg 556

35. Home computers  Refer to Exercise 31. In Jason’s SRS, 41 of the students had a computer at home. (a) Calculate the test statistic. (b) Find the P-value using Table A or technology. Show this result as an area under a standard Normal curve. 36. Walking to school  Refer to Exercise 32. For DeAnna’s survey, 17 students in the sample said they typically walk to school. (a) Calculate the test statistic. (b) Find the P-value using Table A or technology. Show this result as an area under a standard Normal curve. 37. Significance tests  A test of H0 : p = 0.5 versus Ha : p > 0.5 has test statistic z = 2.19. (a) What conclusion would you draw at the 5% significance level? At the 1% level? (b) If the alternative hypothesis were Ha : p ≠ 0.5, what conclusion would you draw at the 5% significance level? At the 1% level?

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Section 9.2 Tests about a Population Proportion 38. Significance tests  A test of H0 : p = 0.65 against Ha : p < 0.65 has test statistic z = –1.78. (a) What conclusion would you draw at the 5% significance level? At the 1% level? (b) If the alternative hypothesis were Ha : p ≠ 0.65, what conclusion would you draw at the 5% significance level? At the 1% level? 39. Better parking  A local high school makes a change that should improve student satisfaction with the parking situation. Before the change, 37% of the school’s students approved of the parking that was provided. After the change, the principal surveys an SRS of 200 of the over 2500 students at the school. In all, 83 students say that they approve of the new parking arrangement. The principal cites this as evidence that the change was effective. Perform a test of the principal’s claim at the a = 0.05 significance level.

pg 559

40. Side effects  A drug manufacturer claims that less than 10% of patients who take its new drug for treating Alzheimer’s disease will experience nausea. To test this claim, researchers conduct an experiment. They give the new drug to a random sample of 300 out of 5000 Alzheimer’s patients whose families have given informed consent for the patients to participate in the study. In all, 25 of the subjects experience nausea. Use these data to perform a test of the drug manufacturer’s claim at the a = 0.05 significance level. 41. Are boys more likely?  We hear that newborn babies are more likely to be boys than girls. Is this true? A random sample of 25,468 firstborn children included 13,173 boys.11 (a) Do these data give convincing evidence that firstborn children are more likely to be boys than girls? (b) To what population can the results of this study be generalized: all children or all firstborn children? Justify your answer. 42. Fresh coffee  People of taste are supposed to prefer fresh-brewed coffee to the instant variety. On the other hand, perhaps many coffee drinkers just want their caffeine fix. A skeptic claims that only half of all coffee drinkers prefer fresh-brewed coffee. To test this claim, we ask a random sample of 50 coffee drinkers in a small city to take part in a study. Each person tastes two unmarked cups— one containing instant coffee and one containing fresh-brewed coffee—and says which he or she prefers. We find that 36 of the 50 choose the fresh coffee. (a) Do these results give convincing evidence that coffee drinkers favor fresh-brewed over instant coffee?

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(b) We presented the two cups to each coffee drinker in a random order, so that some people tasted the fresh coffee first, while others drank the instant coffee first. Why do you think we did this? 43. Bullies in middle school  A University of Illinois study on aggressive behavior surveyed a random sample of 558 middle school students. When asked to describe their behavior in the last 30 days, 445 students said their behavior included physical aggression, social ridicule, teasing, name-calling, and issuing threats. This behavior was not defined as bullying in the questionnaire.12 Is this evidence that more than three-quarters of middle school students engage in bullying behavior? To find out, Maurice decides to perform a significance test. Unfortunately, he made a few errors along the way. Your job is to spot the mistakes and correct them. H0 : p = 0.75 Ha : p^ > 0.797

•

where p = the true mean proportion of middle school students who engaged in bullying. A random sample of 558 middle school students was surveyed.

• 558(0.797) = 444.73 is at least 10. z=

0.75 − 0.797 "0.797(0.203) 445

= −2.46; P-value = 2(0.0069) = 0.0138

The probability that the null hypothesis is true is only 0.0138, so we reject H0. This proves that more than three-quarters of the school engaged in bullying behavior. 44. Is this coin fair?  The French naturalist Count Buffon (1707–1788) tossed a coin 4040 times. He got 2048 heads. That’s a bit more than one-half. Is this evidence that Count Buffon’s coin was not balanced? To find out, Luisa decides to perform a significance test. Unfortunately, she made a few errors along the way. Your job is to spot the mistakes and correct them. H0 : m > 0.5 Ha : x– = 0.5

•

10%: 4040(0.5) = 2020 and 4040(1 – 0.5) = 2020 are both at least 10.

•

Large Counts:  There are at least 40,400 coins in the world.

t=

0.5 − 0.507

= −0.89; P-value = 1 − 0.1867 = 0.8133 0.5(0.5) Å 4040 Reject H0 because the P-value is so large and ­conclude that the coin is fair.

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45. Teen drivers  A state’s Division of Motor Vehicles (DMV) claims that 60% of teens pass their driving test on the first attempt. An investigative reporter examines an SRS of the DMV records for 125 teens; 86 of them passed the test on their first try. Is there convincing evidence at the a = 0.05 significance level that the DMV’s claim is incorrect?

pg 562

46. We want to be rich  In a recent year, 73% of firstyear college students responding to a national survey identified “being very well-off financially” as an important personal goal. A state university finds that 132 of an SRS of 200 of its first-year students say that this goal is important. Is there convincing evidence at the a = 0.05 significance level that the proportion of all first-year students at this university who think being very well-off is important differs from the national value, 73%? 47. Teen drivers  Refer to Exercise 45. (a) Construct and interpret a 95% confidence interval for the proportion of all teens in the state who passed their driving test on the first attempt. (b) Explain what the interval in part (a) tells you about the DMV’s claim. 48. We want to be rich  Refer to Exercise 46. (a) Construct and interpret a 95% confidence interval for the true proportion p of all first-year students at the university who would identify being well-off as an important personal goal.

51. Teens and sex  The Gallup Youth Survey asked a random sample of U.S. teens aged 13 to 17 whether they thought that young people should wait to have sex until marriage.15 The Minitab output below shows the results of a significance test and a 95% confidence interval based on the survey data.

(a) Define the parameter of interest. (b) Check that the conditions for performing the significance test are met in this case. (c) Interpret the P-value in context. (d) Do these data give convincing evidence that the actual population proportion differs from 0.5? Justify your answer with appropriate evidence. 52. Reporting cheating  What proportion of students are willing to report cheating by other students? A student project put this question to an SRS of 172 undergraduates at a large university: “You witness two students cheating on a quiz. Do you go to the professor?” The Minitab output below shows the results of a significance test and a 95% confidence interval based on the survey data.16

(b) Explain what the interval in part (a) tells you about whether the national value holds at this university. 49. Do you Tweet?  In early 2012, the Pew Internet and American Life Project asked a random sample of U.S. adults, “Do you ever . . . use Twitter or another service to share updates about yourself or to see updates about others?” According to Pew, the resulting 95% confidence interval is (0.123, 0.177).13 Does this interval provide convincing evidence that the actual proportion of U.S. adults who would say they use Twitter differs from 0.16? Justify your answer. 50. Losing weight  A Gallup Poll found that 59% of the people in its sample said “Yes” when asked, “Would you like to lose weight?” Gallup announced: “For results based on the total sample of national adults, one can say with 95% confidence that the margin of (sampling) error is ±3 percentage points.”14 Does this interval provide convincing evidence that the actual proportion of U.S. adults who would say they want to lose weight differs from 0.55? Justify your answer.

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(a) Define the parameter of interest. (b) Check that the conditions for performing the significance test are met in this case. (c) Interpret the P-value in context. (d) Do these data give convincing evidence that the actual population proportion differs from 0.15? Justify your answer with appropriate evidence. 53. Better parking  Refer to Exercise 39. (a) Describe a Type I error and a Type II error in this setting, and explain the consequences of each. (b) The test has a power of 0.75 to detect that p = 0.45. Explain what this means. (c) Identify two ways to increase the power in part (b).

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Section 9.2 Tests about a Population Proportion 54. Side effects  Refer to Exercise 40. (a) Describe a Type I error and a Type II error in this setting, and explain the consequences of each. (b) The test has a power of 0.54 to detect that p = 0.07. Explain what this means. (c) Identify two ways to increase the power in part (b). 55. Error probabilities  You read that a statistical test at significance level a = 0.05 has power 0.78. What are the probabilities of Type I and Type II errors for this test? 56. Error probabilities  You read that a statistical test at the a = 0.01 level has probability 0.14 of making a Type II error when a specific alternative is true. What is the power of the test against this alternative? 57. Power  A drug manufacturer claims that fewer than 10% of patients who take its new drug for treating Alzheimer’s disease will experience nausea. To test this claim, a significance test is carried out of H0 : p = 0.10 Ha : p < 0.10 You learn that the power of this test at the 5% significance level against the alternative p = 0.08 is 0.29. (a) Explain in simple language what “power = 0.29” means in this setting.

(a) Explain in simple language what “power = 0.23” means in this setting. (b) You could get higher power against the same alternative with the same a by changing the number of measurements you make. Should you make more measurements or fewer to increase power? (c) If you decide to use a = 0.10 in place of a = 0.05, with no other changes in the test, will the power increase or decrease? Justify your answer. (d) If you shift your interest to the alternative m = 5.2, with no other changes, will the power increase or decrease? Justify your answer. Multiple choice: Select the best answer for Exercises 59 to 62. 59. After once again losing a football game to the archrival, a college’s alumni association conducted a survey to see if alumni were in favor of firing the coach. An SRS of 100 alumni from the population of all living alumni was taken, and 64 of the alumni in the sample were in favor of firing the coach. Suppose you wish to see if a majority of living alumni are in favor of firing the coach. The appropriate test statistic is (a)  z =

(b) You could get higher power against the same ­alternative with the same a by changing the ­number of measurements you make. Should you make more measurements or fewer to increase power? Explain.

(b)  t =

(c) If you decide to use a = 0.01 in place of a = 0.05, with no other changes in the test, will the power increase or decrease? Justify your answer.

(c)  z =

(d) If you shift your interest to the alternative p = 0.07 with no other changes, will the power increase or decrease? Justify your answer. 58. What is power?  You manufacture and sell a liquid product whose electrical conductivity is supposed to be 5. You plan to make six measurements of the conductivity of each lot of product. If the product meets specifications, the mean of many measurements will be 5. You will therefore test H0 : m = 5 Ha : m ≠ 5 If the true conductivity is 5.1, the liquid is not suitable for its intended use. You learn that the power of your test at the 5% significance level against the alternative m = 5.1 is 0.23.

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0.64 − 0.5 Å

0.64(0.36) 100

0.64 − 0.5

0.64(0.36) Å 100 0.64 − 0.5





(d) ​z =

(e) ​z =

0.64 − 0.5 Å

0.64(0.36) 64

0.5 − 0.64 0.5(0.5) Å 100

0.5(0.5) Å 100

60. Which of the following is not a condition for ­performing a significance test about a population ­proportion p? (a) The data should come from a random sample or randomized experiment. (b) Both np0 and n(1 – p0) should be at least 10. (c) If you are sampling without replacement from a finite population, then you should sample no more than 10% of the population. (d) The population distribution should be approximately Normal, unless the sample size is large. (e) All of the above are conditions for performing a significance test about a population proportion.

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61. The z statistic for a test of H0 : p = 0.4 versus Ha : p ≠ 0.4 is z = 2.43. This test is (a) not significant at either a = 0.05 or a = 0.01. (b) significant at a = 0.05 but not at a = 0.01. (c) significant at a = 0.01 but not at a = 0.05. (d) significant at both a = 0.05 and a = 0.01. (e) inconclusive because we don’t know the value of p^ . 62. Which of the following 95% confidence intervals would lead us to reject H0 : p = 0.30 in favor of Ha : p ≠ 0.30 at the 5% significance level? (a) ​(0.19, 0.27)

(c) ​(0.27, 0.31)

(b) ​(0.24, 0.30)

(d)  (0.29, 0.38)

(e) ​None of these

63. Packaging CDs (6.2, 5.3)  A manufacturer of compact discs (CDs) wants to be sure that their CDs will fit inside the plastic cases they have bought for packaging. Both the CDs and the cases are circular. According to the supplier, the plastic cases vary Normally with mean diameter m = 4.2 inches and standard deviation s = 0.05 inches. The CD manufacturer decides to produce CDs with mean diameter m = 4 inches. Their diameters follow a Normal distribution with s = 0.1 inches. (a) Let X = the diameter of a randomly selected CD and Y = the diameter of a randomly selected case.

9.3 What You Will Learn

Describe the shape, center, and spread of the distribution of the random variable X – Y. What is the importance of this random variable to the CD manufacturer? (b) Compute the probability that a randomly selected CD will fit inside a randomly selected case. (c) The production process actually runs in batches of 100 CDs. If each of these CDs is paired with a randomly chosen plastic case, find the probability that all the CDs fit in their cases. 64. Cash to find work? (4.2)  Will cash bonuses speed the return to work of unemployed people? The Illinois Department of Employment Security designed an experiment to find out. The subjects were 10,065 people aged 20 to 54 who were filing claims for unemployment insurance. Some were offered $500 if they found a job within 11 weeks and held it for at least 4 months. Others could tell potential employers that the state would pay the employer $500 for hiring them. A control group got neither kind of bonus.17 (a) Describe a completely randomized design for this experiment. (b) How will you label the subjects for random assignment? Use Table D at line 127 to choose the first 3 subjects for the first treatment. (c) Explain the purpose of a control group in this setting.

Tests about a Population Mean By the end of the section, you should be able to:

• State and check the Random, 10%, and Normal/Large Sample conditions for performing a significance test about a population mean. • Perform a significance test about a population mean.

• Use a confidence interval to draw a conclusion for a two-sided test about a population parameter. • Perform a significance test about a mean difference using paired data.

Confidence intervals and significance tests for a population proportion p are based on z-values from the standard Normal distribution. Inference about a population mean m uses a t distribution with n – 1 degrees of freedom, ­except in the rare case when the population standard deviation s is known. We learned how to construct confidence intervals for a population mean in ­Section 8.3. Now we’ll examine the details of testing a claim about a population mean m.

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575

Carrying Out a Significance Test for m In an earlier example, a company claimed to have developed a new AAA battery that lasts longer than its regular AAA batteries. Based on years of experience, the company knows that its regular AAA batteries last for 30 hours of continuous use, on average. An SRS of 15 new batteries lasted an average of 33.9 hours with a standard deviation of 9.8 hours. Do these data give convincing evidence that the new batteries last longer on average? To find out, we perform a test of H0 : m = 30 hours Ha : m > 30 hours where m is the true mean lifetime of the new deluxe AAA batteries.

Conditions  ​In Chapter 8, we introduced conditions that should be met be-

fore we construct a confidence interval for a population mean: Random, 10% when sampling without replacement, and Normal/Large Sample. These same three conditions must be verified before performing a significance test about a population mean. As in the previous chapter, the Normal/Large Sample condition for means is population distribution is Normal or sample size is large (n ≥ 30) We often don’t know whether the population distribution is Normal. But if the sample size is large (n ≥ 30), we can safely carry out a significance test. If the sample size is small, we should examine the sample data for any obvious departures from Normality, such as strong skewness and outliers.

Conditions for Performing a Significance Test about a Mean • Random: The data come from a well-designed random sample or randomized experiment. 1 ~~ 10%: When sampling without replacement, check that n ≤ N. 10 • Normal/Large Sample: The population has a Normal distribution or the sample size is large (n ≥ 30). If the population distribution has unknown shape and n < 30, use a graph of the sample data to assess the Normality of the population. Do not use t procedures if the graph shows strong skewness or outliers. Here’s an example that shows how to check the conditions.

Example

Better Batteries Checking conditions Figure 9.11 on the next page shows a dotplot, boxplot, and Normal probability plot of the battery lifetimes for an SRS of 15 batteries.

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Expected z-score

576

0

(a)

10

20

30

40

50

Battery life (hours)

60

0

(b)

10

20

30

40

50

60

1.0 0.0 –1.0 –2.0 0

(c)

Battery life (hours)

2.0

10

20

30

40

50

60

Battery life (hours)

FIGURE 9.11  (a) A dotplot, (b) a boxplot, and (c) a Normal probability plot of the lifetimes of a simple random sample of 15 AAA batteries.

Problem:  Check the conditions for carrying out a significance test of the company’s claim about its deluxe AAA batteries. Solution: •  Random:  The company tested a simple random sample of 15 new AAA batteries. ºº 10%:  Because the batteries are being sampled without replacement, we need to check that there are at least 10(15) = 150 new AAA batteries. This seems reasonable to believe. •  Normal/Large Sample:  We don’t know if the population distribution of battery lifetimes for the company’s new AAA batteries is Normal. With such a small sample size (n = 15), we need to graph the data to look for any departures from Normality. The dotplot and boxplot show slight right-skewness but no outliers. The Normal probability plot is fairly linear. Because none of the graphs shows any strong skewness or outliers, we should be safe performing a test about the population mean lifetime m. For Practice Try Exercise There is a small number of real-world situations in which we might know the population standard deviation s. When this is the case, the test statistic z=

x– − m0 s∙!n

will follow a standard Normal distribution if the Normal/Large Sample condition is met. If so, then we can calculate P-values using Table A or technology. The TI-83/84 and TI-89’s Z-Test option in the TESTS menu is designed for this special situation.

65

Calculations: Test Statistic and P-Value  ​When performing a signifi-

cance test, we do calculations assuming that the null hypothesis H0 is true. The test statistic measures how far the sample result diverges from the parameter value specified by H0, in standardized units. As before, test statistic =

statistic − parameter standard deviation of statistic

For a test of H0 : m = m0, our statistic is the sample mean x– . Its standard deviation is s !n In an ideal world, our test statistic would be sx– =

z=

x– − m0 s∙!n

Because the population standard deviation s is usually unknown, we use the sample standard deviation sx in its place. The resulting test statistic has the standard error of x– in the denominator t=

x– − m0 sx∙!n

As we saw earlier, when the Normal/Large Sample condition is met, this statistic has approximately a t distribution with n – 1 degrees of freedom.

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577

In Section 8.3, we used Table B to find critical values from the t distributions when constructing confidence intervals about an unknown population mean m. Once we have calculated the test statistic, we can use Table B to find the P-value for a significance test about m. The following example shows how this works.

Example

Better Batteries Computing the test statistic and P-value The battery company wants to test H0 : m = 30 versus Ha : m > 30 based on an SRS of 15 new AAA batteries with mean lifetime x– = 33.9 hours and standard deviation sx = 9.8 hours. The test statistic is test statistic = t=

statistic − parameter standard deviation of statistic x– − m0 sx∙!n

=

33.9 − 30 = 1.54 9.8∙!15

The P-value is the probability of getting a result this large or larger in the direction indicated by Ha, that is, P(t ≥ 1.54). Figure 9.12 shows this probability as an area under the t distribution curve with df = 15 – 1 = 14. We can find this P-value using Table B.

t distribution with df = 14

Upper-tail probability p

P-value = P(t > 1.54)

–4

–2

0

2

4

t t = 1.54

FIGURE 9.12  The P-value for a one-sided test with t = 1.54.

Go to the df = 14 row. The df .10 .05 .025 t statistic falls between the 13 1.350 1.771 2.160 values 1.345 and 1.761. If 14 1.345 1.761 2.145 you look at the top of the corresponding columns in 15 1.341 1.753 2.131 Table B, you’ll find that the 80% 90% 95% “Upper-tail probability p” is Confidence level C between 0.10 and 0.05. (See the excerpt from Table B at right.) Because we are looking for P(t ≥ 1.54), this is the probability we seek. That is, the P-value for this test is ­between 0.05 and 0.10.

As you can see, Table B gives an interval of possible P-values for a significance test. We can still draw a conclusion from the test in much the same way as if we had a single probability. In the case of the new AAA batteries, for instance, we would fail to reject H0 : m = 30 because the P-value exceeds our default a = 0.05 significance level. We don’t have convincing evidence that the company’s new AAA batteries last longer than 30 hours, on average. Table B has other limitations for finding P-values. It includes probabilities only for t distributions with degrees of freedom from 1 to 30 and then skips to df = 40, 50, 60, 80, 100, and 1000. (The bottom row gives probabilities for df = ∞, which corresponds to the standard Normal distribution.) Also, Table B shows probabilities only for positive values of t. To find a P-value for a negative value of t, we use the symmetry of the t distributions. The next example shows how we deal with both of these issues.

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Example

Two-Sided Tests, Negative t-Values, and More Using Table B wisely What if you were performing a test of H0 : m = 5 versus Ha : m ≠ 5 based on a sample size of n = 37 and obtained t = –3.17? Because this is a two-sided test, you are interested in the probability of getting a value of t less than or equal to –3.17 or greater than or equal to 3.17. Figure 9.13 shows the desired P-value as an area under the t distribution curve with 36 degrees of freedom. Notice that P(t ≤ –3.17) = P(t ≥ 3.17) due to the symmetric shape of the density curve. Table B shows only positive t-values, so we will focus on t = 3.17.

P-value = area in both tails

–4

–2

t = – 3.17

0

2

4

t = 3.17

FIGURE 9.13  The P-value for a two-sided test with t = –3.17. Upper-tail probability p df

.005

.0025

.001

29

2.756

3.038

3.396

30

2.750

3.030

3.385

40

2.704

2.971

3.307

99%

99.5% 99.8%

Confidence level C

Because df = 37 – 1 = 36 is not available on the table, use df = 30. You might be tempted to use df = 40, but doing so would result in a smaller P-value than you are entitled to with df = 36. (In other words, you’d be cheating!) Move across the df = 30 row, and notice that t = 3.17 falls between 3.030 and 3.385. The corresponding “Upper-tail probability p” is between 0.001 and 0.0025. (See the ­excerpt from Table B.) For this two-sided test, the corresponding P-value would be between 2(0.001) = 0.002 and 2(0.0025) = 0.005.

c

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!

n

One point from the example deserves repeating: if the df you need isn’t autio provided in Table B, use the next lower df that is available. It’s no fair “rounding up” to a larger df. This is like pretending that your sample size is larger than it really is. Doing so would give you a smaller P-value than is true and would make you more likely to incorrectly reject H0 when it’s true (make a Type I error). Given the limitations of Table B, our advice is to use technology to find P-values when carrying out a significance test about a population mean.

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19. Technology Corner

579

Computing P-values from t distributions on the calculator TI-Nspire instructions in Appendix B; HP Prime instructions on the book’s Web site.

You can use the tcdf command on the TI-83/84 and TI-89 to calculate areas under a t distribution curve. The syntax is tcdf(lower bound,upper bound,df). Let’s use the tcdf command to compute the P-values from the last two examples. Better batteries: To find P(t ≥ 1.54),

TI-83/84 • Press 2nd VARS (DISTR) and choose tcdf(. OS 2.55 or later: In the dialog box, enter these values: lower:1.54, upper: 10000, df:14, choose Paste, and then press ENTER . Older OS: Complete the command tcdf(1.54,10000,14) and press ENTER .

TI-89 • In the Stats/List Editor, press F5 (Distr) and choose t Cdf.... • In the dialog box, enter these values: Lower value:1.54, Upper value:10000, Deg of Freedom, df:14, and then choose ENTER .

Two-sided test: To find the P-value for the two-sided test with df = 36 and t = –3.17, do tcdf(–10000,–3.17,36) and multiply the result by 2.

Check Your Understanding

The makers of Aspro brand aspirin want to be sure that their tablets contain the right amount of active ingredient (acetylsalicylic acid). So they inspect a random sample of 36 tablets from a batch in production. When the production process is working properly, Aspro tablets have an average of m = 320 milligrams (mg) of active ingredient. The amount of active ingredient in the 36 selected tablets has mean 319 mg and standard deviation 3 mg. 1. State appropriate hypotheses for a significance test in this setting. 2. Check that the conditions are met for carrying out the test. 3. Calculate the test statistic. Show your work. 4. Use Table B to find the P-value. Then use technology to get a more accurate result. What conclusion would you draw?

The One-Sample t Test When the conditions are met, we can test a claim about a population mean m ­using a one-sample t test for a mean. Here are the details.

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One-Sample t Test for a Mean Suppose the conditions are met. To test the hypothesis H0 : m = m0, compute the one-sample t statistic t=

x– − m0 sx∙!n

Find the P-value by calculating the probability of getting a t statistic this large or larger in the direction specified by the alternative hypothesis Ha in a t distribution with df = n – 1: Ha : µ > µ 0

Ha : µ = µ 0

Ha : µ < µ 0

t

 

 

t

t

– t

Now we are ready to test a claim about an unknown population mean. Once again, we follow the four-step process.

Example

Healthy Streams

STEP

Performing a significance test about m

4

The level of dissolved oxygen (DO) in a stream or river is an ­important indicator of the water’s ability to support aquatic life. A researcher measures the DO level at 15 randomly chosen locations along a stream. Here are the results in milligrams per liter (mg/l): 4.53  5.04  3.29  5.23  4.13  5.50  4.83  4.40  5.42  6.38  4.01  4.66  2.87  5.73  5.55

A dissolved oxygen level below 5 mg/l puts aquatic life at risk.

Problem: (a) Do we have convincing evidence at the a = 0.05 significance level that aquatic life in this stream is at risk? (b) Given your conclusion in part (a), which kind of mistake—a Type I error or a Type II error—could you have made? Explain what this mistake would mean in context.

Solution: (a) We will follow the four-step process. State:  We want to test a claim about the true mean dissolved oxygen level m in this stream at the a = 0.05 level. Our hypotheses are H0 : m = 5 Ha : m < 5

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Plan:  If the conditions are met, we should do a one-sample t test for m.

AP® Exam Tip  It is not enough just to make a graph of the data on your calculator when assessing Normality. You must sketch the graph on your paper to receive credit. You don’t have to draw multiple graphs—any appropriate graph will do.

•  Random:  The researcher measured the DO level at 15 randomly chosen locations. ºº 10%: There is an infinite number of possible locations along the stream, so it isn’t necessary to check the 10% condition. •  Normal/Large Sample:  We don’t know whether the population distribution of DO levels at all points along the stream is Normal. With such a small sample size (n = 15), we need to graph the data to see if it’s safe to use t procedures. Figure 9.14 shows our hand sketches of a calculator histogram, boxplot, and Normal probability plot for these data. The histogram looks roughly symmetric; the boxplot shows no outliers; and the Normal probability plot is fairly linear. With no outliers or strong skewness, the t procedures should be pretty accurate. 3 2 1 0

1

–1 1

2

3

4

5

6

7

1

8

2

3

4

5

6

7

8

(a)

3

4

5

6

7

8

–2 –3

DO level (mg/l)

DO level (mg/l)

2

DO level (mg/l)

(b)

(c)

FIGURE 9.14  Sketches of (a) a histogram, (b) a boxplot, and (c) a Normal probability plot for the dissolved oxygen (DO) readings in the sample, in mg/l.

Do:  We entered the data into our calculator and did 1-Var Stats (see screen shot). The sample mean is x– = 4.771 and the sample standard deviation is sx = 0.9396.

–4

–2

0

Values of t t = – 0.94

2

4

•  Test statistic

FIGURE 9.15  ​The P-value for a one-sided test when t = –0.94.

t=

x– − m0 sx∙!n

=

4.771 − 5 = − 0.94 0.9396∙!15

•  P-value The P-value is the area to the left of t = –0.94 under the t distribution curve with degrees of freedom df = 15 – 1 = 14. Figure 9.15 shows this probability. Using the table:  Table B shows only areas in the upper tail of the distribution. Because the t distributions are symmetric, P(t ≤ –0.94) = P(t ≥ 0.94). Search the df = 14 row of Table B for entries that bracket t = 0.94 (see the excerpt at right). Because the observed t lies between 0.868 and 1.076, the P-value lies between 0.15 and 0.20.

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Upper-tail probability p df

.25

.20

.15

13

.694

.870

1.079

14

.692

.868

1.076

15

.691

.866

1.074

50%

60%

70%

Confidence level C

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Using technology:  We can find the exact P-value using a calculator: tcdf(lower: –100, upper: –0.94, df: 14) = 0.1816. Conclude:  Because the P-value, 0.1816, is greater than our a = 0.05 significance level, we fail to reject H0. We don’t have convincing evidence that the mean DO level in the stream is less than 5 mg/l. (b) Because we decided not to reject H0 in part (a), we could have made a Type II error (failing to reject H0 when H0 is false). If we did, then the mean dissolved oxygen level m in the stream is actually less than 5 mg/l, but we didn’t find convincing evidence of that with our significance test. For Practice Try Exercise

73

Because the t procedures are so common, all s­tatistical software packages will do the calculations for you. Figure 9.16 shows the output from Minitab for the one-sample t test in the previous example. Note that the results match! You can also use your calculator to carry out a onesample t test. But be sure to read the AP® exam tip at the end of the Technology Corner. FIGURE 9.16  Minitab output for the one-sample t test from the dissolved oxygen example.

20. Technology Corner

One-sample t test for a mean on the calculator TI-Nspire instructions in Appendix B; HP Prime instructions on the book’s Web site.

You can perform a one-sample t test using either raw data or summary statistics on the TI-83/84 or TI-89. Let’s use the calculator to carry out the test of H0: m = 5 versus Ha: m < 5 from the dissolved oxygen example. Start by entering the sample data in L1/list1. Then, to do the test:

TI-83/84

TI-89

• Press STAT , choose TESTS and T-Test.

• Press 2nd F1 ([F6]) and choose T-Test.

• Adjust your settings as shown.

• Adjust your settings as shown.

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If you select “Calculate,” the following screen appears:

The test statistic is t = –0.94 and the P-value is 0.1809. If you specify “Draw,” you see a t distribution curve (df = 14) with the lower tail shaded.

Note: If you are given summary statistics instead of the original data, you would select the option “Stats” instead of “Data” in the first screen. AP® EXAM TIP ​Remember: if you just give calculator results with no work, and one or more values are wrong, you probably won’t get any credit for the “Do” step. If you opt for the calculator-only method, name the procedure (t test) and report the test statistic (t = –0.94), degrees of freedom (df = 14), and P-value (0.1809).

Check Your Understanding

A college professor suspects that students at his school are getting less than 8 hours of sleep a night, on average. To test his belief, the professor asks a random sample of 28 students, “How much sleep did you get last night?” Here are the data (in hours): 9 6 8 6 8 8 6 6.5 6 7 9 4 3 4 5 6 11 6 3 6 6 10 7 8 4.5 9 7 7 Do these data provide convincing evidence at the a = 0.05 significance level in support of the professor’s suspicion?

Two-Sided Tests and Confidence Intervals Now let’s look at an example involving a two-sided test.

Example

Juicy Pineapples A two-sided test

STEP

4

At the Hawaii Pineapple Company, managers are interested in the sizes of the pineapples grown in the company’s fields. Last year, the mean weight of the pineapples harvested from one large field was 31 ounces. A different irrigation system was installed in this field after the growing season. Managers wonder how this change will affect the mean weight of future pineapples grown in the field. To

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find out, they select and weigh a random sample of 50 pineapples from this year’s crop. The Minitab output below summarizes the data. Descriptive Statistics: Weight (oz) Variable N Weight (oz) 50

Mean SE Mean 31.935 0.339

StDev 2.394

Minimum Q1 Median Q3 26.491 29.990 31.739 34.115

Maximum 35.547

Problem: (a) Do these data give convincing evidence that the mean weight of pineapples produced in the field has changed this year? (b) Can we conclude that the different irrigation system caused a change in the mean weight of pineapples produced? Explain your answer.

Solution: (a) State:  We want to perform a test of H0 : m = 31 Ha : m ≠ 31 where m = the mean weight (in ounces) of all pineapples grown in the field this year. Because no significance level is given, we’ll use a = 0.05. Plan:  If the conditions are met, we should conduct a one-sample t test for m. •  Random:  The data came from a random sample of 50 pineapples from this year’s crop. ºº 10%: There need to be at least 10(50) = 500 pineapples in the field because managers are sampling without replacement. We would expect many more than 500 pineapples in a “large field.” •  Normal/Large Sample:  We don’t know whether the population distribution of pineapple weights this year is Normally distributed. But n = 50 ≥ 30, so the large sample size makes it OK to use t procedures. Do:  From the Minitab output, x– = 31.935 ounces and t distribution, sx = 2.394 ounces. 49 degrees of •  Test statistic freedom x– − m0 31.935 − 31 t= = = 2.762 sx∙!n 2.394∙!50 P-value = 0.0081 •  P-value  Figure 9.17 displays the P-value for this two-sided test as an area under the t distribution curve with 50 – 1 = 49 degrees of freedom. Using the table:  Table B doesn’t have an entry for df = 49, so we have to use the more conservative df = 40. As the excerpt below shows, Values of t the upper-tail probability is between 0.0025 and 0.005. So the t = – 2.762 t = 2.762 desired P-value for this two-sided test is between 2(0.0025) = FIGURE 9.17  The P-value for the two-sided test with t = 2.762. 0.005 and 2(0.005) = 0.01. Upper-tail probability p df

.005

.0025

.001

30

2.750

3.030

3.385

40

2.704

2.971

3.307

50

2.678

2.937

3.261

99%

99.5% 99.8%

Confidence level C

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Using technology:  The calculator’s T-Test command gives t = 2.762 and P-value 0.0081 using df = 49.

Conclude:  Because the P-value, 0.0081, is less than a = 0.05, we reject H0 . We have con-

vincing evidence that the mean weight of the pineapples grown this year is not 31 ounces. (b) No. This was not a comparative experiment, so we cannot infer causation. It is possible that other things besides the irrigation system changed from last year’s growing season. Maybe the weather was different this year, and that’s why the pineapples have a different mean weight than last year. For Practice Try Exercise

77

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Section 9.3 Tests about a Population Mean

The significance test in the previous example gives convincing evidence that the mean weight m of the pineapples grown in the field this year differs from last year’s 31 ounces. Unfortunately, the test doesn’t give us an idea of what the actual value of m is. For that, we need a confidence interval.

Example

Juicy Pineapples Confidence intervals give more information Minitab output for a significance test and confidence interval based on the pineapple data is shown below. The test statistic and P-value match what we got earlier (up to rounding). One-Sample T: Weight (oz) Test of mu = 31 vs not = 31 Variable

N

Mean

StDev SE Mean

Weight (oz) 50 31.935 2.394

0.339

95% CI

T

P

(31.255, 32.616) 2.76 0.008

The 95% confidence interval for the mean weight of all the pineapples grown in the field this year is 31.255 to 32.616 ounces. We are 95% confident that this interval captures the true mean weight m of this year’s pineapple crop. As with proportions, there is a link between a two-sided test at significance level a and a 100(1 – a)% confidence interval for a population mean m. For the pineapples, the two-sided test at a = 0.05 rejects H0 : m = 31 in favor of Ha : m ≠ 31. The corresponding 95% confidence interval does not include 31 as a plausible value of the parameter m. In other words, the test and interval lead to the same conclusion about H0. But the confidence interval provides much more information: a set of plausible values for the population mean.

The connection between two-sided tests and confidence intervals is even stronger for means than it was for proportions. That’s because both inference methods for means use the standard error of x– in the calculations. test statistic: t =

x– − m0

sx   Confidence interval: x– ± t* sx∙!n !n

When the two-sided significance test at level a rejects H0 : m = m0, the 100(1 – a)% confidence interval for m will not contain the hypothesized value m0. And when the test fails to reject the null hypothesis, the confidence interval will contain m0.

THINK ABOUT IT

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Is there a connection between one-sided tests and ­confidence intervals for a population mean?  ​As you might expect, the answer is

yes. But the link is more complicated. Consider a one-sided test of H0 : m = 10 versus Ha : m > 10 based on an SRS of 30 observations. With df = 30 − 1 = 29, Table B says that the test will reject H0 at a = 0.05 if the test statistic t is greater than 1.699. For this to happen, the sample mean x– would have to exceed m0 = 10 by more than 1.699 standardized units. Table B also shows that t* = 1.699 is the critical value for a 90% confidence interval. That is, a 90% confidence interval will extend 1.699 standardized units

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on either side of the sample mean x– . If x– exceeds 10 by more than 1.699 standardized units, the resulting interval will not include 10. And the one-sided test will reject H0 : m = 10. There’s the link: our one-sided test at a = 0.05 gives the same conclusion about H0 as a 90% confidence interval for m.

Check Your Understanding

The health director of a large company is concerned about the effects of stress on the company’s middle-aged male employees. According to the National Center for Health Statistics, the mean systolic blood pressure for males 35 to 44 years of age is 128. The health director examines the medical records of a random sample of 72 male employees in this age group. The Minitab output displays the results of a significance test and a confidence interval. 1. Do the results of the significance test give convincing evidence that the mean blood pressure for all the company’s middle-aged male employees differs from the national average? Justify your answer. 2. Interpret the 95% confidence interval in context. Explain how the confidence interval leads to the same conclusion as in Question 1.

Inference for Means: Paired Data Study designs that involve making two observations on the same individual, or one observation on each of two similar individuals, yield paired data. When paired data result from measuring the same quantitative variable twice, we can make comparisons by analyzing the differences in each pair. If the conditions for inference are met, we can use one-sample t procedures to perform inference about the mean difference md. (These methods are sometimes called paired t procedures.) An example should help illustrate what we mean.

Example

Is Caffeine Dependence Real? Paired data and one-sample t procedures

STEP

4

Researchers designed an experiment to study the effects of caffeine ­withdrawal. They recruited 11 volunteers who were diagnosed as being caffeine dependent to serve as subjects. Each subject was barred from coffee, colas, and other substances with caffeine for the duration of the experiment. During one 2-day period, subjects took capsules containing their normal caffeine intake. ­During another 2-day period, they took placebo capsules. The order in which subjects took caffeine and the placebo was randomized. At the end of each 2-day period, a test for depression was given to all 11 subjects. ­Researchers ­wanted to know whether being deprived of caffeine would lead to an increase in d ­ epression.18

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The table below contains data on the subjects’ scores on the depression test. Higher scores show more symptoms of depression. For each subject, we calculated the difference in test scores following each of the two treatments (placebo – caffeine). We chose this order of subtraction to get mostly positive values. Results of a caffeine-deprivation study Subject

Depression (caffeine)

Depression (placebo)

Difference (placebo – caffeine)

1

5

16

11

2

5

23

18

3

4

5

1

4

3

7

4

5

8

14

6

6

5

24

19

7

0

6

6

8

0

3

3

9

2

15

13

10

11

12

1

11

1

0

–1

Problem: (a) Why did researchers randomly assign the order in which subjects received placebo and caffeine? (b) Carry out a test to investigate the researchers’ question.

Solution: (a) Researchers want to be able to conclude that any statistically significant change in depression score is due to the treatments themselves and not to some other variable. One obvious concern is the order of the treatments. Suppose that caffeine were given to all the subjects during the first 2-day period. What if the weather were nicer on these 2 days than during the second 2-day period when all subjects were given a placebo? Researchers wouldn’t be able to tell if a large increase in the mean depression score is due to the difference in weather or due to the treatments. Random assignment of the caffeine and placebo to the two time periods in the experiment should help ensure that no other variable (like the weather) is systematically affecting subjects’ responses. (b) We’ll follow the four-step process. State:  If caffeine deprivation has no effect on depression, then we would expect the actual mean difference in depression scores to be 0. We therefore want to test the hypotheses H0 : md = 0 Ha : md > 0 It is uncommon for the subjects in an experiment to be randomly selected from some larger population. In fact, most experiments use recruited volunteers as subjects. When there is no sampling, we don’t need to check the 10% condition.

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where md is the true mean difference (placebo – caffeine) in depression score for subjects like these. Because no significance level is given, we’ll use a = 0.05. Plan:  If the conditions are met, we should conduct a paired t test for md. •  Random:  Researchers randomly assigned the treatments—placebo then caffeine, caffeine then placebo—to the subjects. ºº 10%: We aren’t sampling, so it isn’t necessary to check the 10% condition.

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•  Normal/Large Sample:  We don’t know whether the actual distribution of difference in depression scores (placebo – caffeine) for subjects like these is Normal. With such a small sample size (n = 11), we need to graph the data to see if it’s safe to use t procedures. Figure 9.18 shows hand sketches of a calculator histogram, boxplot, and Normal probability plot for these data. The histogram has an irregular shape with so few values; the boxplot shows some right skewness but no outliers; and the Normal probability plot is slightly curved, indicating mild skewness. With no outliers or strong skewness, the t  procedures should be fairly accurate. 3 2 1 –4 –4

(a)

4

8

12

16

Change in depression (placebo - caffeine)

20

–4

(b)

4

8

12

16

Change in depression (placebo - caffeine)

–1

4

8

12

16

20

–2

20

–3

(c)

Change in depression (placebo - caffeine)

FIGURE 9.18  Sketches of (a) a histogram, (b) a boxplot, and (c) a Normal probability plot of the change in depression scores (placebo – caffeine) for the 11 subjects in the caffeine experiment.

Do:  We entered the differences in list1 and then

used the calculator’s t test command with the “Draw” option.

Just by looking at the data, it appears that the true mean change in depression score md is greater than 0. However, it’s possible that there has been no change and we got a result this much larger than md = 0 by the luck of the random assignment. The significance test tells us whether this explanation is plausible.

•  Test statistic  t = 3.53 •  P-value  0.0027, which is the area to the right of t = 3.53 on the t distribution curve with df = 11 – 1 = 10. Note:  The calculator doesn’t report the degrees of freedom, but you should.

Conclude:  Because the P-value of 0.0027 is less than a = 0.05, we reject H0 : md = 0. We

have convincing evidence that the true mean difference (placebo – caffeine) in depression score is positive for subjects like these.

For Practice Try Exercise

85

A few follow-up comments about this example are in order. 1. We could have calculated the test statistic in the example using the formula t=

x–d − m0 sd∙!n

=

7.364 − 0 = 3.53 6.918∙!11

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!

n

and obtained the P-value using Table B or technology. Check with your teacher on whether the calculator-only method is acceptable. Be sure to report autio the degrees of freedom with any t procedure, even if technology doesn’t. 2. The subjects in this experiment were not chosen at random from the population of caffeine-dependent individuals. As a result, we can’t generalize our findings to all caffeine-dependent people—only to people like the ones who took part in this experiment. c



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589

3. Because researchers randomly assigned the treatments, they can make an inference about cause and effect. The data from this experiment provide convincing evidence that depriving caffeine-dependent subjects like these of ­caffeine causes an average increase in depression scores. Until now, we have only used one-sample t procedures in settings involving random sampling. The paired data in the caffeine example came from a matched pairs experiment, in which each subject received both treatments in a random order. A coin toss or some other chance process was used to carry out the random assignment. Why is it legitimate to use a t distribution to perform inference about the parameter m in a randomized experiment? The answer to that question will have to wait until the next chapter.

Check Your Understanding

Refer to the Data Exploration from Chapter 4 on page 257. Do the data give convincing evidence at the a = 0.05 significance level that filling tires with nitrogen instead of air decreases pressure loss?

Using Tests Wisely Significance tests are widely used in reporting the results of research in many fields. New drugs require significant evidence of effectiveness and safety. Courts ask about statistical significance in hearing discrimination cases. Marketers want to know whether a new ad campaign significantly outperforms the old one, and medical researchers want to know whether a new therapy performs significantly better. In all these uses, statistical significance is valued because it points to an effect that is unlikely to occur simply by chance. Carrying out a significance test is often quite simple, especially if you use technology. Using tests wisely is not so simple. Here are some points to keep in mind when using or interpreting significance tests.

Determining Sample Size  How large a sample should researchers take

when they plan to carry out a significance test? The answer depends on three factors: significance level, effect size, and the desired power of the test. Here are the questions that researchers must answer to decide how many observations they need: 1. Significance level. How much risk of a Type I error—rejecting the null hypothesis when H0 is actually true—are we willing to accept? If a Type I error has serious consequences, we might opt for a = 0.01. Otherwise, we should choose a = 0.05 or a = 0.10. Recall that using a higher significance level would decrease the Type II error probability and increase the power.

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2. Effect size. How large a difference between the null parameter value and the actual parameter value is important for us to detect? 3. Power. What chance do we want our study to have to detect a difference of the size we think is important? Let’s illustrate typical answers to these questions using an example.

Example

Developing Stronger Bones Planning a study Can a 6-month exercise program increase the total body bone mineral content (TBBMC) of young women? A team of researchers is planning a study to examine this question. The researchers would like to perform a test of H0 : m = 0 Ha : m > 0 where m is the true mean percent change in TBBMC due to the exercise program. To decide how many subjects they should include in their study, researchers begin by answering the three questions above. 1. Significance level. The researchers decide that a = 0.05 gives enough protection against declaring that the exercise program increases bone mineral content when it really doesn’t (a Type I error). 2. Effect size. A mean increase in TBBMC of 1% would be considered important to detect. 3. Power. The researchers want probability at least 0.9 that a test at the chosen significance level will reject the null hypothesis H0 : m = 0 when the truth is m = 1.

The following Activity gives you a chance to investigate the sample size needed to achieve a power of 0.9 in the bone mineral content study.

Activity Materials: Computer with Internet connection and display capability

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PLET AP

Investigating Power In this Activity, you will use the Statistical Power applet at the book’s Web site to determine the sample size needed for the exercise study of the previous example. Based on the results of a previous study, researchers are willing to assume that s = 2 for the percent change in TBBMC over the 6-month period. We’ll start by seeing whether or not 25 subjects are enough.

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1. ​Go to www.whfreeman.com/tps5e and launch the Statistical Power applet. Enter the values: H0 : m = 0, Ha : m > 0, s = 2, n = 25, a = 0.05, and alternate m = 1. Then click “Update.” What is the power? As a class, discuss what this number means in simple terms. 2. ​Change the significance level to 0.01. What effect does this have on the power of the test to detect m = 1? Why? 3. ​The researchers decide that they are willing to risk a 5% chance of making a Type I error. Change the significance level back to a = 0.05. Now increase the sample size to 30. What happens to the power? Why? 4. ​Keep increasing the sample size until the power is at least 0.90. What minimum sample size should the researchers use for their study? 5. ​Would the researchers need a smaller or a larger sample size to detect a mean increase of 1.5% in TBBMC? A 0.85% increase? Use the applet to investigate. 6. ​Summarize what you have learned about how significance level, effect size, and power influence the sample size needed for a significance test.

Here is a summary of influences on “How large a sample do I need?” from the Activity. • If you insist on a smaller significance level (such as 1% rather than 5%), you have to take a larger sample. A smaller significance level requires stronger evidence to reject the null hypothesis. • If you insist on higher power (such as 0.99 rather than 0.90), you will need a larger sample. Higher power gives a better chance of detecting a difference when it really exists. • At any significance level and desired power, detecting a small difference between the null and alternative parameter values requires a larger sample than detecting a large difference.

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Statistical Significance and Practical Importance  ​When a null hypothesis (“no effect” or “no difference”) can be rejected at the usual levels ­(a = 0.05 or a = 0.01), there is convincing evidence of a difference. But that difference may be very small. When large samples are available, even tiny deviations from the null hypothesis will be significant.

Example

Wound Healing Time Significant doesn’t mean important Suppose we’re testing a new antibacterial cream, “Formulation NS,” on a small cut made on the inner forearm. We know from previous research that with no medication, the mean healing time (defined as the time for the scab to fall off) is 7.6 days. The claim we want to test here is that Formulation NS speeds healing. We will use a 5% significance level. Procedure: We cut a random sample of 250 college students and apply Formulation NS to the wounds. The mean healing time for these subjects is x– = 7.5 days and the standard deviation is sx = 0.9 days. Discussion: We want to test a claim about the mean healing time m in the population of college students whose cuts are treated with Formulation NS. Our hypotheses are H0 : m = 7.6 days Ha : m < 7.6 days An examination of the data reveals no outliers or strong skewness, so the conditions for performing a one-sample t test are met. We carry out the test and find that t = −1.76 and P-value = 0.04 with df = 249. Because 0.04 is less than a = 0.05, we reject H0. We have convincing evidence that Formulation NS reduces the average healing time. However, this result is not practically important. Having your scab fall off one-tenth a day sooner is no big deal.

c

c

!

n

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!

n

Remember the wise saying: Statistical significance is not the same thing autio as practical importance. The remedy for attaching too much importance to statistical significance is to pay attention to the actual data as well as to the P-value. Plot your data and examine them carefully. Are there outliers or other departures from a consistent pattern? A few outlying observations can produce highly significant results if you blindly apply common significance tests. Outliers can also destroy the significance of otherwise-convincing data. The foolish user of statistics who feeds the data to a calculator or computer autio without exploratory analysis will often be embarrassed. Is the difference you are seeking visible in your plots? If not, ask yourself whether the difference is large enough to be practically important. Give a confidence interval for the parameter in which you are interested. A confidence interval gives a set of plausible values for the parameter rather than simply asking if the observed result is too surprising to occur by chance alone when H0 is true. Confidence intervals are not used as often as they should be, whereas significance tests are perhaps overused.

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593

Beware of Multiple Analyses  ​Statistical significance ought to mean that

you have found a difference that you were looking for. The reasoning behind statistical significance works well if you decide what difference you are seeking, design a study to search for it, and use a significance test to weigh the evidence you get. In other settings, significance may have little meaning.

Example

Cell Phones and Brain Cancer Don’t search for significance! Might the radiation from cell phones be harmful to users? Many studies have found little or no connection between using cell phones and various illnesses. Here is part of a news account of one study: A hospital study that compared brain cancer patients and a similar group without brain cancer found no statistically significant difference between brain cancer rates for the two groups. But when 20 distinct types of brain cancer were considered separately, a significant difference in brain cancer rates was found for one rare type. Puzzlingly, however, this risk appeared to decrease rather than increase with greater mobile phone use.19 Think for a moment. Suppose that the 20 null hypotheses for these 20 significance tests are all true. Then each test has a 5% chance of being significant at the 5% level. That’s what a = 0.05 means: results this extreme occur only 5% of the time just by chance when the null hypothesis is true. We expect about 1 of 20 tests to give a significant result just by chance. Running one test and reaching the a = 0.05 level is reasonably good evidence that you have found something; running 20 tests and reaching that level only once is not.

For more on the pitfalls of multiple analyses, do an Internet search for the XKCD comic about jelly beans causing acne.

Searching data for patterns is certainly legitimate. Exploratory data analysis is an important part of statistics. But the reasoning of formal inference does not apply when your search for a striking effect in the data is successful. The remedy is clear. Once you have a hypothesis, design a study to search specifically for the effect you now think is there. If the result of this study is statistically significant, you have real evidence.

case closed

Do You Have a Fever? At the beginning of the chapter, we described a study investigating whether “normal” human body temperature is really 98.6°F. Here is a summary of the details we provided in the chapteropening Case Study (page 537).

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96

97

98

99

Temperature (°F)

100

101

• The mean temperature was x– = 98.25°F. • The standard deviation of the temperature readings was sx = 0.73°F. • 62.3% of the temperature readings were less than 98.6°F. 1. If “normal” body temperature really is 98.6°F, we would expect that about half of all healthy 18- to 40-year-olds will have a body temperature less than 98.6°F. Do the data from this study provide convincing evidence at the a = 0.05 level that this is not the case? 2. The test in Question 1 has power 0.66 to detect that the actual population proportion is 0.60. Describe two changes that could be made to increase the power of the test. Do the data provide convincing evidence that average normal body temperature is not 98.6°F? The computer output below shows the results of a onesample t test and a 95% confidence interval for the population mean m. One-Sample T Test of mu = 98.6 vs not = 98.6 N

Mean

130 98.2500

StDev 0.7300

SE Mean 0.0640

95% CI

T

(98.1233, 98.3767) −5.47

P 0.000

3. What conditions must be satisfied for a one-sample t test to give valid results? Show that these conditions are met in this setting. 4. Explain how the P-value and the confidence interval lead to the same conclusion for the significance test. 5. Based on the conclusion in Question 4, which type of error could have been made: a Type I error or a Type II error? Justify your answer.

Section 9.3

Summary •

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The conditions for performing a significance test of H0: m = m0 are: • Random:  The data were produced by a well-designed random sample or randomized experiment. ~~ 10%: When sampling without replacement, check that the population is at least 10 times as large as the sample.

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595

Normal/Large Sample:  The population distribution is Normal or the sample size is large (n ≥ 30). When the sample size is small (n < 30), examine a graph of the sample data for any possible departures from Normality in the population. You should be safe using a t distribution as long as there is no strong skewness and no outliers are present. The one-sample t test for a mean uses the test statistic •

•

t=

x– − m0 sx∙!n

with P-values calculated from the t distribution with n − 1 degrees of freedom. • Confidence intervals provide additional information that significance tests do not—namely, a set of plausible values for the parameter m. A two-sided test of H0 : m = m0 at significance level a gives the same conclusion as a 100(1 − a)% confidence interval for m. • Analyze paired data by first taking the difference within each pair to produce a single sample. Then use one-sample t procedures. • There are three factors that influence the sample size required for a statistical test: significance level, effect size, and the desired power of the test. • Very small differences can be highly significant (small P-value) when a test is based on a large sample. A statistically significant difference need not be practically important. • Many tests run at once will probably produce some significant results by chance alone, even if all the null hypotheses are true.

9.3 T echnology Corners TI-Nspire Instructions in Appendix B; HP Prime instructions on the book’s Web site. 19. Computing P-values from t distributions on the calculator 20. One-sample t test for a mean on the calculator

Section 9.3

Exercises

65. Attitudes  The Survey of Study Habits and Attitudes (SSHA) is a psychological test that measures students’ attitudes toward school and study habits. Scores range from 0 to 200. Higher scores indicate more positive attitudes. The mean score for U.S. college students is about 115. A teacher suspects that older students have better attitudes toward school. She gives the SSHA to an SRS of 45 of the over 1000 students at her college who are at least

pg 575

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page 579 page 582

30 years of age. Check the conditions for carrying out a ­significance test of the teacher’s suspicion. 66. Anemia  Hemoglobin is a protein in red blood cells that carries oxygen from the lungs to body tissues. People with fewer than 12 grams of hemoglobin per deciliter of blood (g/dl) are anemic. A public health official in Jordan suspects that Jordanian children are at risk of anemia. He measures a random sample of

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50 children. Check the conditions for carrying out a significance test of the official’s suspicion. 67. Ancient air  The composition of the earth’s atmosphere may have changed over time. To try to discover the nature of the atmosphere long ago, we can examine the gas in bubbles inside ancient amber. Amber is tree resin that has hardened and been trapped in rocks. The gas in bubbles within amber should be a sample of the atmosphere at the time the amber was formed. Measurements on 9 specimens of amber from the late Cretaceous era (75 to 95 million years ago) give these percents of nitrogen:20 63.4 65.0 64.4 63.3 54.8 64.5 60.8 49.1 51.0

Explain why we should not carry out a one-sample t test in this setting.

68. Paying high prices?  A retailer entered into an exclusive agreement with a supplier who guaranteed to provide all products at competitive prices. The retailer eventually began to purchase supplies from other vendors who offered better prices. The original supplier filed a lawsuit claiming violation of the agreement. In defense, the retailer had an audit performed on a random sample of 25 invoices. For each audited invoice, all purchases made from other suppliers were examined and compared with those offered by the original supplier. The percent of purchases on each invoice for which an alternative supplier offered a lower price than the original supplier was recorded.21 For example, a data value of 38 means that the price would be lower with a different supplier for 38% of the items on the invoice. A histogram and some computer output for these data are shown below. Explain why we should not carry out a one-sample t test in this setting. 14

10 Frequency

(a) Calculate the test statistic. (b) Find the P-value using Table B. Then obtain a more precise P-value from your calculator. 70. Anemia  For the study of Jordanian children in ­Exercise 66, the sample mean hemoglobin level was 11.3 mg/dl and the sample standard deviation was 1.6 mg/dl. (a) Calculate the test statistic. (b) Find the P-value using Table B. Then obtain a more precise P-value from your calculator. 71. One-sided test  Suppose you carry out a significance test of H0 : m = 5 versus Ha : m < 5 based on a sample of size n = 20 and obtain t = −1.81. (a) Find the P-value for this test using Table B or technology. What conclusion would you draw at the 5% significance level? At the 1% significance level? (b) Redo part (a) using an alternative hypothesis of Ha : m ≠ 5. 72. Two-sided test  The one-sample t statistic from a sample of n = 25 observations for the two-sided test of H0 : m = 64 Ha : m ≠ 64

has the value t = −1.12.

(a) Find the P-value for this test using Table B or technology. What conclusion would you draw at the 5% significance level? At the 1% significance level? (b) Redo part (a) using an alternative hypothesis of Ha : m < 64. 73. Construction zones  Every road has one at some point—construction zones that have much lower speed limits. To see if drivers obey these lower speed limits, a police officer uses a radar gun to measure the speed (in miles per hours, or mph) of a random sample of 10 drivers in a 25 mph construction zone. Here are the data:

pg 580

12

8

27 33 32 21 30 30 29 25 27 34

6

(a) Is there convincing evidence that the average speed of drivers in this construction zone is greater than the posted speed limit?

4 2 0 0

20

40

60

80

100

120

Percent lower

Summary statistics Column

69. Attitudes  In the study of older students’ attitudes from Exercise 65, the sample mean SSHA score was 125.7 and the sample standard deviation was 29.8.

n

Mean Std. Dev. Std. Err. Median Min Max Q1 Q3

pctlower 25 77.76

32.6768

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6.5353603

100

0

100 68 100

(b) Given your conclusion in part (a), which kind of mistake—a Type I error or a Type II error—could you have made? Explain what this mistake would mean in context. 74. Heat through the glass  How well materials conduct heat matters when designing houses, for example. Conductivity is measured in terms of watts of heat

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Section 9.3 Tests about a Population Mean power transmitted per square meter of surface per degree Celsius of temperature difference on the two sides of the material. In these units, glass has conductivity about 1. The National Institute of Standards and Technology provides exact data on properties of materials. Here are measurements of the heat conductivity of 11 randomly selected pieces of a particular type of glass:22 1.11 1.07 1.11 1.07 1.12 1.08 1.08 1.18 1.18 1.18 1.12 (a) Is there convincing evidence that the mean conductivity of this type of glass is greater than 1?

(a) Do these data give convincing evidence to support the lawyer’s case? Justify your answer. (b) Interpret the P-value in context. 77. Pressing pills  A drug manufacturer forms tablets by compressing a granular material that contains the active ingredient and various fillers. The hardness of a sample from each batch of tablets produced is measured to control the compression process. The target value for the hardness is m = 11.5. The hardness data for a random sample of 20 tablets are

pg 583

11.627  11.613  11.493  11.602  11.360 11.374  11.592  11.458  11.552  11.463 11.383  11.715  11.485  11.509  11.429 11.477  11.570  11.623  11.472  11.531

(b) Given your conclusion in part (a), which kind of mistake—a Type I error or a Type II error—could you have made? Explain what this mistake would mean in context. 75. Healthy bones  The recommended daily allowance (RDA) of calcium for women between the ages of 18 and 24 years is 1200 milligrams (mg). Researchers who were involved in a large-scale study of women’s bone health suspected that their participants had significantly lower calcium intakes than the RDA. To test this suspicion, the researchers measured the daily calcium intake of a random sample of 36 women from the study who fell in the desired age range. The Minitab output below displays the results of a significance test.



One-Sample T: Calcium intake (mg)

299.4 297.7 301.0 298.9 300.2 297.0

Variable

N

Mean

StDev

SE Mean

T

P

Calcium

36

856.2

306.7

51.1

−6.73

0.000

(a) Do these data give convincing evidence to support the researchers’ suspicion? Justify your answer. (b) Interpret the P-value in context. 76. Taking stock  An investor with a stock portfolio worth several hundred thousand dollars sued his broker due to the low returns he got from the portfolio at a time when the stock market did well overall. The investor’s lawyer wants to compare the broker’s performance against the market as a whole. He collects data on the broker’s returns for a random sample of 36 weeks. Over the 10-year period that the broker has managed portfolios, stocks in the Standard & Poor’s 500 index gained an average of 0.95% per week. The Minitab output below displays the results of a significance test. One-Sample T: Return (percent)

Test of mu = 0.95 vs < 0.95 N

Return

36 −1.441

(percent)

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Mean

Is there convincing evidence at the 5% level that the mean hardness of the tablets differs from the target value?

78. Filling cola bottles  Bottles of a popular cola are supposed to contain 300 milliliters (ml) of cola. There is some variation from bottle to bottle because the filling machinery is not perfectly precise. An inspector measures the contents of six randomly selected bottles from a single day’s production. The results are

Test of mu = 1200 vs < 1200

Variable

597

StDev

SE Mean

T

P

4.810

0.802

−2.98

0.003



Do these data provide convincing evidence that the mean amount of cola in all the bottles filled that day differs from the target value of 300 ml?

79. Pressing pills  Refer to Exercise 77. Construct and interpret a 95% confidence interval for the population mean m. What additional information does the confidence interval provide? 80. Filling cola bottles  Refer to Exercise 78. Construct and interpret a 95% confidence interval for the population mean m. What additional information does the confidence interval provide? 81. Fast connection?  How long does it take for a chunk of information to travel from one server to another and back on the Internet? According to the site ­internettrafficreport.com, a typical response time is 200 milliseconds (about one-fifth of a second). Researchers collected data on response times of a random sample of 14 servers in Europe. A graph of the data reveals no strong skewness or outliers. The following figure displays Minitab output for a one-sample t interval for the population mean. Is there convincing evidence at the 5% significance level that the site’s claim is incorrect? Justify your answer.

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82. Water!  A blogger claims that U.S. adults drink an average of five 8-ounce glasses of water per day. Skeptical researchers ask a random sample of 24 U.S. adults about their daily water intake. A graph of the data shows a roughly symmetric shape with no outliers. The figure below displays Minitab output for a one-sample t interval for the population mean. Is there convincing evidence at the 10% significance level that the blogger’s claim is incorrect? Justify your answer.

83. Tests and CIs The P-value for a two-sided test of the null hypothesis H0 : m = 10 is 0.06. (a) Does the 95% confidence interval for m include 10? Why or why not?

Subject

Right thread

Left thread

1

113

137

2

105

105

3

130

133

4

101

108

5

138

115

6

118

170

7

 87

103

8

116

145

9

 75

 78

10

 96

107

11

122

 84

12

103

148

13

116

147

14

107

 87

15

118

166

16

103

146

17

111

123

18

104

135

19

111

112

20

 89

 93

21

 78

 76

22

100

116

23

 89

 78

24

 85

101

25

 88

123

(b) Does the 90% confidence interval for m include 10? Why or why not?

(a) Explain why it was important to randomly assign the order in which each subject used the two knobs.

84. Tests and CIs The P-value for a two-sided test of the null hypothesis H0 : m = 15 is 0.03.

(b) The project designers hoped to show that right-­ handed people find right-hand threads easier to use, on average. Carry out a test at the 5% significance level to investigate this claim.

(a) Does the 99% confidence interval for m include 15? Why or why not? (b) Does the 95% confidence interval for m include 15? Why or why not? 85. Right versus left  The design of controls and instruments affects how easily people can use them. A student project investigated this effect by asking 25 right-handed students to turn a knob (with their right hands) that moved an indicator. There were two identical instruments, one with a right-hand thread (the knob turns clockwise) and the other with a left-hand thread (the knob must be turned counterclockwise). Each of the 25 students used both instruments in a random order. The following table gives the times in seconds each subject took to move the indicator a fixed distance.23 Note that smaller times are better.

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86. Floral scents and learning  We hear that listening to Mozart improves students’ performance on tests. Maybe pleasant odors have a similar effect. To test this idea, 21 subjects worked two different but roughly equivalent paper-and-pencil mazes while wearing a mask. The mask was either unscented or carried a floral scent. Each subject used both masks, in a random order. The table below gives the subjects’ times (in seconds) with both masks.24 Note that smaller times are better. Subject

Unscented

Scented

1

30.60

37.97

2

48.43

51.57

3

60.77

56.67

4

36.07

40.47

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Section 9.3 Tests about a Population Mean Subject

Unscented

Scented

5

68.47

49.00

6

32.43

43.23

7

43.70

44.57

8

37.10

28.40

9

31.17

28.23

10

51.23

68.47

11

65.40

51.10

12

58.93

83.50

13

54.47

38.30

14

43.53

51.37

15

37.93

29.33

16

43.50

54.27

17

87.70

62.73

18

53.53

58.00

19

64.30

52.40

20

47.37

53.63

21

53.67

47.00

(a) Explain why it was important to randomly assign the order in which each subject used the two masks. (b) Do these data provide convincing evidence that the floral scent improved performance, on average? 87. Growing tomatoes  Researchers suspect that Variety A tomato plants have a higher average yield than Variety B tomato plants. To find out, researchers randomly select 10 Variety A and 10 Variety B tomato plants. Then the researchers divide in half each of 10 small plots of land in different locations. For each plot, a coin toss determines which half of the plot gets a Variety A plant; a Variety B plant goes in the other half. After harvest, they compare the yield in pounds for the plants at each location. The 10 differences in yield (Variety A − Variety B) are recorded. A graph of the differences looks roughly symmetric and single-peaked with no outliers. A paired t test on the differences yields t = 1.295 and P-value = 0.1138. (a) State appropriate hypotheses for the paired t test. Be sure to define your parameter. (b) What are the degrees of freedom for the paired t test? (c) Interpret the P-value in context. What conclusion should the researchers draw? (d) Describe a Type I error and a Type II error in this setting. Which mistake could researchers have made based on your answer to part (c)? 88. Music and memory  Does listening to music while studying hinder students’ learning? Two AP® Statistics students designed an experiment to find out.

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599

They selected a random sample of 30 students from their medium-sized high school to participate. Each subject was given 10 minutes to memorize two different lists of 20 words, once while listening to music and once in silence. The order of the two word lists was determined at random; so was the order of the treatments. The difference in the number of words recalled (music − silence) was recorded for each subject. A paired t test on the differences yielded t = −3.01 and P-value = 0.0027. (a) State appropriate hypotheses for the paired t test. Be sure to define your parameter. (b) What are the degrees of freedom for the paired t test? (c) Interpret the P-value in context. What conclusion should the students draw? (d) Describe a Type I error and a Type II error in this setting. Which mistake could students have made based on your answer to part (c)? 89. The power of tomatoes  Refer to Exercise 87. Explain two ways that the researchers could have increased the power of the test to detect m = 0.5. 90. Music and memory  Refer to Exercise 88. Which of the following changes would give the test a higher power to detect m = −1: using a = 0.01 or a = 0.10? Explain. 91. Significance and sample size  A study with 5000 subjects reported a result that was statistically significant at the 5% level. Explain why this result might not be particularly large or important. 92. Sampling shoppers  A marketing consultant observes 50 consecutive shoppers at a supermarket, recording how much each shopper spends in the store. Explain why it would not be wise to use these data to carry out a significance test about the mean amount spent by all shoppers at this supermarket. 93. Do you have ESP?  A researcher looking for evidence of extrasensory perception (ESP) tests 500 subjects. Four of these subjects do significantly better (P < 0.01) than random guessing. (a) Is it proper to conclude that these four people have ESP? Explain your answer. (b) What should the researcher now do to test whether any of these four subjects have ESP? 94. Ages of presidents  Joe is writing a report on the backgrounds of American presidents. He looks up the ages of all the presidents when they entered office. Because Joe took a statistics course, he uses these numbers to perform a significance test about the mean age of all U.S. presidents. Explain why this makes no sense.

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CHAPTER 9

Testing a Claim on Friday the 6th and Friday the 13th in the same month. The dotplot and computer output below summarize the data on the difference in the number of shoppers at each store on these two days (subtracting in the order 6th minus 13th).25

Multiple choice: Select the best answer for Exercises 95 to 102. 95. The reason we use t procedures instead of z procedures when carrying out a test about a population mean is that (a) z requires that the sample size be large. (b) z requires that you know the population standard deviation s.

–800

N

(d) z requires that the population distribution be perfectly Normal. 96. You are testing H0 : m = 10 against Ha : m < 10 based on an SRS of 20 observations from a Normal population. The t statistic is t = −2.25. The P-value (a) falls between 0.01 and 0.02. (b) falls between 0.02 and 0.04. (c) falls between 0.04 and 0.05. (d) falls between 0.05 and 0.25. (e) is greater than 0.25. 97. You are testing H0 : m = 10 against Ha : m ≠ 10 based on an SRS of 15 observations from a Normal population. What values of the t statistic are statistically significant at the a = 0.005 level? (a)  t > 3.326

(d)  t < −3.326 or t > 3.326

(b)  t > 3.286

(e)  t < −3.286 or t > 3.286

– 400

–200

0

200

Diff

(c) z requires that the data come from a random sample or randomized experiment.

(e) z can only be used for proportions.

– 600

Mean Median TrMean StDev SEMean

45 -46.5 -11.0



-37.4 178.0

26.1

Min

Max

Q1

Q3

-774.0 302.0 -141.0 53.5

Researchers would like to carry out a test of H0 : md = 0 versus Ha : md ≠ 0, where md is the true mean difference in the number of grocery shoppers on these two days. Which of the following conditions for performing a paired t test are clearly satisfied?

I. Random  II. 10%   III. Normal/Large Sample (a)  I only

(c)  III only

(b)  II only

(d)  I and II only

(e)  I, II, and III

100. The most important condition for sound conclusions from statistical inference is that (a) the data come from a well-designed random sample or randomized experiment. (b) the population distribution be exactly Normal. (c) the data contain no outliers. (d) the sample size be no more than 10% of the population size.

(c)  t > 2.977

(e) the sample size be at least 30.

98. After checking that conditions are met, you perform a significance test of H0 : m = 1 versus Ha : m ≠ 1. You obtain a P-value of 0.022. Which of the following must be true?

(b) A 95% confidence interval for m will include the value 0.

101. Vigorous exercise helps people live several years longer (on average). Whether mild activities like slow walking extend life is not clear. Suppose that the added life expectancy from regular slow walking is just 2 months. A statistical test is more likely to find a significant increase in mean life expectancy if (a) it is based on a very large random sample and a 5% significance level is used.

(c) A 99% confidence interval for m will include the value 1.

(b) it is based on a very large random sample and a 1% significance level is used.

(d) A 99% confidence interval for m will include the value 0.

(c) it is based on a very small random sample and a 5% significance level is used.

(e) None of these is necessarily true.

(d) it is based on a very small random sample and a 1% significance level is used.

(a) A 95% confidence interval for m will include the value 1.

99. Does Friday the 13th have an effect on people’s behavior? Researchers collected data on the number of shoppers at a sample of 45 nearby grocery stores

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(e) the size of the sample doesn’t have any effect on the significance of the test.

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Section 9.3 Tests about a Population Mean 102. A researcher plans to conduct a significance test at the a = 0.01 significance level. She designs her study to have a power of 0.90 at a particular alternative value of the parameter of interest. The probability that the researcher will commit a Type II error for the particular alternative value of the parameter at which she computed the power is (a)  0.01.  (b) 0.10.  (c) 0.89.  (d) 0.90.  (e) 0.99. 103. Is your food safe? (8.1)  “Do you feel confident or not confident that the food available at most grocery stores is safe to eat?” When a Gallup Poll asked this question, 87% of the sample said they were confident.26 Gallup announced the poll’s margin of error for 95% confidence as ±3 percentage points. Which of the following sources of error are included in this margin of error? Explain. (a) Gallup dialed landline telephone numbers at random and so missed all people without landline phones, including people whose only phone is a cell phone.

601

(b) Some people whose numbers were chosen never answered the phone in several calls or answered but refused to participate in the poll. (c) There is chance variation in the random selection of telephone numbers. 104. Spinning for apples (6.3 or 7.3)  In the “Ask Marilyn” column of Parade magazine, a reader posed this question: “Say that a slot machine has five wheels, and each wheel has five symbols: an apple, a grape, a peach, a pear, and a plum. I pull the lever five times. What are the chances that I’ll get at least one apple?” Suppose that the wheels spin independently and that the five symbols are equally likely to appear on each wheel in a given spin. (a) Find the probability that the slot player gets at least one apple in one pull of the lever. Show your method clearly. (b) Now answer the reader’s question. Show your method clearly.

FRAPPY!  Free Response AP® Problem, Yay! The following problem is modeled after actual AP® Statistics exam free response questions. Your task is to generate a complete, concise response in 15 minutes.

Station

Price

5

3.97

6

3.99

Directions: Show all your work. Indicate clearly the methods you use, because you will be scored on the correctness of your methods as well as on the accuracy and completeness of your results and explanations.

7

4.05

8

3.98

9

4.09

10

4.02

Mean

4.038

SD

0.0533

Anne reads that the average price of regular gas in her state is $4.06 per gallon. To see if the average price of gas is different in her city, she selects 10 gas stations at random and records the price per gallon for regular gas at each station. The data, along with the sample mean and standard deviation, are listed in the table below. Station

Price

1

4.13

2

4.01

3

4.09

4

4.05

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Do the data provide convincing evidence that the average price of gas in Anne’s city is different from $4.06 per gallon? After you finish, you can view two example solutions on the book’s Web site (www.whfreeman.com/tps5e). Determine whether you think each solution is “complete,” “substantial,” “developing,” or “minimal.” If the solution is not complete, what improvements would you suggest to the student who wrote it? Finally, your teacher will provide you with a scoring rubric. Score your response and note what, if anything, you would do differently to improve your own score.

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Chapter Review Section 9.1: Significance Tests: The Basics In this section, you learned the basic ideas of significance testing. Start by stating the hypotheses that you want to test. The null hypothesis (H0) is typically a statement of “no difference” and the alternative hypothesis (Ha) describes what we suspect is true. Remember that hypotheses are always about parameters, not statistics. When sample data provide support for the alternative hypothesis, there are two possible explanations: (1) the null hypothesis is true, and data supporting the alternative hypothesis occurred just by chance, or (2) the alternative ­hypothesis is true, and the data are consistent with an alternative value of the parameter. In a significance test, always start with the belief that the null hypothesis is true. If you can rule out chance as a plausible explanation for the observed data, there is convincing evidence that the alternative hypothesis is true. The P-value in a significance test measures how likely it is to get results at least as extreme as the observed results by chance alone, assuming the null hypothesis is true. To determine if the P-value is small enough to reject H0, compare it to a predetermined significance level such as a = 0.05. If P-value < a, reject H0—there is convincing evidence that the alternative hypothesis is true. However, if P-value ≥ a, fail to reject H0—there is not convincing evidence that the alternative hypothesis is true. Because conclusions are based on sample data, there is a possibility that the conclusion will be incorrect. You can make two types of errors in a significance test: a Type I error occurs if you find convincing evidence for the alternative hypothesis when, in reality, the null hypothesis is true. A Type II error occurs when you don’t find convincing evidence that the alternative hypothesis is true when, in reality, the alternative hypothesis is true. The probability of making a Type I error is equal to the significance level (a) of the test. Section 9.2: Tests about a Population Proportion In this section, you learned the details of conducting a significance test for a population proportion p. Whenever you are asked if there is convincing evidence for a claim about a population proportion, you are expected to respond using the familiar four-step process. STATE: Give the hypotheses you are testing in terms of p, state the significance level, and define the parameter p. PLAN: Name the procedure you plan to use (one-sample z test for a population proportion) and check the appropriate conditions (Random, 10%, Large Counts) to see if the procedure is appropriate.

• Random: The data come from a well-designed random sample or randomized experiment. ~~

10%: The sample size should be no larger than 10% of the population when sampling without replacement.

• Large Counts: Both np0 and n(1−p0) must be at least 10, where p0 is the value of p in the null hypothesis.

DO: Calculate the test statistic and P-value. The test statistic z measures how far away the sample statistic is from the hypothesized parameter value in standardized units: z=

p^ − p0 p0(1 − p0) n Å

To calculate the P-value, use Table A or technology. CONCLUDE: Use the P-value to make an appropriate conclusion about the hypotheses in context. Perform a two-sided test when looking for convincing evidence that the true value of the parameter is different from the hypothesized value, in either direction. The P ­ -­value for a two-sided test is calculated by finding the probability of getting a sample statistic at least as extreme as the observed statistic, in either direction, assuming the null hypothesis is true. You can also use a confidence interval to make a conclusion for a two-sided test. If the null parameter value is one of the plausible values in the interval, there isn’t convincing evidence that the alternative hypothesis is true. However, if the null parameter value is not one of the plausible values in the interval, there is convincing evidence that the alternative hypothesis is true. Besides helping you draw a conclusion, the interval tells you which alternative parameter values are plausible. The probability that you avoid making a Type II error when an alternative value of the parameter is true is called the power of the test. Power is good—if the alternative hypothesis is true, we want to maximize the probability of finding convincing evidence that it is true. We can increase the power of a significance test by increasing the sample size or by increasing the significance level. The power of a test will also be greater when the alternative value of the parameter is farther away from the null hypothesis value. Section 9.3: Tests about a Population Mean In this section, you learned the details of conducting a significance test for a population mean. Although some of the details are different, the reasoning and structure of the tests in this section are the same as in Section 9.2. In fact, the “State” and “Conclude” steps are exactly the same, other than the switch from proportions to means.

602

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603

Section 2.1 Scatterplots and Correlations

PLAN: Name the procedure you are using (one-sample t test for a population mean), and check the conditions (Random, 10%, and Normal/Large Sample). The Random and 10% conditions are the same as in Section 9.2. The Normal/Large Sample condition states that the population distribution must be Normal or the sample size must be large (n ≥ 30). If the sample is small and the population shape is unknown, graph the sample data to make sure there is no strong skewness or outliers. DO: Calculate the test statistic and P-value. The test statistic t measures how far away the sample statistic is from the hypothesized parameter value in standardized units: t=

x– − m sx∙!n

To calculate the P-value, determine the degrees of freedom (df = n – 1) and use Table B or technology. Use a paired t test to analyze the results of comparative experiments and observational studies that produce paired data. Start by calculating the difference for each pair and use the set of differences to check the Normal/Large Sample condition and to calculate the test statistic and P-value. Remember to use significance tests wisely. When planning a study, use a large enough sample size so the test will have adequate power. Also, remember that statistically significant results aren’t always “practically” important. Finally, be aware that the probability of making at least one Type I error goes up dramatically when conducting multiple tests.

What Did You Learn? Learning Objective

Section

Related Example on Page(s)

State the null and alternative hypotheses for a significance test about a population parameter.

9.1

540

R9.1

Interpret a P-value in context.

9.1

543, 544

R9.5

Determine if the results of a study are statistically significant and draw an appropriate conclusion using a significance level.

9.1

546

R9.5

Interpret a Type I and a Type II error in context, and give a ­consequence of each.

9.1

548

State and check the Random, 10%, and Large Counts conditions for performing a significance test about a population proportion.

9.2

555

R9.4

Perform a significance test about a population proportion.

9.2

559, 562

R9.4

Interpret the power of a test and describe what factors affect the power of a test.

9.2

565, Discussion on 568

R9.3

Describe the relationship among the probability of a Type I error (significance level), the probability of a Type II error, and the power of a test.

9.2

565

R9.3

State and check the Random, 10%, and Normal/Large Sample conditions for performing a significance test about a population mean.

9.3

575

Perform a significance test about a population mean.

9.3

580, 583

9.2, 9.3

563, 585

9.3

586

Use a confidence interval to draw a conclusion for a two-sided test about a population parameter. Perform a significance test about a mean difference using paired data.

Relevant Chapter Review Exercise(s)

R9.3, R9.4

R9.2, R9.6, R9.7 R9.6 R9.5, R9.6 R9.7

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Chapter 9 Chapter Review Exercises These exercises are designed to help you review the important ideas and methods of the chapter. R9.1 Stating hypotheses  State the appropriate null and alternative hypotheses in each of the following settings. Be sure to define the parameter. (a) The average height of 18-year-old American women is 64.2 inches. You wonder whether the mean height of this year’s female graduates from a large local high school (over 3000 students) differs from the national average. You measure an SRS of 48 female graduates and find that x– = 63.1 inches. (b) Mr. Starnes believes that less than 75% of the students at his school completed their math homework last night. The math teachers inspect the homework assignments from a random sample of students at the school to help Mr. Starnes test his claim. R9.2 Fonts and reading ease  Does the use of fancy type fonts slow down the reading of text on a computer screen? Adults can read four paragraphs of text in the common Times New Roman font in an average time of 22 seconds. Researchers asked a random sample of 24 adults to read this text in the ornate font named Gigi. Here are their times, in seconds: 23.2 21.2 28.9 27.7 29.1 27.3 16.1 22.6 25.6 34.2 23.9 26.8 20.5 34.3 21.4 32.6 26.2 34.1 31.5 24.6 23.0 28.6 24.4 28.1



State and check the conditions for performing a significance test using these data.

R9.3 Strong chairs?  A company that manufactures classroom chairs for high school students claims that the mean breaking strength of the chairs that they make is 300 pounds. One of the chairs collapsed beneath a 220-pound student last week. You wonder whether the manufacturer is exaggerating the breaking strength of the chairs. (a) State appropriate null and alternative hypotheses in this setting. Be sure to define your parameter. (b) Describe a Type I error and a Type II error in this setting, and give the consequences of each. (c) Would you recommend a significance level of 0.01, 0.05, or 0.10 for this test? Justify your choice. (d) The power of the test to detect m = 294 using a = 0.05 is 0.71. Interpret this value in context. (e) Explain two ways that you could increase the power of the test from (d). R9.4 Flu vaccine  A drug company has developed a new vaccine for preventing the flu. The company claims

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that fewer than 5% of adults who use its vaccine will get the flu. To test the claim, researchers give the vaccine to a random sample of 1000 adults. Of these, 43 get the flu. (a) Do these data provide convincing evidence to support the company’s claim? (b) Which kind of mistake—a Type I error or a Type II error—could you have made in (a)? Explain. (c) From the company’s point of view, would a Type I error or Type II error be more serious? Why? R9.5 Roulette  An American roulette wheel has 18 red slots among its 38 slots. To test if a particular roulette wheel is fair, you spin the wheel 50 times and the ball lands in a red slot 31 times. The resulting P-value is 0.0384. (a) Interpret the P-value in context. (b) Are the results statistically significant at the a = 0.05 level? Explain. What conclusion would you make? (c) The casino manager uses your data to produce a 99% confidence interval for p and gets (0.44, 0.80). He says that this interval provides convincing evidence that the wheel is fair. How do you respond? R9.6 Radon detectors  Radon is a colorless, odorless gas that is naturally released by rocks and soils and may concentrate in tightly closed houses. Because radon is slightly radioactive, there is some concern that it may be a health hazard. Radon detectors are sold to homeowners worried about this risk, but the detectors may be inaccurate. University researchers placed a random sample of 11 detectors in a chamber where they were exposed to 105 picocuries per liter of radon over 3 days. A graph of the radon readings from the 11 detectors shows no strong skewness or outliers. The mean reading is 104.82 and the standard deviation of the readings is 9.54. (a) Is there convincing evidence at the 10% level that the mean reading differs from the true value 105? (b) A 90% confidence interval for the true mean reading is (99.61, 110.03). Is this interval consistent with your conclusion from part (a)? Explain. R9.7 Better barley  Does drying barley seeds in a kiln increase the yield of barley? A famous experiment by William S. Gosset (who discovered the t distributions) investigated this question. Eleven pairs of adjacent plots were marked out in a large field. For each pair, regular barley seeds were planted in one plot and kiln-dried seeds were planted in the other. The following table displays the data on yield (lb/acre).27

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AP® Statistics Practice Test Plot

Regular

Kiln

1

1903

2009

2

1935

1915

3

1910

2011

4

2496

2463

5

2108

2180

6

1961

1925

7

2060

2122

8

1444

1482

9

1612

1542

10

1316

1443

11

1511

1535

605

(a) How can the Random condition be satisfied in this study? (b) Assuming that the Random condition has been met, do these data provide convincing evidence that drying barley seeds in a kiln increases the yield of barley, on average? Justify your answer.

Chapter 9 AP® Statistics Practice Test Section I: Multiple Choice  ​Select the best answer for each question. T9.1 An opinion poll asks a random sample of adults whether they favor banning ownership of handguns by private citizens. A commentator believes that more than half of all adults favor such a ban. The null and alternative hypotheses you would use to test this claim are (a) ​H0 : p^ = 0.5; Ha : p^ > 0.5 (b) H0 : p = 0.5; Ha : p > 0.5 (c) H0 : p = 0.5; Ha : p < 0.5 (d) H0 : p = 0.5; Ha : p ≠ 0.5 (e) H0 : p > 0.5; Ha : p = 0.5 T9.2 You are thinking of conducting a one-sample t test about a population mean m using a 0.05 significance level. Which of the following statements is correct? (a) You should not carry out the test if the sample does not have a Normal distribution. (b) You can safely carry out the test if there are no outliers, regardless of the sample size. (c) You can carry out the test if a graph of the data shows no strong skewness, regardless of the sample size. (d) You can carry out the test only if the population standard deviation is known. (e) You can safely carry out the test if your sample size is at least 30.

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T9.3 To determine the reliability of experts who interpret lie detector tests in criminal investigations, a random sample of 280 such cases was studied. The results were Suspect’s True Status Examiner’s Decision “Innocent”

Innocent 131

15

9

125

“Guilty”



Guilty

If the hypotheses are H0: suspect is innocent versus Ha: suspect is guilty, then we could estimate the probability that experts who interpret lie detector tests will make a Type II error as

(a) 15/280. (b) 9/280.

(c) 15/140. (d) 9/140.

(e) 15/146.

T9.4 A significance test allows you to reject a null hypothesis H0 in favor of an alternative Ha at the 5% significance level. What can you say about significance at the 1% level? (a) H0 can be rejected at the 1% significance level. (b) There is insufficient evidence to reject H0 at the 1% significance level. (c) There is sufficient evidence to accept H0 at the 1% significance level.

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(d) Ha can be rejected at the 1% significance level. (e) The answer can’t be determined from the information given. T9.5 A random sample of 100 likely voters in a small city produced 59 voters in favor of Candidate A. The observed value of the test statistic for testing the null hypothesis H0 : p = 0.5 versus the alternative hypothesis Ha : p > 0.5 is (a) z =

(b) z =

(c) z =

0.59 − 0.5 0.59(0.41) Å 100 0.59 − 0.5

0.5(0.5) Å 100



0.5 − 0.59



(d)  z =

(e)  t =

0.5 − 0.59 0.5(0.5) Å 100

0.59 − 0.5 0.5(0.5) Å 100

0.59(0.41) Å 100

T9.6 A researcher claims to have found a drug that causes people to grow taller. The coach of the basketball team at Brandon University has expressed interest but demands evidence. Over 1000 Brandon students volunteer to participate in an experiment to test this new drug. Fifty of the volunteers are randomly selected, their heights are measured, and they are given the drug. Two weeks later, their heights are measured again. The power of the test to detect an average increase in height of 1 inch could be increased by (a) using only volunteers from the basketball team in the experiment. (b) using a = 0.01 instead of a = 0.05. (c) using a = 0.05 instead of a = 0.01. (d) giving the drug to 25 randomly selected students instead of 50. (e) using a two-sided test instead of a one-sided test. T9.7 A 95% confidence interval for a population mean m is calculated to be (1.7, 3.5). Assume that the conditions for performing inference are met. What conclusion can we draw for a test of H0 : m = 2 versus Ha : m ≠ 2 at the a = 0.05 level based on the confidence interval? (a) None. We cannot carry out the test without the original data. (b) None. We cannot draw a conclusion at the a = 0.05 level because this test corresponds to the 97.5% confidence interval.

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(c) None. Confidence intervals and significance tests are unrelated procedures. (d) We would reject H0 at level a = 0.05. (e) We would fail to reject H0 at level a = 0.05. T9.8 In a test of H0 : p = 0.4 against Ha : p ≠ 0.4, a random sample of size 100 yields a test statistic of z = 1.28. The P-value of the test is approximately equal to (a) 0.90. (b) 0.40.

(c) 0.05. (d) 0.20.

(e) 0.10.

T9.9 An SRS of 100 postal employees found that the average time these employees had worked at the postal service was 7 years with standard deviation 2 years. Do these data provide convincing evidence that the mean time of employment m for the population of postal employees has changed from the value of 7.5 that was true 20 years ago? To determine this, we test the hypotheses H0 : m = 7.5 versus Ha : m ≠ 7.5 using a one-­ sample t test. What conclusion should we draw at the 5% significance level? (a) There is convincing evidence that the mean time working with the postal service has changed. (b) There is not convincing evidence that the mean time working with the postal service has changed. (c) There is convincing evidence that the mean time working with the postal service is still 7.5 years. (d) There is convincing evidence that the mean time working with the postal service is now 7 years. (e) We cannot draw a conclusion at the 5% significance level. The sample size is too small. T9.10 Are TV commercials louder than their surrounding programs? To find out, researchers collected data on 50 randomly selected commercials in a given week. With the television’s volume at a fixed setting, they measured the maximum loudness of each commercial and the maximum loudness in the first 30 seconds of regular programming that followed. Assuming conditions for inference are met, the most appropriate method for answering the question of interest is (a) a one-proportion z test. (b) a one-proportion z interval. (c) a paired t test. (d) a paired t interval. (e) None of these.

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Section II: Free Response  ​Show all your work. Indicate clearly the methods you use, because you will be graded on the correctness of your methods as well as on the accuracy and completeness of your results and explanations. T9.11 A software company is trying to decide whether to produce an upgrade of one of its programs. Customers would have to pay $100 for the upgrade. For the upgrade to be profitable, the company needs to sell it to more than 20% of their customers. You contact a random sample of 60 customers and find that 16 would be willing to pay $100 for the upgrade. (a) Do the sample data give good evidence that more than 20% of the company’s customers are willing to purchase the upgrade? Carry out an appropriate test at the a = 0.05 significance level. (b) Which would be a more serious mistake in this setting—a Type I error or a Type II error? Justify your answer. (c) Describe two ways to increase the power of the test in part (a). T9.12 “I can’t get through my day without coffee” is a common statement from many students. Assumed benefits include keeping students awake during lectures and making them more alert for exams and tests. Students in a statistics class designed an experiment to measure memory retention with and without drinking a cup of coffee one hour before a test. This experiment took place on two different days in the same week (Monday and Wednesday). Ten students were used. Each student received no coffee or one cup of coffee one hour before the test on a parti­cular day. The test consisted of a series of words flashed on a screen, after which the student had to write down as many of the words as possible. On the other day, each student received a different amount of coffee (none or one cup).

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(a) One of the researchers suggested that all the subjects in the experiment drink no coffee before Monday’s test and one cup of coffee before Wednesday’s test. Explain to the researcher why this is a bad idea and suggest a better method of deciding when each subject receives the two treatments. (b) The data from the experiment are provided in the table below. Set up and carry out an appropriate test to determine whether there is convincing evidence that drinking coffee improves memory. Student

No cup

One cup

1

24

25

2

30

31

3

22

23

4

24

24

5

26

27

6

23

25

7

26

28

8

20

20

9

27

27

10

28

30

T9.13 A government report says that the average amount of money spent per U.S. household per week on food is about $158. A random sample of 50 households in a small city is selected, and their weekly spending on food is recorded. The sample data have a mean of $165 and a standard deviation of $20. Is there convincing evidence that the mean weekly spending on food in this city differs from the national figure of $158?

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Chapter

10 Introduction 610 Section 10.1 612 Comparing Two Proportions Section 10.2 634 Comparing Two Means Free Response AP® Problem, YAY! 662 Chapter 10 Review 662 Chapter 10 Review Exercises 664 Chapter 10 AP® Statistics Practice Test 666 Cumulative AP® Practice Test 3

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Comparing Two Populations or Groups case study

Fast-Food Frenzy! More than $70 billion is spent each year in the drive-thru lanes of America’s fast-food restaurants. H ­ aving quick, accurate, and friendly service at a drive-thru window translates directly into revenue for the restaurant. According to Jack Greenberg, former CEO of McDonald’s, sales increase 1% for every six seconds saved at the drive-thru. So industry executives, stockholders, and analysts closely follow the ratings of fast-food drive-thru lanes that appear annually in QSR, a publication that reports on the quick-service restaurant industry. The 2012 QSR magazine drive-thru study involved visits to a random sample of restaurants in the 20 largest fast-food chains in all 50 states. During each visit, the researcher ordered a modified main item (for example, a hamburger with no pickles), a side item, and a drink. If any item was not received as o­ rdered, or if the restaurant failed to give the correct change or supply a straw and a napkin, then the order was considered “inaccurate.” Service time, which is the time from when the car stopped at the speaker to when the entire order was received, was measured each visit. Researchers also recorded whether or not each restaurant had an order-confirmation board in its drive-thru.1 Here are some results from the 2012 QSR study: • For restaurants with order-confirmation boards, 1169 of 1327 visits (88.1%) resulted in accurate orders. For restaurants with no order-confirmation board, 655 of 726 visits (90.2%) resulted in accurate orders. • McDonald’s average service time for 362 drive-thru visits was 188.83 seconds with a standard deviation of 17.38 seconds. Burger King’s service time for 318 drive-thru visits had a mean of 201.33 seconds and a standard deviation of 18.85 seconds. Was there a significant difference in accuracy at restaurants with and without order-confirmation boards? How much better was the average service time at McDonald’s than at Burger King restaurants in 2012? By the end of the chapter, you should have acquired the tools to help answer questions like these.

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Introduction Which of two popular drugs—Lipitor or Pravachol—helps lower “bad cholesterol” more? Researchers designed an experiment, called the PROVE-IT Study, to find out. They used about 4000 people with heart disease as subjects. These individuals were randomly assigned to one of two treatment groups: Lipitor or Pravachol. At the end of the study, researchers compared the proportion of subjects in each group who died, had a heart attack, or suffered other serious consequences within two years. For those using Pravachol, the proportion was 0.263; for those using Lipitor, it was 0.224.2 Could such a difference have occurred purely by the chance involved in the random assignment? This is a question about comparing two proportions. Who studies more in college—men or women? Researchers asked separate random samples of 30 males and 30 females at a large university how many minutes they studied on a typical weeknight. The females reported studying an average of 165.17 minutes; the male average was 117.17 minutes. How large is the difference in the corresponding population means? This is a question about comparing two means. Comparing two proportions or means based on random sampling or a randomized experiment is one of the most common situations encountered in statistical practice. In the PROVE-IT experiment, the goal of inference is to determine whether the treatments (Lipitor and Pravachol) caused the observed difference in the proportion of subjects who experienced serious consequences in the two groups. For the college studying survey, the goal of inference is to draw a conclusion about the actual mean study times for all women and all men at the university. The following Activity gives you a taste of what lies ahead in this chapter.

Activity Materials: Set of 50 index cards or standard deck of ­playing cards for each pair of ­students

Is Yawning Contagious? According to the popular TV show Mythbusters, the answer is “Yes.” The Mythbusters team conducted an experiment involving 50 subjects. Each subject was placed in a booth for an extended period of time and monitored by hidden camera. Thirty-four subjects were given a “yawn seed” by one of the experimenters; that is, the experimenter yawned in the subject’s presence before leaving the room. The remaining 16 subjects were given no yawn seed. What ­happened in the experiment? The table below shows the results:3 Yawn Seed? Subject Yawned?

Yes

No

Total

Yes

10

 4

14

No

24

12

36

Total

34

16

50

Ten of the 34 subjects (29.4%) in the yawn-seed group yawned, compared to 4 of the 16 subjects (25.0%) in the no-yawn-seed group. The difference in the proportions who yawned for the

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two groups is 10/34 − 4/16 = 0.044. Adam Savage and Jamie Hyneman, the cohosts of MythBusters, used this difference as evidence that yawning is contagious. But is the evidence convincing? In this Activity, your class will investigate whether the results of the experiment were really statistically significant. Let’s see what would happen just by chance if we randomly reassign the 50 people in this experiment to the two groups (yawn seed and no yawn seed) many times, assuming the treatment received doesn’t affect whether or not a person yawns. 1. ​We need 50 cards to represent the subjects in this study. In the MythBusters experiment, 14 people yawned and 36 didn’t. Because we’re ­assuming that the treatment received won’t change whether each subject yawns, we use 14 cards to represent people who yawn and 36 cards to represent those who don’t. •  Using index cards: Write “Yes” on 14 cards and “No” on 36 cards. •  Using playing cards: Remove the ace of spades and ace of clubs from the deck. All jacks, queens, kings, and aces represent subjects who yawn. All remaining cards represent subjects who don’t yawn. 2. ​Shuffle and deal two piles of cards—one with 34 cards and one with 16 cards. The first pile represents the yawn-seed group and the second pile represents the no-yawn-seed group. The shuffling reflects our assumption that the outcome for each subject is not affected by the treatment. Calculate the difference in the proportions who yawned for the two groups (yawn seed – no yawn seed). For example, if you get 9 yawners in the yawn-seed group and 5 yawners in the no-yawn-seed group, the resulting difference in proportions is 9 5 − = −0.048 34 16 A negative difference would mean that a smaller proportion of people in the yawn-seed group yawned during the experiment than in the no-yawn-seed group. 3.  Your teacher will draw and label axes for a class dotplot. Plot the result you got in Step 2 on the graph. 4.  Repeat Steps 2 and 3 if needed to get a total of at least 40 repetitions of the simulation for your class. 5. ​ Based on the class’s simulation results, how surprising would it be to get a difference in proportions of 0.044 (what the Mythbusters got in their experiment) or larger simply due to the chance involved in the random a­ ssignment? 6. ​ What conclusion would you draw about whether yawning is contagious? Explain.

Here is an example of what the class dotplot in the Activity might look like after 100 trials. In this simulation, 50 of the 100 trials (in red) produced a difference in proportions of at least 0.044, so the approximate P-value is 0.50. It is very likely that a difference this big could occur just due to the chance variation in random assignment! This result is not statistically significant and does not provide convincing evidence that yawning is ­contagious.

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10.1 What You Will Learn

Comparing Two Proportions By the end of the section, you should be able to:

• Describe the shape, center, and spread of the sampling distribution of p^1 − p^2. • Determine whether the conditions are met for doing inference about p1 − p2.

• Construct and interpret a confidence interval to ­compare two proportions. • Perform a significance test to compare two proportions.

Suppose we want to compare the proportions of individuals with a certain characteristic in Population 1 and Population 2. Let’s call these parameters of interest p1 and p2. The ideal strategy is to take a separate random sample from each population and to compare the sample proportions p^ 1 and p^ 2 with that characteristic. What if we want to compare the effectiveness of Treatment 1 and Treatment 2 in a completely randomized experiment? This time, the parameters p1 and p2 that we want to compare are the true proportions of successful outcomes for each treatment. We use the proportions of successes in the two treatment groups, p^ 1 and p^ 2, to make the comparison. Here’s a table that summarizes these two situations: Population or treatment

Parameter

Statistic

Sample size

1

p1

p^ 1

n1

2

p2

p^ 2

n2

We compare the populations or treatments by doing inference about the difference p1 − p2 between the parameters. The statistic that estimates this difference is the difference between the two sample proportions, p^ 1 − p^ 2. To use p^ 1 − p^ 2 for inference, we must know its sampling distribution.

The Sampling Distribution of a Difference between Two Proportions To explore the sampling distribution of p^ 1 − p^ 2, let’s start with two populations having a known proportion of successes. Suppose that there are two large high schools, each with over 2000 students, in a certain town. At School 1, 70% of students did their homework last night. Only 50% of the students at School 2 did their homework last night. The counselor at School 1 takes an SRS of 100 students and records the proportion p^ 1 that did the homework. School 2’s counselor takes an SRS of 200 students and records the proportion p^ 2 that did the homework. What can we say about the difference p^ 1 − p^ 2 in the sample proportions? We used Fathom software to take an SRS of n1 = 100 students from School 1 and a separate SRS of n2 = 200 students from School 2 and to plot the values of p^ 1, p^ 2, and p^ 1 − p^ 2 from each sample. Our first set of simulated samples gave p^ 1 = 0.68 and p^ 2 = 0.505, so dots were placed above each of those values in Figure 10.1(a) and (b).

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Section 10.1 Comparing Two Proportions (a) Approximate sampling distribution of p^1

(b) Approximate sampling distribution of p^2

613

(c) Approximate sampling distribution of p^1 − p^2

FIGURE 10.1  ​Simulated sampling distributions of (a) the sample proportion p^1 of successes in 1000 SRSs of size n1 = 100 from a population with p1 = 0.70, (b) the sample proportion p^2 of ­successes in 1000 SRSs of size n2 = 200 from a population with p2 = 0.50, and (c) the ­difference in sample proportions p^1 − p^2 for each of the 1000 repetitions.

The difference in the sample proportions for this first set of samples is p^ 1 − p^ 2 = 0.68 − 0.505 = 0.175. A dot for this value appears in Figure 10.1(c). The three dotplots in Figure 10.1 show the results of repeating this process 1000 times. These are the approximate sampling distributions of p^ 1, p^ 2, and p^ 1 − p^ 2. In Chapter 7, we saw that the sampling distribution of a sample proportion p^ has the following properties: Shape:  Approximately Normal if np ≥ 10 and n(1 − p) ≥ 10 Center:  mp^ = p Spread:  sp^ =

p(1 − p) 1   if n ≤ N n Å 10

For the sampling distributions of p^ 1 and p^ 2 in this case: Sampling distribution of p^ 1

Sampling distribution of p^2

Shape

Approximately Normal; n1p1 = 100(0.70) = 70 ≥ 10 and n1(1 − p1) = 100(0.30) = 30 ≥ 10

Approximately Normal; n2p2 = 200(0.50) = 100 ≥ 10 and n2(1 − p2) = 200(0.50) = 100 ≥ 10

Center

mp^ 1 = p1 = 0.70

mp^ 2 = p2 = 0.50

Spread

p1(1 − p1) 0.7(0.3) = = 0.0458 Å n1 Å 100 because School 1 has a population of over 10(100) = 1000 students. sp^ 1 =

p2(1 − p2) 0.5(0.5) = = 0.0354 Å n2 Å 200 because School 2 has a population of over 10(200) = 2000 students.

sp^ 2 =

The approximate sampling distributions in Figures 10.1(a) and (b) give similar results. What about the sampling distribution of p^ 1 − p^ 2? Figure 10.1(c) suggests that it has an approximately Normal shape, is centered at about 0.198, and has standard deviation about 0.0572. The shape makes sense because we are combining two independent random variables, p^ 1 and p^ 2, that have approximately Normal distributions. How about the center? The true proportion of students who did last night’s homework at School 1 is p1 = 0.70 and at School 2 is p2 = 0.50. We expect the difference p^ 1 − p^ 2 to center on the actual difference in the population proportions, p1 − p2 = 0.70 − 0.50 = 0.20. The spread, however, is a bit more complicated.

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THINK ABOUT IT

How can we find formulas for the mean and standard deviation of the sampling distribution of p^ 1 − p^ 2?  ​Both p^ 1 and p^ 2 are ran-

dom variables. That is, their values would vary in repeated independent SRSs of size n1 and n2. Independent random samples yield independent random variables p^ 1 and p^ 2. The statistic p^ 1 − p^ 2 is the difference of these two independent random variables. In Chapter 6, we learned that for any two random variables X and Y, mX−Y = mX − mY For the random variables p^ 1 and p^ 2, we have mp^1 −p^2 = mp^1 − mp^2 = p1 − p2 In the school homework survey, mp^1 −p^2 = p1 − p2 = 0.70 − 0.50 = 0.20 We also learned in Chapter 6 that for independent random variables X and Y, s2X−Y = s2X + s2Y For the random variables p^ 1 and p^ 2, we have s2p^1−p^2= s2p^1 + s2p^2= a So sp^1−p^2=

p1(1 − p1) 2 p2(1 − p2) 2 p1(1 − p1) p2(1 − p2) + b +a b = n1 n2 n1 n2 Å Å

p1(1 − p1) p2(1 − p2) + . n1 n2 Å

In the school homework survey, sp^1−p^2=

p1(1 − p1) p2(1 − p2) 0.7(0.3) 0.5(0.5) + = + = 0.058 n1 n2 Å Å 100 200

This is similar to the result from the Fathom simulation. Here are the facts we need.

The Sampling Distribution of p^1 − p^2 Choose an SRS of size n1 from Population 1 with proportion of successes p1 and an independent SRS of size n2 from Population 2 with proportion of successes p2. • Shape: When n1p1, n1(1 − p1), n2p2, and n2(1 − p2) are all at least 10, the sampling distribution of p^ 1 − p^ 2 is approximately Normal. • Center:  The mean of the sampling distribution of p^ 1 − p^ 2 is p1 − p2. • Spread:  The standard deviation of the sampling distribution of p^ 1 − p^ 2 is



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p1(1 − p1) p2(1 − p2) + n1 n2 Å

as long as each sample is no more than 10% of its population.

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Section 10.1 Comparing Two Proportions

615

When conditions are met, the sampling distribution of p^ 1 − p^ 2 will be approximately Normal with mean mp^ 1 −p^ 2 = p1 − p2 and standard deviation p1(1 − p1) p2(1 − p2) . Figure 10.2 displays this distribution. + n1 n2 Å The formula for the standard deviation of the sampling distribution involves the unknown parameters p1 and p2. Just as in Chapters 8 and 9, we must replace these by estimates to do inference. And just as before, we do this a bit differently for confidence intervals and for tests. We’ll get to inference shortly. For now, let’s focus on the sampling distribution of p^ 1 − p^ 2. Standard deviation sp^ 1 −p^ 2 =

p2(1 – p2) p1(1 – p1) + n1 n2 Mean p1 – p2

FIGURE 10.2  ​Select independent SRSs from two populations having proportions of successes p1 and p2. The proportions of successes in the two samples are p^1 and p^2. When the samples are large, the sampling distribution of the difference p^1 − p^2 is ­approximately Normal.

^ –p ^ Values of p 1 2

Example

Yummy Goldfish! Describing the sampling distribution of p^ 1 − p^ 2 Your teacher brings two bags of colored goldfish crackers to class. Bag 1 has 25% red crackers and Bag 2 has 35% red crackers. Each bag contains more than 1000 crackers. Using a paper cup, your teacher takes an SRS of 50 crackers from Bag 1 and a separate SRS of 40 crackers from Bag 2. Let p^ 1 − p^ 2 be the difference in the sample proportions of red crackers.

PROBLEM: (a) What is the shape of the sampling distribution of p^1 − p^2 ? Why? (b) Find the mean of the sampling distribution. Show your work. (c) Find the standard deviation of the sampling distribution. Show your work.

SOLUTION: (a) Because n1p1 = 50(0.25) = 12.5, n1(1 − p1) = 50(0.75) = 37.5, n2p2 = 40(0.35) = 14, and n2(1 − p2) = 40(0.65) = 26 are all at least 10, the sampling distribution of p^1 − p^2 is approximately Normal. (b) The mean is mp^1 −p^2 = p1 − p2 = 0.25 − 0.35 = −0.10. (c) Because there are at least 10(50) = 500 crackers in Bag 1 and 10(40) = 400 crackers in Bag 2, the standard deviation is sp^1−p^2=

p1(1 − p1) p2(1 − p2) 0.25(0.75) 0.35(0.65) + = + = 0.0971 n1 n2 Å Å 50 40 For Practice Try Exercise

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Confidence Intervals for p1 − p2 When data come from two independent random samples or two groups in a randomized experiment (the Random condition), the statistic p^ 1 − p^ 2 is our best guess for the value of p1 − p2. We can use our familiar formula to calculate a confidence interval for p1 − p2: statistic ± (critical value) · (standard deviation of statistic) When the 10% condition is met, the standard deviation of the statistic p^ 1 − p^ 2 is sp^1−p^2=

p1(1 − p1) p2(1 − p2) + n1 n2 Å

If the Large Counts condition is met, we find the critical value z* for the given confidence level from the standard Normal curve.

Conditions for Constructing a Confidence Interval about a Difference in Proportions • Random: The data come from two independent random samples or from two groups in a randomized experiment. 1 ~~ 10%: When sampling without replacement, check that n1 ≤ N1 10 1 and n2 ≤ N2. 10 • Large Counts: The counts of “successes” and “failures” in each sample or group—n1p^1, n1(1 − p^ 1), n2p^ 2, n2(1 − p^ 2)—are all at least 10.

Because we don’t know the values of the parameters p1 and p2, we replace them in the standard deviation formula with the sample proportions. The result is the ­standard error (also called the estimated standard deviation) of the statistic p^ 1 − p^ 2: SEp^1−p^2=

p^1(1 − p^1) p^ 2(1 − p^ 2) + n1 n2 Å

This value tells us how far the difference in sample proportions will typically be from the difference in population proportions if we repeat the random sampling or random assignment many times. When the conditions are met, our confidence interval for p1 − p2 is therefore statistic ± (critical value) · (standard deviation of statistic) (p^ 1 − p^ 2) ± z*

p^1(1 − p^1) p^ 2(1 − p^ 2) + n1 n2 Å

This is often called a two-sample z interval for a difference between two proportions.

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Section 10.1 Comparing Two Proportions

Two-Sample z Interval for a Difference between Two Proportions When the conditions are met, an ­approximate C% confidence interval for p^ 1 − p^ 2 is p^1(1 − p^1) p^ 2(1 − p^ 2) (p^1 − p^ 2) ± z* + n1 n2 Å

where z* is the critical value for the standard Normal curve with C% of its area ­between −z* and z*. The following example shows how to construct and interpret a confidence interval for a difference in proportions. As usual with inference problems, we follow the four-step process. Because you are expected to include these four steps whenever you construct a confidence interval or perform a significance test, we will limit our use of the four-step icon to examples from this point forward.

Example

Teens and Adults on Social Networking Sites

STEP

4

Confidence interval for p1 − p2 As part of the Pew Internet and American Life Project, researchers conducted two surveys in 2012. The first survey asked a random sample of 799 U.S. teens about their use of social media and the ­Internet. A second survey posed similar questions to a random sample of 2253 U.S. adults. In these two studies, 80% of teens and 69% of adults used social-networking sites.

PROBLEM: (a) Calculate the standard error of the sampling distribution of the difference in the sample proportions (teens − adults). What information does this value provide? (b) Construct and interpret a 95% confidence interval for the difference between the proportion of all U.S. teens and adults who use social-networking sites.

SOLUTION: (a) The sample proportions of teens and adults who use social-networking sites are p^ 1 = 0.80 and p^2 = 0.69, respectively. The standard error of the sampling distribution of p^1 − p^2 is SE p^1−p^2=

p^1(1 − p^1) p^ 2(1 − p^ 2) 0.80(0.20) 0.69(0.31) = 0.0172 = + + n n Å 799 2253 Å 1 2

If we were to take many random samples of 799 teens and 2253 adults, the difference in the sample proportions of teens and adults who use social-networking sites will typically be 0.0172 from the true difference in proportions of all teens and adults who use social-networking sites. (b) State:  Our parameters of interest are p1 = the proportion of all U.S. teens who use socialnetworking sites and p2 = the proportion of all U.S. adults who use social-networking sites. We want to estimate the difference p1 − p2 at a 95% confidence level.

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Plan:  We should use a two-sample z interval for p1 − p2 if the conditions are met. •  Random:  The data come from independent random samples of 799 U.S. teens and 2253 U.S. adults. ºº 10%:  The researchers are sampling without replacement, so we must check the 10% condition: there are at least 10(799) = 7990 U.S. teens and at least 10(2253) = 22,530 U.S. adults. •  Large Counts:  We check the counts of “successes” and “failures”: n1p^ 1 = 799(0.80) = 639.2 S 639 n1(1 − p^ 1) = 799(1 − 0.80) = 159.8 S 160 n2p^2 = 2253(0.69) = 1554.57 S 1555  n2(1 − p^2) = 2253(1 − 0.69) = 698.43 S 698 Note that the observed counts have to be whole numbers! Because all four values are at least 10, this condition is met. Do:  We know that n1 = 799, p^1 = 0.80, n2 = 2253, and p^2 = 0.69. For a 95% confidence level, the critical value is z* = 1.96. So our 95% confidence interval for p1 − p2 is (p^1 − p^2) ± z* Å

This interval suggests that more teens than adults in the United States engage in social networking by between about 7.6 and 14.3 percentage points.

p^1(1 − p^ 1) p^2(1 − p^2) 0.80(0.20) 0.69(0.31) + = (0.80 − 0.69) ± 1.96 + n1 n2 Å 799 2253 = 0.11 ± 1.96(0.0172) = 0.11 ± 0.034 = (0.076, 0.144)

Using technology:  Refer to the Technology Corner that follows the example. The calculator’s 2-PropZInt gives (0.07588, 0.14324). Conclude:  We are 95% confident that the interval from 0.07588 to 0.14324 captures the true difference in the proportion of all U.S. teens and adults who use social-networking sites. For Practice Try Exercise

9

The researchers in the previous example selected independent random samples from the two populations they wanted to compare. In practice, it’s common to take one random sample that includes individuals from both populations of interest and then to separate the chosen individuals into two groups. The twosample z procedures for comparing proportions are still valid in such situations, provided that the two groups can be viewed as independent samples from their respective populations of interest. You can use technology to perform the calculations in the “Do” step. Remember that this comes with potential benefits and risks on the AP® exam.

21. Technology Corner

Confidence interval for a difference in proportions TI-Nspire instructions in Appendix B; HP Prime instructions on the book’s Web site.

The TI-83/84 and TI-89 can be used to construct a confidence interval for p1 − p2. We’ll demonstrate using the previous example. Of n1 = 799 teens surveyed, X = 639 said they used social-networking sites. Of n2 = 2253 adults surveyed, X = 1555 said they engaged in social networking. To construct a confidence interval:

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•

TI-83/84

619

TI-89

Press STAT , then choose TESTS and 2-PropZInt.

•

In the Statistics/List Editor, press 2nd F2 ([F7]) and choose 2-PropZInt.

• When the 2-PropZInt screen appears, enter the values shown.





• Highlight “Calculate” and press ENTER .



AP® EXAM TIP ​The formula for the two-sample z interval for p1−p2 often leads to calculation errors by students. As a result, we recommend using the calculator’s 2-PropZInt feature to compute the confidence interval on the AP® exam. Be sure to name the procedure (twoproportion z interval) and to give the interval (0.076, 0.143) as part of the “Do” step.

Check Your Understanding

Are teens or adults more likely to go online daily? The Pew Internet and American Life Project asked a random sample of 799 teens and a separate random sample of 2253 adults how often they use the Internet. In these two surveys, 63% of teens and 68% of adults said that they go online every day. Construct and interpret a 90% confidence interval for p1 − p2.

Significance Tests for p1 − p2 An observed difference between two sample proportions can reflect an actual difference in the parameters, or it may just be due to chance variation in random sampling or random assignment. Significance tests help us decide which explanation makes more sense. The null hypothesis has the general form H0: p1 − p2 = hypothesized value

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C H A P T E R 1 0 C o m pa r i n g T w o P o p u l at i o n s o r G r o u p s We’ll restrict ourselves to situations in which the hypothesized difference is 0. Then the null hypothesis says that there is no difference between the two parameters: H0: p1 − p2 = 0 or, alternatively, H0: p1 = p2 The alternative hypothesis says what kind of difference we expect.

Example

Hungry Children Stating hypotheses Researchers designed a survey to compare the proportions of children who come to school without eating breakfast in two low-income elementary schools. An SRS of 80 students from School 1 found that 19 had not eaten breakfast. At School 2, an SRS of 150 students included 26 who had not had breakfast. More than 1500 students attend each school. Do these data give convincing evidence of a difference in the population proportions?

Problem:  State appropriate hypotheses for a significance test to answer this question. Define any parameters you use. Solution:  We should carry out a test of H0: p1 − p2 = 0 Ha: p1 − p2 ≠ 0 where p1 = the true proportion of students at School 1 who did not eat breakfast and p2 = the true proportion of students at School 2 who did not eat breakfast. For Practice Try Exercise

13

The conditions for performing a significance test about p1 − p2 are the same as for constructing a confidence interval.

Conditions for Performing a Significance Test about a Difference in Proportions • Random: The data come from two independent random samples or from two groups in a randomized experiment. 1 ~~ 10%: When sampling without replacement, check that n1 ≤ N1 10 1 and n2 ≤ N2. 10 • Large Counts: The counts of “successes” and “failures” in each sample or group—n1p^1, n1(1 − p^ 1), n2p^ 2, n2(1 − p^ 2)—are all at least 10.

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If the conditions are met, we can proceed with calculations. To do a test, standardize p^ 1 − p^ 2 to get a z statistic: test statistic = z=

statistic − parameter standard deviation of statistic (p^1 − p^ 2) − 0 standard deviation of statistic

If H0: p1 = p2 is true, the two parameters are the same. We call their common value p. But now we need a way to estimate p, so it makes sense to combine the data from the two samples as if they came from one larger sample. This pooled (or combined) sample proportion is p^ C =

Breakfast?

1

No

19

Yes

61

Total

80

count of successes in both samples combined X1 + X2 = count of individuals in both samples combined n1 + n2

In other words, p^C gives the overall proportion of successes in the combined samples. Let’s look at how to calculate p^C in the hungry children example. The two-way table below summarizes the survey data. We have combined the independent SRSs from the two schools in the right-hand Total column. Because researchers want to compare the proportions of students School at School 1 and School 2 who have not eaten breakfast, we treat 2 Total the individuals in the “No” row as successes. It is easy to see from the table that the overall proportion of successes in the combined  26  45 45 124 185 = 0.1957. We can also get this result using samples is p^C = 230 150 230 the formula above: p^C =

X1 + X2 19 + 26 45 = = = 0.1957 n1 + n2 80 + 150 230

Recall that the standard deviation of p^ 1 − p^ 2 is sp^ 1 −p^ 2 =

p1(1 − p1) p2(1 − p2) + n1 n2 Å

Use p^C in place of both p1 and p2 in this expression for the denominator of the test statistic: We can use a little algebra to rewrite the denominator of the test statistic: p^C (1 − p^ C) p^ C (1 − p^C) + = Å n1 n2 Å

p^C (1 − p^ C)a

1 1 + b= n1 n2

1 1 + Å n1 n2 The final formula looks like the one given on the AP® exam formula sheet. !p^ C (1 − p^ C)

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z=

(p^ 1 − p^ 2) − 0 p^ C(1 − p^ C) p^ C(1 − p^ C) + n1 n2 Å

When the Large Counts condition is met, this will yield a z statistic that has approximately the standard Normal distribution when H 0 is true. Here are the details for the two-sample z test for the difference between two proportions.

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Some people prefer to use p^C to check the Large Counts condition. If the expected counts n1p^C , n1(1 − p^C ), n2p^C , and n2(1 − p^C ) are all at least 10, the sampling distribution of p^1 − p^2 is approximately Normal. Checking the observed counts of successes and failures is more conservative, as the expected counts will always be at least 10 if the observed counts are at least 10.

Two-Sample z Test for the Difference between Two Proportions Suppose the conditions are met. To test the hypothesis H0: p1 − p2 = 0, first find the pooled proportion p^C of ­successes in both samples combined. Then compute the z statistic z=

(p^ 1 − p^ 2) − 0 p^ C(1 − p^ C) p^ C(1 − p^ C) + n1 n2 Å

Find the P-value by calculating the probability of getting a z statistic this large or larger in the direction specified by the alternative hypothesis Ha: Ha : p 1 – p 2 > 0

z

Ha : p 1 – p 2 < 0

z

Ha : p 1 – p 2 = 0

–z

z

Now we can finish the test we started earlier.

Example

Hungry Children Significance test for p1 − p2

STEP

4

Researchers designed a survey to compare the proportions of children who come to school without eating breakfast in two low-income elementary schools. An SRS of 80 students from School 1 found that 19 had not eaten breakfast. At School 2, an SRS of 150 students included 26 who had not had breakfast. More than 1500 students attend each school. Do these data give convincing evidence at the a = 0.05 level of a difference in the population proportions?

State:  Our hypotheses are H0: p1 − p2 = 0 Ha: p1   − p2 ≠ 0 where p1 = the true proportion of students at School 1 who did not eat breakfast and p2 = the true proportion of students at School 2 who did not eat breakfast. Plan:  If conditions are met, we should perform a two-sample z test for p1 − p2. •  Random:  The data were produced using two independent random samples—80 students from School 1 and 150 students from School 2. ºº 10%:  The researchers are sampling without replacement, so we check the 10% condition: there are at least 10(80) = 800 students at School 1 and at least 10(150) = 1500 students at School 2. •  Large Counts:  We check the counts of “successes” and “failures”: n1p^ 1 = 19, n1(1 − p^ 1) = 61, n2p^2 = 26, n2(1 − p^ 2) = 124 All four values are at least 10, so this condition is met.

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623

19 26 = 0.2375, n2 = 150, and p^ 2 = = 0.1733. Our point 80 150 estimate for the difference in population proportions is p^1 − p^ 2 = 0.2375 − 0.1733 = 0.0642. The pooled proportion of students who didn’t eat breakfast in the two samples is 19 + 26 45 p^C = = = 0.1957 80 + 150 230

Do:  We know that n1 = 80, p^ 1 = School Breakfast?

1

2

Total

No

19

 26

45

Yes

61

124

185

Total

80

150

230

z=

See the two-way table in the margin for confirmation. •  Test statistic (p^ 1 − p^ 2) − 0 p^C (1 − p^C) p^C (1 − p^C) + n1 n2 Å

=

0.0642 − 0 0.1957(1 − 0.1957) 0.1957(1 − 0.1957) + Å 80 150

= 1.17

• P-value  Figure 10.3 displays the P-value as an area under the standard Normal curve for this two-tailed test. Using Table A or normalcdf, the desired P-value is 2P(Z ≥ 1.17) = 2(1 − 0.8790) = 0.2420. Using technology:  Refer to the Technology Corner that follows the example. The calculator’s 2-PropZTest gives z = 1.1683 and P-value = 0.2427.

N(0,1)

Area = 0. 2420

0

FIGURE 10.3  ​The P-value for the two-sided test.

z = -1.1 7

z = 1.1 7

Conclude:  Because our P-value, 0.2427, is greater than a = 0.05, we fail to reject H0. There is not convincing evidence that the true proportions of students at the two schools who didn’t eat breakfast are different. For Practice Try Exercise

The two-sample z test and two-sample z interval for the difference between two proportions don’t always give consistent results. That’s because the “standard deviation of the statistic” used in calculating the test statistic is p^ C (1 − p^ C ) p^ C (1 − p^ C) + Å n1 n2

but for the confidence interval, it’s p^ 1 (1 − p^ 1 ) p^ 2 (1 − p^ 2) + Å n1 n2

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15

Exactly what does the P-value in the previous example tell us? If we repeated the random sampling process many times, we’d get a difference in sample proportions as large as or larger than 0.0642 in either direction about 24% of the time when H0: p1 − p2 = 0 is true. With such a high probability of getting a result like this just by chance when the null hypothesis is true, we don’t have enough evidence to reject H0. We can get additional information about the difference between the population proportions at School 1 and School 2 with a confidence interval. The TI-84’s 2-PropZInt gives the 95% confidence interval for p1 − p2 as (−0.047, 0.175). That is, we are 95% confident that the difference in the true proportions of students who ate breakfast at the two schools is between 4.7 percentage points lower at School 1 and 17.5 percentage points higher at School 1. This is consistent with our “fail to reject H0” conclusion in the example because 0 is included in the interval of plausible values for p1 − p2.

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22. TECHNOLOGY CORNER

Significance test for a difference in proportions TI-Nspire instructions in Appendix B; HP Prime instructions on the book’s Web site.

The TI-83/84 and TI-89 can be used to perform significance tests for comparing two proportions. Here, we use the data from the hungry children example. To perform a test of H0: p1 − p2 = 0 versus Ha: p1 − p2 ≠ 0: TI-83/84 TI-89 •

Press STAT , then choose TESTS and 2-PropZTest.

•

I n the Statistics/List Editor, press 2nd F1 ([F6]) and choose 2-PropZTest.

• When the 2-PropZTest screen appears, enter the values x1 = 19, n1 = 80, x2 = 26, n2 = 150. Specify the alternative hypothesis p1 ≠ p2, as shown.





• If you select “Calculate” and press ENTER , you will see that the test statistic is z = 1.168 and the P-value is 0.2427. Do you see the combined proportion of students who didn’t eat breakfast? It’s the value labeled p^ , 0.1957.





• If you select the “Draw” option, you will see the screen shown here.



AP® EXAM TIP ​The formula for the two-sample z statistic for a test about p1 − p2 often leads to calculation errors by students. As a result, we recommend using the calculator’s 2-PropZTest feature to perform calculations on the AP® exam. Be sure to name the procedure (two-proportion z test) and to report the test statistic (z = 1.17) and P-value (0.2427) as part of the “Do” step.

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Section 10.1 Comparing Two Proportions

Inference for Experiments Most of the examples in this section have involved doing inference about p1 − p2 using data that were produced by random sampling. In such cases, the parameters p1 and p2 are the true proportions of successes in the corresponding populations. However, many important statistical results come from randomized ­comparative experiments. Defining the parameters in experimental settings is more ­challenging. The “Is Yawning Contagious?” Activity on page 610 describes an experiment that used 50 volunteer adults as subjects. Researchers randomly assigned 34 subjects to get a yawn seed and 16 subjects to get no yawn seed. Then researchers compared the proportions of people in the two groups who yawned. The parameters in this setting are: p1 = the true proportion of people like these who would yawn when given a yawn seed p2 = the true proportion of people like these who would yawn when no yawn seed is given

Most experiments on people use recruited volunteers as subjects. When subjects are not randomly selected, researchers cannot generalize the results of an experiment to some larger populations of interest. But researchers can draw causeand-effect conclusions that apply to people like those who took part in the experiment. This same logic applies to experiments on animals or things. Also note that unless the experimental units are randomly selected, we don’t need to check the 10% condition when performing inference about an experiment. Here is an example that involves comparing two proportions.

Example

Cholesterol and Heart Attacks Significance test in an experiment

STEP

4

High levels of cholesterol in the blood are associated with higher risk of heart attacks. Will using a drug to lower blood cholesterol reduce heart attacks? The Helsinki Heart Study recruited middle-aged men with high cholesterol but no history of other serious medical problems to investigate this question. The volunteer subjects were assigned at random to one of two treatments: 2051 men took the drug gemfibrozil to reduce their cholesterol levels, and a control group of 2030 men took a placebo. During the next five years, 56 men in the gemfibrozil group and 84 men in the placebo group had heart attacks. Is this difference statistically significant at the a = 0.01 level?

State:  We hope to show that gemfibrozil reduces heart attacks, so we have a one-sided alternative: H0: p1 − p2 = 0       H0: p1 = p2 or, equivalently, Ha: p1 − p2  < 0           Ha: p1  < p2 where p1 is the actual heart attack rate for middle-aged men like the ones in this study who take gemfibrozil, and p2 is the actual heart attack rate for middle-aged men like the ones in this study who take only a placebo. We’ll use a = 0.01.

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Note that we did not need to check the 10% condition because the subjects in the experiment were not sampled without replacement from some larger population.

Plan:  If conditions are met, we will do a two-sample z  test for p1 − p2. • Random:  The data come from two groups in a randomized experiment. ºº 10%:  Don’t need to check because there was no sampling. •  Large Counts:  The number of successes (heart attacks!) and failures in the two groups are 56, 1995, 84, and 1946. These are all at least 10, so this condition is met. Do:  The proportions of men who had heart attacks in each group are p^ 1 =

Drug taken Gemfibrozil

Placebo

Total

Yes

  56

  84

 140

No

1995

1946

3941

Total

2051

2030

4081

Heart attack?

56 84 = 0.0273 (gemfibrozil group) and p^ 2 = = 0.0414 (placebo group) 2051 2030

The pooled proportion of heart attacks for the two groups is p^C =

count of heart attacks in both samples combined count of subjects in both samples combined

=

56 + 84 140 = = 0.0343 2051 + 2030 4081

See the two-way table in the margin. We’ll use the calculator’s 2-PropZTest to perform calculations. •  Test statistic z = −2.47 • P-value  This is the area under the standard Normal curve to the left of z = −2.47, shown in Figure 10.4.

N(0,1)

Area = 0.0068

Conclude:  Because the P-value, 0.0068, is less than

0.01, we can reject H0. The r­ esults are statistically significant at the a = 0.01 level. There is convincing evidence of a lower heart attack rate for middle-aged men like these who take gemfibrozil than for those who take only a placebo.

0 z = -2.47

FIGURE 10.4  ​The P-value for the one-sided test.

For Practice Try Exercise

21

We chose a = 0.01 in the example to reduce the chance of making a Type I ­error—finding convincing evidence that gemfibrozil reduces heart attack risk when it actually doesn’t. This error could have serious consequences if an ineffective drug was given to lots of middle-aged men with high cholesterol! The random assignment in the Helsinki Heart Study allowed researchers to draw a cause-and-effect conclusion. They could say that gemfibrozil reduces the rate of heart attacks for middle-aged men like those who took part in the experiment. Because the subjects were not randomly selected from a larger population, researchers could not generalize the findings of this study any further. No conclusions could be drawn about the effectiveness of gemfibrozil at preventing heart attacks for all middle-aged men, for older men, or for women.

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Why do the inference methods for random sampling work for randomized experiments?  ​Confidence intervals and tests for p1 − p2

THINK ABOUT IT

are based on the ­sampling distribution of p^ 1 − p^ 2. But in experiments, we aren’t sampling at ­random from any larger populations. We can think about what would happen if the random assignment were repeated many times under the assumption that H0: p1 − p2 = 0 is true. That is, we assume that the specific treatment received doesn’t affect an individual subject’s response. Let’s see what would happen just by chance if we randomly reassign the 4081 subjects in the Helsinki Heart Study to the two groups many times, assuming the drug received doesn’t affect whether or not each individual has a heart attack. We used Fathom software to redo the random assignment 500 times. The approximate randomization distribution of p^ 1 − p^ 2 is shown in Figure 10.5. It has an approximately Normal shape with mean 0 and standard deviation 0.0058. These are roughly the same as the shape, center, and spread of the sampling distribution of p^ 1 − p^ 2 that we used to perform calculations in the previous example because p^ C(1 − p^ C) p^ C(1 − p^ C) 0.0343(1 − 0.0343) 0.0343(1 − 0.0343) + = + = 0.0057 n1 n2 Å Å 2051 2030

In 500 random reassignments, there were only 5 times when the difference in sample proportions was as small as or smaller than the observed 20.0141.

Shape: Approximately Normal Center: Mean 5 0 Spread: SD 5 0.0058

FIGURE 10.5  ​Fathom simulation showing the approximate randomization distribution of p^1 − p^2 from 500 random reassignments of subjects to treatment groups in the Helsinki Heart Study.

In the Helsinki Heart Study, the difference in the proportions of subjects who had a heart attack in the gemfibrozil and placebo groups was 0.0273 − 0.0414 = −0.0141. How likely is it that a difference this large or larger would happen just by chance when H0 is true? Figure 10.5 provides a rough answer: 5 of the 500 random reassignments yielded a difference in proportions less than or equal to −0.0141. That is, our estimate of the P-value is 0.01. This is quite close to the 0.0068 P-value that we calculated in the previous example. Figure 10.6 shows the value of the z test statistic for each of the 500 re-­randomizations, calculated using our familiar formula z=

Standard Normal curve

–3

–2

(p^ 1 − p^ 2) − 0 p^ C(1 − p^ C) p^ C(1 − p^ C) + n1 n2 Å

The standard Normal density curve is shown in blue. We can see that the z test statistic has approximately the standard Normal distribution in this case. Whenever the conditions are met, the randomization distribution of p^ 1 − p^ 2 looks much like its sampling distribution. We are therefore safe using two-sample z procedures for comparing two proportions in a randomized experiment.

–1

0

z

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1

2

3

FIGURE 10.6  ​The distribution of the z test statistic for the 500 random reassignments in Figure 10.5.

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Check Your Understanding

To study the long-term effects of preschool programs for poor children, researchers designed an experiment. They recruited 123 children who had never attended preschool from low-income families in Michigan. Researchers randomly assigned 62 of the children to attend preschool (paid for by the study budget) and the other 61 to serve as a control group who would not go to preschool. One response variable of interest was the need for social services as adults. Over a 10-year period, 38 children in the preschool group and 49 in the control group have needed social services.4 Does this study provide convincing evidence that preschool reduces the later need for social services? Justify your answer.

Section 10.1

Summary •

•

•

•

•

Choose independent SRSs of size n1 from Population 1 with proportion of successes p1 and of size n2 from Population 2 with proportion of successes p2. The sampling distribution of p^ 1 − p^ 2 has the following properties: • Shape  Approximately Normal if the samples are large enough that n1p1, n1(1 − p1), n2p2, and n2 (1 − p2) are all at least 10. • Center  The mean is p1 − p2. • Spread  As long as each sample is no more than 10% of its population, p1(1 − p1) p2(1 − p2) the standard deviation is + . n1 n2 Å Confidence intervals and tests to compare the proportions p1 and p2 of successes for two populations or treatments are based on the difference p^ 1 − p^ 2 between the sample proportions. Before estimating or testing a claim about p1 − p2, check that these conditions are met: • Random: The data come from two independent random samples or from two groups in a randomized experiment. ~~ 10%: When sampling without replacement, check that the two populations are at least 10 times as large as the corresponding samples. • Large Counts:  The counts of “successes” and “failures” in each sample or group—n1p^ 1, n1(1 − p^ 1), n2p^ 2, and n2 (1 − p^ 2)—are all at least 10. When conditions are met, an approximate C% confidence interval for p1 − p2 is (p^1 − p^ 2) ± z* Å

where z* is the standard Normal critical value with C% of its area between −z* and z*. This is called a two-sample z interval for p1 ∙ p2. Significance tests of H0: p1 − p2 = 0 use the pooled (combined) sample proportion in the standard error formula: p^ C =

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p^1(1 − p^1) p^ 2(1 − p^ 2) + n1 n2

count of successes in both samples combined X1 + X2 = count of individuals in both samples combined n1 + n2

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Section 10.1 Comparing Two Proportions

When conditions are met, the two-sample z test for p1 ∙ p2 uses the test statistic z=

•

STEP

4 •

(p^ 1 − p^ 2) − 0 p^ C(1 − p^ C) p^ C(1 − p^ C) + n1 n2 Å

with P-values calculated from the standard Normal distribution. Inference about the difference p1 − p2 in the effectiveness of two treatments in a completely randomized experiment is based on the randomization distribution of p^ 1 − p^ 2. When conditions are met, our usual inference procedures based on the sampling distribution of p^ 1 − p^ 2 will be approximately correct. Be sure to follow the four-step process whenever you construct a confidence interval or perform a significance test for comparing two proportions.

10.1 T echnology Corners TI-Nspire Instructions in Appendix B; HP Prime instructions on the book’s Web site. 21.  Confidence interval for a difference in proportions 22. Significance test for a difference in proportions

Section 10.1 STEP

4

1. pg 615

Exercises

Remember: We are no longer reminding you to use the four-step process in exercises that require you to perform inference.

Goldfish  Refer to the example on page 615. Suppose that your teacher decides to take SRSs of 100 crackers from both bags instead.

3.

(a) What is the shape of the sampling distribution of p^1 − p^2? Why? (b) Find the mean of the sampling distribution. Show your work. (c) Find the standard deviation of the sampling distribution. Show your work. 2.

page 618 page 624

Homework  Refer to page 612. Suppose that both school counselors decide to take SRSs of 150 students instead.

(a) What is the shape of the sampling distribution of p^1 − p^2? Why? (b) Find the mean of the sampling distribution. Show your work. (c) Find the standard deviation of the sampling distribution. Show your work.

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I want red!  A candy maker offers Child and Adult bags of jelly beans with different color mixes. The company claims that the Child mix has 30% red jelly beans, while the Adult mix contains 15% red jelly beans. Assume that the candy maker’s claim is true. Suppose we take a random sample of 50 jelly beans from the Child mix and a separate random sample of 100 jelly beans from the Adult mix. Let p^ C and p^ A be the sample proportions of red jelly beans from the Child and Adult mixes, respectively.

(a) What is the shape of the sampling distribution of p^ C − p^ A? Why? (b) Find the mean of the sampling distribution. Show your work. (c) Find the standard deviation of the sampling distribution. Show your work. 4.

Literacy  A researcher reports that 80% of high school graduates, but only 40% of high school dropouts, would pass a basic literacy test.5 Assume that the researcher’s claim is true. Suppose we give

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a basic literacy test to a random sample of 60 high school graduates and a separate random sample of 75 high school dropouts. Let p^ G and p^ D be the sample proportions of graduates and dropouts, respectively, who pass the test. (a) What is the shape of the sampling distribution of p^G − p^D? Why? (b) Find the mean of the sampling distribution. Show your work. (c) Find the standard deviation of the sampling distribution. Show your work. Explain why the conditions for constructing a two-sample z interval for p1 − p2 are not met in the settings of ­Exercises 5 through 8. 5.

6.

7.

8.

Don’t drink the water!  The movie A Civil Action (Touchstone Pictures, 1998) tells the story of a major legal battle that took place in the small town of Woburn, Massachusetts. A town well that supplied water to eastern Woburn residents was contaminated by industrial chemicals. During the period that residents drank water from this well, 16 of the 414 babies born had birth defects. On the west side of Woburn, 3 of the 228 babies born during the same time period had birth defects. In-line skaters  A study of injuries to in-line skaters used data from the National Electronic Injury Surveillance System, which collects data from a random sample of hospital emergency rooms. The researchers interviewed 161 people who came to emergency rooms with injuries from in-line skating. Wrist injuries (mostly fractures) were the most common.6 The interviews found that 53 people were wearing wrist guards and 6 of these had wrist injuries. Of the 108 who did not wear wrist guards, 45 had wrist injuries. Shrubs and fire  Fire is a serious threat to shrubs in dry climates. Some shrubs can resprout from their roots after their tops are destroyed. One study of resprouting took place in a dry area of Mexico.7 The investigators randomly assigned shrubs to treatment and control groups. They clipped the tops of all the shrubs. They then applied a propane torch to the stumps of the treatment group to simulate a fire. All 12 of the shrubs in the treatment group resprouted. Only 8 of the 12 shrubs in the control group resprouted. Broken crackers  We don’t like to find broken crackers when we open the package. How can makers reduce breaking? One idea is to microwave the crackers for 30 seconds right after baking them. Breaks start as hairline cracks called “checking.” ­Randomly

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assign 65 newly baked crackers to the microwave and another 65 to a control group that is not microwaved. After one day, none of the microwave group and 16 of the control group show checking.8 9. pg 617

Who tweets?  Do younger people use Twitter more often than older people? In a random sample of 316 adult Internet users aged 18 to 29, 26% used Twitter. In a separate random sample of 532 adult Internet users aged 30 to 49, 14% used Twitter.9

(a) Calculate the standard error of the sampling distribution of the difference in the sample proportions (younger adults − older adults). What information does this value provide? (b) Construct and interpret a 90% confidence interval for the difference between the true proportions of adult Internet users in these age groups who use Twitter. 10. Listening to rap  Is rap music more popular among young blacks than among young whites? A sample survey compared 634 randomly chosen blacks aged 15 to 25 with 567 randomly selected whites in the same age group. It found that 368 of the blacks and 130 of the whites listened to rap music every day.10 (a) Calculate the standard error of the sampling distribution of the difference in the sample proportions (blacks − whites). What information does this value provide? (b) Construct and interpret a 95% confidence interval for the difference between the proportions of black and white young people who listen to rap every day. 11. Young adults living at home  A surprising number of young adults (ages 19 to 25) still live in their parents’ homes. A random sample by the National Institutes of Health included 2253 men and 2629 women in this age group.11 The survey found that 986 of the men and 923 of the women lived with their parents. (a) Construct and interpret a 99% confidence interval for the difference in the true proportions of men and women aged 19 to 25 who live in their parents’ homes. (b) Does your interval from part (a) give convincing evidence of a difference ­between the population proportions? Explain. 12. Fear of crime  The elderly fear crime more than younger people, even though they are less likely to be victims of crime. One study recruited separate random samples of 56 black women and 63 black men over the age of 65 from Atlantic City, New Jersey. Of the women, 27 said they “felt vulnerable” to crime; 46 of the men said this.12

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Section 10.1 Comparing Two Proportions (a) Construct and interpret a 90% confidence interval for the difference in the true proportions of black women and black men in Atlantic City who would say they felt vulnerable to crime.

age groups were compared on their ability to sort new products into the correct product category (milk or juice).14 Here are some of the data:

(b) Does your interval from part (a) give convincing evidence of a difference between the population proportions? Explain. 13. Who owns iPods?  As part of the Pew Internet and American Life Project, researchers surveyed a random sample of 800 teens and a separate random sample of 400 young adults. For the teens, 79% said that they own an iPod or MP3 player. For the young adults, this figure was 67%. Do the data give convincing evidence of a difference in the proportions of all teens and young adults who would say that they own an iPod or MP3 player? State appropriate hypotheses for a test to answer this question. Define any parameters you use.

pg 622 15.

Who owns iPods?  Refer to Exercise 13. Carry out a significance test at the a = 0.05 level.

16. Steroids in high school  Refer to Exercise 14. Carry out a significance test at the a = 0.05 level. 17. Who owns iPods?  Refer to Exercise 13. Construct and interpret a 95% confidence interval for the difference between the population proportions. Explain how the confidence interval is consistent with the results of the test in Exercise 15.

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Number who sorted correctly

4- to 5-year-olds 6- to 7-year-olds

50 53

10 28

20. Marriage and status  “Would you marry a person from a lower social class than your own?” Researchers asked this question of a random sample of 385 black, never-married college students. Of the 149 men in the sample, 91 said “Yes.” Among the 236 women, 117 said “Yes.”15 Did a significantly higher proportion of the men than the women who were surveyed say “Yes”? Give appropriate evidence to justify your answer. 21. Driving school  A driving school owner believes that Instructor A is more effective than Instructor B at pg 625 preparing students to pass the state’s driver’s license exam. An incoming class of 100 students is randomly assigned to two groups, each of size 50. One group is taught by Instructor A; the other is taught by Instructor B. At the end of the course, 30 of Instructor A’s students and 22 of Instructor B’s students pass the state exam. (a) Do these results give convincing evidence at the a = 0.05 level that Instructor A is more effective? (b) Describe a Type I and a Type II error in this setting. Which error could you have made in part (a)? 22. Preventing strokes  Aspirin prevents blood from clotting and so helps prevent strokes. The Second European Stroke Prevention Study asked whether adding another anticlotting drug, named dipyridamole, would be more effective for patients who had already had a stroke. Here are the data on strokes during the two years of the study:16

18. Steroids in high school  Refer to Exercise 14. Construct and interpret a 95% confidence interval for the difference between the population proportions. Explain how the confidence interval is consistent with the results of the test in Exercise 16. 19. Children make choices  Many new products introduced into the market are targeted toward children. The choice behavior of children with regard to new products is of particular interest to companies that design marketing strategies for these products. As part of one study, randomly selected children in different

N

Did a significantly higher proportion of the 6- to 7-year-olds than the 4- to 5-year-olds sort correctly? Give appropriate evidence to justify your answer.

pg 620

14. Steroids in high school  A study by the National Athletic Trainers Association surveyed random samples of 1679 high school freshmen and 1366 high school seniors in Illinois. Results showed that 34 of the freshmen and 24 of the seniors had used anabolic steroids. Steroids, which are dangerous, are sometimes used in an attempt to improve athletic performance.13 Do the data give convincing evidence of a difference in the proportion of all Illinois high school freshmen and seniors who have used anabolic steroids? State appropriate hypotheses for a test to answer this question. Define any parameters you use.

Age group



Number of patients

Number of strokes

Aspirin alone

1649

206

Aspirin + dipyridamole

1650

157

The study was a randomized comparative experiment.

(a) Is there convincing evidence at the a = 0.05 level that adding dipyridamole helps reduce the risk of stroke? (b) Describe a Type I and a Type II error in this ­setting. Which is more serious? Explain.

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Exercises 23 and 24 involve the following setting. Some women would like to have children but cannot do so for medical reasons. One option for these women is a procedure called in vitro fertilization (IVF), which involves injecting a fertilized egg into the woman’s uterus. 23. Prayer and pregnancy  Two hundred women who were about to undergo IVF served as subjects in an experiment. Each subject was randomly assigned to either a treatment group or a control group. Women in the treatment group were intentionally prayed for by several people (called intercessors) who did not know them, a process known as intercessory prayer. The praying continued for three weeks following IVF. The intercessors did not pray for the women in the control group. Here are the results: 44 of the 88 women in the treatment group got pregnant, compared to 21 out of 81 in the control group.17 Is the pregnancy rate significantly higher for women who received intercessory prayer? To find out, researchers perform a test of H0: p1 = p2 versus Ha: p1 > p2, where p1 and p2 are the actual pregnancy rates for women like those in the study who do and don’t receive intercessory prayer, respectively. (a) Name the appropriate test and check that the conditions for carrying out this test are met.

Is the pregnancy rate significantly higher for women who received acupuncture? To find out, researchers perform a test of H0: p1 = p2 versus Ha: p1 > p2, where p1 and p2 are the actual pregnancy rates for women like those in the study who do and don’t receive ­acupuncture, respectively. (a) Name the appropriate test and check that the conditions for carrying out this test are met. (b) The appropriate test from part (a) yields a P-value of 0.0152. Interpret this P-value in context. (c) What conclusion should researchers draw at the a = 0.05 significance level? Explain. (d) The women in the study knew whether or not they received acupuncture. Explain why this is important. Multiple choice: Select the best answer for Exercises 25 to 28. Exercises 25 to 27 refer to the following setting. A sample survey interviews SRSs of 500 female college students and 550 male college students. Researchers want to determine whether there is a difference in the proportion of male and female college students who worked for pay last summer. In all, 410 of the females and 484 of the males say they worked for pay last summer.

(b) The appropriate test from part (a) yields a P-value of 0.0007. Interpret this P-value in context.

25. Take pM and pF to be the proportions of all college males and females who worked last summer. The hypotheses to be tested are

(c) What conclusion should researchers draw at the a = 0.05 significance level? Explain.

(a) H0: pM − pF = 0 versus Ha: pM − pF ≠ 0.

(d) The women in the study did not know whether they were being prayed for. Explain why this is important.

(c) H0: pM − pF = 0 versus Ha: pM − pF < 0.

24. Acupuncture and pregnancy  A study reported in the medical journal Fertility and Sterility sought to determine whether the ancient Chinese art of acupuncture could help infertile women become pregnant.18 One hundred sixty healthy women who planned to have IVF were recruited for the study. Half of the subjects (80) were randomly assigned to receive acupuncture 25 minutes before embryo transfer and again 25 minutes after the transfer. The remaining 80 women were assigned to a control group and instructed to lie still for 25 minutes after the embryo transfer. Results are shown in the table below.

Pregnant Not pregnant Total

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Acupuncture group

Control group

34 46 80

21 59 80

(b) H0: pM − pF = 0 versus Ha: pM − pF > 0. (d) H0: pM − pF > 0 versus Ha: pM − pF = 0. (e) H0: pM − pF ≠ 0 versus Ha: pM − pF = 0. 26. The researchers report that the results were statistically significant at the 1% level. Which of the following is the most appropriate conclusion? (a) Because the P-value is less than 1%, fail to reject H0. There is not convincing evidence that the proportion of male college students in the study who worked for pay last summer is different from the proportion of female college students in the study who worked for pay last summer. (b) Because the P-value is less than 1%, fail to reject H0. There is not convincing evidence that the proportion of all male college students who worked for pay last summer is different from the proportion of all female college students who worked for pay last summer.

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(d) Because the P-value is less than 1%, reject H0. There is convincing evidence that the proportion of all male college students in the study who worked for pay last summer is different from the proportion of all female college students in the study who worked for pay last summer.

(e) should not be used because the Large Counts condition is violated. Exercises 29 and 30 refer to the following setting. Thirty randomly selected seniors at Council High School were asked to report the age (in years) and mileage of their main vehicles. Here is a scatterplot of the data:

200,000

Mileage

(c) Because the P-value is less than 1%, reject H0. There is convincing evidence that the proportion of all male college students who worked for pay last summer is the same as the proportion of all female college students who worked for pay last summer.

(e) Because the P-value is less than 1%, reject H0. There is convincing evidence that the proportion of all male college students who worked for pay last summer is different from the proportion of all female college students who worked for pay last summer. 27. Which of the following is the correct margin of error for a 99% confidence interval for the difference in the proportion of male and female college students who worked for pay last summer? (a) 2.576 Å

(b) 2.576 Å (c) 2.576 Å (d) 1.960 Å

0.851(0.149) 0.851(0.149) + 550 500 0.851(0.149) 1050

Represents 2 points

100,000

0 0.0

2.5

5.0

7.5

10.0

Age (years)



We used Minitab to perform a least-squares regression analysis for these data. Part of the computer output from this regression is shown below.

Predictor

Coef

Stdev

t-ratio

Constant

−13832

8773

−1.58

0.126

14954

1546

9.67

0.000

Age s = 22723

R-sq = 77.0%

P

R-sq(adj) = 76.1%

0.880(0.120) 0.820(0.180) + 550 500

29. Drive my car (3.2)

0.851(0.149) 0.851(0.149) + 550 500

(a) What is the equation of the least-squares ­regression line? Be sure to define any symbols you use.

0.880(0.120) 0.820(0.180) (e) 1.960 + Å 550 500 28. In an experiment to learn whether Substance M can help restore memory, the brains of 20 rats were treated to damage their memories. First, the rats were trained to run a maze. After a day, 10 rats (determined at random) were given M and 7 of them succeeded in the maze. Only 2 of the 10 control rats were successful. The two-sample z test for “no difference” against “a significantly higher proportion of the M group succeeds”

(b) Interpret the slope of the least-squares line in the context of this problem. (c) One student reported that her 10-year-old car had 110,000 miles on it. Find and interpret the residual for this data value. Show your work. 30. Drive my car (3.2, 4.3) (a) Explain what the value of r2 tells you about how well the least-squares line fits the data.

(a) gives z = 2.25, P < 0.02.

(b) The mean age of the students’ cars in the sample was x– = 8 years. Find the mean mileage of the cars in the sample. Show your work.

(b) gives z = 2.60, P < 0.005.

(c) Interpret the value of s in the context of this ­setting.

(c) gives z = 2.25, P < 0.04 but not < 0.02. (d) should not be used because the Random condition is violated.

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(d) Would it be reasonable to use the least-squares line to predict a car’s mileage from its age for a Council High School teacher? Justify your answer.

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10.2 What You Will Learn

Comparing Two Means By the end of the section, you should be able to:

• Describe the shape, center, and spread of the sampling distribution of x–1 − x–2. • Determine whether the conditions are met for doing inference about m1 − m2. • Construct and interpret a confidence interval to ­compare two means.

• Perform a significance test to compare two means. • Determine when it is appropriate to use two-sample t procedures versus paired t procedures.

In the previous section, we developed methods for comparing two proportions. What if we want to compare the mean of some quantitative variable for the individuals in Population 1 and Population 2? Our parameters of interest are the population means m1 and m2. Once again, the best approach is to take separate random samples from each population and to compare the sample means x–1 and x–2. Suppose we want to compare the average effectiveness of two treatments in a completely randomized experiment. In this case, the parameters m1 and m2 are the true mean responses for Treatment 1 and Treatment 2, respectively. We use the mean response in the two groups, x–1 and x–2, to make the comparison. Here’s a table that summarizes these two situations: Population or treatment

Parameter

1

m1

Statistic x–1

m2

x–2

2

Sample size n1 n2

We compare the populations or treatments by doing inference about the difference m1 − m2 between the parameters. The statistic that estimates this difference is the difference between the two sample means, x–1 − x–2. To use x–1 − x–2 for inference, we must know its sampling distribution. Here is an Activity that gives you a preview of what lies ahead.

Activity Materials: 10 small pieces of card stock (or index cards) per pair of students

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​Does Polyester Decay? How quickly do synthetic fabrics such as polyester decay in landfills? A researcher buried polyester strips in the soil for different lengths of time, then dug up the strips and measured the force required to break them. Breaking strength is easy to measure and is a good indicator of decay. Lower strength means the fabric has decayed. The researcher buried 10 strips of polyester fabric in well-drained soil in the summer. The strips were randomly assigned to two groups: 5 of them were buried for 2 weeks and the other 5 were buried for 16 weeks. Here are the breaking strengths in pounds:19

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Section 10.2 Comparing Two Means

Group 1 (2 weeks): Group 2 (16 weeks):

118 124

126  98

126 110

120 140

129 110

Do the data give convincing evidence that polyester decays more in 16 weeks than in 2 weeks? 1. ​The Fathom dotplot displays the data from the experiment. Discuss what this graph shows with your classmates.

For Group 1, the mean breaking strength was x–1 = 123.8 pounds. For Group 2, the mean breaking strength was x–2 = 116.4 pounds. The observed difference in average breaking strength for the two groups is x–1 − x–2 = 123.8 − 116.4 = 7.4 pounds. Is it plausible that this difference is due to the chance involved in the random assignment and not to the treatments themselves? To find out, your class will perform a simulation. Suppose that the length of time in the ground has no effect on the breaking strength of the polyester specimens. Then each specimen would have the same breaking strength regardless of whether it was assigned to Group 1 or Group 2. In that case, we could examine the results of repeated random assignments of the specimens to the two groups. 2. ​Write each of the 10 breaking-strength measurements on a separate card. Mix the cards well and deal them face down into two piles of 5 cards each. Be sure to decide which pile is Group 1 and which is Group 2 before you look at the cards. Calculate the difference in the mean breaking strength (Group 1 − Group 2). Record this value. 3. ​Your teacher will draw and label axes for a class dotplot. Plot the result you got in Step 2 on the graph. 4. ​Repeat Steps 2 and 3 if needed to get a total of at least 40 repetitions of the simulation for your class. 5.  Based on the class’s simulation results, how surprising would it be to get a difference in means of 7.4 or larger simply due to the chance involved in the random assignment? 6.  What conclusion would you draw about whether polyester decays more when left in the ground for longer periods of time? Explain. In this simulation, 14 of the 100 trials (in red) produced a difference in means of at least 7.4 pounds, so the approximate P-value is 0.14. It is likely that a difference this big could have happened just due to the chance variation in random assignment. The observed difference is not statistically significant and does not provide convincing evidence that polyester decays more in 16 weeks than in 2 weeks.

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The Sampling Distribution of a Difference between Two Means To explore the sampling distribution of x–1 − x–2, let’s start with two Normally distributed populations having known means and standard deviations. Based on information from the U.S. National Health and Nutrition Examination Survey (NHANES), the heights of 10-year-old girls follow a Normal distribution with mean mF = 56.4 inches and standard deviation sF = 2.7 inches. The heights of 10-year-old boys follow a Normal distribution with mean mM = 55.7 inches and standard deviation sM = 3.8 inches.20 Suppose we take independent SRSs of 12 girls and 8 boys of this age and measure their heights. What can we say about the difference x–F − x–M in the average heights of the sample of girls and the sample of boys? We used Fathom software to take an SRS of 12 ten-year-old girls and 8 ten-yearold boys and to plot the values of x–F, x–M, and x–F − x–M for each sample. Our first set of simulated samples gave x–F = 56.09 inches and x–M = 54.68 inches, so dots were placed above each of those values in Figure 10.7(a) and (b). The difference in the sample means is x–F − x–M = 56.09 − 54.68 = 1.41 inches. A dot for this value appears in Figure 10.7(c). The three dotplots in Figure 10.7 show the results of repeating this process 1000 times. These are the approximate sampling distributions of x–F, x–M, and x–F − x–M. (a) Approximate sampling distribution of x– F

Shape:  Approximately Normal Center: Mean = 56.40 inches Spread: SD = 0.80 inches

(b) Approximate sampling distribution of x– M

Shape:  Approximately Normal Center: Mean = 55.73 inches Spread: SD = 1.35 inches

(c) Approximate sampling distribution of x– F − x– M

Shape:  Approximately Normal Center: Mean = 0.67 inches Spread: SD = 1.56 inches

FIGURE 10.7  ​Simulated sampling distributions of (a) the sample mean height x–F in 1000 SRSs of size nF = 12 from the population of 10-year-old girls, (b) the sample mean height x–M in 1000 SRSs of size nM = 8 from the population of 10-year-old boys, and (c) the difference in sample means x–F − x–M for each of the 1000 repetitions.

In Chapter 7, we saw that the sampling distribution of a sample mean x– has the following properties: Shape:  (1) If the population distribution is Normal, then so is the sampling distribution of x– ; (2) if the population distribution isn’t Normal, the sampling distribution of x– will be approximately Normal if the sample size is large enough (say, n ≥ 30) by the central limit theorem (CLT).

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637

Center:  mx– = m s if the sample is no more than 10% of the population !n For the sampling distributions of x–F and x–M in this case:

Spread:  sx– =

Sampling distribution of x–F

Sampling distribution of x–M

Shape

Normal, because the population ­distribution is Normal

Normal, because the population distribution is Normal

Center

mx–F = mF = 56.4 inches

mx–M = mM = 55.7 inches

Spread

sx–F =

sF !nF

=

2.7 = 0.78 inches !12

because there are way more than 10(12) = 120 ten-year-old girls in the United States.

sx–M =

sM !nM

=

3.8 = 1.34 inches !8

because there are way more than 10(8) = 80 ten-year-old boys in the United States.

The approximate sampling distributions in Figures 10.7(a) and (b) give similar results. What about the sampling distribution of x–F − x–M? Figure 10.7(c) suggests that it has a roughly Normal shape, is centered at about 0.67 inches, and has standard deviation about 1.56 inches. The shape makes sense because we are combining two independent Normal random variables, x–F and x–M. How about the center? The actual mean height of 10-year-old girls is mF = 56.4 inches. For 10-year-old boys, the actual mean height is mM = 55.7 inches. We’d expect the difference x–F − x–M to center on the actual difference in the population means, mF − mM = 56.4 − 55.7 = 0.7 inches. The spread, however, is a bit more complicated.

THINK ABOUT IT

How can we find formulas for the mean and standard deviation of the sampling distribution of x–1 − x–2?  ​Both x–1 and x–2 are

random variables. That is, their values would vary in repeated independent SRSs of size n1 and n2. Independent random samples yield independent random variables x–1 and x–2. The statistic x–1 − x–2 is the difference of these two independent random variables. In Chapter 6, we learned that for any two random variables X and Y, mX−Y = mX − mY For the random variables x–1 and x–2, we have mx–1 −x–2 = mx–1 − mx–2 = m1 − m2 In the observational study of the heights of 10-year-olds, mx–F −x–M = mF − mM = 56.4 − 55.7 = 0.70 inches We also learned in Chapter 6 that for independent random variables X and Y, s2X−Y = s2X + s2Y For the random variables x–1 and x–2, we have s2x–1 −x–2 = s2x–1 + s2x– 2 = a

So sx–1 −x–2 =

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s21 s22 + . Å n1 n2

s1 2 s2 2 s21 s22 b +a b = + n1 n2 !n1 !n2

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C H A P T E R 1 0 C o m pa r i n g T w o P o p u l at i o n s o r G r o u p s In the observational study of the heights of 10-year-olds, sx– F −x–M =

s2F s2M 2.72 3.82 + = + = 1.55 Å nF nM Å 12 8

This is similar to the result from the Fathom simulation. Here are the facts we need.

The Sampling Distribution of x–1 − x–2 Choose an SRS of size n1 from Population 1 with mean m1 and standard ­deviation s1 and an independent SRS of size n2 from Population 2 with mean m2 and standard deviation s2. FIGURE 10.8  ​Select independent SRSs from two populations having means m1 and m2 and standard deviations s1 and s2. The two sample means are x–1 and x–2. If the population distributions are both Normal, the sampling distribution of the difference x–1 − x–2 is Normal. The sampling distribution will be approximately Normal in other cases if both samples are large enough (n1 ≥ 30 and n2 ≥ 30).

• Shape:  When the population distributions are Normal, the sampling distribution of x–1 − x–2 is Normal. In other cases, the sampling distribution of x–1 − x–2 will be approximately Normal if the sample sizes are large enough (n1 ≥ 30 and n2 ≥ 30). • Center:  The mean of the sampling distribution of x–1 − x–2 is m1 − m2. • Spread:  The standard deviation of the sampling distribution of x–1 − x–2 is s21 s22 + Å n1 n2

as long as each sample is no more than 10% of its population. Standard deviation 2

2

σ1 σ2 n1 + n2

Mean µ1 – µ 2

Values of x1 – x2

Example

When conditions are met, the sampling distribution of x–1 − x–2 will be approximately Normal with mean mx–1 −x– 2 = m1 − m2 and s21 s22 standard deviation sx–1 −x–2 = + . Figure 10.8 displays this Å n1 n2 distribution. The formula for the standard deviation of the sampling distribution involves the parameters s1 and s2, which are usually unknown. Just as in Chapters 8 and 9, we must replace these by estimates to do inference. We’ll get to confidence intervals and significance tests shortly. For now, let’s focus on the sampling distribution of x–1 − x–2.

Medium or Large Drink? Describing the sampling distribution of x–1 − x–2 A fast-food restaurant uses an automated filling machine to pour its soft drinks. The machine has different settings for small, medium, and large drink cups. According to the machine’s manufacturer, when the large setting is chosen, the amount of liquid L dispensed by the machine follows a Normal distribution with mean 27 ounces and

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639

standard deviation 0.8 ounces. When the medium setting is chosen, the amount of liquid M dispensed follows a Normal distribution with mean 17 ounces and standard deviation 0.5 ounces. To test the manufacturer’s claim, the restaurant manager measures the amount of liquid in each of 20 cups filled with the large setting and 25 cups filled with the medium setting. Let x–L − x–M be the difference in the sample mean amount of liquid under the two settings.

PROBLEM:

(a) What is the shape of the sampling distribution of x–L − x–M ? Why? (b) Find the mean of the sampling distribution. Show your work. (c) Find the standard deviation of the sampling distribution. Show your work.

SOLUTION:

(a) The sampling distribution of x–L − x–M is Normal because both population distributions are Normal. (b) The mean is mx–L −x–M = mL − mM = 27 − 17 = 10 ounces. (0.80)2 (0.50)2 s2L s2M + = + = 0.205 ounces. nM Å 20 Å nL 25 Note that we do not need to check the 10% condition because we are not sampling without replacement from a finite population. (c) The standard deviation is sx–L −x–M =

For Practice Try Exercise

31

The Two-Sample t Statistic When data come from two independent random samples or two groups in a randomized experiment (the Random condition), the statistic x–1 − x–2 is our best guess for the value of m1 − m2. If the 10% condition is met, the standard deviation of the sampling distribution of x–1 − x–2 is

sx– 1 −x–2 =

s21 s22 + Å n1 n2

If the Normal/Large Sample condition is met, we can standardize the observed difference x–1 − x–2 to obtain a z statistic that is modeled well by a standard Normal distribution:

z=

(x–1 − x–2) − (m1 − m2) s21 s22 + Å n1 n2

In the unlikely event that both population standard deviations are known, this two-sample z statistic is the basis for inference about m1 − m2.

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C H A P T E R 1 0 C o m pa r i n g T w o P o p u l at i o n s o r G r o u p s Suppose now that the population standard deviations s1 and s2 are not known. We estimate them by the standard deviations s1 and s2 from our two samples. The result is the standard error (also called the estimated standard deviation) of x–1 − x–2: SEx–1 −x–2 =

s22 s21 + Å n1 n2

Now when we standardize the point estimate x–1 − x–2, the result is the two-sample t statistic: t=

(x–1 − x–2) − (m1 − m2) s22 s21 + Å n1 n2

The statistic t has the same interpretation as any z or t statistic: it says how far – 1 − x 2 is from its mean in standard deviation units. When the Normal/Large Sample condition is met, the two-sample t statistic has approximately a t distribution. It does not have exactly a t distribution even if the populations are both exactly Normal. In practice, however, the approximation is very accurate. x–

Conditions for Performing Inference about m1 − m2 • Random: The data come from two independent random samples or from two groups in a randomized experiment. 1 ~~ 10%: When sampling without replacement, check that n1 ≤ N1 10 1 and n2 ≤ N2. 10 • Normal/Large Sample: Both population distributions (or the true distributions of responses to the two treatments) are Normal or both sample sizes are large (n1 ≥ 30 and n2 ≥ 30). If either population (­treatment) distribution has unknown shape and the corresponding sample size is less than 30, use a graph of the sample data to assess the Normality of the population (treatment) distribution. Do not use two-sample t procedures if the graph shows strong skewness or outliers. There are two practical options for using the two-sample t procedures when the conditions are met. The two options are exactly the same except for the degrees of freedom used for t critical values and P-values. Option 1 (Technology):  Use the t distribution with degrees of freedom calculated from the data by the formula below. Note that the df given by this formula is usually not a whole number.

You can thank statisticians B. L. Welch and F. E. Satterthwaite for discovering this fairly remarkable formula.

df =

a

s21 s22 2 + b n1 n2

s21 2 s22 2 1 1 a b + a b n1 − 1 n1 n2 − 1 n2

Option 2 (Conservative):  Use the t distribution with degrees of freedom equal to the smaller of n1 − 1 and n2 − 1. With this option, the resulting confidence interval has a margin of error as large as or larger than is needed for the desired confidence level. The significance test using this option gives a P-value equal to or

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greater than the true P-value. As the sample sizes increase, confidence levels and P-values from Option 2 become more accurate.21

Confidence Intervals for m1 − m2 If the Random, 10%, and Normal/Large Sample conditions are met, we can use our standard formula to construct a confidence interval for m1 − m2: statistic ± (critical value) ⋅ (standard deviation of statistic)

s22 s21 = (x–1 − x–2) ± t* + Å n1 n2

We can use either technology or the conservative approach with Table B to find the critical value t* for the given confidence level. This method is called a twosample t interval for a difference between two means.

Two-Sample t Interval for a Difference between Two Means When the conditions are met, an ­approximate C% confidence interval for m1 − m2 is s22 s21 (x–1 − x–2) ± t* + Å n1 n2

Here, t* is the critical value with C% of its area between −t* and t* for the t distribution with degrees of freedom using either Option 1 (technology) or Option 2 (the smaller of n1 − 1 and n2 − 1). The following example shows how to construct and interpret a confidence interval for a difference in means. As usual with inference problems, we follow the four-step process.

Example

Big Trees, Small Trees, Short Trees, Tall Trees

STEP

4

Confidence interval for m1 − m2 The Wade Tract Preserve in Georgia is an old-growth forest of longleaf pines that has survived in a relatively undisturbed state for hundreds of years. One question of interest to foresters who study the area is “How do the sizes of longleaf pine trees in the northern and southern halves of the forest compare?” To find out, researchers took random samples of 30 trees from each half and measured the diameter at breast height (DBH) in centimeters.22 Here are comparative boxplots of the data and summary statistics from Minitab. Descriptive Statistics: North, South

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Variable

N

Mean

StDev

North

30

23.70

17.50

South

30

34.53

14.26

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Problem: (a) Based on the graph and numerical summaries, write a few sentences comparing the sizes of longleaf pine trees in the two halves of the forest. (b) Construct and interpret a 90% confidence interval for the difference in the mean DBH of longleaf pines in the northern and southern halves of the Wade Tract Preserve.

Solution: (a) The distribution of DBH measurements in the northern sample is skewed to the right, while the distribution of DBH measurements in the southern sample is skewed to the left. It appears that trees in the southern half of the forest have larger diameters. The mean and median DBH for the southern sample are both much larger than the corresponding measures of center for the northern sample. Furthermore, the boxplots show that more than 75% of the southern trees have diameters that are above the northern sample’s median. There is more variability in the diameters of the northern longleaf pines, as we can see from the larger range, IQR, and standard deviation for this sample. No outliers are present in either sample. (b) State:  Our parameters of interest are m1 = the true mean DBH of all trees in the southern half of the forest and m2 = the true mean DBH of all trees in the northern half of the forest. We want to estimate the difference m1 − m2 at a 90% confidence level. PLAN:  If conditions are met, we’ll construct a two-sample t interval for m1 − m2. • Random:  The data came from independent random samples of 30 trees each from the northern and southern halves of the forest. ºº 10%:  Because sampling without replacement was used, there have to be at least 10(30) = 300 trees in each half of the forest. This is fairly safe to assume.

Upper-tail probability p df

.10

.05

.025

28

1.313

1.701

2.048

29

1.311

1.699

2.045

30

1.310

1.697

2.042

80%

90%

95%

Confidence level C

•  Normal/Large Sample:  The boxplots give us reason to believe that the population distributions of DBH measurements may not be Normal. However, because both sample sizes are at least 30, we are safe using two-sample t procedures. DO:  From the Minitab output, x–1 = 34.53, s1 = 14.26, n1 = 30, x–2 = 23.70, s2 = 17.50, and n2 = 30. We’ll use the conservative df = the smaller of n1 − 1 and n2 − 1, which is 29. For a 90% confidence level the critical value from Table B is t* = 1.699. So a 90% confidence interval for m1 − m2 is ( x–1 − x–1) ± t*

s21 s22 14.262 17.502 + = (34.53 − 23.70) ± 1.699 + Å n1 n2 Å 30 30 = 10.83 ± 7.00 = (3.83, 17.83)

Using technology:  Refer to the Technology Corner that follows the example. The ­calculator’s 2-SampTInt gives (3.9362, 17.724) using df = 55.728. Conclude:  We are 90% confident that the interval from 3.9362 to 17.724 centimeters captures the difference in the actual mean DBH of the southern trees and the actual mean DBH of the northern trees. For Practice Try Exercise

37

The 90% confidence interval in the example does not include 0. This gives convincing evidence that the difference in the mean diameter of northern and southern trees in the Wade Tract Preserve isn’t 0. However, the confidence interval provides more information than a simple reject or fail to reject H0 conclusion. It gives a set of plausible values for m1 − m2. The interval suggests that the mean diameter of the southern trees is between 3.83 and 17.83 cm larger than the mean diameter of the northern trees.

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We chose the parameters in the DBH example so that x–1 − x–2 would be positive. What if we had defined m1 as the mean DBH of the northern trees and m2 as the mean DBH of the southern trees? The 90% confidence interval for m1 − m2 would be (23.70 − 34.53) ± 1.699

17.502 14.262 + = −10.83 ± 7.00 = (−17.83, −3.83) Å 30 30

This interval suggests that the mean diameter of the northern trees is between 3.83 and 17.83 cm smaller than the mean diameter of the southern trees. Changing the order of subtraction doesn’t change the result. As with other inference procedures, you can use technology to perform the calculations in the “Do” step. Remember that technology comes with potential benefits and risks on the AP® exam.

23. Technology Corner

​Two-sample t intervals on the ­calculator TI-Nspire instructions in Appendix B; HP Prime instructions on the book’s Web site.

You can use the two-sample t interval command on the TI-83/84 or TI-89 to construct a confidence interval for the difference between two means. We’ll show you the steps using the summary statistics from the pine trees example. TI-83/84 TI-89 • Press STAT , then choose TESTS •   Press 2nd F2 ([F7]) Ints and and 2-SampTInt....   choose 2-SampTInt.... • Choose Stats as the input method and enter the summary statistics as shown.





• Enter the confidence level: C-level: .90. For Pooled: choose “No.” We’ll discuss pooling later. • Highlight Calculate and press ENTER .



AP® EXAM TIP  The formula for the two-sample t interval for m1 − m2 often leads to calculation errors by students. As a result, we recommend using the calculator’s 2-SampTInt feature to compute the confidence interval on the AP® exam. Be sure to name the procedure (two-sample t interval) and to give the interval (3.9362, 17.724) and df (55.728) as part of the “Do” step.

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C H A P T E R 1 0 C o m pa r i n g T w o P o p u l at i o n s o r G r o u p s The calculator’s 90% confidence interval for m1 − m2 is (3.936, 17.724). This interval is narrower than the one we found by hand earlier: (3.83, 17.83). Why the difference? We used the conservative df = 29, but the calculator used df = 55.73. With more degrees of freedom, the calculator’s critical value is smaller than our t* = 1.699, which results in a smaller margin of error and a narrower interval.

Check Your Understanding

The U.S. Department of Agriculture (USDA) conducted a survey to estimate the average price of wheat in July and in September of the same year. Independent random samples of wheat producers were selected for each of the two months. Here are summary statistics on the reported price of wheat from the selected producers, in dollars per bushel:23 Month

n

x–

sx

July

90

$2.95

$0.22

September

45

$3.61

$0.19

Construct and interpret a 99% confidence interval for the difference in the true mean wheat price in July and in September.

Significance Tests for m1 − m2 An observed difference between two sample means can reflect an actual difference in the parameters m1 and m2, or it may just be due to chance variation in random sampling or random assignment. Significance tests help us decide which explanation makes more sense. The null hypothesis has the general form H0: m1 − m2 = hypothesized value We’re often interested in situations in which the hypothesized difference is 0. Then the null hypothesis says that there is no difference between the two parameters: H0: m1 − m2 = 0  or, alternatively,  H0: m1 = m2 The alternative hypothesis says what kind of difference we expect. If the Random, 10%, and Normal/Large Sample conditions are met, we can proceed with calculations. To do a test, standardize x–1 − x–2 to get a two-sample t statistic: statistic − parameter test statistic = standard deviation of statistic t=

(x–1 − x–2) − (m1 − m2) s22 s21 + Å n1 n2

To find the P-value, use the t distribution with degrees of freedom given by ­Option 1 (technology) or Option 2 (df = smaller of n1 − 1 and n2 − 1). Here are the details for the two-sample t test for the difference between two means.

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Two-Sample t  Test for the Difference between Two Means Suppose the conditions are met. To test the hypothesis H0: m1 − m2 = hypothesized value, compute the two-sample t statistic t=

(x–1 − x–2) − (m1 − m2) s22 s21 + Å n1 n2

Find the P-value by calculating the probability of getting a t statistic this large or larger in the direction specified by the alternative hypothesis Ha. Use the t distribution with degrees of freedom approximated by technology or the smaller of n1 − 1 and n2 − 1. Ha : µ1 – µ 2 > hypothesized value Ha : µ1 – µ 2 < hypothesized value Ha : µ1 – µ 2 = hypothesized value

t

t

– t

t

Here’s an example that shows how to perform a two-sample t test for a randomized experiment.

Example

Calcium and Blood Pressure

STEP

Comparing two means

4

Does increasing the amount of calcium in our diet reduce blood pressure? Examination of a large sample of people revealed a relationship between calcium intake and blood pressure. The relationship was strongest for black men. Such observational studies do not establish causation. Researchers therefore designed a randomized comparative experiment. The subjects were 21 healthy black men who volunteered to take part in the experiment. They were randomly assigned to two groups: 10 of the men received a calcium supplement for 12 weeks, while the control group of 11 men received a placebo pill that looked identical. The experiment was doubleblind. The response variable is the decrease in systolic (top number) blood pressure for a subject after 12 weeks, in millimeters of mercury. An increase appears as a negative number.24 Here are the data: Group 1 (calcium):

 7

−4

18

17

−3

−5

1

10

  11

−2

Group 2 (placebo):

−1

12

−1

−3

  3

−5

5

2

−11

−1

−3

Problem: (a) Do the data provide convincing evidence that a calcium supplement reduces blood pressure more than a placebo? Carry out an appropriate test to support your answer. (b) Interpret the P-value you got in part (a) in the context of this experiment.

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Solution: (a) State:  We want to perform a test of   H0: m1 = m2 H0: m1 − m2 = 0   or, equivalently, Ha: m1 − m2 > 0 Ha: m1 > m2

AP® EXAM TIP When checking the Normal condition on an AP® exam question involving inference about means, be sure to include graphs. Don’t expect to receive credit for describing graphs that you made on your calculator but didn’t put on paper.

where m1 is the true mean decrease in systolic blood pressure for healthy black men like the ones in this study who take a calcium supplement and m2 is the true mean decrease in systolic blood pressure for healthy black men like the ones in this study who take a placebo. No significance level was specified, so we’ll use a = 0.05. plan:  If conditions are met, we will carry out a two sample t test for m1 − m2. •  Random:  The 21 subjects were randomly assigned to the two treatments. ºº 10%:  Don’t need to check because there was no sampling. •  Normal/Large Sample:  With such small sample sizes, we need to graph the data to see if it’s reasonable to believe that the actual distributions of differences in blood pressure when taking calcium or placebo are Normal. Figure 10.9 shows hand sketches of calculator boxplots for these data. The graphs show no strong skewness and no outliers. So we are safe using two-sample t procedures.

Calcium

Placebo

FIGURE 10.9  ​Sketches of boxplots of the changes in blood pressure for the two groups of subjects in the calcium and blood pressure experiment.

-10

-5

0

5

10

15

20

Change in BP

dO:  From the data, we calculated summary statistics: Group

Treatment

n

x–

sx

1

Calcium

10

5.000

8.743

2

Placebo

11

−0.273

5.901

•  Test statistic (x–1 − x–2) − (m1 − m2) [5.000 − (−0.273)] − 0 5.273 t= = = = 1.604 2 2 2 2 3.2878 s1 s2 8.743 5.901 + + Å n1 n2 Å 10 11

•  P-value  By the conservative method, the smaller of n1 − 1 and Upper-tail probability p n2 − 1 gives df = 9. Because Ha counts only positive values of t as df .10 .05 .025 evidence against H0, the P-value is the area to the right of 8 1.397 1.860 2.306 t = 1.604 under the t distribution curve with df = 9. Figure 9 1.383 1.833 2.262 10.10 illustrates this P- value. Table B shows that the P-value lies 10 1.372 1.812 2.228 between 0.05 and 0.10. Using technology:  Refer to the Technology Corner that follows the example. The calculator’s 2-SampTTest gives t = 1.60 and P-value = 0.0644 using df = 15.59.

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Conclude:  Because the P-value is greater than a = 0.05, we

t distribution, 9 degrees of freedom

P -value

fail to reject H0. The experiment does not provide convincing evidence that the true mean decrease in systolic blood pressure is higher for men like these who take calcium than for men like these who take a placebo. (b)  ​Assuming H0: m1 − m2 = 0 is true, there is a 0.0644 probability of getting a ­difference in mean blood pressure reduction for the two groups (calcium − placebo) of 5.273 or greater just by the chance involved in the random assignment.

t = 1.604

FIGURE 10.10  ​The P-value for the one-sided test using the conservative method, which leads to the t distribution with 9 degrees of freedom.

For Practice Try Exercise

41

When a significance test leads to a fail to reject H0 decision, as in the previous example, be sure to interpret the results as “We don’t have convincing evidence to conclude Ha.” Saying anything that sounds like you believe H0 is (or might be) true is incorrect.

THINK ABOUT IT

24. Technology Corner

Why didn’t researchers find a significant difference in the calcium and blood pressure experiment?  ​The difference in mean

systolic blood pressures for the two groups was 5.273 millimeters of mercury. This seems like a fairly large difference. With the small group sizes, however, this difference wasn’t large enough to reject H0: m1 − m2 = 0 in favor of the one-sided alternative. We suspect that larger groups might show a similar difference in mean blood pressure reduction, which would indicate that calcium has a significant ­effect. If so, then the researchers in this experiment made a Type II error—failing to reject a false H0. In fact, later analysis of data from an experiment with more subjects resulted in a P-value of 0.008. Sample size strongly affects the power of a test. It is easier to detect an actual difference in the effectiveness of two treatments if both are applied to large numbers of subjects.

Two-sample t  tests on the calculator TI-Nspire instructions in Appendix B; HP Prime instructions on the book’s Web site.

Technology gives smaller P-values for two-sample t tests than the conservative method. That’s because calculators and software use the more complicated formula on page 640 to obtain a larger number of degrees of freedom. • Enter the Group 1 (calcium) data in L1/list1 and the Group 2 (placebo) data in L2/list2. • To perform the significance test, go to STAT/TESTS (Tests menu in the Statistics/List Editor on the TI-89) and choose 2-SampTTest. • In the 2-SampTTest screen, specify “Data” and adjust your other settings as shown.

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TI-83/84



TI-89



• Highlight “Calculate” and press ENTER . (The Pooled option will be discussed shortly.)





If you select “Draw” instead of “Calculate,” the appropriate t distribution will be displayed, showing the test statistic and the shaded area corresponding to the P-value.



AP® EXAM TIP  The formula for the two-sample t statistic for a test about m1 − m2 often leads to calculation errors by students. As a result, we recommend using the calculator’s 2-­SampTTest feature to perform calculations on the AP® exam. Be sure to name the procedure (two-sample t test) and to report the test statistic (t = 1.60), P-value (0.0644), and df (15.59) as part of the “Do” step.

Inference for Experiments  Confidence intervals and tests for m1 − m2

are based on the sampling distribution of x–1 − x–2. But in experiments, we aren’t sampling at random from any larger populations. We can think about what would happen if the random assignment were repeated many times under the assumption that H0: m1 − m2 = 0 is true. That is, we assume that the specific treatment received doesn’t affect an individual subject’s response. Let’s see what would happen just by chance if we randomly reassign the 21 subjects in the calcium and blood pressure experiment to the two groups many times, assuming the drug received doesn’t affect each individual’s change in systolic blood pressure. We used Fathom software to redo the random assignment 1000 times. The approximate randomization distribution of x–1 − x–2 is shown in Figure 10.11. It has an a­ pproximately Normal shape with mean 0 (no difference) and standard deviation 3.42.

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Section 10.2 Comparing Two Means FIGURE 10.11  ​Fathom simulation showing the approximate randomization distribution of x–1 − x–2 from 1000 random reassignments of subjects to treatment groups in the calcium and blood pressure experiment.

t distribution with df = 15.59

In the actual experiment, the difference in the mean change in blood pressure in the calcium and placebo groups was 5.000 − (−0.273) = 5.273. How likely is it that a difference this large or larger would happen just by chance when H0 is true? Figure 10.11 provides a rough answer: 61 of the 1000 random reassignments yielded a difference in means greater than or equal to 5.273. That is, our estimate of the P-value is 0.061. This is quite close to the 0.0644 P-value that we obtained in the Technology Corner. If Figure 10.11 displayed the results of all possible random reassignments of subjects to treatment groups, it would be the actual randomization distribution of x–1 − x–2. The P-value obtained from this distribution would be exactly correct. Using the two-sample t test to calculate the P-value gives only approximately correct results. Figure 10.12 shows the value of the two-sample t test statistic for each of the 1000 re-randomizations, calculated using our familiar formula t=

–6

–4

–2

0

2

4

6

t

FIGURE 10.12  ​The distribution of the two-sample t test statistic for the 1000 random reassignments in Figure 10.11.

649

(x– 1 − x–2) − (m1 − m2) s22 s21 + Å n1 n2

The density curve for the t distribution with df = 15.59 is shown in blue. We can see that the test statistic follows the t distribution quite closely in this case. Whenever the conditions are met, the randomization distribution of x–1 − x–2 looks much like its sampling distribution. We are therefore safe using two-sample t procedures for comparing two means in a randomized experiment.

Check Your Understanding

How quickly do synthetic fabrics such as polyester decay in landfills? A researcher buried polyester strips in the soil for different lengths of time, then dug up the strips and measured the force required to break them. Breaking strength is easy to measure and is a good indicator of decay. Lower strength means the fabric has decayed. For one part of the study, the researcher buried 10 strips of polyester fabric in welldrained soil in the summer. The strips were randomly assigned to two groups: 5 of them were buried for 2 weeks and the other 5 were buried for 16 weeks. Here are the breaking strengths in pounds:25 Group 1 (2 weeks): Group 2 (16 weeks):

118 124

126  98

126 110

120 140

129 110

Do the data give convincing evidence that polyester decays more in 16 weeks than in 2 weeks?

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Using Two-Sample t Procedures Wisely

c

Example

!

n

In Chapter 9, we used paired t procedures to compare the mean change in ­depression scores for a group of caffeine-dependent individuals when taking caffeine and a placebo. The inference involved paired data because the same 11 subjects received both treatments. In this chapter, we used two-sample t procedures to compare the mean change in blood pressure for a group of healthy autio black men when taking calcium and a placebo. This time, the inference ­involved two distinct groups of subjects. The proper method of analysis depends on the design of the study.

Comparing Tires and Comparing Workers Independent samples versus paired data Problem:  In each of the following settings, decide whether you should use paired t pro-

cedures or two-sample t procedures to perform inference.26 Explain your choice. (a) To test the wear characteristics of two tire brands, A and B, one Brand A tire is mounted on one side of each car in the rear, while a Brand B tire is mounted on the other side. Which side gets which brand is determined by flipping a coin. (b) Can listening to music while working increase productivity? Twenty factory workers agree to take part in a study to investigate this question. Researchers randomly assign 10 workers to do a repetitive task while listening to music and the other 10 workers to do the task in silence.

Solution: (a) Paired t procedures. This is a matched pairs experiment, with the two treatments (Brand A and Brand B) being randomly assigned to the rear pair of wheels on each car. (b) Two-sample t procedures. The data are being produced using two distinct groups of workers in a randomized experiment. For Practice Try Exercise

53

The same logic applies when data are produced by random sampling. If independent random samples are taken from each of two populations, we should use two-sample t procedures to perform inference about m1 − m2 if conditions are met. If one random sample is taken, and two data values are recorded for each individual, we should use paired t procedures to perform inference about the population mean difference mD if conditions are met.

The Pooled Two-Sample t Procedures (Don’t Use Them!)  Most

software ­offers a choice of two-sample t statistics. One is often labeled “unequal” variances; the other, “equal” variances. The “unequal” variance procedure uses our two-sample t statistic. This test is valid whether or not the population variances are equal.

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Section 10.2 Comparing Two Means

c

!

n

Remember, we always use the pooled sample proportion p^C when performing a significance test for comparing two proportions. But we don’t recommend pooling when comparing two means.

The other choice is a special version of the two-sample t statistic that assumes that the two populations have the same variance. This procedure combines (the statistical term is pools) the two sample variances to estimate the common population variance. The resulting statistic is called the pooled two-sample t statistic. The pooled t statistic has exactly the t distribution with n1 + n2 − 2 degrees of freedom if the two population variances really are equal and the population distributions are exactly Normal. This method offers more degrees of freedom than Option 1 (technology), which leads to narrower confidence intervals and smaller P-values. The pooled t procedures were in common use before software made it easy to use Option 1 for our two-sample t statistic. In the real world, distributions are not exactly Normal, and population variances are not exactly equal. In practice, the Option 1 two-sample t procedures are almost always more accurate than the pooled procedures. Our advice: autio Never use the pooled t procedures if you have software that will carry out Option 1.

case closed

Fast-Food Frenzy! Let’s return to the chapter-opening Case Study (page 609) about drivethru service at fast-food restaurants. Here, once again, are some results from the 2012 QSR study. • For restaurants with order-confirmation boards, 1169 of 1327 visits (88.1%) resulted in accurate orders. For restaurants with no orderconfirmation board, 655 of 726 visits (90.2%) resulted in accurate orders. • M  cDonald’s average service time for 362 drive-thru visits was 188.83 seconds with a standard deviation of 17.38 seconds. Burger King’s service time for 318 drive-thru visits had a mean of 201.33 seconds and a standard deviation of 18.85 seconds. You are now ready to use what you have learned about comparing population parameters to perform inference about accuracy and average service time in the drive-thru lane. 1. Is there a significant difference in order accuracy between restaurants with and without order-confirmation boards? Carry out an appropriate test at the a = 0.05 level to help answer this question. A 95% confidence interval for the difference in the population proportions of accurate orders at restaurants with and without order-confirmation boards is (–0.049, 0.00649). 2. Interpret the meaning of “95% confident” in the context of this study.

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3. Explain how the confidence interval is consistent with your conclusion from Question 1. Now turn your attention to the speed-of-service data. 4. Construct and interpret a 99% confidence interval for the difference in the mean service times at McDonald’s and Burger King drive-thrus.

Section 10.2

Summary •

•

•

Choose independent SRSs of size n1 from Population 1 and size n2 from Population 2. The sampling distribution of x–1 − x–2 has the following properties: • Shape:  Normal if both population distributions are Normal; approximately Normal otherwise if both samples are large enough (n1 ≥ 30 and n2 ≥ 30) by the central limit theorem. • Center:  Its mean is m1 − m2. • Spread:  As long as each sample is no more than 10% of its population, s21 s22 its standard deviation is + . Å n1 n2 Confidence intervals and tests for the difference between the means of two populations or the mean responses to two treatments m1 and m2 are based on the difference x–1 − x–2 between the sample means. Because we almost never know the population standard deviations in practice, we use the two-sample t statistic t=



•

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(x– 1 − x–2) − (m1 − m2) s22 s21 + Å n1 n2

This statistic has approximately a t distribution. There are two options for using a t distribution to approximate the distribution of the two-sample t statistic: • Option 1 (Technology)  Use the t distribution with degrees of freedom calculated from the data by a somewhat messy formula. The degrees of freedom probably won’t be a whole number. • Option 2 (Conservative)  Use the t distribution with degrees of freedom equal to the smaller of n1 − 1 and n2 − 1. This method gives wider confidence intervals and larger P-values than Option 1. Before estimating or testing a claim about m1 − m2, check that these conditions are met: • Random:  The data are produced by independent random samples of size n1 from Population 1 and of size n2 from Population 2 or by two groups of size n1 and n2 in a randomized experiment.

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10%: When sampling without replacement, check that the two populations are at least 10 times as large as the corresponding samples. • Normal/Large Sample:  Both population distributions (or the true distributions of responses to the two treatments) are Normal or both ­sample sizes are large (n1 ≥ 30 and n2 ≥ 30). If either population (treatment) distribution has unknown shape and the corresponding sample size is less than 30, use a graph of the sample data to assess the Normality of the population (treatment) distribution. Do not use two-sample t procedures if the graph shows strong skewness or outliers. An approximate C% confidence interval for m1 − m2 is ~~

•

•

•

STEP

4

• •

s22 s21 (x– 1 − x– 2) ± t* + Å n1 n2 where t* is the critical value with C% of its area between −t* and t* for the t distribution with degrees of freedom from either Option 1 (technology) or Option 2 (the s­maller of n1 − 1 and n2 − 1). This is called a two-sample t interval for m1 ∙ m2. To test H0: m1 − m2 = hypothesized value, use a two-sample t test for m1 ∙ m2. The test statistic is (x– 1 − x–2) − (m1 − m2) t= s22 s21 + Å n1 n2

P-values are calculated using the t distribution with degrees of freedom from either Option 1 (technology) or Option 2 (the smaller of n1 − 1 and n2 − 1). Inference about the difference m1 − m2 in the effectiveness of two treatments in a completely randomized experiment is based on the randomization distribution of x–1 − x–2. When the conditions are met, our usual inference procedures based on the sampling distribution of x–1 − x–2 will be approximately correct. Don’t use two-sample t procedures to compare means for paired data. Be sure to follow the four-step process whenever you construct a confidence interval or perform a significance test for comparing two means.

10.2 T echnology Corners TI-Nspire Instructions in Appendix B; HP Prime instructions on the book’s Web site. 23. Two-sample t intervals on the calculator 24. Two-sample t tests on the calculator

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Section 10.2 STEP

4

Exercises

Remember: We are no longer reminding you to use the four-step process in exercises that require you to perform inference.

31. Cholesterol  The level of cholesterol in the blood for all men aged 20 to 34 follows a Normal distribution with mean 188 milligrams per deciliter (mg/dl) and standard deviation 41 mg/dl. For 14-year-old boys, blood cholesterol levels follow a Normal distribution with mean 170 mg/dl and standard deviation 30 mg/dl. Suppose we select independent SRSs of 25 men aged 20 to 34 and 36 boys aged 14 and calculate the sample mean cholesterol levels x–M and x–B .

pg 638

(a) What is the shape of the sampling distribution of x–M − x–B ? Why?

of 20 male students from their school. Then they recorded the number of pairs of shoes that each respondent reported having. The back-to-back stemplot displays the data. 34. Household size  How do the numbers of people living in households in the United Kingdom (U.K.) and South Africa compare? To help answer this question, we used CensusAtSchool’s random data selector to choose independent samples of 50 students from each country. Here is a Fathom dotplot of the household sizes reported by the students in the survey.

(b) Find the mean of the sampling distribution. Show your work. (c) Find the standard deviation of the sampling distribution. Show your work. 32. How tall?  The heights of young men follow a Normal distribution with mean 69.3 inches and standard deviation 2.8 inches. The heights of young women follow a Normal distribution with mean 64.5 inches and standard deviation 2.5 inches. Suppose we select independent SRSs of 16 young men and 9 young women and calculate the sample mean heights x–M and x–W . (a) What is the shape of the sampling distribution of x–M − x–W ? Why? (b) Find the mean of the sampling distribution. Show your work.

35. Literacy rates  Do males have higher average literacy rates than females in Islamic countries? The table below shows the percent of men and women who were literate in the major Islamic nations at the time of this writing.27 (We omitted countries with populations of less than 3 million.) Country

Male (%)

Female (%)

(c) Find the standard deviation of the sampling distribution. Show your work.

Afghanistan

43

13

Algeria

80

60

In Exercises 33 to 36, determine whether or not the ­conditions for using two-sample t procedures are met.

Azerbaijan

99.9

99.7

Bangladesh

61

52

33. Shoes  How many pairs of shoes do teenagers have? To find out, a group of AP® Statistics students conducted a survey. They selected a random sample of 20 female students and a separate random sample

Egypt

80

64

Indonesia

94

86.8

Iran

84

70

Iraq

86

71

Jordan

96

89

100

99

Females

333 95 4332 66 410 8 9 100 7

0 0 1 1 2 2 3 3 4 4 5 5

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Males 4 555677778 0000124 2

58

Kazakhstan

Key: 2|2 represents a male student with 22 pairs of shoes.

Kyrgyzstan

99.3

98.1

Lebanon

93

82

Libya

96

83

Malaysia

92

85

Morocco

69

44

Pakistan

68.6

30.3

Saudi Arabia

90

81

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Section 10.2 Comparing Two Means Country Syria

Male (%)

Female (%)

86

74

100

100

Tunisia

83

65

Turkey

98

90

Turkmenistan

99.3

98.3

Tajikistan

Uzbekistan Yemen

100

99

81

47

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(b) Construct and interpret a 90% confidence interval for the difference in mean percent change in polyphenol levels for the red wine and white wine treatments. (c) Does the interval in part (b) suggest that red wine is more effective than white wine? Explain. 38. Tropical flowers  Different varieties of the tropical flower Heliconia are fertilized by different species of hummingbirds.

36. Long words  Mary was interested in comparing the mean word length in articles from a medical journal and an airline’s in-flight magazine. She counted the number of letters in the first 400 words of an article in the medical journal and in the first 100 words of an article in the airline magazine. Mary then used Minitab statistical software to produce the histograms shown. Note that J is for journal and M is for magazine.

Researchers believe that over time, the lengths of the flowers and the forms of the hummingbirds’ beaks have evolved to match each other. Here are data on the lengths in millimeters for random samples of two color varieties of the same species of flower on the island of Dominica:29 37. Is red wine better than white wine? Observational pg 641 studies suggest that moderate use of alcohol by adults reduces heart attacks and that red wine may have special benefits. One reason may be that red wine contains polyphenols, substances that do good things to cholesterol in the blood and so may reduce the risk of heart attacks. In an experiment, healthy men were assigned at random to drink half a bottle of either red or white wine each day for two weeks. The level of polyphenols in their blood was measured before and after the two-week period. Here are the percent changes in level for the subjects in both groups:28 Red wine:

3.5 8.1

White wine:

3.1 0.5

7.4 4.0

0.7 4.9

−3.8 4.1 −0.6 2.7

8.4

H. caribaea red 41.90

42.01

41.93

43.09

41.17

41.69

39.78

40.57

39.63

42.18

40.66

37.87

39.16

37.40

38.20

38.07

38.10

37.97

38.79

38.23

38.87

37.78

38.01

36.78

37.02

36.52

H. caribaea yellow 36.11 36.03 35.45

38.13

35.17

36.82

36.66

35.68

36.03

34.57

37.10

34.63

(a) A Fathom dotplot of the data is shown below. Write a few sentences comparing the distributions.

7.0 5.5

1.9 −5.9 0.1

(a) A Fathom dotplot of the data is shown below. Write a few sentences comparing the distributions. (b) Construct and interpret a 95% confidence interval for the difference in the mean lengths of these two varieties of flowers. (c) Does the interval support the researchers’ belief that the two flower varieties have different average lengths? Explain.

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39. Paying for college  College financial aid offices expect students to use summer earnings to help pay for college. But how large are these earnings? One large university studied this question by asking a random sample of 1296 students who had summer jobs how much they earned. The financial aid office separated the responses into two groups based on gender. Here are the data in summary form:30 Group

n

x–

sx

Males

675

$1884.52

$1368.37

Females

621

$1360.39

$1037.46

(a) How can you tell from the summary statistics that the distribution of earnings in each group is strongly skewed to the right? The use of two-sample t procedures is still justified. Why? (b) Construct and interpret a 90% confidence interval for the difference between the mean summer earnings of male and female students at this university. (c) Interpret the 90% confidence level in the context of this study. 40. Happy customers  As the Hispanic population in the United States has grown, businesses have tried to understand what Hispanics like. One study interviewed a random sample of customers leaving a bank. Customers were classified as Hispanic if they preferred to be interviewed in Spanish or as Anglo if they preferred English. Each customer rated the importance of several aspects of bank service on a 10-point scale.31 Here are summary results for the importance of “reliability” (the accuracy of account records and so on): Group

n

x–

sx

Anglo Hispanic

92 86

6.37 5.91

0.60 0.93

breeding next year? Researchers randomly assigned 7 pairs of birds to have the natural caterpillar ­supply supplemented while feeding their young and ­another 6 pairs to serve as a control group relying on natural food supply. The next year, they measured how many days after the caterpillar peak the birds produced their nestlings.32 The investigators expected the control group to adjust their breeding date the next year, whereas the well-fed supplemented group had no reason to change. Here are the data (days after caterpillar peak): Control:

 4.6

 2.3

7.7

 6.0

 4.6 −1.2

Supplemented:

15.5

11.3

5.4

16.5

11.3

11.4

7.7

(a) Do the data provide convincing evidence to confirm the researchers’ belief? (b) Interpret the P-value from part (a) in the context of this study. 42. DDT in rats  Poisoning by the pesticide DDT causes convulsions in humans and other mammals. ­Researchers seek to understand how the convulsions are caused. In a randomized comparative experiment, they compared 6 white rats poisoned with DDT with a control group of 6 unpoisoned rats. Electrical measurements of nerve activity are the main clue to the nature of DDT poisoning. When a nerve is stimulated, its electrical response shows a sharp spike followed by a much smaller second spike. The researchers measured the height of the second spike as a percent of the first spike when a nerve in the rat’s leg was stimulated.33 For the poisoned rats, the results were 12.207 16.869 25.050 22.429 8.456 20.589

The control group data were 11.074 9.686 12.064 9.351 8.182 6.642

(a) The distribution of reliability ratings in each group is not Normal. The use of two-sample t procedures is still justified. Why?

(a) Do these data provide convincing evidence that DDT affects the mean relative height of the second spike’s electrical response?

(b) Construct and interpret a 95% confidence interval for the difference between the mean ratings of the importance of reliability for Anglo and Hispanic bank customers.

(b) Interpret the P-value from part (a) in the context of this study.

(c) Interpret the 95% confidence level in the context of this study. 41. Baby birds  Do birds learn to time their breeding? Blue titmice eat caterpillars. The birds would like lots of caterpillars around when they have young to feed, but they must breed much earlier. Do the birds learn from one year’s experience when to time their

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43. Who talks more—men or women? Researchers equipped random samples of 56 male and 56 ­female students from a large university with a small device that secretly records sound for a random 30 seconds during each 12.5-minute period over two days. Then they counted the number of words spoken by each subject during each recording period and, from this, estimated how many words per day each subject speaks. The female

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estimates had a mean of 16,177 words per day with a standard deviation of 7520 words per day. For the male estimates, the mean was 16,569 and the standard deviation was 9108. Do these data provide convincing evidence of a difference in the average number of words spoken in a day by male and female students at this university? 44. Competitive rowers  What aspects of rowing technique distinguish between novice and skilled competitive rowers? Researchers compared two randomly selected groups of female competitive rowers: a group of skilled rowers and a group of novices. The researchers measured many mechanical aspects of rowing style as the subjects rowed on a Stanford Rowing Ergometer. One important variable is the angular velocity of the knee, which describes the rate at which the knee joint opens as the legs push the body back on the sliding seat. The data show no outliers or strong skewness. Here is the SAS computer output:34 TTEST PROCEDURE Variable: KNEE GROUP

N

Mean

Std Dev

Std Error

SKILLED

10

4.182

0.479

0.151

NOVICE

8

3.010

0.959

0.339



The researchers believed that the knee velocity would be higher for skilled rowers. Do the data provide convincing evidence to support this belief?

(a) Based on the graph and numerical summaries, write a few sentences comparing the DRP scores for the two groups. (b) Is the mean DRP score significantly higher for the students who did the reading activities? Give appropriate evidence to justify your answer. (c) Can we conclude that the new reading activities caused an increase in the mean DRP score? Explain. 46. Does breast-feeding weaken bones? Breast-feeding mothers secrete calcium into their milk. Some of the calcium may come from their bones, so mothers may lose bone mineral. Researchers compared a random sample of 47 breast-feeding women with a random sample of 22 women of similar age who were neither pregnant nor lactating. They measured the percent change in the bone mineral content (BMC) of the women’s spines over three months. Comparative ­boxplots and summary statistics for the data from Fathom are shown below.36

45. Teaching reading  An educator believes that new reading activities in the classroom will help elementary school pupils improve their reading ability. She recruits 44 third-grade students and randomly assigns them into two groups. One group of 21 students does these new activities for an 8-week period. A control group of 23 thirdgraders follows the same curriculum without the activities. At the end of the 8 weeks, all students are given the Degree of Reading Power (DRP) test, which measures the aspects of reading ability that the treatment is designed to improve. Comparative boxplots and summary statistics for the data from Fathom are shown below.35

(a) Based on the graph and numerical summaries, write a few sentences comparing the percent changes in BMC for the two groups.

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(b) Is the mean change in BMC significantly lower for the mothers who are breast-feeding? Give appropriate evidence to justify your answer.

(b) Based on your conclusion in part (a), could you have made a Type I error or a Type II error? Justify your answer.

(c) Can we conclude that breast-feeding causes a mother’s bones to weaken? Why or why not?

51. Rewards and creativity  Dr. Teresa Amabile conducted a study involving 47 college students who were randomly assigned to two treatment groups. The 23 students in one group were given a list of statements about external reasons (E) for writing, such as public recognition, making money, or pleasing their parents. The 24 students in the other group were given a list of statements about internal reasons (I) for writing, such as expressing yourself and enjoying playing with words. Both groups were then instructed to write a poem about laughter. Each student’s poem was rated separately by 12 different poets using a creativity scale.37 The 12 poets’ ratings of each student’s poem were averaged to obtain an overall creativity score.

47. Who talks more—men or women?  Refer to Exercise 43. Construct and interpret a 95% confidence interval for the difference in mean number of words spoken in a day. Explain how this interval provides more information than the significance test in Exercise 43. 48. DDT in rats  Refer to Exercise 42. Construct and interpret a 95% confidence interval for the difference in mean relative height of the second spike’s electrical response. Explain how this interval provides more information than the significance test in Exercise 42. 49. A better drug?  In a pilot study, a company’s new cholesterol-reducing drug outperforms the currently available drug. If the data provide convincing evidence that the mean cholesterol reduction with the new drug is more than 10 milligrams per deciliter of blood (mg/dl) greater than with the current drug, the company will begin the expensive process of mass-producing the new drug. For the 14 subjects who were assigned at random to the current drug, the mean cholesterol reduction was 54.1 mg/dl with a standard deviation of 11.93 mg/dl. For the 15 subjects who were randomly assigned to the new drug, the mean cholesterol reduction was 68.7 mg/dl with a standard deviation of 13.3 mg/dl. Graphs of the data reveal no outliers or strong skewness.

We used Fathom software to randomly reassign the 47 subjects to the two groups 1000 times, assuming the treatment received doesn’t affect each individual’s average creativity rating. The dotplot shows the ­approximate randomization distribution of x–I − x–E .

(a) Carry out an appropriate significance test. What conclusion would you draw? (Note that the null hypothesis is not H0: m1 − m2 = 0.) (b) Based on your conclusion in part (a), could you have made a Type I error or a Type II error? Justify your answer. 50. Down the toilet  A company that makes hotel toilets claims that its new pressure-assisted toilet reduces the average amount of water used by more than 0.5 gallon per flush when compared to its current model. To test this claim, the company randomly selects 30 toilets of each type and measures the amount of water that is used when each toilet is flushed once. For the current-model toilets, the mean amount of water used is 1.64 gal with a standard deviation of 0.29 gal. For the new toilets, the mean amount of water used is 1.09 gal with a standard deviation of 0.18 gal. (a) Carry out an appropriate significance test. What conclusion would you draw? (Note that the null hypothesis is not H0: m1 − m2 = 0.)

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(a) Why did researchers randomly assign the subjects to the two treatment groups? (b) In the actual experiment, x–I − x–E = 4.15. This value is marked with a blue line in the figure. What conclusion would you draw? Justify your answer with appropriate evidence. (c) Based on your conclusion in part (b), could you have made a Type I error or a Type II error? Justify your answer. 52. Sleep deprivation  Does sleep deprivation linger for more than a day? Researchers designed a study using 21 volunteer subjects between the ages of 18 and 25. All 21 participants took a computer-based visual discrimination test at the start of the study. Then the subjects were randomly assigned into two groups. The 11 subjects in one group, D, were deprived of sleep for an entire night in a laboratory setting. The

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Section 10.2 Comparing Two Means 10 subjects in the other group, A, were allowed unrestricted sleep for the night. Both groups were allowed as much sleep as they wanted for the next two nights. On Day 4, all the subjects took the same visual discrimination test on the computer. Researchers recorded the improvement in time (measured in milliseconds) from Day 1 to Day 4 on the test for each subject.38 We used Fathom software to randomly reassign the 21 subjects to the two groups 1000 times, assuming the treatment received doesn’t affect each individual’s time improvement on the test. The dotplot shows the approximate randomization distribution of x–A − x–D .

woman was measured before the assigned diet and again after 10 weeks on the diet. 54. Paired or unpaired?  In each of the following settings, decide whether you should use paired t procedures or two-sample t procedures to perform inference. Explain your choice.40 (a) To compare the average weight gain of pigs fed two different rations, nine pairs of pigs were used. The pigs in each pair were littermates. A coin toss was used to decide which pig in each pair got Ration A and which got Ration B. (b) Separate random samples of male and female college professors are taken. We wish to compare the average salaries of male and female teachers. (c) To test the effects of a new fertilizer, 100 plots are treated with the new fertilizer, and 100 plots are treated with another fertilizer. A computer’s random number generator is used to determine which plots get which fertilizer.

(a) Explain why the researchers didn’t let the subjects choose whether to be in the sleep deprivation group or the unrestricted sleep group. (b) In the actual experiment, x–A − x–D = 15.92. This value is marked with a blue line in the figure. What conclusion would you draw? Justify your answer with appropriate evidence. (c) Based on your conclusion in part (b), could you have made a Type I error or a Type II error? Justify your answer.

Exercises 55 and 56 refer to the following setting. Coaching companies claim that their courses can raise the SAT scores of high school students. Of course, students who retake the SAT without paying for coaching generally raise their scores. A random sample of students who took the SAT twice found 427 who were coached and 2733 who were uncoached.41 Starting with their Verbal scores on the first and second tries, we have these summary statistics: Try 1 n Coached Uncoached

–

x

Try 2 sx

–

x

Gain sx

–

sx

x

427

500

 92

529

 97

29

59

2733

506

101

527

101

21

52

55. Coaching and SAT scores  Let’s first ask if students who are coached increased their scores significantly.

53. Paired or unpaired?  In each of the following settings, decide whether you should use paired t procedures or two-sample t procedures to perform inference. Explain your choice.39

(a) You could use the information on the Coached line to carry out either a two-sample t test comparing Try 1 with Try 2 for coached students or a paired t test using Gain. Which is the correct test? Why?

(a) To test the wear characteristics of two tire brands, A and B, each brand of tire is randomly assigned to 50 cars of the same make and model.

(b) Carry out the proper test. What do you conclude?

pg 650

(b) To test the effect of background music on productivity, factory workers are ­observed. For one month, each subject works without music. For another month, the subject works while listening to music on an MP3 player. The month in which each subject listens to music is determined by a coin toss. (c) A study was designed to compare the effectiveness of two weight-reducing diets. Fifty obese women who volunteered to participate were randomly assigned into two equal-sized groups. One group used Diet A and the other used Diet B. The weight of each

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56. Coaching and SAT scores  What we really want to know is whether coached students improve more than uncoached students, and whether any advantage is large enough to be worth paying for. Use the information above to answer these questions: (a) How much more do coached students gain on the average? Construct and interpret a 99% confidence interval. (b) Does the interval in part (a) give convincing evidence that coached students gain more, on ­average, than uncoached students? Explain. (c) Based on your work, what is your opinion: do you think coaching courses are worth paying for?

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Multiple choice: Select the best answer for Exercises 57 to 60. 57. There are two common methods for measuring the concentration of a pollutant in fish tissue. Do the two methods differ, on average? You apply both methods to each fish in a random sample of 18 carp and use (a) the paired t test for md.

(d) Assuming that the true mean road rage score is the same for males and females, there is a 0.002 probability that the null hypothesis is true. (e) Assuming that the true mean road rage score is the same for males and females, there is a 0.002 probability that the alternative hypothesis is true.

(b) the one-sample z test for p.

60. Based on the P-value in Exercise 59, which of the following must be true?

(c) the two-sample t test for m1 − m2.

(a) A 90% confidence interval for mM − mF will contain 0.

(d) the two-sample z test for p1 − p2.

(b) A 95% confidence interval for mM − mF will contain 0.

(e) none of these.

(c) A 99% confidence interval for mM − mF will contain 0.

Exercises 58 to 60 refer to the following setting. A study of road rage asked random samples of 596 men and 523 women about their behavior while driving. Based on their answers, each person was assigned a road rage score on a scale of 0 to 20. The participants were chosen by random digit dialing of phone numbers. The researchers performed a test of the following hypotheses: H0: mM = mF versus Ha: mM ≠ mF.

(d) A 99.9% confidence interval for mM − mF will ­contain 0.

58. Which of the following describes a Type II error in the context of this study? (a) Finding convincing evidence that the true means are different for males and females, when in reality the true means are the same (b) Finding convincing evidence that the true means are different for males and females, when in reality the true means are different (c) Not finding convincing evidence that the true means are different for males and females, when in reality the true means are the same (d) Not finding convincing evidence that the true means are different for males and females, when in reality the true means are different (e) Not finding convincing evidence that the true means are different for males and females, when in reality there is convincing evidence that the true means are different 59. The P-value for the stated hypotheses is 0.002. Interpret this value in the context of this study. (a) Assuming that the true mean road rage score is the same for males and females, there is a 0.002 probability of getting a difference in sample means. (b) Assuming that the true mean road rage score is the same for males and females, there is a 0.002 probability of getting an observed difference at least as extreme as the observed difference. (c) Assuming that the true mean road rage score is different for males and females, there is a 0.002 probability of getting an observed difference at least as extreme as the observed difference.

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(e) It is impossible to determine whether any of these statements is true based only on the P-value. In each part of Exercises 61 and 62, state which inference procedure from Chapter 8, 9, or 10 you would use. Be specific. For example, you might say, “Two-sample z test for the difference between two proportions.” You do not need to carry out any procedures. 61. Which inference method? (a) Drowning in bathtubs is a ­major cause of death in children less than 5 years old. A random sample of parents was asked many questions related to bathtub safety. Overall, 85% of the sample said they used baby bathtubs for infants. Estimate the percent of all parents of young children who use baby bathtubs. (b) How seriously do people view speeding in comparison with other annoying behaviors? A large random sample of adults was asked to rate a number of behaviors on a scale of 1 (no problem at all) to 5 (very severe problem). Do speeding drivers get a higher average rating than noisy neighbors? (c) You have data from interviews with a random sample of students who failed to graduate from a particular college in 7 years and also from a random sample of students who entered at the same time and did graduate. You will use these data to compare the percents of students from rural backgrounds among dropouts and graduates. (d) Do experienced computer game players earn higher scores when they play with someone present to cheer them on or when they play alone? Fifty teenagers with experience playing a particular computer game have volunteered for a study. We randomly assign 25 of them to play the game alone and the other 25 to play the game with a supporter present. Each player’s score is recorded.

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Section 10.2 Comparing Two Means 62. Which inference method? (a) How do young adults look back on adolescent romance? Investigators interviewed 40 couples in their midtwenties. The female and male partners were interviewed separately. Each was asked about his or her current relationship and also about a romantic relationship that lasted at least two months when they were aged 15 or 16. One response variable was a measure on a numerical scale of how much the attractiveness of the adolescent partner mattered. You want to find out how much men and women differ on this measure. (b) Are more than 75% of Toyota owners generally satisfied with their vehicles? Let’s design a study to find out. We’ll select a random sample of 400 Toyota owners. Then we’ll ask each individual in the sample: “Would you say that you are generally satisfied with your Toyota vehicle?” (c) Are male college students more likely to binge drink than female college students? The Harvard School of Public Health surveys random samples of male and female undergraduates at four-year colleges and universities about whether they have engaged in binge drinking. (d) A bank wants to know which of two incentive plans will most increase the use of its credit cards and by how much. It offers each incentive to a group of current credit card customers, determined at random, and compares the amount charged during the following six months. 63. Quality control (2.2, 5.3, 6.3)  Many manufacturing companies use statistical techniques to ensure that the products they make meet standards. One common way to do this is to take a random sample of products at regular intervals throughout the production shift. Assuming that the process is working properly, the mean measurement x– from a random sample varies according to a Normal distribution with mean m x– and standard deviation sx–. For each question that follows, assume that the process is working properly.

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(a) What’s the probability that at least one of the next two sample means will fall more than 2sx– from the target mean m x– ? Show your work. (b) What’s the probability that the first sample mean that is greater than m x– + 2sx– is the one from the fourth sample taken? Plant managers are trying to develop a criterion for determining when the process is not working properly. One idea they have is to look at the 5 most recent sample means. If at least 4 of the 5 fall outside the interval (m x– − sx– , m x– + sx– ), they will conclude that the process isn’t working. (c) Find the probability that at least 4 of the 5 most ­recent sample means fall outside the interval, assuming the process is working properly. Is this a reasonable criterion? Explain. 64. Information online (8.2, 10.1)  A random digit dialing sample of 2092 adults found that 1318 used the Internet.42 Of the users, 1041 said that they expect businesses to have Web sites that give product information; 294 of the 774 nonusers said this. (a) Construct and interpret a 95% confidence interval for the proportion of all adults who use the Internet. (b) Construct and interpret a 95% confidence interval to compare the proportions of users and nonusers who expect businesses to have Web sites that give product information. 65. Coaching and SAT scores: Critique (4.1, 4.3) The data in Exercises 55 and 56 came from a random sample of students who took the SAT twice. The response rate was 63%, which is fairly good for nongovernment surveys. (a) Explain how nonresponse could lead to bias in this study. (b) We can’t be sure that coaching actually caused the coached students to gain more than the uncoached students. Explain briefly but clearly why this is so.

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FRAPPY!  Free Response AP® Problem, Yay! The following problem is modeled after actual AP® Statistics exam free response questions. Your task is to generate a complete, concise response in 15 minutes.

Directions: Show all your work. Indicate clearly the methods you use, because you will be scored on the correctness of your methods as well as on the accuracy and completeness of your results and explanations. Will using name-brand microwave popcorn result in a greater percentage of popped kernels than using storebrand microwave popcorn? To find out, Briana and ­Maggie randomly selected 10 bags of name-brand microwave popcorn and 10 bags of store-brand microwave popcorn. The chosen bags were arranged in a random order. Then each bag was popped for 3.5 minutes, and the percentage of popped kernels was calculated. The results are displayed in the following table.

Name-brand Store-brand

Name-brand Store-brand 90 78

95

91

88

89

97

84

84

82

93

86

94

82

91

86

81

77

86

90

Do the data provide convincing evidence that using namebrand microwave popcorn will result in a greater mean percentage of popped kernels? After you finish, you can view two example solutions on the book’s Web site (www.whfreeman.com/tps5e). Determine whether you think each solution is “complete,” “substantial,” “developing,” or “minimal.” If the solution is not complete, what improvements would you suggest to the student who wrote it? Finally, your teacher will provide you with a scoring rubric. Score your response and note what, if anything, you would do differently to improve your own score.

Chapter Review Section 10.1: Comparing Two Proportions In this section, you learned how to construct confidence intervals and perform significance tests for a difference between two proportions. Inference for a difference in proportions is based on the sampling distribution of p^ 1 − p^ 2. When the conditions are met, the sampling distribution of p^ 1 − p^ 2 is approximately Normal with a mean of mp^ 1 −p^ 2 = p1 − p2 and p1(1 − p1) p2(1 − p2) a standard deviation of sp^ 1 −p^ 2 = + . n1 n2 Å The conditions for inference about a difference in proportions are the same for confidence intervals and significance tests. The Random condition says that the data must be from two independent random samples or two groups in a randomized experiment. The 10% condition says that each sample size should be less than 10% of the corresponding population size when sampling without replacement. The Large Counts condition says that the number of successes and number of failures from each sample/group should be at least 10. That is, n1p^ 1, n1(1 − p^ 1), n2p^ 2, n2(1 − p^ 2) are all ≥10. A confidence interval for a difference between two proportions provides an interval of plausible values for the true difference in proportions. The formula is

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p^ 1(1 − p^ 1) p^ 2(1 − p^ 2) (p^ 1 − p^ 2) ± z* + n1 n2 Å The logic of confidence intervals, including how to interpret the confidence interval and the confidence level, is the same as it was in Chapter 8, when you first learned about confidence intervals. Likewise, a significance test for a difference between two proportions uses the same logic as the significance tests you learned about in Chapter 9. We start by assuming the null hypothesis is true and asking how likely it would be to get results at least as unusual as the results observed in a study by chance alone. If it is plausible that a difference in proportions could be the result of sampling variability or the chance variation due to random assignment, we do not have convincing evidence that the alternative hypothesis is true. However, if the difference is too big to attribute to chance, there is convincing evidence to believe that the alternative hypothesis is true. For a test of H0: p1 − p2 = 0, the test statistic is z=

(p^ 1 − p^ 2) − 0 p^ C(1 − p^ C) p^ C(1 − p^ C) + n1 n2 Å

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where p^ C is the combined (overall) proportion of successes: X1 + X2 p^ C = . n1 + n2 Finally, you learned that the inference techniques used for analyzing a difference in proportions from two ­independent random samples work very well for analyzing a difference in proportions from two groups in a completely randomized experiment. Section 10.2: Comparing Two Means In this section, you learned how to construct confidence intervals and perform significance tests for a difference in two means. Inference for a difference in means is based on the sampling distribution of x–1 − x–2. When the conditions are met, the sampling distribution of x–1 − x–2 is approximately Normal with a mean of mx–1 −x– 2 = m1 − m2 and a standard s21 s22 ­deviation of sx–1 − x–2 = + . Å n1 n2 The conditions for inference about a difference in means are the same for confidence intervals and significance tests. The Random condition says that the data must be from two independent random samples or two groups in a randomized experiment. The 10% condition says that each sample size should be less than 10% of the corresponding population size when sampling without replacement. The Normal/­Large Sample condition says that the two populations are Normal or that the two sample/group sizes are large (n1 ≥ 30, n2 ≥ 30). If the sample/group sizes are small and

the population shapes are unknown, graph both sets of data to make sure there is no strong skewness or outliers. As in Chapters 8 and 9, inference techniques for means are based on the t distributions. There are two options for calculating the number of degrees of freedom to use. The first option is to use technology to calculate the degrees of freedom. The second option is to use the smaller of n1 − 1 and n2 − 1. The technology option is preferred because it produces a larger number of degrees of freedom, resulting in narrower confidence intervals and smaller P-values. If you are using technology, always choose the unpooled option. A confidence interval for a difference between two means provides an interval of plausible values for the true difference in means. The formula is s21 s21 (x–1 − x–2) ± t* + Å n1 n2 Use a significance test to decide between two competing hypotheses about a difference in true means. The test statistic is t=

(x–1 − x–2) − (m1 − m2) s21 s22 + Å n1 n2

where m1 − m2 is the difference specified by the null hypothesis. When constructing confidence intervals or performing significance tests for a difference in means, make sure that the data are not paired. If the data are paired, use the paired t procedures from Chapter 9.

What Did You Learn? Learning Objective

Section

Related Example on Page(s)

Relevant Chapter Review Exercise(s)

Describe the shape, center, and spread of the sampling distribution of p^ 1 − p^ 2 .

10.1

615

R10.2

Determine whether the conditions are met for doing inference about p1 − p 2 .

10.1

617

R10.5, R10.6

Construct and interpret a confidence interval to compare two proportions.

10.1

617

R10.2

Perform a significance test to compare two proportions.

10.1

622, 625

R10.5

Describe the shape, center, and spread of the sampling distribution of x–1 − x–2 .

10.2

638

R10.3

Determine whether the conditions are met for doing inference for m1 − m2 .

10.2

641

R10.3, R10.4, R10.6

Construct and interpret a confidence interval to compare two means.

10.2

641

R10.4

Perform a significance test to compare two means.

10.2

645

R10.7

Determine when it is appropriate to use two-sample t procedures versus paired t procedures.

10.2

650

R10.1, R10.7

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Chapter 10 Chapter Review Exercises These exercises are designed to help you review the important ideas and methods of the chapter. R10.1 Which procedure?  For each of the following settings, say which inference procedure from Chapter 8, 9, or 10 you would use. Be specific. For example, you might say, “Two-sample z test for the difference between two proportions.” You do not need to carry out any procedures.43 (a) Do people smoke less when cigarettes cost more? A random sample of 500 smokers was selected. The number of cigarettes each person smoked per day was recorded over a one-month period before a 30% cigarette tax was imposed and again for one month after the tax was imposed. (b) How much greater is the percent of senior citizens who attend a play at least once per year than the percent of people in their twenties who do so? Random samples of 100 senior citizens and 100 people in their twenties were surveyed. (c) You have data on rainwater collected at 16 locations in the Adirondack Mountains of New York State. One measurement is the acidity of the ­water, measured by pH on a scale of 0 to 14 (the pH of distilled water is 7.0). Estimate the average acidity of rainwater in the Adirondacks. (d) Consumers Union wants to see which of two brands of calculator is easier to use. They recruit 100 volunteers and randomly assign them to two equal-sized groups. The people in one group use Calculator A and those in the other group use Calculator B. Researchers record the time ­required for each volunteer to carry out the same series of routine calculations (such as figuring discounts and sales tax, totaling a bill) on the ­assigned calculator. R10.2 Seat belt use  The proportion of drivers who use seat belts depends on things like age (young people are more likely to go unbelted) and gender (women are more likely to use belts). It also depends on local law. In New York City, police can stop a driver who is not belted. In Boston at the time of the study, police could cite a driver for not wearing a seat belt only if the driver had been stopped for some other violation. Here are data from observing random samples of female Hispanic drivers in these two cities:44 City

Drivers

Belted

New York

220

183

Boston

117

 68

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(a) Calculate the standard error of the sampling distribution of the difference in the proportions of female Hispanic drivers in the two cities who wear seat belts. What information does this value provide? (b) Construct and interpret a 95% confidence interval for the difference in the proportions of female ­Hispanic drivers in the two cities who wear seat belts. R10.3 Expensive ads  Consumers who think a product’s advertising is expensive often also think the product must be of high quality. Can other information undermine this effect? To find out, marketing researchers did an experiment. The subjects were 90 women from the clerical and administrative staff of a large organization. All subjects read an ad that described a fictional line of food products called “Five Chefs.” The ad also described the major TV commercials that would soon be shown, an unusual expense for this type of product. The 45 women who were randomly assigned to the control group read nothing else. The 45 in the “undermine group” also read a news story headlined “No Link between Advertising Spending and New Product Quality.” All the subjects then rated the quality of Five Chefs products on a 7-point scale. The study report said, “The mean quality ratings were significantly lower in the undermine treatment (x–A = 4.56) than in the control treatment (x–C = 5.05; t = 2.64, P < 0.01).”45 (a) The 90 women who participated in the study were not randomly selected from a population. Explain why the Random condition is still satisfied. (b) The distribution of individual responses is not Normal, because there is only a 7-point scale. What is the shape of the sampling distribution of x–C − x–A ? Explain. (c) Interpret the P-value in context. R10.4 Men versus women  The National Assessment of Educational Progress (NAEP) Young Adult Literacy Assessment Survey interviewed a random sample of 1917 people 21 to 25 years old. The sample contained 840 men and 1077 women.46 The mean and standard deviation of scores on the NAEP’s test of quantitative skills were x–1 = 272.40 and s1 = 59.2 for the men in the sample. For the women, the results were x–2 = 274.73 and s2 = 57.5. (a) Construct and interpret a 90% confidence interval for the difference in mean score for male and female young adults. (b) Based only on the interval from part (a), is there convincing evidence of a difference in mean score for male and female young adults?

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Chapter Review Exercises R10.5 Treating AIDS  The drug AZT was the first drug that seemed effective in delaying the onset of AIDS. Evidence for AZT’s effectiveness came from a large randomized comparative experiment. The subjects were 870 volunteers who were infected with HIV, the virus that causes AIDS, but did not yet have AIDS. The study assigned 435 of the subjects at random to take 500 milligrams of AZT each day and another 435 to take a placebo. At the end of the study, 38 of the placebo subjects and 17 of the AZT subjects had developed AIDS. (a) Do the data provide convincing evidence at the a = 0.05 level that taking AZT lowers the proportion of infected people who will develop AIDS in a given period of time? (b) Describe a Type I error and a Type II error in this setting and give a consequence of each error. Based on your conclusion in part (a), which error could have been made in this study? R10.6 Conditions  Explain why it is not safe to use the methods of this chapter to perform inference in each of the following settings. (a) Lyme disease is spread in the northeastern United States by infected ticks. The ticks are infected mainly by feeding on mice, so more mice result in more infected ticks. The mouse population in turn rises and falls with the abundance of acorns, their favored food. Experimenters studied two similar forest areas in a year when the acorn crop failed. They added hundreds of thousands of acorns to one area to imitate an abundant acorn crop, while leaving the other area untouched. The next spring, 54 of the 72 mice trapped in the first area were in breeding condition, versus 10 of the 17 mice trapped in the second area.47 (b) Who texts more—males or females? For their ­final project, a group of AP® Statistics students ­investigated their belief that females text more than males. They asked a random sample of 31 students—15 males and 16 females—from their school to record the number of text messages sent and received over a 2-day period. Boxplots of their data are shown below.

R10.7 Each day I am getting better in math  A “subliminal” message is below our threshold of awareness but may nonetheless influence us. Can subliminal messages help students learn math? A group of 18 students who had failed the mathematics part of the City University of New York Skills Assessment Test agreed to participate in a study to find out. All received a daily subliminal message, flashed on a screen too rapidly to be consciously read. The treatment group of 10 students (assigned at random) was exposed to “Each day I am getting better in math.” The control group of 8 students was exposed to a neutral message, “People are walking on the street.” All 18 students participated in a summer program designed to raise their math skills, and all took the assessment test again at the end of the program. The table below gives data on the subjects’ scores before and after the program.48 Treatment Group

Control Group

Pretest Posttest Difference Pretest Posttest Difference 18 18 21 18 18 20 23 23 21 17

24 25 33 29 33 36 34 36 34 27

 6  7 12 11 15 16 11 13 13 10

18 24 20 18 24 22 15 19

29 29 24 26 38 27 22 31

11  5  4  8 14  5  7 12

(a) Explain why a two-sample t test is more appropriate than a paired t test for analyzing these data. (b) The Fathom boxplots below display the differences in pretest and posttest scores for the students in the control (Cdiff) and treatment (Tdiff) groups. Write a few sentences comparing the performance of these two groups.

Males 0 6 28

83 127

213 214

Females 0

34 0

107 100

191 200

379 300

Number of text messages in 2-day period

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400

(c) Do the data provide convincing evidence that subliminal messages help students learn math? (d) Can we generalize these results to the population of all students who failed the mathematics part of the City University of New York Skills Assessment Test? Why or why not?

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Chapter 10 AP® Statistics Practice Test Section I: Multiple Choice  ​Select the best answer for each question. T10.1 A study of road rage asked separate random samples of 596 men and 523 women about their behavior while driving. Based on their answers, each respondent was assigned a road rage score on a scale of 0 to 20. Are the conditions for performing a two-­ sample t test satisfied? (a) Maybe; we have independent random samples, but we need to look at the data to check Normality. (b) No; road rage scores in a range between 0 and 20 can’t be Normal. (c) No; we don’t know the population standard ­deviations. (d) Yes; the large sample sizes guarantee that the corresponding population distributions will be Normal. (e) Yes; we have two independent random samples and large sample sizes. T10.2 Thirty-five people from a random sample of 125 workers from Company A admitted to using sick leave when they weren’t really ill. Seventeen employees from a random sample of 68 workers from Company B admitted that they had used sick leave when they weren’t ill. A 95% confidence interval for the difference in the proportions of workers at the two companies who would admit to using sick leave when they weren’t ill is 0.03 ± (a)

(0.28)(0.72) (0.25)(0.75) + Å 125 68

(b) 0.03 ± 1.96 Å

(0.28)(0.72) (0.25)(0.75) + 125 68

(c) 0.03 ± 1.645 Å (d) 0.03 ± 1.96 Å

(0.28)(0.72) (0.25)(0.75) + 125 68

(0.269)(0.731) (0.269)(0.731) + 125 68

(e) 0.03 ± 1.645 Å

(0.269)(0.731) (0.269)(0.731) + 125 68

T10.3 The power takeoff driveline on tractors used in agriculture is a potentially serious hazard to operators of farm equipment. The driveline is covered by a shield in new tractors, but for a variety of reasons, the shield is often missing on older tractors. Two types of shields are the bolt-on and the flip-up. It was believed that the bolt-on shield was perceived

as a nuisance by the operators and deliberately removed, but the flip-up shield is easily lifted for inspection and maintenance and may be left in place. In a study initiated by the U.S. National Safety Council, random samples of older tractors with both types of shields were taken to see what proportion of shields were removed. Of 183 tractors designed to have bolt-on shields, 35 had been removed. Of the 136 tractors with flip-up shields, 15 were removed. We wish to perform a test of H0: pb = pf versus Ha: pb > pf, where pb and pf are the proportions of all tractors with the bolt-on and flip-up shields removed, respectively. Which of the following is not a condition for performing the significance test? (a) Both populations are Normally distributed. (b) The data come from two independent samples. (c) Both samples were chosen at random. (d) The counts of successes and failures are large enough to use Normal calculations. (e) Both populations are at least 10 times the corresponding sample sizes. T10.4 A quiz question gives random samples of n = 10 observations from each of two Normally distributed populations. Tom uses a table of t distribution critical values and 9 degrees of freedom to calculate a 95% confidence interval for the difference in the two population means. Janelle uses her calculator’s two-sample t interval with 16.87 degrees of freedom to compute the 95% confidence interval. Assume that both students calculate the intervals correctly. Which of the following is true? (a) Tom’s confidence interval is wider. (b) Janelle’s confidence interval is wider. (c) Both confidence intervals are the same. (d) There is insufficient information to determine which confidence interval is wider. (e) Janelle made a mistake; degrees of freedom has to be a whole number. Exercises T10.5 and T10.6 refer to the following setting. A researcher wished to compare the average amount of time spent in extracurricular activities by high school students in a suburban school district with that in a school district of a large city. The researcher obtained an SRS of 60 high school students in a large suburban school district and found the mean time spent in extracurricular activities per week

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AP® Statistics Practice Test to be 6 hours with a standard deviation of 3 hours. The researcher also obtained an independent SRS of 40 high school students in a large city school district and found the mean time spent in extracurricular activities per week to be 5 hours with a standard deviation of 2 hours. Suppose that the researcher decides to carry out a significance test of H0: msuburban = mcity versus a two-sided alternative. T10.5 The correct test statistic is z= (a)

(b) z=

(6 − 5) − 0 3 2 + Å 60 40 (6 − 5) − 0

32 22 + Å 60 40 (6 − 5) − 0 t= (c) 3 2 + "60 "40 (d) t=

(e) t=

(6 − 5) − 0 3 2 + Å 60 40 (6 − 5) − 0 32 22 + Å 60 40

T10.6 The P-value for the test is 0.048. A correct conclusion is to (a) fail to reject H0 at the a = 0.05 level. There is convincing evidence of a difference in the average time spent on extracurricular activities by students in the suburban and city school districts. (b) fail to reject H0 at the a = 0.05 level. There is not convincing evidence of a difference in the average time spent on extracurricular activities by students in the suburban and city school districts. (c) fail to reject H0 at the a = 0.05 level. There is convincing evidence that the average time spent on extracurricular activities by students in the suburban and city school districts is the same. (d) reject H0 at the a = 0.05 level. There is not convincing evidence of a difference in the average time spent on extracurricular activities by students in the suburban and city school districts. (e) reject H0 at the a = 0.05 level. There is convincing evidence of a difference in the average time spent on extracurricular activities by students in the suburban and city school districts. T10.7 At a baseball game, 42 of 65 randomly selected people own an iPod. At a rock concert occurring at the same time across town, 34 of 52 randomly selected people own an iPod. A researcher wants

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to test the claim that the proportion of iPod owners at the two venues is different. A 90% confidence interval for the difference in population proportions (game − concert) is (−0.154, 0.138). Which of the following gives the correct outcome of the researcher’s test of the claim? (a) Because the confidence interval includes 0, the researcher can conclude that the proportion of iPod owners at the two venues is the same. (b) Because the center of the interval is –0.008, the researcher can conclude that a higher proportion of people at the rock concert own iPods than at the baseball game. (c) Because the confidence interval includes 0, the researcher cannot conclude that the proportion of iPod owners at the two venues is different. (d) Because the confidence interval includes more negative than positive values, the researcher can conclude that a higher proportion of people at the rock concert own iPods than at the baseball game. (e) ​The researcher cannot draw a conclusion about a claim without performing a significance test. T10.8 An SRS of size 100 is taken from Population A with proportion 0.8 of successes. An independent SRS of size 400 is taken from Population B with proportion 0.5 of successes. The sampling distribution for the difference (Population A – ­Population B) in sample proportions has what mean and standard deviation? (a) mean = 0.3; standard deviation = 1.3 (b) mean = 0.3; standard deviation = 0.40 (c) mean = 0.3; standard deviation = 0.047 (d) mean = 0.3; standard deviation = 0.0022 (e) mean = 0.3; standard deviation = 0.0002 T10.9 How much more effective is exercise and drug treatment than drug treatment alone at reducing the rate of heart attacks among men aged 65 and older? To find out, researchers perform a completely randomized experiment involving 1000 healthy males in this age group. Half of the subjects are assigned to receive drug treatment only, while the other half are assigned to exercise regularly and to receive drug treatment. The most appropriate inference method for answering the original research question is (a) one-sample z test for a proportion. (b) two-sample z interval for p1 − p2. (c) two-sample z test for p1 − p2. (d) two-sample t interval for m1 − m2. (e) two-sample t test for m1 − m2.

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T10.10 Researchers are interested in evaluating the effect of a natural product on reducing blood pressure. This will be done by comparing the mean reduction in blood pressure of a treatment (natural product) group and a placebo group using a two-sample t test. The researchers would like to be able to detect whether the natural product reduces blood pressure by at least 7 points more, on average, than the placebo. If groups of size 50 are used in the experiment, a two-sample t test using a = 0.01 will have a power of 80% to detect a 7-point difference in mean blood pressure

r­ eduction. If the researchers want to be able to detect a 5-point difference instead, then the power of the test (a) would be less than 80%. (b) would be greater than 80%. (c) would still be 80%. (d) could be either less than or greater than 80%, depending on whether the natural product is effective. (e) would vary depending on the standard deviation of the data.

Section II: Free Response  ​Show all your work. Indicate clearly the methods you use, because you will be graded on the correctness of your methods as well as on the accuracy and completeness of your results and explanations. T10.11 Researchers wondered whether maintaining a patient’s body temperature close to normal by heating the patient during surgery would affect wound infection rates. Patients were assigned at random to two groups: the normothermic group (patients’ core temperatures were maintained at near normal, 36.5°C, with heating blankets) and the hypothermic group (patients’ core temperatures were allowed to decrease to about 34.5°C). If keeping patients warm during surgery alters the chance of infection, patients in the two groups should have hospital stays of very different lengths. Here are summary statistics on hospital stay (in number of days) for the two groups: Group Normothermic Hypothermic

n

x–

sx

104  96

12.1 14.7

4.4 6.5

(a) Construct and interpret a 95% confidence interval for the difference in the true mean length of hospital stay for normothermic and hypothermic ­patients. (b) Does your interval in part (a) suggest that keeping ­patients warm during surgery affects the average length of patients’ hospital stays? Justify your ­answer. (c) Interpret the meaning of “95% confidence” in the context of this study. T10.12 A random sample of 100 of a certain popular car model last year found that 20 had a certain minor

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defect in the brakes. The car company made an adjustment in the production process to try to reduce the proportion of cars with the brake problem. A random sample of 350 of this year’s model found that 50 had the minor brake ­defect. (a) Was the company’s adjustment successful? Carry out an appropriate test to support your answer. (b) Describe a Type I error and a Type II error in this ­setting, and give a possible consequence of each. T10.13 Pat wants to compare the cost of one- and twobedroom apartments in the area of her college campus. She collects data for a random sample of 10 advertisements of each type. The table below shows the rents (in dollars per month) for the selected apartments. 1 bedroom:

500 650 600 505 450 550 515 495 650 395

2 bedroom:

595 500 580 650 675 675 750 500 495 670



Pat wonders if two-bedroom apartments rent for significantly more, on average, than one-bedroom apartments.

(a) State an appropriate pair of hypotheses for a significance test. Be sure to define any parameters you use. (b) Name the appropriate test and show that the conditions for carrying out this test are met. (c) The appropriate test from part (b) yields a P-value of 0.029. Interpret this P-value in context. (d) What conclusion should Pat draw at the a = 0.05 significance level? Explain.

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Cumulative AP® Practice Test 3

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Cumulative AP® Practice Test 3 Section I: Multiple Choice  Choose the best answer. AP3.1 Suppose the probability that a softball player gets a hit in any single at-bat is 0.300. Assuming that her chance of getting a hit on a particular time at bat is independent of her other times at bat, what is the probability that she will not get a hit until her fourth time at bat in a game? 4 (a) a b(0.3)1(0.7)3 3 4 (b) a b(0.3)3(0.7)1 3 4 (c) a b(0.3)3(0.7)1 1

(d) ​(0.3)3(0.7)1 (e) ​(0.3)1(0.7)3

AP3.6 You can find some interesting polls online. Anyone can become part of the sample just by clicking on a response. One such poll asked, “Do you prefer watching first-run movies at a movie theater, or waiting until they are available to watch at home or on a digital device?” In all, 8896 people responded, with only 12% (1118 people) saying they preferred theaters. You can conclude that (a) American adults strongly prefer watching movies at home or on their digital devices. (b) the high nonresponse rate prevents us from drawing a conclusion.

AP3.2 The probability that Color Me Dandy wins a horse race at Batavia Downs given good track conditions is 0.60. The probability of good track conditions on any given day is 0.85. What is the probability that Color Me Dandy wins or the track conditions are good? (a) 0.94  (b)  0.51  (c) ​0.49   (d) ​0.06 (e) The answer cannot be determined from the given information.

(c) the sample is too small to draw any conclusion. (d) the poll uses voluntary response, so the results tell us little about all American adults. (e) American adults strongly prefer seeing movies at a movie theater.

AP3.3 Sports Illustrated planned to ask a random sample of Division I college athletes, “Do you believe performance-enhancing drugs are a problem in college sports?” How many athletes must be interviewed to estimate the proportion concerned about use of drugs within ±2% with 90% confidence?

AP3.7 A certain candy has different wrappers for various holidays. During Holiday 1, the candy wrappers are 30% silver, 30% red, and 40% pink. During Holiday 2, the wrappers are 50% silver and 50% blue. Forty pieces of candy are randomly selected from the Holiday 1 distribution, and 40 pieces are randomly selected from the Holiday 2 distribution. What are the expected value and standard deviation of the total number of silver wrappers?

(a) 17 (b) 21

(a) 32, 18.4 (b) 32, 6.06

(c) ​1680 (d) ​1702

(e) ​2401

AP3.4 The distribution of grade point averages for a ­certain college is approximately Normal with a mean of 2.5 and a standard deviation of 0.6. Within which of the following intervals would we expect to find approximately 81.5% of all GPAs for students at this college? (a) (0.7, 3.1) (b) (1.3, 3.7)

(c) ​(1.9, 3.7) (d) ​(1.9, 4.3)

(e) ​(0.7, 4.3)

AP3.5 Which of the following will increase the power of a significance test? (a) Increase the Type II error probability. (b) Decrease the sample size. (c) Reject the null hypothesis only if the P-value is smaller than the level of significance. (d) Increase the significance level a. (e) Select a value for the alternative hypothesis closer to the value of the null hypothesis.

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(c) ​32, 4.29 (d) ​80, 18.4

(e) ​80, 4.29

AP3.8 A beef rancher randomly sampled 42 cattle from her large herd to obtain a 95% confidence interval to estimate the mean weight of the cows in the herd. The interval ­obtained was (1010, 1321). If the rancher had used a 98% confidence interval instead, the interval would have been (a) wider and would have less precision than the original estimate. (b) wider and would have more precision than the original estimate. (c) wider and would have the same precision as the original estimate. (d) narrower and would have less precision than the original estimate. (e) narrower and would have more precision than the original estimate.

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AP3.9 School A has 400 students and School B has 2700 students. A local newspaper wants to compare the distributions of SAT scores for the two schools. Which of the following would be the most useful for making this comparison? (a) Back-to-back stemplots for A and B (b) A scatterplot of A versus B (c) Dotplots for A and B drawn on the same scale (d) Two relative frequency histograms of A and B drawn on the same scale (e) Two bar graphs for A and B drawn on the same scale AP3.10 Let X represent the outcome when a fair six-sided die is rolled. For this random variable, mX = 3.5 and sX = 1.71. If the die is rolled 100 times, what is the approximate probability that the total score is at least 375? (a) 0.0000 (b) 0.0017

(c) ​0.0721 (d) ​0.4420

(e) ​0.9279

AP3.11 An agricultural station is testing the yields for six different varieties of seed corn. The station has four large fields available, which are located in four distinctly different parts of the county. The agricultural researchers consider the climatic and soil conditions in the four parts of the county as being unequal but are reasonably confident that the conditions within each field are fairly similar throughout. The researchers divide each field into six sections and then randomly assign one variety of corn seed to each section in that field. This procedure is done for each field. At the end of the growing season, the corn will be harvested, and the yield, measured in tons per acre, will be compared. Which one of the following statements about the design is correct? (a) This is an observational study because the researchers are watching the corn grow. (b) This a randomized block design with fields as blocks and seed types as treatments. (c) This is a randomized block design with seed types as blocks and fields as treatments. (d) This is a completely randomized design because the six seed types were randomly assigned to the four fields. (e) This is a completely randomized design with 24 treatments—6 seed types and 4 fields. AP3.12 The correlation between the heights of fathers and the heights of their (grownup) sons is r = 0.52, both measured in inches. If fathers’ heights were measured in feet instead, the correlation between heights of fathers and heights of sons would be

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(a) much smaller than 0.52. (b) slightly smaller than 0.52. (c) unchanged; equal to 0.52. (d) slightly larger than 0.52. (e) much larger than 0.52. AP3.13 A random sample of 200 New York State voters included 88 Republicans, while a random sample of 300 California voters produced 141 Republicans. Which of the following represents the 95% confidence interval that should be used to estimate the true difference in the proportions of Republicans in New York State and ­California? (0.44 − 0.47) ± 1.96 (a)

(0.44)(0.56) + (0.47)(0.53)

!200 + 300 (0.44)(0.56) (0.47)(0.53) (b) (0.44 − 0.47) ± 1.96 + !200 !300 (0.44 − 0.47) ± 1.96 (c)

(d) (0.44 − 0.47) ± 1.96 (e) (0.44 − 0.47) ± 1.96

(0.44)(0.56) (0.47)(0.53) + Å 200 300 (0.44)(0.56) + (0.47)(0.53) Å 200 + 300

Å

(0.45)(0.55) (0.45)(0.55) + 200 300

AP3.14 Which of the following is not a property of a ­binomial setting? (a) Outcomes of different trials are independent. (b) The chance process consists of a fixed number of trials, n. (c) The probability of success is the same for each trial. (d) Trials are repeated until a success occurs. (e) Each trial can result in either a success or a failure. AP3.15 Mrs. Woods and Mrs. Bryan are avid vegetable gardeners. They use different fertilizers, and each claims that hers is the best fertilizer to use when growing tomatoes. Both agree to do a study using the weight of their tomatoes as the response variable. They had each planted the same v­ arieties of tomatoes on the same day and fertilized the plants on the same schedule throughout the growing season. At harvest time, they each randomly select 15 t­ omatoes from their respective gardens and weigh them. After performing a two-sample t test on the difference in mean weights of t­ omatoes, they get t = 5.24 and P = 0.0008. Can the gardener with the larger mean claim that her fertilizer caused her tomatoes to be heavier? (a) Yes, because a different fertilizer was used on each ­garden.

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Cumulative AP® Practice Test 3 (b) Yes, because random samples were taken from each garden. (c) Yes, because the P-value is so small. (d) No, because the soil conditions in the two gardens is a potential confounding variable. (e) No, because there was no replication. AP3.16 The Environmental Protection Agency is charged with monitoring industrial emissions that pollute the atmosphere and water. So long as emission levels stay within specified guidelines, the EPA does not take action against the polluter. If the polluter is in violation of the regulations, the offender can be fined, forced to clean up the problem, or possibly closed. Suppose that for a particular industry the acceptable emission level has been set at no more than 5 parts per million (5 ppm). The null and alternative ­hypotheses are H0 : m = 5 versus Ha : m > 5. Which of the following describes a Type II error? (a) The EPA fails to find convincing evidence that emissions exceed acceptable limits when, in fact, they are within acceptable limits. (b) The EPA finds convincing evidence that emissions exceed acceptable limits when, in fact, they are within acceptable limits. (c) The EPA finds convincing evidence that emissions exceed acceptable limits when, in fact, they do ­exceed acceptable limits. (d) The EPA takes more samples to ensure that they make the correct decision. (e) The EPA fails to find convincing evidence that emissions exceed acceptable limits when, in fact, they do exceed acceptable limits. AP3.17 Which of the following is false? (a) A measure of center alone does not completely describe the characteristics of a set of data. Some measure of spread is also needed. (b) If the original measurements are in ­inches, converting them to centimeters will not change the mean or standard deviation. (c) One of the disadvantages of a histogram is that it doesn’t show each data value. (d) Between the range and the interquartile range, the IQR is a better measure of spread if there are ­outliers. (e) If a distribution is skewed, the median and interquartile range should be reported rather than the mean and standard deviation. AP3.18 A 96% confidence interval for the proportion of the labor force that is unemployed in a certain city

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671

is (0.07, 0.10). Which of the following statements about this interval is true? (a) The probability is 0.96 that between 7% and 10% of the labor force is unemployed. (b) About 96% of the intervals constructed by this method will contain the true proportion of unemployed in the city. (c) In repeated samples of the same size, there is a 96% chance that the sample proportion will fall between 0.07 and 0.10. (d) The true rate of unemployment lies within this interval 96% of the time. (e) Between 7% and 10% of the labor force is unemployed 96% of the time. AP3.19 A large toy company introduces a lot of new toys to its product line each year. The company wants to predict the demand as measured by y, first-year sales (in millions of dollars) using x, awareness of the product (as measured by the percent of customers who had heard of the product by the end of the second month after its introduction). A random sample of 65 new products was taken, and a correlation of 0.96 was computed. Which of the following is a correct interpretation of this value? (a) Ninety-six percent of the time, the least-squares regression line accurately predicts first-year sales. (b) About 92% of the time, the percent of people who have heard of the product by the end of the second month will correctly predict first-year sales. (c) About 92% of first-year sales can be accounted for by the percent of people who have heard of the product by the end of the second month. (d) For each increase of 1% in awareness of the new product, the predicted sales will go up by 0.96 million dollars. (e) About 92% of the variation in first-year sales can be accounted for by the least-squares regression line with percent of people who have heard of the product by the end of the second month as the explanatory variable. AP3.20 Final grades for a class are approximately Normally distributed with a mean of 76 and a standard deviation of 8. A professor says that the top 10% of the class will receive an A, the next 20% a B, the next 40% a C, the next 20% a D, and the bottom 10% an F. What is the approximate maximum grade a student could attain and still ­receive an F for the course? (a) 70 (b) 69.27

(c) ​65.75 (d) ​62.84

(e) ​57

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AP3.21 National Park rangers keep data on the bears that inhabit their park. Below is a histogram of the weights of 143 bears measured in a recent year. 35

Frequency

30 25 20 15 10 5 0 40

80 120 160 200 240 280 320 360 400 440 480 520

Weight (pounds)



Which statement below is correct?



(a) The median will lie in the interval (140, 180), and the mean will lie in the interval (180, 220).



(b) The median will lie in the interval (140, 180), and the mean will lie in the interval (260, 300).



(c) The median will lie in the interval (100, 140), and the mean will lie in the interval (180, 220).



(d) The mean will lie in the interval (140, 180), and the median will lie in the interval (260, 300).



(e) The mean will lie in the interval (100, 140), and the median will lie in the interval (180, 220).

AP3.22 A random sample of size n will be selected from a population, and the proportion of those in the sample who have a Facebook page will be calculated. How would the margin of error for a 95% confidence interval be affected if the sample size were increased from 50 to 200? (a) It remains the same. (b) It is multiplied by 2. (c) It is multiplied by 4. (d) It is divided by 2. (e) It is divided by 4. AP3.23 A scatterplot and a least-squares regression line are shown in the figure below. What effect does point P have on the slope of the regression line and the correlation?

(a) Point P increases the slope and increases the correlation. (b) Point P increases the slope and decreases the correlation. (c) Point P decreases the slope and decreases the correlation. (d) Point P decreases the slope and increases the ­correlation. (e) No conclusion can be drawn because the other coordinates are unknown. AP3.24 The following dotplots show the average high temperatures (in degrees Fahrenheit) for a sample of tourist cities from around the world. Both the January and July average high temperatures are shown. What is one statement that can be made with certainty from an analysis of the graphical display?

(a) Every city has a larger average high temperature in July than in January. (b) The distribution of temperatures in July is skewed right, while the distribution of temperatures in January is skewed left. (c) The median average high temperature for January is higher than the median average high temperature for July. (d) There appear to be outliers in the average high temperatures for January and July. (e) There is more variability in average high temperatures in January than in July. AP3.25 Suppose the null and alternative hypotheses for a significance test are defined as H0 : m = 40 Ha : m < 40

Which of the following specific values for Ha will give the highest power?

(a) m = 38 (b) m = 39

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(c) ​m = 40 (d) ​m = 41

(e) ​m = 42

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Cumulative AP® Practice Test 3 AP3.26 A large university is considering the establishment of a schoolwide recycling program. To gauge interest in the program by means of a questionnaire, the university takes separate random samples of undergraduate students, graduate students, faculty, and staff. This is an example of what type of sampling design? (a) Simple random sample (b) Stratified random sample (c) Convenience sample (d) Cluster sample (e) Randomized block design AP3.27 Suppose the true proportion of people who use public transportation to get to work in the Washington, D.C., area is 0.45. In a simple random sample of 250 people who work in Washington, about how far do you expect the sample proportion to be from the true proportion? (a) 0.4975 (b) 0.2475

(c) 0.0315 (d) ​0.0009

(e) ​0

Questions 28 and 29 refer to the following setting. According to sleep researchers, if you are between the ages of 12 and 18 years old, you need 9 hours of sleep to be fully functional. A simple random sample of 28 students was chosen from a large high school, and these students were asked how much sleep they got the previous night. The mean of the responses was 7.9 hours, with a standard deviation of 2.1 hours. AP3.28 If we are interested in whether students at this high school are getting too little sleep, which of the ­following represents the appropriate null and alternative ­hypotheses? (a) H0 : m = 7.9 and Ha : m < 7.9 (b) H0 : m = 7.9 and Ha : m ≠ 7.9 (c) H0 : m = 9 and Ha : m ≠ 9 (d) H0 : m = 9 and Ha : m < 9 (e) H0 : m ≤ 9 and Ha : m ≥ 9

673

AP3.29 Which of the following is the test statistic for the hypothesis test? 7.9 − 9 9 − 7.9 7.9 − 9   (b)  t =   (c)  t = 2.1 2.1 2.1 !28 !28 Å 28 7.9 − 9 9 − 7.9 (d) t=   (e)  t = 2.1 2.1 !27 !27 AP3.30 Shortly before the 2012 presidential election, a survey was taken by the school newspaper at a very large state university. Randomly selected students were asked, “Whom do you plan to vote for in the upcoming presidential election?” Here is a twoway table of the ­responses by political persuasion for 1850 students: (a) t=

Candidate of choice Obama Romney Other Undecided Total



Political Persuasion Republican Independent

Democrat

Total

925

 78

 26

1029

 78   2  32 1037

598   8  28 712

 19  11  45 101

 695   21  105 1850

Which of the following statements about these data is true?

(a) The percent of Republicans among the respondents is 41%. (b) The marginal distribution of the variable choice of candidate is given by Obama: 55.6%; Romney: 37.6%; Other: 1.1%; Undecided: 5.7%. (c) About 11.2% of Democrats reported that they planned to vote for Romney. (d) About 44.6% of those who are undecided are ­Independents. (e) The conditional distribution of political persuasion among those for whom Romney is the candidate of choice is Democrat: 7.5%; Republican: 84.0%; Independent: 18.8%

Section II: Free Response  Show all your work. Indicate clearly the methods you use, because you will be graded on the correctness of your methods as well as on the accuracy and completeness of your results and ­explanations. AP3.31 A researcher wants to determine whether or not a five-week crash diet is effective over a long period of time. A random sample of 15 dieters is selected. Each person’s weight is recorded before starting the diet and one year ­after it is concluded. Based on the data shown at right (weight in pounds), can we conclude that the diet has a long-term effect, that is, that dieters manage to not regain the weight they lose? Include appropriate statistical evidence to justify your answer.

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Before After

Before After

1

2

3

4

5

6

7

8

158 163

185 182

176 188

172 150

164 161

234 220

258 235

200 191

9

10

11

12

13

14

15

228 228

246 237

198 209

221 220

236 222

255 268

231 234

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C H A P T E R 1 0 C o m pa r i n g T w o P o p u l at i o n s o r G r o u p s

AP3.32 Starting in the 1970s, medical technology allowed babies with very low birth weight (VLBW, less than 1500 grams, or about 3.3 pounds) to survive without major handicaps. It was noticed that these children nonetheless had difficulties in school and as adults. A long study has followed 242 randomly selected VLBW babies to age 20 years, along with a control group of 233 randomly selected babies from the same population who had normal birth weight.49 (a) Is this an experiment or an observational study? Why? (b) At age 20, 179 of the VLBW group and 193 of the control group had graduated from high school. Is the graduation rate among the VLBW group significantly lower than for the normal-birth-weight controls? Give appropriate statistical evidence to justify your answer.

Distance to nearest fish (meters)

AP3.33 A nuclear power plant releases water into a nearby lake every afternoon at 4:51 p.m. Environmental researchers are concerned that fish are being driven away from the area around the plant. They believe that the temperature of the water discharged may be a factor. The scatterplot below shows the temperature of the water (°C) released by the plant and the measured distance (in meters) from the outflow pipe of the plant to the nearest fish found in the water on eight randomly chosen afternoons. 160 140 120 100 80 60 40 20 0 15

20

25

30

Temperature (ºC)

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35

40



Computer output from a least-squares regression analysis on these data and a residual plot are shown below. Predictor Constant Temperature

Coef SE Coef -73.64 15.48 5.7188 0.5612

T -4.76 -10.19

P 0.003 0.000

S = 11.4175 R-Sq = 94.5% R-Sq(adj) = 93.6% 20

10

Residual

674

0

−10

−20 0

20

40

60

80

100

120

140

160

Fitted value

(a) Write the equation of the least-squares regression line. Define any variables you use. (b) Interpret the slope of the regression line in context. (c) Is a linear model appropriate for describing the relationship between temperature and distance to the nearest fish? Justify your answer. (d) Compute the residual for the point (29, 78). Interpret this residual in context. AP3.34 The Candy Shoppe assembles gift boxes that ­contain 8 chocolate truffles and 2 handmade caramel nougats. The truffles have a mean weight of 2 ounces with a standard deviation of 0.5 ounce, and the nougats have a mean weight of 4 ounces with a standard deviation of 1 ounce. The empty boxes weigh 3 ounces with a standard deviation of 0.2 ounce.

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Cumulative AP® Practice Test 3 (a) Assuming that the weights of the truffles, nougats, and boxes are independent, what are the mean and standard deviation of the weight of a box of candy? (b) Assuming that the weights of the truffles, nougats, and boxes are approximately Normally distributed, what is the probability that a randomly selected box of candy will weigh more than 30 ounces? (c) If five gift boxes are randomly selected, what is the probability that at least one of them will weigh more than 30 ounces? (d) If five gift boxes are randomly selected, what is the probability that the mean weight of the five boxes will be more than 30 ounces? AP3.35 An investor is comparing two stocks, A and B. She wants to know if over the long run, there is a significant difference in the return on investment as measured by the percent increase or decrease in the price of the stock from its date of purchase. The investor takes a random sample of 50 annualized daily returns over the past five years for each stock. The data are summarized below. Stock A B

Mean return 11.8% 7.1%

tions in Stock A significantly exceed those of Stock B, as measured by their standard deviations? Identify an appropriate set of hypotheses that the investor is interested in testing. (c) To measure this, we will construct a test statistic ­defined as F=

large sample variance smaller sample variance

What value(s) of the statistic would indicate that the price fluctuations in Stock A significantly exceed those of Stock B? Explain.

(d) Calculate the value of the F statistic using the information given in the table. (e) Two hundred simulated values of this test statistic, F, were calculated assuming no difference in the standard ­deviations of the returns for the two stocks. The results of the simulation are displayed in the following dotplot.

Standard deviation 12.9% 9.6%

(a) Is there a significant difference in the mean return on investment for the two stocks? Support your answer with appropriate statistical evidence. Use a 5% significance level. (b) The investor believes that although the return on investment for Stock A usually exceeds that of Stock B, Stock A represents a riskier investment, where the risk is measured by the price volatility of the stock. The standard deviation is a statistical measure of the price volatility and indicates how much an investment’s actual performance during a specified period varies from its average performance over a longer period. Do the price fluctua-

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675

0.6

0.9

1.2

1.5

1.8

2.1

Larger variance/smaller variance



Use these simulated values and the test ­statistic that you calculated in part (d) to determine ­whether the observed data provide convincing evidence that Stock A is a riskier investment than Stock B. ­Explain your reasoning.

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Chapter

11 Introduction 678 Section 11.1 Chi-Square Tests for Goodness of Fit

680

Section 11.2 Inference for Two-Way Tables

697

Free Response AP® Problem, Yay! 730 Chapter 11 Review 731 Chapter 11 Review Exercises 732 Chapter 11 AP® Statistics Practice Test 734

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Inference for Distributions of Categorical Data case study

Do Dogs Resemble Their Owners? Some people look a lot like their pets. Maybe they deliberately choose animals that match their ­appearance. Or maybe we’re perceiving similarities that aren’t really there. Researchers at the University of California, San Diego, decided to investigate. They designed an experiment to test whether or not dogs resemble their owners. The researchers believed that resemblance between dog and owner might differ for purebred and mixed-breed dogs. A random sample of 45 dogs and their owners was photographed separately at three dog parks. Then, researchers “constructed triads of pictures, each consisting of one owner, that owner’s dog, and one other dog photographed at the same park.” The subjects in the experiment were 28 undergraduate psychology students. Each subject was presented with the individual sets of photographs and asked to identify which dog belonged to the pictured owner. A dog was classified as resembling its owner if more than half of the 28 undergraduate students matched dog to owner.1 The table below summarizes the results. There is some support for the researchers’ belief that ­resemblance between dog and owner might differ for purebred and mixed-breed dogs. Breed status Resemblance?

Purebred dogs

Mixed-breed dogs

Resemble owner

16

 7

Don’t resemble

 9

13

Do these data provide convincing evidence of an association between dogs’ breed status and whether or not they resemble their owners? By the end of this chapter, you will have developed the tools you need to answer this question.

677

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Introduction In the previous chapter, we discussed inference procedures for ­comparing the proportion of successes for two populations or treatments. Sometimes we want to examine the distribution of a single categorical variable in a ­population. The chi-square test for goodness of fit allows us to determine whether a h ­ ypothesized distribution seems valid. This method is useful in a field like ­genetics, where the laws of probability give the expected proportion of o­ utcomes in each ­category. We can decide whether the distribution of a categorical variable differs for two or more populations or treatments using a chi-square test for homogeneity. In d ­ oing so, we will often organize our data in a two-way table. It is also possible to use the information in a two-way table to study the relationship between two categorical variables. The chi-square test for independence allows us to determine if there is convincing evidence of an association between the variables in the population at large. The methods of this chapter help us answer questions such as these: • Are the birthdays of NHL players evenly distributed throughout the year? • Does background music influence customer purchases? • Is there an association between anger level and heart disease? Of course, we have to do a careful job of describing the data before we proceed to statistical inference. In Chapter 1, we discussed graphical and numerical ­methods of data analysis for categorical variables. You may want to quickly review Section 1.1 now. Here’s an Activity that gives you a taste (pardon the pun) of what lies ahead.

Activity Materials: Large bag of M&M’S® Milk Chocolate Candies for the class; TI-83/84 or TI-89 for each team of 3 to 4 students

The Candy Man Can Mars, Incorporated, which is headquartered in McLean, Virginia, makes milk chocolate candies. Here’s what the company’s Consumer Affairs Department says about the color distribution of its M&M’S Milk Chocolate Candies: On average, M&M’S Milk Chocolate Candies will contain 13 percent of each of browns and reds, 14 percent yellows, 16 percent greens, 20 percent oranges and 24 percent blues. The purpose of this activity is to determine whether the company’s claim is ­believable. 1.  Your teacher will take a random sample of 60 M&M’S from a large bag and give one or more pieces of candy to each student. As a class, count the number of M&M’S® Chocolate Candies of each color. Make a table on the board that summarizes these observed counts. 2.  How can you tell if the sample data give convincing evidence to dispute the company’s claim? Each team of three or four students should discuss this question and devise a formula for a test statistic that measures the difference between the observed and expected color distributions. The test statistic should yield a

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679

single number when the observed and expected values are plugged in. Here are some questions for your team to consider: •  Should we look at the difference between the observed and expected proportions in each color category or between the observed and expected counts in each category? •  Should we use the differences themselves, the absolute value of the differences, or the square of the differences? •  Should we divide each difference value by the sample size, expected count, or nothing at all? 3.  Each team will share its proposed test statistic with the class. Your teacher will then reveal how the chi-square statistic c2 is calculated. 4.  Discuss as a class: If your sample is consistent with the company’s claimed distribution of M&M’S® Chocolate Candies colors, will the value of c2 be large or small? If your sample is not consistent with the company’s claimed color distribution, will the value of c2 be large or small? 5.  Compute the value of the chi-square test statistic for the class’s data. Is this value large or small? To find out, you and your classmates will perform a ­simulation. 6.  Suppose that the company’s claimed color distribution is correct. We’ll use numerical labels from 1 to 100 to represent the color of a randomly chosen M&M’S Milk Chocolate Candy: 1−13 = brown 14−26 = red  27−40 = yellow  41−56 = green  57−76 = orange 77−100 = blue Use the calculator command below to simulate choosing a random sample of 60 candies. TI-83/84: RandInt(1,100,60) S L1 TI-89: tistat.randint(1,100,60) S list1 Sort the list in ascending order. Then record the observed counts in each color category and compute the value of c2 for your simulated sample. 7.  Your teacher will draw and label axes for a class dotplot. Plot the result you got in Step 6 on the graph. 8.  Repeat Steps 6 and 7 if needed to get a total of at least 40 repetitions of the simulation for your class. 9.  Based on the class’s simulation results, how surprising would it be to get a c2 -value as large as or larger than the one you did in Step 5 by chance alone when sampling from the claimed distribution? What conclusion would you draw?

Here is an example of what the class dotplot in the Activity might look like after 100 trials. The graph shows what values of the chi-square statistic are likely to occur by chance alone when sampling from the company’s claimed M&M’S® Chocolate Candies color distribution. Where did your class’s c2-value fall? You will learn more about the sampling distribution of the chi-square statistic shortly.

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11.1 What You Will Learn

Chi-Square Tests for Goodness of Fit By the end of the section, you should be able to:

• State appropriate hypotheses and compute expected counts for a chi-square test for goodness of fit. • Calculate the chi-square statistic, degrees of freedom, and P-value for a chi-square test for goodness of fit.

• Perform a chi-square test for goodness of fit. • Conduct a follow-up analysis when the results of a ­chi-square test are statistically significant.

Jerome’s class did the Candy Man Can Activity. The one-way table below ­summarizes the data from the class’s sample of M&M’S® Milk Chocolate Candies. In general, one-way tables display the distribution of a single categorical variable for the individuals in a sample. Color:

Blue

Orange

Green

Yellow

Red

Brown

Total

Count:

9

8

12

15

10

6

60

9 = 0.15. Because the c­ ompany 60 claims that 24% of all M&M’S Milk Chocolate Candies are blue, Jerome might believe that something fishy is going on. We could use the one-sample z test for a proportion from Chapter 9 to test the hypotheses The sample proportion of blue M&M’S is p^ =

Note that the correct alternative hypothesis Ha is two-sided. A sample proportion of blue M&M’S much higher or much lower than 0.24 would give Jerome reason to be suspicious about the company’s claim. It’s not appropriate to adjust Ha after looking at the sample data!

H0: p = 0.24 Ha: p ≠ 0.24 where p is the true population proportion of blue M&M’S® Chocolate Candies. We could then ­perform additional significance tests for each of the remaining colors. Not only would this method be fairly inefficient, it would also lead to the problem of multiple comparisons, which we’ll discuss in Section 11.2. More important, this approach wouldn’t tell us how likely it is to get a random sample of 60 candies with a color distribution that differs as much from the one claimed by the company as the class’s sample does, taking all the colors into consideration at one time. For that, we need a new kind of significance test, called a chi-square test for goodness of fit.

Comparing Observed and Expected Counts: The Chi-Square Statistic As with any test, we begin by stating hypotheses. The null hypothesis in a chisquare test for goodness of fit should state a claim about the distribution of a single categorical variable in the population of interest. In the case of the ­Candy Man Can Activity, the categorical variable we’re measuring is color and the

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Section 11.1 Chi-Square Tests for Goodness of Fit

­ opulation of interest is all M&M’S® Milk Chocolate Candies. The appropriate p null ­hypothesis is H0: The company’s stated color distribution for all   M&M’S Milk Chocolate Candies is correct. The alternative hypothesis in a chi-square test for goodness of fit is that the c­ ategorical variable does not have the specified distribution. For the M&M’S, our alternative hypothesis is Ha: The company’s stated color distribution for all    M&M’S Milk Chocolate Candies is not correct.

THINK ABOUT IT

Why did we state hypotheses in words for a chi-square test for goodness of fit?  We can also write the hypotheses in symbols as H0: pblue = 0.24, porange = 0.20, pgreen = 0.16,   pyellow = 0.14, pred = 0.13, pbrown = 0.13, Ha:  At least two of the pi’s are incorrect

c

!

n

where pcolor = the true population proportion of M&M’S Milk Chocolate Candies of that color. Why don’t we write the alternative hypothesis as Ha: At least one of the pi’s is incorrect? If the stated proportion in one category is wrong, then the stated proportion in at least one other category must be wrong because the sum of the pi’s must be 1. Don’t state the alternative hypothesis in a way that suggests that all the autio proportions in the hypothesized distribution are wrong. For instance, it would be incorrect to write Ha: pblue ≠ 0.24, porange ≠ 0.20, pgreen ≠ 0.16, pyellow ≠ 0.14, pred ≠ 0.13, pbrown ≠ 0.13 The idea of the chi-square test for goodness of fit is this: we compare the observed counts from our sample with the counts that would be expected if H0 is true. (Remember: we always assume that H0 is true when ­performing a significance test.) The more the observed counts differ from the expected counts, the more evidence we have against the null hypothesis. In general, the expected counts can be obtained by multiplying the sample size by the proportion in each category according to the null hypothesis. Here’s an example that illustrates the process.

Example

Return of the M&M’S® Chocolate Candies Computing expected counts Problem:  Jerome’s class collected data from a random sample of 60 M&M’S Milk Chocolate Candies. ­Calculate the expected counts for each color. Show your work. Solution:  Assuming that the color distribution stated by Mars, Inc., is true, 24% of all M&M’S

Milk Chocolate Candies produced are blue. For random samples of 60 candies, the average number of blue M&M’S should be (60)(0.24) = 14.40. This is our expected count of blue M&M’S® Chocolate Candies. Using this same method, we find the expected counts for the other color categories:

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Orange: (60)(0.20) = 12.00

Color

Observed

Expected

Green: (60)(0.16) = 9.60

Blue

9

14.40

Orange

8

12.00

Green

12

9.60

Yellow

15

8.40

Red

10

7.80

6

7.80

Yellow: (60)(0.14) = 8.40 Red: (60)(0.13) = 7.80 Brown: (60)(0.13) = 7.80

Brown

For Practice Try Exercise

1

Count

Did you notice that the expected count sounds a lot like the expected value of a random variable from Chapter 6? That’s no coincidence. The number of M&M’S® Chocolate Candies of a specific color in a random sample of 60 candies is a binomial random variable. Its expected value is np, the average number of candies of this color in many samples of 60 M&M’S Milk Chocolate Candies. The expected value is not likely to be a whole number. To see if the data give convincing evidence for the alternative hypothesis, we compare the observed counts from our 16 sample with the expected counts. If the observed counts are 14 Observed far from the expected counts, that’s the evidence we were seekExpected 12 ing. The table in the example gives the observed and expected 10 counts for the sample of 60 M&M’S in Jerome’s class. Figure 8 11.1 shows a side-by-side bar graph comparing the observed 6 and expected counts. 4 We see some fairly large differences between the observed 2 and expected counts in several color categories. How likely is it 0 that differences this large or larger would occur just by chance Blue Orange Green Yellow Red Brown in random samples of size 60 from the population distribution Color FIGURE 11.1  ​Bar graph comparing observed and ­expected claimed by Mars, Inc.? To answer this question, we calculate a statistic that measures how far apart the observed and expected counts for Jerome’s class sample of 60 M&M’S® Milk counts are. The statistic we use to make the comparison is the Chocolate Candies. chi-square statistic (Observed − Expected)2 c =∙ Expected 2

(The symbol c is the lowercase Greek letter chi, pronounced “kye.”)

Definition: Chi-square statistic The chi-square statistic is a measure of how far the observed counts are from the expected counts. The formula for the statistic is c2 =

∙

(Observed − Expected)2 Expected

where the sum is over all possible values of the categorical variable.

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Section 11.1 Chi-Square Tests for Goodness of Fit

Let’s use this formula to compare the observed and expected counts for Jerome’s class sample.

Example

Return of the M&M’S® Chocolate Candies Calculating the chi-square statistic The table shows the observed and expected counts for the random sample of 60 M&M’S Milk Chocolate Candies in Jerome’s class.

Color

Observed

Expected

Blue

9

14.40

Orange

8

12.00

Problem:  Calculate the chi-square statistic. Solution:  The formula for the chi-square statistic is

Green

12

9.60

Yellow

15

8.40

Red

10

7.80

6

7.80

2

c =

∙

(Observed − Expected)2

Brown

Expected

For Jerome’s data, we add six terms—one for each color category: c2 =

(9 − 14.40)2 (8 − 12.00)2 (12 − 9.60)2 + + 14.40 12.00 9.60 +

(15 − 8.40)2 (10 − 7.80)2 (6 − 7.80)2 + + 8.40 7.80 7.80

= 2.025 + 1.333 + 0.600 + 5.186 + 0.621 + 0.415 = 10.180 For Practice Try Exercise

THINK ABOUT IT

3

Why do we divide by the expected count when calculating the chi-square statistic?  ​Suppose you obtain a random sample of

60 M&M’S Milk Chocolate Candies. Which would be more surprising: getting 18 blue candies or 12 yellow candies in the sample? Earlier, we computed the expected counts for these two categories as 14.4 and 8.4, respectively. The difference in the observed and expected counts for the two colors would be Blue: 18 − 14.4 = 3.6   Yellow: 12 − 8.4 = 3.6 In both cases, the number of M&M’S® Chocolate Candies in the sample exceeds the expected count by the same amount. But it’s much more surprising to be off by 3.6 out of an expected 8.4 yellow candies (almost a 50% discrepancy) than to be off by 3.6 out of an expected 14.4 blue candies (a 25% discrepancy). For that reason, we want the category with a larger relative difference to contribute more heavily to the evidence against H0 and in favor of Ha measured by the c2 statistic. If we just computed (Observed − Expected)2 for each category instead, the contributions of these two color categories would be the same: Blue: (18 − 14.40)2 = 12.96   Yellow: (12 − 8.40)2 = 12.96

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C H A P T E R 1 1 I n f e r e n c e f o r D i s t r i b u t i o n s o f C at e g o r i c a l Data (Observed − Expected)2 , we guarantee that the color category with the Expected larger relative difference will contribute more heavily to the total:

By using

Blue:

(18 − 14.40)2 (12 − 8.40)2 = 0.90  Yellow: = 1.54 14.40 8.40

Think of c2 as a measure of the distance of the observed counts from the expected counts. Like any distance, it is always zero or positive, and it is zero only when the observed counts are exactly equal to the expected counts. Large values of c2 are stronger evidence for Ha because they say that the observed counts are far from what we would expect if H0 were true. Small values of c2 suggest that the data are consistent with the null hypothesis. Is c2 = 10.180 a large value? You know the drill: compare the observed value 10.180 against the sampling distribution that shows how c2 would vary in repeated random sampling if the null hypothesis were true.

Check Your Understanding

Mars, Inc., reports that their M&M’S® Peanut Chocolate Candies are produced according to the following color distribution: 23% each of blue and orange, 15% each of green and yellow, and 12% each of red and brown. Joey bought a randomly selected bag of Peanut Chocolate Candies and counted the colors of the candies in his sample: 12 blue, 7 orange, 13 green, 4 yellow, 8 red, and 2 brown. 1. State appropriate hypotheses for testing the company’s claim about the color distribution of M&M’S Peanut Chocolate Candies. 2. Calculate the expected count for each color, assuming that the company’s claim is true. Show your work. 3. Calculate the chi-square statistic for Joey’s sample. Show your work.

The Chi-Square Distributions and P-Values We used Fathom software to simulate taking 500 random samples of size 60 from the population distribution of M&M’S Milk Chocolate Candies given by Mars, Inc. Figure 11.2 shows a dotplot of the values of the chi-square statistic for these 500 samples. The blue vertical line is plotted at the value of c2 = 10.180 from Jerome’s class data. Recall that larger values of c2 give more convincing evidence against H0 and in favor of Ha. According to Fathom, 37 of the 500 simulated samples resulted in a chi-square statistic of 10.180 or higher. Our estimated P-value is 37/500 = 0.074. Because the P-value exceeds the default a = 0.05 significance level, we fail to reject H0. We do not have convincFIGURE 11.2  ​Fathom dotplot showing values ing evidence that the company’s claimed color distribution is incorrect. of the chi-square statistic in 500 simulated As Figure 11.2 suggests, the sampling distribution of the chi-square samples of size n = 60 from the population ® statistic is not a Normal distribution. It is a right-skewed distribution distribution of M&M’S Milk Chocolate Candies that allows only nonnegative values because c2 can never be negative. stated by the company.

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685

The sampling distribution of c2 differs depending on the number of possible values for the categorical variable (that is, on the number of categories). When the expected counts are all at least 5, the sampling distribution of the c2 statistic is modeled well by a chi-square distribution with degrees of freedom (df) equal to the number of categories minus 1. As with the t distributions, there is a different chi-square distribution for each possible value of df. Here are the facts.

The Chi-Square Distributions The chi-square distributions are a family of density curves that take only nonnegative values and are skewed to the right. A particular chi-square distribution is specified by giving its degrees of freedom. The chi-square test for goodness of fit uses the chi-square distribution with degrees of freedom = the number of categories − 1. Figure 11.3 shows the density curves for three members of the chi-square family of distributions. As the degrees of freedom (df) increase, the density curves become less skewed, and df = 4 larger values become more probable. Here are two other interesting facts about the chi-square distributions: df = 8 • The mean of a particular chi-square distribution is equal to its degrees of f­reedom. • For df > 2, the mode (peak) of the chi-square density curve is at df − 2. 0 2 4 6 8 10 12 14 16 Chi-square When df = 8, for example, the chi-square distribution has a FIGURE 11.3  ​The density curves for three members of the mean of 8 and a mode of 6. chi-square family of distributions. To get P-values from a chi-square distribution, we can use technology or Table C in the back of the book. The following example shows how to use the table. df = 1

Example

Return of the M&M’S® Chocolate Candies Finding the P-value

Chi-square distribution with df = 5

In the last example, we computed the chi-square statistic for the random sample of 60 M&M’S Milk Chocolate Candies in Jerome’s class: c2 = 10.180. Now let’s find the P-value. Because all the expected counts are at least 5, the c2 statistic will be modeled well by a chi-square distribution when H0 is true. There are 6 color categories for M&M’S Milk Chocolate Candies, so df = 6 − 1 = 5.

P -value

0

2

4

6

8

10 2

12

14

16

18

20

= 10.180

FIGURE 11.4  ​The P-value for a chi-square test for goodness of fit using Jerome’s M&M’S® Chocolate Candies class data.

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The P-value is the probability of getting a value of c2 as large as or larger than 10.180 when H0 is true. Figure 11.4 shows this probability as an area under the chi-square density curve with 5 degrees of freedom.

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To find the P-value using Table C, look in the df = 5 row. The value c2 = 10.180 falls between the critical values 9.24 and 11.07. The corresponding areas in the right tail of the chi-square distribution with 5 degrees of freedom are 0.10 and 0.05. (See the excerpt from Table C on the right.) So the P-value for a test based on Jerome’s data is between 0.05 and 0.10.

P df

.15

.10

.05

4

6.74

7.78

9.49

5

8.12

9.24

11.07

6

9.45

10.64

12.59

Now let’s look at how to find the P-value with your calculator.

25. TECHNOLOGY CORNER

Finding P-values for chi-square tests on the calculator TI-Nspire instructions in Appendix B; HP Prime instructions on the book’s Web site.

To find the P-value in the M&M’S® example with your calculator, use the c2cdf(Chi-square Cdf on the TI-89) command. We ask for the area between c2 = 10.180 and a very large number (we’ll use 10,000) under the chi-square density curve with 5 degrees of freedom. TI-83/84 TI-89 2 • Press 2nd VARS (DISTR) and choose c cdf(. • In the Stats/List Editor, Press F5 (Distr) and choose Chi-square Cdf... . OS 2.55 or later: In the dialog box, enter these values: lower:10.18, upper:10000, df:5, choose • In the dialog box, enter these values: Lower Paste, and then press ENTER . Older OS: Complete ­value:10.18, Upper value:10000, Deg of the command c2cdf(10.180,10000,5) and press Freedom,df:5, and then choose ENTER . ENTER .





As the calculator screen shots show, this method gives a more precise P-value than Table C.

Table C gives us an interval in which the P-value falls. The calculator’s c2cdf (Chi-square Cdf on the TI-89) command gives a result that is consistent with Table C but more precise. For that reason, we recommend using your calculator to compute P-values from a chi-square distribution. Based on Jerome’s sample, what conclusion can we draw about H0: the company’s stated color distribution for all M&M’S® Milk Chocolate Candies is correct? Because our P-value of 0.07 is greater than a = 0.05, we fail to reject H0. We don’t have convincing evidence that the company’s claimed color distribution is incorrect.

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687 autio

!

n

Failing to reject H0 does not mean that the null hypothesis is true! That is, we can’t conclude that the color distribution claimed by Mars, Inc., is correct. All we can say is that the sample data did not provide convincing evidence to reject H0.

c

Section 11.1 Chi-Square Tests for Goodness of Fit

Check Your Understanding

Let’s continue our analysis of Joey’s sample of M&M’S® Peanut Chocolate Candies from the previous Check Your Understanding (page 684). 1. Confirm that the expected counts are large enough to use a chi-square distribution. Which distribution (specify the degrees of freedom) should we use? 2. Sketch a graph like Figure 11.4 on page 685 that shows the P-value. 3. Use Table C to find the P-value. Then use your calculator’s c2cdf ­command. 4. What conclusion would you draw about the company’s claimed color distribution for M&M’S® Peanut Chocolate Candies? Justify your answer.

Carrying Out a Test Like our test for a population proportion, the chi-square test for goodness of fit uses some approximations that become more accurate as we take larger samples. The Large Counts condition says that all expected counts must be at least 5. Before performing a test, we must also check that the Random and 10% conditions are met.

Conditions for Performing a Chi-Square Test for Goodness of Fit • Random: The data come from a well-designed random sample or randomized experiment. 1 ~~ 10%: When sampling without replacement, check that n ≤ N. 10 • Large Counts: All expected counts are at least 5.

c

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!

n

Before we start using the chi-square test for goodness of fit, we have two important cautions to offer. 1. The chi-square test statistic compares observed and expected counts. autio Don’t try to perform calculations with the observed and expected proportions in each category. 2. When checking the Large Counts condition, be sure to examine the expected counts, not the observed counts.

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We can also write these hypotheses symbolically using pi to represent the proportion of individuals in the population that fall in category i : H0: p1 = ___, p2 = ___, …, pk = ___. Ha: At least two of the p i’s are incorrect.

The Chi-Square Test for Goodness of Fit Suppose the conditions are met. To determine whether a categorical variable has a specified distribution in the population of interest, expressed as the proportion of individuals falling into each possible category, perform a test of H0: The stated distribution of the categorical variable in the population of interest is correct. Ha: The stated distribution of the categorical variable in the population of interest is not correct. Start by finding the expected count for each category assuming that H0 is true. Then calculate the chi-square statistic c2 =

P-value

χ2

(Observed − Expected)2 ∙ Expected

where the sum is over the k different categories. The P-value is the area to the right of c2 under the density curve of the chi-square distribution with k − 1 degrees of freedom.

The next example shows the chi-square test for goodness of fit in action. As always, we follow the four-step process when performing inference.

Example

Birthdays and Hockey

STEP

A test for equal proportions

4

In his book Outliers, Malcolm Gladwell suggests that a hockey player’s birth month has a big influence on his chance to make it to the highest levels of the game. Specifically, because ­January 1 is the cut-off date for youth leagues in Canada (where many ­ National Hockey League (NHL) players come from), players born in January will be competing against players up to 12 months younger. The older players tend to be bigger, stronger, and more coordinated and hence get more playing time, more coaching, and have a better chance of being successful. To see if birth date is related to success (judged by whether a player makes it into the NHL), a random sample of 80 NHL players from a recent season was selected and their birthdays were recorded. The one-way table below summarizes the data on birthdays for these 80 players: Birthday: Number of players:

Jan–Mar

Apr–Jun

Jul–Sep

Oct–Dec

32

20

16

12

Do these data provide convincing evidence that the birthdays of NHL players are not uniformly distributed throughout the year?

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Section 11.1 Chi-Square Tests for Goodness of Fit

The null hypothesis says that NHL players’ birthdays are evenly distributed across the four quarters of the year. In that case, all 4 proportions must be 1/4. So we could write the hypotheses in symbols as H0: pJan-Mar = pApr-Jun = pJul-Sep = pOct-Dec = 1/4 Ha: At least two of the proportions are not 1/4

689

State:  We want to perform a test of H0: The birthdays of all NHL players are evenly distributed across the four quarters of the year. Ha: The birthdays of all NHL players are not evenly distributed across the four quarters of the year. No significance level was specified, so we’ll use a = 0.05. Plan:  If the conditions are met, we will perform a chi-square test for goodness of fit. •  Random:  The data came from a random sample of NHL players. ºº 10%:  Because we are sampling without replacement, there must be at least 10(80) = 800 NHL players. In the season when the data were collected, there were 879 NHL players. •  Large Counts:  If birthdays are evenly distributed across the four quarters of the year, then the expected counts are all 80(1/4) = 20. These counts are all at least 5.

Do: Chi-square distribution, df = 3

0

2

•  Test statistic: c2 =

P-value

10

= 11.2

20

=

∙

(Observed − Expected)2 Expected

(32 − 20)2 (20 − 20)2 (16 − 20)2 (12 − 20)2 + + + 20 20 20 20

= 7.2 + 0 + 0.8 + 3.2 = 11.2

FIGURE 11.5  ​The P-value for the chi-square test for goodness of fit with c2 = 11.2 and df = 3.

•  P-value:  Figure 11.5 displays the P-value for this test as an area under the chi-square distribution with 4 − 1 = 3 degrees of freedom. As the excerpt at right shows, c2 = 11.2 corresponds to a P-value between 0.01 and 0.02. Using Technology:  Refer to the Technology Corner that follows p the example. The calculator’s c2 GOF-Test gives df 0.02 0.01 0.005 c2 = 11.2 and P-value = 0.011 using df = 3. 2 7.82 9.21 10.60 Conclude:  Because the P-value, 0.011, is less than 3 9.84 11.34 12.84 a = 0.05, we reject H0. We have convincing evidence that the 4 11.67 13.28 14.86 birthdays of NHL players are not evenly distributed across the four quarters of the year. For Practice Try Exercise

15

You can use your calculator to carry out the “Do” step for a chi-square test for goodness of fit. Remember that this comes with potential benefits and risks on the AP® exam.

26. TECHNOLOGY CORNER

Chi-square test for goodness of fit on the calculator TI-Nspire instructions in Appendix B; HP Prime instructions on the book’s Web site.

You can use the TI-83/84 or TI-89 to perform the calculations for a chi-square test for goodness of fit. We’ll use the data from the hockey and birthdays example to illustrate the steps.

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1. Enter the counts.

Birthday

Observed

Expected

• Enter the observed counts in L1/list1. Enter the expected counts in L2/list2.

Jan–Mar

32

20

Apr–Jun

20

20

2. Perform a chi-square test for goodness of fit.

Jul–Sep

16

20

Oct–Dec Note: Some older TI-83s and TI-84s don’t have this test. TI-84 users can get this functionality by upgrading their operating systems. TI-83/84: Press STAT , arrow over to TESTS and choose c2GOF-Test... .

12

20

TI-89: In the Stats/List Editor APP, press 2nd F1 ([F6]) and choose Chi2GOF... . Enter the inputs shown below. If you choose Calculate, you’ll get a screen with the test statistic, P-value, and df. If you choose the Draw option, you’ll get a picture of the appropriate chi-square distribution with the test statistic marked and shaded area corresponding to the P-value.

 

 

 

 

We’ll discuss the CNTRB and Comp Lst results shortly. AP® EXAM TIP  You can use your calculator to carry out the mechanics of a significance test on the AP® exam. But there’s a risk involved. If you just give the calculator answer with no work, and one or more of your values is incorrect, you will probably get no credit for the “Do” step. We recommend writing out the first few terms of the chi-square calculation followed by “. . .”. This approach might help you earn partial credit if you enter a number incorrectly. Be sure to name the procedure (c2GOF-Test) and to report the test statistic (c2 =11.2), degrees of freedom (df = 3), and P-value (0.011).

Follow-up Analysis  In the chi-square test for goodness of fit, we test the null

hypothesis that a categorical variable has a specified distribution in the population of interest. If the sample data lead to a statistically significant result, we can conclude that our variable has a distribution different from the one stated. When this happens, start by examining which categories of the variable show large deviations between the observed and expected counts. Then look at the individual terms (Observed − Expected)2 that are added together to produce the test statistic c2. Expected These components show which terms contribute most to the chi-square statistic.

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691

Let’s return to the hockey and birthdays example. The table of observed and expected counts for the 80 randomly selected NHL players is repeated below. We have added a column that shows the components of the chi-square test statistic. Looking at the counts, we see that there were many more players born in January through March than expected and far fewer players born in October through December than expected. The component for January to March birthdays made the largest contribution to the chi-square statistic. These results support Malcolm Gladwell’s claim that NHL players are more likely to be born early in the year. Birthday

Observed

Expected

O−E

(O−E)2/E

Jan–Mar

32

20

12

7.2

Apr–Jun

20

20

0

0.0

Jul–Sep

16

20

−4

0.8

Oct–Dec

12

20

−8

3.2

Note: When we ran the chi-square test for goodness of fit on the calculator, a list of these individual components was stored. On the TI-83/84, the list is called CNTRB (for contribution). On the TI-89, it’s called Comp Lst (component list).

Check Your Understanding

Biologists wish to mate pairs of fruit flies having genetic makeup RrCc, indicating that each has one dominant gene (R) and one recessive gene (r) for eye color, along with one dominant (C) and one recessive (c) gene for wing type. Each offspring will receive one gene for each of the two traits from each parent. The following table, known as a Punnett square, shows the possible combinations of genes received by the offspring: Parent 2 passes on: Parent 1 passes on:

RC

Rc

rC

rc

RC

RRCC (x)

RRCc (x)

RrCC (x)

RrCc (x)

Rc

RRCc (x)

RRcc (y)

RrCc (x)

Rrcc (y)

rC

RrCC (x)

RrCc (x)

rrCC (z)

rrCc (z)

rc

RrCc (x)

Rrcc (y)

rrCc (z)

rrcc (w)

Any offspring receiving an R gene will have red eyes, and any offspring receiving a C gene will have straight wings. So based on this Punnett square, the biologists predict a ratio of 9 red-eyed, straight-winged (x):3 red-eyed, curly-winged (y):3 white-eyed, straightwinged (z):1 white-eyed, curly-winged (w) offspring. To test their hypothesis about the distribution of offspring, the biologists mate a random sample of pairs of fruit flies. Of 200 offspring, 99 had red eyes and straight wings, 42 had red eyes and curly wings, 49 had white eyes and straight wings, and 10 had white eyes and curly wings. Do these data differ significantly from what the biologists have predicted? Carry out a test at the a = 0.01 significance level.

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Section 11.1

Summary • • •

•

A one-way table is often used to display the distribution of a single categorical variable for a sample of individuals. The chi-square test for goodness of fit tests the null hypothesis that a categorical variable has a specified distribution in the population of interest. This test compares the observed count in each category with the counts that would be expected if H0 were true. The expected count for any category is found by multiplying the sample size by the proportion in each category according to the null hypothesis. The chi-square statistic is

c2 =

∙

(Observed − Expected)2 Expected

where the sum is over all possible categories. • The conditions for performing a chi-square test for goodness of fit are: • Random: The data were produced by a well-designed random sample or randomized experiment. ~~ 10%: When sampling without replacement, check that the population is at least 10 times as large as the sample. • Large Counts: All expected counts must be at least 5. • When the conditions are met, the sampling distribution of the statistic c2 can be modeled by a chi-square distribution. • Large values of c2 are evidence against H0 and in favor of Ha. The P-value is the area to the right of c2 under the chi-square distribution with degrees of freedom df = number of categories − 1. • If the test finds a statistically significant result, consider doing a follow-up analysis that compares the observed and expected counts and that looks for the largest components of the chi-square statistic.

11.1 T echnology Corners TI-Nspire Instructions in Appendix B; HP Prime instructions on the book’s Web site. 25. Finding P-values for chi-square tests on the calculator 26. Chi-square test for goodness of fit on the calculator

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Section 11.1 Chi-Square Tests for Goodness of Fit

Exercises

Section 11.1 1. pg 681

Aw, nuts!  A company claims that each batch of its deluxe mixed nuts contains 52% cashews, 27% almonds, 13% macadamia nuts, and 8% brazil nuts. To test this claim, a quality-control inspector takes a random sample of 150 nuts from the latest batch. The one-way table below displays the sample data. Nut:

Cashew

Almond

Macadamia

Brazil

83

29

20

18

Count:

(a) State appropriate hypotheses for performing a test of the company’s claim. (b) Calculate the expected counts for each type of nut. Show your work. 2.

6.

(a) Confirm that the expected counts are large enough to use a chi-square distribution to calculate the P-value. What degrees of freedom should you use? (b) Sketch a graph like Figure 11.4 (page 685) that shows the P-value. (c) Use Table C to find the P-value. Then use your calculator’s c2cdf command. (d) What conclusion would you draw about whether or not the roulette wheel is operating correctly? Justify your answer. 7.

Birds in the trees  Researchers studied the behavior of birds that were searching for seeds and insects in an Oregon forest. In this forest, 54% of the trees were Douglas firs, 40% were ponderosa pines, and 6% were other types of trees. At a randomly selected time during the day, the researchers observed 156 ­red-breasted nuthatches: 70 were seen in Douglas firs, 79 in ponderosa pines, and 7 in other types of trees.2 Do these data provide convincing evidence that nuthatches prefer particular types of trees when they’re searching for seeds and insects?

8.

Seagulls by the seashore  Do seagulls show a preference for where they land? To answer this question, biologists conducted a study in an enclosed outdoor space with a piece of shore whose area was made up of 56% sand, 29% mud, and 15% rocks. The biologists chose 200 seagulls at random. Each seagull was released into the outdoor space on its own and observed until it landed somewhere on the piece of shore. In all, 128 seagulls landed on the sand, 61 landed in the mud, and 11 landed on the rocks. Do these data provide convincing evidence that seagulls show a preference for where they land?

9.

No chi-square  A school’s principal wants to know if students spend about the same amount of time on homework each night of the week. She asks a random sample of 50 students to keep track of their homework time for a week. The following table displays the average amount of time (in minutes) students reported per night:

Roulette  Casinos are required to verify that their games operate as advertised. American roulette wheels have 38 slots—18 red, 18 black, and 2 green. In one casino, managers record data from a random sample of 200 spins of one of their American roulette wheels. The one-way table below displays the results. Color:

Red

Black

Green

Count:

85

99

16

(a) State appropriate hypotheses for testing whether these data give convincing evidence that the distribution of outcomes on this wheel is not what it should be. (b) Calculate the expected counts for each color. Show your work. pg 683

3.

Aw, nuts!  Calculate the chi-square statistic for the data in Exercise 1. Show your work.

4.

Roulette  Calculate the chi-square statistic for the data in Exercise 2. Show your work.

5.

Aw, nuts!  Refer to Exercises 1 and 3.

(a) Confirm that the expected counts are large enough to use a chi-square distribution to calculate the P-value. What degrees of freedom should you use? (b) Sketch a graph like Figure 11.4 (page 685) that shows the P-value. (c) Use Table C to find the P-value. Then use your calculator’s c2cdf command. (d) What conclusion would you draw about the company’s claimed distribution for its deluxe mixed nuts? Justify your answer.

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Roulette  Refer to Exercises 2 and 4.

Night: Average time:

Sunday Monday Tuesday Wednesday Thursday Friday Saturday 130

108

115

104

99

37

62

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Explain carefully why it would not be appropriate to perform a chi-square test for goodness of fit using these data.

10. No chi-square  The principal in Exercise 9 also asked the random sample of students to record whether they did all of the homework that was assigned on each of the five school days that week. Here are the data: Monday

Tuesday

Wednesday

Thursday

Friday

No. who did  homework:

34

29

32

28

19

Explain carefully why it would not be appropriate to perform a chi-square test for goodness of fit using these data.

11. Benford’s law  Faked numbers in tax returns, invoices, or expense account claims often display ­patterns that aren’t present in legitimate records. Some patterns are obvious and easily avoided by a clever crook. Others are more subtle. It is a striking fact that the first digits of numbers in legitimate records often follow a model known as Benford’s law.3 Call the first digit of a randomly chosen record X for short. Benford’s law gives this probability model for X (note that a first digit can’t be 0): First digit: Probability:



1

2

3

4

5

6

7

8

9

0.301 0.176 0.125 0.097 0.079 0.067 0.058 0.051 0.046

A forensic accountant who is familiar with Benford’s law inspects a random sample of 250 invoices from a company that is accused of committing fraud. The table below displays the sample data.

First digit:

1

 2

 3

 4

 5

 6

7

8

9

Count:

61

50

43

34

25

16

7

8

6

(a) Are these data inconsistent with Benford’s law? Carry out an appropriate test at the a = 0.05 level to support your answer. If you find a significant result, perform a follow-up analysis. (b) Describe a Type I error and a Type II error in this setting, and give a possible consequence of each. Which do you think is more serious? 12. Housing  According to the Census Bureau, the ­distribution by ethnic background of the New York City population in a recent year was Hispanic: 28%  Black: 24%  White: 35% Asian: 12%  Others: 1%

The manager of a large housing complex in the city wonders whether the distribution by race of the

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Race:

Hispanic

Black

White

Asian

Other

Count:

212

202

270

94

22



School day:



complex’s residents is consistent with the population distribution. To find out, she records data from a random sample of 800 residents. The table below displays the sample data.4

Are these data significantly different from the city’s distribution by race? Carry out an appropriate test at the a = 0.05 level to support your answer. If you find a significant result, perform a follow-up analysis.

13. Skittles  Statistics teacher Jason Molesky contacted Mars, Inc., to ask about the color distribution for Skittles candies. Here is an excerpt from the response he received: “The original flavor blend for the SKITTLES BITE SIZE CANDIES is lemon, lime, orange, strawberry and grape. They were chosen as a result of consumer preference tests we conducted. The flavor blend is 20 percent of each flavor.” (a) State appropriate hypotheses for a significance test of the company’s claim. (b) Find the expected counts for a bag of Skittles with 60 candies. (c) How large a c2 statistic would you need to have significant evidence against the ­company’s claim at the a = 0.05 level? At the a = 0.01 level? (d) Create a set of observed counts for a bag with 60 candies that gives a P-value between 0.01 and 0.05. Show the calculation of your chi-square statistic. 14. Is your random number generator working? Use your calculator’s RandInt function to generate 200 digits from 0 to 9 and store them in a list. (a) State appropriate hypotheses for a chi-square test for goodness of fit to determine whether your ­calculator’s random number generator gives each digit an equal chance to be generated. (b) Carry out a test at the a = 0.05 significance level. For parts (c) and (d), assume that the students’ random number generators are all working properly. (c) What is the probability that a student who does this exercise will make a Type I error? (d) Suppose that 25 students in an AP Statistics class independently do this exercise for homework. Find the probability that at least one of them makes a Type I error. 15. What’s your sign?  The University of Chicago’s General Social Survey (GSS) is the nation’s most important social science sample survey. For reasons known

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Section 11.1 Chi-Square Tests for Goodness of Fit only to social scientists, the GSS regularly asks a random sample of people their astrological sign. Here are the counts of responses from a recent GSS: Sign:

Aries

Taurus

Gemini

Cancer

Leo

Virgo

Count:

321

360

367

374

383

402

Sign:

Libra Scorpio Sagittarius

Count:

392



329

Capricorn Aquarius Pisces

331

354

376

355

If births are spread uniformly across the year, we expect all 12 signs to be equally likely. Do these data provide convincing evidence that all 12 signs are not equally likely? If you find a significant result, perform a follow-up analysis.

16. Munching Froot Loops  Kellogg’s Froot Loops cereal comes in six fruit flavors: orange, lemon, cherry, raspberry, blueberry, and lime. Charise poured out her morning bowl of cereal and methodically counted the number of cereal pieces of each flavor. Here are her data: Flavor:

Orange

Lemon

Count:

28

21



Cherry Raspberry Blueberry 16

25

14

Lime 16

Do these data provide convincing evidence that Kellogg’s Froot Loops do not contain an equal proportion of each flavor? If you find a significant result, perform a follow-up analysis.

17. Mendel and the peas  Gregor Mendel (1822–1884), an Austrian monk, is considered the father of genetics. Mendel studied the inheritance of various traits in pea plants. One such trait is whether the pea is smooth or wrinkled. Mendel predicted a ratio of 3 smooth peas for every 1 wrinkled pea. In one experiment, he observed 423 smooth and 133 wrinkled peas. Assume that the conditions for inference were met. Carry out an appropriate test of the genetic model that Mendel predicted. What do you conclude? 18. You say tomato  The paper “Linkage Studies of the Tomato” (Transactions of the Canadian Institute, 1931) reported the following data on phenotypes resulting from crossing tall cut-leaf tomatoes with dwarf potato-leaf tomatoes. We wish to investigate whether the following frequencies are consistent with genetic laws, which state that the phenotypes should occur in the ratio 9:3:3:1. Phenotype: Frequency:

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Tall cut

Tall potato

Dwarf cut

Dwarf potato

926

288

293

104



Assume that the conditions for inference were met. Carry out an appropriate test of the proposed genetic model. What do you conclude?

Multiple choice: Select the best answer for Exercises 19 to 22. Exercises 19 to 21 refer to the following setting. The manager of a high school cafeteria is planning to offer several new types of food for student lunches in the following school year. She wants to know if each type of food will be equally popular so she can start ordering supplies and making other plans. To find out, she selects a random sample of 100 students and asks them, “Which type of food do you prefer: Asian food, Mexican food, pizza, or hamburgers?” Here are her data: Type of Food: Count:

Asian

Mexican

Pizza

Hamburgers

18

22

39

21

19. An appropriate null hypothesis to test whether the food choices are equally popular is (a) H0 :m = 25, where m = the mean number of students that prefer each type of food. (b) H0 :p = 0.25, where p = the proportion of all students who prefer Asian food. (c) H0 :nA = nM = nP = nH= 25, where nA is the number of students in the school who would choose Asian food, and so on. (d) H0 :pA = pM = pP= pH= 0.25, where pA is the proportion of students in the school who would choose Asian food, and so on. (e) H0:p^ A = p^ M = p^ P = p^ H = 0.25, where p^ A is the proportion of students in the sample who chose Asian food, and so on. 20. The chi-square statistic is (a) (b) (c) (d) (e) 21.

(18 − 25)2 (22 − 25)2 (39 − 25)2 (21 − 25)2 + + + 25 25 25 25 (25 − 18)2 (25 − 22)2 (25 − 39)2 (25 − 21)2 + + + 18 22 39 21 (18 − 25) (22 − 25) (39 − 25) (21 − 25) + + + 25 25 25 25 (18 − 25)2 (22 − 25)2 (39 − 25)2 (21 − 25)2 + + + 100 100 100 100 (0.18 − 0.25)2 (0.22 − 0.25)2 (0.39 − 0.25)2 + + 0.25 0.25 0.25 (0.21 − 0.25)2 + 0.25 The P-value for a chi-square test for goodness of fit is 0.0129. Which of the following is the most appropriate conclusion?

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(a) Because 0.0129 is less than a = 0.05, reject H0. There is convincing evidence that the food choices are equally popular.

23. Reading and grades (1.3)  Write a few sentences comparing the distributions of English grades for light and heavy readers.

(b) Because 0.0129 is less than a = 0.05, reject H0. There is not convincing evidence that the food choices are equally popular.

24. Reading and grades (10.2)  Summary statistics for the two groups from Minitab are provided below.

(c) Because 0.0129 is less than a = 0.05, reject H0. There is convincing evidence that the food choices are not equally popular. (d) Because 0.0129 is less than a = 0.05, fail to reject H0. There is not convincing evidence that the food choices are equally popular. (e) Because 0.0129 is less than a = 0.05, fail to reject H0. There is convincing evidence that the food choices are equally popular. 22. Which of the following is false? (a) A chi-square distribution with k degrees of freedom is more right-skewed than a chi-square distribution with k + 1 degrees of freedom. (b) A chi-square distribution never takes negative values.

Type of reader

N

Mean

StDev

Heavy

47

3.640

0.324

SE Mean 0.047

Light

32

3.356

0.380

0.067

(a) Explain why it is acceptable to use two-sample t procedures in this setting. (b) Construct and interpret a 95% confidence interval for the difference in the mean English grade for light and heavy readers. (c) Does the interval in part (b) provide convincing evidence that reading more causes a difference in students’ English grades? Justify your answer. 25. Reading and grades (3.2)  The Fathom scatterplot below shows the number of books read and the English grade for all 79 students in the study. A leastsquares regression line has been added to the graph.

(c) The degrees of freedom for a chi-square test is determined by the sample size. (d) P(c2 > 10) is greater when df = k + 1 than when df = k. (e) The area under a chi-square density curve is always equal to 1. Exercises 23 through 25 refer to the following setting. Do students who read more books for pleasure tend to earn higher grades in English? The boxplots below show data from a simple random sample of 79 students at a large high school. Students were classified as light readers if they read fewer than 3 books for pleasure per year. Otherwise, they were classified as heavy readers. Each student’s average English grade for the previous two marking periods was converted to a GPA scale where A+ = 4.3, A = 4.0, A− =3.7, B + =3.3, and so on.

(a) Interpret the meaning of the slope and y intercept in ­context. (b) The student who reported reading 17 books for pleasure had an English GPA of 2.85. Find this student’s residual and interpret this value in context. (c) How strong is the relationship between English grades and number of books read? Give appropriate evidence to support your answer. 26. Yahtzee (5.3, 6.3)  In the game of Yahtzee, 5 sixsided dice are rolled simultaneously. To get a Yahtzee, the player must get the same number on all 5 dice. (a) Luis says that the probability of getting a Yahtzee in one 1 5 roll of the dice is a b . Explain why Luis is wrong. 6 (b) Nassir decides to keep rolling all 5 dice until he gets a Yahtzee. He is surprised when he still hasn’t gotten a Yahtzee after 25 rolls. Should he be? Calculate an appropriate probability to support your answer.

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Section 11.2 Inference for Two-Way Tables

11.2 What You Will Learn

697

Inference for Two-Way Tables By the end of the section, you should be able to:

• Compare conditional distributions for data in a two-way table. • State appropriate hypotheses and compute expected counts for a chi-square test based on data in a two-way table. • Calculate the chi-square statistic, degrees of freedom, and P-value for a chi-square test based on data in a two-way table.

• Perform a chi-square test for homogeneity. • Perform a chi-square test for independence. • Choose the appropriate chi-square test.

The two-sample z procedures of Chapter 10 allow us to compare the proportions of successes in two populations or for two treatments. What if we want to compare more than two samples or groups? More generally, what if we want to compare the distributions of a single categorical variable across several populations or treatments? We need a new statistical test. The new test starts by presenting the data in a two-way table. Two-way tables have more general uses than comparing distributions of a single categorical variable. As we saw in Section 1.1, they can be used to describe relationships between any two categorical variables. In this section, we will start by developing a test to determine whether the distribution of a categorical variable is the same for each of several populations or treatments. This test is called a chisquare test for homogeneity. Then we’ll examine a related test to see whether there is convincing evidence of an association between the row and column variables in a two-way table. This test is known as a chi-square test for independence.

Comparing Distributions of a Categorical Variable We’ll start with an example involving a randomized experiment.

Example

Does Background Music Influence What Customers Buy? Comparing conditional distributions Market researchers suspect that background music may affect the mood and buying behavior of customers. One study in a European restaurant compared three randomly assigned treatments: no music, French accordion music, and Italian string music. Under each condition, the researchers recorded the number of

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c­ ustomers who ordered French, Italian, and other entrees.5 Here is a table that summarizes the data: Type of Music None

French

Italian

Total

French

30

39

30

 99

Italian

11

 1

19

 31

Other

43

35

35

113

Total

84

75

84

243

Entree ordered

Problem: (a) Calculate the conditional distribution (in proportions) of entrees ordered for each treatment. (b) Make an appropriate graph for comparing the conditional distributions in part (a). (c) Write a few sentences comparing the distributions of entrees ordered under the three music treatments.

Solution: (a) When no music was playing, the distribution of entree orders was 30 11 43 = 0.357  Italian: = 0.131  Other: = 0.512 84 84 84 When French accordion music was playing, the distribution of entree orders was French:

39 1 35 = 0.520  Italian: = 0.013  Other: = 0.467 75 75 75 When Italian string music was playing, the distribution of entree orders was French:

30 19 35 = 0.357  Italian: = 0.226  Other: = 0.417 84 84 84 (b)  ​The bar graphs in Figure 11.6 compare the distributions of entrees ordered for each of the three music treatments. French:

Music = French

Music = Italian

60

60

50

50

50

40

40

40

30

Percent

60

Percent

Percent

Music = None

30

30

20

20

20

10

10

10

0

French

Italian

Entree (a)

Other

0

French

Italian

Entree (b)

Other

0

French

Italian

Entree

Other

(c)

FIGURE 11.6  ​Bar graphs comparing the distributions of entrees ordered for different music c­ onditions.

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Section 11.2 Inference for Two-Way Tables

699

(c) The type of entree that customers order seems to differ considerably across the three music ­treatments. Orders of Italian entrees are very low (1.3%) when French music is playing but are higher when Italian music (22.6%) or no music (13.1%) is playing. French entrees seem popular in this restaurant, as they are ordered frequently under all music conditions but notably more often when French music is playing. For all three music treatments, the percent of Other entrees ordered was similar. For Practice Try Exercise

27

The researchers in the restaurant study expected that music would influence customer orders, so the type of music played is the explanatory variable and the type of entree ordered is the response variable. A good general strategy is to compare the conditional distributions of the response variable for each value of the explanatory variable. That’s why we compared the conditional distributions of entrees ordered for each type of music played. It is common practice to describe a two-way table by its number of rows and columns (not including totals). For instance, the data in the previous example were given in a 3 × 3 table. The following Check Your Understanding involves a 3 × 2 table.

Check Your Understanding

The Pennsylvania State University has its main campus in the town of State College and more than 20 smaller “commonwealth campuses” around the state. The Penn State Division of Student Affairs polled separate random samples of undergraduates from the main campus and commonwealth campuses about their use of online social networking. Facebook was the most popular site, with more than 80% of students having an account. Here is a comparison of Facebook use by undergraduates at the main campus and commonwealth campuses who have a Facebook account:6 Use Facebook

Main campus

Commonwealth

Several times a month or less

 55

 76

At least once a week

215

157

At least once a day

640

394

Total Facebook users

910

627

1. Calculate the conditional distribution (in proportions) of Facebook use for each campus setting. 2. Why is it important to compare proportions rather than counts in Question 1? 3. Make a bar graph that compares the two conditional distributions. What are the most important differences in Facebook use between the two campus settings?

Stating Hypotheses  The null hypothesis in the restaurant example is H0: There is no difference in the true distributions of entrees ordered at this restaurant when no music, French accordion music, or Italian string music is played.

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C H A P T E R 1 1 I n f e r e n c e f o r D i s t r i b u t i o n s o f C at e g o r i c a l Data If the null hypothesis is true, the observed differences in the distributions of entrees ordered for the three groups are due to the chance involved in the random assignment of treatments. The alternative hypothesis says that there is a difference but does not specify the nature of that difference: Ha: There is a difference in the true distributions of entrees ordered at this restaurant when no music, French accordion music, or Italian string music is played. Any difference among the three true distributions of entrees ordered when no music, French accordion music, or Italian string music is played means that the null hypothesis is false and the alternative hypothesis is true. The alternative hypothesis is not one-sided or two-sided. We might call it “many-sided” because it allows any kind of difference. With only the methods we already know, we might start by comparing the proportions of French entrees ordered when no music and French accordion music are played. We could similarly compare other pairs of proportions, ending up with many tests and many P-values. This is a bad idea. The P-values belong to each test separately, not to the collection of all the tests together. Type of Music Entree ordered French Italian Other Total

None 30 11 43 84

French 39  1 35 75

Italian 30 19 35 84

Total  99  31 113 243

c

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!

n

When we do many individual tests or construct many confidence inter- autio vals, the individual P-values and confidence levels don’t tell us how confident we can be in all the inferences taken together. Because of this, it’s cheating to pick out one large difference from the two-way table and then perform a significance test as if it were the only comparison we had in mind. For example, the proportions of French entrees ordered under the no music and French accordion music treatments are 30/84 = 0.357 and 39/75 = 0.520, respectively. A two-sample z test shows that the difference between the proportions is statistically significant (z = 2.06, P = 0.039) if we make just this one comparison. But we could also pick a comparison that is not significant. For example, the proportions of Italian entrees ordered for the no music and Italian string music treatments are 11/84 = 0.131 and 19/84 = 0.226, respectively. These two proportions do not differ significantly (z = 1.61, P = 0.107). Individual comparisons can’t tell us whether the three distributions of the categorical variable (in this case, type of entree ordered) are significantly different. The problem of how to do many comparisons at once with an overall measure of confidence in all our conclusions is common in statistics. This is the problem of multiple comparisons. Statistical methods for dealing with multiple comparisons usually have two parts: 1. An overall test to see if there is convincing evidence of any differences among the parameters that we want to compare. 2. A detailed follow-up analysis to decide which of the parameters differ and to estimate how large the differences are.

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The overall test uses the familiar chi-square statistic. But in this new setting the test will be used to compare the distribution of a categorical variable for several populations or treatments. The follow-up analysis can be quite elaborate. We will concentrate on the overall test and do a follow-up analysis only when the observed results are statistically significant.

Expected Counts and the Chi-Square Statistic A chi-square test for homogeneity begins with the hypotheses It would also be correct to state the null hypothesis as H0: The distribution of a categorical variable is the same for each of several populations or treatments. We prefer the “no difference” wording because it’s more consistent with the language we used in the significance tests of Chapter 10.

Example

H0: There is no difference in the distribution of a categorical variable for ­several populations or treatments. Ha: There is a difference in the distribution of a categorical variable for ­several populations or treatments. To perform a test, we compare the observed counts in a two-way table with the counts we would expect if H0 were true. Finding the expected counts is not that difficult, as the ­following example illustrates.

Does Background Music Influence What Customers Buy? Computing expected counts The null hypothesis in the restaurant experiment is that there’s no difference in the distribution of entrees ordered when no music, French accordion music, or Italian string music is played. To find the expected counts, we start by assuming that H0 is true. We can see from the two-way table that 99 of the 243 entrees ordered during the study were French.

Entree ordered French Italian Other Total

Observed Counts Type of Music None French Italian 30 39 30 11  1 19 43 35 35 84 75 84

Total  99  31 113 243

If the specific type of music that’s playing has no effect on entree orders, the proportion of French entrees ordered under each music condition should be 99/243 = 0.4074. For instance, there were 84 total entrees ordered when no music was playing. We would expect 99 = 84(0.4074) = 34.22 84 # 243

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Although any count of entrees ordered must be a whole number, an expected count need not be. The expected count gives the average number of entrees ordered if H0 is true and the random assignment process is repeated many times.

of those entrees to be French, on average. The expected counts of French entrees ordered under the other two music conditions can be found in a similar way: French music: 75(0.4074) = 30.56  Italian music: 84(0.4074) = 34.22 We repeat the process to find the expected counts for the other two types of entrees. The overall proportion of Italian entrees ordered during the study was 31/243 = 0.1276. So the expected counts of Italian entrees ordered under each treatment are No music: 84(0.1276) = 10.72   French music: 75(0.1276) = 9.57 Italian music: 84(0.1276) = 10.72 The overall proportion of Other entrees ordered during the experiment was 113/243 = 0.465. So the expected counts of Other entrees ordered for each treatment are No music: 84(0.465) = 39.06   French music: 75(0.465) = 34.88 Italian music: 84(0.465) = 39.06 The following table summarizes the expected counts for all three treatments. Note that the values for no music and Italian music are the same because 84 total entrees were ordered under each condition. We can check our work by adding the expected counts to obtain the row and Expected Counts column totals, as in the table. Type of Music These should be the same as None French Italian Total those in the table of observed Entree ordered French 34.22 30.56 34.22  99 counts except for small roundItalian 10.72  9.57 10.72  31 off errors, such as 75.01 rather 39.06 34.88 39.06 113 than 75 for the total number of Other Total 84 75 84 243 French entrees ordered.

Let’s take a look at the two-way table from the restaurant study one more time. In the example, we found the expected count of French entrees ordered when no music was playing as follows: 99 84 # = 34.22 243

Entree ordered French Italian Other Total

None 30 11 43 84

Observed Counts Type of Music French 39  1 35 75

Italian 30 19 35 84

Total  99  31 113 243

We marked the three numbers used in this calculation in the table. These values are the row total for French entrees ordered, the column total for entrees ordered

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when no music was playing, and the table total of entrees ordered during the experiment. We can rewrite the original calculation as 84 # 99 99 # 84 = = 34.22 243 243 This suggests a more general formula for the expected count in any cell of a twoway table: row total # column total table total

Finding Expected Counts When H0 is true, the expected count in any cell of a two-way table is expected count =

row total # column total table total

All the expected counts in the restaurant study are at least 5. This satisfies the Large Counts condition. The Random condition is met because the treatments were assigned at random. We don’t need to check the 10% condition because the researchers were not sampling without replacement from some population of interest. They just performed an experiment using customers who happened to be in the restaurant at the time.

Conditions for Performing a Chi-Square Test for Homogeneity • Random: The data come from independent random samples or from the groups in a randomized experiment. 1 ~~ 10%: When sampling without replacement, check that n ≤ N. 10 • Large Counts: All expected counts are at least 5.

Just as we did with the chi-square test for goodness of fit, we compare the observed counts with the expected counts using the statistic c2 =

(Observed − Expected)2 ∙ Expected

This time, the sum is over all cells (not including the totals!) in the two-way table.

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Example

Does Background Music Influence What Customers Buy? The chi-square statistic Problem:  The tables below show the observed and expected counts for the restaurant experiment. Calculate the chi-square statistic. Show your work. Observed Counts Type of Music None French Italian

Entree ordered French Italian Other Total

AP® EXAM TIP ​​In the “Do” step, you aren’t required to show every term in the chi-square statistic. Writing the first few terms of the sum followed by “. . .” is considered as “showing work.” We suggest that you do this and then let your calculator tackle the computations.

30 11 43 84

39  1 35 75

30 19 35 84

Total

Entree ordered

Expected Counts Type of Music None French Italian

 99  31 113 243

French Italian Other Total

34.22 10.72 39.06 84

30.56   9.57 34.88 75

34.22 10.72 39.06 84

Total  99  31 113 243

Solution:  For French entrees with no music, the observed count is 30 orders and the expected count is 34.22. The contribution to the c2 statistic for this cell is (Observed − Expected)2 Expected

=

(30 − 34.22)2 = 0.52 34.22

The c2 statistic is the sum of nine such terms: c =∙ 2

(Observed − Expected)2 Expected

(30 − 34.22)2 (39 − 30.56)2 c (35 − 39.06)2 + + + 34.22 30.56 39.06 c = 0.52 + 2.33 + + 0.42 = 18.28

=

For Practice Try Exercise

29

As in the test for goodness of fit, you should think of the chi-square statistic c2 as a measure of how much the observed counts deviate from the expected counts. Once again, large values of c2 are evidence against H0 and in favor of Ha. The P-value measures the strength of this evidence. When conditions are met, P-values for a chi-square test for homogeneity come from a chi-square distribution with df = (number of rows − 1) × (number of columns − 1).

Example

Does Background Music Influence What Customers Buy? P-value and conclusion Earlier, we started a significance test of H0: There is no difference in the true distributions of entrees ordered at this restaurant when no music, French accordion music, or Italian string music is played.

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Ha: There is a difference in the true distributions of entrees ordered at this restaurant when no music, French accordion music, or Italian string music is played.

Entree ordered French Italian Other Total

Observed Counts Type of Music None French Italian 30 39 30 11  1 19 43 35 35 84 75 84

Total  99  31 113 243

We already checked that the conditions are met. Our calculated test statistic is c2 = 18.28.

Problem: (a) Use Table C to find the P-value. Then use your calculator’s c2cdf command. (b) Interpret the P-value from the calculator in context. (c) What conclusion would you draw? Justify your answer.

Solution: (a) Because the two-way table that summarizes the data from the study has three rows and three columns, we use a chi-square distribution with df = (3 − 1)(3 − 1) = 4 to find the P-value. •  Table:  Look at the df = 4 row in Table C. The calculated value P c2 = 18.28 lies between the critical values 16.42 and 18.47. The df .0025 .001 corresponding P-value is between 0.001 and 0.0025. 4 16.42 18.47 •  Calculator:  The command c2cdf(18.28,10000,4) gives 0.0011. (b)  ​Assuming that there is no difference in the true distributions of entrees ordered in this restaurant when no music, French accordion music, or Italian string music is played, there is a 0.0011 probability of observing a difference in the distributions of entrees ordered among the three treatment groups as large as or larger than the one in this study. (c)  Because the P-value, 0.0011, is less than our default a = 0.05 significance level, we reject H0. We have convincing evidence of a difference in the true distributions of entrees ordered at this restaurant when no music, French accordion music, or Italian string music is played. Furthermore, the random assignment allows us to say that the difference is caused by the music that’s played. For Practice Try Exercise

31

Check Your Understanding

In the previous Check Your Understanding (page 699), we presented data on the use of Facebook by two randomly selected groups of Penn State students. Here are the data once again. Use Facebook

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Main campus

Commonwealth

Several times a month or less

 55

 76

At least once a week

215

157

At least once a day

640

394

Total Facebook users

910

627

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C H A P T E R 1 1 I n f e r e n c e f o r D i s t r i b u t i o n s o f C at e g o r i c a l Data Do these data provide convincing evidence of a difference in the distributions of Facebook use among students in the two campus settings? 1. State appropriate null and alternative hypotheses for a significance test to help answer this question. 2. Calculate the expected counts. Show your work. 3. Calculate the chi-square statistic. Show your work. 4. Use Table C to find the P-value. Then use your calculator’s c2cdf command. 5. Interpret the P-value from the calculator in context. 6. What conclusion would you draw? Justify your answer.

Calculating the expected counts and then the chi-square statistic by hand is a bit time-consuming. As usual, technology saves time and gets the arithmetic right.

27. Technology Corner

Chi-square tests for two-way tables on the calculator TI-Nspire instructions in Appendix B; HP Prime instructions on the book’s Web site.

You can use the TI-83/84 or TI-89 to perform calculations for a chi-square test for homogeneity. We’ll use the data from the restaurant study to illustrate the process. 1. Enter the observed counts in matrix [A].

TI-83/84

• Press 2nd X choose A.

-1

(MATRIX), arrow to EDIT, and

• Enter the dimensions of the matrix: 3 × 3.

TI-89 • Press APPS , select Data/Matrix Editor and then New. . . . • Adjust your settings to match those shown.

• Enter the observed counts from the two-way table in the same locations in the matrix.

2. Specify the chi-square test, the matrix where the observed counts are found, and the matrix where the expected counts will be stored.

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• Press STAT , arrow to TESTS, and choose c2-Test.

• In the Statistics/List Editor, press 2nd F1 ([F6]), and choose Chi2 2-way. . . .

• Adjust your settings as shown.

• Adjust your settings as shown.



707



3. Choose “Calculate” or “Draw” to carry out the test. If you choose “Calculate,” you should get the test statistic, P-value, and df shown below. If you specify “Draw,” the chi-square distribution with 4 degrees of freedom will be drawn, the area in the tail will be shaded, and the P-value will be displayed.

4. To see the expected counts, go to the home screen and ask for a display of the matrix [B]. • Press 2nd X-1 (MATRIX), arrow to EDIT, and choose [B].



•

Press 2nd − (Var-LINK) and choose B.

AP® EXAM TIP  You can use your calculator to carry out the mechanics of a significance test on the AP® exam. But there’s a risk involved. If you just give the calculator answer with no work, and one or more of your values is incorrect, you will probably get no credit for the “Do” step. We recommend writing out the first few terms of the chi-square calculation followed by “. . . ”. This approach might help you earn partial credit if you enter a number incorrectly. Be sure to name the procedure (c2-Test for homogeneity) and to report the test statistic (c2 =18.279), degrees of freedom (df = 4), and P-value (0.0011).

The Chi-Square Test for Homogeneity In Section 11.1, we used a chi-square test for goodness of fit to test a hypothesized model for the distribution of a categorical variable. Our P-values came from a chi-square distribution with df = the number of categories −1. When the Random, 10%, and Large Counts conditions are met, the c2 statistic calculated from

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C H A P T E R 1 1 I n f e r e n c e f o r D i s t r i b u t i o n s o f C at e g o r i c a l Data a two-way table can be used to perform a test of H0: There is no difference in the distribution of a categorical variable for several populations or treatments. This new procedure is known as a chi-square test for homogeneity.

This test is also known as a chi-square test for homogeneity of proportions. We prefer the simpler name.

Chi-Square Test for Homogeneity Suppose the conditions are met. You can use the chi-square test for homogeneity to test H0: There is no difference in the distribution of a categorical variable for several populations or treatments. Ha: There is a difference in the distribution of a categorical variable for several populations or treatments. Start by finding the expected counts. Then calculate the chi-square statistic c2 =

P-value

χ2

∙

(Observed − Expected)2 Expected

where the sum is over all cells (not including totals) in the two-way table. If H0 is true, the c2 statistic has approximately a chi-square distribution with degrees of freedom = (number of rows − 1)(number of columns − 1). The P-value is the area to the right of c2 under the corresponding chi-square ­density curve. Let’s look at an example of a chi-square test for homogeneity from start to f­inish. As usual, we follow the four-step process when performing a significance test.

Example

Are Cell-Only Telephone Users Different?

STEP

The chi-square test for homogeneity

4

Random digit dialing telephone surveys used to exclude cell phone numbers. If the opinions of people who have only cell phones differ from those of people who have landline service, the poll results may not represent the entire adult population. The Pew Research Center interviewed separate random samples of cell-only and landline telephone users who were less than 30 years old. Here’s what the Pew survey found about how these people describe their political party affiliation:7

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Cell-only sample

Landline sample

Democrat or lean Democratic

49

 47

Refuse to lean either way

15

 27

Republican or lean Republican

32

 30

Total

96

104

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709

Problem: (a) Compare the distributions of political party affiliation for cell-only and landline phone users. (b) Do these data provide convincing evidence at the a = 0.05 level that the distribution of party affiliation differs in the under-30 cell-only and landline user populations?

Solution: (a) Because the sample sizes are different, we should compare the proportions of individuals in each political affiliation category in the two samples. The table below shows the conditional distributions of political party affiliation for cell-only and landline phone users. We made a segmented bar graph to compare these two distributions. Cell-only users appear slightly more likely to declare themselves as Democrats or Republicans than people who have landlines. People with landlines seem much more likely to say they don’t lean Democratic or Republican than those who use only cell phones.

Proportion

Phone Status Political affiliation

1.00 0.90 0.80 0.70 0.60 0.50 0.40 0.30 0.20 0.10 0.00

Cell only

Landline

Democrat

0.51

0.45

No lean

0.16

0.26

Republican

0.33

0.29

(b)  ​StatE: We want to perform a test of Cell-only

H0: There is no difference in the distribution of party affiliation in the under-30 cell-only and landline populations.

Landline

Phone status Democrat

No lean

Ha: There is a difference in the distribution of party affiliation in the under-30 cell-only and landline populations.

Republican

at the a = 0.05 level.

Plan:  If conditions are met, we should use a chi-square test for homogeneity. • Random:  The data came from independent random samples of 96 cell-only and 104 landline users. ºº 10%:  Sampling without replacement was used, so there need to be at least 10(96) = 960 cell-only users under age 30 and at least 10(104) = 1040 landline users under age 30. This is safe to assume. •  Large Counts:  We followed the steps in the Technology Corner on page 706 to get the expected counts. The calculator screen shot confirms that all expected counts are at least 5.

Do:  A chi-square test on the calculator gave •  Test statistic: 2

c =

∙

(Observed − Expected)2 Expected

(49 − 46.08)2 (47 − 49.92)2 c = + + = 3.22 46.08 49.92 •  P-value:  Using df = (number of rows − 1)(number of columns − 1) = (3 − 1)(2 − 1) = 2, the P-value is 0.1999.

Conclude:  Because our P-value, 0.1999, is greater than a = 0.05, we fail to reject H0. There is not convincing evidence that the distribution of party affiliation differs in the under-30 cell-only and landline user populations. For Practice Try Exercise

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Follow-up Analysis  The chi-square test for homogeneity allows us to compare the distribution of a categorical variable for any number of populations or treatments. If the test allows us to reject the null hypothesis of no difference, we may want to do a follow-up analysis that examines the differences in detail. Start by examining which cells in the two-way table show large deviations between the observed and ex(Observed − Expected)2 pected counts. Then look at the individual components Expected to see which terms contribute most to the chi-square statistic. Our earlier restaurant study found significant differences among the true distributions of entrees ordered under each of the three music conditions. We entered the two-way table for the study into Minitab software and requested a chi-square test. The output appears in Figure 11.7. Minitab repeats the two-way table of observed counts and puts the expected count for each cell below the observed count. Finally, the software prints the 9 individual components that ­contribute to the c2 statistic. Chi-Square Test: None, French, Italian Expected counts are printed below observed counts Chi-Square contributions are printed below expected counts None 30 34.22 0.521

French 39 30.56 2.334

Italian 30 34.22 0.521

Total 99

2

11 10.72 0.008

1 9.57 7.672

19 10.72 6.404

31

3

43 39.06 0.397

35 34.88 0.000

35 39.06 0.422

113

84

75

84

243

1

FIGURE 11.7  ​Minitab output for the two-way table in the restaurant study. The output gives the observed counts, the expected counts, and the individual components of the chi-square statistic.

Total

Chi-Sq = 18.279, DF = 4, P-Value = 0.001

Looking at the output, we see that just two of the nine components that make up the chi-square statistic contribute about 14 (almost 77%) of the total c2 = 18.28. Comparing the observed and expected counts in these two cells, we see that orders of Italian entrees are much below expectation when French music is playing and well above expectation when Italian music is playing. We are led to a specific conclusion: orders of Italian entrees are strongly affected by Italian and French music. More advanced methods provide tests and confidence intervals that make this follow-up analysis more complete.

THINK ABOUT IT

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What if we want to compare several proportions?  Many studies involve comparing the proportion of successes for each of several populations or treatments. The two-sample z test from Chapter 10 allows us to test the null hypothesis H0: p1 = p2, where p1 and p2 are the true proportions of successes for the two populations or treatments. The chi-square test for homogeneity allows us to test H0: p1 = p2 = c = pk. This null hypothesis says that there is no difference in the proportions of successes for the k populations or treatments. The alternative hypothesis is Ha: at least two of the pi’s are different. Many students incorrectly state Ha as “all the proportions are different.” Think about it this way: the opposite of “all the proportions are equal” is “some of the proportions are not equal.”

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Check Your Understanding

Canada has universal health care. The United States does not but often offers more elaborate treatment to patients with access. How do the two systems compare in treating heart attacks? Researchers compared random samples of U.S. and Canadian heart attack patients. One key outcome was the patients’ own assessment of their quality of life relative to what it had been before the heart attack. Here are the data for the patients who survived a year: Quality of life

Canada

United States

Much better

 75

541

Somewhat better

 71

498

About the same

 96

779

Somewhat worse

 50

282

Much worse

 19

 65

Total

311

2165

1. Construct an appropriate graph to compare the distributions of opinion about quality of life among heart attack patients in Canada and the United States. 2. Is there a significant difference between the two distributions of quality-of-life ratings? Carry out an appropriate test at the a = 0.01 level.

Relationships between Two Categorical Variables Two-way tables can arise in several ways. The restaurant experiment compared entrees ordered for three music treatments. The phone use and political party affiliation observational study compared independent random samples from the cell-only and landline user populations. In both cases, we are comparing the distributions of a categorical variable for several populations or treatments. We use the chi-square test for homogeneity to perform inference in such settings. Another common situation that leads to a two-way table is when a single random sample of individuals is chosen from a single population and then classified based on two categorical variables. In that case, our goal is to analyze the relationship between the variables. The next example describes a study of this type.

Example

Angry People and Heart Disease Relationships between two categorical variables A study followed a random sample of 8474 people with normal blood pressure for about four years.8 All the individuals were free of heart disease at the beginning of the study. Each person took the Spielberger Trait Anger Scale test, which measures how prone a person is to sudden anger. Researchers also recorded whether each individual developed coronary heart disease (CHD). This includes people who had

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heart attacks and those who needed medical treatment for heart disease. Here is a two-way table that summarizes the data: CHD

Low anger

Moderate anger

High anger

Total

  53

 110

  27

 190

No CHD

3057

4621

606

8284

Total

3110

4731

633

8474

Problem: (a) Is this an observational study or an experiment? Justify your answer. (b) Make a well-labeled bar graph that compares CHD rates for the different anger levels. Describe what you see.

Solution: (a) This is an observational study. Researchers did not deliberately impose any treatments. They just recorded data about two variables—anger level and whether or not the person developed CHD—for each randomly chosen individual. (b) In this setting, anger level is the explanatory variable and whether or not a person gets heart disease is the response variable. So we compare the percents of people who did and did not get heart disease in each of the three anger categories: CHD no CHD 53 3057 = 0.0170 = 1.70%   = 0.9830 = 98.30% Low anger:   3110 3110 110 4621 Moderate anger:   = 0.0233 = 2.33%   = 0.9767 = 97.67% 4731 4731 27 606 High anger:   = 0.0427 = 4.27%    = 0.9573 = 95.73% 633 633 The bar graph in Figure 11.8 shows the percent of people in each of the three anger categories who developed CHD. ​There is a clear trend: as the anger score increases, so does the percent who suffer heart disease. A much higher percent of people in the high anger category developed CHD (4.27%) than in the moderate (2.33%) and low (1.70%) anger categories.

Percent with CHD

5

FIGURE 11.8  ​Bar graph comparing the percents of people in each anger category who got coronary heart disease (CHD).

4 3 2 1 0

Low

Moderate

High

Anger rating

For Practice Try Exercise

41

Anger rating on the Spielberger scale is a categorical variable that takes three possible values: low, medium, and high. Whether or not someone gets heart disease is also a categorical variable. The two-way table in the example shows the relationship between anger rating and heart disease for a random sample of

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713

8474 people. Do these data provide convincing evidence of an association between the variables in the larger population? To answer that question, we work with a new significance test.

The Chi-Square Test for Independence We often gather data from a random sample and arrange them in a two-way table to see if two categorical variables are associated. The sample data are easy to investigate: turn them into percents and look for a relationship between the variables. Is the association in the sample evidence of an association between these variables in the entire population? Or could the sample association easily arise just from the luck of random sampling? This is a question for a significance test. Our null hypothesis is that there is no association between the two categorical variables in the population of interest. The alternative hypothesis is that there is an association between the variables. For the observational study of anger level and coronary heart disease, we want to test the hypotheses H0: There is no association between anger level and heart-disease status in the population of people with normal blood pressure. Ha: There is an association between anger level and heart-disease status in the population of people with normal blood pressure. We could substitute the word “dependent” in place of “not independent” in the alternative hypothesis. We’ll avoid this practice, however, because saying that two variables are dependent sounds too much like saying that changes in one variable cause changes in the other.

No association between two variables means that knowing the value of one variable does not help us predict the value of the other. That is, the variables are independent. An equivalent way to state the hypotheses is therefore H0: Anger and heart-disease status are independent in the population of people with normal blood pressure. Ha: Anger and heart-disease status are not independent in the population of people with normal blood pressure. As with the two previous types of chi-square tests, we begin by comparing the observed counts in a two-way table with the expected counts if H0 is true.

Example

Angry People and Heart Disease Finding expected counts The null hypothesis is that there is no association between anger level and heart-­disease status in the population of interest. If we assume that H0 is true, then anger level and CHD status are independent. We can find the expected cell counts in the two-way table using the definition of independent events from Chapter 5: P(A | B) = P(A). The chance process here is randomly selecting a person and recording his or her anger level and CHD status.

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Low anger

Moderate anger

High anger

Total

CHD

  53

 110

 27

 190

No CHD

3057

4621

606

8284

Total

3110

4731

633

8474

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Let’s start by considering the events “CHD” and “low anger.” We see from the two-way table that 190 of the 8474 people in the study had CHD. If we imagine choosing one of these people at random, P(CHD) = 190/8474. Because anger level and CHD status are independent, knowing that the selected individual is low anger does not change the probability that this person develops CHD. That is to say, P(CHD | low anger) = P(CHD) = 190/8474 = 0.02242. Of the 3110 low-anger people in the study, we’d expect 190 3110 # = 3110(0.02242) = 69.73 8474 to get CHD. You can see that the general formula we developed earlier for a test for homo­geneity applies in this situation also: expected count =

row total # column total 190 # 3110 = = 69.73 table total 8474

To find the expected count in the “low anger, no CHD” cell, we begin by noting that P(no CHD) = 8284/8474 = 0.97758 for a randomly selected person in the study. Of the 3110 low-anger people in the study, we would expect 8284 3110 # = 3110(0.97758) = 3040.27 8474 to not develop CHD. We find the expected counts for the remaining cells in the two-way table in a similar way. CHD, Low CHD, Moderate CHD, High 3110(0.02242) = 69.73 no CHD, Low

4731(0.02242) = 106.08 no CHD, Moderate

633(0.02242) = 14.19 no CHD, High

3110(0.97758) = 3040.27 4731(0.97758) = 4624.92 633(0.97758) = 618.81

The 10% and Large Counts conditions for the chi-square test for independence are the same as for the homogeneity test. There is a slight difference in the Random condition for the two tests: a test for independence uses data from one sample but a test for homogeneity uses data from two or more samples/groups.

Conditions for Performing a Chi-Square Test for Independence • Random: The data come from a well-designed random sample or randomized experiment. 1 ~~ 10%: When sampling without replacement, check that n ≤ N. 10 • Large Counts: All expected counts are at least 5.

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Section 11.2 Inference for Two-Way Tables

If the Random, 10%, and Large Counts conditions are met, the c2 statistic calculated from a two-way table can be used to perform a test of H0: There is no association between two categorical variables in the population of interest. P-values for this test come from a chi-square distribution with df = (number of rows − 1) × (number of columns − 1). This new procedure is known as a chi-square test for independence.

Chi-Square Test for Independence The chi-square test for independence is also known as the chi-square test for association.

Suppose the conditions are met. You can use the chi-square test for independence to test H0: There is no association between two categorical variables in the population of interest. Ha: There is an association between two categorical variables in the population of interest. Or, alternatively, H0: Two categorical variables are independent in the population of interest. Ha: Two categorical variables are not independent in the population of interest. Start by finding the expected counts. Then calculate the chi-square statistic c2 =

P-value

χ2

(Observed − Expected)2 ∙ Expected

where the sum is over all cells in the two-way table. If H0 is true, the c2 statistic has approximately a chi-square distribution with degrees of freedom = (number of rows − 1)(number of columns − 1). The P-value is the area to the right of c2 under the corresponding chi-square density curve.

Now we’re ready to complete the significance test for the anger and heart disease study.

Example

Angry People and Heart Disease Chi-square test for independence

STEP

4

Here is the complete table of observed and expected counts for the CHD and anger study side by side: Observed

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Expected

Low

Moderate

High

Low

Moderate

High

CHD

  53

 110

 27

  69.73

 106.08

 14.19

No CHD

3057

4621

606

3040.27

4624.92

618.81

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C H A P T E R 1 1 I n f e r e n c e f o r D i s t r i b u t i o n s o f C at e g o r i c a l Data

Do the data provide convincing evidence of an association between anger level and heart disease in the population of interest?

State:  We want to perform a test of H0: There is no association between anger level and heart-disease status in the population of people with normal blood pressure. Ha: There is an association between anger level and heart-disease status in the population of people with normal blood pressure. Because no significance level was stated, we’ll use a = 0.05.

Plan:  If conditions are met, we should carry out a chi-square test for independence. •  Random:  The data came from a random sample of 8474 people with normal blood pressure. ºº 10%:  Because the researchers sampled without replacement, we need to check that the total number of people in the population with normal blood pressure is at least 10(8474) = 84,740. This seems reasonable to assume. •  Large Counts:  All the expected counts are at least 5 (the smallest is 14.19), so this condition is met.

Do:  We perform calculations assuming H0 is true. •  Test statistic: 2

c =

∙

(Observed − Expected)2

Expected (53 − 69.73)2 (110 − 106.08)2 c (606 − 618.81)2 = + + + 69.73 106.08 618.81 = 4.014 + 0.145 + c+ 0.265 = 16.077

•  P-value:  The two-way table of anger level versus heart disease has 2 rows and 3 columns. We will use the chi-square distribution with df = (2 − 1)(3 − 1) = 2 to find the P-value. Look at the df = 2 line in Table C. The observed statistic c2 = 16.077 is larger than the critical value 15.20 for a = 0.0005. So the P-value is less than 0.0005. Using Technology:  The calculator’s c2-Test gives c2 = 16.077 and P-value = 0.00032 using df = 2.

Conclude:  Because the P-value of 0.00032 is less than a = 0.05, we reject H0 . We have convincing evidence of an association between anger level and heart-disease status in the population of people with normal blood pressure. For Practice Try Exercise

45

c

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!

n

A follow-up analysis reveals that two cells contribute most of the chi-square statistic: Low anger, CHD (4.014) and High anger, CHD (11.564). A much ­smaller number of low-anger people developed CHD than expected. And a much larger number of high-anger people got CHD than expected. Can we conclude that proneness to anger causes heart disease? No. autio The anger and heart-disease study is an observational study, not an experiment. It isn’t surprising that some other variables are confounded with anger level. For example, people prone to anger are more likely than others to be men who drink and smoke. We don’t know whether the increased rate of heart disease among those with higher anger levels in the study is due to their anger or perhaps to their drinking and smoking or maybe even to gender.

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Check Your Understanding

Many popular businesses are franchises—think of McDonald’s. The owner of a local franchise benefits from the brand recognition, national advertising, and detailed guidelines provided by the franchise chain. In return, he or she pays fees to the franchise firm and agrees to follow its policies. The relationship between the local owner and the franchise firm is spelled out in a detailed contract. One clause that the contract may or may not contain is the entrepreneur’s right to an exclusive territory. This means that the new outlet will be the only representative of the franchise in a specified territory and will not have to compete with other outlets of the same chain. How does the presence of an exclusive-territory clause in the contract relate to the survival of the business? A study designed to address this question collected data from a random sample of 170 new franchise firms. Two categorical variables were measured for each franchisor. First, the franchisor was classified as successful or not based on whether or not it was still offering franchises as of a certain date. Second, the contract each franchisor offered to franchisees was classified according to whether or not there was an exclusive-territory clause. Here are the count data, arranged in a two-way table:9

Exclusive Territory Success

Yes

No

Total

Yes

108

15

123

No

 34

13

 47

Total

142

28

170

Do these data provide convincing evidence at the a = 0.01 level of an association between an exclusive-territory clause and business survival for new franchise firms?

Using Chi-Square Tests Wisely

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c

Both the chi-square test for homogeneity and the chi-square test for autio independence start with a two-way table of observed counts. They even calculate the test statistic, degrees of freedom, and P-value in the same way. The questions that these two tests answer are different, however. A chi-square test for homogeneity tests whether the distribution of a categorical variable is the same for each of several populations or treatments. The chi-square test for independence tests whether two categorical variables are associated in some population of interest. Unfortunately, it is quite common to see questions asking about association when a test for homogeneity applies and questions asking about differences between proportions or the distribution of a variable when a test of independence applies. Sometimes, people avoid the distinction altogether and pose questions about the “relationship” between two variables. Instead of focusing on the question asked, it’s much easier to look at how the data were produced. If the data come from two or more independent random samples or treatment groups in a randomized experiment, then do a chi-square test for homogeneity. If the data come from a single random sample, with the individuals classified according to two categorical variables, use a chi-square test for independence.

!

n

AP® EXAM TIP ​If you have trouble distinguishing the two types of chi-square tests for two-way tables, you’re better off just saying “chi-square test” than choosing the wrong type. Better yet, learn to tell the difference!

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Example

Scary Movies and Fear Choosing the right type of chi-square test Are men and women equally likely to suffer lingering fear from watching scary movies as children? Researchers asked a random sample of 117 college students to write narrative accounts of their exposure to scary movies before the age of 13. More than one-fourth of the students said that some of the fright symptoms are still present when they are awake.10 The following table breaks down these results by gender. Gender Male

Female

Total

Yes

 7

29

 36

No

31

50

 81

Total

38

79

117

Fright symptoms?

Minitab output for a chi-square test using these data is shown below. Chi-Square Test: Male, Female Expected counts are printed below observed counts Chi-Square contributions are printed below expected counts Male Female Total 1 7 29 36 11.69 24.31 1.883 0.906 2 31 50 81 26.31 54.69 0.837 0.403 Total 38

79 117

Chi-Sq = 4.028, DF = 1, P-Value = 0.045

Problem:  Assume that the conditions for performing inference are met. (a) Explain why a chi-square test for independence and not a chi-square test for ­homogeneity should be used in this setting. (b) State an appropriate pair of hypotheses for researchers to test in this setting. (c) Which cell contributes most to the chi-square statistic? In what way does this cell differ from what the null hypothesis suggests? (d) Interpret the P-value in context. What conclusion would you draw at a = 0.01?

Solution: (a) The data were produced using a single random sample of college students, who were then ­classified by gender and whether or not they had lingering fright symptoms. The chi-square test for homogeneity requires independent random samples from each population. (b) The null hypothesis is H0: There is no association between gender and ongoing fright symptoms in the population of college students. The alternative hypothesis is Ha: There is an association between gender and ongoing fright symptoms in the population of college students.

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Section 11.2 Inference for Two-Way Tables

(c) Men who admit to having lingering fright symptoms account for the largest component of the chi-square statistic: 1.883 of the total 4.028. Far fewer men in the sample admitted to fright symptoms (7) than we would expect if H0 were true (11.69). (d) If gender and ongoing fright symptoms really are independent in the population of interest, there is a 0.045 chance of obtaining a random sample of 117 students that gives a chi-square statistic of 4.028 or higher. Because the P-value, 0.045, is greater than 0.01, we would fail to reject H0. We do not have convincing evidence that there is an association between gender and fright symptoms in the population of college students. For Practice Try Exercise

47

What if we want to compare two proportions?  Shopping at second-

hand stores is becoming more popular and has even attracted the attention of business schools. A study of customers’ attitudes toward secondhand stores interviewed separate random samples of shoppers at two secondhand stores of the same chain in two cities. The two-way table shows the breakdown of respondents by gender.11 City 1

City 2

Men

 38

 68

Women

203

150

Total

241

218

Do the data provide convincing evidence of a difference in the true gender distributions of shoppers at the two stores? To answer this question, we could perform a chi-square test for homogeneity. Our hypotheses are H0: There is no difference in the true gender distributions of shoppers at the two stores. Ha: There is a difference in the true gender distributions of shoppers at the two stores. But a difference in gender distributions would mean that there is a difference in the true proportions of female shoppers at the two stores. So we could also use a two-sample z test from Section 10.1 to compare two proportions. The hypotheses for this test are H0: p1 − p2 = 0 Ha: p1 − p2 ∙ 0

c

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!

n

where p1 and p2 are the true proportions of women shoppers at Store 1 and Store 2, respectively. The TI-84 screen shots in the margin show the results from a two-sample z test for p1 − p2 and from a chi-square test for homogeneity. (We checked that the Random, 10%, and Large Counts conditions are met before carrying out the calculations.) Note that the P-values from the two tests are the same except for rounding errors. You can also check that the chi-square statistic is the square of the two‑sample z statistic: (3.915…)2 = 15.334. As the previous example suggests, the chi-square test for homogeneity autio based on a 2 × 2 two-way table is equivalent to the two-sample z test for p1 − p2 with a two-sided alternative hypothesis. We cannot use a chi-square test for a one-sided alternative hypothesis. The two-sample z procedures allow us

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C H A P T E R 1 1 I n f e r e n c e f o r D i s t r i b u t i o n s o f C at e g o r i c a l Data to perform one-sided tests and to construct confidence intervals for the difference between proportions. For that reason, we recommend the Chapter 10 methods for comparing two proportions whenever you are given a choice.

Grouping quantitative data into categories  As we mentioned in Chapter 1, it is possible to convert a quantitative variable to a categorical variable by grouping together intervals of values. Here’s an example. Researchers surveyed independent random samples of shoppers at two secondhand stores of the same chain in two cities. The two-way table below summarizes data on the incomes of the shoppers in the two samples. Income Under $10,000 $10,000 to $19,999 $20,000 to $24,999 $25,000 to $34,999 $35,000 or more

City 1 70 52 69 22 28

City 2 62 63 50 19 24

Personal income is a quantitative variable. But by grouping the values of this variable, we create a categorical variable. We could use these data to carry out a chi-square test for homogeneity because the data came from independent random samples of shoppers at the two stores. Comparing the distributions of income for shoppers at the two stores would give more information than simply comparing their mean incomes.

What can we do if the expected cell counts aren’t all at least 5?  ​Let’s look at a situation where this is the case. A sample survey asked a

random sample of young adults, “Where do you live now? That is, where do you stay most often?” A two-way table of all 2984 people in the sample (both men and women) classified by their age and by where they lived is shown below.12 Living arrangement is a categorical variable. Even though age is quantitative, the two-way table treats age as dividing the young adults into four categories. The table gives the observed counts for all 20 combinations of age and living ­arrangement. Living arrangement Parents’ home Another person’s home Your own place Group quarters Other Total

19 324  37 116  58   5 540

Age (years) 20 21 378 337  47  40 279 372  60  49   2   3 766 801

Total 1357  162 1254  192   19 2984

22 318  38 487  25   9 877

Our null hypothesis is H0: There is no association between age and living arrangement in the population of young adults. The table below shows the expected counts assuming H0 is true. We can see that two of the expected counts (circled in red) are less than 5. This violates the Large Counts condition. Age (years) Living arrangement

20

21

22

Total

245.57

348.35

364.26

398.82

1357

29.32

41.59

43.49

47.61

162

Your own place

226.93

321.90

336.61

368.55

1254

Group quarters

34.75

49.29

51.54

56.43

192

Other

3.44

4.88

5.10

5.58

19

Total

540

766

801

877

2984

Parents’ home Another person’s home

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Section 11.2 Inference for Two-Way Tables

721

A clever strategy is to “collapse” the table by combining two or more rows or columns. In this case, it might make sense to combine the Group quarters and Other living arrangements. Doing so and then running a chi-square test in Minitab gives the following output. Notice that the Large Counts condition is now met. Chi-Square Test: 19, 20, 21, 22 Expected counts are printed below observed counts Chi-Square contributions are printed below expected counts

19

1

20

21

22 Total

324 378 337 318 1357

245.57 348.35 364.26 398.82

25.049 2.525 2.040 16.379 2

37 47 40 38 162



29.32 41.59 43.49 47.61



2.014 0.705 0.279 1.940

3

116 279 372 487 1254

226.93 321.90 336.61 368.55

54.226 5.719 3.720 38.068 4

63 62 52 34 211

38.18 54.16 56.64 62.01

16.129 1.134 0.380 12.654 Total 540

766

801

877

2984

Chi-Sq = 182.961, DF = 9, P-Value = 0.000

case closed

Do Dogs Resemble Their Owners? In the chapter-opening Case Study (page 677), we described a study that investigated whether or not dogs resemble their owners. The researchers who conducted the experiment believe that resemblance between dog and owner might differ for purebred and mixed-breed dogs. Here is a two-way table summarizing the results of the experiment: Breed status

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Resemblance?

Purebred dogs

Mixed-breed dogs

Resemble owner

16

 7

Don’t resemble

 9

13

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C H A P T E R 1 1 I n f e r e n c e f o r D i s t r i b u t i o n s o f C at e g o r i c a l Data

1. Why did researchers photograph a random sample of dogs and their owners in this study? Do the data from this study provide convincing evidence of an association between dogs’ breed status and whether or not they resemble their owners? Questions 2 through 5 address this issue. 2. Which type of chi-square test should be used to help answer the question of interest? State an appropriate pair of hypotheses for the test you choose. 3. The table shows the expected counts for the appropriate chisquare test in Question 2. Breed status Resemblance?

Purebred dogs

Mixed-breed dogs

Resemble owner

12.78

10.22

Don’t resemble

12.22

 9.78

(a) Show how the expected count for the cell “purebred dogs, resemble owner” was computed. (b) Explain why the Large Counts condition is met. 4. Find the test statistic and P-value. Be sure to state the degrees of freedom you are using. 5. What conclusion would you draw?

Section 11.2

Summary •

•

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We can use a two-way table to summarize data involving two categorical variables. To analyze the data, we compare the conditional distributions of one variable for each value of the other variable. Then we turn to formal inference. Two different ways of producing data for two-way tables lead to two different types of chi-square tests. Some studies aim to compare the distribution of a single categorical variable for each of several populations or treatments. In such cases, researchers should take independent random samples from the populations of interest or use the groups in a randomized experiment. The null hypothesis is that there is no difference in the distribution of the categorical variable for each of the populations or treatments. We use the chi-square test for homogeneity to test this hypothesis.

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Section 11.2 Inference for Two-Way Tables •

•

•

•

The conditions for performing a chi-square test for homogeneity are: • Random: The data come from independent random samples or the groups in a randomized experiment. ~~ 10%: When sampling without replacement, check that the population is at least 10 times as large as the sample. • Large Counts: All expected counts must be at least 5. Other studies are designed to investigate the relationship between two categorical variables. In such cases, researchers take a random sample from the population of interest and classify each individual based on the two categorical variables. The chi-square test for independence tests the null hypothesis that there is no association between the two categorical variables in the population of interest. Another way to state the null hypothesis is H0: The two categorical variables are independent in the population of interest. The conditions for performing a chi-square test for independence are: • Random: The data come from a well-designed random sample or randomized experiment. ~~ 10%: When sampling without replacement, check that the population is at least 10 times as large as the sample. • Large Counts: All expected counts must be at least 5. The expected count in any cell of a two-way table when H0 is true is expected count =

•

P-value

χ2

•

row total # column total table total

The chi-square statistic is (Observed − Expected)2 Expected where the sum is over all cells in the two-way table. Both types of chi-square tests for two-way tables compare the value of the statistic c2 with critical values from the chi-square distribution with df = (number of rows − 1)(number of columns − 1). Large values of c2 are evidence against H0 and in favor of Ha, so the P-value is the area under the chi-square density curve to the right of c2. If the test finds a statistically significant result, consider doing a follow-up analysis that compares the observed and expected counts and that looks for the largest components of the chi-square statistic. c2 =

•

723

∙

11.2 T echnology Corner TI-Nspire Instructions in Appendix B; HP Prime instructions on the book’s Web site. 27. Chi-square tests for two-way tables on the calculator

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Exercises

Section 11.2

27. Why men and women play sports  Do men and women participate in sports for the same reasons? One goal for sports participants is social comparison— the desire to win or to do better than other people. Another is mastery—the desire to improve one’s skills or to try one’s best. A study on why students participate in sports collected data from independent random samples of 67 male and 67 female undergraduates at a large university.13 Each student was classified into one of four categories based on his or her responses to a questionnaire about sports goals. The four categories were high social comparison–high mastery (HSC-HM), high social comparison–low mastery (HSC-LM), low social comparison–high mastery (LSC-HM), and low social comparison–low mastery (LSC-LM). One purpose of the study was to compare the goals of male and female students. Here are the data displayed in a two-way table:

(a) Calculate the conditional distribution (in proportions) of responses for each group of parents.

pg 697

(b) Make an appropriate graph for comparing the conditional distributions in part (a). (c) Write a few sentences comparing the distributions of responses for the three groups of parents. 29. Why women and men play sports  Refer to Exercise 27. Do the data provide convincing evidence of a difference in the distributions of sports goals for male and female undergraduates at the university?

pg 704

(a) State appropriate null and alternative hypotheses for a significance test to help answer this question. (b) Calculate the expected counts. Show your work. (c) Calculate the chi-square statistic. Show your work. 30. How are schools doing?  Refer to Exercise 28. Do the data provide convincing evidence of a difference in the distributions of opinions about how high schools are doing among black, Hispanic, and white parents?

Gender Goal

Female

Male

14  7 21 25

31 18  5 13

HSC-HM HSC-LM LSC-HM LSC-LM

(a) State appropriate null and alternative hypotheses for a significance test to help answer this question. (b) Calculate the expected counts. Show your work.

(a) Calculate the conditional distribution (in proportions) of the reported sports goals for each gender. (b) Make an appropriate graph for comparing the conditional distributions in part (a). (c) Write a few sentences comparing the distributions of sports goals for male and female undergraduates. 28. How are schools doing?  The nonprofit group Public Agenda conducted telephone interviews with three randomly selected groups of parents of high school children. There were 202 black parents, 202 Hispanic parents, and 201 white parents. One question asked was “Are the high schools in your state doing an excellent, good, fair, or poor job, or don’t you know enough to say?” Here are the survey results:14

Excellent Good Fair Poor Don’t know Total

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Black parents

Hispanic parents

White parents

12 69 75 24 22 202

34 55 61 24 28 202

22 81 60 24 14 201

(c) Calculate the chi-square statistic. Show your work. pg 704

31. Why women and men play sports  Refer to Exercises 27 and 29. (a) Check that the conditions for performing the ­chi-square test are met. (b) Use Table C to find the P-value. Then use your calculator’s c2cdf command. (c) Interpret the P-value from the calculator in ­context. (d) What conclusion would you draw? Justify your ­answer. 32. How are schools doing?  Refer to Exercises 28 and 30. (a) Check that the conditions for performing the chi-square test are met. (b) Use Table C to find the P-value. Then use your calculator’s c2cdf command. (c) Interpret the P-value from the calculator in context. (d) What conclusion would you draw? Justify your answer.

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Section 11.2 Inference for Two-Way Tables 33. Python eggs  How is the hatching of water python eggs influenced by the temperature of the snake’s nest? Researchers randomly assigned newly laid eggs to one of three water temperatures: hot, neutral, or cold. Hot duplicates the extra warmth provided by the mother python, and cold duplicates the absence of the mother. Here are the data on the number of eggs that hatched and didn’t hatch:15

pg 708

36. Going Nuts  The UR Nuts Company sells Deluxe and Premium nut mixes, both of which contain only cashews, brazil nuts, almonds, and peanuts. The Premium nuts are much more expensive than the Deluxe nuts. A consumer group suspects that the two nut mixes are really the same. To find out, the group took separate random samples of 20 pounds of each nut mix and recorded the weights of each type of nut in the sample. Here are the data:18

Water Temperature Hatched? Yes No

Cold

Neutral

Hot

16 11

38 18

75 29

Type of mix Type of nut

(a) Compare the distributions of hatching status for the three treatments. (b) Are the differences between the three groups statistically significant? Give appropriate evidence to support your answer. 34. Don’t do drugs!  Cocaine addicts need cocaine to feel any pleasure, so perhaps giving them an antidepressant drug will help. A three-year study with 72 chronic cocaine users compared an antidepressant drug called desipramine with lithium (a standard drug to treat cocaine addiction) and a placebo. One-third of the subjects were randomly assigned to receive each treatment. Here are the results:16



Desipramine

Lithium

Placebo

Yes

10

18

20

No

14

6

4

Deluxe

6 lb

5 lb

Brazil nut

3 lb

4 lb

Almond

5 lb

6 lb

Peanut

6 lb

5 lb

Explain why we can’t use a chi-square test to determine whether these two distributions differ significantly.

37. How to quit smoking  It’s hard for smokers to quit. Perhaps prescribing a drug to fight depression will work as well as the usual nicotine patch. Perhaps combining the patch and the drug will work better than either treatment alone. Here are data from a randomized, double-blind trial that compared four treatments.19 A “success” means that the subject did not smoke for a year following the beginning of the study.

Drug administered Relapsed?

Premium

Cashew

Group

Treatment

1 2 3 4

Nicotine patch Drug Patch plus drug Placebo

Subjects

Successes

244 244 245 160

40 74 87 25

(a) Compare the distributions of relapse status for the three treatments.

(a) Summarize these data in a two-way table. Then compare the success rates for the four treatments.

(b) Are the differences among the three groups statistically significant? Give appropriate evidence to support your answer.

(b) Explain in words what the null hypothesis H0:  p1 = p2 = p3 = p4 says about subjects’ smoking habits.

35. Sorry, no chi-square  How do U.S. residents who travel overseas for leisure differ from those who travel for business? The following is the breakdown by occupation:17 Occupation Professional/technical Manager/executive Retired Student Other Total



Leisure travelers (%)

Business travelers (%)

36 23 14  7 20 100

39 48  3  3  7 100

Explain why we can’t use a chi-square test to learn whether these two distributions differ significantly.

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(c) Do the data provide convincing evidence of a difference in the effectiveness of the four treatments at the a = 0.05 significance level? 38. Preventing strokes  Aspirin prevents blood from clotting and so helps prevent strokes. The Second European Stroke Prevention Study asked whether adding another anticlotting drug named dipyridamole would be more effective for patients who had already had a stroke. Here are the data on strokes during the two years of the study:20 Group

Treatment

1 2 3 4

Placebo Aspirin Dipyridamole Both

Number of patients Number of strokes 1649 1649 1654 1650

250 206 211 157

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C H A P T E R 1 1 I n f e r e n c e f o r D i s t r i b u t i o n s o f C at e g o r i c a l Data

(a) Summarize these data in a two-way table. Then compare the stroke rates for the four treatments.

(b) Compute the expected counts assuming that H0 is true. Show your work.

(b) Explain in words what the null hypothesis H0: p1 = p2 = p3 = p4 says about the incidence of strokes.

(c) Calculate the chi-square statistic, df, and P-value.

(c) Do the data provide convincing evidence of a difference in the effectiveness of the four treatments at the a = 0.05 significance level? 39. How to quit smoking  Perform a follow-up analysis of the test in Exercise 37 by finding the individual components of the chi-square statistic. Which cell(s) contributed most to the final result and in what direction? 40. Preventing strokes  Perform a follow-up analysis of the test in Exercise 38 by finding the individual components of the chi-square statistic. Which cell(s) contributed most to the final result and in what direction? 41. Attitudes toward recycled products  Some people believe recycled products are lower in quality than other products, a fact that makes recycling less practical. Here are data on attitudes toward coffee filters made of recycled paper from a random sample of adults:21

pg 711

(d) What conclusion would you draw? 44. Is astrology scientific?  Refer to Exercise 42. (a) State appropriate hypotheses for performing a chisquare test of independence in this setting. (b) Compute the expected counts assuming that H0 is true. Show your work. (c) Calculate the chi-square statistic, df, and P-value. (d) What conclusion would you draw? 45. Regulating guns  The National Gun Policy Survey asked a random sample of adults, “Do you think there should be a law that would ban possession of handguns except for the police and other authorized persons?” Here are the responses, broken down by the respondent’s level of education:23

pg 715

Education Less than High school Some College high school grad college grad

Recycled coffee filter status Quality rating

Buyers

Nonbuyers

20  7  9

29 25 43

Higher Same Lower



Yes No



Make a well-labeled bar graph that compares buyers’ and nonbuyers’ opinions about recycled filters. Describe what you see.

42. Is astrology scientific?  The General Social Survey asked a random sample of adults their opinion about whether astrology is very scientific, sort of scientific, or not at all scientific. Here is a two-way table of counts for people in the sample who had three levels of higher education:22 Degree Held

Not at all scientific Very or sort of scientific



Associate’s

Bachelor’s

Master’s

169  65

256  65

114  18

Make a well-labeled bar graph that compares opinions about astrology for the three education categories. Describe what you see.

43. Attitudes toward recycled products  Refer to Exercise 41. (a) State appropriate hypotheses for performing a chisquare test of independence in this setting.

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58 58

 84 129

169 294

 98 135

Postgrad degree 77 99

Does the sample provide convincing evidence of an association between education level and opinion about a handgun ban in the adult ­p opulation?

46. Market research  Before bringing a new product to market, firms carry out extensive studies to learn how consumers react to the product and how best to advertise its advantages. Here are data from a study of a new laundry detergent.24 The participants are a random sample of people who don’t currently use the established brand that the new product will compete with. Give subjects free samples of both detergents. After they have tried both for a while, ask which they prefer. The answers may depend on other facts about how people do laundry. Laundry Practices Soft water, Soft water, Hard water, Hard water, warm wash hot wash warm wash hot wash Prefer  standard  product Prefer new  product

53

27

42

30

63

29

68

42

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Section 11.2 Inference for Two-Way Tables

Does the sample provide convincing evidence of an association between laundry practices and product preference in the population of interest?

47. Where do young adults live?  A survey by the ­National Institutes of Health asked a random sample of young adults (aged 19 to 25 years), “Where do you live now? That is, where do you stay most often?” Here is the full two-way table (omitting a few who refused to answer and one who claimed to be homeless):25

The answer may differ for different groups of students. Here are results for separate random samples of American and Asian students at a large mid­ western university:26

pg 718

Parents’ home Another person’s home Own place Group quarters

Female

Male

923 144 1294 127

986 132 1129 119

(a) Should we use a chi-square test for homogeneity or a chi-square test for independence in this setting? Justify your answer. (b) State appropriate hypotheses for performing the type of test you chose in part (a). Minitab output from a chi-square test is shown below. Chi-Square Test: Female, Male Expected counts are printed below observed counts Chi-Square contributions are printed below expected counts Female Male Total 1 923 986 1909 978.49 930.51 3.147 3.309 2

144 132 276 141.47 134.53 0.045 0.048

3

1294 1129 2423 1241.95 1181.05 2.181 2.294

4

127 119 246 126.09 119.91 0.007 0.007

Total 2488 2366 4854 Chi-Sq = 11.038, DF = 3, P-Value = 0.012

(c) Check that the conditions for carrying out the test are met. (d) Interpret the P-value in context. What conclusion would you draw? 48. Students and catalog shopping  What is the most important reason that students buy from catalogs?

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727

Save time Easy Low price Live far from stores No pressure to buy

American

Asian

29 28 17 11 10

10 11 34  4  3

(a) Should we use a chi-square test for homogeneity or a chi-square test for independence in this setting? Justify your answer. (b) State appropriate hypotheses for performing the type of test you chose in part (a). Minitab output from a chi-square test is shown below. Chi-Square Test: American, Asian Expected counts are printed below observed counts Chi-Square contributions are printed below expected counts American Asian Total 1 29 10 39 23.60 15.40 1.236 1.894 2

28 11 39 23.60 15.40 0.821 1.258

3

17 34 51 30.86 20.14 6.225 9.538

4

11 4 15 9.08 5.92 0.408 0.625

5

10 3 13 7.87 5.13 0.579 0.887

Total 95 62 157 Chi-Sq = 23.470, DF = 4, P-Value = 0.0001

(c) Check that the conditions for carrying out the test are met. (d) Interpret the P-value in context. What conclusion would you draw? 49. Treating ulcers  Gastric freezing was once a recommended treatment for ulcers in the upper intestine. Use of gastric freezing stopped after experiments showed it had no effect. One randomized comparative experiment found that 28 of the 82 gastric-­freezing patients improved, while 30 of

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C H A P T E R 1 1 I n f e r e n c e f o r D i s t r i b u t i o n s o f C at e g o r i c a l Data

the 78 patients in the placebo group improved.27 We can test the hypothesis of “no difference” in the effectiveness of the treatments in two ways: with a two-sample z test or with a chi-square test. (a) Minitab output for a chi-square test is shown below. State appropriate hypotheses and interpret the Pvalue in context. What conclusion would you draw? Chi-Square Test: Gastric freezing, Placebo Expected counts are printed below observed counts Chi-Square contributions are printed below expected counts Gastric freezing Placebo Total 1 28 30 58 29.73 28.27 0.100 0.105 2 54 48 102 52.27 49.73 0.057 0.060 Total 82 78 160 Chi-Sq = 0.322, DF = 1, P-Value = 0.570

(b) Minitab output for a two-sample z test is shown below. Explain how these results are consistent with the test in part (a). Test for Two Proportions Sample X N Sample p 1 28 82 0.341463 2 30 78 0.384615 Difference = p (1) − p (2) Estimate for difference: −0.0431520 Test for difference = 0 (vs not = 0): Z = −0.57 P-Value = 0.570

50. Opinions about the death penalty  The General Social Survey asked separate random samples of people with only a high school degree and people with a bachelor’s degree, “Do you favor or oppose the death penalty for persons convicted of murder?” The following table gives the ­responses of people whose highest education was a high school degree and of people with a bachelor’s degree: Highest education level High school

Bachelor’s degree

1010  369

319 185

Favor Oppose



We can test the hypothesis of “no difference” in support for the death penalty among people in these educational categories in two ways: with a twosample z test or with a chi-square test. (a) Minitab output for a chi-square test is shown below. State appropriate hypotheses and interpret the Pvalue in context. What conclusion would you draw?

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Chi-Square Test: C1, C2 Expected counts are printed below ­observed counts Chi-Square contributions are printed below expected counts C1 C2 Total 1 1010 319 1329 973.28 355.72 1.385 3.790 2 369 185 554 405.72 148.28 3.323 9.092 Total 1379 504 1883 Chi-Sq = 17.590, DF = 1, P-Value = 0.000

(b) Minitab output for a two-sample z test is shown below. Explain how these results are consistent with the test in part (a). Test for Two Proportions Sample X N Sample p 1 1010 1379 0.732415 2 319 504 0.632937 Difference = p (1) − p (2) Estimate for difference: 0.0994783 Test for difference = 0 (vs not = 0): Z = 4.19 P-Value = 0.000

Multiple choice: Select the best answer for Exercises 51 to 56. Exercises 51 to 55 refer to the following setting. The National Longitudinal Study of Adolescent Health interviewed a random sample of 4877 teens (grades 7 to 12). One question asked was “What do you think are the chances you will be married in the next ten years?” Here is a two-way table of the responses by gender:28

Almost no chance Some chance, but probably not A 50–50 chance A good chance Almost certain

Female

Male

119 150 447 735 1174

103 171 512 710 756

51. Which of the following would be the most appropriate type of graph for these data? (a) A bar chart showing the marginal distribution of opinion about marriage (b) A bar chart showing the marginal distribution of gender (c) A bar chart showing the conditional distribution of gender for each opinion about marriage (d) A bar chart showing the conditional distribution of opinion about marriage for each gender (e) Dotplots that display the number in each opinion category for each gender

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Section 11.2 Inference for Two-Way Tables 52. The appropriate null hypothesis for performing a chi-square test is that

For Exercises 57 and 58, you may find the inference summary chart inside the back cover helpful.

(a) equal proportions of female and male teenagers are almost certain they will be married in 10 years.

57. Inference recap (8.1 to 11.2)  In each of the following settings, state which inference procedure from Chapter 8, 9, 10, or 11 you would use. Be specific. For example, you might say “two-sample z test for the difference between two proportions.” You do not need to carry out any procedures.29

(b) there is no difference between the distributions of female and male teenagers’ opinions about marriage in this sample. (c) there is no difference between the distributions of female and male teenagers’ opinions about marriage in the population. (d) there is no association between gender and opinion about marriage in the sample. (e) there is no association between gender and opinion about marriage in the population. 53. The expected count of females who respond “almost certain” is (a) 487.7.

(c) 965.

(b) 525.

(d) 1038.8.

(e) 1174.

54. The degrees of freedom for the chi-square test for this two-way table are (a) 4.

(c) 10.

(b) 8.

(d) 20.

(e) 4876.

55. For these data, c2 = 69.8 with a P-value of approximately 0. Assuming that the researchers used a significance level of 0.05, which of the following is true? (a) A Type I error is possible. (b) A Type II error is possible. (c) Both a Type I and a Type II error are possible. (d) There is no chance of making a Type I or Type II error because the P-value is approximately 0. (e) There is no chance of making a Type I or Type II error because the calculations are correct. 56. When analyzing survey results from a two-way table, the main distinction between a test for independence and a test for homogeneity is (a) how the degrees of freedom are calculated. (b) how the expected counts are calculated.

(a) What is the average voter turnout during an election? A random sample of 38 cities was asked to report the percent of registered voters who actually voted in the most recent election. (b) Are blondes more likely to have a boyfriend than the rest of the single world? Independent random samples of 300 blondes and 300 nonblondes were asked whether they have a boyfriend. 58. Inference recap (8.1 to 11.2)  In each of the following settings, state which inference procedure from Chapter 8, 9, 10, or 11 you would use. Be specific. For example, you might say “two-sample z test for the difference between two proportions.” You do not need to carry out any procedures.30 (a) Is there a relationship between attendance at religious services and alcohol consumption? A random sample of 1000 adults was asked whether they regularly attend religious services and whether they drink alcohol daily. (b) Separate random samples of 75 college students and 75 high school students were asked how much time, on average, they spend watching television each week. We want to estimate the difference in the average amount of TV watched by high school and college students. Exercises 59 to 60 refer to the following setting. For their final project, a group of AP® Statistics students investigated the following question: “Will changing the rating scale on a survey affect how people answer the question?” To find out, the group took an SRS of 50 students from an alphabetical roster of the school’s just over 1000 students. The first 22 students chosen were asked to rate the cafeteria food on a scale of 1 (terrible) to 5 (excellent). The remaining 28 students were asked to rate the cafeteria food on a scale of 0 (terrible) to 4 (excellent). Here are the data: 1 to 5 scale Rating Frequency

1 2

(c) the number of samples obtained. (d) the number of rows in the two-way table. (e) the number of columns in the two-way table.

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2 3

3 1

4 13

5 3

0 to 4 scale Rating Frequency

0 0

1 0

2 2

3 18

4 8

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C H A P T E R 1 1 I n f e r e n c e f o r D i s t r i b u t i o n s o f C at e g o r i c a l Data

59. Design and analysis (4.2) (a) Was this an observational study or an experiment? Justify your answer. (b) Explain why it would not be appropriate to perform a chi-square test in this setting. 60. Average ratings (1.3, 10.2)  The students decided to compare the average ratings of the cafeteria food on the two scales.

(b) For the students who were given the 0-to-4 scale, the ratings have a mean of 3.21 and a standard deviation of 0.568. Since the scales differ by one point, the group decided to add 1 to each of these ratings. What are the mean and standard deviation of the adjusted ratings? (c) Would it be appropriate to compare the means from parts (a) and (b) using a two-sample t test? Justify your answer.

(a) Find the mean and standard deviation of the ratings for the students who were given the 1-to-5 scale.

FRAPPY!  Free Response AP® Problem, Yay! The following problem is modeled after actual AP® Statistics exam free response questions. Your task is to generate a complete, concise response in 15 minutes.

Directions: Show all your work. Indicate clearly the methods you use, because you will be scored on the correctness of your methods as well as on the accuracy and completeness of your results and explanations. Two statistics students wanted to know if including additional information in a survey question would change the distribution of responses. To find out, they randomly selected 30 teenagers and asked them one of the following two questions. Fifteen of the teenagers were randomly assigned to answer Question A, and the other 15 students were assigned to answer Question B. A: When choosing a college, how important is a good athletic program: very important, important, somewhat important, not that important, or not important at all? B: It’s sad that some people choose a college based on its athletic program. When choosing a college, how important is a good athletic program: very important, important, somewhat important, not that important, or not important at all? The table below summarizes the responses to both questions. For these data, the chi-square test statistic is c2 = 6.12. Question A

Question B

Total

Very important

7

2

9

Important

4

3

7

Somewhat important

2

3

5

Not that important

1

2

3

Not important at all

1

5

6

15

15

30

Total

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(a) State the hypotheses that the students are interested in testing. (b) Describe a Type I error and a Type II error in the context of the hypotheses stated in part (a). (c) For these data, explain why it would not be appropriate to use a chi-square distribution to calculate the P-value. (d) To estimate the P-value, 100 trials of a simulation were conducted, assuming that the additional information didn’t have an effect on the response to the question. In each trial of the simulation, the value of the chi-square statistic was calculated. These simulated chi-square statistics are displayed in the dotplot below.

Based on the results of the simulation, what conclusion would you make about the hypotheses stated in part (a)? After you finish, you can view two example solutions on the book’s Web site (www.whfreeman.com/tps5e). Determine whether you think each solution is “complete,” “substantial,” “developing,” or “minimal.” If the solution is not complete, what improvements would you suggest to the student who wrote it? Finally, your teacher will provide you with a scoring rubric. Score your response and note what, if anything, you would do differently to improve your own score.

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Chapter Review Section 11.1: Chi-Square Tests for Goodness of Fit In this section, you learned the details for performing a chisquare test for goodness of fit. The null hypothesis is that a single categorical variable follows a specified distribution. The alternative hypothesis is that the variable does not follow the specified distribution. The chi-square statistic measures the difference between the observed distribution of a categorical variable and its hypothesized distribution. To calculate the chi-square statistic, use the following formula that involves the observed and expected counts for each value of the categorical variable: c2 =

∙

(Observed − Expected)2 Expected

To calculate the expected counts, multiply the total sample size by the hypothesized proportion for each category. Larger values of the chi-square statistic provide more convincing evidence that the categorical variable does not have the hypothesized distribution. When the Random, 10%, and Large Counts conditions are satisfied, we can accurately model the sampling distribution of a chi-square statistic using a chi-square distribution (density curve). The Random condition says that the data are from a well-designed random sample or a randomized experiment. The 10% condition says that the sample size should be at most 10% of the population size when sampling without replacement. The Large Counts condition says that the expected counts for each category must be at least 5. In a test for goodness of fit, use a chi-square distribution with degrees of freedom = number of categories – 1. When the results of a test for goodness of fit are significant, consider doing a follow-up analysis. Identify which categories of the variable had the largest contributions to the chi-square statistic and whether the observed values in those categories were larger or smaller than expected. Section 11.2: Inference for Two-Way Tables In this section, you learned how to perform inference for categorical data that are summarized in a two-way table. To begin the analysis, compare the conditional distributions of the response variable for each value of the explanatory variable. Displaying these distributions with a bar graph will help you make an effective comparison.

There are two types of chi-square tests that could apply when data are summarized in a two-way table. A test for homogeneity compares the distribution of a single categorical variable for two or more populations or treatments. A test for independence looks for an association between two categorical variables in a single population. In a chi-square test for homogeneity, the null hypothesis is that there is no difference between the true distributions of a categorical variable for two or more populations or treatments. The alternative hypothesis is that there is a difference in the distributions. The Random condition is that the data come from independent random samples or groups in a randomized experiment. The 10% condition applies when sampling without replacement, but not in experiments. Finally, the Large Counts condition remains the same—the expected counts must be at least 5 in each cell of the two-way table. To calculate the expected counts for a test for homogeneity, use the following formula: expected count =

row total # column total table total

To calculate the P-value, compute the chi-square statistic and use a chi-square distribution with degrees of freedom = (number of rows – 1)(number of columns – 1). In a chi-square test for independence, the null hypothesis is that there is no association between two categorical variables in one population. The alternative hypothesis is that there is an association between the two variables. For this test, the Random condition says that the data must come from a single random sample. The 10% condition applies when sampling without replacement. The Large Counts condition is still the same—the expected counts must all be at least 5. The method for calculating expected counts, the chi-square statistic, the degrees of freedom, and the P-value are exactly the same in a test for independence and a test for homogeneity. As with tests for goodness of fit, when the results of a test for homogeneity or independence are significant, consider doing a follow-up analysis. Identify which cells in the twoway table had the largest contributions to the chi-square statistic and whether the observed values in those cells were larger or smaller than expected.

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What Did You Learn? Learning Objective

Section

Related Example on Page(s)

Relevant Chapter Review Exercise(s)

State appropriate hypotheses and compute expected counts for a chi-square test for goodness of fit.

11.1

681

R11.1

Calculate the chi-square statistic, degrees of freedom, and P-value for a chi-square test for goodness of fit.

11.1

683, 685

R11.1

Perform a chi-square test for goodness of fit.

11.1

688

R11.1

11.1,11.2

Discussion on 690–691

R11.4

Compare conditional distributions for data in a two-way table.

11.2

697, 711

R11.3, R11.5

State appropriate hypotheses and compute expected counts for a chi-square test based on data in a two-way table.

11.2

701, 713

R11.2, R11.3, R11.4, R11.5

Calculate the chi-square statistic, degrees of freedom, and P-value for a chi-square test based on data in a two-way table.

11.2

704

R11.3, R11.5

Conduct a follow-up analysis when the results of a chi-square test are statistically significant.

Perform a chi-square test for homogeneity.

11.2

708

R11.3

Perform a chi-square test for independence.

11.2

715

R11.5

Choose the appropriate chi-square test.

11.2

718

R11.4

Chapter 11 Chapter Review Exercises These exercises are designed to help you review the important ideas and methods of the chapter. R11.1 Testing a genetic model  Biologists wish to cross pairs of tobacco plants having genetic makeup Gg, indicating that each plant has one dominant gene (G) and one recessive gene (g) for color. Each offspring plant will receive one gene for color from each parent. The Punnett square below shows the possible combinations of genes received by the offspring: Parent 2 passes on: Parent 1 passes on:



G

g

G

GG

Gg

g

Gg

gg

The Punnett square suggests that the expected ratio of green (GG) to yellow-green (Gg) to albino (gg) tobacco plants should be 1:2:1. In other words, the biologists predict that 25% of the off-

spring will be green, 50% will be yellow-green, and 25% will be albino. To test their hypothesis about the distribution of offspring, the biologists mate 84 randomly selected pairs of yellow-green parent plants. Of 84 offspring, 23 plants were green, 50 were yellow-green, and 11 were albino. Do the data provide convincing evidence at the a = 0.01 level that the true distribution of offspring is different from what the biologists predict? R11.2 Sorry, no chi-square  We would prefer to learn from teachers who know their subject. Perhaps even preschool children are affected by how knowledgeable they think teachers are. Assign 48 three- and four-year-olds at random to be taught the name of a new toy by either an adult who claims to know about the toy or an adult who claims not to know about it. Then ask the children to pick out a picture of the new toy in a set of pictures of other toys and say its name. The response variable is the count of right answers in four tries. Here are the data:31

732

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Chapter Review Exercises

733

Correct Answers Knowledgeable teacher Ignorant teacher



 0

1

2

3

4

 5 20

1 0

6 3

3 0

9 1

The researchers report that children taught by the teacher who claimed to be knowledgeable did significantly better (c2 = 20.24, P < 0.05). Explain why this result isn’t valid.

R11.3 Stress and heart attacks  You read a newspaper article that describes a study of whether stress management can help reduce heart attacks. The 107 subjects all had reduced blood flow to the heart and so were at risk of a heart attack. They were assigned at random to three groups. The article goes on to say:

One group took a four-month stress management program, another underwent a fourmonth exercise program, and the third received usual heart care from their personal physicians. In the next three years, only three of the 33 people in the stress management group suffered “cardiac events,” defined as a fatal or non-fatal heart attack or a surgical procedure such as a bypass or angioplasty. In the same period, 7 of the 34 people in the exercise group and 12 out of the 40 patients in usual care suffered such events.32

(a) Use the information in the news article to make a two-way table that describes the study results. (b) Compare the success rates of the three treatments in preventing cardiac events. (c) Do the data provide convincing evidence that the true success rates are not the same for the three treatments? R11.4 Sexy magazine ads?  Researchers looked at 1509 full-page ads that show a model. The two-way table below shows the main audience of the magazines in which the ads were found (young men, young women, or young adults in general) and whether or not the ad was “sexual.” This was determined based on how the model was dressed (or not dressed).33

(a) Describe how these data could have been collected so that a test for homogeneity is appropriate. (b) Describe how these data could have been collected so that a test for independence is appropriate. (c) Show how each of the numbers 60.94, 424.8, and 12.835 was obtained for the “notsexy, Women” cell. (d) Which cell contributes most to the chi-square statistic? How do the observed and expected counts compare for this cell? R11.5 Popular kids  Who were the popular kids at your elementary school? Did they get good grades or have good looks? Were they good at sports? A study was performed to examine the factors that determine social status for children in grades 4, 5, and 6. Researchers administered a questionnaire to a random sample of 478 students in these grades. One of the questions they asked was “What would you most like to do at school: make good grades, be good at sports, or be popular?” The two-way table below summarizes the students’ responses.34 Gender Goal Grades Popular Sports

Female

Male

130  91  30

117  50  60

Readers Sexual Not sexual



Men

Women

General

105 514

225 351

 66 248

The following figure displays Minitab output for a chi-square test using these data.

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(a) Construct an appropriate graph to compare male and female responses. Write a few sentences ­describing the relationship between gender and goals. (b) Is there convincing evidence at the a = 0.05 level of an association between gender and goals for elementary school students?

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C H A P T E R 1 1 I n f e r e n c e f o r D i s t r i b u t i o n s o f C at e g o r i c a l Data

Chapter 11 AP® Statistics Practice Test Section I: Multiple Choice  ​Select the best answer for each question. T11.1 A chi-square test is used to test whether a 0 to 9 spinner is “fair” (that is, the outcomes are all equally likely). The spinner is spun 100 times, and the r­ esults are recorded. The degrees of freedom for the test will be

II. All expected counts are at least 5. III. The population sizes are at least 10 times the sample sizes.

(a)  8.  (b) 9.  (c) ​10.   (d) ​99.   (e) ​None of these.

(a) I only (b) II only

Exercises T11.2 and T11.3 refer to the following setting. Recent revenue shortfalls in a midwestern state led to a reduction in the state budget for higher education. To offset the reduction, the largest state university proposed a 25% tuition increase. It was determined that such an increase was needed simply to compensate for the lost support from the state. Separate random samples of 50 freshmen, 50 sophomores, 50 juniors, and 50 seniors from the university were asked whether they were strongly opposed to the increase, given that it was the minimum increase necessary to maintain the university’s budget at current levels. Here are the results. Year

Strongly Opposed?

Yes No

Freshman

Sophomore

Junior

Senior

39 11

36 14

29 21

18 32

T11.2 Which hypotheses would be appropriate for performing a chi-square test? (a) The null hypothesis is that the closer students get to graduation, the less likely they are to be opposed to tuition increases. The alternative is that how close students are to graduation makes no difference in their opinion. (b) The null hypothesis is that the mean number of students who are strongly opposed is the same for each of the 4 years. The alternative is that the mean is different for at least 2 of the 4 years. (c) The null hypothesis is that the distribution of student opinion about the proposed tuition increase is the same for each of the 4 years at this university. The alternative is that the distribution is different for at least 2 of the 4 years. (d) The null hypothesis is that year in school and student opinion about the tuition increase in the sample are independent. The alternative is that these variables are dependent. (e) The null hypothesis is that there is an association between year in school and opinion about the ­tuition increase at this university. The alternative hypothesis is that these variables are not associated. T11.3 ​The conditions for carrying out the chi-square test in exercise T11.2 are I. Independent random samples from the populations of interest.

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Which of the conditions is (are) satisfied in this case? (c)  I and II only (d) ​II and III only

(e) ​I, II, and III

Exercises T11.4 to T11.6 refer to the following setting. A random sample of traffic tickets given to motorists in a large city is examined. The tickets are classified according to the race of the driver. The results are summarized in the following table. Race:

White

Black

Hispanic

Other

69

52

18

9

Number of tickets:

The proportion of this city’s population in each of the racial categories listed above is as follows: Race:

White

Black

Hispanic

Other

Proportion:

0.55

0.30

0.08

0.07

We wish to test H0: The racial distribution of traffic tickets in the city is the same as the racial distribution of the city’s population. T11.4 Assuming H0 is true, the expected number of Hispanic drivers who would receive a ticket is (a) 8.  (b) 10.36.  (c) ​11.   (d) ​11.84.   (e) ​12. T11.5 We compute the value of the c2 statistic to be 6.58. Assuming that the conditions for inference are met, the P-value of our test is (a) greater than 0.20. (d) between 0.01 and 0.05. (b) between 0.10 and 0.20. (e) less than 0.01. (c) between 0.05 and 0.10. T11.6 The category that contributes the largest component to the c2 statistic is (a) White. (c) ​Hispanic. (b) Black. (d) ​Other. (e) The answer cannot be determined because this is only a sample. Exercises T11.7 to T11.10 refer to the following setting. All currentcarrying wires produce electromagnetic (EM) radiation, including the electrical wiring running into, through, and out of our homes. High-frequency EM radiation is thought to be a cause of cancer. The lower frequencies associated with household current are generally assumed to be harmless. To investigate the relationship between current configuration and type of cancer, researchers visited the addresses of a random sample of children who had died of some form of cancer (leukemia, lymphoma, or some other type) and classified the wiring configuration outside the dwelling as either a high-current configuration (HCC) or a low-current configuration (LCC). Here are the data:

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AP® Statistics Practice Test

HCC LCC

Leukemia

Lymphoma

Other cancers

52 84

10 21

17 31

Computer software was used to analyze the data. The output included the value c2 = 0.435. T11.7 The appropriate degrees of freedom for the c2 statistic is (a) 1.  (b) ​2.   (c) ​3.   (d) ​4.   (e) ​5. T11.8 The expected count of cases with lymphoma in homes with an HCC is 79 # 31 10 # 21 79 # 31 136 # 31 (a) .  (b)  .  (c)  .  (d)  . 215 215 10 215 (e) None of these. T11.9 Which of the following may we conclude, based on the test results? (a) There is convincing evidence of an association between wiring configuration and the chance that a child will develop some form of cancer. (b) HCC either causes cancer directly or is a major contributing factor to the development of cancer in ­children.

735

(c) Leukemia is the most common type of cancer among children. (d) There is not convincing evidence of an association between wiring configuration and the type of cancer that caused the deaths of children in the study. (e) There is convincing evidence that HCC does not cause cancer in children. T11.10 A Type I error would occur if we found convincing evidence that (a) HCC wiring caused cancer when it actually didn’t. (b) HCC wiring didn’t cause cancer when it actually did. (c) there is no association between the type of wiring and the form of cancer when there actually is an ­association. (d) there is an association between the type of wiring and the form of cancer when there actually is no ­association. (e) the type of wiring and the form of cancer have a positive correlation when they actually don’t.

Section II: Free Response  ​Show all your work. Indicate clearly the methods you use, because you will be graded on the correctness of your methods as well as on the accuracy and completeness of your results and explanations. T11.11 A large distributor of gasoline claims that 60% of all cars stopping at their service stations choose regular unleaded gas and that premium and supreme are each s­ elected 20% of the time. To investigate this claim, ­researchers collected data from a random sample of d ­ rivers who put gas in their vehicles at the distributor’s service stations in a large city. The results were as follows: Gasoline Selected



Regular

Premium

Supreme

261

51

88

Carry out a test of the distributor’s claim at the 5% significance level.

T11.12 A study conducted in Charlotte, North Carolina, tested the effectiveness of three police responses to spouse abuse: (1) advise and possibly separate the couple, (2) issue a citation to the offender, and (3) arrest the offender. ­Police officers were trained to recognize eligible cases. When presented with an eligible case, a police officer called the dispatcher, who would randomly assign one of the three available treatments to be administered. There were a total of 650 cases in the study. Each case was classified according to Treatment Subsequent arrest? No Yes

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Advise and separate

Citation

Arrest

187  25

181  43

175  39

whether the abuser was subsequently arrested within six months of the original incident.35 (a) Explain the purpose of the random assignment in the design of this study. (b) Construct a well-labeled graph that is suitable for comparing the effectiveness of the three treatments. (c) State an appropriate pair of hypotheses for performing a chi-square test in this setting. (d) Assume that all the conditions for performing the test in part (b) are met. The test yields c2 = 5.063 and a P-value of 0.0796. Interpret this P-value in context. What conclusion should we draw from the study? T11.13 In the United States, there is a strong relationship between education and smoking: well-educated people are less likely to smoke. Does a similar relationship hold in France? To find out, researchers recorded the level of education and smoking status of a random sample of 459 French men ­ elow aged 20 to 60 years.36 The two-way table b displays the data. Education Smoking Status Primary School Secondary School University Nonsmoker Former Moderate Heavy



56 54 41 36

37 43 27 32

53 28 36 16

Is there convincing evidence of an association between smoking status and educational level among French men aged 20 to 60 years?

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Chapter

12 Introduction 738 Section 12.1 739 Inference for Linear Regression Section 12.2 765 Transforming to Achieve Linearity Free Response AP® Problem, Yay! 793 Chapter 12 Review 794 Chapter 12 Review Exercises 795 Chapter 12 AP® Statistics Practice Test 797 Cumulative AP® Practice Test 4 800

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More about Regression case study

Do Longer Drives Mean Lower Scores on the PGA Tour?

Mean score

Recent advances in technology have led to golf balls that fly farther, clubs that generate more speed at impact, and swings that have been perfected through computer video analysis. Moreover, today’s professional golfers are fitter than ever. The net result is many more players who routinely hit drives traveling 300 yards or more. Does greater distance off the tee translate to better (lower) scores? We collected data on mean drive distance (in yards) and mean score per round from an SRS of 72.0 19 of the 197 players on the Professional Golfers Association (PGA) Tour in a recent year. Figure 12.1 71.5 is a scatterplot of the data with results from a least71.0 squares regression analysis added. The graph shows that there is a moderately weak negative linear re70.5 lationship between mean drive distance and mean 1 score for the 19 players in the sample. 70.0 270

280

290

300

310

Mean distance (yards)

FIGURE 12.1  ​Scatterplot and least-squares regression line of mean score versus mean drive distance for a random sample of 19 players on the PGA Tour. The slope of the least-squares regression line for the sample data is about −0.02. This line predicts a 0.02 ­decrease in mean score per round for each 1-yard ­increase in mean driving distance. But a slope of −0.02 is very close to 0. Do these data give c ­ onvincing evidence that the slope of the true regression line for all 197 PGA Tour golfers is negative? By the end of this chapter, you’ll have developed the tools you need to a ­ nswer this question.

737

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C H A P T E R 1 2 More about Regression

Introduction

It is conventional to refer to a scatterplot of the points (x, y ) as a graph of y versus x. So a scatterplot of life span versus temperature uses life span as the response variable and temperature as the explanatory variable.

Activity Materials: 50 copies of the helicopter template from the Teacher’s Resource Materials, scissors, tape measures, stopwatches

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When a scatterplot shows a linear relationship between a quantitative explanatory variable x and a quantitative response variable y, we can use the least-squares line calculated from the data to predict y for a given value of x. If the data are a random ­sample from a larger population, we need statistical inference to answer questions like these: • Is there really a linear relationship between x and y in the population, or could the pattern we see in the scatterplot plausibly happen just by chance? • In the population, how much will the predicted value of y change for each increase of 1 unit in x? What’s the margin of error for this estimate? If the data come from a randomized experiment, the values of the explanatory variable correspond to the levels of some factor that is being manipulated by the researchers. For instance, researchers might want to investigate how temperature affects the life span of mosquitoes. They could set up several tanks at each of several different temperatures and then randomly assign hundreds of mosquitoes to each of the tanks. The response variable of interest is the average time (in days) from hatching to death. Suppose that a scatterplot of average life span versus temperature has a linear form. We need statistical inference to decide if it’s plausible that there is no linear relationship between the variables, and that the pattern observed in the graph is due simply to the chance involved in the random assignment. In Section 12.1, we will show you how to estimate and test claims about the slope of the population (true) regression line that describes the relationship between two quantitative variables. The following Activity gives you a preview of inference for linear regression.

​The Helicopter Experiment Is there a linear relationship between the height from which a paper helicopter is released and the time it takes to hit the ground? In this Activity, your class will perform an experiment to investigate this question.2 1. ​Follow the directions provided with the template to construct 50 long-rotor helicopters. 2. ​Randomly assign 10 helicopters to each of five different drop heights. (The experiment works best for drop heights of 5 feet or more.) 3. ​Work in teams to release the helicopters from their assigned drop heights and record the descent times. 4. ​Make a scatterplot of the data in Step 3. Find the least-squares regression line for predicting descent time from drop height. 5. ​Interpret the slope of the regression line from Step 4 in context. What is your best guess for the increase in descent time for each additional foot of drop height? 6. ​Does it seem plausible that there is really no linear relationship between descent time and drop height and that the observed slope happened just by chance due to the random assignment? Discuss this as a class.

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Section 12.1 Inference for Linear Regression

739

Sometimes a scatterplot reveals that the relationship between two quantitative variables has a strong curved form. One strategy is to transform one or both variables so that the graph shows a linear pattern. Then we can use least-squares regression to fit a linear model to the data. Section 12.2 examines methods of transforming data to achieve linearity.

12.1 What You Will Learn

Inference for Linear Regression By the end of the section, you should be able to:

• Check the conditions for performing inference about the slope b of the population (true) regression line. • Interpret the values of a, b, s, SEb, and r 2 in context, and determine these values from computer output.

• Construct and interpret a confidence interval for the slope b of the population (true) regression line. • Perform a significance test about the slope b of the population (true) regression line.

In Chapter 3, we examined data on eruptions of the Old Faithful geyser. Figure 12.2 is a scatterplot of the duration and interval of time until the next eruption for all 222 recorded eruptions in a single month. The least-squares regression line for this population of data has been added to the graph. Its equation is predicted interval = 33.97 + 10.36 (duration) We call this the population regression line (or true regression line) because it uses all the observations that month. 100 predicted interval = 33.97 + 10.36 (duration)

Interval (minutes)

90 80 70 60 50 40 1

2

3

4

Duration (minutes)

The sample regression line is sometimes called the estimated regression line.

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5

FIGURE 12.2  ​Scatterplot of the duration and interval between eruptions of Old Faithful for all 222 eruptions in a single month. The population least-squares line is shown in blue.

Suppose we take an SRS of 20 eruptions from the population and calculate the least-squares regression line y^ = a + bx for the sample data. How does the slope b of the sample regression line relate to the slope of the population regression line? Figure 12.3 on the next page shows the results of taking three different SRSs of 20 Old Faithful eruptions in this month. Each graph displays the selected points and the least-squares regression line for that sample. Notice that the slopes of the sample regression lines (10.2, 7.7, and 9.5) vary quite a bit from the slope of the population regression line, 10.36. The pattern of variation in the slope b is described by its sampling distribution.

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C H A P T E R 1 2 More about Regression 100 90 80 70 60 50 40 1

2

3

4

100 90 80 70 60 50 40

5

Interval (minutes)

FIGURE 12.3  ​Scatterplots and least-squares regression lines for three different SRSs of 20 Old Faithful eruptions.

Interval (minutes)



Interval (minutes)

740

1

Duration (minutes)

2

3

4

100 90 80 70 60 50 40

5

1

Duration (minutes)

(a) Sample 1: y^ = 32.8 + 10.2x

2

3

4

5

Duration (minutes)

(b) Sample 2: y^ = 44.0 + 7.7x

(c) Sample 3: y^ = 36.0 + 9.5x

Sampling Distribution of b

Approximate sampling distribution of b (n = 20)

Confidence intervals and significance tests about the slope of the population regression line are based on the sampling distribution of b, the slope of the sample regression line. We used Fathom software to simulate choosing 1000 SRSs of n = 20 from the Old Faithful data, each time calculating the equation y^ = a + bx of the least-squares regression line for the sample. Figure 12.4 displays the values of the slope b for the 1000 sample regression lines. We have added a vertical line at 10.36 corresponding to the slope of the population regression line. Let’s describe this approximate sampling distribution of b. Shape:  We can see that the distribution of b-values is roughly symmetric and unimodal. Figure 12.5(a) is a Normal proba6 8 10 12 14 bility plot of these sample regression line slopes. The strong Slope linear pattern in the graph tells us that the approximate samFIGURE 12.4  ​Dotplot of the slope b of the least-squares pling distribution of b is close to Normal. regression line in 1000 simulated SRSs by Fathom software. Center:  The mean of the 1000 b-values is 10.35. This value is quite close to the slope of the population (true) regression line, 10.36. Spread:  The standard deviation of the 1000 b-values is 1.29. Soon, we will see that the standard deviation of the sampling distribution of b is actually 1.27. Figure 12.5(b) is a histogram of the b-values from the 1000 simulated SRSs. We have superimposed the density curve for a Normal distribution with mean 10.36 and standard deviation 1.27. This curve models the approximate sampling distribution of the slope quite well. 4

Expected z-score

3 2 1 0 –1 –2 –3 –4 0

(a)

2

4

6

8

Slope

10

12

14

6

16

(b)

8

10

12

14

Slope

FIGURE 12.5  ​(a) Normal probability plot and (b) histogram of the 1000 sample regression line slopes from Figure 12.4. The red density curve in Figure 12.5(b) is for a Normal distribution with mean 10.36 (marked by the yellow line) and standard deviation 1.27.

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Section 12.1 Inference for Linear Regression 100

Let’s do a quick recap. For all 222 eruptions of Old Faithful in a single month, the population regression line is: predicted interval = 33.97 + 10.36 (duration). We use the symbols a  = 33.97 and b = 10.36 to represent the y intercept and slope parameters. The standard deviation of the residuals for this line is the parameter s = 6.131. Figure 12.5(b) shows the approximate sampling distribution of the slope b of the sample regression line for samples of 20 eruptions. If we take all possible SRSs of size n = 20 from the population, we get the sampling distribution of b. Can you guess its shape, center, and spread? 3 4 5 Shape: Approximately Normal Duration (minutes) Center: mb = b = 10.36 (b is an unbiased estimator of b) s 6.131 = = 1.27 where sx is the standard deviation of Spread: sb = sx !n 1.0815 !20 the 222 eruption durations. We interpret sb just like any other standard deviation: the slopes of the sample regression lines typically differ from the slope of the population regression line by about 1.27. Here’s a summary of the important facts about the sampling distribution of b.

predicted interval = 33.97 = 10.36 (duration)

90

Interval (minutes)

741

80 70 60 50 40 1

2

Note that the symbols a and b here refer to the intercept and slope, respectively, of the population regression line. They are in no way related to Type I and Type II error probabilities, which are sometimes designated by these same symbols.

Sampling Distribution of a Slope Choose an SRS of n observations (x, y) from a population of size N with leastsquares regression line predicted y = a + bx Let b be the slope of the sample regression line. Then: • The mean of the sampling distribution of b is mb = b. • The standard deviation of the sampling distribution of b is s sb = sx !n

1 N. 10 • The sampling distribution of b will be approximately Normal if the values of the response variable y follow a Normal distribution for each value of the explanatory variable x (the Normal condition). as long as the 10% condition is satisfied: n ≤

We’ll say more about the Normal condition in a moment.

THINK ABOUT IT

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What’s with that formula for s b?  There are three factors that affect the standard deviation of the sampling distribution of b: • s, the standard deviation of the residuals for the population regression line. Because s is in the numerator of the formula, when s is larger, so is sb. When the points are more spread out around the population (true) regression line, we should expect more variability in the slopes b of sample regression lines from repeated random sampling or random assignment.

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C H A P T E R 1 2 More about Regression • sx, the standard deviation of the explanatory variable. Because sx is in the denominator of the formula, when sx is larger, sb is smaller. More variability in the values of the explanatory variable leads to a more precise estimate of the slope of the true regression line. • n, the sample size. Just like every other formula for the standard deviation of a statistic, the variability of the statistic gets smaller as the sample size increases. A larger sample size will lead to a more precise estimate of the true slope.

Conditions for Regression Inference We can fit a least-squares line to any data relating two quantitative variables, but the results are useful only if the scatterplot shows a linear pattern. Inference about regression involves more detailed conditions. Figure 12.6 shows the regression model when the conditions are met in picture form. The regression model requires that for each possible value of the explanatory variable x: 1. The mean value of the response variable my falls on the population (true) regression line my = a + bx. 2. The values of the response variable y follow a Normal distribution with common standard deviation s. For any fixed x, the responses y follow a Normal distribution with standard deviation σ .

σ σ σ

y μy = α + βx x1

x2

x3

x

FIGURE 12.6  ​The regression model when the conditions for inference are met. The line is the population (true) regression line, which shows how the mean response my changes as the explanatory variable x changes. For any fixed value of x, the observed response y varies according to a Normal distribution having mean my and standard deviation s.

THINK ABOUT IT

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What does the regression model in Figure 12.6 tell us?  ​Consider

the population of all eruptions of the Old Faithful geyser in a given year. For each eruption, let x be the duration (in minutes) and y be the interval of time (in minutes) until the next eruption. Suppose that the conditions for regression inference are met for this data set, that the population regression line is my = 34 + 10.4x, and that the spread around the line is given by s = 6. Let’s focus on the eruptions that lasted x = 2 minutes. For this “subpopulation”: • The average amount of time until the next eruption is my = 34 + 10.4(2) = 54.8 minutes.

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Section 12.1 Inference for Linear Regression

743

σ

=

6

σ

=

6

σ

=

6

• The amounts of time until the next eruption follow a Normal distribution with mean 54.8 minutes and standard deviation 6 minutes. • For about 95% of these eruptions, the amount of time y until the next eruption is between 54.8 – 2(6) = 42.8 minutes and 54.8 + 2(6) = 66.8 minutes. That is, if the previous eruption lasted 2 minutes, 95% of the time the next eruption will occur in 42.8 to 66.8 minutes.

Interval (minutes)

.8

.8

54

.8

66

42

μy = α + βx

1.5

2 2.5 Duration (minutes)

Here are the conditions for performing inference about the linear regression model.

Conditions for Regression Inference Suppose we have n observations on an explanatory variable x and a response variable y. Our goal is to study or predict the behavior of y for given values of x.

The acronym LINER should help you remember the conditions for inference about regression.

• Linear:  The actual relationship between x and y is linear. For any fixed value of x, the mean response my falls on the population (true) regression line my = a + bx. • Independent:  Individual observations are independent of each other. When sampling without replacement, check the 10% condition. • Normal:  For any fixed value of x, the response y varies according to a ­Normal distribution. • Equal SD:  The standard deviation of y (call it s) is the same for all values of x. • Random:  The data come from a well-designed random sample or randomized experiment.

Although the conditions for regression inference are a bit complicated, it is not hard to check for major violations. Most of the conditions involve the population (true) regression line and the deviations of responses from this line. We usually can’t observe the population line, but the sample regression line estimates it. The residuals from the sample regression line estimate the deviations from the population line. We can check several of the conditions for regression inference by looking at graphs of the residuals. Start by making a residual plot and a histogram or Normal probability plot of the residuals. Here’s a summary of how to check the conditions one by one.

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C H A P T E R 1 2 More about Regression • Linear: Examine the scatterplot to see if the overall pattern is roughly linear. Make sure there are no curved patterns in the residual plot. Check to see that the residuals center on the “residual = 0” line at each x-value in the residual plot. 







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Good: Scatterplot has a linear form.

 

  

  

  

   0LOHVGULYHQ   

  

  

  

Bad: Residual plot shows a curved pattern.

• Independent: Look at how the data were produced. Random sampling and random assignment help ensure the independence of individual observations. If sampling is done without replacement, remember to check that the population is at least 10 times as large as the sample (10% condition). But there are other issues that can lead to a lack of independence. One example is measuring the same variable at intervals over time, yielding what is known as timeseries data. Knowing that a young girl’s height at age 6 is 48 inches would definitely give you additional information about her height at age 7. You should avoid doing inference about the regression model for time-series data. • Normal: Make a stemplot, histogram, or Normal probability plot of the residuals and check for clear skewness or other major departures from Normality. Ideally, we would check the distribution of residuals for Normality at each possible value of x. Because we rarely have enough observations at each x-value, however, we make one graph of all the residuals to check for Normality. • Equal SD: Look at the scatter of the residuals above and below the “residual = 0” line in the residual plot. The vertical spread of the residuals should be roughly the same from the smallest to the largest x-value. 4 3 2

Residual

1 0 –1 –2 –3

   Good: Residuals have roughly equal spread at all x-values in the data set.

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–4 x

Bad: The response variable y has greater spread for larger values of the explanatory variable x.

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Section 12.1 Inference for Linear Regression

745

• Random: See if the data came from a well-designed random sample or randomized experiment. If not, we can’t make inferences about a larger population or about cause and effect. Let’s look at an example that illustrates the process of checking conditions.

Example

The Helicopter Experiment Checking conditions Mrs. Barrett’s class did a variation of the helicopter experiment on page 738. Students randomly assigned 14 helicopters to each of five drop heights: 152 centimeters (cm), 203 cm, 254 cm, 307 cm, and 442 cm. Teams of students released the 70 helicopters in a predetermined random order and measured the flight times in seconds. The class used Minitab to carry out a least-squares regression analysis for these data. A scatterplot and residual plot, plus a histogram and Normal probability plot of the residuals are shown below.

Problem:  Check whether the conditions for performing inference about the regression model

are met.

Solution:  We’ll use our LINER acronym!

Note that we do not have to check the 10% condition here because there was no random sampling.

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• Linear:  The scatterplot shows a clear linear form. The residual plot shows a random scatter about the horizontal line. For each drop height used in the experiment, the residuals are centered on the horizontal line at 0. • Independent:  Because the helicopters were released in a random order and no helicopter was used twice, knowing the result of one observation should not help us predict the value of another observation. • Normal:  The histogram of the residuals is single-peaked and somewhat bell-shaped. In addition, the Normal probability plot is very close to linear.

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C H A P T E R 1 2 More about Regression

•  Equal SD:  The residual plot shows a similar amount of scatter about the residual = 0 line for the 152, 203, 254, and 442 cm drop heights. Flight times (and the corresponding residuals) seem to vary a little more for the helicopters that were dropped from a height of 307 cm. • Random:  The helicopters were randomly assigned to the five possible drop heights. Except for a slight concern about the equal-SD condition, we should be safe performing inference about the regression model in this setting. For Practice Try Exercise

3

You will always see some irregularity when you look for Normality and equal standard deviation in the residuals, especially when you have few observations. Don’t overreact to minor issues in the graphs when checking these two conditions.

AP® EXAM TIP ​The AP® exam formula sheet gives y^ = b0 + b1x for the equation of the sample regression line. We will stick with our simpler notation, y^ = a + bx , which is also used by TI calculators. Just remember: the coefficient of x is always the slope, no matter what symbol is used.

Because s is estimated from data, it is sometimes called the regression standard error or the root mean squared error.

Estimating the Parameters When the conditions are met, we can do inference about the regression model my = a + bx. The first step is to estimate the unknown parameters. If we calculate the sample regression line y^ = a + bx, the slope b is an unbiased estimator of the true slope b, and the y intercept a is an unbiased estimator of the true y intercept a. The remaining parameter is the standard deviation s, which describes the variability of the response y about the population (true) regression line. The least-squares regression line computed from the sample data estimates the population (true) regression line. So the residuals estimate how much y varies about the population line. Because s is the standard deviation of responses about the population (true) regression line, we estimate it by the standard deviation of the residuals s=

∙residuals

2

Å

n−2

=

∙(y − y^ ) i

Å

i

2

n−2

Recall from Chapter 3 that s describes the size of a “typical” prediction error. It is possible to do inference about any of the three parameters in the regression model: a, b, or s. However, the slope b of the population (true) regression line is usually the most important parameter in a regression problem. So we’ll restrict our attention to inference about the slope. When the conditions are met, the sampling distribution of the slope b is approximately Normal with mean mb = b and standard deviation s sb = sx !n

In practice, we don’t know s for the true regression line. So we estimate it with the standard deviation of the residuals, s. We also don’t know the standard deviation sx for the population of x-values. For reasons beyond the scope of this text, we replace the denominator with sx !n − 1. So we estimate the spread of the sampling distribution of b with the standard error of the slope s SEb = sx !n − 1

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What happens if we transform the values of b by standardizing? Because the sampling distribution of b is approximately Normal, the statistic z=

b−b sb

is modeled well by the standard Normal distribution. Replacing the standard deviation sb of the sampling distribution with its standard error gives the statistic t=

b−b SEb

which has a t distribution with n − 2 degrees of freedom. (The explanation of why df = n − 2 is beyond the scope of this book.) Let’s return to the Old Faithful eruption data. Figure 12.7(a) displays the simulated sampling distribution of the slope b from 1000 SRSs of n = 20 eruptions. Figure 12.7(b) shows the result of standardizing the b-values from these 1000 samples. The superimposed curve is a t distribution with df = 20 – 2 = 18.

(a)

(b) t distribution with 18 degrees of freedom

6

8

10

12

14

–4

–2

0

2

4

Standardized slope

Slope

FIGURE 12.7  (a) The approximate sampling distribution of the slope b for samples of size n = 20 eruptions. This distribution has a roughly Normal shape with mean about 10.36 and standard deviation about 1.27. (b) The sampling distribution of the standardized slope values has approximately a t distribution with df = n – 2.

Constructing a Confidence Interval for the Slope In a regression setting, we often want to estimate the slope b of the population (true) regression line. The slope b of the sample regression line is our point estimate for b. A confidence interval is more useful than the point estimate because it gives a set of plausible values for b. The confidence interval for b has the familiar form statistic ± (critical value) · (standard deviation of statistic) Because we use the statistic b as our point estimate, the confidence interval is b ± t*SEb

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C H A P T E R 1 2 More about Regression We call this a t interval for the slope. Here are the details.

AP® EXAM TIP ​The AP® exam formula sheet gives the formula for the standard error of the slope as

∙(y − y^ ) i

sb1 =

Å

∙

i

t  Interval for the Slope When the conditions for regression inference are met, a C% confidence interval for the slope b of the population (true) regression line is

2

n−2

" (xi − x– )2

The numerator is just a fancy way of writing the standard deviation of the residuals s. Can you show that the denominator of this formula is the same as ours?

b ± t*SEb In this formula, the standard error of the slope is SEb =

s sx !n − 1

and t* is the critical value for the t distribution with df = n − 2 having C% of its area between −t* and t*.

Although we give the formula for the standard error of b, you should rarely have to calculate it by hand. Computer output gives the standard error SEb along with b itself. However we get it, SEb estimates how much the slope of the sample regression line typically varies from the slope of the population (true) regression line if we repeat the data production process many times.

Example

The Helicopter Experiment A confidence interval for b Earlier, we used Minitab to perform a least-squares regression analysis on the helicopter data for Mrs. Barrett’s class. Recall that the data came from dropping 70 paper helicopters from various heights and measuring the flight times. Some computer output from this regression is shown below. We checked conditions for performing inference earlier. Regression Analysis: Flight time versus Drop height Predictor

Coef

SE Coef

T

P

Constant

−0.03761

0.05838

−0.64

0.522

0.0002018

28.37

0.000

Drop height (cm)

0.0057244

S = 0.168181  R-Sq = 92.2%  R-Sq(adj) = 92.1%

Problem: (a) Give the standard error of the slope, SEb. Interpret this value in context. (b) Find the critical value for a 95% confidence interval for the slope of the true regression line. Then calculate the confidence interval. Show your work. (c) Interpret the interval from part (b) in context. (d) Explain the meaning of “95% confident” in context.

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Solution: When we compute the leastsquares regression line based on a random sample of data, we can think about doing inference for the population regression line. When our least-squares regression line is based on data from a randomized experiment, as in this example, the resulting inference is about the true regression line relating the explanatory and response variables. We’ll follow this convention from now on.

(a) We got the value of the standard error of the slope, 0.0002018, from the “SE Coef” column in the computer output. If we repeated the random assignment many times, the slope of the sample regression line would typically vary by about 0.0002 from the slope of the true regression line for predicting flight time from drop height. (b) Because the conditions are met, we can calculate a t interval for the slope b based on a t distribution with df = n − 2 = 70 − 2 = 68. Using the more conservative df = 60 from Table B gives t* = 2.000. The 95% confidence interval is b ± t*SEb = 0.0057244 ± 2.000(0.0002018) = 0.0057244 ± 0.0004036 = (0.0053208, 0.0061280) using technology:  From invT(.025,68), we get t* = 1.995. The resulting 95% confidence interval is 0.0057244 ± 1.995(0.0002018) = 0.0057244 ± 0.0004026 = (0.0053218, 0.0061270) This interval is slightly narrower due to the more precise t* critical value. (c) We are 95% confident that the interval from 0.0053218 to 0.0061270 seconds per cm captures the slope of the true regression line relating the flight time y and drop height x of paper helicopters. (d) If we repeat the experiment many, many times, and use the method in part (b) to construct a confidence interval each time, about 95% of the resulting intervals will capture the slope of the true regression line relating flight time y and drop height x of paper helicopters. For Practice Try Exercise

7

c

Example

How Much Is That Truck Worth? Confidence interval for a slope

STEP

!

n

The values of t given in the computer regression output are not the autio critical values for a confidence interval. They come from carrying out a significance test about the y intercept or slope of the population (true) regression line. We’ll discuss tests in more detail shortly. You can find a confidence interval for the y intercept a of the population (true) regression line in the same way, using a and SEa from the “Constant” row of the Minitab output. However, we are usually interested only in the point estimate for a that’s provided in the computer output. Here is an example using a familiar context that illustrates the four-step process for calculating and interpreting a confidence interval for the slope.

4

Everyone knows that cars and trucks lose value the more they are driven. Can we predict the price of a used Ford F-150 SuperCrew 4 × 4 if we know how many miles it has on the odometer? A random sample of 16 used Ford F-150 SuperCrew 4 × 4s was selected from among those listed for sale on autotrader.com. The number of miles driven and price (in dollars) were recorded for each of the trucks.3

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C H A P T E R 1 2 More about Regression

Here are the data: Miles driven:

70,583

129,484

29,932

29,953

24,495

75,678

8359

4447

Price (in dollars):

21,994

9500

29,875

41,995

41,995

28,986

31,891

37,991

Miles driven:

34,077

58,023

44,447

68,474 144,162 140,776

29,397 131,385

Price (in dollars):

34,995

29,988

22,896

33,961

27,495

16,883

20,897

13,997

Minitab output from a least-squares regression analysis for these data is shown below. (a)

(b)

45,000

10,000

35,000

5000

30,000

Residual

Price (in dollars)

40,000

25,000 20,000 15,000

0

5000

10,000 10,000

5000 0

0

20,000 40,000 60,000 80,000 100,000 120,000 140,000 160,000

Miles driven

20,000 40,000 60,000 80,000 100,000 120,000 140,000 160,000

Miles driven



Regression Analysis: Price (dollars) versus Miles driven Predictor

Coef

SE Coef

T

P

Constant

38257

Miles driven

−0.16292

2446

15.64

0.000

0.03096

−5.26

0.000

S = 5740.13  R-Sq = 66.4%  R-Sq(adj) = 64.0%

Problem:  Construct and interpret a 90% confidence interval for the slope of the population

regression line.

Solution:  We will follow the familiar four-step process. STATE:  We want to estimate the slope b of the population regression line relating miles driven to

price with 90% confidence. PLAN:  If the conditions are met, we will use a t interval for the slope of a regression line. •  Linear:  The scatterplot shows a clear linear pattern. Also, the residual plot shows a random scatter of points about the residual = 0 line. •  Independent:  Because we sampled without replacement to get the data, there have to be at least 10(16) = 160 used Ford F-150 SuperCrew 4 × 4s listed for sale on autotrader.com. This seems reasonable to believe. •  Normal:  The histogram of the residuals is roughly symmetric and single-peaked, so there are no obvious departures from Normality. •  Equal SD:  The scatter of points around the residual = 0 line appears to be about the same at all x-values. •  Random:  We randomly selected the 16 pickup trucks in the sample.

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751

DO:  We use the t distribution with 16 – 2 = 14 degrees of freedom to find the critical value. For a 90% confidence level, the critical value is t* = 1.761. So the 90% confidence interval for b is b ± t*SEb = −0.16292 ± 1.761(0.03096) = −0.16292 ± 0.05452 = (−0.21744, −0.10840) Using technology:  Refer to the Technology Corner that follows the example. The calculator’s LinRegTInt gives (−0.2173, −0.1084) using df = 14. CONCLUDE:  We are 90% confident that the interval from −0.2173 to −0.1084 captures the slope of the population regression line relating price to miles driven for used Ford F-150 SuperCrew 4 × 4s listed for sale on autotrader.com. For Practice Try Exercise

9

The predicted change in price of a used Ford F-150 is quite small for a 1-mile increase in miles driven. What if miles driven increased by 1000 miles? We can just multiply both endpoints of the confidence interval in the example by 1000 to get a 90% confidence interval for the corresponding predicted change in average price. The resulting interval is (−217.3, −108.4). That is, the population regression line predicts a decrease in price of between $108.40 and $217.30 for every additional 1000 miles driven. So far, we have used computer regression output when performing inference about the slope of a population (true) regression line. The TI-83/84, TI-89, and TINspire can do the calculations for inference when the sample data are provided.

28. TECHNOLOGY CORNER

Confidence interval for slope on the calculator TI-Nspire instructions in Appendix B; HP Prime instructions on the book’s Web site.

Let’s use the data from the previous example to construct a confidence interval for the slope of a population (true) regression line on the TI-83/84 and TI-89. Enter the x-values (miles driven) into L1/list1 and the y-values (price) into L2/list2.



TI-83/84 with recent OS

TI-89

• Press STAT , then choose TESTS and LinRegTInt. . . .

• Press 2nd F2 ([F7]) and choose LinRegTInt. . . .

• In the LinRegTInt screen, adjust the inputs as shown. Then highlight “Calculate” and press ENTER .

• In the LinRegTInt screen, adjust the inputs as shown and press ENTER .



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752

C H A P T E R 1 2 More about Regression

• The linear regression t interval results are shown below. The TI-84 Plus C fits the results on one screen. The TI-83/84 and TI-89 require you to arrow down to see the rest of the output.





Note that s is the standard deviation of the residuals, not the standard error of the slope. AP® EXAM TIP ​The formula for the t interval for the slope of a population (true) regression line often leads to calculation errors by students. As a result, we recommend using the calculator’s LinRegTInt feature to compute the confidence interval on the AP® Exam. Be sure to name the procedure (t interval for slope) and to give the interval (–0.217, –0.108) and df (14) as part of the “Do” step.

Check Your Understanding

Does fidgeting keep you slim? Some people don’t gain weight even when they overeat. Perhaps fidgeting and other “nonexercise activity” (NEA) explain why—some people may spontaneously increase nonexercise activity when fed more. Researchers deliberately overfed a random sample of 16 healthy young adults for 8 weeks. They measured fat gain (in kilograms) as the response variable and change in energy use (in calories) from activity other than deliberate exercise—fidgeting, daily living, and the like—as the explanatory variable. Here are the data:4 NEA change (cal):

−94

−57

−29

135

143

151

245

355

Fat gain (kg):

4.2

3.0

3.7

2.7

3.2

3.6

2.4

1.3

NEA change (cal):

392

473

486

535

571

580

620

690

Fat gain (kg):

3.8

1.7

1.6

2.2

1.0

0.4

2.3

1.1

Minitab output from a least-squares regression analysis for these data is shown below.

  

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Regression Analysis: Fat gain versus NEA change Predictor

Coef

SE Coef

T

P

Constant

 3.5051

0.03036

11.54

0.000

NEA change

−0.0034415

0.0007414

−4.64

0.000

S = 0.739853  R-Sq = 60.6%  R-Sq(adj) = 57.8%

Construct and interpret a 95% confidence interval for the slope of the population (true) regression line.

Performing a Significance Test for the Slope When the conditions for inference are met, we can use the slope b of the sample regression line to construct a confidence interval for the slope b of the population (true) regression line. We can also perform a significance test to determine whether a specified value of b is plausible. The null hypothesis has the general form H0 : b = b0. To do a test, standardize b to get the test statistic: statistic − parameter test statistic = standard deviation of statistic b − b0 t= SEb To find the P-value, use a t distribution with n − 2 degrees of freedom. Here are the details for the t test for the slope.

t  Test for the Slope Suppose the conditions for inference are met. To test the hypothesis H0 : b = b0, compute the test statistic t=

b − b0 SEb

Find the P-value by calculating the probability of getting a t statistic this large or larger in the direction specified by the alternative hypothesis Ha. Use the t distribution with df = n − 2. Ha : β > hypothesized value

t

Ha : β < hypothesized value

t

Ha : β = hypothesized value

– t

t

If sample data suggest a linear relationship between two variables, how can we determine whether this happened just by chance or whether there is actually a linear relationship between x and y in the population? By performing a test of H0 : b = 0. A regression line with slope 0 is horizontal. That is, the mean of y does not change at all when x changes. So H0 : b = 0 says that there is no linear relationship between x and y in the population. Put another way, H0 says that linear regression of y on x is of no value for predicting y.

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Most technology will only do a test with H0 : b = 0.

Example

Regression output from statistical software usually gives t and its two-sided P-value for a test of H0 : b = 0. For a one-sided test in the proper direction, just divide the P-value in the output by 2. The following example shows what we mean.

Crying and IQ

STEP

Significance test for b

4

Infants who cry easily may be more easily stimulated than others. This may be a sign of higher IQ. Child development researchers explored the relationship between the crying of infants 4 to 10 days old and their later IQ test scores. A snap of a rubber band on the sole of the foot caused the infants to cry. The researchers recorded the crying and measured its intensity by the number of peaks in the most active 20 seconds. They later measured the children’s IQ at age three years using the Stanford-Binet IQ test. The table below contains data from a random sample of 38 infants.5

®

AP EXAM TIP  When you see a list of data values on an exam question, don’t just start typing the data into your calculator. Read the question first. Often, additional information is provided that makes it unnecessary for you to enter the data at all. This can save you valuable time on the AP® exam.

Crycount

IQ

Crycount

IQ

Crycount

IQ

Crycount

10

87

12

97

20

90

16

100

 9

103

23

16

106

18

109

15

IQ

17

94

12

94

19

103

12

103

103

13

104

14

106

27

108

18

109

10

109

15

112

18

112

23

113

114

21

114

16

118

 9

119

12

119

12

120

19

120

16

124

20

132

15

133

22

135

31

135

16

136

17

141

30

155

22

157

33

159

13

162

Some computer output from a least-squares regression analysis on these data is shown below.

Regression Analysis: IQ versus Crycount Predictor

Coef

SE Coef

T

P

Constant

91.268

8.934

10.22

0.000

3.07

0.004

Crycount

1.4929

0.4870

S = 17.50  R-Sq = 20.7%  R-Sq(adj) = 18.5%

  

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755

PROBLEM: (a) What is the equation of the least-squares regression line for predicting IQ at age 3 from the number of crying peaks (crycount)? Interpret the slope and y intercept of the regression line in context. (b) Explain what the value of s means in this setting. (c) Do these data provide convincing evidence of a positive linear relationship between crying counts and IQ in the population of infants?

Solution: (a) The equation of the least-squares line is predicted  IQ  score = 91.268 + 1.4929 (crycount) Because there were no infants who recorded fewer than 9 crying peaks in their most active 20 seconds, it is a risky extrapolation to use this line to predict the value of y when x = 0.

Slope:  For each additional crying peak in the most active 20 seconds, the regression line predicts an increase of about 1.5 IQ points. y intercept:  The model predicts that an infant who doesn’t cry when flicked with a rubber band will have a later IQ score of about 91. (b) The size of a typical prediction error when using the regression line in part (a) is 17.50 IQ points. (c) We’ll follow the four-step process. State:  We want to perform a test of H0 : b = 0 Ha : b > 0 where b is the slope of the population regression line relating crying count to IQ score. No significance level was given, so we’ll use a = 0.05.

Plan:  If the conditions are met, we will do a t test for the slope b. •  Linear:  The scatterplot suggests a moderately weak positive linear relationship between crying peaks and IQ. The residual plot shows a random scatter of points about the residual = 0 line. •  Independent:  Due to sampling without replacement, there have to be at least 10(38) = 380 infants in the population from which these children were selected. •  Normal:  The Normal probability plot of the residuals shows slight curvature, but no strong skewness or obvious outliers that would prevent use of t procedures. •  Equal SD:  The residual plot shows a fairly equal amount of scatter around the horizontal line at 0 for all x-values. •  Random:  We are told that these 38 infants were randomly selected.

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Do:  We can get the test statistic and P-value from the Minitab output.

Our usual formula for the test statistic confirms the value in the computer output: b − b0 1.4929 − 0 t= = = 3.07 SEb 0.4870

•  Test statistic:  t = 3.07 (look in the “T” column of the computer output across from “Crycount”) •  P-value:  Figure 12.8 displays the P-value for this one-sided test as an area under the t distribution curve with 38 − 2 = 36 degrees of freedom. The Minitab output gives P = 0.004 as the P-value for a two-sided test. The P-value for the one-sided test is half of this, P = 0.002. Using technology:  Refer to the Technology Corner that follows the example. The calculator’s LinRegTTest gives t =3.065 and P-value = 0.002 using df = 36. Conclude:  Because the P-value, 0.002, is less than a = 0.05, we reject H0. There P -value = 0.002 is convincing evidence of a positive linear relationship between intensity of crying and IQ score in the population of infants.

t distribution, 36 degrees of freedom

Values of t

t = 3.07

For Practice Try Exercise

FIGURE 12.8  ​The P-value for the one-sided test.

13

Based on the results of the crying and IQ study, should we ask doctors and parents to make infants cry more so that they’ll be smarter later in life? Hardly. This observational study gives statistically significant evidence of a positive linear relationship between the two variables. However, we can’t conclude that more intense crying as an infant causes an increase in IQ. Maybe infants who cry more are more alert to begin with and tend to score higher on intelligence tests.

29. Technology Corner

Significance test for slope on the calculator TI-Nspire instructions in Appendix B; HP Prime instructions on the book’s Web site.

Let’s use the data from the crying and IQ study to perform a significance test for the slope of the population regression line on the TI-83/84 and TI-89. Enter the x-values (crying count) into L1/list1 and the y-values (IQ score) into L2/list2. TI-83/84 TI-89 • Press STAT , then choose TESTS and • Press 2nd F1 ([F6]) and choose LinRegTTest. . . . LinRegTTest. . . . • In the LinRegTTest screen, adjust the inputs as shown. Then highlight “Calculate” and press ENTER .



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• In the LinRegTTest screen, adjust the inputs as shown and press ENTER .



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Section 12.1 Inference for Linear Regression

• The linear regression t test results take two screens to present. We show only the first screen.



AP® EXAM TIP ​ The formula for the test statistic in a t test for the slope of a population (true) regression line often leads to calculation errors by students. As a result, we recommend using the calculator’s LinRegTTest feature to perform calculations on the AP® exam. Be sure to name the procedure (t test for slope) and to report the test statistic (t = 3.065), P-value (0.002), and df (36) as part of the “Do” step.

THINK ABOUT IT

What’s with that r > 0 in the LinRegTTest screen?  ​The slope b

of the least-squares regression line is closely related to the correlation r between sy the explanatory and response variables x and y. (Recall that b = r ). In the same sx way, the slope b of the population regression line is closely related to the correlation r (the lowercase Greek letter rho) between x and y in the population. In particular, the slope is 0 when the correlation is 0. Testing the null hypothesis H0 : b = 0 is, therefore, exactly the same as testing that there is no correlation between x and y in the population from which we drew our data. You can use the test for zero slope to test the hypothesis H0 : r = 0 of zero correlation between any two quantitative variables. That’s a useful trick. Because correlation also makes sense when there is no explanatory-response distinction, it is handy to be able to test correlation without doing regression.

Check Your Understanding

The previous Check Your Understanding (page 752) described some results from a study of nonexercise activity (NEA) and fat gain. Here, again, is the Minitab output from a leastsquares regression analysis for these data. Regression Analysis: Fat gain versus NEA change Predictor Constant NEA change

Coef 3.5051 −0.0034415

SE Coef

T

P

0.3036

11.54

0.000

0.0007414

−4.64

0.000

S = 0.739853   R-Sq = 60.6%   R-Sq(adj) = 57.8%

Do these data provide convincing evidence at the a = 0.05 significance level of a negative linear relationship between fat gain and NEA change in the population of healthy young adults? Assume that the conditions for regression inference are met.

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Section 12.1

Summary •

•

•

• •

•

Least-squares regression fits a straight line of the form y^ = a + bx to data to predict a response variable y from an explanatory variable x. Inference in this setting uses the sample regression line to estimate or test a claim about the population (true) regression line. The conditions for regression inference are • Linear:  The actual relationship between x and y is linear. For any fixed value of x, the mean response my falls on the population (true) regression line my = a + bx. • Independent: Individual observations are independent. When sampling is done without replacement, check the 10% condition. • Normal:  For any fixed value of x, the response y varies according to a Normal distribution. • Equal SD:  The standard deviation of y (call it s) is the same for all values of x. • Random:  The data are produced from a well-designed random sample or randomized experiment. The slope b and intercept a of the sample regression line estimate the slope b and intercept a of the population (true) regression line. Use the standard deviation of the residuals, s, to estimate s. Confidence intervals and significance tests for the slope b of the population regression line are based on a t distribution with n − 2 degrees of freedom. The t interval for the slope b has the form b ± t*SEb, where the standard s . error of the slope is SEb = sx !n − 1 To test the null hypothesis H0 : b = b0, carry out a t test for the slope. This b − b0 test uses the statistic t = . The most common null hypothesis is SEb H0 : b  = 0, which says that there is no linear relationship between x and y in the population.

12.1 T echnology CornerS TI-Nspire Instructions in Appendix B; HP Prime instructions on the book’s Web site. 28.  Confidence interval for slope on the calculator 29. Significance test for slope on the calculator

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Section 12.1 Inference for Linear Regression

Exercises

Section 12.1 1.

759

Oil and residuals  Exercise 53 on page 194 (Chapter 3) examined data on the depth of small defects in the Trans-Alaska Oil Pipeline. Researchers compared the results of measurements on 100 defects made in the field with measurements of the same defects made in the laboratory.6 The figure below shows a residual plot for the least-squares regression line based on these data. Explain why the conditions for performing ­inference about the slope b of the population regression line are not met.

3. pg 745

Beer and BAC  How well does the number of beers a person drinks predict his or her blood alcohol content (BAC)? Sixteen volunteers aged 21 or older with an initial BAC of 0 took part in a study to find out. Each volunteer drank a randomly assigned number of cans of beer. Thirty minutes later, a police officer measured their BAC. Least-squares regression was performed on the data. A residual plot and a histogram of the residuals are shown below. Check whether the ­conditions for performing inference about the regression model are met.

Residual of field measurement

30

20

10

0

–10

–20

–30 0

20

40

60

80

Laboratory measurement

2.

SAT Math scores  In Chapter 3, we examined data on the percent of high school graduates in each state who took the SAT and the state’s mean SAT Math score in a recent year. The figure below shows a residual plot for the least-squares regression line based on these data. Explain why the conditions for performing ­inference about the slope b of the population ­regression line are not met.

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4.

Prey attracts predators  Here is one way in which nature regulates the size of animal populations: high population density attracts predators, which remove a higher proportion of the population than when the density of the prey is low. One study looked at kelp perch and their common predator, the kelp bass. The researcher set up four large circular pens on sandy ocean bottoms off the coast of southern California. He chose young perch at random from a large group and placed 10, 20, 40, and 60 perch in the four pens. Then he dropped the nets protecting the pens, allowing bass to swarm in, and counted the

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C H A P T E R 1 2 More about Regression

perch left after two hours. Here are data on the proportions of perch eaten in four repetitions of this setup:7 Number of Perch 10 20 40 60



6.

Proportion Killed 0.0 0.2 0.075 0.517

0.1 0.3 0.3 0.55

0.3 0.3 0.6 0.7

0.3 0.6 0.725 0.817

Predictor Constant Perch

We used Minitab software to carry out a least-squares regression analysis for these data. A residual plot and a histogram of the residuals are shown below. Check whether the conditions for performing inference about the regression model are met.

Coef 0.12049 0.008569

Stdev. 0.09269 0.002456

t-ratio 1.30 3.49

p 0.215 0.004

S = 0.1886  R-Sq = 46.5%  R-Sq(adj) = 42.7%

The explanatory variable is the number of perch (the prey) in a confined area. The response variable is the proportion of perch killed by bass (the predator) in two hours when the bass are allowed access to the perch. A scatterplot of the data shows a linear relationship.



The model for regression inference has three ­parameters: a, b, and s. Explain what each parameter represents in context. Then provide an estimate for each. Prey attracts predators  Refer to Exercise 4. ­Computer output from the least-squares regression analysis on the perch data is shown below.

pg 748



The model for regression inference has three parameters: a, b, and s. Explain what each parameter represents in context. Then provide an estimate for each.

7.

Beer and BAC  Refer to Exercise 5.

(a) Give the standard error of the slope, SEb. Interpret this value in context. (b) Find the critical value for a 99% confidence ­interval for the slope of the true regression line. Then calculate the confidence interval. Show your work. (c) Interpret the interval from part (b) in context. (d) Explain the meaning of “99% confident” in context. 8.

Prey attracts predators  Refer to Exercise 6.

(a) Give the standard error of the slope, SEb. Interpret this value in context. (b) Find the critical value for a 90% confidence interval for the slope of the true regression line. Then calculate the confidence interval. Show your work. (c) Interpret the interval from part (b) in context. (d) Explain the meaning of “90% confident” in context. 9. pg 749

5.

Beer and BAC  Refer to Exercise 3. Computer output from the least-squares regression analysis on the beer and blood alcohol data is shown below.

Dependent variable is:

BAC

No Selector R squared = 80.0% R squared (adjusted) = 78.6% s = 0.0204 with 16  −  2 = 14 degrees of freedom Variable Coefficient s.e. of Coeff Constant Beers

t-ratio

prob

−0.012701

0.0126

−1.00

0.3320

0.017964

0.0024

7.84

≤0.0001

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Beavers and beetles  Do beavers benefit beetles? Researchers laid out 23 circular plots, each 4 ­meters in diameter, at random in an area where beavers were cutting down cottonwood trees. In each plot, they counted the number of stumps from trees cut by beavers and the number of clusters of beetle larvae. Ecologists think that the new sprouts from stumps are more tender than other cottonwood growth, so that beetles prefer them. If so, more stumps should produce more beetle larvae.8 Minitab output for a regression analysis on these data is shown below. Construct and interpret a 99% confidence interval for the slope of the population regression line. Assume that the conditions for performing inference are met.

Regression Analysis: Beetle larvae versus Stumps Predictor Coef SE Coef T P −1.286 2.853 −0.45 0.657 Constant Stumps 11.894 1.136 10.47 0.000 S = 6.41939  R-Sq = 83.9%  R-Sq(adj) = 83.1%

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Section 12.1 Inference for Linear Regression 10. Ideal proportions  The students in Mr. Shenk’s class measured the arm spans and heights (in inches) of a random sample of 18 students from their large high school. Some computer output from a least-squares regression analysis on these data is shown below. Construct and interpret a 90% confidence interval for the slope of the population regression line. Assume that the conditions for performing inference are met. Predictor Coef Stdev t-ratio p Constant 11.547 5.600 2.06 0.056 Armspan 0.84042 0.08091 10.39 0.000 S = 1.613  R-Sq = 87.1%  R-Sq(adj) = 86.3%

11. Beavers and beetles  Refer to Exercise 9. (a) How many clusters of beetle larvae would you predict in a circular plot with 5 tree stumps cut by beavers? Show your work. (b) About how far off do you expect the prediction in part (a) to be from the actual number of clusters of beetle larvae? Justify your answer. 12. Ideal proportions  Refer to Exercise 10. (a) What height would you predict for a student with an arm span of 76 inches? Show your work. (b) About how far off do you expect the prediction in part (a) to be from the student’s actual height? Justify your answer.

Predictor Coef Constant 166.483 Weeds per -1.0987  meter

SE Coef 2.725 0.5712

T 61.11 -1.92

P 0.000 0.075

S = 7.97665 R-Sq = 20.9%   R-Sq(adj) = 15.3%

(a) What is the equation of the least-squares regression line for predicting corn yield from the number of lamb’s quarter plants per meter? Interpret the slope and y intercept of the regression line in context. (b) Explain what the value of s means in this settting. (c) Do these data provide convincing evidence at the a = 0.05 level that more weeds reduce corn yield? Assume that the conditions for performing inference are met. 14. Time at the table  Does how long young children remain at the lunch table help predict how much they eat? Here are data on a random sample of 20 toddlers observed over several months.10 “Time” is the average number of minutes a child spent at the table when lunch was served. “Calories” is the average number of calories the child consumed during lunch, calculated from careful observation of what the child ate each day.

Some computer output from a least-squares regression analysis on these data is shown below.

13. Weeds among the corn  Lamb’s-quarter is a common weed that interferes with the growth of corn. An agriculture researcher planted corn at the same rate in 16 small plots of ground and then weeded the plots by hand to allow a fixed number of lamb’squarter plants to grow in each meter of corn row. The decision of how many of these plants to leave in each plot was made at random. No other weeds were allowed to grow. Here are the yields of corn (bushels per acre) in each of the plots:9

pg 754



Some computer output from a least-squares regression analysis on these data is shown below.

Predictor Constant Time

Coef 560.65 −3.0771

SE Coef 29.37 0.8498

T 19.09 −3.62

P 0.000 0.002

S = 23.3980 R-Sq = 42.1% R-Sq(adj) = 38.9%

(a) What is the equation of the least-squares ­regression line for predicting calories consumed from time at the table? Interpret the slope of the regression line in context. Does it make sense to interpret the y intercept in this case? Why or why not? (b) Explain what the value of s means in this setting. (c) Do these data provide convincing evidence at the a = 0.01 level of a linear relationship between time

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C H A P T E R 1 2 More about Regression

at the table and calories consumed in the population of toddlers? Assume that the conditions for performing inference are met.

150,000

120,000

Alcohol from wine (liters/year)

Heart disease death rate (per 100,000)

Alcohol from wine (liters/year)

Heart disease death rate (per 100,000)

2.5 3.9 2.9 2.4 2.9 0.8 9.1 2.7 0.8 0.7

211 167 131 191 220 297 71 172 211 300

7.9 1.8 1.9 0.8 6.5 1.6 5.8 1.3 1.2

107 167 266 227 86 207 115 285 199

Mileage

15. Is wine good for your heart?  A researcher from the University of California, San Diego, collected data on average per capita wine consumption and heart disease death rate in a random sample of 19 countries for which data were available. The following table displays the data.11

90,000

60,000

30,000

0 0



Is there statistically significant evidence of a ­negative linear relationship between wine consumption and heart disease deaths in the population of countries? Carry out an appropriate significance test at the a = 0.05 level.

16. The professor swims  Here are data on the time (in minutes) Professor Moore takes to swim 2000 yards and his pulse rate (beats per minute) after swimming on a random sample of 23 days:



Time: Pulse:

34.12 152

35.72 124

34.72 140

34.05 152

34.13 146

35.72 128

Time: Pulse:

36.17 136

35.57 144

35.37 148

35.57 144

35.43 136

36.05 124

Time: Pulse:

34.85 148

34.70 144

34.75 140

33.93 156

34.60 136

34.00 148

Time: Pulse:

34.35 148

35.62 132

35.68 124

35.28 132

35.97 139

Is there statistically significant evidence of a negative linear relationship between Professor Moore’s swim time and his pulse rate in the population of days on which he swims 2000 yards? Carry out an appropriate significance test at the a = 0.05 level.

17. Stats teachers’ cars  A random sample of AP® Statistics teachers was asked to report the age (in years) and mileage of their primary vehicles. A scatterplot of the data is shown at top right.

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5.0

7.5

10.0

Car age

Computer output from a least-squares regression analysis of these data is shown below (df = 19). Assume that the conditions for regression inference are met.

Variable Coef



2.5

Constant Car age

SE Coef t-ratio prob

7288.54 6591 11630.6

1249

1.11

0.2826

9.31

0. (b) H0 : b = 0 versus Ha : b < 0.

(c) Compute the test statistic and P-value for the test in part (b). What conclusion would you draw at the a = 0.01 significance level?

(c) H0 : b = 0 versus Ha : b ≠ 0.

(d) Does the confidence interval in part (a) lead to the same conclusion as the test in part (c)? Explain.

(e) H0 : b = 1 versus Ha : b > 1.

Multiple choice: Select the best answer for Exercises 19 to 24, which are based on the following information. To determine property taxes, Florida reappraises real estate every year, and the county appraiser’s Web site lists the current “fair market value” of each piece of property. Property usually sells for somewhat more than the appraised market value. We collected data on the appraised market values x and actual selling prices y (in thousands of dollars) of a random sample of 16 condominium units in Florida. We checked that the conditions for inference about the slope of the population regression line are met. Here is part of the Minitab output from a least-squares regression analysis using these data.13 Predictor Coef Constant 127.27 Appraisal 1.0466

SE Coef 79.49 0.1126

T 1.60 9.29

P 0.132 0.000

S = 69.7299 R-Sq = 86.1% R-Sq(adj) = 85.1%

19. The equation of the least-squares regression line for predicting selling price from appraised value is

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(d) H0 : b > 0 versus Ha : b = 0. 22. Which of the following is the best interpretation for the value 0.1126 in the computer output? (a) For each increase of $1000 in appraised value, the average selling price increases by about 0.1126. (b) When using this model to predict selling price, the predictions will typically be off by about 0.1126. (c) 11.26% of the variation in selling price is accounted for by the linear relationship between selling price and appraised value. (d) There is a weak, positive linear relationship between selling price and appraised value. (e) In repeated samples of size 16, the sample slope will typically vary from the population slope by about 0.1126. 23. A 95% confidence interval for the population slope b is (a) 1.0466 ± 1.046.

(d) 1.0466 ± 0.2207.

(b) 1.0466 ± 0.2415.

(e) 1.0466 ± 0.2400.

(c) 1.0466 ± 0.2387.

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C H A P T E R 1 2 More about Regression

24. Which of the following would have resulted in a violation of the conditions for inference? (a) If the entire sample was selected from one neighborhood (b) If the sample size was cut in half (c) If the scatterplot of x = appraised value and y = ­selling price did not show a perfect linear relationship (d) If the histogram of selling prices had an outlier (e) If the standard deviation of appraised values was different from the standard deviation of selling prices Exercises 25 to 28 refer to the following setting. Does the color in which words are printed affect your ability to read them? Do the words themselves affect your ability to name the color in which they are printed? Mr. Starnes d ­ esigned a study to investigate these questions using the 16 students in his AP® Statistics class as subjects. Each student performed two tasks in a random order while a partner timed: (1) read 32 words aloud as quickly as possible, and (2) say the color in which each of 32 words is printed as quickly as possible. Try both tasks for yourself using the word list below. YELLOW RED GREEN YELLOW BLUE RED BLUE GREEN

RED GREEN RED BLUE YELLOW BLUE GREEN YELLOW

BLUE YELLOW BLUE GREEN RED YELLOW GREEN RED

GREEN YELLOW BLUE RED RED GREEN BLUE YELLOW

27. Color words (9.3)  Explain why it is not safe to use paired t procedures to do inference about the difference in the mean time to complete the two tasks. 28. Color words (3.1, 3.2, 12.1)  Can we use a student’s word task time to predict his or her color task time? (a) Make an appropriate scatterplot to help answer this question. Describe what you see. (b) Use your calculator to find the equation of the leastsquares regression line. Define any symbols you use. (c) Find and interpret the residual for the student who completed the word task in 9 seconds. (d) Assume that the conditions for performing inference about the slope of the true regression line are met. The P-value for a test of H0 : b = 0 versus Ha : b > 0 is 0.0215. Explain what this value means in context.

25. Color words (4.2)  Let’s review the design of the study. (a) Explain why this was an experiment and not an observational study. (b) Did Mr. Starnes use a completely randomized design or a randomized block design? Why do you think he chose this experimental design? (c) Explain the purpose of the random assignment in the context of the study. The data from Mr. Starnes’s experiment are shown below. For each subject, the time to perform the two tasks is given to the nearest second. Subject

Words

Colors

Subject

Words

Colors

1 2 3 4 5 6 7 8

13 10 15 12 13 11 14 16

20 21 22 25 17 13 32 21

9 10 11 12 13 14 15 16

10 9 11 17 15 15 12 10

16 13 11 26 20 15 18 18

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26. Color words (1.3)  Do the data provide evidence of a difference in the average time required to perform the two tasks? Include an appropriate graph and numerical summaries in your answer.

Note:  John Ridley Stroop is often credited with the discovery in 1935 of the fact that the color in which “color words” are printed interferes with people’s ability to identify the color. The so-called Stroop Effect, though, was originally published by German researchers in 1929.

Exercises 29 and 30 refer to the following setting. Yellowstone National Park surveyed a random sample of 1526 winter visitors to the park. They asked each person whether he or she owned, rented, or had never used a snowmobile. Respondents were also asked whether they belonged to an environmental organization (like the Sierra Club). The two-way table summarizes the survey responses. Environmental Clubs Never used Snowmobile renter Snowmobile owner Total

No

Yes

Total

445 497 279 1221

212 77 16 305

657 574 295 1526

29. Snowmobiles (5.2, 5.3) (a) If we choose a survey respondent at random, what’s the probability that this individual (i)  is a snowmobile owner? (ii)  belongs to an environmental organization or owns a snowmobile? (iii)  has never used a snowmobile given that the person belongs to an environmental organization?

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Section 12.2 Transforming to Achieve Linearity (b) Are the events “is a snowmobile owner” and ­“belongs to an environmental organization” ­independent for the members of the sample? Justify your answer. (c) If we choose two survey respondents at random, what’s the probability that (i)  both are snowmobile owners? (ii)  at least one of the two belongs to an environmental organization?

12.2 What You Will Learn

765

30. Snowmobiles (11.2)  Do these data provide convincing evidence at the 5% significance level of an association between environmental club membership and snowmobile use for the population of visitors to Yellowstone National Park? Justify your answer.

Transforming to Achieve Linearity By the end of the section, you should be able to:

• Use transformations involving powers and roots to find a power model that describes the relationship between two variables, and use the model to make predictions. • Use transformations involving logarithms to find a power model or an exponential model that describes

the relationship between two variables, and use the model to make predictions. • Determine which of several transformations does a ­better job of producing a linear relationship.

In Chapter 3, we learned how to analyze relationships between two quantitative variables that showed a linear pattern. When two-variable data show a curved relationship, we must develop new techniques for finding an appropriate model. This section describes several simple transformations of data that can straighten a nonlinear pattern. Once the data have been transformed to achieve linearity, we can use least-squares regression to generate a useful model for making predictions. And if the conditions for regression inference are met, we can estimate or test a claim about the slope of the population (true) regression line using the transformed data.

Example

Health and Wealth Straightening out a curved pattern The Gapminder Web site, www.gapminder.org, provides loads of data on the health and well-being of the world’s inhabitants. Figure 12.9 on the next page is a scatterplot of data from Gapminder.14 The individuals are all the world’s nations for which data are available. The explanatory variable is a measure of how rich a country is: income per person. The response variable is life expectancy at birth.

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We expect people in richer countries to live longer because they have better access to medical care and typically lead healthier lives. The overall pattern of the ­scatterplot does show this, but the relationship is not linear. Life expectancy rises very quickly as income per person increases and then levels off. People in very rich countries such as the United States live no longer than people in poorer but not extremely poor nations. In some less wealthy countries, people live longer than in the United States. Four African nations are outliers. Their life expectancies are similar to those of their neighbors, but their income per person is higher. Gabon and Equatorial Guinea produce oil, and South Africa and Botswana produce diamonds. It may be that income from mineral exports goes mainly to a few people and so pulls up income per person without much effect on either the income or the life expectancy of ordinary citizens. That is, income per person is a mean, and we know that mean income can be much higher than median income.

80

Life expectancy

United States

Qatar

America

70

East Asia & Pacific Europe & Central Asia Gabon

Middle East & North Africa

60

South Asia

South Africa

Sub-Saharan Africa Equatorial Guinea 50

0 90

,0 0

0 80

,0 0

0 70

,0 0

0 60

,0 0

0 50

,0 0

0 ,0 0 40

30

,0 0

0

0 ,0 0 20

10

,0 0

0

Botswana

Income per person in 2012

FIGURE 12.9  ​Scatterplot of the life expectancy of people in many nations against each nation’s income per person. The color of each circle indicates the geographic region in which that country is located. The size of each circle is based on the population of the country—bigger circles indicate larger populations.

The scatterplot in Figure 12.9 shows a curved pattern. We can straighten things out using logarithms. Figure 12.10 (on the facing page) plots the logarithm of income per person against life expectancy for these same countries. The effect is almost magical. This graph has a clear, linear pattern.

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Section 12.2 Transforming to Achieve Linearity

767

80

Life expectancy

America 70

East Asia & Pacific Europe & Central Asia Middle East & North Africa

60

South Asia Sub-Saharan Africa

50

500

1,000

5,000

10,000

50,000

Income per person in 2012

FIGURE 12.10  ​Scatterplot of life expectancy against income per person (on a logarithm scale) for many nations.

Applying a function such as the logarithm or square root to a quantitative variable is called transforming the data. We will see in this section that understanding how simple functions work helps us choose and use transformations to straighten nonlinear patterns. Transforming data amounts to changing the scale of measurement that was used when the data were collected. We can choose to measure temperature in degrees Fahrenheit or in degrees Celsius, distance in miles or in kilometers. These changes of units are linear transformations, discussed in Chapter 2. Linear transformations cannot straighten a curved relationship between two variables. To do that, we resort to functions that are not linear. The logarithm function, applied in the “Health and Wealth” example, is a nonlinear function. We’ll return to transformations involving logarithms later.

Transforming with Powers and Roots When you visit a pizza parlor, you order a pizza by its diameter—say, 10 inches, 12 inches, or 14 inches. But the amount you get to eat depends on the area of the pizza. The area of a circle is p times the square of its radius r. So the area of a round pizza with diameter x is x 2 x2 p area = pr2 = pa b = pa b = x2 2 4 4

This is a power model of the form y = axp with a = p/4 and p = 2.

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C H A P T E R 1 2 More about Regression When we are dealing with things of the same general form, whether circles or fish or people, we expect area to go up with the square of a dimension such as diameter or height. Volume should go up with the cube of a linear dimension. That is, geometry tells us to expect power models in some settings. There are other physical relationships between two variables that are described by power models. Here are some examples from science. • The distance that an object dropped from a given height falls is related to time since release by the model distance = a(time)2 • The time it takes a pendulum to complete one back-and-forth swing (its ­period) is related to its length by the model period = a"length = a(length)1∙2 • The intensity of a light bulb is related to distance from the bulb by the model intensity =

a = a(distance)−2 2 distance

Although a power model of the form y = axp describes the relationship between x and y in each of these settings, there is a linear relationship between xp and y. If we transform the values of the explanatory variable x by raising them to the p power, and graph the points (xp, y), the scatterplot should have a linear form. The following example shows what we mean.

Example

Go Fish! Transforming with powers Imagine that you have been put in charge of organizing a fishing tournament in which prizes will be given for the heaviest Atlantic Ocean rockfish caught. You know that many of the fish caught during the tournament will be measured and released. You are also aware that using delicate scales to try to weigh a fish that is flopping around in a moving boat will probably not yield very accurate results. It would be much easier to measure the length of the fish while on the boat. What you need is a way to convert the length of the fish to its weight. You contact the nearby marine research laboratory, and they provide reference data on the length (in centimeters) and weight (in grams) for Atlantic Ocean rockfish of several sizes.15

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Length:

5.2

8.5

11.5

14.3

16.8

19.2

21.3

23.3

25.0

26.7

Weight:

2

8

21

38

69

117

148

190

264

293

Length:

28.2

29.6

30.8

32.0

33.0

34.0

34.9

36.4

37.1

37.7

Weight:

318

371

455

504

518

537

651

719

726

810

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Section 12.2 Transforming to Achieve Linearity

769

Figure 12.11 is a scatterplot of the data. Note the clear curved shape. Because length is one-dimensional and weight (like volume) is three-dimensional, a power model of the form weight = a (length)3 should describe the relationship. What happens if we cube the lengths in the data table and then graph weight versus length3? Figure 12.12 gives us the answer. This transformation of the explanatory variable helps us produce a graph that is quite linear. 800

800

700

600

Weight (g)

Weight (g)

700 500 400 300 200

600 500 400 300 200

100

100

0

0 8

16

24

32

40

0

10,000

20,000

30,000

40,000

50,000

Length3

Length (cm)

FIGURE 12.12  ​The scatterplot of weight versus length3 is linear.

FIGURE 12.11  ​Scatterplot of Atlantic Ocean rockfish weight versus length.

There’s another way to transform the data in the example to achieve linearity. 3 We can take the cube root of the weight values and graph ! weight versus length. Figure 12.13 shows that the resulting scatterplot has a linear form. Why does this transformation work? Start with weight = a(length)3 and take the cube root of both sides of the equation: 3 3 ! weight = " a(length)3 3 3 ! weight = ! a(length)

3 That is, there is a linear relationship between length and ! weight. 10 9

3

Weight

8 7 6 5 4 3 2 1

FIGURE 12.13  ​The scatterplot of !  3 weight versus length is linear.

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5

10

15

20

25

30

35

40

Length (cm)

Once we straighten out the curved pattern in the original scatterplot, we fit a least-squares line to the transformed data. This linear model can be used to predict values of the response variable y. As in Chapter 3, a residual plot tells us if the linear model is appropriate. The values of s and r2 tell us how well the regression line fits the data.

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C H A P T E R 1 2 More about Regression

Example

Go Fish! Transforming with Powers and Roots Here is Minitab output from separate regression analyses of the two sets of transformed Atlantic Ocean rockfish data. Transformation 1: (length3, weight) Predictor

Coef

SE Coef

Constant

4.066

6.902

Length^3

0.0146774

0.0002404

T

P

0.59

0.563

61.07

0.000

S = 18.8412  R-Sq = 99.5%  R-Sq(adj) = 99.5%

Predictor Constant Length

3 Transformation 2: (length,  !weight)

Coef

SE Coef

T

P

−0.02204

0.07762

−0.28

0.780

0.002868

86.00

0.000

0.246616

S = 0.124161  R-Sq = 99.8%  R-Sq(adj) = 99.7%

Problem:  Do each of the following for both transformations. (a) Give the equation of the least-squares regression line. Define any variables you use. (b) Suppose a contestant in the fishing tournament catches an Atlantic Ocean rockfish that’s 36 centimeters long. Use the model from part (a) to predict the fish’s weight. Show your work.

Solution: (a) Transformation 1: weight = 4.066 + 0.0146774 (length3)  3 = − 0.02204 + 0.246616 (length) Transformation 2: !weight

(b) Transformation 1: weight = 4.066 + 0.0146774(363) = 688.9 grams  3 = − 0.02204 + 0.246616(36) = 8.856 Transformation 2: !weight weight = 8.8563 = 694.6 grams

For Practice Try Exercise

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771

When experience or theory suggests that the relationship between two variables is described by a power model of the form y = axp, you now have two strategies for transforming the data to achieve linearity. 1. Raise the values of the explanatory variable x to the p power and plot the points (xp, y). 2. Take the pth root of the values of the response variable y and plot the points p (x, !y)  

What if you have no idea what power to choose? You could guess and test until you find a transformation that works. Some technology comes with built-in sliders that allow you to dynamically adjust the power and watch the scatterplot change shape as you do. It turns out that there is a much more efficient method for linearizing a curved pattern in a scatterplot. Instead of transforming with powers and roots, we use logarithms. This more general method works when the data follow an unknown power model or any of several other common mathematical models.

Transforming with Logarithms Not all curved relationships are described by power models. For instance, in the “Health and Wealth” example (page 765), a graph of life expectancy versus the logarithm (base 10) of income per person showed a linear pattern. We used Fathom software to fit a least-squares regression line to the transformed data and to make a residual plot. Figure 12.14 shows the results. The regression line is predicted life expectancy = 19.5 + 13.2 log(income) How well does this model fit the data? The residual plot shows a random scatter of prediction errors about the residual = 0 line. Also, because r2 = 0.61, about 61% of the variation in life expectancy is accounted FIGURE 12.14  ​Scatterplot with least-squares for by the linear model using log(income) as the explanatory variable. line added and residual plot from Fathom for the transformed data about the health and wealth of The relationship between life expectancy and income per person is nations. described by a logarithmic model of the form y = a + b log x. We can use this model to predict how long a country’s c­ itizens will live from how much money they make. For the United States, which has income per person of $42,296.20, predicted life expectancy = 19.5 + 13.2 log(42,296.20) = 80.567 years The actual U.S. life expectancy in 2012 was 78.80 years. Taking the logarithm of the income per person values straightened out the curved pattern in the original scatterplot. The logarithm transformation can also help achieve linearity when the relationship between two variables is described by a power model or an exponential model.

Power Models Biologists have found that many characteristics of living

things are described quite closely by power models. There are more mice than elephants, and more flies than mice—the abundance of species follows a power model with body weight as the explanatory variable. So do pulse rate, length of life, the number of eggs a bird lays, and so on.

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C H A P T E R 1 2 More about Regression Sometimes the powers can be predicted from geometry, but sometimes they are mysterious. Why, for example, does the rate at which animals use energy go up as the 3/4 power of their body weight? Biologists call this relationship Kleiber’s law. It has been found to work all the way from bacteria to whales. The search goes on for some physical or geometrical explanation for why life follows power laws. To achieve linearity from a power model, we apply the logarithm transformation to both variables. Here are the details: 1. A power model has the form y = axp, where a and p are constants. 2. Take the logarithm of both sides of this equation. Using properties of log­ arithms, we get log y = log(axp) = log a + log(xp) = log a + p log x The equation log y = log a + p log x shows that taking the logarithm of both variables results in a linear relationship between log x and log y. 3. ​Look carefully: the power p in the power model becomes the slope of the straight line that links log y to log x. If a power model describes the relationship between two variables, a scatterplot of the logarithms of both variables should produce a linear pattern. Then we can fit a least-squares regression line to the transformed data and use the linear model to make predictions. Here’s an example.

EXAMPLE

Go Fish!

Weight (g)

Transforming with logarithms Let’s return to the fishing tournament from the previous example. Our goal remains the same: to find a model for predicting the weight of an Atlantic Ocean rockfish from its length. We still expect a power model of the form weight = a(length)3 based on geometry. Here once again is a scatterplot of the data from the local marine research lab.

800 700 600 500 400 300 200 100 0 8

16

24

32

40

Length (cm)

Earlier, we transformed the data in two ways to try to achieve linearity: (1) cubing the length values and (2) taking the cube root of the weight values. This time we’ll use logarithms.

We took the logarithm (base 10) of the values for both variables. Some computer output from a linear regression analysis on the transformed data is shown below. 3.0 0.050

0.025

2.0 Residual

log(Weight)

2.5

1.5

-0.025

0.5 0.0

-0.050

0.6

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0.000

1.0

0.8

1.0 1.2 log(Length)

1.4

1.6

0.6

 

0.8

1.0 1.2 log(Length)

1.4

1.6

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773

Regression Analysis: log(Weight) versus log(Length) Predictor Constant log(Length)

Coef −1.89940 3.04942

SE Coef

T

P

0.03799 0.02764

−49.99 110.31

0.000 0.000

S = 0.0281823  R-Sq = 99.9%  R-Sq(adj) = 99.8%

PROBLEM: (a) Based on the output, explain why it would be reasonable to use a power model to describe the relationship between weight and length for Atlantic Ocean rockfish. (b) Give the equation of the least-squares regression line. Be sure to define any variables you use.

SOLUTION: (a) If a power model describes the relationship between two variables x and y, then a linear model should describe the relationship between log x and log y. The scatterplot of log(weight) versus log(length) has a linear form, and the residual plot shows a fairly random scatter of points about the residual = 0 line. So a power model seems reasonable here. (b) log(weight) = −1.89940 + 3.04942 log(length) For Practice Try Exercise

On the TI-83/84, you can “undo” the logarithm using the 2nd function keys. To solve log y = 2, press 2nd    LOG    2    ENTER . To solve ln y = 2, press 2nd    LN    2    ENTER .

35

If we fit a least-squares regression line to the transformed data, we can find the predicted value of the logarithm of y for any value of the explanatory variable x by substituting our x-value into the equation of the line. To obtain the corresponding prediction for the response variable y, we have to “undo” the logarithm transformation to return to the original units of measurement. One way of doing this is to use the definition of a logarithm as an exponent: logb a = x 1 bx = a For instance, if we have log y = 2, then log y = 2 1 log10 y = 2 1 102 = y 1 100 = y If instead we have ln y = 2, then ln y = 2 1 loge y = 2 1 e2 = y 1 7.389 = y

EXAMPLE

Go Fish! Making predictions PROBLEM:  Suppose a contestant in the fishing tournament catches an Atlantic Ocean rockfish that’s 36 centimeters long. Use the model from part (b) of the previous example to predict the fish’s weight. Show your work. SOLUTION:  For a length of 36 centimeters, we have log(weight) = −1.89940 + 3.04942 log(36) = 2.8464

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To find the predicted weight, we use the definition of a logarithm as an exponent: log10 (weight) = 2.8464 weight = 102.8464 ≈ 702.1 This model predicts that a 36-centimeter-long rockfish will weigh about 702 grams. For Practice Try Exercise

37

Your calculator and most statistical software will calculate the logarithms of all the values of a variable with a single command. The important thing to remember is this: if the relationship between two variables is described by a power model, then we can linearize the relationship by taking the logarithm of both the explanatory and response variables.

How do we find the power model for predicting y from x?   The

THINK ABOUT IT

least-squares line for the transformed rockfish data is

log(weight) = −1.89940 + 3.04942 log(length) If we use the definition of the logarithm as an exponent, we can rewrite this equation as weight = 10−1.89940 + 3.04942log(length) Using properties of exponents, we can simplify this as follows:

800

weight = 10−1.89940 # 103.04942log(length)

using the fact that bmbn = bm+n

3.04942 weight = 10−1.89940 # 10log(length)

using the fact that p log x = log xp

weight = 0.0126(length)3.04942

using the fact that 10log x = x

This equation is now in the familiar form of a power model y = axp with a = 0.0126 and b = 3.04942. Notice how close the power is to 3, as expected from geometry. We could use the power model to predict the weight of a 36-centimeter-long Atlantic Ocean rockfish:

weight = 0.0126(length)3.04942

700

Weight (g)

600 500 400

weight = 0.0126(36)3.04942 ≈ 701.76 grams

300 200 100 0 0

5

10

15

20

25

Length (cm)

30

35

40

This is the same prediction we got earlier (up to rounding). The scatterplot of the original rockfish data with the p ­ ower model added appears in Figure 12.15. Note how well this ­model fits the data!

FIGURE 12.15  ​Rockfish data with power model.

Exponential Models  A linear model has the form y = a + bx. The value

of y increases (or decreases) at a constant rate as x increases. The slope b describes the constant rate of change of a linear model. That is, for each 1 unit increase in x, the model predicts an increase of b units in y. You can think of a linear model as describing the repeated addition of a constant amount. Sometimes the relationship between y and x is based on repeated multiplication by a constant factor. That

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is, each time x increases by 1 unit, the value of y is multiplied by b. An exponential model of the form y = abx describes such multiplicative growth. Populations of living things tend to grow exponentially if not restrained by outside limits such as lack of food or space. More pleasantly (unless we’re talking about credit card debt!), money also displays exponential growth when interest is compounded each time period. Compounding means that last period’s income earns income in the next period.

EXAMPLE

Money, Money, Money Understanding exponential growth Suppose that you invest $100 in a savings account that pays 6% interest ­compounded annually. After a year, you will have earned $100(0.06) = $6.00 in interest. Your new account balance is the initial deposit plus the interest earned: $100 + ($100)(0.06), or $106. We can rewrite this as $100(1 + 0.06), or more simply as $100(1.06). That is, 6% annual interest means that any amount on deposit for the entire year is multiplied by 1.06. If you leave the money invested for a second year, your new balance will be [$100(1.06)](1.06) = $100(1.06)2 = $112.36. Notice that you earn $6.36 in interest during the second year. That’s another $6 in ­interest from your initial $100 deposit plus the interest on your $6 interest earned for Year 1. After x years, your account balance y is given by the exponential model y = 100(1.06)x. The table below shows the balance in your savings account at the end of each of the first six years. Figure 12.16 shows the growth in your investment over 100 years. It is characteristic of exponential growth that the increase appears slow for a long period and then seems to explode.

FIGURE 12.16  ​Scatterplot of the growth of a $100 investment in a savings account paying 6% interest, compounded annually.    

Time x (years)

Account ­balance y

0

$100.00

1

$106.00

2

$112.36

3

$119.10

4

$126.25

5

$133.82

6

$141.85

…

…

 

If an exponential model of the form y = abx describes the relationship between x and y, we can use logarithms to transform the data to produce a l­inear relationship. Start by taking the logarithm (we’ll use base 10, but the natural logarithm ln

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C H A P T E R 1 2 More about Regression using base e would work just as well). Then use algebraic properties of logarithms to simplify the resulting expressions. Here are the details: log y = log (abx)    

taking the logarithm of both sides

log y = log a + log (bx)   using the property log(mn) = log m + log n log y = log a + x log b  

using the property log mp = p log m

We can then rearrange the final equation as log y = log a + (log b)x. Notice that log a and log b are constants because a and b are constants. So the equation gives a linear model relating the explanatory variable x to the transformed variable log y. Thus, if the relationship between two variables follows an exponential model, and we plot the logarithm (base 10 or base e) of y against x, we should observe a straight-line pattern in the transformed data.

EXAMPLE

Moore’s Law and Computer Chips Logarithm transformations and exponential models Gordon Moore, one of the founders of Intel Corporation, predicted in 1965 that the number of transistors on an integrated circuit chip would double every 18 months. This is Moore’s law, one way to measure the revolution in computing. Here are data on the dates and number of transistors for Intel microprocessors:16

FIGURE 12.17  ​Scatterplot of the number of transistors on a computer chip from 1971 to 2010.

Processor

Date

Transistors

4004

1971

2,250

8008

1972

2,500

8080

1974

5,000

8086

1978

29,000

286

1982

120,000

386

1985

275,000

486 DX

1989

1,180,000

Pentium

1993

3,100,000

Pentium II

1997

7,500,000

Pentium III

1999

24,000,000

Pentium 4

2000

42,000,000

Itanium 2

2003

220,000,000

Itanium 2 w/9MB cache

2004

592,000,000

Dual-core Itanium 2

2006

1,700,000,000

Six-core Xeon 7400

2008

1,900,000,000

8-core Xeon Nehalem-EX 2010

2,300,000,000

Figure 12.17 shows the growth in the number of transistors on a computer chip from 1971 to 2010. Notice that we used “years since 1970” as the explanatory variable. We’ll explain this later. If Moore’s law is correct, then an exponential model should describe the relationship between the variables.

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Problem: (a) A scatterplot of the natural logarithm (log base e or ln) of the number of transistors on a ­computer chip versus years since 1970 is shown. Based on this graph, explain why it would be reasonable to use an exponential model to describe the relationship between number of transistors and years since 1970. (b) Minitab output from a linear regression analysis on the transformed data is shown below. Give the equation of the least-squares regression line. Be sure to define any variables you use. Predictor

Coef

SE Coef

T

P

Constant

7.0647

0.2672

26.44

0.000

Years since 1970

0.36583

0.01048

34.91

0.000

S = 0.544467  R-Sq = 98.9%  R-Sq(adj) = 98.8%

(c) Use your model from part (b) to predict the number of transistors on an Intel computer chip in 2020. Show your work. (d) A residual plot for the linear regression in part (b) is shown at left. Discuss what this graph tells you about the appropriateness of the model.

Solution: (a) If an exponential model describes the relationship between two variables x and y, then we expect a scatterplot of (x, ln y) to be roughly linear. The scatterplot of ln(transistors) versus years since 1970 has a fairly linear pattern, especially through the year 2000. So an exponential model seems reasonable here. (b) In(transistors) = 7.0647 + 0.36583(years since 1970) (c) Because 2020 is 50 years since 1970, we have In(transistors) = 7.0647 + 0.36583(50) = 25.3562 To find the predicted number of transistors, we use the definition of a logarithm as an exponent: In(transistors) = 25.3562 1 loge (transistors) = 25.3562 transistors = e25.362 ≈ 1.028 # 1011 This model predicts that an Intel chip made in 2020 will have about 100 billion transistors. (d) The residual plot shows a distinct pattern, with the residuals going from positive to negative to positive as we move from left to right. But the residuals are small in size relative to the transformed y-values. Also, the scatterplot of the transformed data is much more linear than the original scatterplot. We feel reasonably comfortable using this model to make predictions about the number of transistors on a computer chip. For Practice Try Exercise

41

Make sure that you understand the big idea here. The necessary transformation is carried out by taking the logarithm of the response variable. The crucial property of the logarithm for our purposes is that if a variable grows exponentially, its logarithm grows linearly.

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THINK ABOUT IT

How do we find the exponential model for predicting y from x? ​ The least-squares line for the transformed data in the computer chip example is ln (transistors) = 7.0647 + 0.36583 (years since 1970) If we use the definition of the logarithm as an exponent, we can rewrite this ­equation as transistors = e7.0647+0.36583(years since 1970) Using properties of exponents, we can simplify this as follows: transistors = e7.0647 # e0.36583(years since 1970)

using the fact that bmbn = bm+n

transistors = e7.0647 # (e0.36583)(years since 1970)

using the fact that (bm)n = bmn

transistors = 1169.93 # (1.44171)(years since 1970) simplifying This equation is now in the familiar form of an exponential model y = abx with a = 1169.93 and b = 1.44171. We could use the exponential model to predict the number of transistors on an Intel chip in 2020: transistors = 1169.93(1.44171)50 ≈ 1.0281 # 1011. This is the same prediction we got earlier. How does this compare with the prediction from Moore’s law? Suppose the number of transistors on an Intel computer chip doubles every 18 months (1.5 years). Then in the 49 years from 1971 to 2020, the number of transistors would double 49/1.5 = 32.67 times. So the predicted number of transistors on an Intel chip in 2020 would be transistors = 2250(2)32.67 = 1.54 # 1013 Moore’s law predicts more rapid exponential growth than our model does.

The calculation at the end of the Think about It feature might give you some idea of why we used years since 1970 as the explanatory variable in the example. To make a prediction, we substituted the value x = 50 into the equation for the exponential model. This value is the exponent in our calculation. If we had used years as the explanatory variable, our exponent would have been 2020. Such a large exponent can lead to overflow errors on a calculator.

Putting It All Together: Which Transformation Should We Choose? Suppose that a scatterplot shows a curved relationship between two quantitative variables x and y. How can we decide whether a power model or an exponential model better describes the relationship? The following example shows the strategy we should use.

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EXAMPLE

What’s a Planet, Anyway? Power models and logarithm transformations On July 31, 2005, a team of astronomers announced that they had discovered what appeared to be a new planet in our solar system. They had first observed this object almost two years earlier using a telescope at Caltech’s Palomar Observatory in California. Originally named UB313, the potential planet is bigger than Pluto and has an average distance of about 9.5 billion miles from the sun. (For reference, Earth is about 93 million miles from the sun.) Could this new astronomical body, now called Eris, be a new planet?

Period (yr)

At the time of the discovery, there were nine known planets in our solar system. Here are data on the distance from the sun and period of revolution of those planets. Note that distance is measured in astronomical units (AU), the number of Earth distances the object is from the sun.17

250

Planet

Distance from sun (astronomical units)

Period of revolution (Earth years)

200

Mercury

0.387

0.241

150

Venus

0.723

0.615

100

Earth

1.000

1.000

Mars

1.524

1.881

Jupiter

5.203

11.862

Saturn

9.539

29.456

Uranus

19.191

84.070

Neptune

30.061

164.810

Pluto

39.529

248.530

50 0 0

10

20

30

40

Distance (AU)

FIGURE 12.18  ​Scatterplot of planetary distance from the sun and period of revolution.

Figure 12.18 is a scatterplot of the planetary data. There appears to be a strong curved relationship between distance from the sun and period of revolution.

6 5 4 3 2 1 0 –1 –2

Problem:  The graphs below show the results of two different transformations of the data. Figure 12.19(a) plots the natural logarithm of period against distance from the sun for all nine planets. Figure 12.19(b) plots the natural logarithm of period against the natural logarithm of distance from the sun for the nine planets.

In (period)

In (period)

In August 2006, the International Astronomical Union agreed on a new definition of “planet.” Both Pluto and Eris were classified as “dwarf planets.”

0

10

20

30

40

Distance (AU) (a)

6 5 4 3 2 1 0 –1 –2

–1

0

1

2

3

4

In (distance) (b)

FIGURE 12.19  ​(a) A scatterplot of ln(period) versus distance. (b) A scatterplot of ln(period) versus ln(distance).

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(a) Explain why a power model would provide a more appropriate description of the relationship between period of revolution and distance from the sun than an exponential model. (b) Minitab output from a linear regression analysis on the transformed data in Figure 12.19(b) is shown below. Give the equation of the least-squares regression line. Be sure to define any variables you use. Predictor

Coef

SE Coef

Constant

0.0002544

0.0001759

ln(distance) 1.49986

0.00008

T

P 1.45

0.191

18598.27

0.000

S = 0.000393364  R-Sq = 100.0%  R-Sq(adj) = 100.0%

(c) Use your model from part (b) to predict the period of revolution for Eris, which is 9,500,000,000/93,000,000 = 102.15 AU from the sun. Show your work. (d) A residual plot for the linear regression in part (b) is shown at left. Do you expect your prediction in part (c) to be too high, too low, or about right? Justify your answer.

0.00075 0.00050

Residual

0.00025 0.00000

Solution:

– 0.00025

(a) The scatterplot of ln(period) versus distance is clearly curved, so an exponential model would not be appropriate. However, the graph of 1 2 3 4 0 –1 ln(period) versus ln(distance) has a strong linear pattern, indicating that In (distance) a power model would be more appropriate. (b) In(period) = 0.0002544 + 1.49986 In(distance) (c) Eris’s average distance from the sun is 102.15 AU. Using this value for distance in our model from part (b) gives Starnes/Yates/Moore: The Practice of Statistics, 4E In(period) = 0.0002544 + 1.49986 In(102.15) = 6.939 New Fig.: 12_UN43 Perm. Fig.: 12077 To predict the period, we have to undo the logarithm transformation: First Pass: 2010-09-22 –0.00050

period = e6.939 ≈ 1032 years We wouldn’t want to wait for Eris to make a full revolution to see if our prediction is accurate! (d) Eris’s value for ln(distance) is ln(102.15) = 4.626, which would fall at the far right of the residual plot, where all the residuals are positive. Because residual = actual y − predicted y seems likely to be positive, we would expect our prediction to be too low. For Practice Try Exercise 250

Period (yr)

200 150 100 50 0 0

5

10

15

20

25

30

35

Distance (AU)

FIGURE 12.20  ​Planetary data with power model.

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40

43

The scatterplot of the original data with the power model added appears in Figure 12.20. It seems remarkable that period of revolution is closely related to the 1.5 power of distance from the sun. Johannes Kepler made this fascinating discovery about 400 years ago without the aid of modern technology—a result known as Kepler’s third law. What if the scatterplots of (log x, log y) and (x, log y) both look linear? Fit a least-squares regression line to both sets of transformed data. Then compare residual plots and look for the one with the most random scatter. If the residual plots look roughly the same, use the values of s and r2 to decide whether a power model or an exponential model is a better choice.

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781

We have used statistical software to do all the transformations and linear regression analysis in this section so far. Now let’s look at how the process works on a graphing calculator.

30. Technology Corner

Transforming to achieve linearity on the calculator TI-Nspire instructions in Appendix B; HP Prime instructions on the book’s Web site.

We’ll use the planet data to illustrate a general strategy for performing transformations with logarithms on the TI-83/84 and TI-89. A similar approach could be used for transforming data with powers and roots. TI-83/84 TI-89 • Enter the values of the explanatory variable in L1/list1 and the values of the response variable in L2/list2. Make a scatterplot of y versus x and confirm that there is a curved pattern.





• Define L3/list3 to be the natural logarithm (ln) of L1/list1 and L4/list4 to be the natural logarithm of L2/list2. To see whether a power model fits the original data, make a plot of ln y (L4/list4) versus ln x (L3/list3) and look for linearity. To see whether an exponential model fits the original data, make a plot of ln y (L4/list4) versus x (L1/list1) and look for linearity.

• If a linear pattern is present, calculate the equation of the least-squares ­regression line and store it in Y1. For the planet data, we executed the command LinReg(a+bx)L3,L4,Y1.





Construct a residual plot to look for any departures from the linear pattern. For Xlist, enter the list you used as the explanatory variable in the linear regression calculation. For Ylist, use the RESID list stored in the calculator. For the planet data, we used L3/list3 as the Xlist.

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• To make a prediction for a specific value of the explanatory variable, compute log x or ln x, if appropriate. Then use Y1(k) to obtain the predicted value of log y or ln y. To get the predicted value of y, use 10^Ans or e^Ans to undo the logarithm transformation. Here’s our prediction of the period of revolution for Eris, which is at a distance of 102.15 AU from the sun:





Check Your Understanding

One sad fact about life is that we’ll all die someday. Many adults plan ahead for their eventual passing by purchasing life insurance. Many different types of life insurance policies are available. Some provide coverage throughout an individual’s life (whole life), while others last only for a specified number of years (term life). The policyholder makes regular payments (premiums) to the insurance company in return for the coverage. When the insured person dies, a payment is made to designated family members or other beneficiaries. How do insurance companies decide how much to charge for life insurance? They rely on a staff of highly trained actuaries—people with expertise in probability, statistics, and advanced mathematics—to establish premiums. For an individual who wants to buy life ­insurance, the premium will depend on the type and amount of the policy as well as on personal characteristics like age, sex, and health status. The table shows monthly premiums for a 10-year term-life insurance policy worth $1,000,000.18

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Age (years)

Monthly premium

40

$29

45

$46

50

$68

55

$106

60

$157

65

$257

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Section 12.2 Transforming to Achieve Linearity

The Fathom screen shots below show three possible models for predicting monthly premium from age. Option 1 is based on the original data, while Options 2 and 3 involve transformations of the original data. Each screen shot includes a scatterplot with a leastsquares regression line added and a residual plot.

Option 1

Option 2

Option 3

1. Use each model to predict how much a 58-year-old would pay for such a policy. Show your work. 2. What type of function—linear, power, or exponential—best describes the relationship between age and monthly premium? Explain.

case closed!

Do Longer Drives Mean Lower Scores on the PGA Tour? In the chapter-opening Case Study (page 737), we examined data on the mean drive distance (in yards) and mean score per round for an SRS of 19 of the 197 players on the PGA Tour in a recent year. Here is some Minitab output from a least-squares regression analysis on these data:

72.0

Coef

SE Coef

T

P

Constant

76.904

Avg. distance

−0.02016

3.808

20.20

0.000

0.01319

−1.53

0.145

71.5 Mean score

Predictor

S = 0.618396  R-Sq = 12.1%  R-Sq(adj) = 6.9%

71.0

70.5

70.0 270

 

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280

290 300 Mean distance (yards)

310

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C H A P T E R 1 2 More about Regression

1. Calculate the residual for the player with a mean drive distance of 275.4 yards and a mean score per round of 72.1. Show your work. 2. Interpret the value of s in this setting and explain what parameter s is estimating. 3. Do these data give convincing evidence at the a = 0.05 level that the slope of the population regression line is negative? 4. Which kind of mistake—a Type I error or a Type II error—could you have made in Question 3? Justify your answer.

Section 12.2

Summary •

•

•

•

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Curved relationships between two quantitative variables can sometimes be changed into linear relationships by transforming one or both of the variables. Once we transform the data to achieve linearity, we can fit a leastsquares regression line to the transformed data and use this linear model to make predictions. When theory or experience suggests that the relationship between two variables follows a power model of the form y = axp, there are two transformations involving powers and roots that can linearize a curved pattern in a scatterplot. Option 1: Raise the values of the explanatory variable x to the power p, then look at a graph of (xp, y). Option 2: Take the pth root of the values of the p response variable y, then look at a graph of (x, !y) Another useful strategy for straightening a curved pattern in a scatterplot is to take the logarithm of one or both variables. When a power model describes the relationship between two variables, a plot of log y (ln y) versus log x (ln x) should be linear. In a linear model of the form y = a + bx, the values of the response variable are predicted to increase by a constant amount b for each increase of 1 unit in the explanatory variable. For an exponential model of the form y = abx, the predicted values of the response variable are multiplied by a factor of b for each increase of 1 unit in the explanatory variable. When an exponential model describes the relationship between two variables, a plot of log y (ln y) versus x should be linear.

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Section 12.2 Transforming to Achieve Linearity

785

12.2 T echnology Corner TI-Nspire Instructions in Appendix B; HP Prime instructions on the book’s Web site. 30. Transforming to achieve linearity on the calculator

Section 12.2

page 781

Exercises

31. The swinging pendulum  Mrs. Hanrahan’s ­precalculus class collected data on the length (in ­centimeters) of a pendulum and the time (in seconds) the ­pendulum took to complete one back-andforth swing (called its period). Here are their data: Length (cm)

Period (s)

16.5

0.777

17.5

0.839

19.5

0.912

22.5

0.878

28.5

1.004

31.5

1.087

34.5

1.129

37.5

1.111

43.5

1.290

46.5

1.371

106.5

2.115

(c) Use the following graph to identify the transformation that was used to linearize the curved pattern in part (a).

(a) Make a reasonably accurate scatterplot of the data by hand, using length as the explanatory variable. Describe what you see. (b) The theoretical relationship between a ­pendulum’s length and its period is period =

2p "length !g

where g is a constant representing the acceleration due to gravity (in this case, g = 980 cm/s2). Use the following graph to identify the transformation that was used to linearize the curved pattern in part (a).

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32. Boyle’s law  If you have taken a chemistry or physics class, then you are probably familiar with Boyle’s law: for gas in a confined space kept at a constant temperature, pressure times volume is a constant (in symbols, PV = k). Students collected the following data on pressure and volume using a syringe and a pressure probe.

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C H A P T E R 1 2 More about Regression

Volume (cubic centimeters)

Pressure (atmospheres)

6

2.9589

8

2.4073

10

1.9905

12

1.7249

14

1.5288

16

1.3490

18

1.2223

20

1.1201

(a) Make a reasonably accurate scatterplot of the data by hand using volume as the explanatory variable. Describe what you see. (b) If the true relationship between the pressure and volume of the gas is PV = k, we can divide both sides of this equation by V to obtain the theoretical model P = k/V, or P = k(1/V). Use the graph below to ­identify the transformation that was used to linearize the curved pattern in part (a).

Transformation 2: (length, period2) Predictor Constant Length (cm)

SE Coef 0.05802

T P -2.67 0.026

0.042836 0.001320 32.46 0.000

S = 0.105469  R-Sq = 99.2%  R-Sq(adj) = 99.1%



(c) Use the graph below to identify the transformation that was used to linearize the curved pattern in part (a).

Coef -0.15465

Do each of the following for both transformations.

(a) Give the equation of the least-squares regression line. Define any variables you use. (b) Use the model from part (a) to predict the period of a pendulum with length 80 centimeters. Show your work. 34. Boyle’s law  Refer to Exercise 32. Here is Minitab output from separate regression analyses of the two sets of transformed pressure data: 1 , pressureb volume Predictor Coef SE Coef T P Constant 0.36774 0.04055 9.07 0.000 1/V 15.8994 0.4190 37.95 0.000 S = 0.044205  R-Sq = 99.6%  R-Sq(adj) = 99.5% Transformation 1: a

33. The swinging pendulum  Refer to Exercise 31. Here is Minitab output from separate regression analyses of the two sets of transformed pendulum data:

pg 770

Transformation 1: ( !length, period)

Predictor Coef Constant -0.08594 sqrt  (length)

SE Coef 0.05046

T P -1.70 0.123

0.209999 0.008322 25.23 0.000

S = 0.0464223 R-Sq = 98.6% R-Sq(adj) = 98.5%

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Section 12.2 Transforming to Achieve Linearity 1 b pressure Predictor Coef SE Coef T P Constant 0.100170 0.003779 26.51 0.000 Volume 0.0398119 0.0002741 145.23 0.000 S = 0.003553  R-Sq = 100.0%  R-Sq(adj) = 100.0% Transformation 2: avolume,

(a) Based on the output, explain why it would be reasonable to use a power model to describe the relationship between the length and period of a pendulum. (b) Give the equation of the least-squares regression line. Be sure to define any variables you use. 36. Boyle’s law  Refer to Exercise 32. We took the logarithm (base 10) of the values for both variables. Some computer output from a linear regression analysis on the transformed data is shown below. Regression Analysis: log(Pressure) versus log(Volume) Predictor

Coef

Constant

1.11116 log(Volume) -0.81344

SE Coef

T

0.01118

99.39 0.000 -79.78 0.000

0.01020

P

S = 0.00486926   R-Sq = 99.9%  R-Sq(adj) = 99.9%



Do each of the following for both transformations.

(a) Give the equation of the least-squares regression line. Define any variables you use.

0.5

0.4

log(Pressure)

(b) Use the model from part (a) to predict the ­pressure in the syringe when the volume is 17 cubic centimeters. Show your work.

0.3

0.2

0.1

35. The swinging pendulum  Refer to Exercise 31. We pg 772 took the logarithm (base 10) of the values for both variables. Some computer output from a linear regression analysis on the transformed data is shown below.

0.0

0.8

0.9

1.0 1.1 log(Volume)

1.2

1.3

0.0050

Regression Analysis: log(Period) versus log(Length) Coef -0.73675

Constant log(Length)

0.51701

0.0025

SE Coef T P 0.03808 -19.35 0.000 0.02511

Residual

Predictor

20.59 0.000

0.0000 -0.0025 -0.0050

S = 0.0185568   R-Sq = 97.9%   R-Sq(adj) = 97.7%

-0.0075 0.8

0.9

1.0 1.1 log(Volume)

1.2

1.3

0.3

(a) Based on the output, explain why it would be reasonable to use a power model to describe the relationship between pressure and volume.

log(Period)

0.2

0.1

(b) Give the equation of the least-squares regression line. Be sure to define any variables you use.

0.0

-0.1 1.2

1.3

1.4

1.5

1.6 1.7 log(Length)

1.8

1.9

2.0

2.1

0.03

38. Boyle’s law  Use your model from Exercise 36 to predict the pressure in the syringe when the volume is 17 cubic centimeters. Show your work.

0.02 0.01 Residual

37. The swinging pendulum  Use your model from Exercise 35 to predict the period of a pendulum with length 80 centimeters. Show your work.

pg 773

0.00 -0.01 -0.02 -0.03 1.2

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1.3

1.4

1.5

1.6 1.7 log(Length)

1.8

1.9

2.0

2.1

39. Brawn versus brain  How is the weight of an animal’s brain related to the weight of its body? ­Researchers collected data on the brain weight (in grams) and body weight (in kilograms) for 96 species of ­mammals.19 The following figure is a scatterplot of

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C H A P T E R 1 2 More about Regression

the ­logarithm of brain weight against the logarithm of body weight for all 96 species. The least-squares regression line for the transformed data is

(b) A scatterplot of the natural logarithm of the number of surviving bacteria versus time is shown below. Based on this graph, explain why it would be reasonable to use an exponential model to describe the relationship between count of bacteria and time.

log y = 1.01 + 0.72 log x

Logarithm of brain weight

(a) Make a reasonably accurate scatterplot of the data by hand, using time as the explanatory variable. Describe what you see.

3

2

1

0

–1

0

1

2

3

Logarithm of body weight



Based on footprints and some other sketchy evidence, some people believe that a large apelike animal, called Sasquatch or Bigfoot, lives in the Pacific Northwest. His weight is estimated to be about 280 pounds, or 127 kilograms. How big is Bigfoot’s brain? Show your method clearly.

40. Determining tree biomass  It is easy to measure the “diameter at breast height” of a tree. It’s hard to measure the total “aboveground biomass” of a tree, because to do this you must cut and weigh the tree. The biomass is important for studies of ecology, so ecologists commonly estimate it using a power model. Combining data on 378 trees in tropical rain forests gives this relationship between biomass y measured in kilograms and diameter x measured in centimeters:20

(c) Minitab output from a linear regression analysis on the transformed data is shown below. Predictor Coef Constant Time

SE Coef

T

P

5.97316 0.05978 99.92 0.000 -0.218425 0.006575 -33.22 0.000

S = 0.110016  R-Sq = 98.8%  R-Sq(adj) = 98.7%

ln y = −2.00 + 2.42 ln x

Use this model to estimate the biomass of a tropical tree 30 centimeters in diameter. Show your work.

41. Killing bacteria  Expose marine bacteria to X-rays for time periods from 1 to 15 minutes. Here are the ­number of surviving bacteria (in hundreds) on a culture plate after each exposure time:21

pg 776

Time t

Count y

Time t

Count y

1 2 3 4 5 6 7 8

355 211 197 166 142 106 104 60

9 10 11 12 13 14 15

56 38 36 32 21 19 15

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Give the equation of the least-squares regression line. Be sure to define any variables you use.

(d) Use your model to predict the number of ­surviving bacteria after 17 minutes. Show your work. 42. Light through the water  Some college students ­collected data on the intensity of light at various depths in a lake. Here are their data: Depth (m)

Light intensity (lumens)

5 6 7 8 9 10 11

168.00 120.42 86.31 61.87 44.34 31.78 22.78

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Section 12.2 Transforming to Achieve Linearity (a) Make a reasonably accurate scatterplot of the data by hand, using depth as the explanatory variable. Describe what you see.



789

Here is a scatterplot of the data:

(b) A scatterplot of the natural logarithm of light intensity versus depth is shown below. Based on this graph, explain why it would be reasonable to use an exponential model to describe the relationship between light intensity and depth.

(a) The following graphs show the results of two different transformations of the data. Would an exponential model or a power model provide a ­better ­description of the relationship between bounce ­number and height? Justify your answer.

(c) Minitab output from a linear regression analysis on the transformed data is shown below.

Predictor Coef

SE Coef

T

P

Constant

6.78910 0.00009 78575.46 0.000 Depth (m) -0.333021 0.000010 -31783.44 0.000 S = 0.000055  R-Sq = 100.0%  R-Sq(adj) = 100.0%



Give the equation of the least-squares regression line. Be sure to define any variables you use.

(d) Use your model to predict the light intensity at a depth of 12 meters. Show your work. 43. Follow the bouncing ball  Students in Mr. Handford’s class dropped a kickball beneath a motion detector. The detector recorded the height of the ball as it bounced up and down several times. Here are the heights of the ball at the highest point on the first five bounces:

pg 779

(b) Minitab output from a linear regression analysis on the transformed data of log(height) versus bounce number is shown below. Give the equation of the least-squares regression line. Be sure to define any variables you use. Predictor Constant Bounce

Coef SE Coef T P 0.45374 0.01385 32.76 0.000 -0.117160 0.004176 -28.06 0.000

S = 0.0132043  R-Sq = 99.6%  R-Sq(adj) = 99.5% Bounce number

Height (ft)

1

2.240

2

1.620

3

1.235

4

0.958

5

0.756

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(c) Use your model from part (b) to predict the ­highest point the ball reaches on its seventh bounce. Show your work. (d) A residual plot for the linear regression in part (b) is shown on the next page. Do you expect your prediction

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C H A P T E R 1 2 More about Regression

in part (c) to be too high, too low, or about right? Justify your answer.

44. Counting carnivores  Ecologists look at data to learn about nature’s patterns. One pattern they have found relates the size of a carnivore (body mass in ­kilograms) to how many of those carnivores there are in an area. A good measure of “how many” is to count carnivores per 10,000 kilograms (kg) of their prey in the area. The table below gives data for 25 carnivore species.22

Carnivore species

Body mass (kg)

Least weasel Ermine Small Indian mongoose Pine marten Kit fox Channel Islands fox Arctic fox Red fox Bobcat Canadian lynx European badger Coyote Ethiopian wolf Eurasian lynx Wild dog Dhole Snow leopard Wolf Leopard Cheetah Puma Spotted hyena Lion Tiger Polar bear

0.14 0.16 0.55 1.3 2.02 2.16 3.19 4.6 10.0 11.2 13.0 13.0 14.5 20.0 25.0 25.0 40.0 46.0 46.5 50.0 51.9 58.6 142.0 181.0 310.0

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Here is a scatterplot of the data.

(a) The following graphs show the results of two different transformations of the data. Would an exponential model or a power model provide a better description of the relationship between body mass and abundance? Justify your answer.

Abundance (per 10,000 kg of prey) 1656.49 406.66 514.84 31.84 15.96 145.94 21.63 32.21 9.75 4.79 7.35 11.65 2.70 0.46 1.61 0.81 1.89 0.62 6.17 2.29 0.94 0.68 3.40 0.33 0.60

(b) Minitab output from a linear regression analysis on the transformed data of log(abundance) versus log(body mass) is shown below. Give the equation of the least-squares regression line. Be sure to define any variables you use. Predictor Coef Constant log(body  mass)

SE Coef T

P

1.9503 0.1342 14.53 0.000 -1.04811 0.09802 -10.69 0.000

S = 0.423352  R-Sq = 83.3%  R-Sq(adj) = 82.5%

(c) Use your model from part (b) to predict the ­abundance of black bears, which have a body mass of 92.5 kilograms. Show your work. (d) A residual plot for the linear regression in part (b) is shown at top right. Explain what this graph tells you about the appropriateness of the model.

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Section 12.2 Transforming to Achieve Linearity

791

(a) Make an appropriate scatterplot for predicting horizontal distance traveled from ramp height. ­Describe what you see. (b) Use transformations to linearize the relationship. Does the relationship between distance and height seem to follow an exponential model or a power model? Justify your answer.

45. Heart weights of mammals  Here are some data on the hearts of various mammals.23 Mammal

Length of cavity of left ventricle (cm)

Mouse Rat Rabbit Dog Sheep Ox Horse

0.55 1.0 2.2 4.0 6.5 12.0 16.0

Heart weight (g) 0.13 0.64 5.8 102 210 2030 3900

(c) Perform least-squares regression on the transformed data. Give the equation of your regression line. Define any variables you use. (d) Use your model from part (c) to predict the ­horizontal distance a ball would travel if the ramp height was 700. Show your work. Multiple Choice: Select the best answer for ­Exercises 47 to 50. 47. Suppose that the relationship between a response variable y and an explanatory variable x is modeled by y = 2.7(0.316)x. Which of the following scatterplots would approximately follow a straight line? (a) A plot of y against x

(a) Make an appropriate scatterplot for predicting heart weight from length. Describe what you see.

(b) A plot of y against log x

(b) Use transformations to linearize the relationship. Does the relationship between heart weight and length seem to follow an exponential model or a power model? Justify your answer.

(d) A plot of log y against log x

(c) Perform least-squares regression on the transformed data. Give the equation of your regression line. ­Define any variables you use. (d) Use your model from part (c) to predict the heart weight of a human who has a left ventricle 6.8 centimeters long. Show your work. 46. Galileo’s experiment  Galileo studied motion by rolling balls down ramps. He rolled a ball down a ramp with a horizontal shelf at the end of it so that the ball was moving horizontally when it started to fall off the shelf. The top of the ramp was placed at different heights above the floor (that is, the length of the ramp varied), and Galileo measured the horizontal distance the ball traveled before it hit the floor. Here are Galileo’s data. (We won’t try to describe the obsolete seventeenth-century units Galileo used to measure distance and height.)24

Starnes-Yates5e_c12_736-811hr3.indd 791

Height

Distance

1000 828 800 600 300

1500 1340 1328 1172 800

(c) A plot of log y against x (e) A plot of !y against x.

48. Some high school physics students dropped a ball and ­measured the distance fallen (in centimeters) at various times (in seconds) after its release. If you have studied physics, then you probably know that the ­theoretical relationship ­between the variables is distance = 490(time)2. A scatterplot of the students’ data showed a clear curved pattern. At 0.68 seconds after release, the ball had fallen 220.4 centimeters. How much more or less did the ball fall than the theoretical model predicts?

(a) More by 226.576 centimeters (b) More by 6.176 centimeters (c) No more and no less (d) Less by 226.576 centimeters (e) Less by 6.176 centimeters 49. A scatterplot of x = Super Bowl number and y = cost of a 30-second advertisement on the Super Bowl broadcast (in dollars) shows a strong, positive, nonlinear association. A scatterplot of ln(cost) versus Super Bowl number is roughly linear. The least-squares regression line for this association is ln(cost) = 10.97 + 0.0971 (Super Bowl number). Predict the cost of a 30-second advertisement for Super Bowl 40.

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C H A P T E R 1 2 More about Regression

(a) $3

(d) $83,132

(b) $15

(e) $2,824,947

Do you have a tattoo?

Yes 14%

(c) $58,153 50. A scatterplot of y versus x shows a positive, nonlinear association. Two different transformations are attempted to try to linearize the association: using the logarithm of the y values and using the square root of the y values. Two least-squares regression lines are calculated, one that uses x to predict log(y) and the other that uses x to predict !y. Which of the following would be the best reason to prefer the least-squares regression line that uses x to predict log(y)? (a) The value of r2 is smaller. (b) The standard deviation of the residuals is smaller. (c) The slope is greater. (d) The residual plot has more random scatter. (e) The distribution of residuals is more Normal. 51. Shower time (1.3, 2.2, 6.3, 7.3)  Marcella takes a shower every morning when she gets up. Her time in the shower varies according to a Normal distribution with mean 4.5 minutes and standard deviation 0.9 minutes. (a) Find the probability that Marcella’s shower lasts between 3 and 6 minutes on a randomly selected day. Show your work. (b) If Marcella took a 7-minute shower, would it be classified as an outlier by the 1.5IQR rule? Justify your answer. (c) Suppose we choose 10 days at random and record the length of Marcella’s shower each day. What’s the probability that her shower time is 7 minutes or higher on at least 2 of the days? Show your work. (d) Find the probability that the mean length of her shower times on these 10 days exceeds 5 minutes. Show your work. 52. Tattoos (8.2)  What percent of U.S. adults have one or more tattoos? The Harris Poll conducted an online survey of 2302 adults during January 2008. According to the published report, “Respondents for this survey were selected from among those who have agreed to participate in Harris Interactive surveys.”25 The pie chart at top right summarizes the responses from those who were surveyed. Explain why it would not be appropriate to use these data to construct a 95% ­confidence interval for the proportion of all U.S. adults who have tattoos.

Starnes-Yates5e_c12_736-811hr3.indd 792

No 86%

Exercises 53 and 54 refer to the following setting. About 1100 high school teachers attended a weeklong summer institute for teaching AP® classes. After hearing about the survey ® Starnes/Yates/Moore: The Practice 4E class in Exercise 52, the teachers in theof APStatistics, Statistics New Fig.:whether 12_UN84the results Perm. Fig.: 120121 wondered of the tattoo survey would First Pass: 2010-09-22 be similar for teachers. They designed a survey to find out. 2nd Pass: 2010-10-05 The class opted to take a random sample of 100 teachers at the institute. One of the questions on the survey was

Do you have any tattoos on your body? (Circle one) YES   NO

53. Tattoos (8.2, 9.2)  Of the 98 teachers who responded, 23.5% said that they had one or more tattoos. (a) Construct and interpret a 95% confidence ­interval for the actual proportion of teachers at the AP® institute who would say they had tattoos. (b) Does the interval in part (a) provide convincing evidence that the proportion of teachers at the institute with tattoos is not 0.14 (the value cited in the Harris Poll report)? Justify your answer. (c) Two of the selected teachers refused to respond to the survey. If both of these teachers had responded, could your answer to part (b) have changed? Justify your answer. 54. Tattoos (4.1)  One of the first decisions the class had to make was what kind of sampling method to use. (a) They knew that a simple random sample was the “preferred” method. With 1100 teachers in 40 different sessions, the class decided not to use an SRS. Give at least two reasons why you think they made this decision. (b) The AP® Statistics class believed that there might be systematic differences in the proportions of teachers who had tattoos based on the subject areas that they taught. What sampling method would you recommend to account for this possibility? Explain a statistical advantage of this method over an SRS.

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Section 12.2 Transforming to Achieve Linearity

FRAPPY!  Free Response AP® Problem, Yay!

Directions: Show all your work. Indicate clearly the methods you use, because you will be scored on the correctness of your methods as well as on the accuracy and completeness of your results and explanations. A random sample of 14 golfers was selected from the 147 players on the Ladies Professional Golf Association (LPGA) tour in a recent year. The total amount of money won during the year (in dollars) and the scoring average for each player in the sample was recorded. Lower scoring averages are better in golf. The scatterplot below displays the relationship between money and scoring average for these 14 players.

14

13

12

11

10 70

71

72 73 Scoring average

Predictor Coef

SE Coef

T

P

Constant

7.035

11.02 -9.35

0.000

Scoring average

1800000 1600000

77.537 -0.90470

0.09679

74

75

0.000

S = 0.475059  R-Sq = 87.9%  R-Sq(adj) = 86.9%

1400000 1200000 Money

puter output for a least-squares regression using the transformed data.

ln(Money)

The following problem is modeled after actual AP® Statistics exam free response questions. Your task is to generate a complete, concise response in 15 minutes.

(b) Predict the amount of money won for an LPGA golfer with a scoring average of 70. (c) Calculate and interpret a 95% confidence interval for the slope of the least-squares regression line relating ln(money) to scoring average. Assume that the conditions for inference have been met.

1000000 800000 600000 400000 200000 0 70

71

72 73 Scoring average

74

75

(a) Explain why it would not be appropriate to construct a confidence interval for the slope of the least-squares regression line relating money to scoring average. A scatterplot of the natural logarithm of money versus scoring average is shown at top right along with some com-

Starnes-Yates5e_c12_736-811hr3.indd 793

After you finish, you can view two example solutions on the book’s Web site (www.whfreeman.com/tps5e). Determine whether you think each solution is “complete,” “substantial,” “developing,” or “minimal.” If the solution is not complete, what improvements would you suggest to the student who wrote it? Finally, your teacher will provide you with a scoring rubric. Score your response and note what, if anything, you would do differently to improve your own score.

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Chapter Review Section 12.1: Inference for Linear Regression In this section, you learned how to conduct inference about the slope of a population (true) least-squares regression line. The sampling distribution of the sample slope b is the foundation for doing inference about the population (true) slope b. When the conditions are met, the sampling distribution of b has an approximately Normal distribution with s mean mb = b and standard deviation sb = . sx !n There are five conditions for performing inference about a population (true) slope. Remember them with the acronym LINER. • The Linear condition says that the mean value of the response variable my falls on the population (true) regression line my = a + bx. To check the Linear condition, verify that there are no leftover patterns in the residual plot. • The Independent condition says that individual observations are independent of each other. To check the Independent condition, verify that the sample size is less than 10% of the population size. Also, convince yourself that knowing the response for one individual won’t help you predict the response for another individual. • The Normal condition says that the distribution of y values is approximately Normal for each value of x. To check the Normal condition, graph a dotplot, histogram, or Normal probability plot of the residuals and verify that there are no outliers or strong skewness. • The Equal SD condition says that for each value of x, the distribution of y should have the same standard deviation. To check the Equal SD condition, verify that the residuals have roughly the same amount of scatter around the residual = 0 line for each value of x on the residual plot. • The Random condition says that the data are from a random sample or a randomized experiment. To check the Random condition, verify that randomness was properly used in the data collection process.

To construct and interpret a confidence interval for the slope of the population (true) least-squares regression line, follow the familiar four-step process. The formula for the confidence interval is b ± t*SEb, where t* is the t critical value with df = n – 2. The standard error of the slope SEb describes how far the sample slope typically varies from the

population (true) slope in repeated random samples or random assignments. The formula for the standard error of the s slope is SEb = . The standard error of the slope is sx !n − 1 typically provided with standard computer output for leastsquares regression. In most cases, when you conduct a significance test for the slope of the population (true) least-squares regression line, the null hypothesis is H0 :b = 0. This hypothesis says that a straight-line relationship between x and y is of no value for predicting y. To do the calculations, use the test b − b0 statistic t = with df = n – 2. The value of the test staSEb tistic, along with a two-sided P-value, is typically provided with standard computer output for least-squares regression. Section 12.2: Transforming to Achieve Linearity When the association between two variables is nonlinear, transforming one or both of the variables can result in a linear association. If the association between two variables follows a power model in the form y = axp, there are several transformations that will result in a linear association. • Raise the values of x to the power of p and plot y versus xp. p • Calculate the pth root of the y values and plot !y   versus x.

• Calculate the logarithms of the x values and the y values and plot log(y) versus log(x). You can use base 10 logarithms (log) or base e logarithms (ln).

If the association between two variables follows an exponential model in the form y = abx, transform the data by computing the logarithms of the y values and plot log(y) versus x (or ln(y) versus x). Once you have achieved linearity, calculate the equation of the least-squares regression line using the transformed data. Remember to include the transformed variables when you are writing the equation of the line. Likewise, when using the line to make predictions, make sure that the prediction is in the original units of y. If you transformed the y variable, you will need to undo the transformation after using the least-squares regression line. To decide between two or more transformations, look at the residual plots and choose the one with the most random scatter.

794

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What Did You Learn? Learning Objective

Section

Related Example on Page(s)

Relevant Chapter Review Exercise(s)

12.1

745

R12.2, R12.3, R12.4

Interpret the values of a, b, s, SEb, and r in context, and determine these values from computer output.

12.1

748, 754

R12.1

Construct and interpret a confidence interval for the slope b of the population (true) regression line.

12.1

749

R12.3

Perform a significance test about the slope b of the population (true) regression line.

12.1

754

R12.2

Use transformations involving powers and roots to find a power model that describes the relationship between two variables, and use the model to make predictions.

12.2

768, 770

R12.5

Use transformations involving logarithms to find a power model or an exponential model that describes the relationship between two variables, and use the model to make predictions.

12.2

772, 773, 776

R12.6

Determine which of several transformations does a better job of producing a linear relationship.

12.2

779

R12.6

Check the conditions for performing inference about the slope b of the population (true) regression line. 2

Chapter 12 Chapter Review Exercises These exercises are designed to help you review the important ideas and methods of the chapter. Exercises R12.1 to R12.3 refer to the following setting. In the casting of metal parts, molten metal flows through a “gate” into a die that shapes the part. The gate ­velocity (the speed at which metal is forced through the gate) plays a critical role in die casting. A firm that casts ­cylindrical ­aluminum pistons examined a random sample of 12 ­pistons formed from the same alloy of metal. What is the relationship ­between the cylinder wall thickness (inches) and the gate velocity (feet per second) chosen by the

skilled workers who do the casting? If there is a clear pattern, it can be used to direct new workers or to automate the process. A scatterplot of the data is shown below.26 A least-squares regression analysis was performed on the data. Some computer output and a residual plot are shown below. A Normal probability plot of the residuals (not shown) is roughly linear. Predictor Coef SE Coef T Constant 70.44 52.90 1.33 Thickness 274.78 88.18 *** S = 56.3641  R-Sq = 49.3%  R-Sq(adj) =

P 0.213 *** 44.2%

795

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796



C H A P T E R 1 2 More about Regression

R12.1 Casting aluminum (a) Describe what the scatterplot tells you about the relationship between cylinder wall thickness and gate velocity. (b) What is the equation of the least-squares regression line? Define any variables you use. (c) One of the cylinders in the sample had a wall thickness of 0.4 inches. The gate velocity chosen for this cylinder was 104.8 feet per second. Does the regression line in part (b) overpredict or ­underpredict the gate velocity for this cylinder? By how much? Show your work. (d) Is a linear model appropriate in this setting? Justify your answer with appropriate evidence. (e) Interpret each of the following in context: (i) The slope (ii) s (iii) r2 (iv) The standard error of the slope R12.2 Casting aluminum  Do the data provide convincing evidence at the a = 0.05 level of a linear relationship between thickness and gate velocity in the population of pistons formed from this alloy of metal? R12.3 Casting aluminum  Construct and interpret a 95% confidence interval for the slope of the population regression line. Explain how this interval is consistent with the results of Exercise R12.2. R12.4 SAT essay—is longer better?  Following the debut of the new SAT Writing test in March 2005, Dr. Les Perelman from the Massachusetts Institute of Technology recorded the number of words and score for each essay in a sample provided by the College Board. A least-squares regression analysis

was ­performed on these data. The two graphs at bottom left display the results of that analysis. Explain why the conditions for ­performing inference are not met in this setting. R12.5 Light intensity  In a physics class, the intensity of a 100-watt lightbulb was measured by a sensor at ­various distances from the light source. A scatterplot of the data is shown below. Note that a candela is a unit of luminous intensity in the International System of Units.



Physics textbooks suggest that the relationship between light intensity y and distance x should follow an “inverse square law,” that is, a power law model 1 of the form y = ax−2 = a 2 . We transformed the x distance measurements by squaring them and then taking their reciprocals. Some computer output and a residual plot from a least-squares regression analysis on the transformed data are shown below. Note that the horizontal axis on the residual plot displays ­predicted light intensity.

Predictor Constant Distance^(-2)

Coef SE Coef T P -0.000595 0.001821 -0.33 0.751 0.299624 0.003237 92.56 0.000

S = 0.00248369  R-Sq = 99.9%  R-Sq(adj) = 99.9%

(a) Did this transformation achieve linearity? Give appropriate evidence to justify your answer. (b) What is the equation of the least-squares regression line? Define any variables you use. (c) What would you predict for the intensity of a 100-watt bulb at a distance of 2.1 meters? Show your work.

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797

AP® Statistics Practice Test R12.6 An experiment was conducted to determine the effect of practice time (in seconds) on the percent of ­unfamiliar words recalled. Here is a Fathom scatterplot of the results with a least-squares regression line superimposed.

(a) Sketch a residual plot. Be sure to label your axes. (b) Explain why a linear model is not appropriate for describing the relationship between practice time and percent of words recalled. (c) We used Fathom to transform the data in hopes of achieving linearity. The screen shots on the right show the results of two different transformations. Would an exponential model or a power model describe the relationship better? Justify your answer.

(d) Use each model to predict the word recall for 25 seconds of practice. Show your work. Which prediction do you think will be better?

Chapter 12 AP® Statistics Practice Test Section I: Multiple Choice  ​Select the best answer for each question.

(a) For each value of x, the population of y-values is ­Normally distributed. (b) The standard deviation s of the population of y-values corresponding to a particular value of x is always the same, regardless of the specific value of x. (c) The sample size—that is, the number of paired observations (x, y)—exceeds 30. (d) There exists a straight line y = a + bx such that, for each value of x, the mean my of the corresponding population of y-values lies on that straight line. (e) The data come from a random sample or a randomized experiment.

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T12.2 Students in a statistics class drew circles of varying diameters and counted how many Cheerios could be placed in the circle. The scatterplot shows the results. 180 160 140 120

Cheerios

T12.1 Which of the following is not one of the conditions that must be satisfied in order to perform inference about the slope of a least-squares regression line?

100 80 60 40 20 0 0

2

4

6

8

10

12

14

16

Diameter

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C H A P T E R 1 2 More about Regression

The students want to determine an appropriate equation for the relationship between diameter and the number of Cheerios. The students decide to transform the data to make it appear more linear before computing a leastsquares regression line. Which of the following transformations would be reasonable for them to try? I. Plot the square root of the number of Cheerios against diameter. II. Plot the cube of the number of Cheerios against diameter. III. Plot the log of the number of Cheerios against the log of the diameter. IV. Plot the number of Cheerios against the log of the diameter. (a) I and II (c) ​II and III (e) ​I and IV (b) I and III (d) ​II and IV T12.3 Inference about the slope b of a least-squares regres­sion line is based on which of the following distributions? (a) The t distribution with n − 1 degrees of freedom (b) The standard Normal distribution (c) The chi-square distribution with n − 1 degrees of ­freedom (d) The t distribution with n − 2 degrees of freedom (e) The Normal distribution with mean m and standard deviation s Exercises T12.4 through T12.8 refer to the following s­etting. An old saying in golf is “You drive for show and you putt for dough.” The point is that good putting is more important than long driving for shooting low scores and hence ­winning money. To see if this is the case, data from a r­ andom sample of 69 of the nearly 1000 players on the PGA Tour’s world money list are examined. The average number of putts per hole and the player’s total winnings for the previous season are recorded. A least-squares regression line was fitted to the data. The following results were obtained from statistical software. Predictor Coef Constant 7897179 Avg. Putts -4139198

SE Coef 3023782 1698371

T P 6.86 0.000 **** ****

S = 281777  R-Sq = 8.1%  R-Sq(adj) = 7.8%

T12.4 The correlation between total winnings and average number of putts per hole for these players is (a) −0.285. (c)  −0.007. (e) ​0.285. (b) −0.081. (d) ​0.081. T12.5 Suppose that the researchers test the hypotheses H0 : b = 0, Ha : b < 0. The value of the t statistic for this test is (a) 2.61. (c) ​0.081. (e) ​ −20.24. (b) 2.44. (d) ​ −2.44. T12.6 The P-value for the test in Question T12.5 is 0.0087. A correct interpretation of this result is that

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(a) the probability that there is no linear relationship ­between average number of putts per hole and total ­winnings for these 69 players is 0.0087. (b) the probability that there is no linear relationship ­between average number of putts per hole and total ­winnings for all players on the PGA Tour’s world money list is 0.0087. (c) if there is no linear relationship between average number of putts per hole and total winnings for the players in the sample, the probability of getting a random sample of 69 players that yields a least-squares regression line with a slope of −4139198 or less is 0.0087. (d) if there is no linear relationship between average ­number of putts per hole and total winnings for the ­players on the PGA Tour’s world money list, the probability of ­getting a random sample of 69 players that yields a least-squares regression line with a slope of −4139198 or less is 0.0087. (e) the probability of making a Type I error is 0.0087. T12.7 A 95% confidence interval for the slope b of the population regression line is (a) 7,897,179 ± 3,023,782. (b) 7,897,179 ± 6,047,564. (c) −4,139,198 ± 1,698,371. (d) −4,139,198 ± 3,328,807. (e) −4,139,198 ± 3,396,742. T12.8 A residual plot from the least-squares regression is shown below. Which of the following statements is supported by the graph?

Residuals

798

1,000,000 800,000 600,000 400,000 200,000 0 −200,000 −400,000 −600,000 −800,000 −1,000,000 1.700

1.725

1.750

1.775

1.800

1.825

1.850

Average Putts per Round

(a) The residual plot contains dramatic evidence that the standard deviation of the response about the population ­regression line increases as the average number of putts per round increases. (b) The sum of the residuals is not 0. Obviously, there is a major error present. (c) Using the regression line to predict a player’s total ­winnings from his average number of putts almost always results in errors of less than $200,000. (d) For two players, the regression line underpredicts their total winnings by more than $800,000. (e) The residual plot reveals no correlation between average putts per round and prediction errors from the least-squares line for these players.

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AP® Statistics Practice Test T12.9 Which of the following would provide evidence that a power law model of the form y = axb, where b ≠ 0 and b ≠ 1, describes the relationship between a response variable y and an explanatory variable x? (a) A scatterplot of y versus x looks approximately linear. (b) A scatterplot of ln y versus x looks approximately linear. (c) A scatterplot of y versus ln x looks approximately linear. (d) A scatterplot of ln y versus ln x looks approximately ­linear. (e) None of these

799

T12.10 We record data on the population of a ­particular country from 1960 to 2010. A scatterplot reveals a clear curved relationship between population and year. ­However, a different scatterplot reveals a strong linear relationship between the logarithm (base 10) of the population and the year. The leastsquares regression line for the transformed data is log (population) = −13.5 + 0.01(year)

Based on this equation, the population of the country in the year 2020 should be about

(a) 6.7.

(c) ​5,000,000.

(b) 812.

(d) ​6,700,000.

(e) ​8,120,000.

Section II: Free Response  ​Show all your work. Indicate clearly the methods you use, because you will be graded on the correctness of your methods as well as on the accuracy and completeness of your results and explanations. T12.11 Growth hormones are often used to increase the weight gain of chickens. In an experiment using 15 chickens, 3 chickens were randomly assigned to each of 5 different doses of growth hormone (0, 0.2, 0.4, 0.8, and 1.0 milligrams). The subsequent weight gain (in ounces) was recorded for each chicken. A researcher plots the data and finds that a linear relationship appears to hold. Computer output from a least-squares regression analysis for these data is shown below. Assume that the conditions for performing inference about the slope b of the true regression line are met. Predictor Coef SE Coef T P Constant 4.5459 0.6166 7.37 58. The P-value for the resulting onesample t test is 0.035. Which of the following best describes what the P-value measures?

AP4.35 In a clinical trial, 30 patients with a certain blood disease are randomly assigned to two groups. One group is then randomly assigned the currently marketed medicine, and the other group receives the experimental medicine. Each week, patients report to the clinic where blood tests are conducted. The lab technician is unaware of the kind of medicine the patient is taking, and the patient is also unaware of which medicine he or she has been given. This design can be described as

(a) The probability that the new integrated circuit has the same life span as the current model is 0.035. (b) The probability that the test correctly rejects the null ­hypothesis in favor of the alternative hypothesis is 0.035. (c) The probability that a single new integrated circuit will not last as long as one of the earlier circuits is 0.035. (d) The probability of getting a sample statistic as far or farther from 58 if there really is no difference between the new and the old circuits is 0.035. (e) The probability of getting a sample mean for the new integrated circuit that is lower than the mean for the earlier model is 0.035.

(a) a double-blind, completely randomized experiment, with the currently marketed medicine and the experimental medicine as the two treatments. (b) a single-blind, completely randomized experiment, with the currently marketed medicine and the experimental medicine as the two treatments. (c) a double-blind, matched pairs design, with the currently marketed medicine and the experimental medicine forming a pair. (d) a double-blind, block design that is not a matched pairs design, with the currently marketed medicine and the experimental medicine as the two blocks. (e) a double-blind, randomized observational study.

Cumulative relative frequency

AP4.36 A local investment club that meets monthly has 200 members ranging in age from 27 to 81. A cumulative relative frequency graph is shown below. Approximately how many members of the club are more than 60 years of age? 100 80 60 40

Questions AP4.38 and AP4.39 refer to the following situation. Do children’s fear levels change over time and, if so, in what ways? Little research has been done on the prevalence and persistence of fears in children. Several years ago, two researchers surveyed a randomly selected group of 94 thirdand fourth-grade children, asking them to rate their level of fearfulness about a variety of situations. Two years later, the children again completed the same survey. The ­researchers computed the overall fear rating for each child in both years and were interested in the relationship between these ratings. They then assumed that the true regression line was mlater rating = a + b (initial rating) and that the assumptions for regression inference were satisfied. This model was fitted to the data using least-squares regression. The following results were obtained from statistical software. Predictor

Coefficient

St. Dev.

Constant

0.877917

0.1184

Initial Rating

0.397911

0.0676

S = 0.2374

R-Sq = 0.274

Here is a scatterplot of the later ratings versus the initial ratings and a plot of the residuals versus the initial ratings.

20 0 20

30

40

50

60

70

80

Age of members (years)

(c) ​78 (d) ​90

(e) ​110

AP4.37 A manufacturer of electronic components is testing the durability of a newly designed integrated circuit to determine whether its life span is longer than that of the earlier model, which has a mean life span of 58 months. The company takes a simple random sample of 120 integrated ­circuits

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Later ratings

(a) 20 (b) 44

2.1

1.8

1.5

1.2 1.2

1.6

2.0

2.4

Initial ratings

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808



C H A P T E R 1 2 More about Regression AP4.39 George’s initial fear rating was 0.2 higher than Jonny’s. What does the model predict about their final fear ratings? (a) George’s will be about 0.96 higher than Jonny’s. (b) George’s will be about 0.40 higher than Jonny’s. (c) George’s will be about 0.20 higher than Jonny’s (d) George’s will be about 0.08 higher than Jonny’s. (e) George’s will be about the same as Jonny’s.

Residuals

0.4

0.0

– 0.4

1.2

1.6

2.0

2.4

Initial ratings

AP4.38 Which of the following statements is supported by these plots?

AP4.40 The table below provides data on the political affiliation and opinion about the death penalty of 850 randomly selected voters from a congressional district.

(a) There is no striking evidence that the assumptions for regression inference are violated.

Republican

(d) These plots call into question the validity of the assumption that the later ratings vary Normally about the least-squares line for each value of the initial ratings. (e) A linear model isn’t appropriate here because the residual plot shows no association.

Oppose

Total

299

98

397

77

171

248

Other

118

87

205

Total

494

356

850

Democrat

(b) The abundance of outliers and influential observations in the plots means that the assumptions for regression are clearly violated. (c) These plots contain dramatic evidence that the standard deviation of the response about the true regression line is not approximately the same for each x-value.

Favor



Which of the following does not support the conclusion that being a Republican and favoring the death penalty are not independent?

299 98 (a) ∙ 494 356

(d)

494 397 ∙ 850 850

299 397 (b) ∙ 494 850

(e)

(397)(494) ∙ 299 850

494 299 (c) ∙ 850 397

Section II: Free Response  Show all your work. Indicate clearly the methods you use, because you will be graded on the correctness of your methods as well as on the accuracy and completeness of your results and ­explanations. AP4.41 The body’s natural electrical field helps wounds heal. If diabetes changes this field, it might explain why people with diabetes heal more slowly. A study of this idea compared randomly selected normal mice and randomly selected mice bred to spontaneously develop diabetes. The investigators attached sensors to the right hip and front feet of the mice and measured the difference in electrical potential (in millivolts) between these locations. Graphs of the data for each group reveal no outliers or strong skewness. The following computer output provides numerical summaries of the data.27 Variable

N

Mean

StDev

Diabetic mice

24

13.090

4.839

Minimum 1.050

Normal mice

18

10.022

2.915

4.950

Q1

Median

Q3

Maximum

10.038

12.650

17.038

22.600

8.238

9.250

12.375

16.100

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The researchers want to know whether the difference in mean electrical potentials between normal mice and mice with diabetes is statistically significant at the a = 0.05 level. Carry out a test and report your conclusion.

AP4.42 Can physical activity in youth lead to mental sharpness in old age? A 2010 study investigating this question ­involved 9344 randomly selected, mostly white women over age 65 from four U.S. states. These women were asked about their levels of physical activity during their teenage years, thirties, fifties, and later years. Those who reported being physically active as teens enjoyed the lowest level of cognitive decline—only 8.5% had cognitive impairment—compared with 16.7% of women who reported not being ­physically active at that time. (a) State an appropriate pair of hypotheses that the ­researchers could use to test whether the proportion of women who suffered a cognitive decline was

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Cumulative AP® Practice Test 4

AP4.43 In a recent poll, randomly selected New York State residents at various fast-food restaurants were asked if they supported or opposed a “fat tax” on nondiet sugared soda. Thirty-one percent said that they were in favor of such a tax and 66% were opposed. But when asked if they would support such a tax if the money raised were used to fund health care given the high incidence of obesity in the United States, 48% said that they were in favor and 49% were opposed. (a) In this situation, explain how bias may have been ­introduced based on the way the questions were worded and suggest a way that they could have been worded differently in order to avoid this bias. (b) In this situation, explain how bias may have been introduced based on the way the sample was taken and suggest a way that the sample could have been obtained in order to avoid this bias. (c) This poll was conducted only in New York State. Suppose the pollsters wanted to ensure that estimates for the proportion of people who would support a tax on nondiet sugared soda were available for each state as well as an overall estimate for the nation as a whole. Identify a sampling method that would achieve this goal and briefly ­describe how the sample would be taken. AP4.44 ​ Each morning, coffee is brewed in the school workroom by one of three faculty members, depending on who arrives first at work. Mr. Worcester arrives first 10% of the time, Dr. Currier arrives first 50% of the time, and Mr. Legacy arrives first on the remaining mornings. The probability that the coffee is strong when brewed by Dr. Currier is 0.1, while

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the corresponding probabilities when it is brewed by Mr. Legacy and Mr. Worcester are 0.2 and 0.3, respectively. Mr. Worcester likes strong coffee! (a) What is the probability that on a randomly selected morning the coffee will be strong? Show your work. (b) If the coffee is strong on a randomly selected morning, what is the probability that it was brewed by Dr. Currier? Show your work. AP4.45 The following table gives data on the mean ­number of seeds produced in a year by several common tree species and the mean weight (in milligrams) of the seeds produced. Two species appear twice because their seeds were counted in two locations. We might ­expect that trees with heavy seeds produce fewer of them, but what mathematical model best describes the relationship?28 Tree species Paper birch Yellow birch White spruce Engelmann spruce Red spruce Tulip tree Ponderosa pine White fir Sugar maple Sugar pine American beech American beech Black oak Scarlet oak Red oak Red oak Pignut hickory White oak Chestnut oak

Seed count 27,239 12,158 7202 3671 5051 13,509 2667 5196 1751 1159 463 1892 93 525 411 253 40 184 107

Seed weight (mg) 0.6 1.6 2.0 3.3 3.4 9.1 37.7 40.0 48.0 216 247 247 1851 1930 2475 2475 3423 3669 4535

(a) Based on the scatterplot below, is a linear model appropriate to describe the relationship between seed count and seed weight? Explain. 5000

Seed weight (mg)

significantly lower for women who were physically active in their youth than for women who were not physically active at that time. Be sure to define any parameters you use. (b) Assuming the conditions for performing inference are met, what inference method would you use to test the ­hypotheses you identified in part (b)? Do not carry out the test. (c) Suppose the test in part (b) shows that the proportion of women who suffered a cognitive decline was significantly lower for women who were physically active in their youth than for women who were not physically active at that time. Can we generalize the results of this study to all women aged 65 and older? Justify your answer. (d) We cannot conclude that being physically active as a teen causes a lower level of cognitive decline for women over 65, due to possible confounding with other variables. Explain the concept of confounding and give an example of a potential confounding variable in this study.

809

4000 3000 2000 1000 0 0

5000

10,000

15,000

20,000

25,000

30,000

Seed count

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C H A P T E R 1 2 More about Regression

(b) Two alternative models based on transforming the original data are proposed to predict the seed weight from the seed count. Graphs and computer output from a least-squares r­egression analysis on the transformed data are shown below. Model A: 8

2

1

Residual



810

0

−1

ln(weight)

6

−2 0

4

2

4

6

8

10

Fitted value 2

Predictor Coef Constant 0

5000

10,000

15,000

20,000

25,000

30,000

1

Residual

P

14.33 -10.35

0.000 0.000

Which model, A or B, is more appropriate for predicting seed weight from seed count? Justify your answer.

2

(c) Using the model you chose in part (b), predict the seed weight if the seed count is 3700. (d) Interpret the value of r2 for your model.

0 −1 −2 −3 −4 −4

−2

0

2

4

6

Fitted value

Predictor Coef

SE Coef

T

Constant

0.5726

10.72 0.000

6.1394

P

-0.00033869 0.00007187 -4.71 0.000

Seed Count

S = 2.08100  R-Sq = 56.6%  R-Sq(adj) = 54.1%

AP4.46 Suppose a company manufactures plastic lids for disposable coffee cups. When the manufacturing process is working correctly, the diameters of the lids are approximately Normally distributed with a mean diameter of 4 inches and a standard deviation of 0.02 inches. To make sure the machine is not producing lids that are too big or too small, each hour a random sample of 25 lids is selected and the sample mean is calculated. (a) Describe the shape, center, and spread of the sampling distribution of the sample mean diameter, assuming the machine is working properly. The company decides that it will shut down the ­machine if the sample mean diameter is less than 3.99 inches or greater than 4.01 inches, because this indicates that some lids will be too small or too large for the cups. If the sample mean is less than 3.99 or greater than 4.01, all the lids from that hour are thrown away because the company does not want to sell bad products. (b) Assuming that the machine is working properly, what is the probability that a random sample of 25 lids will have a mean diameter less than 3.99 inches or greater than 4.01 inches? Show your work.

Model B: 8 6

ln(weight)

T

S = 1.16932  R-Sq = 86.3%  R-Sq(adj) = 85.5%

Seed count 3

4 2 0 3

SE Coef

15.491 1.081 ln(count) -1.5222 0.1470

0

4

5

6

7

ln(count)

Starnes-Yates5e_c12_736-811hr3.indd 810

8

9

10

11

Also, to look for any trends, each hour the company records the value of the sample mean on a chart, like the one at top right.

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Sample mean diameter (inches)

Cumulative AP® Practice Test 4



4.01

Hour 2 Hour 3

4.00

Hour 1

3.99

One benefit of using this type of chart is that out-of-control production trends can be noticed before it is too late and lids have to be thrown away. For example, if the sample mean Starnes/Yates/Moore: The Practice of Statistics, 4E increased in 3 consecutive samples, this would New Fig.: CumTest04_11 Perm. Fig.: 4011 suggest that something might be wrong with the First Pass: 2010-09-30 machine. If this trend can be noticed before the sample mean gets larger than 4.01, then the machine can be fixed without having to throw away any lids.

(c) Assuming that the manufacturing process is working correctly, what is the probability that the sample mean diameter will be above the desired mean of 4.00 but ­below the upper boundary of 4.01? Show your work.

Starnes-Yates5e_c12_736-811hr3.indd 811

811

(d) Assuming that the manufacturing process is working correctly, what is the probability that in 5 consecutive samples, 4 or 5 of the sample means will be above the desired mean of 4.00 but below the upper boundary of 4.01? Show your work. (e) Which of the following results gives more convincing evidence that the machine needs to be shut down? Explain. 1. Getting a single sample mean below 3.99 or above 4.01 or 2. Taking 5 consecutive samples and having at least 4 of the sample means be between 4.00 and 4.01. (f) Suggest a different rule (other than 1 and 2 stated in part (e)) for stopping the machine before it starts producing lids that have to be thrown away. Assuming that the machine is working properly, calculate the probability that the m ­ achine will be shut down when using your rule.

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Index

Photo Credits Cover: Joseph Devenney/Getty Images Flamingos used in interior design: © Smileus/ istockphoto p. vi p. vi p. xxi p. xxi p. xxii p. xxiii p. xxiii

Judy Starnes Anne Tabor istockphoto Bloomberg via Getty Images istockphoto istockphoto © Leah Warkentin/Design Pics/Corbis

Chapter 1 p. xxxii Thinkstock p. 2 istockphoto p. 3 Courtesy RSSCSE p. 5 istockphoto p. 9 DILBERT: © Scott Adams/Dist. by United Feature Syndicate, Inc. p. 10 istockphoto p. 12 istockphoto p. 15 Thinkstock p. 19 The Kobal Collection/20th Century Fox/Paramount p. 25 Chuck Myers/MCT via Getty Images p. 26 Photo by Bentley Motors via Getty Images p. 30 istockphoto p. 31 Thinkstock p. 33 Thinkstock p. 50 istockphoto p. 51 W. H. Freeman and Company p. 52 istockphoto p. 57 Photo by Sara Wolfram/Getty Images p. 62 © 2008 Zits Partnership. Distributed by King Features Syndicate. p. 65 Getty Images/OJO Images RF p. 67 Thinkstock Chapter 2 p. 82 PRNewsFoto/Chrysler Group, Diane Bondareff p. 84 Thinkstock p. 86 Fotosearch p. 88 AP Photo/Jae C. Hong p. 91 istockphoto p. 95 istockphoto p. 98 NASA and The Hubble Heritage Team p. 106 FOXTROT©2005 Bill Amend. Universal Press Syndicate, Inc. p. 111 istockphoto p. 115 istockphoto p. 124 istockphoto p. 126 PRNewsFoto/Chrysler Group, Diane Bondareff Chapter 3 p. 140 Thinkstock p. 142 istockphoto p. 144 istockphoto

p. 145 Chris Graythen/Staff, Getty Images p. 148 istockphoto/Thinkstock p. 149 Mother Goose & Grimm-Grimmy, Inc. King Features Syndicate p. 152 Cal Sport Media via AP Images p. 156 W. H. Freeman and Company p. 157 AP Photo/Eric Gay p. 165 Ford Images p. 168 Thinkstock p. 173 Ford Images p. 180 Sports Illustrated/Getty Images p. 181 © Fancy/Alamy p. 183 Thinkstock p. 185 © Laqhill|Dreamstime.com p. 190 istockphoto p. 191 Thinkstock Chapter 4 p. 206 Eric O’Connell/Getty Images p. 208 © MBI/Alamy p. 209 istockphoto p. 210 © Charles O. Cecil/Alamy p. 212 Thinkstock p. 214 Cartoons By Landers p. 215 Urbaniak, G. C., & Plous, S. (2011). Research Randomizer (Version 3.0) [Computer software]. Retrieved on April 22, 2011, from http://www.randomizer. org/ p. 217 istockphoto p. 218 MPI/Getty Images p. 219 © Li Ding/Alamy p. 223 © Richard Levine/Alamy p. 223 © David Hodges/Alamy p. 225 istockphoto p. 226 GSS p. 227 © 1995 Watterson. Universal Press Syndicate. Reprinted with permission. All rights reserved. p. 235 istockphoto p. 237 © Golden Pixels LLC/Alamy p. 238 istockphoto p. 239 istockphoto p. 240 istockphoto p. 241 © moodboard/Alamy p. 243 istockphoto p. 250 istockphoto p. 266 istockphoto/Thinkstock p. 269 istockphoto p. 271 W. H. Freeman and Company p. 272 Eric O’Connell/Getty Images Chapter 5 p. 286 Hemera/Thinkstock p. 288 © Jason Stitt|Dreamstime.com p. 289 Creatas/Thinkstock p. 293 © Eyebyte/Alamy p. 294 Mitchell Layton/Getty Images p. 295 W. H. Freeman and Company p. 298 Drew Hallowell/Getty Images

I-1

p. 306 p. 308 p. 309 p. 313 p. 318 p. 319 p. 320 p. 322 p. 324 p. 327 p. 330

© Judith Collins/Alamy © Murray Lee/Getty Images © Big Cheese Photo/SuperStock © Flint/Corbis istockphoto istockphoto istockphoto Thomas Starke/Bongarts/Getty Images istockphoto Graeme Harris/Getty Images © 2008 Chris Browne 2010 King Features Syndicate, Inc. p. 332 Hemera/Thinkstock Chapter 6 p. 344 Comstock/Thinkstock p. 346 istockphoto p. 347 istockphoto p. 349 Thinkstock p. 351 Thinkstock p. 352 Scott Adams/Dist. by United Feature Syndicate, Inc. p. 357 Alamy p. 365 Alamy p. 368 istockphoto p. 372 moodboard/SuperStock p. 374 (top) istockphoto/Thinkstock p. 374 (inset) Alamy p. 380 istockphoto p. 381 Thinkstock p. 383 istockphoto p. 387 © Alex Clark/Alamy p. 388 istockphoto p. 390 Exactostock/SuperStock p. 396 Thinkstock p. 399 Dreamstime p. 401 © Chen Wee Lip|Dreamstime.com p. 402 istockphoto p. 403 istockphoto p. 405 istockphoto/thinkstock p. 409 Comstock/Thinkstock Chapter 7 p. 420 © Arekmalang|Dreamstime.com p. 423 istockphoto p. 425 istockphoto p. 426 © Kevin Woodrow|Dreamstime.com p. 430 Alamy p. 432 CBS/Landov p. 435 istockphoto p. 440 FOXTROT©2005 Bill Amend. Universal Press Syndicate. p. 445 Dreamstime p. 450 istockphoto p. 455 Alamy p. 460 © Arekmalang|Dreamstime.com Chapter 8 p. 474 Brooke Fasani/age fotostock p. 476 istockphoto p. 478 istockphoto

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PC-2 P h o t o C r e d i t s p. 480 Garfield ©1999 Paws, Inc. Dist. by Universal Uclick. Reprinted by permission. All rights reserved. p. 481 Digital Vision/Thinkstock p. 484 © Jan Haas/dpa/Corbis p. 485 istockphoto p. 493 istockphoto p. 498 istockphoto p. 500 Alamy p. 502 istockphoto p. 509 © Robert Kneschke|Dreamstime.com p. 519 Alamy p. 520 istockphoto p. 522 istockphoto p. 524 Phototake/Alamy p. 525 Brooke Fasani/age footstock Chapter 9 p. 536 istockphoto p. 540 Superstock p. 542 © Otnaydur|Dreamstime.com p. 543 istockphoto p. 544 istockphoto p. 545 istockphoto p. 546 istockphoto p. 549 fotosearch p. 555 fotosearch p. 559 istockphoto p. 562 istockphoto

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p. 577 p. 580 p. 583 p. 586 p. 589 p. 590 p. 592 p. 593

istockphoto istockphoto Alamy Alamy Courtesy Judy Starnes istockphoto istockphoto istockphoto

Chapter 10 p. 608 Jamie Rector/Getty Images p. 610 © TravelStockCollection—Homer Sykes/Alamy p. 612 Alamy p. 615 Envision/Corbis p. 617 Photo by Peter Macdiarmid/Getty Images p. 620 Alamy p. 622 istockphoto p. 625 Photolibrary p. 635 iStock p. 636 Jeff Greenberg/Photo Edit p. 639 Rich Legg/Getty Images p. 639 istockphoto p. 645 Alamy p. 650 Clerkenwell/Getty Images p. 651 Jamie Rector/Getty Images p. 655 istockphoto

Chapter 11 p. 676 Alex-Takes-Photos/Flickr/Getty p. 678 R Rivera Moret p. 681 R Rivera Moret p. 683 R Rivera Moret p. 687 R Rivera Moret p. 688 Christian Petersen/Getty Images p. 698 RF Pictures/Corbis p. 701 Alamy p. 708 Alamy p. 712 Alamy p. 715 Alamy p. 718 Alamy p. 721 Alex-Takes-Photos/Flickr/Getty Chapter 12 p. 736 Mike Ehrmann/Getty Images p. 738 fotosearch p. 748 istockphoto p. 749 Ford Images p. 754 Fotosearch p. 767 istockphoto p. 768 Alamy p. 773 © RGB Ventures LLC dba SuperStock/ Alamy p. 775 istockphoto p. 776 © Nandana De Silva/Alamy p. 779 NASA/JPL-Caltech p. 783 Mike Ehrmann/Getty Images

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Notes and Data Sources Overview: What Is Statistics?

  1. ​Patrizia Frei et al., “Use of mobile phones and risk of brain tumours: update of Danish cohort study,” BMJ 2011;343:d6387 (Published 20 October 2011).   2. ​Nikhil Swaminathan, “Gender Jabber: Do Women Talk More than Men?” Scientific American, July 6, 2007.   3.  Stephen Moss, “Do women really talk more?” The Guardian, November 26, 2006.  4. ​Data from gapminder.org, accessed October 19, 2013.  5. ​ Data from the Josephson Institute’s “2012 Report Card of American Youth,” accessed online at charactercounts.org, March 16, 2013.

Chapter 1

 1. ​These are some of the data from the EESEE story “Stress among Pets and Friends.” The study results appear in K. Allen, J. Blascovich, J. Tomaka, and R. M. Kelsey, “Presence of human friends and pet dogs as moderators of autonomic responses to stress in women,” Journal of Personality and Social Psychology, 83 (1988), pp. 582–589.   2. ​Roller coaster data for 2009 from www.rcdb.com.  3. ​Arbitron, Radio Today, www.arbitron.com.  4. ​Arbitron, Internet and Multimedia, www.arbitron.com.   5. ​We got the idea for this example from David Lane’s case study “Who is buying iMacs?” which we found online at ­onlinestatbook. com/case_studies_rvls/.   6. ​The National Longitudinal Study of Adolescent Health interviewed a stratified random sample of 27,000 adolescents, then reinterviewed many of the subjects six years later, when most were aged 19 to 25. These data are from the Wave III reinterviews in 2000 and 2001, found at the Web site of the Carolina Population Center, www.cpc.unc.edu.   7. ​The U.K. data were obtained using the Random Sampler tool at www.censusatschool.com. The U.S. data were obtained using the Random Sampler tool at www.amstat.org/censusatschool.   8. ​2011 DuPont Automotive Color Survey, at www2.dupont.com/ Media_Center/en_US/color_popularity/.  9. ​Found at spam-filter-review.toptenreviews.com, which claims to have compiled data “from a number of different reputable sources.” 10. ​Centers for Disease Control and Prevention, Births: Final Data for 2005, National Vital Statistics Reports, 56, No. 6, 2007, at www. cdc.gov. 11. ​ National Center for Health Statistics, Deaths: Preliminary Data for 2010, at www.cdc.gov/nchs. 12. ​The Hispanic Population: 2010, at www.census.gov, based on the data collected from the 2010 U.S. Census. 13. ​Data for 2005 from the 2007 Statistical Abstract of the United States at the Census Bureau Web site, www.census.gov. 14. ​Arbitron, Internet and Media, at www.arbitron.com. 15. ​Statistical Abstract of the United States, 2004–2005, Table 1233.

16. ​Lien-Ti Bei, “Consumers’ purchase behavior toward recycled products: an acquisition-transaction utility theory perspective,” MS thesis, Purdue University, 1993. 17. ​S. V. Zagona (ed.), Studies and Issues in Smoking Behavior, University of Arizona Press, 1967, pp. 157–180. 18. ​ From the Web site of the Carolina Population Center, www.cpc.unc.edu. 19. ​Data on the U.S. Women’s National Soccer Team are from the United States Soccer Federation Web site, www.ussoccer.com. 20. ​2012 Fuel Economy Guide, from the U.S. Environmental Protection Agency’s Web site at www.fueleconomy.gov. 21. ​From the American Community Survey, at factfinder2.census. gov. 22. ​James T. Fleming, “The measurement of children’s perception of difficulty in reading materials,” Research in the Teaching of English, 1 (1967), pp. 136–156. 23. ​See Note 13. 24. ​The original paper is T. M. Amabile, “Motivation and creativity: effects of motivational orientation on creative writers,” Journal of Personality and Social Psychology, 48, No. 2 (February 1985), pp. 393–399. The data for Exercise 43 came from Fred L. Ramsey and Daniel W. Schafer, The Statistical Sleuth, 3rd ed., Brooks/Cole Cengage Learning, 2013. 25. ​The cereal data came from the Data and Story Library, http:// lib.stat.cmu.edu/DASL/. 26. ​From a plot in K. Krishna Kumar et al., “Unraveling the mystery of Indian monsoon failure during El Niño,” Science, 314 (2006), pp. 115–119. 27. ​Basketball scores from the Los Angeles Times. 28. ​Monthly stock returns from the Web site of Professor Kenneth French of Dartmouth, mba.tuck.dartmouth.edu/pages/faculty/ken. french. A fine point: the data are the “excess returns” on stocks, the actual returns less the small monthly returns on Treasury bills. 29. ​C. B. Williams, Style and Vocabulary: Numerological Studies, Griffin, 1970. 30. ​From the American Community Survey, at factfinder2.census. gov. 31. ​ International Energy Agency, Key World Energy Statistics 2007, at www.iea.org. 32. ​ Maribeth Cassidy Schmitt, “The effects of an elaborated directed reading activity on the metacomprehension skills of third graders,” PhD thesis, Purdue University, 1987. 33. ​These data were collected by students as a class project. 34. ​The chest size data came from the Data and Story Library, lib.stat.cmu.edu/DASL/. 35. ​Advanced Placement exam results from AP Central Web site, apcentral.collegeboard.com. 36. ​Arbitron, Internet and Media 2006, www.arbitron.com. 37. ​H. Lindberg, H. Roos, and P. Gardsell, “Prevalence of coxarthritis in former soccer players,” Acta Orthopedica Scandinavica, 64 (1993), pp. 165–167. 38. ​From the American Community Survey, at the Census Bureau Web site, www.census.gov. The data are a subsample of the 13,194 individuals in the ACS North Carolina sample who had travel times greater than zero.

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Notes and Data Sources

39. ​Pets data set from Guidelines for Assessment and Instruction in Statistics Education: Pre-K–12 Report, American Statistical Asso­ ciation, 2007. 40. ​This isn’t a mathematical theorem. The mean can be less than the median in right-skewed distributions that take only a few values, many of which lie exactly at the median. The rule almost never fails for distributions taking many values, and most counterexamples don’t appear clearly skewed in graphs even though they may be slightly skewed according to technical measures of skewness. See Paul T. von Hippel, “Mean, median, and skew: correcting a textbook rule,” Journal of Statistics Education, 13, No. 2 (2005), online journal, www.amstat.org/publications/jse/v13n2/­vonhippel.html. 41. ​From the National Association of Realtors, www.realtor.org. 42. ​Tom Lloyd et al., “Fruit consumption, fitness, and cardiovascular health in female adolescents: the Penn State Young Women’s Health Study,” American Journal of Clinical Nutrition, 67 (1998), pp. 624–630. 43. ​The daily returns for the year ending November 8, 2007, for the CREF Equity Index Fund and the TIAA Real Estate Fund. Daily price data for these funds are at www.tiaa-cref.org. Returns can be easily calculated because dividends are incorporated in the daily prices rather than given separately. 44. ​Data from the most recent Annual Demographic Supplement can be found at www.census.gov/cps. 45. ​Ethan J. Temeles and W. John Kress, “Adaptation in a planthummingbird association,” Science, 300 (2003), pp. 630–633. We thank Ethan J. Temeles for providing the data. 46. ​2007 CIRP Freshman Survey; data from the Higher Edu­cation Research Institute’s report “The American freshman: national norms for fall 2007,” published January 2008. 47. ​From Internet Movie Database, www.imdb.com. 48. ​Movie ratings data found at informationplease.com. 49. ​The report “Cell phones key to teens’ social lives, 47% can text with eyes closed” is published online by Harris Interactive, July 2008, www.marketingcharts.com. 50. ​A. Karpinski and A. Duberstein, “A description of Facebook use and academic performance among undergraduate and graduate students,” paper presented at the American Educational Research Association annual meeting, April 2009. Thanks to Aryn Karpinski for providing us with some original data from the study. 51. ​S. M. Stigler, “Do robust estimators work with real data?” Annals of Statistics, 5 (1977), pp. 1055–1078. 52. ​T. Bjerkedal, “Acquisition of resistance in guinea pigs infected with different doses of virulent tubercle bacilli,” American Journal of Hygiene, 72 (1960), pp. 130–148. 53. ​Data from the Bureau of Labor Statistics, Annual Demographic Supplement www.bls.census.gov. 54. ​Data from the report “Is our tuna family-safe?” prepared by Defenders of Wildlife, 2006.

Chapter 2

  1. ​Will Shortz, “A few words about Sudoku, which has none,” New York Times, August 28, 2005.   2. ​Data found online at exploringdata.net.  3. ​S. M. Stigler, “Do robust estimators work with real data?” Annals of Statistics, 5 (1977), pp. 1055–1078.  4. ​Sheldon Ross, Introduction to Probability and Statistics for Engineers and Scientists, 3rd ed., Academic Press, 2004.   5. ​Stephen Jay Gould, “Entropic homogeneity isn’t why no one hits .400 anymore,” Discover, August 1986, pp. 60–66. Gould does not standardize but gives a speculative discussion instead.

Starnes-Yates5e_nds_001_011hr.indd 2

  6. ​Information on bone density in the reference populations was found at www.courses.washington.edu/bonephys/opbmd.html.  7. ​Data from USA Today sports salary data base at www.usatoday. com. 8. ​Data from Gary Community School Corporation, courtesy of Celeste Foster, Department of Education, Purdue University. 9. ​Detailed data appear in P. S. Levy et al., Total Serum Cholesterol Values for Youths 12–17 Years, Vital and Health Sta­tistics, Series 11, No. 155, National Center for Health Statistics, 1976. 10. ​Data found at Bureau of Labor Statistics, www.bls.gov. 11. ​Data provided by Chris Olsen, who found the information in Scuba News and Skin Diver magazines. 12. ​See Note 3. 13. ​The data set was constructed based on information provided in P. D. Wood et al., “Plasma lipoprotein distributions in male and female runners,” in P. Milvey (ed.), The Marathon: Physiological, Medical, Epidemiological, and Psychological Studies, New York Academy of Sciences, 1977. 14. ​Data found online at www.earthtrends.wri.org. 15. ​R. Shine, T. R. L. Madsen, M. J. Elphick, and P. S. Harlow, “The influence of nest temperature and maternal brooding on hatchling phenotypes in water python,” Ecology, 78 (1997), pp. 1713–1721. 16. ​We found the information on birth weights of Norwegian children on the National Institute of Environmental Health Sciences Web site. The relevant article can be accessed here: http://www. ncbi.nlm.nih.gov/pubmed/1536353. 17. ​From a graph in L. Partridge and M. Farquhar, “Sexual activity reduces lifespan of male fruit flies,” Nature, 294 (1981), pp. 580–582. Provided by Brigitte Baldi.

Chapter 3

  1. ​SEC football data from espn.go.com.   2. ​The Florida Boating Accident Statistical Report for each year (at myfwc.com/boating/safety-education/boating-accidents) gives the number of registered vessels. The Florida Wildlife Commission maintains a manatee death data base at research.myfwc.com/ manatees/.  3. ​ Michael Winerip, “On education: SAT essay test rewards length and ignores errors,” New York Times, May 4, 2005.  4. ​Data from The College Board Web site, www.collegeboard. com.   5. ​Data provided by Darlene Gordon, Purdue University.  6. ​ From a graph in Bernt-Erik Saether, Steiner Engen, and Erik Mattysen, “Demographic characteristics and population dynamical patterns of solitary birds,” Science, 295 (2002), pp. 2070–2073.  7. ​ Based on T. N. Lam, “Estimating fuel consumption from engine size,” Journal of Transportation Engineering, 111 (1985), pp.  339–357. The data for 10 to 50 km/h are measured; those for 60 and higher are calculated from a model given in the paper and are therefore smoothed.   8. ​From a graph in Christer G. Wiklund, “Food as a mechanism of density-dependent regulation of breeding numbers in the merlin Falco columbarius,” Ecology, 82 (2001), pp. 860–867.   9. ​From a graph in Naomi I. Eisenberger, Matthew D. Lieberman, and Kipling D. Williams, “Does rejection hurt? An fMRI study of social exclusion,” Science, 302 (2003), pp. 290–292. 10. ​M. A. Houck et al., “Allometric scaling in the earliest fossil bird, Archaeopteryx lithographica,” Science, 247 (1990), pp. 195–198.

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Notes and Data Sources The authors conclude from a variety of evidence that all specimens represent the same species. 11. ​Consumer Reports, June 1986, pp. 366–367. 12. ​G. A. Sacher and E. F. Staffelt, “Relation of gestation time to brain weight for placental mammals: implications for the theory of vertebrate growth,” American Naturalist, 108 (1974), pp. 593–613. We found the data in Fred L. Ramsey and Daniel W. Schafer, The Statistical Sleuth: A Course in Methods of Data Analysis, Duxbury, 1997, p. 228. 13. ​Table 1 of E. Thomassot et al., “Methane-related diamond crystallization in the earth’s mantle: stable isotopes evidence from a single diamond-bearing xenolith,” Earth and Planetary Science Letters, 257 (2007), pp. 362–371. 14. ​The information for exercise 34 came from libertystreeteconomics.newyorkfed.org/2012/03/grading-student-loans.html 15.  Data on used car prices from autotrader.com, September 8, 2012. We searched for F-150 4 x 4’s on sale within 50 miles of College Station, Texas. 16.  N. R. Draper and J. A. John, “Influential observations and outliers in regression,” Technometrics, 23(1981), pp. 21–26. 17.  Frank J. Anscombe, “Graphs in statistical analysis,” American Statistician, 27 (1973), pp. 17–21. 18. ​P. Goldblatt (ed.), Longitudinal Study: Mortality and Social Organization, Her Majesty’s Stationery Office, 1990. At least, so claims Richard Conniff, in The Natural History of the Rich, Norton, 2002, p. 45. We have not been able to access the Goldblatt report. 19. ​W. M. Lewis and M. C. Grant, “Acid precipitation in the western United States,” Science, 207 (1980), pp. 176–177. 20. ​Data from National Institute of Standards and Technology, Engineering Statistics Handbook, www.itl.nist.gov/div898/­ handbook. The analysis there does not comment on the bias of field measurements. 21. ​We obtained the data for exercise 55 from i.nbcolympics.com/ figure-skating/resultsandschedules/event=FSW010000/index.html. 22. ​Gary Smith, “Do statistics test scores regress toward the mean?” Chance, 10, No. 4 (1997), pp. 42–45. 23. ​We got the data for exercise 66 from www.baseball-reference. com. 24. ​From a graph in G. D. Martinsen, E. M. Driebe, and T. G. Whitham, “Indirect interactions mediated by changing plant chemistry: beaver browsing benefits beetles,” Ecology, 79 (1998), pp. 192–200. 25. ​ We got the data for exercise 68 from http://nutrition. mcdonalds.com/getnutrition/nutritionfacts.pdf. 26. The data for exercise 69 came from www.pro-football-reference. com/teams/jax/2011.htm. 27. ​Debora L. Arsenau, “Comparison of diet management instruction for patients with non–insulin dependent diabetes mellitus: learning activity package vs. group instruction,” MS thesis, Purdue University, 1993. 28. ​A. K. Yousafzai et al., “Comparison of armspan, arm length and tibia length as predictors of actual height of disabled and nondisabled children in Dharavi, Mumbai, India,” European Journal of Clinical Nutrition, 57 (2003), pp. 1230–1234. In fact, r2 = 0.93. 29. ​David M. Fergusson and L. John Horwood, “Cannabis use and traffic accidents in a birth cohort of young adults,” Accident Analysis and Prevention, 33 (2001), pp. 703–711. 30. ​The World Almanac and Book of Facts (2009).

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31. ​G. L. Kooyman et al., “Diving behavior and energetics during foraging cycles in king penguins,” Ecological Monographs, 62 (1992), pp. 143–163. 32. ​ We found the data on cherry blossoms in the paper “Linear equations and data analysis,” which was posted on the North Carolina School of Science and Mathematics Web site, www.ncssm.edu. 33. ​From a graph in Craig Packer et al., “Ecological change, group territoriality, and population dynamics in Serengeti lions,” Science, 307 (2005), pp. 390–393.

Chapter 4

  1. ​Carlos Vallbona et al., “Response of pain to static magnetic fields in postpolio patients, a double blind pilot study,” Archives of Physical Medicine and Rehabilitation, 78 (1997), pp.1200–1203.   2. ​“Traffic Safety Facts 2012,” published by the National Highway Traffic Safety Administration on their Web site, www.nhtsa.gov.  3. ​ Results from the Cox Communications Teen Online and Wireless Safety Survey, in Partnership with the National Center for Missing and Exploited Children® (NCMEC) and John Walsh, published May 2009, www.cox.com/takecharge/­​safe_teens_2009.   4. ​Sheldon Cohen, William J. Doyle, Cuneyt M. Alper, Denise Janicki-Deverts, and Ronald B. Turner, “Sleep habits and susceptibility to the common cold,” Archives of Internal Medicine, 169, No. 1 (2009), pp. 62–67.  5. ​ Rebecca Smith and Alison Wessner, The Lawrenceville School, “Does listening to music while studying affect performance on learning assessments?” May 2009.   6. ​Based on a suggested Science Fair project entitled “Do the Eyes Have It?” at the Science Buddies Web site, www.sciencebuddies. org.   7. ​Frederick Mosteller and David L. Wallace. Inference and Disputed Authorship: The Federalist. Addison-Wesley, Reading, MA, 1964.  8. ​ According to the Web site http://en.wikipedia.org/wiki/ Federalist_papers.  9. ​Excerpt obtained from http://www.constitution.org/fed/ federa51.htm. 10. ​ For information on the American Community Survey of households (there is a separate sample of group quarters), go to www.census.gov/acs. 11. ​The Pew press release and the full report “Polls face growing resistance, but still representative” (April 20, 2004) are at peoplepress.org/reports. 12. ​Cynthia Crossen, “Margin of error: studies galore support products and positions, but are they reliable?” Wall Street Journal, November 14, 1991. 13. ​Robert C. Parker and Patrick A. Glass, “Preliminary results of double-sample forest inventory of pine and mixed stands with highand low-density LiDAR,” in Kristina F. Connoe (ed.), Proceedings of the 12th Biennial Southern Silvicultural Research Conference, U.S. Department of Agriculture, Forest Service, Southern Research Station, 2004. The researchers actually sampled every tenth plot. This is a systematic sample. 14. ​Gary S. Foster and Craig M. Eckert, “Up from the grave: a socio-historical reconstruction of an African American community from cemetery data in the rural Midwest,” Journal of Black Studies, 33 (2003), pp. 468–489. 15. ​Bryan E. Porter and Thomas D. Berry, “A nationwide survey of self-reported red light running: measuring prevalence, predictors, and perceived consequences,” Accident Analysis and Prevention, 33 (2001), pp. 735–741.

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Notes and Data Sources

16. ​Giuliana Coccia, “An overview of non-response in Italian telephone surveys,” Proceedings of the 99th Session of the Inter­national Statistics Institute, 1993, Book 3, pp. 271–272. 17. ​Mario A. Parada et al., “The validity of self-reported seatbelt use: Hispanic and non-Hispanic drivers in El Paso,” Accident Analysis and Prevention, 33 (2001), pp. 139–143. 18. ​These figures are cited by Dr. William Dement at “The Sleep Well” Web site, www.stanford.edu/~dement/. 19. ​See, for example, Martin Enserink, “The vanishing promises of hormone replacement,” Science, 297 (2002), pp. 325–326; and Brian Vastag, “Hormone replacement therapy falls out of favor with expert committee,” Journal of the American Medical Association, 287 (2002), pp. 1923–1924. A National Institutes of Health panel’s comprehensive report is International Position Paper on Women’s Health and Menopause, NIH Publication 02-3284. 20. ​“Family dinner linked to better grades for teens: survey finds regular meal time yields additional benefits,” written by John Mackenzie for ABC News’s World News Tonight, September 13, 2005. 21. ​ Information found online at http//ssw.unc.edu/about/news/ careerstart_1-13-09. 22. ​Simplified from Arno J. Rethans, John L. Swasy, and Lawrence J. Marks, “Effects of television commercial repetition, receiver knowledge, and commercial length: a test of the two-­factor model,” Journal of Marketing Research, 23 (February 1986), pp. 50–61. 23. ​Steering Committee of the Physicians’ Health Study Research Group, “Final report on the aspirin component of the ongoing Physicians’ Health Study,” New England Journal of Medicine, 321 (1989), pp. 129–135. 24. ​Martin J. Bergee and Lecia Cecconi-Roberts, “Effects of smallgroup peer interaction on self-evaluation of music performance,” Journal of Research in Music Education, 50 (2002), pp. 256–268. 25. ​ The placebo effect examples are from Sandra Blakeslee, “Placebos prove so powerful even experts are surprised,” New York Times, October 13, 1998. 26. ​The “three-quarters” estimate is cited by Martin Enserink, “Can the placebo be the cure?” Science, 284 (1999), pp. 238–240. An extended treatment is Anne Harrington (ed.), The Placebo Effect: An Interdisciplinary Exploration, Harvard University Press, 1997. 27. ​ The flu trial quotation is from Kristin L. Nichol et al., “Effectiveness of live, attenuated intranasal influenza virus vaccine in healthy, working adults,” Journal of the American Medical Association, 282 (1999), pp. 137–144. 28. ​From the Electronic Encyclopedia of Statistical Examples and Exercises (EESEE), “Anecdotes of Placebos.” 29. ​David L. Strayer, Frank A. Drews, and William A. Johnston, “Cell phone–induced failures of visual attention during simulated driving,” Journal of Experimental Psychology: Applied, 9 (2003), pp. 23–32. 30. ​We obtained the tire pressure loss data from the Consumer Reports Web site: http://news.consumerreports.org/cars/2007/10/ nitrogen-tires-.html. 31. ​J. E. Muscat et al., “Handheld cellular telephone use and risk of brain cancer,” Journal of the American Medical Association, 284 (2000), pp. 3001–3007. 32. ​Early Human Development, 76, No. 2 (February 2004), pp. 139–145. 33. ​National Institute of Child Health and Human Development, Study of Early Child Care and Youth Development. The article

Starnes-Yates5e_nds_001_011hr.indd 4

appears in the July 2003 issue of Child Development. The quotation is from the summary on the NICHD Web site, www.nichd.nih.gov. 34. ​Naomi D. L. Fisher, Meghan Hughes, Marie Gerhard-Herman, and Norman K. Hollenberg, “Flavonol-rich cocoa induces nitricoxide-dependent vasodilation in healthy humans,” Journal of Hypertension, 21, No. 12 (2003), pp. 2281–2286. 35. ​Joel Brockner et al., “Layoffs, equity theory, and work performance: further evidence of the impact of survivor guilt,” Academy of Management Journal, 29 (1986), pp. 373–384. 36. ​Details of the Carolina Abecedarian Project, including references to published work, can be found online at abc.fpg.unc.edu. 37. ​ K. B. Suttle, Meredith A. Thomsen, and Mary E. Power, “Species interactions reverse grassland responses to changing ­climate,” Science, 315 (2007), pp. 640–642. 38. ​Christopher Anderson, “Measuring what works in health care,” Science, 263 (1994), pp. 1080–1082. 39. ​Esther Duflo, Rema Hanna, and Stephan Ryan, “Monitoring works: getting teachers to come to school,” November 21, 2007, at economics.mit.edu/files/2066. 40. ​See Note 28. 41. ​Marielle H. Emmelot-Vonk et al., “Effect of testosterone supplementation on functional mobility, cognition, and other parameters in older men,” Journal of the American Medical Asso­ciation, 299 (2008), pp. 39–52. 42. ​Linda Stern et al., “The effects of low-carbohydrate versus conventional weight loss diets in severely obese adults: one-year follow up of a randomized trial,” Annals of Internal Medicine, 140, No. 10 (May 2004), pp. 778–785. 43. ​W. E. Paulus et al., “Influence of acupuncture on the pregnancy rate in patients who undergo assisted reproductive therapy,” Fertility and Sterility, 77, No. 4 (2002), pp. 721–724. 44. ​Mary O. Mundinger et al., “Primary care outcomes in patients treated by nurse practitioners or physicians,” Journal of the American Medical Association, 238 (2000), pp. 59–68. 45. ​Sterling C. Hilton et al., “A randomized controlled experiment to assess technological innovations in the classroom on student outcomes: an overview of a clinical trial in education,” manuscript, no date. A brief report is Sterling C. Hilton and Howard B. Christensen, “Evaluating the impact of multimedia lectures on student learning and attitudes,” Proceedings of the 6th International Conference on the Teaching of Statistics, at www.stat.auckland.ac.nz. 46. ​Study conducted by cardiologists at Athens Medical School, Greece, and announced at a European cardiology conference in February 2004. 47. ​Evan H. DeLucia et al., “Net primary production of a forest ecosystem with experimental CO2 enhancement,” Science, 284 (1999), pp. 1177–1179. The investigators used the block design. 48. ​Niels Juel-Nielsen, Individual and Environment: Monozygotic Twins Reared Apart, International Universities Press, 1980. 49. ​The sleep deprivation study is described in R. Stickgold, L. James, and J. Hobson, “Visual discrimination learning requires post-training sleep,” Nature Neuroscience, 2000, pp. 1237–1238. We obtained the data from Allan Rossman, who got it courtesy of the authors. 50. ​The idea for this chart came from Fred L. Ramsey and Daniel W. Schafer, The Statistical Sleuth: A Course in Methods of Data Analysis, 2nd edition, Duxbury Press, 2002. 51. ​Thanks to Dan Teague for providing this example. 52. ​Thanks to Josh Tabor for contributing this Activity. 53. ​The Health Consequences of Smoking: 1983, U.S. Health Service, 1983.

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Notes and Data Sources 54. ​See the details on the Web site of the Office for Human Research Protections of the Department of Health and Human Services, hhs.gov/ohrp. 55. ​Charles A. Nelson III et al., “Cognitive recovery in socially deprived young children: the Bucharest Early Intervention Project,” Science, 318 (2007), pp. 1937–1940. 56. ​ The article describing this study is Nikhil Swaminathan, “Gender jabber: do women talk more than men?” Scientific American, July 6, 2007. 57. ​Marilyn Ellis, “Attending church found factor in longer life,” USA Today, August 9, 1999. 58. ​ Scott DeCarlo with Michael Schubach and Vladimir Naumovski, “A decade of new issues,” Forbes, March 5, 2001, www. forbes.com. 59. ​ “Antibiotics no better than placebo for sinus infections” http://in.news.yahoo.com/antibiotics-no-better-placebo-sinusinfections-074240018.html 60. ​Warren McIsaac and Vivek Goel, “Is access to physician services in Ontario equitable?” Institute for Clinical Evaluative Sciences in Ontario, October 18, 1993. 61. ​L. E. Moses and F. Mosteller, “Safety of anesthetics,” in J. M. Tanur et al. (eds.), Statistics: A Guide to the Unknown, 3rd ed., Wadsworth, 1989, pp. 15–24. 62. ​Arizona Daily Star 10-29-2008. 63. ​R. C. Shelton et al., “Effectiveness of St. John’s wort in major depression,” Journal of the American Medical Association, 285 (2001), pp. 1978–1986. 64. ​Based on a news item “Bee off with you,” Economist, November 2, 2002, p. 78. 65. ​Jay Schreiber and Megan Thee, “Fans concerned about steroid use and believe it’s widespread, poll shows,” New York Times, March 31, 2008. 66. ​From the Electronic Encyclopedia of Statistical Examples and Exercises (EESEE) case study “Is Caffeine Dependence Real?”

Addendum for Activity “Sampling sunflowers” (page 219) Sampling sunflowers A

B

C

D

E

F

G

H

I

J

1 106 104 106 105 105 105 108 108 107 106 2 103 102 103 104 103 103 103 104 103 105 3 101

99 102 101 100 100 102 101 102 101 99

4

98

99

98 101

5

98 100 100 100

99

99

99

98 100

98

6

97

99

99

99

99

98

98

99 98

99 100 98

99 101 97

7 103 102 102 104 101 102 102 102 104 102 8 104 103 103 103 102 102 103 104 102 102 9 106 106 104 106 102 107 104 103 106 106 10 107 108 109 110 106 107 109 107 106 107

Chapter 5

  1. ​Note that pennies have rims that make spinning more stable. The probability of a head in spinning a coin depends on the type of coin and also on the surface.

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  2. ​R. Vallone and A. Tversky, “The hot hand in basketball: on the misperception of random sequences,” Cognitive Psychology, 17 (1985), pp. 295–314.   3.  Gur Yaari and Shmuel Eisenmann, The Hot (Invisible?) Hand: Can Time Sequence Patterns of Success/Failure in Sports Be Modeled as Repeated Random Independent Trials? (2011) PLoS ONE 6(10): e24532. doi:10.1371/journal.pone.0024532   4. ​The excerpts in Exercise 22 are from http://marilynvossavant. com/game-show-problem/.   5. ​From the Web site of the Gallup Organization, www.gallup.com. Individual poll reports remain on this site for only a limited time.   6. ​Information for the 1999–2000 academic year is from the 2003 Statistical Abstract of the United States, Table 286.   7. ​Data for 2006 from the Web site of Statistics Canada, www. statcan.ca.   8. ​Gail Burrill, “Two-way tables: introducing probability using real data,” paper presented at the Mathematics Education into the Twentyfirst Century Project, Czech Republic, September 2003. Burrill cites as her source H. Kranendonk, P. Hopfensperger, and R. Scheaffer, Exploring Probability, Dale Seymour Publications, 1999.   9. ​From the EESEE story “Is It Tough to Crawl in March?” 10. ​Pierre J. Meunier et al., “The effects of strontium ranelate on the risk of vertebral fracture in women with postmenopausal osteoporosis,” New England Journal of Medicine, 350 (2004), pp. 459–468. 11. ​The table closely follows the grade distributions for these three schools at the University of New Hampshire in the fall of 2000, found in a self-study document at www.unh.edu/academic-affairs/ neasc/. The counts of grades mirror the proportions of UNH students in these schools. The table is simplified to refer to a university with only these three schools. 12. ​Information about Internet users comes from sample surveys carried out by the Pew Internet and American Life Project, at www. pewinternet.org. 13. ​ Data on Roger Federer’s serve percentages from www. atpworldtour.com. 14. ​Thanks to Michael Legacy for providing these data. 15. ​This is one of several tests discussed in Bernard M. Branson, “Rapid HIV testing: 2005 update,” a presentation by the Centers for Disease Control and Prevention, at www.cdc.gov. The Malawi clinic result is reported by Bernard M. Branson, “Point-of-care rapid tests for HIV antibody,” Journal of Laboratory Medicine, 27 (2003), pp. 288–295. 16. ​The National Longitudinal Study of Adolescent Health interviewed a stratified random sample of 27,000 adolescents, then reinterviewed many of the subjects six years later, when most were aged 19 to 25. These data are from the Wave III reinterviews in 2000 and 2001, found at the Web site of the Carolina Population Center, www.cpc.unc.edu. 17. ​Information about Internet users comes from sample surveys carried out by the Pew Internet and American Life Project, found online at www.pewinternet.org. The music-downloading data were collected in 2003. 18. ​We got these data from the Energy Information Administration on their Web site at http://www.eia.gov/dnav/pet/pet_sum_mkt_ dcu_sct_m.htm. 19. ​From the National Institutes of Health’s National Digestive Dis­eases Information Clearinghouse, found at http://digestive.niddk. nih.gov/. 20. ​The probabilities given are realistic, according to the fundraising firm SCM Associates, at scmassoc.com.

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Notes and Data Sources

21. ​Probabilities from trials with 2897 people known to be free of HIV antibodies and 673 people known to be infected are reported in J. Richard George, “Alternative specimen sources: methods for confirming positives,” 1998 Conference on the Laboratory Science of HIV, found online at the Centers for Disease Control and Prevention, www.cdc.gov. 22. ​Robert P. Dellavalle et al., “Going, going, gone: lost Internet references,” Science, 302 (2003), pp. 787–788. 23. ​Margaret A. McDowell et al., “Anthropometric reference data for children and adults: U.S. population, 1999–2002,” National Center for Health Statistics, Advance Data from Vital and Health Statistics, No. 361 (2005), at www.cdc.gov/nchs. 24. ​The General Social Survey exercises in this chapter present tables constructed using the search function at the GSS archive, sda.berkeley.edu/archive.htm. 25. ​National population estimates for July 1, 2006, at the Census Bureau Web site www.census.gov. The table omits people who consider themselves to belong to more than one race. 26. ​Data provided by Patricia Heithaus and the Department of Biology at Kenyon College. 27. ​Thanks to Tim Brown, The Lawrenceville School, for providing the idea for this exercise.

Chapter 6

  1. ​U.S. Supreme Court, Strauder v. West Virginia, 100 U.S. 303 (1879) https://supreme.justia.com/cases/federal/us/100/303/case. html.   2. ​U.S. Supreme Court, Berghuis v. Smith, Docket No. 08-1402 http://www.law.cornell.edu/supct/html/08-1402.ZO.html.   3. ​In most applications, X takes a finite number of possible values. The same ideas, implemented with more advanced mathematics, apply to random variables with an infinite but still countable collection of values.   4. ​The Apgar score data came from National Center for Health Statistics, Monthly Vital Statistics Reports, Vol. 30, No. 1, Supplement, May 6, 1981.  5. ​Information from www.ncsu.edu.  6. ​The mean of a continuous random variable X with density function f(x) can be found by integration: mX = ∫ xf(x) dx This integral is a kind of weighted average, analogous to the ­discrete-case mean mX =

∙ xipi

The variance of a continuous random variable X is the average squared deviation of the values of X from their mean, found by the integral s2X

= ∫ (x − m) f(x) dx 2

  7. ​You can find a mathematical explanation of Benford’s law in Ted Hill, “The first-digit phenomenon,” American Scientist, 86 (1996), pp. 358–363; and Ted Hill, “The difficulty of faking data,” Chance, 12, No. 3 (1999), pp. 27–31. Applications in fraud detection are discussed in the second paper by Hill and in Mark A. Nigrini, “I’ve got your number,” Journal of Accountancy, May 1999,

Starnes-Yates5e_nds_001_011hr.indd 6

available online at http://www.journalofaccountancy.com/issues/ 1999/may/nigrini.   8. ​The National Longitudinal Study of Adolescent Health interviewed a stratified random sample of 27,000 adolescents, then reinterviewed many of the subjects six years later, when most were aged 19 to 25. These data are from the Wave III reinterviews in 2000 and 2001, found at the Web site of the Carolina Population Center, www.cpc.unc.edu.  9. ​ Data from the Census Bureau’s 1998 American Housing Survey. 10. ​ Thomas K. Cureton et al., Endurance of Young Men, Monographs of the Society for Research in Child Development, Vol. 10, No. 1, 1945. 11. ​The survey question is reported in Trish Hall, “Shop? Many say ‘Only if I must,’” New York Times, November 28, 1990. In fact, 66% (1650 of 2500) in the sample said “Agree.” 12. ​ Office of Technology Assessment, Scientific Validity of Polygraph Testing: A Research Review and Evaluation, Govern­ment Printing Office, 1983. 13. ​ John Schwartz, “Leisure pursuits of today’s young men,” New York Times, March 29, 2004. The source cited is comScore Media Metrix. 14. ​ Population base from the 2006 Labor Force Survey, at www.statistics.gov.uk. Smoking data from Action on Smoking and Health, Smoking and Health Inequality, at www.ash.org.uk.

Chapter 7

 1. ​Data on income and education from the March 2012 CPS supplement, obtained from the Census Bureau Web site.   2. ​This and similar results of Gallup polls are from the Gallup Organization Web site, www.gallup.com.   3. ​From a graph in Stan Boutin et al., “Anticipatory reproduction and population growth in seed predators,” Science, 314 (2006), pp. 1928–1930.   4. ​See Note 2.  5. ​Amanda Lenhart and Mary Madden, “Music downloading, file-sharing and copyright,” Pew Internet and American Life Project, 2003, at www.pewinternet.org.   6. ​This Activity is suggested in Richard L. Schaeffer, Ann Watkins, Mrudulla Gnanadesikan, and Jeffrey A. Witmer, Activity-Based Statistics, Springer, 1996.   7. ​See Note 1.   8. ​We found the information on birth weights of Norwegian children on the National Institute of Environmental Health Sciences Web site: The relevant article can be accessed here: http://www. ncbi.nlm.nih.gov/pubmed/1536353.

Chapter 8

  1. ​Thanks to Floyd Bullard for sharing the idea for this Activity.   2. ​This and similar results from the Pew Internet and American Life Project can be found at www.pewinternet.org.   3. ​Michele L. Head, “Examining college students’ ethical values,” Consumer Science and Retailing honors project, Purdue University, 2003.   4. ​Francisco L. Rivera-Batiz, “Quantitative literacy and the likelihood of employment among young adults,” Journal of Human Resources, 27 (1992), pp. 313–328.

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Notes and Data Sources   5. ​This and similar results of Gallup polls are from the Gallup Organization Web site, www.gallup.com.  6. ​ Amanda Lenhart and Mary Madden, “Teens, privacy and online social networks,” Pew Internet and American Life Project, 2007, at www.pewinternet.org.   7. ​E. W. Campion, “Editorial: power lines, cancer, and fear,” New England Journal of Medicine, 337, No. 1 (1997), pp. 44–46. The study report is M. S. Linet et al., “Residential exposure to magnetic fields and acute lymphoblastic leukemia in children,” pp. 1–8 in the same issue. See also G. Taubes, “Magnetic field–cancer link: will it rest in peace?” Science, 277 (1997), p. 29.   8. ​Karl Pearson and A. Lee, “On the laws of inheritance in man,” Biometrika, 2 (1902), p. 357. These data also appear in D. J. Hand et al., A Handbook of Small Data Sets, Chapman & Hall, 1994. This book offers more than 500 data sets that can be used in statistical exercises.   9. ​Data from the College Alcohol Study Web site, www.hsph.harvard.edu/cas/. 10. ​Linda Lyons, “Teens: sex can wait,” December 14, 2004, from the Gallup Poll Web site, www.gallup.com. 11. ​Eric Sanford et al., “Local selection and latitudinal variation in a marine predator–prey interaction,” Science, 300 (2003), pp. 1135–1137. 12. ​See Note 3. 13. ​Substance Abuse and Mental Health Services Administration, Results from the 2011 National Survey on Drug Use and Health: Summary of National Findings, NSDUH Series H-44, HHS Publication No. (SMA) 12-4713. Rockville, MD: Substance Abuse and Mental Health Services Administration, 2012. 14. ​Wayne J. Camera and Donald Powers, “Coaching and the SAT I,” TIP, July 1999 (online journal at www.siop.org/tip). 15. ​“Poll: men, women at odds on sexual equality,” Associated Press dispatch appearing in the Lafayette (Ind.) Journal and Courier, October 20, 1997. 16. ​ Linda Lyons, “Most teens have in-room entertainment,” February 22, 2005, www.gallup.com/poll/14989/Most-TeensInRoom-Entertainment.aspx. 17. ​Based on information in “NCAA 2003 national study of collegiate sports wagering and associated health risks,” which can be found on the NCAA Web site, www.ncaa.org. 18. ​T. J. Lorenzen, “Determining statistical characteristics of a vehicle emissions audit procedure,” Technometrics, 22 (1980), pp. 483–493. 19. ​Data provided by Drina Iglesia, Purdue University. The data are part of a larger study reported in D. D. S. Iglesia, E. J. Cragoe, Jr., and J. W. Vanable, “Electric field strength and epithelization in the newt (Notophthalmus viridescens),” Journal of Experimental Zoology, 274 (1996), pp. 56–62. 20. ​Harry B. Meyers, “Investigations of the life history of the velvetleaf seed beetle, Althaeus folkertsi Kingsolver,” MS thesis, Purdue University, 1996. The 95% t interval is 1227.9 to 2507.6. A 95% bootstrap BCa interval is 1444 to 2718, confirming that t inference is inaccurate for these data. 21. ​Lisa M. Baril et al., “Willow-bird relationships on Yellowstone’s northern range,” Yellowstone Science, 17, No. 3 (2009), pp. 19–26. 22. ​Helen E. Staal and D. C. Donderi, “The effect of sound on visual apparent movement,” American Journal of Psychology, 96 (1983), pp. 95–105. 23. ​M. Ann Laskey et al., “Bone changes after 3 mo of lactation: influence of calcium intake, breast-milk output, and vitamin

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D–receptor genotype,” American Journal of Clinical Nutrition, 67 (1998), pp. 685–692. 24. ​TUDA results for 2003 from the National Center for Education Statistics, at nces.ed.gov/nationsreportcard. 25. ​Chi-Fu Jeffrey Yang, Peter Gray, and Harrison G. Pope, Jr., “Male body image in Taiwan versus the West,”American Journal of Psychiatry, 162 (2005), pp. 263–269. 26. ​Alan S. Banks et al., “Juvenile hallux abducto valgus association with metatarsus adductus,” Journal of the American Podiatric Medical Association, 84 (1994), pp. 219–224. 27. ​The vehicle is a 1997 Pontiac Transport Van owned by George P. McCabe. 28. ​These data are from “Results report on the vitamin C pilot program,” prepared by SUSTAIN (Sharing United States Technology to Aid in the Improvement of Nutrition) for the U.S. Agency for International Development. The report was used by the Committee on International Nutrition of the National Academy of Sciences/ Institute of Medicine (NAS/IOM) to make recommendations on whether or not the vitamin C content of food commodities used in U.S. food aid programs should be increased. The program was directed by Peter Ranum and Françoise Chomé. 29. ​R. D. Stichler, G. G. Richey, and J. Mandel, “Measurement of treadware of commercial tires,” Rubber Age, 73, No. 2 (May 1953). 30. ​ Data from Pennsylvania State University Stat 500 Applied Statistics online course, https://onlinecourses.science.psu.edu/ stat500/. 31. ​ Based on interviews in 2000 and 2001 by the National Longitudinal Study of Adolescent Health. Found at the Web site of the Carolina Population Center, www.cpc.unc.edu. 32. ​Simplified from Sanjay K. Dhar, Claudia Gonzalez-Vallejo, and Dilip Soman, “Modeling the effects of advertised price claims: tensile versus precise pricing,” Marketing Science, 18 (1999), pp. 154–177. 33. ​From program 19, “Confidence Intervals,” in the Against All Odds video series. 34. ​James T. Fleming, “The measurement of children’s perception of difficulty in reading materials,” Research in the Teaching of English, 1 (1967), pp. 136–156. 35. ​Bryan E. Porter and Thomas D. Berry, “A nationwide survey of self-reported red light running: measuring prevalence, predictors, and perceived consequences,” Accident Analysis and Prevention, 33 (2001), pp. 735–741.

Chapter 9

  1. ​P. A. Mackowiak, S. S. Wasserman, and M. M. Levine, “A critical appraisal of 98.6 degrees F, the upper limit of the normal body temperature, and other legacies of Carl Reinhold August Wunderlich,” Journal of the American Medical Association, 268 (1992), pp. 1578–1580.   2. ​Thanks to Josh Tabor for suggesting the idea for this example.   3. ​R. A. Fisher, “The arrangement of field experiments,” Journal of the Ministry of Agriculture of Great Britain, 33 (1926), p. 504, quoted in Leonard J. Savage, “On rereading R. A. Fisher,” Annals of Statistics, 4 (1976), p. 471. Fisher’s work is described in a biography by his daughter: Joan Fisher Box, R. A. Fisher: The Life of a Scientist, Wiley, 1978.   4. ​For a discussion of statistical significance in the legal setting, see D. H. Kaye, “Is proof of statistical significance relevant?”

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N/DS-8

Notes and Data Sources

Washington Law Review, 61 (1986), pp. 1333–1365. Kaye argues: “Presenting the P-value without characterizing the evidence by a significance test is a step in the right direction. Interval estimation, in turn, is an improvement over P-values.”   5. ​Julie Ray, “Few teens clash with friends,” May 3, 2005, on the Gallup Organization Web site, www.gallup.com.   6. ​The idea for this exercise was provided by Michael Legacy and Susan McGann.   7. ​Projections from the 2011 Digest of Education Statistics, found online at nces.ed.gov.   8. ​From the report “Sex and tech: results from a study of teens and young adults,” published by the National Campaign to Prevent Teen and Unplanned Pregnancy, www.thenationalcampaign.org/ sextech.   9. ​National Institute for Occupational Safety and Health, Stress at Work, 2000, available online at http://www.cdc.gov/niosh/docs/99101/. Results of this survey were reported in Restaurant Business, September 15, 1999, pp. 45–49. 10. ​See Note 5. 11. ​ Matthew A. Carlton and William D. Stansfield, “Making babies by the flip of a coin?” The American Statistician, 59 (2005), pp. 180–182. 12. ​ Dorothy Espelage et al., “Factors associated with bullying behavior in middle school students,” Journal of Early Adolescence, 19, No. 3 (August 1999), pp. 341–362. 13.  Aaron Smith and Joanna Brenner, “Twitter Use 2012,” published by the Pew Internet and American Life Project, May 31, 2012. 14. ​This and similar results of Gallup polls are from the Gallup Organization Web site, www.gallup.com. 15. ​Linda Lyons, “Teens: sex can wait,” December 14, 2004, from the Gallup Organization Web site, www.gallup.com. 16. ​Michele L. Head, “Examining college students’ ethical values,” Consumer Science and Retailing honors project, Purdue University, 2003. 17. ​Based on Stephen A. Woodbury and Robert G. Spiegelman, “Bonuses to workers and employers to reduce unemployment: randomized trials in Illinois,” American Economic Review, 77 (1987), pp. 513–530. 18. ​E. C. Strain et al., “Caffeine dependence syndrome: evidence from case histories and experimental evaluation,” Journal of the American Medical Association, 272 (1994), pp. 1604–1607. 19. ​Warren E. Leary, “Cell phones: questions but no answers,” New York Times, October 26, 1999. 20. ​R. A. Berner and G. P. Landis, “Gas bubbles in fossil amber as possible indicators of the major gas composition of ancient air,” Science, 239 (1988), pp. 1406–1409. The 95% t confidence interval is 54.78 to 64.40. A bootstrap BCa interval is 55.03 to 62.63. So t is reasonably accurate despite the skew and the small sample. 21. ​This exercise is based on events that are real. The data and details have been altered to protect the privacy of the individuals involved. 22. ​From the NIST Web site, www.nist.gov/srd. 23. ​Data provided by Timothy Sturm. 24. ​A. R. Hirsch and L. H. Johnston, “Odors and learning,” Journal of Neurological and Orthopedic Medicine and Surgery, 17 (1996), pp. 119–126. 25. ​From the story “Friday the 13th,” at the Data and Story Library, lib.stat.cmu.edu/DASL.

Starnes-Yates5e_nds_001_011hr.indd 8

26. ​ From the Gallup Organization Web site, www.gallup.com. The poll was taken in December 2006. 27.  W. S. Gosset, “The probable error of a mean,” Biometrika, 6 (1908), 1–25.

Chapter 10

 1. ​ This Case Study is based on the story “Drive-Thru Competition” in the Electronic Encyclopedia of Statistical Examples and Exercises (EESEE). Updated data were obtained from the QSR magazine Web site: www.qsrmagazine.com/ drive-thru.   2. ​C. P. Cannon et al., “Intensive versus moderate lipid lowering with statins after acute coronary syndromes,” New England Journal of Medicine, 350 (2004), pp. 1495–1504.  3. ​This Activity is based on an experiment conducted on the March 9, 2005, episode of the TV show MythBusters. You may be able to find a video clip from this program at the Discovery Channel Web site, www.dsc.discovery.com.  4. ​The study is reported in William Celis III, “Study suggests Head Start helps beyond school,” New York Times, April 20, 1993. See www.highscope.org.   5. ​The idea for this exercise was inspired by an example in David M. Lane’s Hyperstat Online textbook at http://davidmlane.com/ hyperstat.   6. ​Modified from Richard A. Schieber et al., “Risk factors for injuries from in-line skating and the effectiveness of safety gear,” New England Journal of Medicine, 335 (1996). Internet summary at www.nejm.org.   7. ​Francisco Lloret et al., “Fire and resprouting in Mediterranean ecosystems: insights from an external biogeographical region, the Mexican shrubland,” American Journal of Botany, 88 (1999), pp. 1655–1661.   8. ​Saiyad S. Ahmed, “Effects of microwave drying on checking and mechanical strength of low-moisture baked products,” MS thesis, Purdue University, 1994.  9. Aaron Smith and Joanna Brenner, “Twitter Use 2012,” Pew Internet and American Life Project, May 31, 2012. 10. ​ From the Web site of the Black Youth Project, blackyouthproject.uchicago.edu. 11. ​The National Longitudinal Study of Adolescent Health interviewed a stratified random sample of 27,000 adolescents, then reinterviewed many of the subjects six years later, when most were aged 19 to 25. These data are from the Wave III reinterviews in 2000 and 2001, found at the Web site of the Carolina Population Center, www.cpc.unc.edu. 12. ​Janice Joseph, “Fear of crime among black elderly,” Journal of Black Studies, 27 (1997), pp. 698–717. 13. ​ National Athletic Trainers Association, press release dated, September 30, 1994. 14. ​ Based on Deborah Roedder John and Ramnath LakshmiRatan, “Age differences in children’s choice behavior: the impact of available alternatives,” Journal of Marketing Research, 29 (1992), pp. 216–226. 15. ​Louie E. Ross, “Mate selection preferences among African American college students,” Journal of Black Studies, 27 (1997), pp. 554–569. 16. ​Martin Enserink, “Fraud and ethics charges hit stroke drug trial,” Science, 274 (1996), pp. 2004–2005.

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Notes and Data Sources 17. ​Kwang Y. Cha, Daniel P. Wirth, and Rogerio A. Lobo, “Does prayer influence the success of in vitro fertilization–embryo transfer?” Journal of Reproductive Medicine, 46 (2001), pp. 781–787. 18. ​W. E. Paulus et al., “Influence of acupuncture on the pregnancy rate in patients who undergo assisted reproductive therapy,” Fertility and Sterility, 77, No. 4 (2002), pp. 721–724. 19. ​Sapna Aneja, “Biodeterioration of textile fibers in soil,” MS thesis, Purdue University, 1994. 20. ​We obtained the National Health and Nutrition Examination Survey data from the Centers for Disease Control and Prevention Web site at www.cdc.gov/nchs/nhanes.htm. 21. ​Detailed information about the conservative t procedures can be found in Paul Leaverton and John J. Birch, “Small sample power curves for the two sample location problem,” Technometrics, 11 (1969), pp. 299–307; Henry Scheffe, “Practical solutions of the Behrens-Fisher problem,” Journal of the American Statistical Association, 65 (1970), pp. 1501–1508; and D. J. Best and J. C. W. Rayner, “Welch’s approximate solution for the Behrens-Fisher problem,” Technometrics, 29 (1987), pp. 205–210. 22. ​Data for this example from Noel Cressie, Statistics for Spatial Data, Wiley, 1993. 23. ​Based on R. G. Hood, “Results of the prices received by farmers for grain quality assurance project,” U.S. Department of Agriculture Report SRB-95-07, 1995. 24. ​This study is reported in Roseann M. Lyle et al., “Blood pressure and metabolic effects of calcium supplementation in normotensive white and black men,” Journal of the American Medical Association, 257 (1987), pp. 1772–1776. The data were provided by Dr. Lyle. 25. ​See Note 19. 26. ​The idea for this example was provided by Robert Hayden. 27. ​ United Nations data on literacy were found at www. earthtrends.wri.org. 28. ​Shailija V. Nigdikar et al., “Consumption of red wine polyphenols reduces the susceptibility of low-density lipoproteins to oxidation in vivo,” American Journal of Clinical Nutrition, 68 (1998), pp. 258–265. 29. ​Ethan J. Temeles and W. John Kress, “Adaptation in a plant– hummingbird association,” Science, 300 (2003), pp. 630–633. We thank Ethan J. Temeles for providing the data. 30. ​ Data for 1982, provided by Marvin Schlatter, Division of Financial Aid, Purdue University. 31. ​ Gabriela S. Castellani, “The effect of cultural values on Hispanics’ expectations about service quality,” MS thesis, Purdue University, 2000. 32. ​From a graph in Fabrizio Grieco, Arie J. van Noordwijk, and Marcel E. Visser, “Evidence for the effect of learning on timing of reproduction in blue tits,” Science, 296 (2002), pp. 136–138. 33. ​This example is loosely based on D. L. Shankland, “Involvement of spinal cord and peripheral nerves in DDT poisoning syndrome in albino rats,” Toxicology and Applied Pharmacology, 6 (1964), pp. 197–213. 34. ​Based on W. N. Nelson and C. J. Widule, “Kinematic analysis and efficiency estimate of intercollegiate female rowers,” unpublished manuscript, 1983. 35. ​Adapted from Maribeth Cassidy Schmitt, “The effects of an elaborated directed reading activity on the metacomprehension skills of third graders,” PhD dissertation, Purdue University, 1987. 36. ​M. Ann Laskey et al., “Bone changes after 3 mo of lactation: influence of calcium intake, breast-milk output, and vitamin

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D–receptor genotype,” American Journal of Clinical Nutrition, 67 (1998), pp. 685–692. 37. ​See Note 24 in Chapter 1. 38. ​The data for this exercise came from Rossman, Cobb, Chance, and Holcomb’s National Science Foundation project shared at JMM 2008 in San Diego. Their original source was Robert Stickgold, LaTanya James, and J. Allan Hobson, “Visual discrimination learning requires sleep after training,” Nature Neuroscience, 3 (2000), pp. 1237–1238. 39. ​The idea for this exercise was provided by Robert Hayden. 40. ​Pew Internet Project, “Counting on the Internet,” December 29, 2002, www.pewinternet.org. 41. ​ Wayne J. Camera and Donald Powers, “Coaching and the SAT I,” TIP (online journal at www.siop.org/tip), July 1999. 42. ​See Note 40. 43. ​Thanks to Larry Green, Lake Tahoe Community College, for allowing us to use some of the contexts from his “Classifying Statistics Problems” Web site, www.ltcconline.net/greenl/java/ Statistics/catStatProb/categorizingStatProblems12.html. 44. ​JoAnn K. Wells, Allan F. Williams, and Charles M. Farmer, “Seat belt use among African Americans, Hispanics, and whites,” Accident Analysis and Prevention, 34 (2002), pp. 523–529. 45. ​Based on Amna Kirmani and Peter Wright, “Money talks: perceived advertising expense and expected product quality,” Journal of Consumer Research, 16 (1989), pp. 344–353. 46. ​Francisco L. Rivera-Batiz, “Quantitative literacy and the likelihood of employment among young adults,” Journal of Human Resources, 27 (1992), pp. 313–328. 47. ​Clive G. Jones et al., “Chain reactions linking acorns to gypsy moth outbreaks and Lyme disease risk,” Science, 279 (1998), pp. 1023–1026. 48. ​ Data provided by Warren Page, New York City Technical College, from a study done by John Hudesman. 49. ​Maureen Hack et al., “Outcomes in young adulthood for verylow-birth-weight infants,” New England Journal of Medicine, 346 (2002), pp. 149–157. Exercise AP3.32 is simplified, in that the measures reported in this paper have been statistically adjusted for “sociodemographic status.”

Chapter 11

 1. ​M. M. Roy and N. J. S. Christenfeld, “Do dogs resemble their owners?” Psychological Science, 15, No. 5 (May 2004), pp. 361–363.   2. ​R. W. Mannan and E. C. Meslow, “Bird populations and vegetation characteristics in managed and old-growth forests, northwestern Oregon,” Journal of Wildlife Management, 48 (1984), pp. 1219–1238.   3. ​You can find a mathematical explanation of Benford’s law in Ted Hill, “The first-digit phenomenon,” American Scientist, 86 (1996), pp. 358–363; and Ted Hill, “The difficulty of faking data,” Chance, 12, No. 3 (1999), pp. 27–31. Applications in fraud detection are discussed in the second paper by Hill and in Mark A. Nigrini, “I’ve got your number,” Journal of Accountancy, May 1999, available online at http://www.journalofaccountancy.com/issues/1999/may/nigrini.   4. ​The idea for this exercise came from a post to the AP Statistics electronic discussion group by Joshua Zucker.

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N/DS-10

Notes and Data Sources

  5. ​The context of this example was inspired by C. M. Ryan et al., “The effect of in-store music on consumer choice of wine,” Proceedings of the Nutrition Society, 57 (1998), p. 1069A.   6. ​Pennsylvania State University Division of Student Affairs, “Net behaviors, November 2006,” Penn State Pulse, at http://­studentaffairs. psu.edu/.   7. ​Pew Research Center for the People and the Press, “The cell phone challenge to survey research,” news release for May 15, 2006, at www.people-press.org.   8. ​Janice E. Williams et al., “Anger proneness predicts coronary heart disease risk,” Circulation, 101 (2000), pp. 63–95.  9. ​ P. Azoulay and S. Shane, “Entrepreneurs, contracts, and the failure of young firms,” Management Science, 47 (2001), pp. 337–358. 10. ​J. Cantor, “Long-term memories of frightening media often include lingering trauma symptoms,” poster paper presented at the Association for Psychological Science Convention, New York, May 26, 2006. 11. ​William D. Darley, “Store-choice behavior for preowned merchandise,” Journal of Business Research, 27 (1993), pp. 17–31. 12. ​The National Longitudinal Study of Adolescent Health interviewed a stratified random sample of 27,000 adolescents, then reinterviewed many of the subjects six years later, when most were aged 19 to 25. These data are from the Wave III reinterviews in 2000 and 2001, found at www.cpc.unc.edu. 13. ​Joan L. Duda, “The relationship between goal perspectives, persistence, and behavioral intensity among male and female recreational sports participants,” Leisure Sciences, 10 (1988), pp. 95–106. 14. ​Data compiled from a table of percents in “Americans view higher education as key to the American dream,” press release by the National Center for Public Policy and Higher Education, www. highereducation.org, May 3, 2000. 15. ​R. Shine et al., “The influence of nest temperatures and maternal brooding on hatchling phenotypes in water pythons,” Ecology, 78 (1997), pp. 1713–1721. 16. ​D. M. Barnes, “Breaking the cycle of addiction,” Science, 241 (1988), pp. 1029–1030. 17. ​U.S. Department of Commerce, Office of Travel and Tourism Industries, in-flight survey, 2007, at http://tinet.ita.doc.gov. 18. ​The idea for this exercise came from Bob Hayden. 19. ​Douglas E. Jorenby et al., “A controlled trial of sustainedrelease bupropion, a nicotine patch, or both for smoking cessation,” New England Journal of Medicine, 340 (1990), pp. 685–691. 20. ​Martin Enserink, “Fraud and ethics charges hit stroke drug trial,” Science, 274 (1996), pp. 2004–2005. 21. ​Lien-Ti Bei, “Consumers’ purchase behavior toward recycled products: an acquisition-transaction utility theory perspective,” MS thesis, Purdue University, 1993. 22. ​All General Social Survey exercises in this chapter present tables constructed using the search function at the GSS archive, http://sda.berkeley.edu/archive.htm. 23. ​Based closely on Susan B. Sorenson, “Regulating firearms as a consumer product,” Science, 286 (1999), pp. 1481–1482. Because the results in the paper were “weighted to the U.S. population,” we have changed some counts slightly for consistency. 24. ​ Data produced by Ries and Smith, found in William D. Johnson and Gary G. Koch, “A note on the weighted least squares analysis of the Ries-Smith contingency table data,” Technometrics, 13 (1971), pp. 438–447. 25. ​See Note 12.

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26. ​Mei-Hui Chen, “An exploratory comparison of American and Asian consumers’ catalog patronage behavior,” MS thesis, Purdue University, 1994. 27. ​Lillian Lin Miao, “Gastric freezing: an example of the evaluation of medical therapy by randomized clinical trials,” in John P. Bunker, Benjamin A. Barnes, and Frederick Mosteller (eds.), Costs, Risks, and Benefits of Surgery, Oxford University Press, 1977, pp. 198–211. 28. ​See Note 12. 29. ​Thanks to Larry Green, Lake Tahoe Community College, for giving us permission to use several of the contexts from his Web site at www.ltcconline.net/greenl/java/Statistics/catStatProb/categorizingStatProblems12.html. 30. ​See Note 29. 31. ​Mark A. Sabbagh and Dare A. Baldwin, “Learning words from knowledgeable versus ignorant speakers: links between preschoolers’ theory of mind and semantic development,” Child Development, 72 (2001), pp. 1054–1070. Many statistical software packages offer “exact tests” that are valid even when there are small expected counts. 32. ​Brenda C. Coleman, “Study: heart attack risk cut 74% by stress management,” Associated Press dispatch appearing in the Lafayette (Ind.) Journal and Courier, October 20, 1997. 33. ​Tom Reichert, “The prevalence of sexual imagery in ads targeted to young adults,” Journal of Consumer Affairs, 37 (2003), pp. 403–412. 34. ​Based on the EESEE story “What Makes Pre-teens Popular?” 35. ​Based on the EESEE story “Domestic Violence.” 36. ​Karine Marangon et al., “Diet, antioxidant status, and smoking habits in French men,” American Journal of Clinical Nutrition, 67 (1998), pp. 231–239.

Chapter 12

 1. ​Data obtained from the PGA Tour Web site, www.pgatour. com, by Kevin Stevick, The Lawrenceville School.  2. ​The idea for this Activity came from Gloria Barrett, Floyd Bullard, and Dan Teague at the North Carolina School of Science and Math.   3. ​See Note 15 from Chapter 3.   4. ​Data from a plot in James A. Levine, Norman L. Eberhart, and Michael D. Jensen, “Role of non-exercise activity thermogenesis in resistance to fat gains in humans,” Science, 283 (1999), pp. 212–214.  5. ​Samuel Karelitz et al., “Relation of crying activity in early infancy to speech and intellectual development at age three years,” Child Development, 35 (1964), pp. 769–777.  6. ​Data from National Institute of Standards and Technology, Engineering Statistics Handbook, www.itl.nist.gov/div898/ handbook. The analysis there does not comment on the bias of field measurements.   7. ​Todd W. Anderson, “Predator responses, prey refuges, and density-dependent mortality of a marine fish,” Ecology, 81 (2001), pp. 245–257.  8. ​ Based on a plot in G. D. Martinsen, E. M. Driebe, and T. G. Whitham, “Indirect interactions mediated by changing plant chemistry: beaver browsing benefits beetles,” Ecology, 79 (1998), pp. 192–200.   9. ​Data provided by Samuel Phillips, Purdue University. 10. ​Based on Marion E. Dunshee, “A study of factors affecting the amount and kind of food eaten by nursery school children,” Child Development, 2 (1931), pp. 163–183. This article gives the means,

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Notes and Data Sources standard deviations, and correlation for 37 children but does not give the actual data. 11. ​M. H. Criqui, University of California, San Diego, reported in the New York Times, December 28, 1994. 12. ​See Note 29 in Chapter 8. 13. ​Data for the building at 1800 Ben Franklin Drive, Sarasota, Florida, starting in March 2003. From the Web site of the Sarasota County Property Appraiser, http://www.sc-pa.com. 14. ​Information about the sources used to obtain the data can be found under “Documentation” at the Gapminder Web site, www. gapminder.org. 15. ​Gordon L. Swartzman and Stephen P. Kaluzny, Ecological Simulation Primer, Macmillan, 1987, p. 98. 16.  From the Web site, http://en.wikipedia.org/wiki/Transistor_count. 17. ​Planetary data from http://hyperphysics.phy-astr.gsu.edu/hbase/ solar/soldata2.html. 18. ​Sample Pennsylvania female rates provided by Life Quotes, Inc., in USA Today, December 20, 2004. 19. ​G. A. Sacher and E. F. Staffelt, “Relation of gestation time to brain weight for placental mammals: implications for the theory of vertebrate growth,” American Naturalist, 108 (1974), pp. 593–613. We found these data in Fred L. Ramsey and Daniel W. Schafer, The Statistical Sleuth: A Course in Methods of Data Analysis, Duxbury, 1997.

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20. ​Jérôme Chave, Bernard Riéra, and Marc-A. Dubois, “Estimation of biomass in a neotropical forest of French Guiana: spatial and temporal variability,” Journal of Tropical Ecology, 17 (2001), pp. 79–96. 21. ​S. Chatterjee and B. Price, Regression Analysis by Example, Wiley, 1977. 22. ​Chris Carbone and John L. Gittleman, “A common rule for the scaling of carnivore density,” Science, 295 (2002), pp. 2273–2276. 23. ​Data originally from A. J. Clark, Comparative Physiology of the Heart, Macmillan, 1927, p. 84. Obtained from Frank R. Giordano and Maurice D. Weir, A First Course in Mathematical Modeling, Brooks/Cole, 1985, p. 56. 24. ​Stillman Drake, Galileo at Work, University of Chicago Press, 1978. We found these data in D. A. Dickey and J. T. Arnold, “Teaching statistics with data of historic significance,” Journal of Statistics Education, 3 (1995), www.amstat.org/publications/jse/. 25. ​ Tattoo survey data from the Harris Poll®, No. 15, www. harrisinteractive.com, February 12, 2008. 26. ​ Peter H. Chen, Neftali Herrera, and Darren Christiansen, “Relationships between gate velocity and casting features among aluminum round castings,” no date. Provided by Darren Christiansen. 27. ​Data provided by Corinne Lim, Purdue University, from a student project supervised by Professor Joseph Vanable. 28. ​ Data from many studies compiled in D. F. Greene and E. A. Johnson, “Estimating the mean annual seed production of trees,” Ecology, 75 (1994), pp. 642–647.

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Solutions Chapter 1

30 25 20

5

th

w n/ be ig Ye e llo w /g ol d G re en O th er

Br o

Re d

Bl ue

r

y ra G

1.11  (a) A bar graph is given below. A pie chart would also be appropriate because the numbers in the table refer to parts of a single whole. (b) Perhaps induced or C-section births are scheduled for weekdays so doctors don’t have to work as much on the weekend. 

y sda y Th urs da y Fri da y Sa tur da y

sda

We d

Tu e

ne

ay nd

Mo

nd

ay

14,000 12,000 10,000 8000 6000 4000 2000 0

Day

1.13  About 63% are Mexican and 9% are Puerto Rican. 1.15  (a) The given percents represent fractions of different age groups, rather than parts of a single whole. (b) A bar graph is given below. 25 20 15 10 5 0

7y ear 18– s 24 yea rs 25– 34 yea rs 35– 44 yea rs 45– 54 yea rs 55– 64 yea rs 65 an year do s ver

ep

12– 1

ng

Si lv e

Color

Te l

Superpower preference

k

hi te

0

at hy

ty

tre rs

Su pe

isi bi li

In v

Fr

ee

ze

Fl

tim

e

y

0

Te le pa th y

10

Percent

5

e In vi sib ili ty Su pe rs tre ng th

15

15 10

ee ze tim

Fr 20

Su

page 14:  1.  Fly: 99/415 = 23.9%, Freeze time: 96/415 = 23.1%, Invisibility: 67/415 = 16.1%, Superstrength: 43/415 = 10.4%, Telepathy: 110/415 = 26.5%.  2.  A bar graph is shown below. It appears that telepathy, ability to fly, and ability to freeze time were the most popular choices, with about 25% of students choosing each one. Invisibility was the 4th most popular and superstrength was the least popular.

Percent

25

Births

Section 1.1 Answers to Check Your Understanding

1.9  (a) 1%  (b) A bar graph is given below. (c) Yes, because the numbers in the table refer to parts of a single whole.

Bl ac

1.1  Type of wood, type of water repellent, and paint color are categorical. Paint thickness and weathering time are quantitative. 1.3  (a) AP® Statistics students who completed a questionnaire on the first day of class. (b) Categorical: gender, handedness, and favorite type of music. Quantitative: height, homework time, and the total value of coins in a student’s pocket. (c) The individual is a female who is right-handed. She is 58 inches tall, spends 60 minutes on homework, prefers Alternative music, and has 76 cents in her pocket. 1.5  Student answers will vary. For example, quantitative variables could be graduation rate and student-faculty ratio, and categorical variables could be region of the country and type of institution (2-year college, 4-year college, university). 1.7  b

Answers to Odd-Numbered Section 1.1 Exercises

W

Answers to Odd-Numbered Introduction Exercises

Superpower preference

Percent

page 4:  1.  The cars in the student parking lot.  2. He measured the car’s model (categorical), year (quantitative), color (categorical), number of cylinders (quantitative), gas mileage (quantitative), weight (quantitative), and whether it has a navigation system (categorical).

U.K. U.S.

Fl y

Percent

Introduction Answers to Check Your Understanding

35 30 25 20 15 10 5 0

Age group

1.17  (a) The areas of the pictures should be proportional to the numbers of students they represent. (b) A bar graph is given below. Number of students

page 18:  1.  For the U.K. students: 54/200 = 27% said fly, 52/200 = 26% said freeze time, 30/200 = 15% said invisibility, 20/200 = 10% said superstrength, and 44/200 = 22% said telepathy. For the U.S. students: 45/215 = 20.9% said fly, 44/215 = 20.5% said freeze time, 37/215 = 17.2% said invisibility, 23/215 = 10.7% said superstrength, and 66/215 = 30.7% said telepathy. 2. A bar graph is shown in the next column. 3. There is an association between country of origin and superpower preference. Students in the U.K. are more likely to choose flying and freezing time, while students in the U.S. are more likely to choose invisibility or telepathy. Superstrength is about equally unpopular in both countries.

14 12 10 8 6 4 2 0 Cycle

Car

Bus

Walk

Mode of transportation

S-1

Starnes-Yates5e_Answer_S1-S24hr2.indd 1

12/2/13 10:36 AM

Solutions

1.19  (a) 133 people; 36 buyers of coffee filters made of recycled paper. (b) 36.8% said “higher,” 24.1% said “the same,” and 39.1% said “lower.” Overall, 60.9% of the members of the sample think the quality is the same or higher. 1.21  For buyers, 55.6% said higher, 19.4% said the same, and 25% said lower. For the nonbuyers, 29.9% said higher, 25.8% said the same, and 44.3% said lower. We see that buyers are much more likely to consider recycled filters higher in quality and much less likely to consider them lower in quality than nonbuyers. 1.23  Americans are much more likely to choose white/pearl and red, while Europeans are much more likely to choose silver, black, or gray. Preferences for blue, beige/brown, green, and yellow/gold are about the same for both groups. 1.25  A table and a side-by-side bar graph comparing the ­distributions of snowmobile use for environmental club members and nonmembers are shown below. There appears to be an association between environmental club membership and snowmobile use. The visitors who are members of an environmental club are much more likely to have never used a snowmobile and less likely to have rented or owned a snowmobile than visitors who are not in an environmental club.

page 32: 1. Both males and females have distributions that are skewed to the right, though the distribution for the males is more heavily skewed.  The midpoint for the males (9 pairs) is less than the midpoint for the females (26 pairs). The number of shoes owned by females varies more (from 13 to 57) than for males (from 4 to 38). The male distribution has three likely outliers at 22, 35, and 38. The females do not have any likely outliers.  2. ​b  3. ​e  4. ​c  page 38:  1.  One possible histogram is shown below.  2.  The distribution is roughly symmetric and bell-shaped. The typical IQ appears to be between 110 and 120 and the IQs vary from 80 to 150. There do not appear to be any outliers.

Frequency

S-2

18 16 14 12 10 8 6 4 2 0 80 90 100 110 120 130 140 150

IQ

page 39:  1. This is a bar graph because field of study is a categorical variable.  2.  No, because the variable is categorical and the categories could be listed in any order on the horizontal axis. 

Not a member

Member

Never used

445/1221 = 36.4%

212/305 = 69.5%

Snowmobile renter

497/1221 = 40.7%

77/305 = 25.2%

Answers to Odd-Numbered Section 1.2 Exercises

Snowmobile owner

279/1221 = 22.9%

16/305 = 5.2%

1.37  (a) The graph is shown below. (b) The distribution is roughly symmetric with a midpoint of 6 hours. The hours of sleep vary from 3 to 11. There do not appear to be any outliers.

70 60 50 40 30 20 10 0 Club

Percent

3

5

6

7

8

9

10

11

Hours of sleep

No Yes

Snowmobile

Never used

No Yes

No Yes

Snowmobile Snowmobile renter owner

1.27  d 1.29  d 1.31  b 1.33  d 1.35  Answers will vary.  Two possible tables are given below. 10

40

30

20

50

0

30

20

Section 1.2 Answers to Check Your Understanding page 29: 1. This distribution is skewed to the right and unimodal.  2. The midpoint of the 28 values is between 1 and 2.  3.  The number of siblings varies from 0 to 6.  4.  There are two potential outliers at 5 and 6 siblings. 

Starnes-Yates5e_Answer_S1-S24hr1.indd 2

4

1.39  (a) This dot represents a game where the opposing team won by 1 goal. (b) All but 4 of the 25 values are positive, which indicates that the U.S. women’s soccer team had a very good season. They won 21/25 = 84% of their games.  1.41  As coins get older, they are taken out of circulation and new coins are introduced, meaning that most coins will be from recent years with a few from previous years. 1.43  Both distributions are roughly symmetric and have about the same amount of variability. The center of the internal distribution is greater than the center of the external distribution, indicating that external rewards do not promote creativity. Neither distribution appears to have outliers.  1.45  (a) Otherwise, most of the data would appear on just a few stems, making it hard to identify the shape of the distribution. (b) Key: 12 | 1 means that 12.1% of that state’s residents are aged 25 to 34. (c) The distribution of percent of residents aged 25−34 is roughly symmetric with a possible outlier at 16.0%. The center is around 13%. Other than the outlier at 16.0%, the values vary from 11.4% to 15.1%. 1.47  (a) The stemplots are given in the next column. The stemplot with split stems makes it easier to see the shape of the distribution. (b) The distribution is slightly skewed to the right with a center near 780 mm, and values that vary from around 600 mm to 960 mm. There do not appear to be any outliers. (c) In El Niño years, there is typically less rain than in other years (18 of 23 years). 

11/21/13 2:44 PM

Solutions

03557

6

03

7

0124488999

6

557

8

113667

7

01244

9

06

7

88999

8

113

8

667

9

0

9

6

Key: 6 | 3 = 630 mm of rain

1.49  (a) Most people will round their answers to the nearest 10 minutes (or 30 or 60). The students who claimed 300 and 360 minutes of studying on a typical weeknight may have been exaggerating. (b) The stemplots suggest that women (claim to) study more than men.  The center for women (about 175 minutes) is greater than the center for men (about 120 minutes). Women

Men 0 0 3

3

3

3

9 6 0 5 6

6

6

8

9

9

2

2

2

2

4

4

2 2 2 2 2 2 2 2 1 0 2

2

8 8 8 8 8 8 8 8 8 8 7 5 5 5 5 1 5 5

8

4 4 4 0 2 0 0

3

3 0 6 3

20 15 10 5 0 2

4

6

8

10 12 14

Length of words (number of letters)

1.59  The scale on the horizontal axis is very different from one graph to the other. 1.61  A bar graph should be used because birth month is a categorical variable. A possible bar graph is given below.

9

2 Key: 2 | 3 = 230 minutes

Percent of words

6

1.57  (a) The histogram is given below. The distribution of word lengths is skewed to the right and single-peaked. The center is around 4 letters, with words that vary from 1 to 15 letters. There do not appear to be any outliers. (b) There are more short words in Shakespeare’s plays and more very long words in Popular Science articles.

Percent

With splitting stems

9 8 7 6 5 4 3 2 1 0

Jan u Fe ary bru ar Ma y rch Ap ril Ma y Jun e Ju A ly Se ugus pte t mb Oc er t o No b ve er De mber cem be r

Without splitting stems

S-3

Month

14 12 10 8 6 4 2 0

Frequency

1.63  (a) The percents for women sum to 100.1% due to rounding errors. (b) Relative frequency histograms are shown below because there are considerably more men than women. (c) Both histograms are skewed to the right. The center of the women’s distribution of salaries is less than the men’s. The distributions of salaries are about equally variable, and the table shows that there are some outliers in each distribution who make between $65,000 and $70,000. Percent of men

1.51  (a) The distribution is slightly skewed to the left and unimodal. (b) The center is between 0% and 2.5%. (c) The highest return was between 10% and 12.5%. Ignoring the low outliers, the lowest return was between −12.5% and −10%. (d) About 37% of these months (102 out of 273) had negative returns.  1.53  (a) The histogram is given below. (b) The distribution of travel times is roughly symmetric. The center is near 23 minutes and the values vary from 15.5 to 30.9 minutes.  There do not appear to be any outliers.

35 30 25 20 15 10 5 0 10

Frequency

40

50

60

70

35 30 25 20 15 10 5 0 10

20

30

40

50

60

70

Salary ($ thousands)

9 8 7 6 5 4 3 2 1 0 12 18 24 30 36 42 48 54 60

DRP scores

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Percent of women

1.55  The histogram is given below. The distribution of DRP scores is roughly symmetric with the center around 35. The DRP scores vary from 14 to 54. There do not appear to be any outliers.

30

Salary ($ thousands)

14 16 18 20 22 24 26 28 30 32

Travel time (minutes)

20

1.65  The distribution of age is skewed to the right for both males and females, meaning that younger people outnumber older people. Among the younger Vietnamese, there are more males than females. After age 35, however, females seem to outnumber the males, making the center of the female distribution a little greater than the male ­distribution. Both distributions have about the same amount of ­variability and no outliers. 

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Solutions

1.67  (a) Amount of studying. We would expect some students to study very little, but most students to study a moderate amount. Any outliers would likely be high outliers, leading to a right-skewed distribution. (b) Right- versus left-handed. About 90% of the population is right-handed (represented by the bar at 0). (c) Gender. We would expect a more similar percentage of males and females than for the right-handed and left-handed students. (d) Heights. We expect many heights near the average and a few very short or very tall people.  1.69  a 1.71  c 1.73  d 1.75  (a) Major League Baseball players who were on the roster on opening day of the 2012 season. (b) 6. Two variables are categorical (team, position) and the other 4 are quantitative (age, height, weight, and salary).  1.77  (a) 71/858 = 8.3% were elite soccer players and 43/858 = 5.0% of the people had arthritis. (b) 10/71 = 14.1% of elite soccer players had arthritis and 10/43 = 23.3% of those with arthritis were elite soccer players.

Section 1.3 Answers to Check Your Understanding page 53: 1. Because the distribution is skewed to the right, we would expect the mean to be larger than the median.  2.  Yes. The mean is 31.25 minutes, which is greater than the median of 22.5 minutes.  3.  Because the distribution is skewed, the median would be a better measure of the center of the distribution.  page 59:  1.  The data in order are: 290, 301, 305, 307, 307, 310, 324, 345. The 5-number summary is 290, 303, 307, 317, 345.  2. The IQR is 14 pounds. The range of the middle half of the data is 14 pounds.  3.  Any outliers occur below 303 − 1.5(14) = 282 or above 317 + 1.5(14) = 338, so 345 pounds is an outlier.  4.  The boxplot is given below.

290

300

310

320

330

340

350

Weight

page 63:  1.  The mean is 75.  2.  The table is given below. Observation

Deviation

Squared deviation

67

67 − 75 = −8

(−8)2 = 64

72

72 − 75 = −3

(−3)2

76

76 − 75 = 1

12 = 1

76

76 − 75 = 1

12 = 1

84

84 − 75 = 9

92

Total

0

3.  The variance is s2x =

=9

= 81 156

156 = 39 inches squared and the stan5−1

dard deviation is sx = "39 = 6.24 inches. 4.  The players’ heights typically vary by about 6.24 inches from the mean height of 75 inches.

Answers to Odd-Numbered Section 1.3 Exercises 1.79  x = 85

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1.81  (a) median = 85 (b) x = 79.33 and median = 84. The median did not change much but the mean did, showing that the median is more resistant to outliers than the mean.  1.83  The mean is $60,954 and the median is $48,097. The distribution of salaries is likely to be quite right skewed because of a few people who have a very large income, making the mean larger than the median. 1.85  The team’s annual payroll is 1.2(25) = 30 or $30 million. No, because the median only describes the middle value in the distribution. It doesn’t provide specific information about any of the other values. 1.87  (a) Estimating the frequencies of the bars (from left to right) as 10, 40, 42, 58, 105, 60, 58, 38, 27, 18, 20, 10, 5, 5, 1, and 3, the 3504 mean is x = = 7.01. The median is the average of the 250th 500 and 251st values, which is 6. (b) Because the median is less than the mean, we would use the median to argue that shorter domain names are more popular.  1.89  (a)  IQR = 91 − 78 = 13. The middle 50% of the data have a range of 13 points. (b) Any outliers are below 78 − 1.5(13) = 58.5 or above 91 + 1.5(13) = 110.5. There are no outliers. 1.91  (a) Outliers are anything below 3 − 1.5(40) = −57 or above 43 + 1.5(40) = 103, so 118 is an outlier. The boxplot is shown below. (b) The article claims that teens send 1742 texts a month, which is about 58 texts a day. Nearly all of the members of the class (21 of 25) sent fewer than 58 texts per day, which seems to contradict the claim in the article.  120

Number of texts

S-4

*

100 80 60 40 20 0

1.93  (a) Positive numbers indicate students who had more text messages than calls. Because the 1st quartile is about 0, roughly 75% of the students had more texts than calls, which supports the article’s conclusion. (b) No. Students in statistics classes tend to be upperclassmen and their responses might differ from those of underclassmen.  1.95  (a) About 3% and −3.5%. (b) About 0.1%. (c) The stock fund is much more variable. It has higher positive returns, but also higher negative returns. 2.06 = 0.6419 mg∙dl. (b) The phosphate level Å6 − 1 ­typically varies from the mean by about 0.6419 mg/dl.  1.97  (a) sx =

1.99  (a) Skewed to the right, because the mean is much larger than the median and Q3 is much further from the median than Q1. (b) The amount of money spent typically varies from the mean by $21.70. (c) Any points below 19.06 − 1.5(26.66) = −20.93 or above 45.72 + 1.5(26.66) = 85.71 are outliers. Because the maximum of 93.34 is greater than 85.71, there is at least one outlier. 1.101  Yes. For example, in data set 1, 2, 3, 4, 5, 6, 7, 8 the IQR is 4. If 8 is changed to 88, the IQR will still be 4. 1.103  (a) One possible answer is 1, 1, 1, 1. (b) 0, 0, 10, 10. (c) For part (a), any set of four identical numbers will have sx = 0. For part (b), however, there is only one possible answer. We want the values to be as far from the mean as possible, so our best choice is two values at each extreme.

11/21/13 2:44 PM

Solutions 1.105  State: Do the data indicate that men and women differ in their study habits and attitudes toward learning? Plan: We will draw side-by-side boxplots of the data about men and women; compute summary statistics; and compare the shape, center, and spread of both distributions. Do: The boxplots are given below, as is a table of summary statistics.  Female

S-5

R1.3  (a) The “bars” are different widths. For example, the bar for “send/receive text messages” should be roughly twice the size of the bar for “camera” when it is actually about 4 times as large. (b) No, because they do not describe parts of a whole. Students were free to answer in more than one category. (c) A bar graph is given below. 50 40

Percent

*

Male

30 20 10

80

100

120

140

160

180

200

220

Ma ke /r

60

e ce iv ca l e ls Se tex nd/r t m e ce ess ive age s Ca me ra Se nd /re pic ceive tur es

0

SSHA score

StDev

Minimum

Women

18 141.06

N

Mean

26.44

101.00 126.00  138.50 154.00 200.00

Men

20 121.25

32.85

70.00

Q1

Median

Q3

Maximum

98.00 114.50 143.00 187.00

Frequency

Both distributions are slightly skewed to the right. Both the mean and median are higher for women than for men. The scores for men are more variable than the scores for women. There are no outliers in the male distribution and a single outlier at 200 in the female distribution. Conclude: Men and women differ in their study habits and attitudes toward learning. The typical score for females is about 24 greater than the typical score for males. Female scores are also more consistent than male scores.  1.107  d 1.109  e 1.111  A histogram is given below. This distribution is roughly ­symmetric with a center around 170 cm and values that vary from 145.5 cm to 191 cm. There do not appear to be any outliers. 12 10 8 6 4 2 0 160

170

180

190

Heights (cm)

1.113  Women appear to be more likely to engage in behaviors that are indicative of good “habits of mind.” They are especially more likely to revise papers to improve their writing. The difference is a little smaller for seeking feedback on their work, although the ­percentage is still higher for females.

Answers to Chapter 1 Review Exercises R1.1  (a) Movies. (b) Quantitative: Year, time, box office sales. Categorical: Rating, genre. Note: Year might be considered categorical if we want to know how many of these movies were made each year rather than the average year. (c) This movie is Avatar, released in 2009. It was rated PG-13, runs 162 minutes, is an action film, and had box office sales of $2,781,505,847. R1.2  A bar chart is given below. 60 50

Percent

R1.4  (a) 148/219 = 67.6%. Marginal distribution, because it is part of the distribution of one variable for all categories of the other variable. (b) 78/82 = 95.1% of the younger students were Facebook users. 78/148 = 52.7% of the Facebook users were younger. R1.5  There does appear to be an association between age and Facebook status. From both the table and the graph given below, we can see that as age increases, the percent of Facebook users decreases. For younger students, about 95% are members. That drops to 70% for middle students and drops even further to 31.3% for older students. Facebook user? Age

Yes

No

Younger (18–22)

95.1%

4.9%

Middle (23–27)

70.0%

30.0%

Older (28 and up)

31.3%

68.7%

100 150

40 30 20 10 0 G

PG PG-13

Rating

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Features

R

NC-17

80

Percent

Variable

60 40 20

0 Facebook

Yes No

Yes No

Yes No

Age

Younger

Middle

Older

R1.6  (a) A stemplot is given below. (b) The distribution is roughly symmetric with one possible outlier at 4.88. The center of the distribution is between 5.4 and 5.5. The densities vary from 4.88 to 5.85. (c) Because the distribution is roughly symmetric, we can use the mean to estimate the Earth’s density to be about 5.45 times the density of water. 48 8 49  50 7 51 0 52 6799 53 04469 54 2467 55 03578 56 12358 57 59 58 5

Key: 48 | 8 = 4.88

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Solutions

35 30 25 20 15 10 5 0

T1.9  e  T1.10  b  T1.11  d  T1.12  (a) A histogram is given below. (b) Any point below 30 − 1.5(47) = −40.5 or above 77 + 1.5(47) = 147.5 is an outlier. So 151 minutes is an outlier. (c) Median and IQR, because the distribution is skewed and has a high outlier. 10

240

360

480

600

Survival time (days)

6 4 2 0

Answers to Chapter 1 T1.1  d  T1.2  e  T1.3  b  T1.4  b  T1.5  c   T1.6  c  T1.7  b  T1.8  c 

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100 80 60 40 20 0 Birth defects

160

e o None rm ore On c ab

eti

e o None rm ore On tic

Di

Diabetic status

be

(c)  Use the median and IQR to summarize the distribution because the outliers will have a big effect on the mean and standard deviation.  R1.8  (a) About 20% of low-income and 33% of high-income households. (b) The shapes of both distributions are skewed to the right; however, the skewness is much stronger in the distribution for low-income households. On average, household size is larger for high-income households. One-person households might have less income because they would include many young single people who have no job or retired single people with a fixed income. R1.9  (a) The amount of mercury per can of tuna will typically vary from the mean by about 0.3 ppm. (b) Any point below 0.071 − 1.5(0.309) = −0.393 or above 0.38 + 1.5(0.309) = 0.8435 would be considered an outlier. There are no low outliers, but there are several high outliers. (c) The distribution of the amount of mercury in cans of tuna is highly skewed to the right. The median is 0.18 ppm and the IQR is 0.309 ppm. R1.10  The distribution for light tuna is skewed to the right with several high outliers, while the distribution for albacore tuna is more symmetric with just a couple of high outliers. Because it has a greater center, the albacore tuna generally has more mercury. However, the light tuna has a much bigger spread of values, with some cans having as much as twice the amount of mercury as the largest amount in the albacore tuna. 

dia

0

Pre

100

e o None rm ore

200

AP®

120

T1.13  (a) Row totals are 1154, 53, and 1207. Column totals are 785, 375, 47, and 1207. (b) Nondiabetic: 96.1% none and 3.9% one or more. Prediabetic: 96.5% none and 3.5% one or more. Diabetic: 80.9% none and 19.1% one or more. (c) The graph is given below.

On

** *

300

iab

400

80

nd

* **

500

40

Time on Internet (minutes)

No

600

0

Percent

Survival time (days)

(b)  The boxplot is given below.

c

120

8

eti

Frequency

R1.7  (a) A histogram is given below. The survival times are rightskewed, as expected. The median survival time is 102.5 days and the range of survival times is 598 − 43 = 555 days. There are several high outliers with survival times above 500.

Frequency

S-6

(d)  Yes. Nondiabetics and prediabetics appear to have babies with birth defects at about the same rate. However, those with diabetes have a much higher rate of babies with birth defects. T1.14 (a) Between 550 and 559 hours. (b) Because it has a higher minimum lifetime or because its lifetimes are more consistent (less variable). (c) Because it has a higher median lifetime. T1.15  Side-by-side boxplots and descriptive statistics for both leagues are given below. Both distributions are roughly symmetric, although there are two low outliers in the NL. The data suggest that the number of home runs is somewhat less in the NL. All 5 numbers in the 5-number summary are less for the NL teams than for the AL teams. However, there is more variability among the AL teams. American League National League

Statistics Practice Test

**

20

30

40

50

60

70

80

Home runs

Variable N American 14 League National 14 League

Mean StDev Minimum Q1 Median Q3 Maximum 56.93 12.69 35.00 49.00 57.50 68.00 77.00 50.14 11.13 29.00

46.00 50.50 55.00 67.00

11/21/13 2:44 PM

Solutions

Chapter 2 Section 2.1 Answers to Check Your Understanding page 89: 1. c 2.  Her daughter weighs more than 87% of girls her age and she is taller than 67% of girls her age. 3.  About 65% of calls lasted less than 30 minutes, which means that about 35% of calls lasted 30 minutes or longer. 4.  Q1 = 13 minutes, Q3 = 32 minutes, and IQR = 19 minutes. page 91:  1. z = −0.466. Lynette’s height is 0.466 standard deviations below the mean height of the class. 2.  z 5 1.63. Brent’s height is 1.63 standard deviations above the mean height of the 74 − 76 , so s = 2.35 inches. class.  3.  −0.85 = s page 97:  1.  Shape will not change. However, it will multiply the center (mean, median) and spread (range, IQR, standard ­deviation) by 2.54.  2.  Shape and spread will not change. It will, ­however, add 6 inches to the center (mean, median).  3. Shape will not change. However, it will change the mean to 0 and the ­standard deviation to 1.

Answers to Odd-Numbered Section 2.1 Exercises 2.1  (a) She is at the 25th percentile, meaning that 25% of the girls had fewer pairs of shoes than she did. (b) He is at the 85th percentile, meaning that 85% of the boys had fewer pairs of shoes than he did. (c) The boy is more unusual because only 15% of the boys have as many or more than he has. The girl has a value that is closer to the center of the distribution. 2.3  A percentile only describes the relative location of a value in a distribution. Scoring at the 60th percentile means that Josh’s score is better than 60% of the students taking this test. His correct percentage could be greater than 60% or less than 60%, depending on the difficulty of the test. 2.5  The girl weighs more than 48% of girls her age, but is taller than 78% of the girls her age.  2.7  (a) The student sent about 205 text messages in the 2-day period and sent more texts than about 78% of the students in the sample. (b) Locate 50% on the y-axis, read over to the points, and then go down to the x-axis. The median is approximately 115 text messages. 2.9  (a) IQR ≈ $46 − $19 = $27 (b) About the 26th percentile. (c) The histogram is below. 30

Percent

25 20

S-7

2.15  (a) He is at the 76th percentile, meaning his salary is higher than 76% of his teammates.  (b)  z = 0.79. Lidge’s salary was 0.79 standard deviations above the mean salary. 2.17  Multiply each score by 4 and add 27. 2.19  (a) mean = 87.188 inches and median = 87.5 inches. (b) The standard deviation (3.20 inches) and IQR (3.25 inches) do not change because adding a constant to each value in a distribution does not change the spread.  2.21  (a) mean = 5.77 feet and median = 5.79 feet. (b) Standard deviation = 0.267 feet and IQR = 0.271 feet. 9 2.23  Mean = (25) + 32 = 77°F and standard deviation = 5 9 (2) = 3.6°F. 5 2.25  c 2.27  c 2.29  c 2.31  The distribution is skewed to the right with a center around 20 minutes and the range close to 90 minutes. The two largest ­values appear to be outliers.

Section 2.2 Answers to Check Your Understanding page 107:  1.  It is legitimate because it is positive everywhere and it has total area under the curve = 1.  2. 12% 3.  Point A in the graph below is the approximate median. About half of the area is to the left of A and half of the area is to the right of A.  4.  Point B in the graph below is the approximate mean (balance point). The mean is less than the median in this case because the distribution is skewed to the left. Total area under curve = 1 Area = 0.12

7

8

BA

page 112: 1. The graph is given below. 2. Approximately 100% − 68% 100% − 68% = 16%.  3. Approximately = 16% 2 2 have heights below 62 inches and approximately 100% − 99.7% = 0.15% of young women have heights above 2 72 inches, so the remaining 83.85% have heights between 62 and 72 inches.

15 10 5 0 5

25

45

65

85

Amount spent ($)

2.11  Eleanor. Her standardized score (z = 1.8) is higher than Gerald’s (z = 1.5).  2.13  (a) Your bone density is far below average—about 1.5 times farther below average than a typical below-average density. 948 − 956 gives s = 5.52 g∙cm2 .  (b) Solving −1.45 = s

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57.0 59.5 62.0 64.5 67.0 69.5 72.0

Heights

page 116:  (All graphs are shown on the following page.)  1. The proportion is 0.9177.  2.  The proportion is 0.9842.  3.  The proportion is 0.9649 − 0.2877 = 0.6772.  4. The z-score for the 20th percentile is z = −0.84.  5.  45% of the observations are greater than z = 0.13.

11/21/13 2:44 PM

S-8

Solutions Answers to Odd-Numbered Section 2.2 Exercises 2.33  Sketches will vary, but here is one example:

0

1.39

−2.15

 

z

0

2.35  (a) It is on or above the horizontal axis everywhere, and the 1 1 1 area beneath the curve is × 3 = 1. (b) × 1 = . (c) Because 3 3 3 1 1.1 − 0.8 = 0.3, the proportion is × 0.3 = 0.1. 3 2.37  Both are 1.5.  2.39  (a) Mean is C, median is B. (b) Mean is B, median is B. 2.41  The graph is shown below.

z

0.2

−0.56 0

1.81

z

0

 

z

0.45 61.5 64.0 66.5 69.0 71.5 74.0 76.5

Men’s height (inches)

0

z

page 121:   1.  For 14-year-old boys, the amount of cholesterol follows a N(170, 30) distribution and we want to find the percent of boys with cholesterol of more than 240 (see graph below).  240 − 170 z= = 2.33. From Table A, the proportion of z-scores 30 above 2.33 is 1 − 0.9901 = 0.0099. Using technology: normalcdf (lower:240,upper:1000,μ:170,σ:30) = 0.0098. About 1% of 14-year-old boys have cholesterol above 240 mg/dl.  2. For 14-year-old boys, the amount of cholesterol follows a N(170, 30) distribution and we want to find the percent of boys with cholesterol between 200 and 240 (see graph below). 200 − 170 240 − 170 z= = 1 and z = = 2.33. From Table A, the 30 30 ­proportion of z-scores between 1 and 2.33 is 0.9901 − 0.8413 = 0.1488. Using technology: normalcdf(lower:200,upper: 240,μ:170,σ:30) = 0.1488. About 15% of 14-year-old boys have cholesterol between 200 and 240 mg/dl.  3.  For Tiger Woods, the distance his drives travel follows an N (304, 8) distribution and the 80th percentile is the boundary value x with 80% of the ­distribution to its left (see graph below). A z-score of 0.84 gives the x − 304 gives x = area closest to 0.80 (0.7995). Solving 0.84 = 8 310.7. Using technology: invNorm(area:0.8,μ:304,σ:8) = 310.7. The 80th percentile of Tiger Woods’s drive lengths is about 310.7 yards. 

2.43  (a) Between 69 − 2(2.5) = 64 and 69 + 2(2.5) = 74 inches. 100% − 95% 100% − 68% = 2.5%. (c) About = 16% of (b) About 2 2 100% − 95% = 2.5% are men are shorter than 66.5 inches and 2 shorter than 64 inches, so approximately 16% − 2.5% = 13.5% of men have heights between 64 inches and 66.5 inches. (d) Because 100% − 68% = 16% of the area is to the right of 71.5, 71.5 is at the 2 84th percentile. 2.45  Taller curve: standard deviation ≈ 0.2. Shorter curve: standard deviation ≈ 0.5.  2.47  (a) 0.9978. (b) 1 − 0.9978 = 0.0022 (c) 1 − 0.0485 = 0.9515 (d) 0.9978 − 0.0485 = 0.9493 2.49  (a) 0.9505 − 0.0918 = 0.8587 (b) 0.9633 − 0.6915 = 0.2718 2.51  (a) z = −1.28  (b)  z = 0.41 2.53  (a) The length of pregnancies follows a N(266, 16) distribution and we want the proportion of pregnancies that last less than 240 − 266 = − 1.63. From Table 240 days (see graph below). z = 16 A, the proportion of z-scores less than −1.63 is 0.0516. Using technology: normalcdf(lower:–1000,upper:240,μ:266, σ:16) = 0.0521. About 5% of pregnancies last less than 240 days, so 240 days is at the 5th percentile of pregnancy lengths.

240

170

240

Cholesterol levels

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170 200

240

Cholesterol levels

266

Length of pregnancy (days)

(b)  The length of pregnancies follows a N(266, 16) distribution and we want the proportion of pregnancies that last between 240 240 − 266 = −1.63 and 270 days (see the following graph). z = 16 270 − 266 = 0.25. From Table A, the proportion of and z = 16 z-scores between −1.63 and 0.25 is 0.5987 − 0.0516 = 0.5471. Using ­technology: normalcdf(lower:240,upper:270,

12/2/13 10:34 AM

Solutions μ:266,σ:16) = 0.5466. About 55% of pregnancies last between 240 and 270 days.

240

266 270

Length of pregnancy (days)

(c)  The length of pregnancies follows a N(266, 16) distribution and we are looking for the boundary value x that has an area of 0.20 to the right and 0.80 to the left (see graph below). A z-score of 0.84 x − 266 gives gives the area closest to 0.80 (0.7995). Solving 0.84 = 16 x = 279.44. Using technology: invNorm(area:0.8,μ:266, σ:16) = 279.47. The longest 20% of pregnancies last longer than 279.47 days. 

0.8

S-9

2.57  (a) For large lids, the diameter follows a N(m, 0.02) ­ istribution and we want to find the value of m that will result in d only 1% of lids that are too small to fit (see graph below). A z-score of −2.33 gives the value closest to 0.01 (0.0099). 3.95 − m gives m = 4.00. Using technology: Solving −2.33 = 0.02 invNorm(area:0.01,μ:0,σ:1) gives z = −2.326. Solving 3.95 − m gives m = 4.00. The manufacturer should set −2.326 = 0.02 the mean diameter to approximately m = 4.00 to ensure that only 1% of lids are too small.  (b)  For large lids, the diameter follows a N(3.98, s) distribution and we want to find the value of s that will result in only 1% of lids that are too small to fit (see graph below). A z-score of −2.33 gives the value closest to 0.01 (0.0099). 3.95 − 3.98 gives s = 0.013. Using technology: Solving −2.33 = s invNorm(area:0.01,μ:0,σ:1) gives z = −2.326. Solving 3.95 − 3.98 gives s = 0.013. A standard deviation of at −2.326 = s most 0.013 will result in only 1% of lids that are too small to fit.

0.2

266

Length of pregnancy (days)

Area = 0.01

2.55  (a) For large lids, the diameter follows a N(3.98, 0.02) ­distribution and we want to find the percent of lids that have diam3.95 − 3.98 = −1.5. eters less than 3.95 (see graph below).  z = 0.02 From Table A, the proportion of z-scores below −1.5 is 0.0668. Using technology: normalcdf(lower:–1000,upper:3.95, μ:3.98,σ:0.02) = 0.0668. About 7% of the large lids are too small to fit.

µ

Diameter (inches)

(c)  Reduce the standard deviation. This will reduce the number of lids that are too small and the number of lids that are too big. If we make the mean a little larger as in part (a), we will reduce the number of lids that are too small, but we will increase the number of lids that are too big.

Area = 0.01

3.98

Diameter (inches)

3.95

3.98

Lid width (inches)

2.59  (a) z = −1.28 and z = 1.28 (b) Solving −1.28 =

(b)  For large lids, the diameter follows a N(3.98, 0.02) distribution and we want to find the percent of lids that have diameters greater 4.05 − 3.98 = 3.5. From Table A, than 4.05 (see graph below).  z = 0.02 the proportion of z-scores above 3.50 is approximately 0. Using technology: normalcdf(lower:4.05,upper:1000,μ:3.98, σ:0.02) = 0.0002. Approximately 0% of the large lids are too big to fit.

3.98

gives x = 61.3 inches and solving 1.28 = inches. 

x − 64.5 2.5

x − 64.5 gives x = 67.7 2.5

60 − m 75 − m and 1.88 = gives m = 41.43 s s minutes and s = 17.86 minutes. 

2.61  Solving 1.04 =

2.63  (a) A histogram is given below. The distribution of shark lengths is roughly symmetric and somewhat bell-shaped, with a mean of 15.586 feet and a standard deviation of 2.55 feet.  (b) 30/44 = 68.2% , 42/44 = 95.5%, and 44/44 = 100%. These are very close to the 68−95−99.7 rule. 

4.05

Lid width (inches)

(c)  Make a larger proportion of lids too small. If lids are too small, customers will just try another lid. But if lids are too large, the customer may not notice and then spill the drink.

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S-10

Solutions

Expected z-score

(c)  A Normal probability plot is given below. Except for one small shark and one large shark, the plot is fairly linear, indicating that the distribution of shark lengths is approximately Normal. 2

R2.5  (a) Between 336 − 3(3) = 327 days and 336 + 3(3) = 345 100% − 68% = 16%. days. (b) About 2 R2.6  (a) 0.9616 − 0.0122 = 0.9494  (b)  If 35% of all values are greater than a particular z-value, then 65% are lower. A z-score of 0.39 gives the value closest to 0.65 (0.6517). Using technology: invNorm(area:0.65,μ:0,σ:1) gives z = 0.385.

1 0 –1 –2 10 12 14 16 18 20 22 24

Length of shark (feet)

(d)  All indicate that shark lengths are approximately Normal. 2.65  The distribution is close to Normal because the plot is nearly linear. There is a small “wiggle” between 120 and 130, with several values a little larger than would be expected in a Normal distribution. Also, the smallest value and the two largest values are a little farther from the mean than would be expected in a Normal distribution. 2.67  No. If it was Normal, then the minimum value should be around 2 or 3 standard deviations below the mean. However, the actual minimum has a z-score of just z = −1.09. Also, if the distribution was Normal, the minimum and maximum should be about the same distance from the mean. However, the maximum is much farther from the mean (20,209) than the minimum (8741).  2.69  b 2.71  b 2.73  a 2.75  For both kinds of cars, we see that the highway mileage is greater than the city mileage. The two-seater cars have a more variable distribution, both on the highway and in the city. Also the mileage values are slightly lower for the two-seater cars than for the minicompact cars, both on the highway and in the city, with a greater difference on the highway. All four distributions are roughly symmetric.

0.65

−2.25

0

z

 

R2.7  (a) Birth weights follow a N(3668, 511) distribution and we want to find the percent of babies with weights less than 2500 2500 − 3668 = − 2.29. From grams (see graph below). z = 511 Table A, the proportion of z-scores below −2.29 is 0.0110. Using technology: normalcdf(lower:–1000,upper:2500,μ: 3668,σ:511) = 0.0111. About 1% of babies will be identified as low birth weight.

2500

3668

Birth weights (grams)

Answers to Chapter 2 Review Exercises R2.1  (a) z = 1.20. Paul’s height is 1.20 standard deviations above the average male height for his age. (b) 85% of boys Paul’s age are shorter than Paul. R2.2  (a) 58th percentile (b) IQR = 11 − 2.5 = 8.5 hours per week. R2.3  (a) The shape of the distribution would not change. 43.7 42 Mean = = 13.32 meters, median = = 12.80 meters, 3.28 3.28 12.5 = 3.81 meters, standard deviation = 3.28 12.5 IQR = = 3.81 meters. (b) Mean = 43.7 − 42.6 = 1.1 feet; 3.28 standard deviation = 12.5 feet, because subtracting a constant from each observation does not change the spread. R2.4  (a) The median (line A in the graph below) should be slightly to the right of the main peak, with half of the area to the left and half to the right. (b) The mean (line B in the graph below) should be slightly to the right of the line for the median at the balancing point.

0

1.77

z

(b)  Birth weights follow a N(3668, 511) distribution. The 1st quartile is the boundary value with 25% of the area to its left. The 3rd quartile is the boundary value with 75% of the area to its left (see graph below). A z-score of −0.67 gives the value closest to 0.25 x − 3668 gives Q1 = 3325.63. A (0.2514). Solving − 0.67 = 511 z-score of 0.67 gives the value closest to 0.75 (0.7486). Solving x − 3668 0.67 = gives Q3 = 4010.37. Using technology: 511 invNorm(area:0.25,μ:3668,σ:511)gives Q1 = 3323.34 and invNorm(area:0.75,μ:3668,σ:511)gives Q3 = 4012.66. The quartiles are Q1 = 3323.34 grams and Q3 = 4012.66 grams.

Area = 0.25

2135

2646

Area = 0.25

3157

3668

4179

4690

5201

Birth weight (grams)

AB

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R2.8  (a) The amount of ketchup dispensed follows a N(1.05, 0.08) distribution and we want to find the percent of times that the amount of ketchup dispensed will be between 1 and 1.2 ounces (see

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1.2 − 1.05 1 − 1.05 = 1.88 and z = = − 0.63. 0.08 0.08 From Table A, the proportion of z-scores between −0.63 and 1.88 is 0.9699 − 0.2643 = 0.7056. Using technology: normalcdf (lower:1,upper:1.2,μ:1.05,σ:0.08) = 0.7036.  About 70% of the time the dispenser will put between 1 and 1.2 ounces of ketchup on a burger.  graph below). z =

Expected z-score

Solutions

S-11

2 1 0 –1 –2 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95

Length of thorax (mm)

R2.11  The steep, nearly vertical portion at the bottom and the clear bend to the right indicate that the distribution of the data is right-skewed with several outliers and not approximately Normally distributed. 1 1.05

1.2

(b)  The amount of ketchup dispensed follows a N(1.1, s) distribution and we want to find the value of s that will result in at least 99% of burgers getting between 1 and 1.2 ounces of ketchup (see graph below). Because the mean of 1.1 is in the middle of the ­interval from 1 to 1.2, we are looking for the middle 99% of the distribution. This leaves 0.5% in each tail. A z-score of −2.58 gives 1 − 1.1 gives the value closest to 0.005 (0.0049). Solving − 2.58 = s s = 0.039. Using technology: invNorm(area:0.005,μ:0,σ:1) 1 − 1.1 gives s = 0.039. A gives z = −2.576. Solving − 2.576 = s standard deviation of at most 0.039 ounces will result in at least 99% of burgers getting between 1 and 1.2 ounces of ketchup.  

Area = 0.99

1.10

R2.9  If the distribution is Normal, the 10th and 90th percentiles must be equal distances above and below the mean. Thus, the 25 + 475 = 250 points. The 10th percentile in a standard mean is 2 25 − 250 , we Normal distribution is z = −1.28. Solving − 1.28 = s get s = 175.8. Using technology: invNorm(area:0.10,μ:0, 25 − 250 and s = 175.5. σ:1) gives z = −1.282, so − 1.282 = s R2.10  A histogram and Normal probability plot are given below. The histogram is roughly symmetric but not very bell-shaped. The Normal probability plot, however, is roughly linear. For these data, x = 0.8004 and sx = 0.0782. Although the percentage within 1 standard deviation of the mean (55.1%) is less than expected (68%), the percentage within 2 (93.9%) and 3 s­ tandard deviations (100%) match the 68−95−99.7 rule quite well. It is reasonable to say that these data are approximately Normally distributed. Frequency

12 10 8 6 4 2 0 0.60

0.70

0.80

0.90

Length of thorax (mm)

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Answers to Chapter 2 AP® Statistics Practice Test T2.1  e  T2.2  d  T2.3  b  T2.4  b  T2.5  a  T2.6  e  T2.7  c  T2.8  e  T2.9  e  T2.10  c  T2.11  (a) Jane’s performance was better. Because her performance (40) exceeded the standard for the Presidential award (39), she performed above the 85th percentile. Matt’s performance (40) met the standard for the National award (40), meaning he performed at the 50th percentile. (b) Because Jane’s score has a higher percentile than Matt’s score, she is farther to the right in her distribution than Matt is in his. Therefore, Jane’s standardized score will likely be greater than Matt’s. T2.12  (a) For male soldiers, head circumference follows a N(22.8, 1.1) distribution and we want to find the percent of soldiers with head circumference less than 23.9 inches (see graph below). 23.9 − 22.8 = 1. From Table A, the proportion of z-scores z= 1.1 below 1 is 0.8413. Using ­technology: normalcdf(lower: −1000,upper:23.9,μ:22.8,σ:1.1) = 0.8413. About 84% of soldiers have head circumferences less than 23.9 inches. Thus, 23.9 inches is at the 84th percentile. N(22.8, 1.1)

22.8 23.9

Head circumference (inches)

(b)  For male soldiers, head circumference follows a N(22.8, 1.1) distribution and we want to find the percent of soldiers with head circumferences less than 20 inches or greater than 26 inches (see 20 − 22.8 26 − 22.8 = − 2.55 and z = = 2.91. graph below). z = 1.1 1.1 From Table A, the proportion of z-scores below z = −2.55 is 0.0054 and the proportion of z-scores above 2.91 is 1 − 0.9982 = 0.0018, for a total of 0.0054 + 0.0018 = 0.0072. Using technology: 1 − normalcdf(lower:20,upper:26,μ:22.8,σ:1.1) = 1 − 0.9927 = 0.0073. A little less than 1% of soldiers have head

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S-12

Solutions

circumferences less than 20 inches or greater than 26 inches and require custom helmets. 

19.5 20.6 21.7 22.8 23.9 25.0 26.1

Florida and the number of manatees killed.  (b)  r ≈ 0.5. This indicates that there is a moderate, positive linear relationship between the number of named storms predicted and the actual number of named storms. (c)  r ≈ 0.3. This indicates that there is a weak, positive linear relationship between the healing rate of the two front limbs of the newts.  (d)  r ≈ − 0.1. This indicates that there is a weak, negative linear relationship between last year’s percent return and this year’s percent return in the stock market.

Head circumference (inches)

Answers to Odd-Numbered Section 3.1 Exercises (c)  For male soldiers, head circumference follows a N(22.8, 1.1) distribution. The 1st quartile is the boundary value with 25% of the area to its left. The 3rd quartile is the boundary value with 75% of the area to its left (see graph below). A z-score of −0.67 gives the Starnes/Yates/Moore: The Practice of Statistics, 4E x − 22.8 value closestNew to 0.25 (0.2514). Solving − 0.67 = Fig.: 2056 gives Fig.: BM_UNSol_2-T2.12b Perm. 1.1 First Pass: 2010-05-24 z-score of 0.67 gives the value closest to 0.75 Q1 = 22.063. 2ndAPass: 2010-06-29 x − 22.8 3rd Pass: 2010-11-22 (0.7486). Solving 0.67 = gives Q3 = 23.537. Using tech1.1 nology: invNorm(area:0.25,μ:22.8,σ:1.1) gives Q1 = 22.058 and invNorm(area:0.75,μ:22.8,σ:1.1) gives Q3 = 23.542. Thus, IQR = 23.542 − 22.058 = 1.484 inches. 

19.5

20.6

Area = 0.25

21.7

22.8

23.9

25.0

26.1

Head circumference (inches)

T2.13  No. First, there is a large difference between the mean and the median. In a Normal distribution, the mean and median are the same, but in this distribution the mean is 48.25 and the median is 37.80. Second, the distance between the minimum and the median is 35.80 but the distance between the median and the maximum is 167.10. In a Normal distribution, these distances should be about the same. Both of these facts suggest that the distribution is skewed to the right.

Chapter 3 Section 3.1 Answers to Check Your Understanding page 144: 1. Explanatory: number of cans of beer. Response: blood alcohol level.  2.  Explanatory: amount of debt and income. Response: stress caused by college debt.  page 149:  1.  Positive. The longer the duration of the eruption, the longer we should expect to wait between eruptions because long eruptions use more energy and it will take longer to build up the energy needed to erupt again.  2.  Roughly linear with two clusters. The clusters indicate that, in general, there are two types of eruptions—shorter eruptions that last around 2 minutes and longer eruptions that last around 4.5 minutes.  3.  Fairly strong. The points don’t deviate much from the linear form.  4.  There are a few possible outliers around the clusters. However, there aren’t many and potential outliers are not very distant from the main clusters of points.  5.  How long the previous eruption was. page 153:  (a)  r ≈ 0.9. This indicates that there is a strong, positive linear relationship between the number of boats registered in

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3.7  (a) There is a positive association between backpack weight and body weight. For students under 140 pounds, there seems to be a linear pattern in the graph. However, for students above 140 pounds, the association begins to curve. Because the points vary somewhat from the linear pattern, the relationship is only moderately strong. (b) The hiker with body weight 187 pounds and pack weight 30 pounds. This hiker makes the form appear to be nonlinear for weights above 140 pounds. Without this hiker, the association would look very linear for all body weights.  3.9  (a) A scatterplot is shown below. (b) The relationship is curved. Large amounts of fuel were used for low and high values of speed and smaller amounts of fuel were used for moderate speeds. This makes sense because the best fuel efficiency is obtained by driving at moderate speeds. (c) Both directions are present in the scatterplot. The association is negative for lower speeds and positive for higher speeds. (d) The relationship is very strong, with little ­deviation from a curve that can be drawn through the points. Fuel used (liters/100 km)

Area = 0.25

3.1  Explanatory: water temperature (quantitative). Response: weight change (quantitative). 3.3  (a) Positive. Students with higher IQs tend to have higher GPAs and vice versa because both IQ and GPA are related to ­mental ability. (b) Roughly linear, because a line through the ­scatterplot of points would provide a good summary. Moderately strong, because most of the points would be close to the line. (c) IQ ≈ 103 and GPA ≈ 0.4. 3.5  A scatterplot is shown below.

22.5 20.0 17.5 15.0 12.5 10.0 7.5 5.0 0

20 40 60 80 100 120 140 160

Speed (km/h)

3.11  (a) Most of the southern states fall in the same pattern as the rest of the states. However, southern states typically have lower mean SAT math scores than other states with a similar percent of students taking the SAT. (b) West Virginia has a much lower mean

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Solutions

3.31  b 3.33  A histogram is shown below. The distribution is right-skewed, with several possible high outliers. Because of the skewness and outliers, we should use the median (5.4 mg) and IQR (5.5 mg) to describe the center and spread. 

90

25

80

20

Frequency

Percent return

SAT Math score than the other states that have a similar percent of students taking the exam. 3.13  A scatterplot is shown below. There is a negative, linear, moderately strong relationship between the percent returning and the number of breeding pairs.

70 60 50 40 27

28

29

30

31

32

33

Breeding pairs

Humerus length (cm)

90 80 70 60 50 40 45

50

55

60

65

70

75

Femur length (cm)

Mileage (mpg)

3.21  (a) There is a strong, positive linear association between sodium and calories. (b) It increases the correlation. It falls in the linear pattern of the rest of the data and observations with unusually small or unusually large values of x have a big influence on the correlation.  3.23  (a) The correlation would not change, because correlation is not affected by a change of units for either variable. (b) The correlation would not change, because it does not distinguish between explanatory and response variables.  3.25  (a) A scatterplot is shown below. (b)  r = 0 (c) The correlation measures the strength of a linear association, but this plot shows a nonlinear relationship between speed and mileage. 30 29 28 27 26 25 24 20

30

40

50

Speed (mph)

3.27  a 3.29  d

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15 10 5 0 0

8

16

24

32

Weights (mg)

3.15  (a) r = 0.9 (b)  r = 0 (c) r = 0.7 (d) r = − 0.3 (e) r = − 0.9 3.17  (a) Gender is a categorical variable and correlation r is for two quantitative variables.  (b)  The largest possible value of the correlation is r = 1.  (c)  The correlation r has no units. 3.19  (a) The scatterplot below shows a strong, positive linear relationship between the two measurements. It appears that all five specimens come from the same species. (b) The femur measurements have x = 58.2 and sx = 13.2. The humerus measurements have y = 66 and sy = 15.89. The sum of the z-score products is 3.97620, so the correlation coefficient is r = (1∙4)(3.97620) = 0.9941. The very high value of the correlation confirms the strong, positive linear association between femur length and humerus length in the scatterplot from part (a). 

40

S-13

60

Section 3.2 Answers to Check Your Understanding page 168:  1.  40. For each additional week, we predict that a rat will gain 40 grams of weight.  2.  100. The predicted weight for a newborn rat is 100 grams. 3.  y^ = 100 + 40(16) = 740 grams  4.  2 years = 104 weeks, so y^ = 100 + 40(104) = 4260 grams. This is equivalent to 9.4 pounds (about the weight of a large newborn human). This is unreasonable and is the result of extrapolation. page 172:  The answer is given in the text. page 174:  1.  y − y^ = 31,891 − 36,895 = −$5004  2.  The actual price of this truck is $5004 less than predicted based on the number of miles it has been driven.  3.  The truck with 44,447 miles and a price of $22,896. This truck has a residual of −$8120, which means that the line overpredicted the price by $8120. No other truck had a residual that was farther below 0 than this one.  page 176: 1. The backpack for this hiker was almost 4 pounds heavier than expected based on the weight of the hiker.  2. Because there appears to be a negative-positive-negative pattern in the residual plot, a linear model is not appropriate for these data. 

Answers to Odd-Numbered Section 3.2 Exercises 3.35  predicted weight = 80 − 6 (days) 3.37  (a) 1.109. For each 1-mpg increase in city mileage, the predicted highway mileage will increase by 1.109 mpg. (b) 4.62 mpg. This would represent the highway mileage for a car that gets 0 mpg in the city, which is impossible. (c) 22.36 mpg 3.39  (a) −0.0053. For each additional week in the study, the predicted pH decreased by 0.0053 units. (b) 5.43. The predicted pH level at the beginning of the study (weeks = 0) is 5.43. (c) 4.635 3.41  No. 1000 months is well outside the observed time period and we can’t be sure that the linear relationship continues after 150 weeks. 3.43  The line y^ = 1 − x is a much better fit. The sum of squared residuals for this line is only 3, while the sum of squared residuals for y^ = 3 − 2x is 18.  3.45  residual = 5.08 − 5.165 = −0.085. The actual pH value for that week was 0.085 less than predicted. 3.47  (a) The scatterplot (with regression line) is shown below. (b) y^ = 31.9 − 0.304x. (c) For each increase of 1 in the percent of returning birds, the predicted number of new adult birds will decrease by 0.304. (d) residual = 11 − 16.092 = −5.092. In this colony, there were 5.092 fewer new adults than expected based on the percent of returning birds. 

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Solutions 22.5 17.5 15.0 12.5 10.0 7.5 5.0 40

50

60

70

80

Percent return

3.49  (a) Because there is no obvious leftover pattern in the residual plot shown below, a line is an appropriate model to use for these data. (b) The point with the largest residual (66% returning) has a residual of about −6. This means that the colony with 66% returning birds has about 6 fewer new adults than predicted based on the percent returning.  5.0

0.0 –2.5 –5.0 –7.5 40

50

60

70

80

Percent return

3.51  No. Because there is an obvious negative-positive-negative pattern in the residual plot, a linear model is not appropriate for these data.  3.53  (a) There is a positive, linear association between the two variables. There is more variation in the field measurements for larger laboratory measurements. (b) No. The points for the larger depths fall systematically below the line y = x, showing that the field measurements are too small compared to the laboratory measurements. (c) The slope would be closer to 0 and the y intercept would be larger.  3.55  (a) residual = 150.06 − 146.295 = 3.765. Yu-Na Kim’s free skate score was 3.765 points higher than predicted based on her short program score. (b) Because there is no leftover pattern in the residual plot, a linear model is appropriate for these data. (c) When using the least-squares regression line with x = short program score to predict y = free skate score, we will typically be off by about 10.2 points. (d) About 73.6% of the variation in free skate scores is accounted for by the linear model relating free skate scores to short program scores. 3.57  r2 : About 56% of the variation in the number of new adults is accounted for by the linear model relating number of new adults to the percent returning. s: When using the least-squares regression line with x = percent returning to predict y = number of new adults, we will typically be off by 3.67 adults. 3.59  (a) y^ = 266.07 − 6.650x, where y = percent of males that return the next year and x = number of breeding pairs. When x = 30, y^ = 66.57. (b) R-Sq = 74.6% (c) r = − !0.746 = −0.864. The sign is negative because the slope is negative. (d) When using the least-squares regression line with x = number of breeding pairs to predict y = percent returning, we will typically be off by 7.76%. 3.61  (a) y^ = 33.67 + 0.54x. (b) If the value of x is 1 standard deviation below x, the predicted value of y will be r standard deviations of y below y. So the predicted value for the husband is 68.5 − 0.5(2.7) = 67.15 inches. 3.63  (a) r2 = 0.25. About 25% of the variation in husbands’ heights is accounted for by the linear model relating husband’s height to

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Number of clusters of beetle larvae

Residual

2.5

wife’s height. (b) When using the least-squares regression line with x = wife’s height to predict y = husband’s height, we will typically be off by 1.2 inches.  3.65  (a) y^ = x where y = final and x = midterm (b) If x = 50, y^ = 67.1. If x = 100, y^ = 87.6. (c) The student who did poorly on the midterm (50) is predicted to do better on the final (closer to the mean), while the student who did very well on the midterm (100) is predicted to do worse on the final (closer to the mean). 3.67  State: Is a linear model appropriate for these data? If so, how well does the least-squares regression line fit the data? Plan: We will look at the scatterplot and residual plot to see if the association is linear or nonlinear. Then, if a linear model is appropriate, we will use s and r2 to measure how well the line fits the data. Do: The scatterplot below shows a moderately strong, positive linear association between the number of stumps and the number of clusters of beetle larvae. The residual plot doesn’t show any obvious leftover pattern, confirming that a linear model is appropriate.  60 50 40 30 20 10 0 1

2

3

4

5

Number of stumps 10 5

Residual

New adults

20.0

0 –5 – 10 – 15 1

2

3

4

5

Number of stumps

y^ = −1.29 + 11.89x,   where y = number of clusters of beetle larvae and x = number of stumps. s = 6.42, meaning that our predictions will typically be off by about 6.42 clusters when we use the line to predict the number of clusters of beetle larvae from the number of stumps. Finally, r 2 = 0.839, meaning 83.9% of the variation in the number of clusters of beetle larvae is accounted for by the linear model relating number of clusters of beetle larvae to the number of stumps. Conclude: The linear model relating number of clusters of beetle larvae to the number of stumps is appropriate and fits the data well, accounting for more than 80% of the variation in number of clusters of beetle larvae. 3.69  (a) A scatterplot is shown below. There is a moderate, positive linear association between HbA and FBG. There are possible outliers to the far right (subject 18) and near the top of the plot (subject 15). 400 350

FBG mg/ml

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300 250 200 150 100 50 5.0

7.5

10.0

12.5

15.0

17.5

20.0

HbA %

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Solutions

18.7

25.0

3.81  (a) A bar graph is given below. The people who use marijuana more are more likely to have caused accidents. (b) Association does not imply causation. For example, it could be that drivers who use marijuana more often are more willing to take risks than other drivers and that the willingness to take risks is what is causing the higher accident rate.  Accident rate (percent)

35 30 25 20 15 10 5 0 Never

1–10 times

11–50 times

51+ times

Marijuana use

Answers to Chapter 3 Review Exercises R3.1  (a) There is a moderate, positive linear association between gestation and life span. Without the outliers at the top and in the upper right, the association appears moderately strong, positive, and curved. (b) It makes r closer to 0 because it decreases the strength of what would otherwise be a moderately strong positive association. Because this point is close to x but far above y, it won’t affect the slope much but will increase the y intercept. Because it has such a large residual, it increases s. (c) Because it is in the positive, linear pattern formed by most of the data values, it will make r closer to 1. Also, because the point is likely to be above the leastsquares regression line, it will “pull up” the line on the right side,

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dicted based on the age of the car. (c)  r = +"0.837 = 0.915. This shows that there is a strong, positive linear association between the age of cars and their mileage. (d) Yes, because there is no leftover pattern in the residual plot. (e) s = 20,870.5: When using the leastsquares regression line with x = car’s age to predict y = number of miles it has been driven, we will typically be off by about 20,870.5 miles. r2 = 83.7%: About 83.7% of the variability in mileage is accounted for by the linear model relating mileage to age.  R3.4  (a) The scatterplot is shown below. Average March temperature, because changes in March temperature probably have an effect on the date of first bloom. Days in April to first bloom

N(18.7, 4.3)

making the slope larger and the intercept smaller. Because this point is likely to have a small residual, it decreases s.  R3.2  (a) 0.0138. For each increase of 1 meter in dive depth, the predicted duration increases by 0.0138 minutes. (b) The y intercept suggests that a dive of 0 depth would last an average of 2.69 minutes; this obviously does not make any sense. (c) 5.45 minutes (d) If the variables are reversed, the correlation will remain the same.  However, the slope and y intercept will be different.  R3.3  (a) y^ = 3704 + 12,188x, where y represents the mileage of the cars and x represents the age. (b) residual = 65,000 − 76,832 = −11,832. This teacher has driven 11,832 fewer miles than pre-

30 25 20 15 10 5 0

Predicted number of days = 33.12 – 4.69 × temperature 2

3

4

5

6

Temperature in Celsius

(b)  r = −0.85 and y^ = 33.12 − 4.69x, where y represents the number of days and x represents the temperature. r: There is a strong, negative linear association between the average March temperature and the days in April until first bloom. Slope: For every 1° increase in average March temperature, the predicted number of days in April until first bloom decreases by 4.69.  y intercept: If the average March temperature was 0°C, the predicted number of days in April to first bloom is 33.12 (May 3). (c) No, x = 8.2 is well beyond the values of x we have in the data set. (d) residual = 10 − 12.015 = −2.015. In this year, the actual date of first bloom occurred about 2 days earlier than predicted based on the average March temperature. (e) There is no leftover pattern in the residual plot shown below, indicating that a linear model is appropriate.  6 4

Residual

(b)  Because the point is in the positive, linear pattern formed by most of the data values, it makes r closer to 1. Also, because the point is likely to be below the least-squares regression line, it will “pull down” the line on the right side, making the slope closer to 0. Without the outlier, r decreases from 0.4819 to 0.3837 as expected. Likewise, the equation changes from y^ = 66.4 + 10.4x to y^ = 52.3 + 12.1x. (c) The point makes r closer to 0 because it is out of the linear pattern formed by most of the data values. Because this point’s x coordinate is very close to x but the y coordinate is far above y, it won’t influence the slope very much but will increase the y intercept. Without the outlier, r increases from 0.4819 to 0.5684, as expected. Likewise, the equation changes from y^ = 66.4 + 10.4x to y^ = 69.5 + 8.92x.  3.71  a 3.73  c 3.75  d 3.77  b 3.79  For these vehicles, the combined mileage follows a N(18.7, 4.3) distribution and we want to find the percent of cars with lower 25 − 18.7 mileage than 25 (see graph below). z = = 1.47. From 4.3 Table A, the p ­ roportion of z-scores below 1.47 is 0.9292. Using technology: normalcdf(lower:–1000,upper:25,μ:18.7,σ: 4.3) = 0.9286. About 93% percent of vehicles get worse combined mileage than the Chevrolet Malibu. 

S-15

2 0 –2 –4 –6 2

3

4

5

6

Temperature in Celsius

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S-16

Solutions

R3.5  (a) y^ = 30.2 + 0.16x, where y = final exam score and x = total score before the final examination. (b) 78.2 (c) Of all the lines that the professor could use to summarize the relationship between final exam score and total points before the final exam, the leastsquares regression line is the one that has the smallest sum of squared residuals. (d) Because r2 = 0.36, only 36% of the variability in the final exam scores is accounted for by the linear model relating final exam scores to total score before the final exam. More than half (64%) of the variation in final exam scores is not accounted for, so Julie has reason to question this estimate. R3.6  Even though there is a high correlation between number of calculators and math achievement, we shouldn’t conclude that increasing the number of calculators will cause an increase in math achievement. It is possible that students who are more serious about school have better math achievement and also have more calculators. 

Answers to Chapter 3 AP® Statistics Practice Test

Height (centimeters)

T3.1  d  T3.2  e  T3.3  c  T3.4  a  T3.5  a  T3.6  c  T3.7  b  T3.8  e  T3.9  b  T3.10  c  T3.11  (a) A scatterplot with regression line is shown below. (b) y^ = 71.95 + 0.3833x, where y = height and x = age. (c) 255.934 cm, or 100.76 inches (d) This was an extrapolation. Our data were based only on the first 5 years of life and the linear trend will not continue forever. 95.0 92.5 90.0 87.5 85.0 35

40

45

50

55

60

Age (months)

T3.12  (a) The point in the upper-right-hand corner has a very high silicon value for its isotope value.  (b) (i) r would get closer to −1 because it does not follow the linear pattern of the other points. (ii) Because this point is “pulling up” the line on the right side of the plot, removing it will make the slope steeper (more negative) and the y intercept smaller (note that the y axis is to the right of the points in the scatterplot). (iii) Because this point has a large residual, removing it will make s a little smaller. T3.13  (a) y^ = 92.29 − 0.05762x, where y is the percent of the grass burned and x is the number of wildebeest. (b) For every increase of 1000 wildebeest, the predicted percent of grassy area burned decreases by about 0.058. (c) r = −!0.646 = −0.804. There is a strong, negative linear association between the percent of grass burned and the number of wildebeest. (d) Yes, because there is no obvious leftover pattern in the residual plot.

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Chapter 4 Section 4.1 Answers to Check Your Understanding page 213:  1.  Convenience sampling. This could lead the inspector to overestimate the quality of the oranges if the farmer puts the best oranges on top.  2.  Voluntary response sampling. In this case, those who are happy that the UN has its headquarters in the U.S. already have what they want and so are less likely to respond. The proportion who answered “No” in the sample is likely to be higher than the true proportion in the U.S. who would answer “No.” page 223:  1.  You would have to identify 200 different seats, go to those seats in the arena, and find the people who are sitting there, which would take a lot of time.  2.  It is best to create strata where the people within a stratum are very similar to each other but different than the people in other strata. In this case, it would be better to take the lettered rows as the strata because each lettered row is the same distance from the court and so would contain only seats with the same (or nearly the same) ticket price.  3.  It is best if the people in each cluster reflect the variability found in the population. In this case, it would be better to take the numbered sections as the clusters because they include all different seat prices. page 228:  1.  (a) Undercoverage (b) Nonresponse (c) Undercoverage 2.  By making it sound like they are not a problem in the landfill, this question will result in fewer people suggesting that we should ban disposable diapers. The proportion who would say “Yes” to this survey question is likely to be smaller than the proportion who would say “Yes” to a more fairly worded question.

Answers to Odd-Numbered Section 4.1 Exercises 4.1  Population: all local businesses. Sample: the 73 businesses that return the questionnaire. 4.3  Population: the 1000 envelopes stuffed during a given hour. Sample: the 40 randomly selected envelopes. 4.5  This is a voluntary response sample. In this case, it appears that people who strongly support gun control volunteered more often, causing the proportion in the sample to be greater than the proportion in the population. 4.7  This is a voluntary response sample and overrepresents the opinions of those who feel most strongly about the issue being surveyed. 4.9  (a) A convenience sample (b) The first 100 students to arrive at school likely had to wake up earlier than other students, so 7.2 hours is probably less than the true average. 4.11  (a) Number the 40 students from 01 to 40. Pick a starting point on the random number table. Record two-digit numbers, skipping numbers that aren’t between 01 and 40 and any repeated numbers, until you have 5 unique numbers between 01 and 40. Use the 5 students corresponding to these numbers. (b) Using line 107, skip the numbers not in bold: 82 73 95 78 90 20  80  74 75 11 81 67 65 53 00 94 38  31 48 93 60 94 07. Select Johnson (20), Drasin (11), Washburn (38), Rider (31), and Calloway (07). 4.13  (a) Using calculator: Number the plots from 1 to 1410. Use the command randInt(1,1410) to select 141 different integers from 1 to 1410 and use the corresponding 141 plots. (b) Answers will vary. 4.15  (a) False—although, on average, there will be four 0s in every set of 40 digits, the number of 0s can be less than 4 or greater than 4 by chance. (b) True—there are 100 pairs of digits 00 through 99,

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Solutions and all are equally likely. (c) False—0000 is just as likely as any other string of four digits. 4.17  (a) It might be difficult to locate the 20 phones from among the 1000 produced that day. (b) The quality of the phones produced may change during the day, so that the last phones manufactured are not representative of the day’s production. (c) Because each sample of 20 phones does not have the same probability of being selected. In an SRS, it is possible for 2 consecutive phones to be selected in a sample, but this is not possible with a systematic random sample.  4.19  Assign numbers 01 to 30 to the students. Pick a starting point on the random digit table. Record two-digit numbers, skipping any that aren’t between 01 and 30 and any repeated numbers, until you have 4 unique numbers between 01 and 30. Use the corresponding four students. Then assign numbers 0 to 9 to the faculty members. Continuing on the table, record one-digit numbers, skipping any repeated numbers, until you have 2 unique numbers between 0 and 9. Use the corresponding faculty members. Starting on line 123 gives 08-Ghosh, 15-Jones, 07-Fisher, and 27-Shaw for the students and 1-Besicovitch and 0-Andrews for the faculty.  4.21  (a) Use the three types of seats as the strata because people who can afford more expensive tickets probably have different ­opinions about the concessions than people who can afford only the cheaper tickets. (b) A stratified random sample will include seats from all over the stadium, which would make it very timeconsuming to obtain. A cluster sample of numbered sections would be easier to obtain, because the people selected for the sample would be sitting close together.   4.23  No. In an SRS, each possible sample of 250 engineers is equally likely to be selected, including samples that aren’t exactly 200 males and 50 females.  4.25  (a) Cluster sampling. (b) To save time and money. In an SRS, the company would have to visit individual homes all over the rural subdivision instead of only 5 locations.  4.27  (a) It is unlikely, because different random samples will include different students and produce different estimates of the proportion of students who use Twitter. (b) An SRS of 100 students. Larger random samples give us better information about the population than smaller random samples. 4.29  Because you  are sampling only from the lower-priced ticket holders, this will likely produce an estimate that is too small, as fans in the club seats and box seats probably spend more money at the game than fans in cheaper seats.  4.31  (a) 89.1% (b) Because the people who have long commutes are less likely to be at home and be included in the sample, this will likely produce an estimate that is too small. 4.33  We would not expect very many people to claim they have run red lights when they haven’t, but some people will deny running red lights when they have. Thus, we expect that the sample proportion underestimates the true proportion of drivers who have run a red light. 4.35  (a) The wording is clear, but the question is slanted in favor of warning labels because of the first sentence stating that some cell phone users have developed brain cancer. (b) The question is clear, but it is slanted in favor of national health insurance by asserting it would reduce administrative costs and not providing any counterarguments. (c) The wording is too technical for many people to understand. For those who do understand the question, it is slanted because it suggests reasons why one should support recycling. 4.37  c

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S-17

4.39  d 4.41  d 4.43  (a) For each additional day, the predicted sleep debt increases by about 3.17 hours. (b) The predicted sleep debt for a 5-day school week is 2.23 + 3.17(5) = 18.08 hours. This is about 3 hours more than the researcher claimed for a 5-day week, so the students have reason to be skeptical of the research study’s reported results.

Section 4.2 Answers to Check Your Understanding page 237: 1. Experiment, because a treatment (brightness of screen) was imposed on the laptops. 2. Observational study, because students were not assigned to eat a particular number of meals with their family per week.  3.  Explanatory: number of meals per week eaten with their family. Response: GPA. 4. There are probably other variables that are influencing the response variable. For example, students who have part-time jobs may not be able to eat many meals with their families and may not have much time to study, leading to lower grades.  page 247:  1.  Randomly assign the 29 students to two treatments: evaluating the performance in small groups or evaluating the performance alone. The response variable will be the accuracy of their final performance evaluations. To implement this design, use 29 equally sized slips of paper. Label 15 of them “small group” and 14 of them “alone.” Then shuffle the papers and hand them out at random to the 29 students, assigning them to a treatment.  2. The purpose of the control group is to provide a baseline for comparison. Without a group to compare to, it is impossible to determine if the small group treatment is more effective. page 249:  1.  No. Perhaps seeing the image of their unborn child encouraged the mothers who had an ultrasound to eat a better diet, resulting in healthier babies.  2.  No. While the people weighing the babies at birth may not have known whether that particular mother had an ultrasound or not, the mothers knew. This might have affected the outcome because the mothers knew whether they had received the treatment or not.  3.  Treat all mothers as if they had an ultrasound, but for some mothers the ultrasound machine wouldn’t be turned on. To avoid having mothers know the machine was turned off, the ultrasound screen would have to be turned away from all the mothers. 

Answers to Odd-Numbered Section 4.2 Exercises 4.45  Experiment, because students were randomly assigned to the different teaching methods.  4.47  (a) Observational study, because mothers weren’t assigned to eat different amounts of chocolate. (b) Explanatory: the mother’s chocolate consumption. Response: the baby’s temperament. (c) No, this study is an observational study so we cannot draw a cause-and-effect conclusion. It is possible that women who eat chocolate daily have less stressful lives and the lack of stress helps their babies to have better temperaments. 4.49  Type of school. For example, private schools tend to have smaller class sizes and students that come from families with higher socioeconomic status. If these students do better in the future, we wouldn’t know if the better performance was due to smaller class sizes or higher socioeconomic status.  4.51  Experimental units: pine seedlings. Explanatory variable: light intensity. Response variable: dry weight at the end of the study. Treatments: full light, 25% light, and 5% light. 

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S-18

Solutions

4.53  Experimental units: the individuals who were called. Explanatory variables: (1) information provided by interviewer; (2) whether caller offered survey results. Response variable: whether or not the call was completed. Treatments: (1) name/no offer; (2) university/no offer; (3) name and university/no offer; (4) name/ offer; (5) university/offer; (6) name and university/offer.  4.55  Experimental units: 24 fabric specimens. Explanatory variables: (1) roller type; (2) dyeing cycle time; (3) temperature. Response variable: a quality score. Treatments: (1) metal, 30 min, 150º; (2) natural, 30 min, 150º; (3) metal, 40 min, 150º; (4) natural, 40 min, 150º; (5) metal, 30 min, 175º; (6) natural, 30 min, 175º; (7) metal, 40 min, 175º; (8) natural, 40 min, 175º.  4.57  There was no control group. We don’t know if the improvement was due to the placebo effect or if the flavonols actually affected the blood flow.  4.59  (a) Write all names on slips of paper, put them in a container, and mix thoroughly. Pull out 40 slips of paper and assign these ­subjects to Treatment 1. Then pull out 40 more slips of paper and assign these subjects to Treatment 2. The remaining 40 subjects are assigned to Treatment 3. (b) Assign the students numbers from 1 to 120. Using the command RandInt (1,120) on the calculator, assign the students corresponding to the first 40 unique numbers chosen to Treatment 1, the students corresponding to the next 40 unique numbers chosen to Treatment 2, and the remaining 40 students to Treatment 3. (c) Assign the students numbers from 001 to 120. Pick a spot on Table D and read off the first 40 unique numbers between 001 and 120. The students corresponding to these numbers are assigned to Treatment 1. The students corresponding to the next 40 unique numbers between 001 and 120 are assigned to Treatment 2. The remaining 40 students are assigned to Treatment 3.  4.61  Random assignment. If players are allowed to choose which treatment they get, perhaps the more motivated players will choose the new method. If they improve more by the end of the study, the coach can’t be sure if it was the exercise program or player motivation that caused the improvement.  4.63  Comparison: Researchers used a design that compared a lowcarbohydrate diet with a low-fat diet. Random assignment: Subjects were randomly assigned to one of the two diets. Control: The experiment used subjects who were all obese at the beginning of the study and who all lived in the same area. Replication: There were 66 subjects in each treatment group. 4.65  Write the names of the patients on 36 identical slips of paper, put them in a hat, and mix them well. Draw out 9 slips. The corresponding patients will receive the antidepressant. Draw out 9 more slips. Those patients will receive the antidepressant plus stress management. The patients corresponding to the next 9 slips drawn will receive the placebo, and the remaining 9 patients will receive the placebo plus stress management. At the end of the experiment, record the number and severity of chronic tension-type headaches for each of the 36 subjects and compare the results for the 4 groups.  4.67  (a) Other variables include expense and condition of the patient. For example, if a patient is in very poor health, a doctor might choose not to recommend surgery because of the added complications. Then we won’t know if a higher death rate is due to the treatment or the initial health of the subjects. (b) Write the names of all 300 patients on identical slips of paper, put them in a hat, and mix them well. Draw out 150 slips and assign the corresponding subjects to receive surgery. The remaining 150 subjects receive the new method. At the end of the study, count how many patients survived in each group.

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4.69  The subjects developed rashes on the arm exposed to the placebo (a harmless leaf) simply because they thought they were being exposed to a poison ivy leaf. Likewise, most of the subjects didn’t develop rashes on the arm that was exposed to poison ivy because they didn’t think they were being exposed to the real thing.  4.71  Because the experimenter knew which subjects had learned the meditation techniques, he is not blind. If the experimenter believed that meditation was beneficial, he may subconsciously rate subjects in the meditation group as being less anxious. 4.73  (a) To make sure that the two groups were as similar as possible before the treatments were administered. (b) The difference in weight loss was larger than would be expected due to the chance variation created by the random assignment to treatments. (c) Even though the low-carb dieters lost 2 kg more over the year than the low-fat group, a difference of 2 kg could be due just to chance variation created by the random assignment.  4.75  (a) The different diagnoses, because the treatments were randomly assigned to patients within each diagnosis. (b) Using a randomized block design allows us to account for the variability in response due to differences in diagnosis by initially comparing the results within each block. In a completely randomized design, this variability will be unaccounted for, making it harder to determine if there is a difference in health and satisfaction due to the difference between doctors and nurse-practitioners.  4.77  (a) A randomized block design would help us account for the variability in yield that is due to the differences in fertility in the field, making it easier to determine if one variety is better than the others. (b) The rows. There should be a stronger association between row number and yield than column number and yield. (c) Let the digits 1 to 5 correspond to the five corn varieties A to E. Begin with line 111 on the random digit table, and assign the letters to the top row from left to right, ignoring numbers 0 and 6−9 and repeated numbers. Use a different line (111, 112, 113, 114, and 115) for each row. Top row (left to right): ADECB, second row: ECDAB, third row: BEDCA, fourth row: DEACB, bottom row: ADCBE.  4.79  (a) If all rats from litter 1 were fed Diet A and if these rats gained more weight, we would not know if this was because of the diet or because of genetics and initial health. (b) Use a randomized block design with the litters as blocks. For each of the litters, randomly assign half of the rats to receive Diet A and the other half to receive Diet B. This will allow researchers to account for the differences in weight gain caused by the differences in genetics and initial health. 4.81  (a) Matched pairs design. (b) In a completely randomized design, the differences between the students will add variability to the response, making it harder to detect if there is a difference caused by the treatments. In a matched pairs design, each student is compared with himself (or herself), so the differences between students are accounted for. (c) If all the students used the handsfree phone during the first session and performed worse, we wouldn’t know if the better performance during the second session is due to the lack of phone or to learning from their mistakes the first time. By randomizing the order, some students will use the hands-free phone during the first session and others during the second session. (d) The simulator, route, driving conditions, and traffic flow were all kept the same for both sessions, preventing these variables from adding variability to the response variable.

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Solutions 4.83  (a) Randomly assign the 20 subjects into two groups of 10. Write the name of each subject on a note card, shuffle the cards, and select 10 to be assigned to the 70º environment.  The remaining 10 subjects will be assigned to the 90º environment. Then the number of correct insertions will be recorded for each subject and the two groups compared. (b) All subjects will perform the task twice, once in each temperature condition. Randomly choose the order by flipping a coin. Heads: 70º, then 90º. Tails: 90º, then 70º. For each subject, compare the number of correct insertions in each environment. 4.85  (a) If the students find a difference between the two groups, they will not know if the difference is due to gender or the deodorant. (b) Each student should have one armpit randomly assigned to receive Deodorant A and the other Deodorant B. Because each gender uses both deodorants, there is no longer any confounding between gender and deodorant. 4.87  c 4.89  b 4.91  c 4.93  b 4.95  (a) For these seeds, the weights follow a N(525, 110) distribution and we want the proportion of seeds that weigh more than 500 mg 500 − 525 = − 0.23. From Table A, the (see graph below). z = 110 ­proportion of z-scores greater than −0.23 is 1 − 0.4090 = 0.5910. Using technology normalcdf(lower:500,upper:10000, μ:525,σ:110) = 0.5899. About 59% of seeds will weigh more than 500 mg.

N(525,110)

S-19

4.99  Because this study involved random assignment to the treatments, we can infer that the difference between foster care or institutional care caused the difference in response.  4.101  Because this study did not involve random assignment to a treatment, we cannot infer cause and effect. Also, because the individuals were not randomly chosen, we cannot generalize to a larger population. 4.103  As daytime running lights become more common, they may be less effective at catching the attention of other drivers. Also, a driving simulator might not be very realistic.  4.105  Answers will vary.  4.107  Answers will vary.  4.109  Confidential. The person taking the survey knows who is answering the questions, but will not share the results of individuals with anyone else. 4.111  The subjects were not able to give informed consent. They did not know what was happening to them and they were not old enough to understand the ramifications. 4.113  The conditional distributions for males and females are displayed in the table and graph below. Men are more likely to view animal testing as justified if it might save human lives: over twothirds of men agree or strongly agree with this statement, compared to slightly less than half of the women. The percentages who disagree or strongly disagree tell a similar story: 16% of men versus 30% of women. Response

Male

Female

Strongly agree

14.7%

9.3%

Agree

52.3%

38.8%

Neither

16.9%

21.9%

Disagree

11.8%

19.3%

Strongly disagree

4.3%

10.7%

500 525

60

Weight (mg)

40 30 20 10

ro ng l

y

ag r A ee g N ree ro D eith ng is e ly ag r di re sa e gr St ee ro ng ly ag r A ee g N ree St ro D eith ng is e ly ag r di re sa e gr ee

0

St

St

(b)  For these seeds, the weights follow a N(525, 110) distribution and we are looking for the boundary value x that has an area of 0.10 to the left (see graph below). A z-score of −1.28 gives the closest x − 525 gives x = 384.2. value to 0.10 (0.1003). Solving − 1.28 = 110 Using technology: invNorm(area:0.10,μ:525,σ:110) = 384.0. The smallest weight among the remaining seeds should be about 384 mg.  

Percent

50

Male

Female

Answers to Chapter 4 Review Exercises N(525,110)

Area = 0.10

525

Weight (mg)

Section 4.3 Answers to Odd-Numbered Section 4.3 Exercises 4.97  If the study involves random sampling, we can make inferences about the population from which we sampled. If the study involves random assignment, we can make inferences about cause and effect.

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R4.1  (a) Population: all Ontario residents. Sample: the 61,239 people interviewed. (b) Because different samples will produce different estimates, it is unlikely that the percentages in the entire population would be exactly the same as the percentages in the sample. However, they should be fairly close.  R4.2  (a) Announce in a daily bulletin that there is a survey concerning student parking available in the main office for students who want to respond. Because those who feel strongly are more likely to respond, their opinions will be overrepresented. (b) Interview a group of students as they come in from the parking lot. People who already can park on campus might have different opinions about the parking situation than those who cannot. 

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Solutions

R4.3  (a) Number the players from 01 to 25 in alphabetical order. Move from left to right, reading pairs of digits until you find three different pairs between 01 and 25, and select the corresponding players. (b) 17 (Musselman), 09 (Fuhrmann), and 23 (Smith). R4.4  Stratified, because it is likely that the opinions of professors will vary based on which type of institution they are at. Then a stratified random sample will provide a more precise estimate than the other methods. Furthermore, the other methods might miss faculty from one particular type of institution.  R4.5  (a) People may not remember how many movies they watched in a movie theater in the past year.  So shorten the amount of time that they ask about, perhaps 3 or 6 months. (b) This will underrepresent younger adults who use only cell phones. If younger adults go to movies more often than older adults, the estimated mean will be too small. (c) Because the frequent moviegoers will not be at home to respond, the estimated mean will be too small.  R4.6  (a) Different anesthetics were not randomly assigned to the subjects. (b) Type of surgery. If Anesthesia C is used more often with a type of surgery that has a higher death rate, we wouldn’t know if the death rate was higher because of the anesthesia or the type of surgery. R4.7  (a) Units: potatoes. Explanatory: storage method and time from slicing until cooking. Response:  ratings of color and flavor. Treatments: (1) fresh/immediately, (2) fresh/after an hour, (3) room temperature/immediately, (4) room temperature/after an hour, (5) refrigerator/immediately, (6) refrigerator/after an hour. (b) Using 300 identical slips of paper, write “1” on 50 of them, “2” on 50 of them, and so on. Put the papers in a hat and mix well. Then select a potato and randomly select a slip from the hat to determine which treatment that potato will receive. Repeat this process for the remaining 299 potatoes, making sure not to replace the slips of paper into the hat. (c) Use a randomized block design with regular potatoes in one block and sweet potatoes in the other block. Randomly assign the 6 treatments within each block as in part (b). R4.8  (a) No. The 1000 students were not randomly selected from any larger population. (b) Yes. The students were randomly assigned to the three treatments. R4.9  (a) By giving some patients a treatment that should have no effect at all, but appears like the Saint-John’s-wort, the researchers can account for the expectations of patients (the placebo effect) by comparing the results for the two groups. (b) To create two groups of subjects that are roughly equivalent at the beginning of the experiment. (c) The subjects should not know which treatment they are getting so that the researchers can account for the placebo effect. The researchers should be unaware of which subjects received which treatment so that they cannot influence how the results are measured. (d) The difference in improvement between the two groups wasn’t large enough to rule out the chance variation caused by the random assignment to treatments. R4.10  (a) Randomly assign 15 students to easy mazes and the other 15 to hard mazes. Use 30 identical slips of paper and write the name of each subject on a slip. Mix the slips in a hat, select 15 of them at random, and assign these subjects to hard mazes. The remaining 15 will be assigned to easy mazes. After the experiment, compare the time estimates of the two groups. (b) Each student does the activity twice, once with each type of maze. Randomly determine which set of mazes is used first by flipping a coin for each subject. Heads: easy, then hard. Tails: hard, then easy. After the experiment, compare each student’s easy maze and hard maze time estimate. (c) The matched pairs design would be more likely to

Starnes-Yates5e_Answer_S1-S24hr1.indd 20

detect a difference because it accounts for the variability between subjects. R4.11  (a) This does not meet the requirements of informed consent because the subjects did not know the nature of the experiment before they agreed to participate. (b) All individual data should be kept confidential and the experiment should go before an institutional review board before being implemented. 

Answers to Chapter 4 AP® Statistics Practice Test T4.1  c  T4.2  e  T4.3  d  T4.4  c  T4.5  b  T4.6  b  T4.7  d  T4.8  d  T4.9  d  T4.10  b  T4.11  d  T4.12  (a) Experimental units: acacia trees. Treatments: placing either active beehives, empty beehives, or nothing in the trees. Response: damage to the trees caused by elephants. (b) Assign the trees numbers from 01 to 72 and use a random number table to pick 24 different two-digit numbers in this range. Those trees will get the active beehives. The trees corresponding to the next 24 different two-digit numbers from 01 to 72 will get the empty beehives, and the remaining 24 trees will remain empty. Compare the damage caused by elephants to the three groups of trees. T4.13  (a) Not all possible samples of size 1067 were possible. For example, using their method, they could not have had all respondents from the east coast. (b) If the household members who typically answer the phone have a different opinion than those who don’t typically answer the phone, their opinions will be overrepresented. (c) If people without phones or with cell phones only have different opinions than the group of people with residential lines, these opinions will be underrepresented.  T4.14  (a) Each of the 11 individuals will be a block in this matched pairs design, with the order of treatments randomly assigned. This was to help account for the variability in tapping speed caused by the differences in subjects. (b) If all the subjects got caffeine the second time, the researchers wouldn’t know if the increase was due to the caffeine or due to practice with the task. (c) Yes. Neither the subjects nor the people who come in contact with them during the experiment (including those who record the number of taps) need to know the order in which the caffeine or placebo was administered.

Answers to Cumulative AP® Practice Test 1 AP1.1  d AP1.2  e AP1.3  b AP1.4  c AP1.5  a AP1.6  c AP1.7  e AP1.8  e AP1.9  d AP1.10  d AP1.11  d

11/21/13 2:45 PM

Solutions AP1.12  b AP1.13  b AP1.14  a AP1.15  (a) The distribution of gains for subjects using Machine A is roughly symmetric, while the distribution of gains for subjects using Machine B is skewed to the left. The center is greater for Machine B than for Machine A. The distribution for Machine B is more variable than the distribution for Machine A. (b) B. The typical gain using Machine B is greater than the typical gain using Machine A.  (c) A. The spread for Machine A is less than the spread for Machine B. (d) Volunteers from one fitness center were used and these volunteers may be different in some way from the general population of those who are interested in cardiovascular fitness. To broaden their scope of inference, they should randomly select people from the population they would like to draw an inference about. AP1.16  (a) Number the 60 retail sales districts with a two-digit number from 01 to 60. Using a table of random digits, read twodigit numbers until 30 unique numbers from 01 to 60 have been selected. The corresponding 30 districts are assigned to the monetary incentives group and the remaining 30 to the tangible incentives group. After a specified period of time, record the change in sales for each district and compare the two groups. (b) The districts labeled 07, 51, and 18 are the first three to be assigned to the monetary incentives group. (c) Pair the two districts with the largest sales, the next two largest, down to the two smallest districts. For each pair, pick one of the districts and flip a coin. If the flip is “heads,” this district is assigned to the monetary incentives group. If it is “tails,” this district is assigned to the tangible incentives group. The other district in the pair is assigned to the other group. After a specified period of time, record the change in sales for each district and compare within each pair.  AP1.17  (a) There is a very strong, positive, linear association between sales and shelf length.  (b) y^ = 317.94 + 152.68x, where y = weekly sales (in dollars) and x = shelf length (in feet). (c) $1081 (d) When using the least-squares regression line with x = shelf space to predict y = sales, we will typically be off by about s = $23. (e) $$$  About 98.2% of the variation in weekly sales revenue can be accounted for by the linear model relating sales to shelf length. (f) It would be inappropriate to interpret the intercept, because the data represent sales based on shelf lengths of 3 to 6 feet and 0 feet falls substantially outside that domain.

Chapter 5 Section 5.1 Answers to Check Your Understanding page 292:  1. (a) If you asked a large sample of U.S. adults whether they usually eat breakfast, about 61% of them will answer yes. (b) The exact number of breakfast eaters will vary from sample to sample.  2. (a) 0. If an outcome can never occur, then it will occur in 0% of the trials. (b) 1. If an outcome will occur on every trial, then it will occur in 100% of the trials. (c) 0.01. An outcome that occurs in 1% of the trials is very unlikely, but will occur every once in a while. (d) 0.6. An outcome that occurs in 60% of the trials will happen more than half of the time. page 299:  1.  Assign the members of the AP® Statistics class the numbers 01−28 and the rest of the students numbers 29−95. Ignore the numbers 96−99 and 00. In Table D, read off 4 two-digit numbers, making sure that the second number is different than the first and that the fourth number is different than the third. Record

Starnes-Yates5e_Answer_S1-S24hr1.indd 21

S-21

whether all four numbers are between 01 and 28 or not.  2. Assign the numbers 1−10 to Jeff Gordon, 11−40 to Dale Earnhardt, Jr., 41−60 to Tony Stewart, 61−85 to Danica Patrick, and 86−100 to Jimmie Johnson. Then proceed as in the example. 

Answers to Odd-Numbered Section 5.1 Exercises 5.1  (a) If we use a polygraph machine on many, many people who are all telling the truth, the machine will say about 8% of the people are lying. (b) Answers will vary. A false positive would mean that a person telling the truth would be found to be lying. A false negative would mean that a person lying would be found to be telling the truth. 5.3  (a) If we look at many families like this, approximately 25% of them will have a first-born child that develops cystic fibrosis. (b) No. The number of children with cystic fibrosis could be smaller or larger than 4 by random chance.  5.5  (a) Answers will vary. (b) Spin the coin many more times. 5.7  In the short run, there was quite a bit of variability in the percentage of made free throws. However, this percentage became less variable and approached 0.30 as the number of shots increased. 5.9  No, he is incorrectly applying the law of large numbers to a small number of at-bats. 5.11  (a) There are 10,000 four-digit numbers (0000, 0001, . . . , 2873, . . . , 9999), and each is equally likely. (b) 2873. To many, 2873 “looks” more random than 9999—we don’t “expect” to get the same number four times in a row. It would be best to choose a number that others would avoid so you don’t have to split the pot with many other people. 5.13  (a) Let diamonds, spades, and clubs represent making a free throw and hearts represent missing. Deal one card from the deck. (b) Let 00−74 represent making the free throw and 75−99 represent missing. Read a two-digit number from Table D. (c) Let 1−3 represent the player making the free throw and 4 represent a miss. Generate a random integer from 1−4. 5.15  (a) There are 19 (not 18) numbers from 00 to 18, 19 (not 18) numbers from 19 to 37, and 3 (not 2) numbers from 38 to 40. (b) Repeats should not be skipped. For example, if the first number selected was 08, then the probability of selecting a left-hander on the next selection would be 9% (instead of 10%). 5.17  (a) Valid. The chance of rolling a 1, 2, or 3 is 75% on a 4-sided die and the 100 rolls represent the 100 randomly selected U.S. adults. (b) Not valid. The probability of heads is 50% rather than 60%. This method will underestimate the number of times she hits the center of the target. 5.19  (a) What is the probability that, in a random selection of 10 passengers, none from first class are chosen? (b) Number the firstclass passengers 01−12 and the other passengers 13−76.  Look up two-digit numbers in Table D until you have 10 unique numbers from 01 to 76.  Count the numbers between 01 and 12. (c) 71 48 70 99 84 29 07 71 48 63 61 68 34 70 52.  There is one person selected who is in first class. (d) It seems plausible that the actual selection was random, because 15/100 is not very small.  5.21  (a) Use a random integer generator to select 30 numbers from 1 to 365. Record whether or not there were any repeats in the sample. (b) Answers will vary. (c) Answers will vary. 5.23  (a) Obtaining a sample percentage of 55% or higher is not particularly unusual (probability ≈ 43∙200) when 50% of all students recycle. (b) Obtaining a sample percentage of at least 63% is very unlikely (probability ≈ 1∙200) when 50% of all students recycle.

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Solutions

5.25  State: What is the probability that, in a sample of 4 randomly selected U.S. adult males, at least one of them is red-green colorblind? Plan: Let 00−06 denote a colorblind man and 07−99 denote a non-colorblind man. Read 4 two-digit numbers from Table D for each sample and record whether or not the sample had at least one red-green colorblind man in it. Do: In our 50 samples, 15 had at least one colorblind man in them. Conclude: The probability that a sample of 4 men would have at least one colorblind man is approximately 15/50 = 0.30. 5.27  State: What is the probability that it takes 20 or more selections in order to find one man who is red-green colorblind? Plan: Let 0−6 denote a colorblind man and 7−99 denote a non-colorblind man. Use technology to pick integers from 0 to 99 until we get a number between 0 and 6. Count how many numbers there are in the sample. Do: In 16 of our 50 samples, it took 20 or more ­selections to get one colorblind man. Conclude: Not surprised. The probability of needing 20 or more selections to get one colorblind man is fairly large (approximately 16/50 = 0.32). 5.29  State: What is the probability that the random assignment will result in at least 6 men in the same group? Plan: Number the men 1−8 and the women 9−20. Use technology to pick 10 unique integers between 1 and 20 for one group. Record if there are at least 6 numbers between 1 and 8 in either group. Do: In our 50 repetitions, 9 had one group with 6 or more men in it. Conclude: Not surprised. The probability of getting 6 or more men in one group is fairly large (approximately 9/50 = 0.18).  5.31  c 5.33  b 5.35  c 5.37  (a) Population: adult U.S. residents. Sample: the 353,564 adults who were interviewed. (b) The people who do not have a telephone were excluded. This would lead to an underestimate of the proportion in the population who experienced stress a lot of the day yesterday if the people without phones are poorer and consequently experience more stress. 

page 309:  1.  A person cannot have a cholesterol level of both 240 or above and between 200 and 239 at the same time.  2.  A person has either a cholesterol level of 240 or above, or they have a cholesterol level between 200 and 239. P(A or B) = 0.16 + 0.29 = 0.45.  3.  P(C) = 1 − 0.45 = 0.55. page 311:  1.  Non-face card

B

Not B

Total

10

10

20

Not E

8

10

18

Total

18

20

38

E

18 20 = 0.474; P(E) = = 0.526. (c) The ball lands in a 38 38 10 spot that is black and even. P(B and E) = = 0.263. (d) If we add 38 the probabilities of B and E, the spots that are black and even will 18 20 10 28 + − = = 0.737. be double-counted. P(B or E) = 38 38 38 38 5.53  (a) (b)  P(B) =

B

M

110

190

130

165

Section 5.2 Answers to Check Your Understanding

Face card

1 1 1 1 4 + + + = = 0.25. 16 16 16 16 16 5.43  (a) Legitimate. (b) Not legitimate: the total is more than 1. (c) Legitimate. 5.45  (a) P(type AB) = 1 − 0.96 = 0.04 (b) P(not type AB) = 1 − P(type AB) = 1 − 0.04 = 0.96 (c) P(type O or B) = 0.49 + 0.20 = 0.69 5.47  (a) 1 − 0.13 − 0.29 − 0.30 = 0.28 (b) Using the complement rule, 1 − 0.13 = 0.87. 275 5.49  (a) P(Female) = = 0.462 (b) P(Eats breakfast regularly) 595 300 110 =  = 0.504. (c) P(Female and breakfast) = = 0.185. 595 595 275 300 110 465 (d) P(Female or breakfast) = + − = = 0.782. 595 595 595 595 5.51  (a)  5.41  P(A) =

430 = 0.723. There is a 0.723 probability that we 595 select a person who is a breakfast eater, a male, or both.  165 (c) P(BC d MC) = = 0.277. There is a 0.277 probability that 595 we select a female who is not a breakfast eater. 5.55  (a) 

(b) P(B c M) =

Total

Heart

3

10

13

Non-heart

9

30

39

Total

12

40

52

2.  P(F and H) = 3/52 = 0.058.  3.  The face cards that are hearts will be double-counted because F and H are not mutually exclu12 13 3 22 sive.  P(F or H) = + − = = 0.423. 52 52 52 52

FB

Not FB

Total

YT

0.66

0.07

0.73

Not YT

0.19

0.08

0.27

Total

0.85

0.15

1

(b) FB

YT

0.19

0.66

0.07

Answers to Odd-Numbered Section 5.2 Exercises 5.39  (a) (1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4), (4,1), (4,2), (4,3), (4,4).  (b) Each outcome has 1 probability . 16

Starnes-Yates5e_Answer_S1-S24hr1.indd 22

0.08

(c)  FB c YT (d)  P(FB c YT) = 0.85 + 0.73 − 0.66 = 0.92.

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Solutions

Average crawling age (weeks)

5.57  c 5.59  c 5.61  The scatterplot for the average crawling age and average temperature is given below.  34 33 32 31

13 = 0.7647. Given that a senator is female, 17 there is a 0.7647 probability that she is a Democrat. 13 (b) P(F 0 D) = = 0.2167. Given that a senator is a Democrat, 60 there is a 0.2167 probability that she is a female.  5.67  (a) P(not English) = 1 − 0.59 = 0.41.

5.65  (a) P(D 0 F) =

(b) P(Spanish 0 other than English) =

30 29 30

40

50

60

70

Average temperature (F )

8

In this scatterplot, there appears to be a moderately strong, negative linear relationship between average temperature and average crawling age. The equation for the least-squares regression line is age = 35.7 − 0.077 (temp). We predict that babies will walk 0.077 weeks earlier for every degree warmer it gets.

Section 5.3 Answers to Check Your Understanding 3656 = 0.3656. There is a 0.3656 probabil10,000 ity of selecting a course grade that is lower than a B. 2.  800 800 P(E 0 L) = = 0.219. P(L 0 E) = = 0.50. P(L 0 E) gives 1600 3656 the probability of getting a lower grade given that the student is studying engineering or physical science. Because this probability (0.50) is greater than P(L) = 0.3656, we can conclude that grades are lower in engineering and physical sciences. 

13 19

35

0.

2.  P(laptop) = 0.30 + 0.175 + 0.175 = 0.65. 0.30 3.  P(made in CA 0 laptop) = = 0.462. 0.65 page 328: 1. Independent. Because we are replacing the cards, knowing what the first card was will not help us predict what the second card will be.  2. Not independent. Once we know the suit of the first card, then the probability of getting a heart on the second card will change depending on what the first card was.  3. Independent. P(right-handed) = 24/28 = 6/7 is the same as P(right-handed | female) = 18/21 = 6/7.  page 331:  1.  P(returned safely) = 0.95. So P(safe return on all 20 missions) = 0.9520 = 0.3585.  2.  No.  Being a college student and being 55 or older are not independent events.

Answers to Odd-Numbered Section 5.3 Exercises 5.63  (a) P(almost certain 0 M) = (b) P(F 0 Some chance) =

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597 = 0.2428. 2459

426 = 0.5983. 712

6 20

14 19

Hard center

Soft center 6 19

Hard center Soft center

5 19

Hard center

88

6 6 14 168 (b) P(one soft d one hard) = A 14 20 B A 19 B + A 20 B A 19 B = 380 = 0.4421  5.77  (a) A tree diagram is below.

Regular gasoline

0.28 Credit card 0.72 No credit card

Mid-grade gasoline

0.34 Credit card 0.66 No credit card

Premium gasoline

0.42 Credit card 0.58 No credit card

0.

40 New York

0.50 Laptops 0.50 Desktops

0. Computers

Candies

Customer

0.02

10

Texas

0.70 Laptops 0.30 Desktops

0.25

Soft center

14 20

0.

California

0.75 Laptops 0.25 Desktops

0.26 = 0.6341. 0.41

5.69  P(B) < P(B 0 T) < P(T) < P(T 0 B). There are very few pro basketball players, so P(B) should be smallest. If you are a pro basketball player, it is quite likely that you are tall, so P(T | B) should be largest. Finally, it’s much more likely to be over 6 feet tall than it is to be a pro basketball player if you’re over 6 feet tall. 0.66 5.71  P(YT 0 FB) = = 0.7765. 0.85 5.73  P(download music) = 0.29, P(don’t care | download music) = 0.67. P(download music d don’t care) = (0.29)(0.67)=0.1943 = 19.43%. 5.75  (a) A tree diagram is below.

page 321:  1. P(L) =

page 326:  1.

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(b)  P(credit card) = (0.88)(0.28) + (0.02)(0.34) + (0.10)(0.42) = 0.0420 0.2952  (c) P(premium gasoline 0 credit card) = = 0.142. 0.2952

5.79  (a) P(lactose intolerant) = (0.82)(0.15) + (0.14)(0.70) + (0.04)(0.90) = 0.257.  0.036 (b)  P(Asian 0 lactose intolerant) = = 0.1401.  0.257 5.81 P(antibody 0 positive) = (0.01)(0.9985) = 0.6270 (0.01)(0.9985) + (0.99)(0.006)

663 1421 = 0.2801  (b) = 0.2944  (c) The events are 2367 4826 not independent because the probabilities in parts (a) and (b) are not the same.  5.85  Not independent. From Exercise 5.65, we saw that 60 P(D 0 F) = 0.7647, which is not the same as P(D) = = 0.60.   100

5.83  (a)

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Solutions

5.87  Independent. P(sum of 7 0 green is 4) = 1∙6 = 0.1667, which equals P(sum of 7) = 6∙36 = 0.1667.  5.89  P(all remain bright) = (0.98)20 = 0.6676 5.91  P(at least one universal donor) = 1 − (0.928)10 = 0.5263  5.93  No, because the events are not independent. If one show starts late, we can predict that the next show will start late as well.  6 1 = = 0.167 5.95  (a) P(doubles) = 36 6 5 1 5 (b) P(no doubles first d doubles second) = a b = = 0.139 6 6 36 5 5 1 25 (c) P(first doubles on third roll) = a b a b = = 0.116 6 6 6 216 5 3 1 5 4 1 (d) 4th: a b a b.  5th: a b a b.  The probability that the first 6 6 6 6 5 k−1 1 doubles are rolled on the kth roll is a b a b. 6 6 5.97  c 5.99  e 5.101 P(at least one is underweight) = 1 − (1 − 0.131)2 = 0.2448

Answers to Chapter 5 Review Exercises R5.1  When the weather conditions are like those seen today, it has rained on the following day about 30% of the time. R5.2  (a) Let the numbers 00−14 represent not wearing a seat belt and 15−99 represent wearing a seat belt. Read 10 sets of two-digit numbers. For each set of 10 two-digit numbers, record whether there are two consecutive numbers between 00−14 or not. (b) The first sample is 29 07 71 48 63 61 68 34 70 52 (not two consecutive). The second sample is 62 22 45 10 25 95 05 29 09 08 (two consecutive). The third sample is 73 59 27 51 86 87 13 69 57 61 (not two consecutive).  R5.3  (a) Difference

1

5

–3

Probability

18 36

6 36

12 36

18 6 24 + = 36 36 36 R5.4  (a) Legitimate. All probabilities are between 0 and 1 and they add up to 1. (b) P(Hispanic) = 0.001 + 0.006 + 0.139 + 0.003 = 0.149 (c) P(not a non-Hispanic white) = 1 − 0.674 = 0.326 (d) People who are white and Hispanic will be double-counted.  P(white or Hispanic) = 0.813 + 0.149 − 0.139 = 0.823. R5.5  (a) (b)  Die A. P(A > B) = P(positive difference) =

Won

R5.6  (a)

Athlete

0 0.1 0.90

Steroids

0.95 + 0.05

No

0.03 + 0.97

–

–

(b) P(+) = (0.10)(0.95) + (0.90)(0.03) = 0.122.  (c) P(steroids 0 +) =

0.095 = 0.7787. 0.122

R5.7  (a) 

Thick

Thin

Total

Mushrooms

2

2

4

No mushrooms

1

2

3

Total

3

4

7

(b) Not independent: P(mushrooms) = P(mushrooms 0 thick crust) =

2 = 0.667. 3

(c) Independent: P(mushrooms) = P(mushrooms 0 thick crust) =

R5.8   (a) P(damage) =

4 = 0.571 does not equal 7

4 1 = = 0.50 is equal to 8 2

2 1 = = 0.50. 4 2

209 = 0.24. 871

(b) P(damage 0 no cover) =

60 = 0.2844, 211

1 76 Padamage 0 < b = = 0.3248, 3 234 Padamage 0

1 2 44 to b = = 0.1991, and 3 3 221

2 29 Padamage 0 > b = = 0.1415. (c) Yes. It appears that deer do 3 205

much more damage when there is no cover or less than 1/3 cover than when there is more cover. R5.9  (a) P(up three consecutive years) = (0.65)3 = 0.274625. 

(b) P(same direction for 3 years) = (0.65)3 + (0.35)3 = 0.3175.

R5.10  (a) (A, A), (A, B), (B, A), (B, B)

Tacos

Blood type

A

AB

B

Probability

0.25

0.5

0.25

(b) (A, A), (A, B), (O, A), (O, B) 15

26

4

36

(b) P(lost and no tacos) = 36/81 = 0.444 (c) P(won or tacos) = 41 30 26 45 + − = = 0.556. 81 81 81 81

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Blood type

A

AB

B

Probability

0.5

0.25

0.25

P(at least 1 type B) = 1 − P(neither are type B) = 1 − (0.75)2 = 0.4375. 

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Solutions

S-25

Answers to Chapter 5 AP® Statistics Practice Test

(b) P(smokes c gets cancer) = 0.25 + 0.12 − 0.08 = 0.29. 

T5.1  e  T5.2  d  T5.3  c  T5.4  b  T5.5  b   T5.6  c  T5.7  e  T5.8  e  T5.9  b  T5.10  c  T5.11  (a) Here is a completed table, with T indicating that the teacher 27 wins and Y indicating that you win. P(teacher wins) = = 0.5625. 48

(c) P(cancer) = 0.12, so P(at least one gets cancer) = 1 −

2

3

4

5

6

7

8

—

T

T

T

T

T

T

T

2

Y

—

T

T

T

T

T

T

3

Y

Y

—

T

T

T

T

T

4

Y

Y

Y

—

T

T

T

T

5

Y

Y

Y

Y

—

T

T

T

6

Y

Y

Y

Y

Y

—

T

T

27 8 5 30 (b) P(A c B) = P(A) + P(B) − P(A d B) = + − = . 48 48 48 48 27 (c) Not independent. P(A) = = 0.5625 does not equal 48 5 P(A 0 B) = = 0.625. 8 T5.12  (a)

B

0.30 Defective 0.70 OK

C

0.40 Defective 0.60 OK

0. Part

0.30

0. 10

0.06 = 0.3158. 0.19 0.09 0.04 P(B 0 defective) = = 0.4737. P(C 0 defective) = = 0.2105. 0.19 0.19

T5.13  (a) Here is a two-way table that summarizes this information: Smokes

Does not smoke

Total

Cancer

0.08

0.04

0.12

No cancer

0.17

0.71

0.88

Total

0.25

0.75

1.00

Starnes-Yates5e_Answer_S25-S34hr.indd 25

page 350:  1.  P(X ≥ 3) is the probability that the student got either an A or a B. P(X ≥ 3) = 0.68. 2.  P(X < 2) = 0.02 + 0.10 = 0.12 3.  The histogram below is skewed to the left. Higher grades are more likely, but there are a few lower grades. 0.4 0.3 0.2 0.1 0.0 0

1

2

3

4

X

page 355: 1.  mX = 1.1. If many, many Fridays are randomly selected, the average number of cars sold will be about 1.1.  2.  sX = !0.89 = 0.943. The number of cars sold on a randomly selected Friday will typically vary from the mean (1.1) by about 0.943 cars. 6.1  (a)

0.19. (c)  Machine B. P(A 0 defective) =

0.08 = 0.32. 0.25

Section 6.1 Answers to Check Your Understanding

Answers to Odd-Numbered Section 6.1 Exercises

(b)  P(defective) = (0.60)(0.10) + (0.30)(0.30) + (0.10)(0.40) =

P(gets cancer 0 smoker) =

Chapter 6

Value

0

1

2

3

4

Probability

1/16

4/16

6/16

4/16

1/16

(b)  The histogram below shows that this distribution is symmetric with a center at 2. 0.4

Probability

0.10 Defective 0.90 OK

60

A

T5.14  (a) Let 00−16 represent out-of-state and 17−99 represent in-state. Read two-digit numbers until you have found two numbers between 00 and 16. Record how many 2-digit numbers you had to read. (b) The first sample is 41 05 09 (it took three cars). The second sample is 20 31 06 44 90 50 59 59 88 43 18 80 53 11 (it took 14 cars). The third sample is 58 44 69 94 86 85 79 67 05 81 18 45 14 (it took 13 cars).

Probability

1 1

P(neither gets cancer) = 1 − 0.882 = 0.2256

0.3 0.2 0.1 0.0 0

1

2

3

4

X

(c)  P(X ≤ 3) = 15∙16 = 0.9375. There is a 0.9375 probability that you will get three or fewer heads in 4 tosses of a fair coin. 6.3  (a) P(X ≥ 1) = 0.9. (b) The event X ≤ 2 is “at most two nonword errors.” P(X ≤ 2) = 0.6. P(X < 2) = 0.3. 6.5  (a) All of the probabilities are between 0 and 1 and they sum to 1. (b) The histogram below is unimodal and skewed to the right.

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S-26

Solutions 0.4

0.25

Probability

Probability

0.30 0.20 0.15 0.10 0.05 0.00 2

4

6

0.1 1 2 3 4 5 6 7 8 9 10

Number of rooms in renter-occupied units

(c)  The event X ≥ 6 is the event that “the first digit in a randomly chosen record is a 6 or higher.” P(X ≥ 6) = 0.222. (d) P(X ≤ 5) = 0.778. 6.7  (a) The outcomes that make up the event A are 7, 8, and 9. P(A) = 0.155. (b) The outcomes that make up the event B are 1, 3, 5, 7, and 9. P(B) = 0.609. (c) The outcomes that make up the event “A or B” are 1, 3, 5, 7, 8, and 9. P(A or B) = 0.660. This is not the same as P(A) + P(B) because the outcomes 7 and 9 are included in both events. 6.9  (a) X

−$1

$2

Probability

0.75

0.25

(b)  E(X) = −$0.25. If the player makes many $1 bets, he will lose about $0.25 per $1 bet, on average. 6.11  mX = 2.1. If many, many undergraduates performed this task, they would make about 2.1 nonword errors, on average. 6.13  (a) This distribution is symmetric and 5 is located at the center. (b) According to Benford’s law, E(X) = 3.441. To detect a fake expense report, compute the sample mean of the first digits. A value closer to 5 suggests a fake report and a value near 3.441 is consistent with a truthful report. (c) P(Y > 6) = 3∙9 = 0.333. Under Benford’s law, P(X > 6) = 0.155. To detect a fake expense report, compute the proportion of first digits that begin with 7, 8, or 9. A value closer to 0.333 suggests a fake report and a value closer to 0.155 is consistent with a truthful report. 6.15  sX = !1.29 = 1.1358. The number of nonword errors in a randomly selected essay will typically differ from the mean (2.1) by about 1.14 words. 6.17  (a) sY = !6.667 = 2.58. (b) sX = !6.0605 = 2.4618. This would not be the best way to tell the difference between a fake and a real expense report because the standard deviations are similar. 6.19  (a) See the following histograms. The distribution of the number of rooms is roughly symmetric for owners and skewed to the right for renters. Renter-occupied units tend to have fewer rooms than owner-occupied units. There is more variability in the number of rooms for owner-occupied units. 0.25

Probability

0.2

0.0

8

X

0.20 0.15 0.10 0.05 0.00 1 2 3 4 5 6 7 8 9 10

Number of rooms in owner-occupied units

Starnes-Yates5e_Answer_S25-S34hr.indd 26

0.3

(b) Owner: mX = 6.284 rooms. Renter: mY = 4.187 rooms. Single people and younger people are more likely to rent and need less space than people with families. (c) sX = !2.68934 = 1.6399. The number of rooms in a randomly selected owner-occupied unit will typically differ from the mean (6.284) by about 1.6399 rooms. sY = !1.71003 = 1.3077. The number of rooms in a randomly selected renter-occupied unit will typically differ from the mean (4.187) by about 1.3077 rooms. 6.21  (a) P(X > 0.49) = 0.51. (b) P(X ≥ 0.49) = 0.51. (c) P(0.19 ≤ X < 0.37 or 0.84 < X ≤ 1.27) = 0.18 + 0.16 = 0.34 6.23  The time Y of a randomly chosen student has the N(7.11, 0.74) 6 − 7.11 distribution. We want to find P(Y < 6). z = = −1.50 and 0.74 P(Z > −1.50) = 0.0668. Using technology: normalcdf(lower: -1000,upper:6,μ:7.11,σ:0.74) = 0.0668. There is about a 7% chance that this student will run the mile in under 6 minutes. 6.25  (a) The speed Y of a randomly chosen serve has the N(115, 6) 120 − 115 distribution. We want to find P(Y > 120). z = = 0.83 6 and P(Z > 0.83) = 0.2033. Using technology: normalcdf (lower:120,upper:1000,μ:115,σ:6) = 0.2023. There is a 0.2023 probability of selecting a serve that is greater than 120 mph. (b) The line above 120 has no area, so P(Y ≥ 120) = P(Y > 120) = 0.2023. (c) We want to find c such c − 115 that P(Y ≤ c) = 0.15. Solving −1.04 = gives c = 108.76. 6 Using technology: invNorm(area:0.15,μ:115,σ:6) = 108.78. Fifteen percent of Nadal’s serves will be less than or equal to 108.78 mph. 6.27  b 6.29  c 6.31  Yes. The mean difference (post − pre) was 5.38 and the median difference was 3. This means that at least half of the students improved their reading scores. 6.33  predicted post-test = 17.897 + 0.78301(pretest).

Section 6.2 Answers to Check Your Understanding page 367: 1.  Y = 500X. mY = 500(1.1) = $550. sY = 500(0.943) = $471.50. 2.  T = Y − 75. mT = 550 − 75 = $475. sT = $471.50. page 376:  1.  mT = 1.1 + 0.7 = 1.8. Over many Fridays, this dealership sells or leases about 1.8 cars in the first hour of business, on average.  2.  s2T = (0.943)2 + (0.64)2 = 1.2988, so sT = "1.2988 = 1.14. 3.  mB = 500(1.1) + 300(0.7) = $760. s2B = (500)2(0.943)2 + (300)2(0.64)2 = 259,176.25, so sB = "259,176.25 = $509.09. page 378: 1.  mD = 1.1 − 0.7 = 0.4. Over many Fridays, this dealer­ship sells about 0.4 cars more than it leases during the first hour of business, on average. 

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Solutions 2.  s2D = (0.943)2 + (0.64)2 = 1.2998, so sD = "1.2998 = 1.14. 3.  mB = 500(1.1) − 300(0.7) = $340. s2B = (500)2(0.943)2 + (300)2(0.64)2 = 259,176.25, so sB = "259,176.25 = $509.09.

Answers to Odd-Numbered Section 6.2 Exercises

6.35  mY = 2.54(1.2) = 3.048 cm and sY = 2.54(0.25) = 0.635 cm. 6.37  (a) The distribution shown below is skewed to the left. Most of the time, the ferry makes $20 or $25.

Probability

0.4 0.3 0.2 0.1 0.0 0

5

10

15

20

25

M

(b)  mM = $19.35. If many ferry trips were selected at random, the ferry would collect about $19.35 per trip, on average. (c) sM = $6.45. The amounts collected on randomly selected ferry trips will typically vary by about $6.45 from the mean ($19.35). 6.39  (a) mG = 5(7.6) + 50 = 88. (b) sG = 5(1.32) = 6.6. (c) s2G = (5sX)2 = 25s2X. The variance of G is 25 times the variance of X. 6.41  (a) mY = − $0.65. If many ferry trips were selected at random, the ferry would lose about $0.65 per trip, on average. (b) sY = $6.45. The amount of profit on randomly selected ferry trips will typically vary by about $6.45 from the mean (−$0.65). 6.43  mY = 6(3.87) − 20 = $3.22. sY = 6(1.29) = $7.74. 6.45  (a) mY = 47.3°F. sY = 4.05 °F. (b) Y has the N(47.3, 4.05) dis40 − 47.3 tribution. We want to find P(Y < 40). z = = − 1.80 4.05 and P(Z < −1.80) = 0.0359. Using technology: normalcdf (lower:-1000,upper:40,μ:47.3,σ:4.05) = 0.0357. There is a 0.0357 probability that the midnight temperature in the cabin is below 40ºF. 6.47  (a) Yes. The mean of a sum is always equal to the sum of the means. (b) No, because it is not reasonable to assume that X and Y are independent. 6.49  mY1 +Y2 = (−0.65) + (−0.65) = − $1.30. s2Y1 +Y2 = 6.452 + 6.452 = 83.205, so sY1 +Y2 = "83.205 = $9.12. 6.51  m3X = 3(2.1) = 6.3 and s3X = 3(1.136) = 3.408. m2Y = 2(1.0) = 2.0 and s2Y = 2(1.0) = 2.0. Thus, m3X+2Y = 6.3 + 2.0 = 8.3 and s23X+2Y = 3.4082 + 2.02 = 15.6145, so s3X+2Y = "15.6145 = 3.95. 6.53  (a) mY−X = 1.0 − 2.1 = − 1.1. If you were to select many essays, there would be about 1.1 fewer word errors than nonword errors, on average. s2Y−X = (1.0)2 + (1.136)2 = 2.2905, so sY−X = !2.2905 = 1.51. The difference in the number errors will typically vary by about 1.51 from the mean (−1.1). (b) The outcomes that make up this event are 1 − 0 = 1, 2 − 0 = 2, 2 − 1 = 1, 3 − 0 = 3, 3 − 1 = 2, 3 − 2 = 1. There is a 0.15 probability that a randomly chosen student will have more word errors than nonword errors. 6.55  The difference in score deductions for a randomly selected essay is 3X − 2Y. m3X = 3(2.1) = 6.3 and s3X = 3(1.136) = 3.408. m2Y = 2(1.0) = 2.0 and s2Y = 2(1.0) = 2.0. Thus, m3X−2Y = 6.3 − 2.0 = 4.3 and s23X−2Y = 3.4082 + 2.02 = 15.6145, so s3X−2Y = !15.6145 = 3.95.

Starnes-Yates5e_Answer_S25-S34hr.indd 27

S-27

6.57  mX1 +X2 = 303.35 + 303.35 = $606.70 and s2X1 +X2 = 9707.572 + 9707.572 = 188,473,830.6, so 1 sX1 +X2 = "188,473,830.6 = $13,728.58. W = (X1 + X2), so 2 1 1 mW = (606.70) = $303.35 and sW = (13,728.58) = $6864.29. 2 2 6.59  (a) Normal with mean = 11 + 20 = 31 seconds and standard deviation = "22 + 42 = 4.4721 seconds. (b)  We want to find the probability that the total time is less than 30 seconds. 30 − 31 z= = − 0.22 and P(Z < −0.22) = 0.4129. Using 4.4721 technology: normalcdf(lower:-1000,upper:30,μ:31,σ: 4.4721) = 0.4115. There is a 0.4115 probability of completing the process in less than 30 seconds for a randomly selected part. 6.61  Let T = the total team swim time. mT = 55.2 + 58.0 + 56.3 + 54.7 = 224.2 seconds and s2T = (2.8)2 + (3.0)2 + (2.6)2 + (2.7)2 = 30.89, so sT = "30.89 = 5.56 seconds. Thus, T has the N(224.2, 5.56) distribution. We want to find P(T < 220). 220 − 224.2 z= = −0.76 and P(Z < −0.76) = 0.2236. Using 5.56 technology: normalcdf(lower:-1000,upper:220,μ: 224.2,σ:5.56) = 0.2250. There is a 0.2250 probability that the total team time is less than 220 seconds in a randomly selected race. 6.63  Let D = X1 − X2 = the difference in NOX levels. mD = 1.4 − 1.4 = 0 and s2X1 −X2 = s2X1 + s2X2 = 0.32 + 0.32 = 0.18, so sX1 −X2 = "0.18 = 0.4243. Thus, D has the N(0, 0.4243) distribution. We want to find P(D > 0.8 or D < −0.8) = P(D > 0.8) + 0.8 − 0 − 0.8 − 0 P(D < −0.8). z = = 1.89 and z = = − 1.89 0.4243 0.4243 and P(Z < −1.89 or Z > 1.89) = 0.0588. Using technology: 1 - normalcdf(lower:-0.8,upper:0.8,μ:0,σ: 0.4243) = 0.0594. There is a 0.0594 probability that the ­difference is at least as large as the attendant observed. 6.65  c 6.67  (a) Fidelity Technology Fund, because its correlation is larger. (b) No, the correlation doesn’t tell us anything about the values of the variables, only about the strength of the linear relationship between them.

Section 6.3 Answers to Check Your Understanding

page 389:  1.  Binomial. Binary? “Success” = get an ace. “Failure” = don’t get an ace. Independent? Because you are replacing the card in the deck and shuffling each time, the result of one trial does not tell you anything about the outcome of any other trial. Number? n = 10. Success? The probability of success is p = 4/52 for each trial.  2.  Not binomial. Binary? “Success” = over 6 feet. “Failure” = not over 6 feet. Independent? Because we are selecting without replacement from a small number of students, the observations are not independent. Number? n = 3. Success? The probability of success will not change from trial to trial.  3.  Not binomial. Binary? “Success” = roll a 5. “Failure” = don’t roll a 5. Independent? Because you are rolling a die, the outcome of any one trial does not tell you anything about the outcome of any other trial. Number? n = 100. Success? No. The probability of success changes when the corner of the die is chipped off. page 397: 1. Binary? “Success” = question answered correctly. “Failure” = question not answered correctly. Independent? The computer randomly assigned correct answers to the questions, so

11/20/13 4:19 PM

S-28

Solutions

knowing the result of one trial (question) should not tell you ­anything about the result on any other trial. Number? n = 10. Success? The probability of success is p = 0.20 for each trial. 10 2.  P(X = 3) = a b (0.2)3(0.8)7 = 0.2013. There is a 20% chance 3 that Patti will answer exactly 3 questions correctly. 3.  P(X ≥ 6) = 1 − P(X ≤ 5) = 1 − 0.9936 = 0.0064. There is only a 0.0064 probability that a student would get 6 or more correct, so we would be quite surprised if Patti was able to pass. page 400:  1.  mX = 10(0.20) = 2. If many students took the quiz, we would expect students to get about 2 answers correct, on average.  2.  sX = "10(0.20)(0.80) = 1.265. If many students took the quiz, we would expect individual students’ scores to typically vary from the mean of 2 correct answers by about 1.265 correct answers.  3.  P(X > 2 + 2(1.265)) = P(X > 4.53) = 1 − P(X ≤ 4) = 1 − 0.9672 = 0.0328. page 408:  1.  Die rolls are independent, the probability of getting doubles is the same on each roll (1/6), and we are repeating the chance process until we get a success (doubles). 5 2 1 2.  P(T = 3) = a b a b = 0.1157. There is a 0.1157 probability 6 6 that you will get the first set of doubles on the third roll of the 1 5 1 5 2 1 dice.  3.  P(T ≤ 3) = + a b a b + a b a b = 0.4213. 6 6 6 6 6

Answers to Odd-Numbered Section 6.3 Exercises

6.69  Binomial. Binary? “Success” = seed germinates and “Failure” = seed does not germinate. Independent? Yes, because the seeds were randomly selected, knowing the outcome of one seed shouldn’t tell us anything about the outcomes of other seeds. Number? n = 20 seeds. Success? p = 0.85. 6.71  Not binomial. Binary? “Success” = person is left-handed and “Failure” = person is right-handed. Independent? Because students are selected randomly, their handedness is independent. Number? There is not a fixed number of trials for this chance process because you continue until you find a left-handed student. Success? p = 0.10. 6.73  (a) Binomial. Binary? “Success” = reaching a live person and “Failure” = any other outcome. Independent? Knowing whether or not one call was completed tells us nothing about the outcome on any other call. Number? n = 15. Success? p = 0.2. (b) This is not a binomial setting because there are not a fixed number of attempts. The Binary, Independent, and Success conditions are satisfied, however, as in part (a). 7 6.75  P(X = 4) = a b(0.44)4(0.56)3 = 0.2304. There is a 0.2304 4 probability that exactly 4 of the 7 elk survive to adulthood. 7 6.77  P(X > 4) = a b(0.44)5(0.56)2 + c= 0.1402. Because this 5 probability isn’t very small, it is not surprising for more than 4 elk to survive to adulthood. 20 b(0.85)17(0.15)3 = 0.2428. 17 20 20 (b) P(X ≤ 12) = a b(0.85)0(0.15)20 + c+ a b(0.85)12(0.15)8 0 12 = 0.0059. Because this is such a low probability, Judy should be suspicious. 6.79  (a) P(X = 17) = a

Starnes-Yates5e_Answer_S25-S34hr.indd 28

6.81  (a) mX = 15(0.20) = 3. If we watched the machine make many sets of 15 calls, we would expect about 3 calls to reach a live person, on average. (b) sX = !15(0.20)(0.80) = 1.55. If we watched the machine make many sets of 15 calls, we would expect the number of calls that reach a live person to typically vary by about 1.55 from the mean (3). 6.83  (a) mY = 15(0.80) = 12. Notice that mX = 3 and 12 + 3 = 15 (the total number of calls). (b) sY = !15(0.80)(0.20) = 1.55. This is the same value as sX , because Y = 15 − X and adding a constant to a random variable doesn’t change the spread. 6.85  (a) Binary? “Success” = win a prize and “Failure” = don’t win a prize. Independent? Knowing whether one bottle wins or not should not tell us anything about the caps on other bottles. Number? n = 7. Success? p = 1∙16. (b) mX = 1.167. If we were to buy many sets of 7 bottles, we would get 1.167 winners per set, on average. sX = 0.986. If we were to buy many sets of 7 bottles, the number of winning bottles would typically differ from the mean (1.167) by 0.986. (c) P(X ≥ 3) = 1 − P(X ≤ 2) = 0.0958. Because 0.0958 isn’t a very small probability, the clerk shouldn’t be surprised. It is p ­ lausible to get 3 or more winners in a sample of 7 bottles by chance alone. 6.87  No. Because we are sampling without replacement and the sample size (10) is more than 10% of the population size (76), we should not treat the observations as independent. 6.89  If the sample is a small fraction of the population (less than 10%), the make-up of the population doesn’t change enough to make the lack of independent trials an issue. 6.91  (a) Binary? “Success” =  visit an auction site at least once a month and “Failure” =  don’t visit an auction site at least once a month. Independent? We are sampling without replacement, but the sample size (500) is far less than 10% of all males aged 18 to 34. Number? n = 500. Success? p = 0.50. (b) np = 250 and n(1 − p) = 250 are both at least 10. (c) mX = 250 and sX = 11.18. Thus, X has approximately the N(250, 11.18) distribution. We want 235 − 250 = −1.34 and P(Z $−1.34) to find P(X ≥ 235). z = 11.18 = 0.9099 Using technology: normalcdf(lower:235, upper:1000,μ:250,σ:11.18) = 0.9102. There is a 0.9102 probability that at least 235 of the men in the sample visit an online auction site. 6.93  Let X be the number of 1s and 2s. Then X has a binomial distribution with n = 90 and p = 0.477 (in the absence of fraud). P(X ≤ 29) = 0.0021. Because the probability of getting 29 or fewer invoices that begin with the digits 1 or 2 is quite small, we have reason to be suspicious that the invoice amounts are not genuine. 6.95  (a) Not geometric. We can’t classify the possible outcomes on each trial (card) as “success” or “failure” and we are not selecting cards until we get a single success. (b) Games of 4-Spot Keno are independent, the probability of winning is the same in each game (p = 0.259), and Lola is repeating a chance process until she gets a success. X = number of games needed to win once is a geometric random variable with p = 0.259. 6.97  (a) Let X = the number of bottles Alan purchases to find one winner. P(X = 5) = (5∙6)4(1∙6) = 0.0804. (b) P(X ≤ 8) = (1∙6) + c + (5∙6)7(1∙6) = 0.7674. 1 6.99  (a) mX = = 10.31. 0.097 (b) P(X ≥ 40) = 1 − P(X ≤ 39) = 0.0187. Because the probability of not getting an 8 or 9 before the 40th invoice is small, we may begin to worry that the invoice amounts are fraudulent. 6.101  b

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Solutions 6.103  d 6.105  c 6.107  (a)

S-29

0−3 = − 2 and P(Z < −2) = 0.0228. 1.5 Using technology: normalcdf(lower:-1000,upper:0, μ:3,σ:1.5) = 0.0228. There is a 0.0228 probability that a randomly selected cap will break when being fastened by the machine.

Managerial & professional

0.20 Smoke 0.80 Don’t smoke

Intermediate

0.29 Smoke 0.71 Don’t smoke

Routine & manual

0.38 Smoke 0.62 Don’t smoke

0.

43

P(C − T < 0). z =

Men

0.34

23

0.

P(smoke) = 0.43(0.20) + 0.34(0.29) + 0.23(0.38) = 0.272 = 27.2% (b) P(routine and manual 0 smoke) =

(0.23)(0.38)  = 0.321 = 32.1% 0.272

Answers to Chapter 6 Review Exercises R6.1  (a) P(X = 5) = 1 − 0.1 − 0.2 − 0.3 − 0.3 = 0.1. (b) Discrete, because it takes a fixed set of values with gaps in between. (c) P(X ≤ 2) = 0.3. P(X < 2) = 0.1. These are not the same because the outcome X = 2 is included in the first­ calculation but not the second. (d) mX = 1(0.1) + c+ 5(0.1) = 3.1. s2X = (1 − 3.1)2(0.1) + c+ (5 − 3.1)2(0.1) = 1.29, so sX = "1.29 = 1.136. R6.2  (a) Temperature is a continuous random variable because it takes all values in an interval of numbers—there are no gaps between possible temperatures. (b) P(X < 540) = P(X ≤ 540) because X is a continuous random variable. In this case, P(X = 540) = 0 because the line segment above X = 540 has no area. (c) Mean = 550 − 550 = 0°C. The standard deviation stays the same, 5.7°C, because subtracting a constant does not change the variability. (d) In degrees Fahrenheit, the mean is 9 mY = (550) + 32 = 1022°F and the standard deviation is 5 9 sY = a b(5.7) = 10.26°F. 5 R6.3  (a) If you were to play many games of 4-Spot Keno, you would get a payout of about $0.70 per game, on average. If you were to play many games of 4-Spot Keno, the payout amounts would typically vary by about $6.58 from the mean ($0.70). (b) Let Y be the amount of Jerry’s payout. mY = 5(0.70) = $3.50 and sY = 5(6.58) = $32.90. (c) Let W be the amount of Marla’s payout. mW = 0.70 + 0.70 + 0.70 + 0.70 + 0.70 = $3.50 and s2W = 6.582 + 6.582 + 6.582 + 6.582 + 6.582 = 216.482, so sW = "216.482 = $14.71. (d) Even though their expected values are the same, the casino would probably prefer Marla since there is less variability in her strategy and her winnings are more predictable. R6.4  (a) C follows a N(10, 1.2) distribution and we want to find 11 − 10 = 0.83 and P(Z > 0.83) = 0.2033. P(C > 11). z = 1.2 Using technology: normalcdf(lower:11,upper:1000, μ:10,σ:1.2) = 0.2023. There is a 0.2023 probability that a randomly selected cap has a strength greater than 11 inch-pounds. (b) The machine that makes the caps and the machine that applies the torque are not the same. (c) C − T is Normal with mean 10 − 7 = 3 inch-pounds and standard deviation "0.92 + 1.22 = 1.5 inch-pounds. (d) We want to find

Starnes-Yates5e_Answer_S25-S34hr.indd 29

R6.5  (a) Binary? “Success” = orange and “Failure” = not orange. Independent? The sample of size n  =  8 is less than 10% of the large bag, so we can assume the outcomes of trials are independent. Number? n  =  8. Success? p  =  0.20. (b) mX = 8(0.2) = 1.6. If we were to select many samples of size 8, we would expect to get about 1.6 orange M&M’S, on average. (c) sX = !8(0.2)(0.8) = 1.13. If we were to select many samples of size 8, the number of orange M&M’S would typically vary by about 1.13 from the mean (1.6). 8 R6.6  (a) P(X = 0) = a b(0.2)0(0.8)8 = 0.1678. Because the prob0 ability is not that small, it would not be surprising to get no orange M&M’S in a sample of size 8. (b) P(X ≥ 5) = 8 a b(0.2)5(0.80)3 + c= 0.0104 Because the probability is small, 5 it would be surprising to find 5 or more orange M&M’S in a sample of size 8. R6.7  Let Y be the number of spins to get a “wasabi bomb.” Y is a 3 geometric random variable with p = = 0.25. P(Y ≤ 3) = 12 (0.75)2(0.25) + (0.75)(0.25) + 0.25 = 0.5781. R6.8  (a) Let X be the number of heads in 10,000 tosses. mX = 10,000(0.5) = 5,000 and sX = "10,000(0.5)(0.5) = 50. (b) np = 10,000(0.5) = 5,000 and n(1 − p) = 10,000(0.5) = 5000 are both at least 10. (c) We want to find P(X ≤ 4933 or X ≥ 5067). 4933 − 5000 5067 − 5000 z= = − 1.34 and z = = 1.34 and 50 50 P(Z ≤ − 1.34) + P(Z ≥ 1.34) = 0.1802. Using technology: 1 - normalcdf(lower:4933,upper:5067,μ:5000, σ:50) = 0.1802. Because this probability isn’t small, we don’t have convincing evidence that Kerrich’s coin was unbalanced—a difference this far from 5000 could be due to chance alone.

Answers to Chapter 6 AP® Statistics Practice Test T6.1  b T6.2  d T6.3  d T6.4  e T6.5  d T6.6  b T6.7  c T6.8  b T6.9  b T6.10  c T6.11  (a) P(Y ≤ 2) = 0.96. (b) mY = 0(0.78) + c= 0.38. If we were to randomly select many cartons of eggs, we would expect about 0.38 to be broken, on average. (c) s2Y = (0 − 0.38)2(0.78) + ... = 0.6756. So sY = "0.6756 = 0.8219. If we were to randomly select many cartons of eggs, the number of broken eggs would typically vary by about 0.6756 from the mean (0.38). (d) Let X stand for the number of cartons inspected to find one carton with at least 2 broken eggs. X is a geometric random variable with p = 0.11. P(X ≤ 3) = (0.11) + (0.89)(0.11) + (0.89)2(0.11) = 0.2950.

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Solutions

T6.12  (a) Binary? “Success” = dog first and “Failure” = not dog first. Independent? We are sampling without replacement, but 12 is less than 10% of all dog owners. Number? n = 12. 12 Success? p = 0.66. (b) P(X ≤ 4) = a b(0.66)0(0.34)12 + c+ 0 12 4 8 a b(0.66) (0.34) = 0.0213. Because this probability is small, it 4 is unlikely to have only 4 or fewer owners greet their dogs first by chance alone. This gives convincing evidence that the claim by the Ladies Home Journal is incorrect. T6.13  (a) mD = 50 − 25 = 25 minutes, s2D = 100 + 25 = 125, and sD = "125 = 11.18 minutes. (b) D follows a N(25, 11.18) dis0 − 25 = − 2.24 tribution and we want to find P(D < 0). z = 11.18 and P(Z < − 2.24) = 0.0125. Using technology: normalcdf (lower:-1000,upper:0,μ:25,σ:11.18) = 0.0127. There is a 0.0127 probability that Ed spent longer on his assignment than Adelaide did on hers. T6.14  (a) Let X stand for the number of Hispanics in the sample. mX = 1200(0.13) = 156 and sX = "1200(0.13)(0.87) = 11.6499. (b) 15% of 1200 is 180, so we want to find P(X ≥ 180) = 1200 a b(0.13)180(0.87)1020 + c= 0.0235. Because this probability 180 is small, it is unlikely to select 180 or more Hispanics in the sample just by chance. This gives us reason to be suspicious about the sampling process.

0.20. The first graph shows the distribution of the colors for one sample and the third graph is centered at 0.40 rather than 0.20. page 434:  1.  No. The mean of the approximate sampling distribution of the sample median (73.5) is not equal to the median of the population (75).  2.  Smaller. Larger samples provide more precise estimates because larger samples include more information about the population distribution.  3.  Skewed to the left and unimodal.

Answers to Odd-Numbered Section 7.1 Exercises 7.1  (a) Population: all people who signed a card saying that they intend to quit smoking. Parameter: the proportion of the population who actually quit smoking. Sample: a random sample of 1000 people who signed the cards. Statistic: the proportion of the sample who actually quit smoking; p^ = 0.21. (b) Population: all the turkey meat. Parameter: minimum temperature in all of the turkey meat. Sample: four randomly chosen locations in the turkey. Statistic: minimum temperature in the sample of four locations; sample minimum = 170°F. 7.3  m = 2.5003 is a parameter and x = 2.5009 is a statistic. 7.5  p^ = 0.48 is a statistic and p = 0.52 is a parameter. 7.7  (a) 2 and 6 (x = 4), 2 and 8 (5), 2 and 10 (6), 2 and 10 (6), 2 and 12 (7), 6 and 8 (7), 6 and 10 (8), 6 and 10 (8), 6 and 12 (9), 8 and 10 (9), 8 and 10 (9), 8 and 12 (10), 10 and 10 (10), 10 and 12 (11), 10 and 12 (11). (b) The sampling distribution of x is skewed to the left and unimodal. The mean of the sampling distribution is 8, which is equal to the mean of the population. The values of x vary from 4 to 11.

Chapter 7 Section 7.1 Answers to Check Your Understanding page 425: 1. Parameter: m = 20 ounces. Statistic: x = 19.6 ounces.  2. Parameter: p = 0.10, or 10% of passengers. Statistic: p^ = 0.08, or 8% of the sample of passengers. page 428:  1.  Individuals: M&M’S Milk Chocolate Candies; variable: color; and parameter of interest: proportion of orange M&M’S. The graph below shows the population distribution. 25

Percent

20 15 10

7.9  (a) In one simulated SRS of 100 students, there were 73 students who did all their assigned homework. (b) The distribution is reasonably symmetric and bell-shaped. It is centered at about 0.60. Values vary from about 0.47 to 0.74. There don’t appear to be any outliers. (c) Yes, because there were no values of p^ less than or equal to 0.45 in the simulation. (d) Because it would be very surprising to get a sample proportion of 0.45 or less in an SRS of size 100 when p = 0.60, we should be skeptical of the newspaper’s claim. 7.11  (a) A graph of the population distribution is shown below.

5 Blue Orange Green Yellow Red

Brown

Color

14 12 10 8 6 4 2 0

0.40 0.20 0 Yes

No

Did homework

(b) Answers will vary. An example bar graph is given. 60

Blue Orange Green Yellow Red Brown

Color

3.  The middle graph is the approximate sampling distribution of p^ because the center of the distribution should be at approximately

Frequency

Frequency

2.  The graph below shows a possible distribution of sample data. 11 = 0.22. For this sample there are 11 orange M&M’S, so p^ = 50

Proportion

0.60

0

50 40 30 20 10 0 Yes

No

Did homework

Starnes/Yates/Moore: The Practice of Statistics, 4E New Fig.: BM_UNCYU_7.1-2.2 Perm. Fig.: 7002 First Pass: 2010-08-13 2nd Pass: 2010-09-23

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Solutions 7.13  (a) Skewed to the right with a center at 9(°F)2 . The values vary from about 2 to 27.5(°F)2. (b) A sample variance of 25(°F)2 provides convincing evidence that the manufacturer’s claim is false and that the thermostat actually has more variability than claimed because a value this large was rare in the simulation. 7.15  If we chose many SRSs and calculated the sample mean x for each sample, we will not consistently underestimate m or consistently overestimate m. 7.17  A larger random sample will provide more information and, therefore, more precise results. 7.19  (a) Statistics ii and iii, because the means of their sampling distributions appear to be equal to the population parameter. (b) Statistic ii, because it is unbiased and has very little variability. 7.21  c 7.23  a 7.25  (a) We are looking for the percentage of values that are 2.5 standard deviations or farther below the mean in a Normal distribution. In other words, we are looking for P(Z #−2.5). Using Table A, P(Z #−2.5) = 0.0062. Using technology: normalcdf(lower: -1000,upper:-2.50,μ:0,σ:1) = 0.0062. Less than 1% of healthy young adults have osteoporosis. (b) Let X be the BMD for women aged 70 − 79 on the standard scale. Then X follows a N(−2, 1) distribution and we want to find P(X #−2.5). − 2.5 − ( − 2) = −0.5 and P(Z #−0.5) = 0.3085. Using 1 technology: normalcdf(lower:-1000,upper:-2.5, z=

μ:-2,σ:1) = 0.3085. About 31% of women aged 70 − 79 have osteoporosis.

Section 7.2 Answers to Check Your Understanding

page 445: 1.  mp^ = p = 0.75.  2. The standard deviation of the sampling distribution of p^ is s p^ =

p(1 − p) Å

n

0.75(0.25) = = Å 1000

0.0137. There are more than 10(1000) = 10,000 young adult Internet users, so the 10% condition has been met.  3.  Yes. Both np = 1000(0.75) = 750 and n(1 − p) = 1000(0.25) = 250 are at least 10. 4.  The sampling distribution would still be approximately Normal with mean 0.75. However, the standard deviation would be smaller by a factor of 3: sp^ =

p(1 − p) Å

n

=

Å

0.75(0.25) = 0.0046. 9000

Answers to Odd-Numbered Section 7.2 Exercises

7.27  (a) We would not be surprised to find 8 (32%) orange candies because values this small happened fairly often in the simulation. However, there were few samples in which there were 5 (20%) or fewer orange candies. So getting 5 orange candies would be surprising. (b) A sample of 50, because we expect to be closer to p = 0.45 in larger samples. p(1 − p) 0.45(0.55) = = 7.29  (a) mp^ = p = 0.45. (b) sp^ = n Å Å 25 0.0995. The 10% condition is met because there are more than 10(25)  =  250 candies in the large machine. (c) Yes, because np = 25(0.45) = 11.25 and n(1 − p) = 25(0.55) = 13.75 are both at least 10. (d) The sampling distribution would still be approximately Normal with a mean of mp^ = 0.45. However, the standard deviation decreases to sp^ =

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p(1 − p) Å

n

=

Å

0.45(0.55) = 0.0497. 100

S-31

7.31  (a) No, because more than 10% of the population (10/76  =  13%) was selected. (b) No, because the sample size was only n =  10. Neither np nor n(1 − p) will be at least 10. 7.33  The Large Counts condition is not met because np = 15(0.3) = 4.5 < 10. 0.7(0.3) 7.35  (a) mp^ = p = 0.70. (b) sp^ = = 0.0144. The 10% Å 1012 condition is met because the sample of size 1012 is less than 10% of the population of all U.S. adults. (c) Yes, because np = 1012(0.70) = 708.4 and n(1 − p) = 1012(0.30) = 303.6 are both at least 10. (d) We want to find P(p^ ≤ 0.67). z = 0.67 − 0.70 = −2.08 and P(Z #−2.08) = 0.0188. Using technol0.0144 ogy: normalcdf(lower:-1000,upper:0.67,μ:0.70,σ: 0.0144) = 0.0186. There is a 0.0186 probability of obtaining a sample in which 67% or fewer say they drink the milk. Because this is a small probability, there is convincing evidence against the claim. 7.37  4048, because using 4n for the sample size halves the standard deviation ("4n = 2"n). 7.39  mp^ = 0.70. Because 267 is less than 10% of the population of 0.7(0.3) = 0.0280. Because np = Å 267 267(0.7) = 186.9 and n(1 − p) = 267(0.3) = 80.1 are both at least 10, the sampling distribution of p^ can be approximated by a Normal 0.75 − 0.7 distribution. We want to find P(p^ ≥ 0.75). z = = 1.79 0.0280 and P(Z ≥ 1.79) = 0.0367. Using technology: normalcdf (lower:0.75,upper:1000,μ:0.7,σ:0.0280) = 0.0371. There is a 0.0371 probability that 75% or more of the women in the sample have been on a diet within the last 12 months. 7.41  (a) mp^ = 0.90. Because 100 is less than 10% of the population college

sp^ =

women,

of orders, sp^ =

0.90(0.10) = 0.03. Because np = 100(0.90) = 90 Å 100

and n(1 − p) = 100(0.10) = 10 are both at least 10, the sampling ­distribution of p^ can be approximated by a Normal distri0.86 − 0.90 bution. We want to find P(p^ ≤ 0.86). z = = −1.33 0.03 and P(Z #−1.33) = 0.0918. Using technology: normalcdf (lower:-1000,upper:0.86,μ:0.90,σ:0.03) = 0.0912. There is a 0.0912 probability that 86% or fewer of orders in an SRS of 100 were shipped within 3 working days. (b) Because the probability isn’t very small, it is plausible that the 90% claim is correct and that the lower than expected percentage is due to chance alone. 7.43  a 7.45  b 7.47  The Venn diagram is shown below.

D

S

17%

12%

9%

62% 62% neither download nor share music files.

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Solutions

Section 7.3 Answers to Check Your Understanding page 456:  1.  X = length of pregnancy follows a N(266, 16) distri270 − 266 bution and we want to find P(X > 270). z = = 0.25 and 16 P(Z > 0.25) = 0.4013. Using technology: normalcdf(lower: 270,upper:1000,μ:266,σ:16) = 0.4013. There is a 0.4013 probability of selecting a woman whose pregnancy lasts for more than 270 days.  2.  m −x = m = 266 days  3.  The sample of size 6 is s 16 less than 10% of all pregnant women, so s −x = = = 6.532 "n "6 days.  4.  x follows a N(266, 6.532) distribution and we want to find 270 − 266 P(x > 270). z = = 0.61 and P(Z > 0.61) = 0.2709. 6.532 Using technology: normalcdf(lower:270,upper:1000, μ:266,σ:6.532) = 0.2701. There is a 0.2701 probability of selecting a sample of 6 women whose mean pregnancy length exceeds 270 days.

Answers to Odd-Numbered Section 7.3 Exercises 7.49  m −x = m = 255 seconds. Because the sample size (10) is less than 10% of the population of songs on David’s iPod, s 60 s −x = = = 18.974 seconds. "n "10 60 60 S "n = = 2 S n = 4. 7.51  30 = 30 "n 7.53  (a) Normal with m −x = m = 188 mg/dl. Because the sample size (100) is less than 10% of all men aged 20 to 34, s −x = s 41 = = 4.1 mg/dl. (b) We want to find P(185 ≤ x ≤ 191). "n "100 185 − 188 191 − 188 z= = − 0.73 and z = = 0.73 4.1 4.1

P(−0.73 ≤ Z ≤ 0.73) = 0.5346. Using technology: normalcdf (lower:185,upper:191,μ:188,σ:4.1) = 0.5357. There is a 0.5357 probability that x estimates m within ±3 mg/dl. s 41 (c) s −x = = = 1.30 mg/dl. So x follows a N(188, 1.30) "n "1000 185 − 188 distribution and we want to find P(185 ≤ x ≤ 191). z = 1.30 191 − 188 = −2.31 and z = = 2.31. P(−2.31 ≤ Z ≤ 2.31) 1.30 = 0.9792. Using technology: normalcdf(lower:185, upper:191,μ:188,σ:1.30) = 0.9790. There is a 0.9790 probability that x estimates m within ±3 mg/dl. The larger sample is better because it is more likely to produce a sample mean within 3 mg/dl of the population mean. 7.55  (a) Let X = amount of cola in a randomly selected bottle. X follows the N(298, 3) distribution and we want to find 295 − 298 P(X < 295). z = = −1 and P(Z < − 1) = 0.1587. 3 Using technology: normalcdf(lower:-1000,upper:295, μ:298,σ:3) = 0.1587. There is a 0.1587 probability that a randomly selected bottle contains less than 295 ml. (b) m −x = m = 298 ml. Because 6 is less than 10% of all bottles pros 3 duced, s −x = = = 1.2247 ml. We want to find P(x < 295) "n "6

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295 − 298 = − 2.45 1.2247 and P(Z < − 2.45) = 0.0071. Using ­technology: normalcdf (lower:-1000,upper:295,μ:298,σ:1.2247) = 0.0072. There is a 0.0072 probability that the mean contents of six randomly selected bottles are less than 295 ml. 7.57  No. The histogram of the sample values will look like the population distribution. The CLT says that the histogram of the sampling distribution of the sample mean will look more and more Normal as the sample size increases. 7.59  (a) Because the distribution of the play times of the population of songs is heavily skewed to the right and n = 10 < 30. (b) Because n = 36 ≥ 30, the CLT applies. m −x = m = 225 seconds. Because 36 is less than 10% of all songs on David’s iPod, s 60 s −x = = = 10 seconds. We want to find P(x > 240) "n "36 240 − 225 using the N(225, 10) distribution. z = = 1.50 and 10 P(Z > 1.50) = 0.0668. Using technology: normalcdf(lower: 240,upper:1000,μ:225,σ:10) = 0.0668. There is a 0.0668 probability that the mean play time is more than 240 seconds. 7.61  (a) We do not know the shape of the distribution of passenger weights. (b) We want to find P(x > 6000∙30) = P(x > 200). Because the sample size is large (n = 30 ≥ 30), the distribution of x is approximately Normal with m −x = m = 190 pounds. Because n = 30 is less than 10% of all possible passengers, using the N(298, 1.2247) distribution. z =

s −x =

s

=

35

= 6.3901 pounds. z =

200 − 190 = 1.56 and 6.3901

"n "30 P(Z > 1.56) = 0.0594. Using ­technology: normalcdf(lower: 200,upper:1000,μ:190,σ:6.3901) = 0.0588. There is a 0.0588 probability that the mean weight exceeds 200 pounds. 7.63  Because the sample size is large (n = 10,000 ≥ 30), the sampling distribution of x is approximately Normal. m −x = m = $250. Assuming 10,000 is less than 10% of all homeowners with fire insurs 1000 = = $10. We want to find P(x ≤ 275) ance, s −x = "n "10,000 275 − 250 using the N(250, 10) distribution. z = = 2.50 and 10 P(Z ≤ 2.50) = 0.9938. Using technology: normalcdf(lower: -1000,upper:275,μ:250,σ:10) = 0.9938. There is a 0.9938 probability that the mean annual loss from a sample of 10,000 policies is no greater than $275. 7.65  b 7.67  b 1062 = 0.0852; high school but 7.69  Didn’t finish high school: 12,470 1977 = 0.0523, less than a bachelor’s degree: no college: 37,834 1462 1097 = 0.0425, college graduate: = 0.0272. The unem34,439 40,390 ployment rate decreases with additional education. 7.71  P(in labor force 0 college graduate) =

40,390 = 0.7830. 51,582

Answers to Chapter 7 Review Exercises R7.1  The population is the set of all eggs shipped in one day. The sample consists of the 200 eggs examined. The parameter is the proportion p = 0.03 of eggs shipped that day that had salmonella.

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Solutions 9 The statistic is the proportion p^ = = 0.045 of eggs in the sam200 ple that had salmonella. R7.2  (a) A sketch of the population distribution is given below.

N(3668, 511)

2135

2646

3157

3668

4179

4690

5201

Birth weight (grams)

(b) Answers will vary. An example dotplot is given. (c) The dot at 2750 represents one SRS of size 5 from this population where the sample range was 2750 grams. 2600

3000

3400

3800

4200

4600

Birth weight (grams)

R7.3  (a) No, because sample range is always less than the actual range (3417). If it were unbiased, the distribution would be c­ entered at 3417. (b) Take larger samples. R7.4  (a) mp^ = p = 0.15. (b) Because the sample size of n = 1540 is less than 10% of the population of all adults, 0.15(0.85) sp^ = = 0.0091. (c) Yes, because np = 1540(0.15) Å 1540 = 231 and n(1 − p) = 1540(0.85) = 1309 are both at least 10.

(d) We want to find P(0.13 ≤ p^ ≤ 0.17). z =

0.13 − 0.15 = −2.20 0.0091

0.17 − 0.15 = 2.20. The desired probability is 0.0091 P(−2.20 ≤ Z ≤ 2.20) = 0.9722. Using technology: normalcdf (lower:0 . 1 3 , u p p e r : 0 . 1 7 , μ : 0 . 1 5 , σ : 0 . 0 0 9 1 ) = 0.9720. There is a 0.9720 probability of obtaining a sample in which between 13% and 17% are joggers. R7.5  (a) mp^ = p = 0.30. Because 100 is less than 10% of the popu0.30(0.70) lation of travelers, sp^ = = 0.0458. Because Å 100 np = 100(0.30) = 30 and n(1 − p) = 100(0.70) = 70 are both at least 10, the sampling distribution of p^ can be approximated by a Normal distribution. We want to find P(p^ ≤ 0.20). 0.20 − 0.30 = −2.18 and P(Z ≤ −2.18) = 0.0146. Using z= 0.0458 technology: normalcdf(lower:-1000,upper:0.20,μ: 0.30,σ:0.0458) = 0.0145. There is a 0.0145 probability that 20% or fewer of the travelers get a red light. (b) Because this is a small probability, there is convincing evidence against the agents’ claim—it isn’t plausible to get a sample proportion of travelers with a red light this small by chance alone. R7.6  (a) X = WAIS score for a randomly selected individual follows a N(100, 15) distribution and we want to find 105 − 100 P(X ≥ 105). z = = 0.33 and P(Z ≥ 0.33) = 0.3707. 15 Using technology: normalcdf(lower:105,upper:1000, μ:100,σ:15) = 0.3694. There is a 0.3694 probability of selecting an individual with a WAIS score of at least 105. (b) m −x = m = 100. Because the sample of size 60 is less than s 15 = = 1.9365. (c) x follows a 10% of all adults, s −x = "n "60 and

z=

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N(100, 1.9365) distribution and we want to find P(x ≥ 105). 105 − 100 z= = 2.58 and P(Z ≥ 2.58) = 0.0049. Using technol1.9365 ogy: normalcdf(lower:105,upper:1000,μ:100,σ: 1.9365) = 0.0049. There is a 0.0049 probability of selecting a sample of 60 adults whose mean WAIS score is at least 105. (d) The answer to part (a) could be quite different depending on the shape of the population distribution. The answer to part (b) would be the same because the mean and standard deviation do not depend on the shape of the population distribution. Because of the large sample size (60 ≥ 30), the answer for part (c) would still be fairly reliable due to the central limit theorem. R7.7  (a) Because n = 50 ≥ 30. (b) m −x = m = 0.5. Because 50 is less than 10% of all traps, the standard deviation is s 0.7 = = 0.0990. Thus, x follows a N(0.5, 0.0990) s −x = "n "50 0.6 − 0.5 ­distribution and we want to find P(x ≥ 0.6). z = = 1.01 0.0990 and P(Z ≥ 1.01) = 0.1562. Using technology: normalcdf (lower:0.6,upper:1000,μ:0.5,σ:0.0990) = 0.1562. There is a 0.1562 probability that the mean number of moths is greater than or equal to 0.6. (c) No. Because this probability is not small, it is plausible that the sample mean number of moths is this high by chance alone.

Answers to Chapter 7 AP® Statistics Practice Test T7.1  c T7.2  c T7.3  c T7.4  a T7.5  b T7.6  b T7.7  b T7.8  e T7.9  c T7.10  e T7.11  A. Both A and B appear to be unbiased, and A has less variability than B. T7.12  (a) We do not know the shape of the population distribution of monthly fees. (b) m −x = m = $38. Because the sample of size 500 is less than 10% of all households with Internet access, s 10 = = 0.4472. (c) Because the sample size is large s −x = "n "500 (n = 500 ≥ 30), the distribution of x will be approximately 39 − 38 = 2.24 and Normal. (d) We want to find P(x > 39). z = 0.4472 P(Z > 2.24) = 0.0125. Using technology: normalcdf(lower: 39,upper:1000,μ:38,σ:0.4472) = 0.0127. There is a 0.0127 probability that the mean monthly fee exceeds $39. T7.13  mp^ = p = 0.22. Because 300 is less than 10% of children 0.22(0.78) = 0.0239. Because under the age of 6, sp^ = Å 300 np = 300(0.22) = 66 and n(1 − p) = 300(0.78) = 234 are both at least 10, the sampling distribution of p^ can be approximated by a Normal distribution. We want to find P(p^ > 0.20). z = 0.20 − 0.22 = − 0.84 and P(Z > − 0.84) = 0.7995. Using tech0.0239 nology: normalcdf(lower:0.20,upper:1000,μ:0.22,σ: 0.0239) = 0.7987. There is a 0.7987 probability that more than 20% of the sample are from poverty-level households.

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Solutions

Answers to Cumulative AP® Practice Test 2

n = 50 is less than 10% of all novels in the library,

AP2.1  a AP2.2  d AP2.3  e AP2.4  b AP2.5  c AP2.6  e AP2.7  c AP2.8  a AP2.9  d AP2.10  c AP2.11  b AP2.12  c AP2.13  d AP2.14  c AP2.15  d AP2.16  c AP2.17  e AP2.18  a AP2.19  c AP2.20  b AP2.21  a AP2.22  (a) Observational study, because no treatments were imposed on the subjects. (b) Two variables are confounded when their effects on the cholesterol level cannot be distinguished from one another. For example, people who take omega-3 fish oil might also exercise more. Researchers would not know whether it was the omega-3 fish oil or the exercise that was the real explanation for lower cholesterol. (c) No. Even though the difference was statistically significant, this wasn’t an experiment and taking fish oil is possibly confounded with exercise. AP2.23  (a) P(type O or Hawaiian-Chinese) = 65,516/145,057 = 0.452. (b) P(type AB | Hawaiian) = 99/4670 = 0.021. (c) P(Hawaiian) = 4670/145,057 = 0.032; P(Hawaiian | type B) = 178/17,604 = 0.010. Because these probabilities are not equal, the two events are not independent. (d) P(type A and white) = 50,008/145,057 = 0.345. P(at least one type A and white) = 1 − P(neither are type A and white) = 1 − (1 − 0.345)2 = 0.571. AP2.24  (a) The distribution of seed mass for the cicada plants is roughly symmetric, while the distribution for the control plants is skewed to the left. The median seed mass is the same for both groups. The cicada plants had a bigger range in seed mass, but the control plants had a bigger IQR. Neither group had any outliers. (b) The cicada plants. The distribution of seed mass for the cicada plants is roughly symmetric, which suggests that the mean should be about the same as the median. However, the distribution of seed mass for the control plants is skewed to the left, which will pull the mean of this distribution below its median toward the lower values. Because the medians of both distributions are equal, the mean for the cicada plants is greater than the mean for the control plants. (c) The purpose of the random assignment is to create two groups of plants that are roughly equivalent at the beginning of the experiment. (d) Benefit: controlling a source of variability. Different types of flowers will have different seed masses, making the response more variable if other types of plants were used. Drawback: we can’t make inferences about the effect of cicadas on other types of plants, because other plants might respond differently to cicadas. AP2.25  (a) We want to find P(x < 25,000∙50) = P(x < 500). Because the sample size is large (n = 50 ≥ 30), the distribution of x is approximately Normal with m −x = m = 525 pages. Because

s 200 500 − 525 = = 28.28 pages. z = = − 0.88 and 28.28 "n "50 P(Z < −0.88) = 0.1894. Using technology: normalcdf(lower: -1000,upper:500,μ:525,σ:28.28) = 0.1883. There is a 0.1883 probability that the total number of pages in 50 novels is fewer than 25,000. (b) Let X be the number of novels that have fewer than 400 pages. X is a binomial random variable with n = 50 and p = 0.30. We want to find P(X ≥ 20). Using technology: P(X ≥ 20) = 1 − P(X ≤ 19) = 1 − binomcdf(trials:50, p:0.30, x value:19) = 0.0848. There is a 0.0848 probability of selecting at least 20 novels that have fewer than 400 pages. Note: Using the Normal approximation, P(X ≥ 20) = 0.0614.

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s −x =

Chapter 8 Section 8.1 Answers to Check Your Understanding page 485:  1.  We are 95% confident that the interval from 2.84 to 7.55 g captures the population standard deviation of the fat content of Brand X hot dogs. 2. If this sampling process were repeated many times, approximately 95% of the resulting confidence intervals would capture the population standard deviation of the fat content of Brand X hot dogs.  3.  False. Once the interval is calculated, it either contains s or it does not contain s.

Answers to Odd-Numbered Section 8.1 Exercises 8.1  Sample mean, x = 30.35. 36 8.3  Sample proportion, p^ = = 0.72. 50 8.5  (a) Approximately Normal with mean m −x = 280 and standard 60 = 2.1. (b) See graph below. (c) About 95% deviation s −x = "840 of the x values will be within 2 standard deviations of the mean. Therefore, m = 2(2.1) = 4.2. (d) About 95%.

273.7 275.8 277.9 280.0 282.1 284.2 286.3

x

8.7  The sketch is given below. The interval with the value of x in the shaded region will contain the population mean (280), while the interval with the value of x outside the shaded region will not contain the population mean (280).

273.7 275.8 277.9 280.0 282.1 284.2 286.3

x

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Solutions 8.9  (a) We are 95% confident that the interval from 0.63 to 0.69 captures the true proportion of those who favor an amendment to the Constitution that would permit organized prayer in public 0.63 + 0.69 = 0.66 and margin schools. (b) Point estimate = p^ = 2 of error = 0.69 − 0.66 = 0.03. (c) Because the value 2/3 = 0.667 (and values less than 2/3) are in the interval of plausible values, there is not convincing evidence that more than two-thirds of U.S. adults favor such an amendment. 8.11  Because only 84% of the intervals actually contained the true parameter, these were probably 80% or 90% confidence intervals. 8.13  Answers will vary. One practical difficulty is response bias: people might answer “yes” because they think they should, even if they don’t really support the amendment. 8.15  Interval: We are 95% confident that the interval from 10.9 to 26.5 captures the true difference (girls − boys) in the mean number of pairs of shoes owned by girls and boys. Level: If this sampling process were repeated many times, approximately 95% of the resulting confidence intervals would capture the true difference (girls − boys) in the mean number of pairs of shoes owned by girls and boys. 8.17  Yes. Because the interval does not include 0 as a plausible value, there is convincing evidence of a difference in the mean number of shoes for boys and girls. 8.19  (a) Incorrect. The interval provides plausible values for the mean BMI of all women, not plausible values for individual BMI measurements. (b) Incorrect. We shouldn’t use the results of one sample to predict the results for future samples. (c) Correct. A confidence interval provides an interval of plausible values for a parameter. (d) Incorrect. The population mean doesn’t change and will either be a value between 26.2 and 27.4 100% of the time or 0% of the time. (e) Incorrect. We are 95% confident that the population mean is between 26.2 and 27.4, but that does not absolutely rule out any other possibility. 8.21  b 8.23  e 8.25  (a) Observational study, because there was no treatment imposed on the pregnant women or the children. (b) No. We cannot make any conclusions about cause and effect because this was not an experiment.

Section 8.2 Answers to Check Your Understanding page 496:  1.  Random: not met because this was a convenience sample. 10%: met because the sample of 100 is less than 10% of the population at a large high school. Large Counts: met because 17 successes and 83 failures are both at least 10. 2. Random: met because the inspector chose an SRS of bags. 10%: met because the sample of 25 is less than 10% of the thousands of bags filled in an hour. Large Counts: not met because there were only 3 successes, which is less than 10. page 499:  1.  p = the true proportion of all U.S. college students who are classified as frequent binge drinkers. 2. Random: met because the statement says that the students were chosen randomly. 10%: met because the sample of 10,904 is less than 10% of all U.S. college students. Large Counts: met because 2486 successes and 1 − 0.99 = 0.005 and the 8418 failures are both at least 10. 3.  2

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S-35

closest area in Table A is 0.0051 (or 0.0049), corresponding to a critical value of z* = 2.57 (or 2.58). Using technology: invNorm(area:0.005, μ:0, σ:1) = −2.576, so z* = 0.228(1 − 0.228) 0.228 ± 2.576 = 0.228 ± 0.010 = Å 10904 (0.218, 0.238).  4. We are 99% confident that the interval from 0.218 and 0.238 captures the true proportion of all U.S. college students who are classified as frequent binge drinkers. 0.80(0.20) ≤ 0.03 for n gives page 503: 1. Solving 1.96 n Å n ≥ 682.95. We should select a sample of at least 683 customers.  2.  The required sample size will be larger because the critical value is larger for 99% confidence (2.576) versus 95% confidence (1.96). The company would need to select at least 1180 customers. 2.576.

Answers to Odd-Numbered Section 8.2 Exercises 8.27  Random: met because Latoya selected an SRS of students. 10%: not met because the sample size (50) is more than 10% of the population of seniors in the dormitory (175). Large Counts: met because np^ = 14 ≥ 10 and n(1 − p^ ) = 36 ≥ 10. 8.29  Random: may not be met because we do not know if the people who were contacted were a random sample. 10%: met because the sample size (2673) is less than 10% of the population of adult heterosexuals. Large Counts: not met because np^ = 2673(0.002) ≈ 5 is not at least 10. 1 − 0.98 8.31  = 0.01, and the closest area is 0.0099, correspond2 ing to a critical value of z* = 2.33. Using technology: invNorm (area:0.01, μ:0, σ:1) = − 2.326, so z* = 2.326. 8.33  (a) Population: seniors at Tonya’s high school. Parameter: true proportion of all seniors who plan to attend the prom. (b) Random: the sample is a simple random sample. 10%: The sample size (50) is less than 10% of the population size (750). Large Counts: np^ = 36 ≥ 10 and n(1 − p^ ) = 14 ≥ 10. 0.72(0.28) (c) 0.72 ± 1.645 = 0.72 ± 0.10 = (0.62, 0.82). Å 50 (d) We are 90% confident that the interval from 0.62 to 0.82 captures the true proportion of all seniors at Tonya’s high school who plan to attend the prom. 8.35  (a) S: p = the true proportion of all full-time U.S. college students who are binge drinkers. P: One-sample z interval for p. Random: the students were selected randomly. 10%: the sample size (5914) is less than 10% of the population of all college students. Large Counts: np^ = 2312 ≥ 10 and n(1 − p^ ) = 3602 ≥ 10. D: (0.375, 0.407). C: We are 99% confident that the interval from 0.375 to 0.407 captures the true proportion of full-time U.S. college students who are binge drinkers. (b) Because the value 0.45 does not appear in our 99% confidence interval, it isn’t plausible that 45% of full-time U.S. college students are binge drinkers. 8.37  Answers will vary. Response bias is one possibility. 8.39  S: p = the true proportion of all students retaking the SAT who receive coaching. P: One-sample z interval for p. Random: the students were selected randomly. 10%: the sample size (3160) is less

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Solutions

than 10% of the population of all students taking the SAT twice. Large Counts: np^ = 427 ≥ 10 and n(1 − p^ ) = 2733 ≥ 10. D: (0.119, 0.151). C: We are 99% confident that the interval from 0.119 to 0.151 captures the true proportion of students retaking the SAT who receive coaching. 8.41  (a) We do not know the sample sizes for the men and for the women. (b) The margin of error for women alone would be greater than 0.03 because the sample size for women alone is smaller than 1019. 0.75(0.25) 8.43  (a) Solving 1.645 ≤ 0.04 gives n ≥ 318. n Å

(b) Solving 1.645 Å

0.5(0.5) ≤ 0.04 gives n ≥ 423. In this case, the n

sample size needed is 105 people larger. 8.45  Solving 1.96 Å

0.5(0.5) ≤ 0.03 gives n ≥ 1068. n

0.64(0.36) gives z* = 2.00. The con8.47  (a) Solving 0.03 = z* Å 1028 fidence level is likely 95%, because 2.00 is very close to 1.96. (b) Teens are hard to reach and often unwilling to participate in surveys, so nonresponse is a major “practical difficulty” for this type of poll. 8.49  a 8.51  c 8.53  (a) A histogram of the number of accidents per hour is given below.

Frequency

7 6 5 4 3 2 1 0 0

5

10

15

20

25

30

Number of accidents/hr

Number of accidents

(b) A graph of the number of accidents is given below. 35 30 25 20 15 10 5 0 2 4 6 8 10 12 14 16 18 20 22 24

Hour of the day

(c) The histogram in part (a) shows that the number of accidents has a distribution that is skewed to the right. (d) The graph in part (b) shows that there is a cyclical nature to the number of accidents.

Section 8.3 Answers to Check Your Understanding page 514: 1. df = 21, t* = 2.189. Using technology: invT (area:0.02, df = 21) = −2.189, so t* = 2.189.  2. df = 70, t* = 2.660 (using df = 60). Using technology: invT(area: 0.005, df = 70) = −2.648, so t* = 2.648. page 522:  S: We are trying to estimate m = the true mean healing rate at a 95% confidence level. P: One-sample t interval for m.

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Random: The description says that the newts were randomly chosen. 10%: The sample size (18) is less than 10% of the population of newts. Normal/Large Sample: The histogram below shows no strong skewness or outliers, so this condition is met. 4

Frequency

S-36

3 2 1 0 10

15

20

25

30

35

40

Healing rate (micrometers/hr)

8.32

b = 25.67 ± 4.14 = (21.53,29.81). C: We "18 are 95% confident that the interval from 21.53 to 29.81 micro­ meters per hour captures the true mean healing time for newts.

D: 25.67 ± 2.110a

page 524: Using s = 154 and z* = 1.645 for 90% confidence, 1.645(154) 2 154 30 ≥ 1.645 . Thus, n ≥ a b = 71.3, so take a sam30 "n ple of 72 students.

Answers to Odd-Numbered Section 8.3 Exercises

8.55  (a) t* = 2.262. (b) t* = 2.861. (c) t* = 1.671 (using technology: t* = 1.665). 8.57  Because the sample size is small (n = 20 < 30) and there are outliers in the data. 8.59  (a) No, because we are trying to estimate a population proportion, not a population mean. (b) No, because the 15 team members are not a random sample from the population. (c) No, because the sample size is small (n = 25 < 30) and there are outliers in the sample. 9.3 = 1.7898. If we take many samples of size 27, 8.61  SE −x = "27 the sample mean blood pressure will typically vary by about 1.7898 from the population mean blood pressure. sx , sx = 91.26 cm. (b) They are using 8.63  (a) Because 19.03 = "23 a critical value of t* = 1. With df = 22, the area between t = −1 and t = 1 is approximately tcdf(lower: -1, upper: 1, df: 22) = 0.67. So, the confidence level is 67%. 8.65  (a) S:m = the true mean percent change in BMC for breastfeeding mothers. P: One-sample t interval form. Random: the mothers were randomly selected. 10%: 47 is less than 10% of all breast-feeding mothers. Normal/Large Sample: n = 47 ≥ 30. D: Using df = 40, (−4.575, −2.605). Using technology: (−4.569, −2.605) with df = 46. C: We are 99% confident that the interval from −4.569 to −2.605 captures the true mean percent change in BMC for breast-feeding mothers. (b) Because all of the plausible values in the interval are negative (indicating bone loss), the data give convincing evidence that breast-feeding mothers lose bone mineral, on average. 8.67  (a) S:m = the true mean size of the muscle gap for the population of American and European young men. P: One-sample t interval for m. Random: the young men were randomly selected. 10%: 200 is less than 10% of young men in America and Europe. Normal/Large Sample: n = 200 ≥ 30. D: Using df = 100, (1.999, 2.701). Using technology: (2.001, 2.699) with df = 199. C: We are

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Solutions 95% confident that the interval from 2.001 to 2.699 captures the true mean size of the muscle gap for the population of American and European young men. (b) The large sample size (n = 200 ≥ 30) allows us to use a t interval for m. 8.69  S:m = the true mean fuel efficiency for this vehicle. P: Onesample t interval for m. Random: the records were selected at random. 10%: it is reasonable to assume that 20 is less than 10% of all records for this vehicle. Normal/Large Sample: the histogram does not show any strong skewness or outliers.

Frequency

5 4 3 2 1 0 13.5

15.5

17.5

19.5

21.5

23.5

Mpg

D: Using df = 19, (17.022, 19.938). C: We are 95% confident that the interval from 17.022 to 19.938 captures the true mean fuel efficiency for this vehicle. 8.71  (a) S: m = the true mean difference in the estimates from these two methods in the population of tires. P: One-sample t interval for m. Random: A random sample of tires was selected. 10%: the sample size (16) is less than 10% of all tires. Normal/Large Sample: The histogram of differences shows no strong skewness or outliers.

Frequency

4 3 2 1 0 0

2

4

6

8

10

Difference (weight – groove)

D: Using df = 15, (2.837, 6.275). C: We are 95% confident that the interval from 2.837 to 6.275 thousands of miles captures the true mean difference in the estimates from these two methods in the population of tires. (b) Because 0 is not included in the confidence interval, there is convincing evidence of a difference in the two methods of estimating tire wear. 7.5 ≤ 1 gives n ≥ 374. 8.73  Solving 2.576 "n 8.75  b 8.77  b 8.79  (a) Because the sum of the probabilities must be 1, P(X = 7) = 0.57. (b) mX = 5.44. If we were to randomly select many young people, the average number of days they watched television in the past 7 days would be about 5.44. (c) Because the sample size is large (n = 100 ≥ 30), we expect the mean number of days x for 100 randomly selected young people (aged 19 to 25) to be approximately Normally distributed with mean m −x = m = 5.44. Because the sample size (100) is less than 10% of all young people aged s 2.14 = = 0.214. 19 to 25, the standard deviation is s −x = "n "100 4.96 − 5.44 = − 2.24 and We want to find P(x ≤ 4.96). z = 0.214

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S-37

P(Z ≤ −2.24) = 0.0125. Using technology: normalcdf(lower: −1000, upper:4.96, μ:5.44, σ:0.214) = 0.0124. There is a 0.0124 probability of getting a sample mean of 4.96 or smaller. Because this probability is small, a sample mean of 4.96 or smaller would be surprising.

Answers to Chapter 8 Review Exercises 1 − 0.94 = 0.03, and the closest area is 0.0301, corre2 sponding to a critical value of z* = 1.88. Using technology: invNorm(area:0.03, μ:0, σ:1) = −1.881, so z* = 1.881. (b) Using Table B and 50 degrees of freedom, t* = 2.678. Using technology: invT(area:0.005, df:57) = −2.665, so t* = 2.665. 430 + 470 R8.2  (a) x = = 450 minutes. Margin of error = 470 2 − 450 = 20 minutes. Because n = 30, df = 29 and t* = 2.045. sx sx Because 20 = 2.045 = 9.780 minutes , standard error = "30 "30 and sx = 53.57 minutes. (b) The confidence interval provided gives an interval estimate for the mean lifetime of batteries produced by this company, not individual lifetimes. (c) No. A confidence interval provides a statement about an unknown population mean, not another sample mean. (d) If we were to take many samples of 30 batteries and compute 95% confidence intervals for the mean lifetime, about 95% of these intervals will capture the true mean lifetime of the batteries. R8.3  (a) p = the proportion of all adults aged 18 and older who would say that football is their favorite sport to watch on television. It may not equal 0.37 because the proportion who choose football will vary from sample to sample. (b) Random: The sample was random. 10%: The sample size (1000) is less than 10% of all adults. Large Counts: np^ = 370 ≥ 10 and n(1 − p^ ) = 630 ≥ 10. 0.37(0.63) = (0.3401, 0.3999). (d) We are 95% (c) 0.37 ± 1.96 Å 1000 confident that the interval from 0.3401 to 0.3999 captures the true proportion of all adults who would say that football is their favorite sport to watch on television. R8.4  (a) m = mean IQ score for the 1000 students in the school. (b) Random: the data are from an SRS. 10%: the sample size (60) is less than 10% of the 1000 students at the school. Normal/Large Sample: n = 60 ≥ 30. (c) Using df = 50, 114.98 14.8 ± 1.676a b = (111.778,118.182). Using technology: (111.79, "60 118.17) with df = 59. (d) We are 90% confident that the interval from 111.79 to 118.17 captures the true mean IQ score for the 1000 students in the school. 0.5(0.5) R8.5  Solving 2.576 ≤ 0.01 gives n ≥ 16,590. Å n R8.6  (a) S: p = the true proportion of all drivers who have run at least one red light in the last 10 intersections they have entered. P: One-sample z interval for p. Random: the drivers were selected at random. 10%: The sample size (880) is less than 10% of all drivers. Large Counts: np^ = 171 ≥ 10 and n(1 − p^ ) = 709 ≥ 10. D: (0.168,0.220). C: We are 95% confident that the interval from 0.168 to 0.220 captures the true proportion of all drivers who have run at least one red light in the last 10 intersections they have R8.1  (a)

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Solutions

Frequency

entered. (b) It is likely that more than 171 respondents have run red lights because some people may lie and say they haven’t run a red light. The margin of error does not account for these sources of bias; it accounts only for sampling variability. R8.7  (a) S: m = the true mean measurement of the critical dimension for the engine crankshafts produced in one day. P: One-sample t interval for m. Random: The data come from an SRS. 10%: the sample size (16) is less than 10% of all crankshafts produced in one day. Normal/Large Sample: the histogram shows no strong skewness or outliers. 9 8 7 6 5 4 3 2 1 0

Forest Service cannot conclude that more than half of visitors to Yellowstone National Park favor the proposal. T8.12  (a) Because the sample size is large (n = 48 ≥ 30), the Normal/Large Sample condition is met. (b) Maurice’s interval uses a z critical value instead of a t critical value. Also, Maurice used the wrong value in the square root—it should be n = 48. Correct: 2.576 Using df = 40, 6.208 ± 2.021a b = (5.457,6.959). Using "48 technology: (5.46, 6.956) with df = 47. T8.13  S: m = the true mean number of bacteria per milliliter of raw milk received at the factory. P: One-sample t interval for m. Random: The data come from a random sample. 10%: the sample size (10) is less than 10% of all 1-ml specimens that arrive at the factory. Normal/Large Sample: the dotplot shows that there is no strong skewness or outliers. 4560 4680 4800 4920 5040 5160 5280 5400

223.90 223.95 224.00 224.05 224.10

Bacteria/ml

Crankshaft measurement (mm)

D: Using df = 15, (223.969, 224.035). C: We are 95% confident that the interval from 223.969 to 224.035 mm captures the true mean measurement of the critical dimension for engine crankshafts produced on this day. (b) Because 224 is a plausible value in this interval, we don’t have convincing evidence that the process mean has drifted. 3000 R8.8 Solving 1.96a b ≤ 1000 gives n ≥ 35. "n R8.9 (a) The margin of error must get larger to increase the capture rate of the intervals. (b) If we quadruple the sample size, the margin of error will decrease by a factor of 2. R8.10 (a) When we use the sample standard deviation sx to estimate the population standard deviation s. (b) The t distributions are wider than the standard Normal distribution and they have a slightly different shape with more area in the tails. (c) As the degrees of freedom increase, the spread and shape of the t distributions become more like the standard Normal distribution.

Answers to Chapter 8 AP® Statistics Practice Test T8.1  a T8.2  d T8.3  c T8.4  d T8.5  b T8.6  a T8.7  c T8.8  d T8.9  e T8.10  d T8.11  (a) S: p = the true proportion of all visitors to Yellowstone who would say they favor the restrictions. P: One-sample z interval for p. Random: the visitors were selected randomly. 10%: the sample size (150) is less than 10% of all visitors to Yellowstone National Park. Large Counts: np^ = 89 ≥ 10 and n(1 − p^ ) = 61 ≥ 10. D: (0.490, 0.696). C: We are 99% confident that the interval from 0.490 to 0.696 captures the true proportion of all visitors who would say that they favor the restrictions. (b) Because there are ­values smaller than 0.50 in the confidence interval, the U.S.

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D: Using df = 9, (4794.37,5105.63). C: We are 90% confident that the interval from 4794.37 to 5105.63 bacterial/ml captures the true mean number of bacteria in the milk received at this factory.

Chapter 9 Section 9.1 Answers to Check Your Understanding

page 541: 1. (a) p = proportion of all students at Jannie’s high school who get less than 8 hours of sleep at night. (b) H0 :p = 0.85 and Ha :p ∙ 0.85.  2. (a) m = true mean amount of time that it takes to complete the census form. (b) H0:m = 10 and Ha:m > 10. page 549:  1.  Finding convincing evidence that the new batteries last longer than 30 hours on average, when in reality their true mean lifetime is 30 hours. 2.  Not finding convincing evidence that the new batteries last longer than 30 hours on average, when in reality their true mean lifetime >30 hours.  3.  Answers will vary. A consequence of a Type I error would be that the company spends the extra money to produce these new batteries when they aren’t any better than the older, cheaper type. A consequence of a Type II error would be that the company would not produce the new batteries, even though they were better.

Answers to Odd-Numbered Section 9.1 Exercises

9.1  H0:m = 115; Ha:m > 115, where m is the true mean score on the SSHA for all students at least 30 years of age at the teacher’s college. 9.3  H0:p = 0.12; Ha:p ∙ 0.12, where p is the true proportion of lefties at his large community college. 9.5  H0: s = 3; Ha: s > 3, where s is the true standard deviation of the temperature in the cabin. 9.7  The null hypothesis is always that there is “no difference” or “no change” and the alternative hypothesis is what we suspect is true. Correct: H0:p = 0.37; Ha:p > 0.37. 9.9  Hypotheses are always about population parameters. Correct: H0:m = 1000 grams; Ha:m < 1000 grams. 9.11  (a) The attitudes of older students do not differ from other students, on average. (b) Assuming the mean score on the SSHA for students at least 30 years of age at this school is really 115, there is a 0.0101 probability of getting a sample mean of at least 125.7 just by chance in an SRS of 45 older students.

11/20/13 5:22 PM

Solutions 9.13  a = 0.10: Because the P-value of 0.2184 > a = 0.10, we fail to reject H0 . We do not have convincing evidence that the proportion of left-handed students at Simon’s college is different from the national proportion. a = 0.05: Because the P-value of 0.2184 > a = 0.05, we fail to reject H0 . We do not have convincing evidence that the proportion of left-handed students at Simon’s college is different from the national proportion. 9.15  a = 0.05: Because the P-value of 0.0101 < a = 0.05, we reject H0 . We have convincing evidence that the true mean score on the SSHA for all students at least 30 years of age at the teacher’s college >115. a = 0.01: Because the P-value of 0.0101 > a = 0.01, we fail to reject H0 . We do not have convincing evidence that the true mean score on the SSHA for all students at least 30 years of age at the teacher’s college >115. 9.17  Either H0 is true or H0 is false—it isn’t true some of the time and not true at other times. 9.19  The P-value should be compared with a significance level (such as a = 0.05), not the hypothesized value of p. Also, the data never “prove” that a hypothesis is true, no matter how large or small the P-value. 9.21  (a) H0:m = 6.7; Ha:m < 6.7, where m represents the mean response time for all accidents involving life-threatening injuries in the city. (b) I: Finding convincing evidence that the mean response time has decreased when it really hasn’t. A consequence is that the city may not investigate other ways to reduce the mean response time and more people could die. II: Not finding convincing evidence that the mean response time has decreased when it really has. A consequence is that the city spends time and money investigating other methods to reduce the mean response time when they aren’t necessary. (c) Type I, because people may end up dying as a result. 9.23  (a) H0:m = $85,000; Ha:m > $85,000, where m = the mean income of all residents near the restaurant. (b) I: Finding convincing evidence that the mean income of all residents near the restaurant exceeds $85,000 when in reality it does not. The consequence is that you will open your restaurant in a location where the residents will not be able to support it. II: Not finding convincing evidence that the mean income of all residents near the restaurant exceeds $85,000 when in reality it does. The consequence of this error is that you will not open your restaurant in a location where the residents would have been able to support it and you lose potential income. 9.25  d 9.27  c 9.29  (a) P(woman) = 0.4168, so (24,611)(0.4168) = 10,258 degrees were awarded to women. (b) No. P(woman) = 0.4168, which is not equal to P(woman 0 bachelors) = 0.43. (c) P(at least 1 of the 2 degrees earned by a woman) = 1 − P(neither degree is earned by a woman) = 14,353 14,352 1− a ba b = 0.6599 24,611 24,610

Section 9.2 Answers to Check Your Understanding

page 560:  S: H0:p = 0.20 versus Ha:p > 0.20, where p is the true proportion of all teens at the school who would say they have electronically sent or posted sexually suggestive images of themselves. P: One-sample z test for p. Random: Random sample. 10%: The sample size (250) < 10% of the 2800 students. Large

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S-39

Counts: 250(0.2) = 50 ≥ 10 and 250(0.8) = 200 ≥ 10. D: 0.252 − 0.20 = 2.06 and P(Z ≥ 2.06) = 0.0197. C: Because z= 0.20(0.80) Å 250 the P-value of 0.0197 < a = 0.05, we reject H0 . We have convincing evidence that more than 20% of the teens in her school would say they have electronically sent or posted sexually suggestive images of themselves. page 563:  S: H0:p = 0.75 versus Ha:p ∙ 0.75, where p is the true proportion of all restaurant employees at this chain who would say that work stress has a negative impact on their personal lives. P: One-sample z test for p. Random: Random sample. 10%: The sample size (100) < 10% of all employees. Large Counts: 100(0.75) = 75 ≥ 10 and 100(0.25) = 25 ≥ 10. 0.68 − 0.75 = − 1.62 and 2P(Z #− 1.62) = 0.1052. C: D: z = 0.75(0.25) Å 100

Because the P-value of 0.1052 > a = 0.05, we fail to reject H0 . We do not have convincing evidence that the true proportion of all restaurant employees at this large restaurant chain who would say that work stress has a negative impact on their personal lives is different from 0.75. page 564:  The confidence interval given in the output includes 0.75, which means that 0.75 is a plausible value for the population proportion that we are seeking. So both the significance test (which didn’t rule out 0.75 as the proportion) and the confidence interval give the same conclusion. The confidence interval, however, gives a range of plausible values for the population proportion instead of only making a decision about a single value. page 569:  1.  A Type II error. If a Type I error occurred, they would reject a good shipment of potatoes and have to wait to get a new delivery. However, if a Type II error occurred, they would accept a bad batch and make potato chips with blemishes. This might upset consumers and decrease sales. To minimize the probability of a Type II error, choose a large significance level such as a = 0.10  2. (a) Increase. Increasing a to 0.10 makes it easier to reject the null hypothesis, which increases power. (b) Decrease. Decreasing the sample size means we don’t have as much information to use when making the decision, which makes it less likely to correctly reject H0 . (c) Decrease. It is harder to detect a difference of 0.02 (0.10 − 0.08) than a difference of 0.03 (0.11 − 0.08).

Answers to Odd-Numbered Section 9.2 Exercises 9.31  Random: Random sample. 10%: The sample size (60) < 10% of all students. Large Counts: 60 (0.80) = 48 ≥ 10 and 60 (0.20) = 12 ≥ 10. 9.33  np0 = 10(0.5) = 5 and n(1 − p0) = 10(0.5) = 5 are both 0.5. 1%: Because the P-value of 0.0143 > a = 0.01, we fail to reject H0 . There is not convincing evidence that p > 0.5. (b) P-value = 0.0286. Because this P-value is still less than a = 0.05 and greater than a = 0.01, we would again reject H0 at the 5% significance level and fail to reject H0 at the 1% significance level. 9.39  S: H0:p = 0.37 versus Ha:p > 0.37, where p = true proportion of all students who are satisfied with the parking situation after the change. P: One-sample z test for p. Random: Random sample. 10%: The sample size (200) < 10% of the population of size 2500. Large Counts: 200(0.37) = 74 ≥ 10 and 200(0.63) = 126 ≥ 10. D: z = 1.32, P-value = 0.0934. C: Because the P-value of 0.0934 > a = 0.05, we fail to reject H0 . We do not have convincing evidence that the true proportion of all students who are satisfied with the parking situation after the change > 0.37. 9.41  (a) S: H0:p = 0.50 versus Ha:p > 0.50, where p is the true proportion of boys among first-born children. P: One-sample z test for p. Random: Random sample. 10%: The sample size (25,468) < 10% of all first-borns. Large Counts: 25,468(0.50) = 12,734 ≥ 10 and 25,468(0.50) = 12,734 ≥ 10. D: z = 5.49, P-value≈0. C: Because the P-value of approximately 0 < a = 0.05, we reject H0 . There is convincing evidence that first-born children are more likely to be boys. (b) First-born children, because that is the group that we sampled from. 9.43  Here are the corrections: Ha : p > 0.75; p = the true proportion of middle school students who engage in bullying behavior; 10%: the sample size (558) < 10% of the population of middle school students; np0 = 558(0.75) = 418.5 ≥ 10 and n(1 − p0) = 0.7975 − 0.75 558(0.25) = 139.5 ≥ 10; z = = 2.59; (0.75)(0.25) Å 558 P-value = 0.0048. Because the P-value of 0.0048 < a = 0.05, we reject H0 . We have convincing evidence that more than three-quarters of middle school students engage in bullying behavior. 9.45  S: H0:p = 0.60 versus Ha:p ∙ 0.60, where p is the true proportion of teens who pass their driving test on the first attempt. P: One-sample z test for p. Random: Random sample. 10%: The sample size (125) < 10% of all teens. Large Counts: 125(0.60) = 75 ≥ 10 and 125(0.40) = 50 ≥ 10. D: z = 2.01, P-value = 0.0444. C: Because our P-value of 0.0444 < a = 0.05, we reject H0 . There is convincing evidence that the true proportion of teens who pass the driving test on their first attempt is different from 0.60. 9.47  (a) D: (0.607,0.769). C: We are 95% confident that the interval from 0.607 to 0.769 captures the true proportion of teens who pass the driving test on the first attempt. (b) Because 0.60 is not in the interval, we have convincing evidence that the true proportion of teens who pass the driving test on their first attempt is different from 0.60. 9.49  No. Because the value 0.16 is included in the interval, we do not have convincing evidence that the true proportion of U.S. adults who would say they use Twitter differs from 0.16.

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9.51  (a) p = the true proportion of U.S. teens aged 13 to 17 who think that young people should wait to have sex until marriage. (b) Random: Random sample. 10%: The sample size (439) < 10% of the population of all U.S. teens. Large Counts: 439(0.5) = 219.5 ≥ 10 and 439(0.5) = 219.5 ≥ 10. (c) Assuming that the true proportion of U.S. teens aged 13 to 17 who think that young people should wait to have sex until marriage is 0.50, there is a 0.011 probability of getting a sample proportion that is at least as different from 0.5 as the proportion in the sample. (d) Yes. Because the P-value of 0.011 < a = 0.05, we reject H0 . There is convincing evidence that the true proportion of U.S. teens aged 13 to 17 who think that young people should wait to have sex until marriage differs from 0.5. 9.53  (a) I: Finding convincing evidence that more than 37% of students were satisfied with the new parking arrangement, when in reality only 37% were satisfied. Consequence: The principal believes that students are satisfied and takes no further action. II: Failing to find convincing evidence that more than 37% are satisfied with the new parking arrangement, when in reality more than 37% are satisfied. Consequence: The principal takes further action on parking when none is needed. (b) If the true proportion of students that are satisfied with the new arrangement is really 0.45, there is a 0.75 probability that the survey provides convincing evidence that the true proportion > 0.37. (c) Increase the sample size or significance level. 9.55  P(Type I) = a = 0.05 and P(Type II) = 0.22. 9.57  (a) If the true proportion of Alzheimer’s patients who would experience nausea is really 0.08, there is a 0.29 probability that the results of the study would provide convincing evidence that the true proportion < 0.10. (b) Increase the number of measurements taken (n) to get more information. (c) Decrease. If a is smaller, it becomes harder to reject the null hypothesis. This makes it harder to correctly reject H0 . (d) Increase. Because 0.07 is further from the null hypothesis value of 0.10, it will be easier to detect a difference between the null value and actual value. 9.59  c 9.61  b 9.63  (a)X − Y has a Normal distribution with mean mX−Y = − 0.2 and standard deviation sX−Y = "(0.1)2 + (0.05)2 = 0.112. To fit in a case, X – Y must take on a negative number. (b) We want to find P(X − Y < 0) using the N( − 0.2,0.112) distribution. 0 − ( − 0.2) z= = 1.79 and P(Z < 1.79) = 0.9633. Using tech0.112 nology: 0.9629. There is a 0.9629 probability that a randomly selected CD will fit in a randomly selected case. (c) P(all fit) = (0.9629)100 = 0.0228. There is a 0.0228 probability that all 100 CDs will fit in their cases.

Section 9.3 Answers to Check Your Understanding

page 579: 1.  H0 :m = 320 versus Ha :m ∙ 320, where m = the true mean amount of active ingredient (in milligrams) in Aspro tablets from this batch of production.  2.  Random: Random sample. 10%: The sample of size 36 < 10% of the population of all tablets in this batch. Normal/Large Sample: n = 36 ≥ 30. 319 − 320 3.  t = = − 2  4.   For this test, df = 35. Using Table B 3∙"36 and df = 30, the tail area is between 0.025 and 0.05. Thus, the P-value for the two-sided test is between 0.05 and 0.10. Using technology: 2tcdf(lower:—1000,upper:—2,df:35) =

11/20/13 5:23 PM

Solutions

Frequency

9 8 7 6 5 4 3 2 1 0 3.0 4.5 6.0 7.5 9.0 10.5 12.0

Sleep (hours)

D: x = 6.643 and sx = 1.981. t = −3.625 and the P-value is between 0.0005 and 0.001. Using technology: P-value = 0.0006. C: Because our P-value of 0.0006 < a = 0.05, we reject H0 . There is convincing evidence that students at this university get less than 8 hours of sleep, on average. page 586: 1.  S: H0:m = 128 versus Ha:m ≠ 128, where m is the true mean systolic blood pressure for the company’s middle-aged male employees. P: One-sample t test for m. Random: Random sample. 10%: The sample size (72) < 10% of the population of middle-aged male employees. Normal/Large Sample: n = 72 ≥ 30. D: t = 1.10 and P-value = 0.275. C: Because our P-value of 0.275 > a = 0.05, we fail to reject H0 . There is not convincing evidence that the mean systolic blood pressure for this company’s middle-aged male employees differs from the national average of 128.  2.  We are 95% confident that the interval from 126.43 to 133.43 captures the true mean systolic blood pressure for the company’s middle-aged male employees. The value of 128 is in this interval and therefore is a plausible mean systolic blood pressure for the males 35 to 44 years of age. page 589: S: H0:md = 0 versus Ha:md > 0, where md is the true mean difference (air – nitrogen) in pressure lost. P: Paired t test for md . Random: Treatments were assigned at random to each pair of tires. Normal/Large Sample: n = 31 ≥ 30. D: x = 1.252 and sx = 1.202. t = 5.80 and P-value ≈ 0. C: Because the P-value of approximately 0 < a = 0.05, we reject H0 . We have convincing evidence that the true mean difference in pressure (air – nitrogen) > 0. In other words, we have convincing evidence that tires lose less pressure when filled with nitrogen than when filled with air, on average.

Answers to Odd-Numbered Section 9.3 Exercises 9.65  Random: Random sample. 10%: The sample size (45) < 10% of the population size of 1000. Normal/Large Sample: n = 45 ≥ 30. 9.67  The Random condition may not be met, because we don’t know if this is a random sample of the atmosphere in the Cretaceous era. Also, the Normal/Large Sample condition is not met. The sample size < 30 and the histogram below shows that the data are strongly skewed to the left.

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5

Frequency

2(0.0267) = 0.0534. Because the P-value of 0.0534 > a = 0.05, we fail to reject H0 . There is not convincing evidence that the true mean amount of the active ingredient in Aspro tablets from this batch of production differs from 320 mg. page 583:  1.  S: H0:m = 8 versus Ha:m < 8, where m is the true mean amount of sleep that students at the professor’s school get each night. P: One-sample t test for m. Random: Random sample. 10%: The sample size (28) < 10% of the population of students. Normal/Large Sample: The histogram below indicates that there is not much skewness and no outliers.

S-41

4 3 2 1 0 48

52

56

60

Nitrogen (percent)

64

125.7 − 115 = 2.409. (b) For this test, df = 44. Using 29.8!45 Table B and df = 40, we have 0.01 < P-value < 0.02. Using technology: tcdf(lower:2.409, upper:1000, df:44) = 0.0101. 9.71  (a) Using Table B and df = 19, we have 0.025 < P-value < 0.05. Using technology: P-value = 0.043. 5%: Because the P-value of 0.043 < a = 0.05, we reject H0 . There is convincing evidence that m < 5. 1%: Because the P-value of 0.043 > a = 0.01, we fail to reject H0 . There is not convincing evidence that m < 5. (b) Using technology: P-value = 0.086. 5%: Because the P-value of 0.086 > a = 0.05, we fail to reject H0 . There is not convincing evidence that m ≠ 5. 1%: same as part (a). 9.73  (a) S: H0 : m = 25 versus Ha : m > 25, where m is the true mean speed of all drivers in a construction zone. P: One-sample t test for m. Random: Random sample. 10%: The sample size (10) < 10% of all drivers. Normal/Large Sample: There is no strong skewness or outliers in the sample. 9.69  (a) t =

20

22

24

26

28

30

32

34

36

S pe e d

D: x = 28.8 and sx = 3.94. t = 3.05, df = 9, and the P-value is between 0.005 and 0.01 (0.0069). C: Because the P-value of 0.0069 < a = 0.05, we reject H0 . We have convincing evidence that the true mean speed of all drivers in the construction zone > 25 mph. (b) Because we rejected H0 , it is possible we made a Type I error—finding convincing evidence that the true mean speed > 25 mph when it really isn’t. 9.75  (a) S: H0:m = 1200 versus Ha:m < 1200, where m is the true mean daily calcium intake of women 18 to 24 years of age. P: Onesample t test for m. Random: Random sample. 10%: The sample size (36) < 10% of all women aged 18 to 24. Normal/Large Sample: n = 36 ≥ 30. D: t = −6.73 and P-value = 0.000. C: Because the P-value of approximately 0 < a = 0.05, we reject H0 . There is convincing evidence that women aged 18 to 24 are getting less than 1200 mg of calcium daily, on average. (b) Assuming that women aged 18 to 24 get 1200 mg of calcium per day, on average, there is about a 0 probability that we would observe a sample mean ≤ 856.2 mg by chance alone. 9.77  S: H0:m = 11.5 versus Ha:m ≠ 11.5, where m is the true mean hardness of the tablets. P: One-sample t test for m. Random: The tablets were selected randomly. 10%: The sample size (20) < 10% of all tablets in the batch. Normal/Large Sample: There is no strong skewness or outliers in the sample.

12/2/13 10:47 AM

Solutions

Frequency

4 3 2 1 0

11.35 11.40 11.45 11.50 11.55 11.60 11.65 11.70

Hardness

D: x = 11.5164 and sx = 0.0950. t = 0.77, df = 19, and the P-value is between 0.40 and 0.50 (0.4494). C: Because our P-value of 0.4494 > a = 0.05, we fail to reject H0 . We do not have convincing evidence that the true mean hardness of these tablets is different from 11.5. 9.79  D: With df = 19, (11.472,11.561). C: We are 95% confident that the interval from 11.472 to 11.561 captures the true mean hardness measurement for this type of pill. The confidence interval gives 11.5 as a plausible value for the true mean hardness μ, but it gives other plausible values as well. 9.81  S: H0:m = 200 versus Ha:m ∙ 200, where m is the true mean response time of European servers. P: One-sample t interval to help us perform a two-sided test for m. Random: The servers were selected randomly. 10%: The sample size (14) < 10% of all servers in Europe. Normal/Large Sample: The sample size is small, but a graph of the data reveals no strong skewness or outliers. D: (158.22, 189.64). C: Because our 95% confidence interval does not contain 200 milliseconds, we reject H0 at the a = 0.05 significance level. We have convincing evidence that the mean response time of European servers is different from 200 milliseconds. 9.83  (a) Yes. Because the P-value of 0.06 > a = 0.05, we fail to reject H0 : m = 10 at the 5% level of significance. Thus, the 95% confidence interval will include 10. (b) No. Because the P-value of 0.06 < a = 0.10, we reject H0:m = 10 at the 10% level of significance. Thus, the 90% confidence interval would not include 10 as a plausible value. 9.85  (a) If all the subjects used the right thread first and they were tired when they used the left thread, then we wouldn’t know if the difference in times was because of tiredness or because of the direction of the thread. (b) S: H0: md = 0 versus Ha: md > 0, where md is the true mean difference (left − right) in the time (in seconds) it takes to turn the knob with the left-hand thread and the right-hand thread. P: Paired t test for md . Random: The order of treatments was determined at random. Normal/Large Sample: There is no strong skewness or outliers.

9.87  (a) H0:md = 0 versus Ha:md > 0, where md is the true mean difference in tomato yield (A − B). (b) df = 9. (c) Interpretation: Assuming that the average yield for both varieties is the same, there is a 0.1138 probability of getting a mean difference as large or larger than the one observed in this experiment. Conclusion: Because the P-value of 0.1138 > a = 0.05, we fail to reject H0. We do not have convincing evidence that the true mean difference in tomato yield (A − B) > 0. (d) I: Finding convincing evidence that Variety A tomato plants have a greater mean yield, when in reality there is no difference. II: Not finding convincing evidence that Variety A tomato plants have a higher mean yield, when in reality Variety A does have a greater mean yield. They might have made a Type II error. 9.89  Increase the significance level a or increase the sample size n. 9.91  When the sample size is very large, rejecting the null hypothesis is very likely, even if the actual parameter is only slightly different from the hypothesized value. 9.93  (a) No, in a sample of size n = 500, we expect to see about (500)(0.01) = 5 people who do better than random guessing, with a significance level of 0.01. (b) The researcher should repeat the procedure on these four to see if they again perform well. 9.95  b 9.97  d 9.99  c 9.101  a 9.103  (a) Not included. The margin of error does not account for undercoverage. (b) Not included. The margin of error does not account for nonresponse. (c) Included. The margin of error is calculated to account for sampling variability.

Answers to Chapter 9 Review Exercises R9.1  (a) H0: m = 64.2; Ha: m ∙ 64.2, where m = the true mean height of this year’s female graduates from the local high school. (b) H0: p = 0.75; Ha: p < 0.75, where p = the true proportion of all students at Mr. Starnes’s school who completed their math homework last night. R9.2  Random: Random sample. 10%: The sample size (24) < 10% of the population of adults. Normal/Large Sample: The histogram below shows that the distribution is roughly symmetric with no outliers. 6

Frequency

S-42

5 4 3 2

Frequency

1 9 8 7 6 5 4 3 2 1 0

0 15 18 21 24 27 30 33 36

Time (seconds)

–50 –35 –20 –5

10 25 40 55

Difference in times (left thread – right thread)

D: x = 13.32 and sx = 22.94. t = 2.903, df = 24, and the P-value is between 0.0025 and 0.005 (0.0039). C: Because the P-value of 0.0039 < a = 0.05, we reject H0 . We have convincing evidence that the true mean difference (left − right) in time it takes to turn the knob >0.

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R9.3  (a) H0 : m = 300 versus Ha: m < 300, where m = the true mean breaking strength of these chairs. (b) I: Finding convincing evidence that the mean breaking strength a = 0.05, we fail to reject H0 . We do not have convincing evidence that the true mean difference (regular − kiln) in yield a = 0.05, we fail to reject H0 . We do not have convincing evidence that fewer than 5% of adults who receive this vaccine will get the flu. (b) Because we failed to reject the null hypothesis, we could have made a Type II error—not finding convincing evidence that the true proportion of adults get the flu after using this vaccine a = 0.10, we fail to reject H0 . We do not have convincing evidence that the true mean reading from the radon detectors is different than 105. (b) Yes. Because 105 is in the interval from 99.61 to 110.03, both the confidence interval and the significance test agree that 105 is a plausible value for the true mean reading from the radon detectors. R9.7  (a) The random condition can be satisfied by randomly allocating which plot got the regular barley seeds and which one got the kiln-dried seeds within each pair of adjacent plots. (b) S: H0: md = 0 versus Ha: md < 0, where md is the true mean difference (regular − kiln) in yield between regular barley seeds and kilndried barley seeds. P: Paired t test for md. Random: Assumed. Normal/Large Sample: The histogram below shows no strong skewness or outliers.

S-43

D: x = − 1 and sx = 0.816. t = −3.873, df = 9, and the P-value is between 0.001 and 0.0025 (0.0019). C: Because the P-value of 0.0019 < a = 0.05, we reject H0 . We have convincing evidence that the mean difference (no coffee − coffee) in word recall < 0. T9.13  S: H0: m = $158 versus Ha: m ∙ $158, where m is the true mean amount spent on food by households in this city. P: Onesample t test for m. Random: Random sample. 10%: The sample size (50) < 10% of households in this small city. Normal/Large

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S-44

Solutions

Sample: n = 50 ≥ 30. D: t = 2.47; using df = 40, the P-value is between 0.01 and 0.02 (using df = 49, 0.0168). C: Because the P-value of 0.0168 < a = 0.05, we reject H0 . We have convincing evidence that the true mean amount spent on food per household in this city is different from the national average of $158.

Chapter 10 Section 10.1 Answers to Check Your Understanding page 619:  S: p1 = true proportion of teens who go online every day and p2 = true proportion of adults who go online every day. P: Twosample z interval for p1 − p2 . Random: Independent random samples. 10%: n1 = 799 < 10% of teens and n2 = 2253 < 10% of adults. Large Counts: 503, 296, 1532, and 721 are all ≥ 10. D: 0.63(0.37) 0.68(0.32) (0.63 − 0.68) ± 1.645 + = Å 799 2253 (−0.0824,−0.0176). C: We are 90% confident that the interval from −0.0824 to −0.0176 captures the true difference in the proportion of U.S. adults and teens who go online every day.  page 628:  S: H0: p1 − p2 = 0 versus Ha:p1 − p2 > 0, where p1 is the true proportion of children like the ones in the study who do not attend preschool that use social services later and p2 is the true proportion of children like the ones in the study who attend preschool that use social services later. P: Two-sample z test for p1 − p2. Random: Two groups in a randomized experiment. Large Counts: 49, 12, 38, (0.8033− 0.6129)−0 = 2.32 0.7073(0.2927) 0.7073(0.2927) + Å 61 62 and P-value = 0.0102. C: Because the P-value of 0.0102 < a = 0.05, we reject H0 . There is convincing evidence that the true proportion of children like the ones in the study who do not attend preschool that use social services later is greater than the true proportion of children like the ones in the study who attend preschool that use social services later. 24 are all ≥ 10. D: z =

Answers to Odd-Numbered Section 10.1 Exercises 10.1  (a) Approximately Normal because 100(0.25) = 25, 100(0.75) = 75, 100(0.35) = 35, and 100(0.65) = 65 are all at least (c) Because mp^1 −p^2 = 0.25 − 0.35 = −0.10. 10. (b) n1 = 100 < 10% of the first bag and n2 = 100 < 10% of the sec0.25(0.75) 0.35(0.65) + = 0.0644. Å 100 100 10.3  (a) Approximately Normal because 50(0.30) = 15, 50(0.7) = 35, 100(0.15) = 15, and 100(0.85) = 85 are all at least 10. (b) mp^C −p^ A = 0.30 − 0.15 = 0.15. (c) Because nC = 50 < 10% of the jelly beans in the Child mix and nA = 100 < 10% of the jelly ond bag, sp^1 −p^2 =

0.3(0.7) 0.15(0.85) beans in the Adult mix, sp^C −p^ A = + = Å 50 100 0.0740. 10.5  The data do not come from independent random samples or two groups in a randomized experiment. Also, there were less than 10 successes (3) in the group from the west side of Woburn. 10.7  There were less than 10 failures (0) in the treatment group, less than 10 successes (8) in the control group, and less than 10 failures in the control group (4).

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0.26(1 − 0.26) 0.14(1 − 0.14) + = 0.0289. Å 316 532 If we were to take many random samples of 316 young adults and 532 older adults, the difference in the sample proportions of young adults and older adults who use Twitter will typically be 0.0289 from the true difference. (b) S: p1 = true proportion of young adults who use Twitter and p2 = true proportion of older adults who use Twitter. P: Two-sample z interval for p1 − p2 . Random: Two ­independent random samples. 10%: n1 = 316 < 10% of all young adults and n2 = 532 < 10% of all older adults. Large Counts: 82, 234, 74, 458 are all at least 10. D: (0.072,0.168). C: We are 90% confident that the interval from 0.072 to 0.168 captures the true difference in the proportions of young adults and older adults who use Twitter. 10.11  (a) S: p1 = true proportion of young men who live in their parents’ home and p2 = true proportion of young women who live in their parents’ home. P: Two-sample z interval for p1 − p2 . Random: Reasonable to consider these independent random samples. 10%: n1 = 2253 < 10% of the population of young men and n2 = 2629 < 10% of the population of young women. Large Counts: 986, 1267, 923, 1706 are all at least 10. D: (0.051,0.123). C: We are 99% confident that the interval from 0.051 to 0.123 captures the true difference in the proportions of young men and young women who live in their parents’ home. (b) Because the interval does not contain 0, there is convincing evidence that the true proportion of young men who live in their parents’ home is different from the true proportion of young women who live in their parents’ home. 10.13  H0 : p1 − p2 = 0 versus Ha : p1 − p2 ∙ 0, where p1 is the true proportion of all teens who would say that they own an iPod or MP3 player and p2 is the true proportion of all young adults who would say that they own an iPod or MP3 player. 10.15  P: Two-sample z test for p1 − p2 . Random: Independent random samples. 10%: n1 = 800 < 10% of all teens and n2 = 400 < 10% of all young adults. Large Counts: 632, 168, 268, and 132 are all at least 10. D : z = 4.53 and P-value ≈ 0. C: Because the P-value of close to 0 < a = 0.05, we reject H0 . There is convincing evidence that the true proportion of teens who would say that they own an iPod or MP3 player is different from the true proportion of young adults who would say that they own an iPod or MP3 player. 10.17  D : (0.066,0.174). C : We are 95% confident that the interval from 0.066 to 0.174 captures the true difference in proportions of teens and young adults who own iPods or MP3 players. Because 0 is not included in the interval, it is consistent with the results of Exercise 15. 10.19  S : H0 : p1 − p2 = 0 versus Ha : p1 − p2 > 0, where p1 is the true proportion of 6- to 7-year-olds who would sort correctly and p2 is the true proportion of 4- to 5-year-olds who would sort correctly. P: Two-sample z test for p1 − p2 . Random: Independent random samples. 10%: n1 = 53 < 10% of all 6- to 7-year olds and n2 = 50 < 10% of all 4- to 5-year-olds. Large Counts: 28, 25, 10, 40 are all ≥ 10. D: z = 3.45 and P-value = 0.0003. C: Because the P-value of 0.0003 < a = 0.05, we reject H0 . We have convincing evidence that the true proportion of 6- to 7-year-olds who would sort correctly is greater than the true proportion of 4- to 5-year-olds who would sort correctly. 10.21  (a) S: H0 : pA − pB = 0 versus Ha : pA − pB > 0, where pA is the true proportion of students like these who would pass the driver’s license exam when taught by instructor A and pB is the true 10.9  (a) SEp^1 −p^2 =

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Solutions proportion of students like these who would pass the driver’s license exam when taught by instructor B. P: Two-sample z test for pA − pB. Random: Two groups in a randomized experiment. Large Counts: 30, 20, 22, 28 are all ≥ 10. D: z = 1.60 and P-value = 0.0547. C: Because the P-value of 0.0547 > a = 0.05, we fail to reject H0 . There is not convincing evidence that the true proportion of students like these who would pass using instructor A is greater than the true proportion who would pass using instructor B. (b) I: Finding convincing evidence that instructor A is more effective than instructor B, when in reality the instructors are equally ­effective. II: Not finding convincing evidence that instructor A is better, when in reality instructor A is more effective. It is possible we made a Type II error. 10.23  (a) Two-sample z test for p1 − p2 . Random: Two groups in a randomized experiment. Large Counts: 44, 44, 21, 60 are all ≥ 10. (b) If no difference exists in the true pregnancy rates of women who are being prayed for and those who are not, there is a 0.0007 probability of getting a difference in pregnancy rates as large or larger than the one observed in the experiment by chance alone. (c) Because the P-value of 0.0007 < a = 0.05, we reject H0 . There is convincing evidence that the pregnancy rates among women like these who are prayed for are higher than the pregnancy rates for those who are not prayed for. (d) Knowing they were being prayed for might have affected their behavior in some way that would have affected whether they became pregnant or not. Then we wouldn’t know if it was the prayer or the other behaviors that caused the higher pregnancy rate. 10.25  a 10.27  c 10.29  (a) y^ = −13,832 + 14,954x, where y^ = the predicted mileage and x = the age in years of the cars. (b) For each year older the car is, the predicted mileage will increase by 14,954 miles. (c) Residual = −25,708. The student’s car had 25,708 fewer miles than expected, based on its age.

Section 10.2 Answers to Check Your Understanding

page 644:  S:m1 = the true mean price of wheat in July and m2 = the true mean price of wheat in September. P: Two-sample t ­interval for m1 − m2 . Random: Independent random samples. 10%: n1 = 90 < 10% of all wheat producers in July and n2 = 45 < 10% of all wheat producers in September. Normal/Large Sample: n1 = 90 ≥ 30 and n2 = 45 ≥ 30. D: Using df = 40, (−0.759, −0.561). Using df = 100.45, (−0.756, −0.564). C: We are 99% confident that the interval from −0.756 to −0.564 captures the true difference in mean wheat prices in July and September.  page 649:  S: H0:m1 − m2 = 0 versus Ha:m1 − m2 > 0, where m1 is the true mean breaking strength for polyester fabric buried for 2 weeks and m2 is the true mean breaking strength for polyester fabric buried for 16 weeks. P: Two-sample t test. Random: Two groups in a randomized experiment. Normal/Large Sample: The dotplots below show no strong skewness or outliers in either group. 2 weeks 16 weeks 102

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D: x1 = 123.8, s1 = 4.60, x2 = 116.4, s2 = 16.09. t = 0.989. Using df = 4, the P-value is between 0.15 and 0.20. Using df = 4.65,

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P-value = 0.1857. C: Because the P-value of 0.1857 > a = 0.05, we fail to reject H0 . We do not have convincing evidence that the true mean breaking strength of polyester fabric that is buried for 2 weeks is greater than the true mean breaking strength for polyester fabric that is buried for 16 weeks.

Answers to Odd-Numbered Section 10.2 Exercises 10.31  (a) Because the distributions of M and B are Normal, the distribution of xM − xB is also Normal. (b) m −x − −x = 188 − 170 = 18 mg∙dl. (c) Because 25 < 10% of all 20- to 34-year-old males and 36 < 10% of all 14-year-old boys, s −x − −x = (41)2 (30)2 + = 9.60 mg∙dl. Å 25 36 10.33  Random: Two independent random samples. 10%: 20 < 10% of all males at the school and 20 < 10% of all females at the school. Normal/Large Sample: not met because there are fewer than 30 observations in each group and the stemplot for Males shows several outliers. 10.35  Random: not met because these data are not from two independent random samples. Knowing the literacy percent for females in a country helps us predict the literacy percent for males in that country. 10%: not met because 24 is more than 10% of Islamic countries. Normal/Large Sample: not met because the samples sizes are both small and both distributions are skewed to the left and have an outlier (see boxplots below). M

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10.37  (a) The distributions of percent change are both slightly skewed to the left. People drinking red wine generally have more polyphenols in their blood, on average. The distribution of percent change for the white wine drinkers is a little bit more variable. (b) S: m1 = the true mean change in polyphenol level in the blood of people like those in the study who drink red wine and m2 = the true mean polyphenol level in the blood of people like those in the study who drink white wine. P: Two-sample t interval for m1 − m2 . Random: Two groups in a randomized experiment. Normal/Large Sample: The dotplots given in the problem do not show strong skewness or outliers. D: x1 = 5.5, s1 = 2.517, x2 = 0.23, s2 = 3.292. Using df = 8, (2.701, 7.839). Using df = 14.97, (2.845, 7.689). C: We are 90% confident that the interval from 2.845  to 7.689 captures the true difference in mean change in polyphenol level for men like these who drink red wine and men like these who drink white wine. (c) Because all of the plausible values in the interval are positive, this interval supports the researcher’s belief that red wine is more effective than white wine. 10.39  (a) Earnings amounts cannot be negative, yet the standard deviation is almost as large as the distance between the mean and 0. However, the sample sizes are both very large (675 ≥ 30 and 621 ≥ 30). (b) S: m1 = the true mean summer earnings of male students and m2 = the true mean summer earnings of female students. P: Two-sample t interval for m1 − m2 . Random:

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Solutions

Reasonable to consider these independent random samples. 10%: n1 = 675 < 10% of male students at a large university and n2 = 621 < 10% of female students at a large university. Normal/Large Sample: n1 = 675 ≥ 30 and n2 = 621 ≥ 30. Using D: Using df = 100, (412.68, 635.58). df = 1249.21, (413.62, 634.64). C: We are 90% confident that the interval from $413.62 to $634.64 captures the true difference in mean summer earnings of male students and female students at this large university. (c) If we took many random samples of 675 males and 621 females from this university and each time constructed a 90% confidence interval in this same way, about 90% of the resulting intervals would capture the true difference in mean earnings for males and females. 10.41  (a) S: H0:m1 − m2 = 0 versus Ha:m1 − m2 < 0, where m1 is the true mean time to breeding for the birds relying on natural food supply and m2 is the true mean time to breeding for birds with food supplementation. P: Two-sample t test. Random: Two groups in a randomized experiment. Normal/Large Sample: Neither distribution displays strong skewness or outliers. Control Supplemented 0.0

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Time to breeding (days behind caterpillar peak)

D: x1 = 4.0, s1 = 3.11, x2 = 11.3, s2 = 3.93. t = −3.74. Using df = 5, the P-value is between 0.005 and 0.01. Using df = 10.95, P-value = 0.0016. C: Because the P-value of 0.0016 < a = 0.05, we reject H0 . We have convincing evidence that the true mean time to breeding is less for birds relying on natural food supply than for birds with food supplements. (b) Assuming that the true mean time to breeding is the same for birds relying on natural food supply and birds with food supplements, there is a 0.0016 probability that we would observe a difference in sample means of −7.3 or smaller by chance alone. 10.43  S: H0: m1 − m2 = 0 versus Ha: m1 − m2 ∙ 0, where m1 is the true mean number of words spoken per day by female students and m2 is the true mean number of words spoken per day by male students. P: Two-sample t test. Random: Independent random samples. 10%: n1 = 56 < 10% of females at a large university and n2 = 56 < 10% of males at a large university. Normal/Large Sample: n1 = 56 ≥ 30 and n2 = 56 ≥ 30. D: t = −0.248. Using df = 50, P-value > 0.50. Using df = 106.20, P-value = 0.8043. C: Because the P-value of 0.8043 > a = 0.05, we fail to reject H0 . We do not have convincing evidence that the true mean number of words spoken per day by female students is different than the true mean number of words spoken per day by male students at this university. 10.45  (a) The distribution for the activities group is slightly skewed to the left, while the distribution for the control group is slightly skewed to the right. The center of the activities group is higher than the center of the control group. The scores in the activities group are less variable than the scores in the control group. (b) S: H0: m1 − m2 = 0 versus Ha: m1 − m2 > 0, where m1 is the true mean DRP score for third-grade students like the ones in the experiment who do the activities and m2 is the true mean DRP score for thirdgrade students like the ones in the experiment who don’t do the activities. P: Two-sample t test. Random: Two groups in a randomized experiment. Normal/Large Sample: No strong skewness or outliers in either boxplot. D: t = 2.311. Using df = 20, the P-value

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is between 0.01 and 0.02. Using df = 37.86, P-value = 0.0132. C: Because the P-value of 0.0132 < a = 0.05, we reject H0 . We have convincing evidence that the true mean DRP score for third-grade students like the ones in the experiment who do the activities is greater than the true mean DRP score for third-grade students like the ones in the experiment who don’t do the activities. (c) Because this was a randomized controlled experiment, we can conclude that the activities caused the increase in the mean DRP score. 10.47  D: Using df = 50, (−3563, 2779). Using df = 106.2, (−3521, 2737). C: We are 95% confident that the interval from −3521 to 2737 captures the true difference between mean number of words spoken per day by female students and the mean number of words spoken per day by male students. This interval allows us to determine if 0 is a plausible value for the difference in means and also provides other plausible values for the difference in mean words spoken per day. 10.49  (a) S: H0: m1 − m2 = 10 versus Ha: m1 − m2 > 10, where m1 is the true mean cholesterol reduction for people like the ones in the study when using the new drug and m2 is the true mean cholesterol reduction for people like the ones in the study when using the current drug. P: Two-sample t test. Random: Two groups in a randomized experiment. Normal/Large Sample: No strong skewness or outliers. D: t = 0.982. Using df = 13, the P-value is between 0.15 and 0.20. Using df = 26.96, P-value = 0.1675. C: Because the P-value of 0.1675 > a = 0.05, we fail to reject H0 . We do not have convincing evidence that the true mean cholesterol reduction is more than 10 mg/dl greater for the new drug than for the current drug. (b) Type II error. It is possible that the difference in mean cholesterol reduction is more than 10 mg/dl greater for the new drug than the current drug, but we didn’t find convincing evidence that it was. 10.51  (a) The researchers randomly assigned the subjects to create two groups that were roughly equivalent at the beginning of the experiment. (b) Only about 5 out of the 1000 differences were ≥ 4.15, P-value ≈ 0.005. Because the P-value of 0.005 < a = 0.05, we have convincing evidence that the true mean rating for students like these that are provided with internal reasons is higher than the true mean rating for students like these that are provided with external reasons. (c) Because we found convincing evidence that the mean is higher for students with internal reasons when it is possible that there is no difference in the means, we could have made a Type I error. 10.53  (a) Two-sample. Two distinct groups of cars in a randomized experiment. (b) Paired. Both treatments are applied to each subject. (c) Two-sample. Two distinct groups of women. 10.55  (a) Paired, because we have two scores for each student. (b) S: H0: md = 0 versus Ha: md > 0, where md is the true mean increase in SAT verbal scores of students who were coached. P: Paired t test for md. Random: Random sample. 10%: nd = 427 < 10% of students who are coached. Normal/Large Sample: 427 ≥ 30. D: t = 10.16. Using df = 426, P-value ≈ 0. C: Because the P-value of approximately 0 < a = 0.05, we reject H0 . There is convincing evidence that students who are coached increase their scores on the SAT verbal test, on average. 10.57  a 10.59  b 10.61  (a) One-sample z interval for a proportion. (b) Paired t test for the mean difference. (c) Two-sample z interval for the difference in proportions. (d) Two-sample t test for a difference in means.

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Solutions 10.63  (a) P(at least one mean outside interval) = 1 − P(neither mean outside interval) = 1−(0.95)2 = 1−0.9025 = 0.0975. (b) Let X = the number of samples that must be taken to observe one falling above m x_ + 2s x_. Then X is a geometric random variable with p = 0.025. P(X = 4) = (1 − 0.025)3(0.025) = 0.0232. (c) Let X = the number of sample means out of 5 that fall outside this interval. X is a binomial random variable with n = 5 and p = 0.32. We want P(X ≥ 4) = 1 − P(X ≤ 3) = 1 − binomcdf (trials:5, p:0.32,x value:3) = 1 − 0.961 = 0.039. This is a reasonable criterion because when the process is under control, we would only get a “false alarm” about 4% of the time. 10.65  (a) Perhaps the people who responded are prouder of their improvements and are more willing to share. This could lead to an overestimate of the true mean improvement. (b) This was an observational study, not an experiment. The students (or their parents) chose whether or not to be coached; students who choose coaching might have other motivating factors that help them do better the second time.

Answers to Chapter 10 Review Exercises R10.1  (a) Paired t test for the mean difference. (b) Two-sample z interval for the difference in proportions. (c) One-sample t interval for the mean. (d) Two-sample t interval for the difference between two means. 0.832(1 − 0.832) 0.581(1 − 0.581) + Å 220 117 = 0.0521. If we were to take many random samples of 220 Hispanic female drivers in New York and 117 Hispanic female drivers in Boston, the difference in the sample proportions who wear seatbelts will typically be 0.0521 from the true difference in proportions of all Hispanic female drivers in New York and Boston who wear seat belts. (b) S: p1 = proportion of all Hispanic female drivers in New York who wear seat belts and p2 = proportion of all Hispanic female drivers in Boston who wear seat belts. P: Two-sample z interval for p1 − p2 . Random: Independent random samples. 10%: n1 = 220 < 10% of all Hispanic female drivers in New York and n2 = 117 < 10% of all Hispanic female drivers in Boston. Large Counts: 183, 37, 68, 49 are all ≥ 10. D: (0.149, 0.353). C: We are 95% confident that the interval from 0.149 to 0.353 captures the true difference in the proportions of Hispanic women drivers in New York and Boston who wear their seat belts. R10.3  (a) The women in the study were randomly assigned to one of the two treatments. (b) Because both groups are large (nC = 45 ≥ 30 and nA = 45 ≥ 30), the sampling distribution of xC − xA should be approximately Normal. (c) Assuming no difference exists in the true mean ratings of the product for women like these who read or don’t read the news story, there is less than a 0.01 probability of observing a difference as large as or larger than 0.49 by chance alone. R10.4  (a) S:m1 = the true mean NAEP quantitative skills test score for young men and m2 = the true mean NAEP quantitative skills test score for young women. P: Two-sample t interval for m1 − m2 . Random: Reasonable to consider these independent random samples. 10%: n1 = 840 < 10% of all young men and n2 = 1077 < 10% of all young women. Normal/Large Sample: n1 = 840 ≥ 30 and n2 = 1077 ≥ 30. D: Using df = 100, (−6.80,2.14). Using df = 1777.52, (−6.76,2.10). C: We are 90% confident that the interval from −6.76 to 2.10 captures the true difference in the mean NAEP quantitative skills test score for young men and the R10.2  (a) SEp^1 −p^2 =

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mean NAEP quantitative skills test score for young women. (b) Because 0 is in the interval, we do not have convincing evidence of a difference in mean score for male and female young adults. R10.5  (a) S: H0: p1 − p2 = 0 versus Ha: p1 − p2 < 0, where p1 is the true proportion of patients like these who take AZT and develop AIDS and p2 is the true proportion of patients like these who take placebo and develop AIDS. P: Two-sample z test for p1 − p2. Random: Two groups in a randomized experiment. Large Counts: 17, 418, 38, 397 are all ≥ 10. D: z = −2.91, P-value = 0.0018. C: Because the P-value of 0.0018 < a = 0.05, we reject H0. We have convincing evidence that taking AZT lowers the proportion of patients like these who develop AIDS compared to a placebo. (b) I: Finding convincing evidence that AZT lowers the risk of developing AIDS, when in reality it does not. Consequence: patients will pay for a drug that doesn’t help. II: Not finding convincing evidence that AZT lowers the risk of developing AIDS, when in reality it does. Consequence: patients won’t take the drug when it could actually delay the onset of AIDS. It is possible that we made a Type I error. R10.6  (a) The Large Counts condition is not met because there are only 7 failures in the control area. (b) The Normal/Large Sample condition is not met because both sample sizes are small and there are outliers in the male distribution. R10.7  (a) Even though each subject has two scores (before and after), the two groups of students are independent. (b) The distribution for the control group is slightly skewed to the right, while the distribution for the treatment group is roughly symmetric. The center for the treatment group is greater than the center for the control group. The differences in the control group are more variable than the differences in the treatment group. (c) S: H0: m1 − m2 = 0 versus Ha: m1 − m2 > 0, where m1 = the true mean difference in test scores for students like these who get the treatment message and m2 = the true mean difference in test scores for students like these who get the neutral message. P: Two-sample t test for m1 − m2. Random: Two groups in a randomized experiment. Normal/Large Sample: Neither boxplot showed strong skewness or any outliers. D: Using the differences, x1 = 11.4, s1 = 3.169, x2 = 8.25, s2 = 3.69. t = 1.91. Using df = 7, the P-value is between 0.025 and 0.05. Using df = 13.92, P-value = 0.0382. C: Because the P-value of 0.0382 < a = 0.05, we reject H0. There is convincing evidence that the true mean difference in test scores for students like these who get the treatment message is greater than the true mean difference in test scores for students like these who get the neutral message. (d) We cannot generalize to all students who failed the test because our sample was not a random sample of all students who failed the test.

Answers to Chapter 10 AP® Statistics Practice Test T10.1  e T10.2  b T10.3  a T10.4  a T10.5  e T10.6  e T10.7  c T10.8  c T10.9  b T10.10  a T10.11  (a) S: m1 = the true mean hospital stay for patients like these who get heating blankets during surgery and m2 = the true mean hospital stay for patients like these who have core temperatures reduced during surgery. P: Two-sample t interval for m1 − m2.

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Solutions

Random: Two groups in a randomized experiment. Normal/Large Sample: n1 = 104 ≥ 30 and n2 = 96 ≥ 30. D: Using df = 80, (−4.17, −1.03). Using df = 165.12, (−4.16, −1.04). C: We are 95% confident that the interval from −4.16 to −1.04 captures the true difference in mean length of hospital stay for patients like these who get heating blankets during surgery and those who have their core temperatures reduced during surgery. (b) Yes. Because 0 is not in the interval, we have convincing evidence that the true mean hospital stay for patients like these who get heating blankets during surgery is different than the true mean hospital stay for patients like these who have core temperatures reduced during surgery. (c) If we were to repeat this experiment many times and calculate 95% confidence intervals for the difference in means each time, about 95% of the intervals would capture the true difference in mean hospital stay for patients like these who get heating blankets during surgery and mean hospital stay for patients like these who have core temperatures reduced during surgery. T10.12  (a) S: H0: p1 − p2 = 0 versus Ha: p1 − p2 > 0, where p1 is the true proportion of cars that have the brake defect in last year’s model and p2 is the true proportion of cars that have the brake defect in this year’s model. P: Two-sample z test for p1 − p2 . Random: Independent random samples. 10%: n1 = 100 < 10% of last year’s model and n2 = 350 < 10% of this year’s model. Large Counts: 20, 80, 50, 300 are all ≥ 10. D: z = 1.39, P-value = 0.0822. C: Because the P-value of 0.0822 > a = 0.05, we fail to reject H0 . We do not have convincing evidence that the true proportion of brake defects is smaller in this year’s model compared to last year’s model. (b) I: Finding convincing evidence that there is a smaller proportion of brake defects in this year’s car model, when in reality there is not. This might result in more accidents because people think that their brakes are safe. II: Not finding convincing evidence that there is a smaller proportion of brake defects in this year’s model, when in reality there is a smaller proportion. This might result in reduced sales of this year’s model. T10.13  (a) H0 : m1 − m2 = 0 versus Ha : m1 − m2 < 0, where m1 = the true mean rent for one-bedroom apartments in the area of her college campus and m2 = the true mean rent for two-bedroom apartments in the area of her college campus. (b) Two-sample t test for m1 − m2. Random: Independent random samples. 10%: n1 = 10 < 10% of all one-bedroom apartments in this area and n2 = 10 < 10% of all two-bedroom apartments in this area. Normal/Large Sample: The dotplots below show no strong skewness or outliers in either distribution. Two bedroom One bedroom 400

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750

Rent (dollars)

(c) Assuming the true mean rent of the two types of apartments is really the same, there is a 0.029 probability of getting an observed difference in mean rents as large as or larger than the one in this study. (d) Because the P-value of 0.029 < a = 0.05, Pat should reject H0 . She has convincing evidence that the true mean rent of two-bedroom apartments is greater than the true mean rent of onebedroom apartments in the area of her college campus.

Answers to Cumulative AP® Practice Test 3 AP3.1  e AP3.2  e AP3.3  d

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AP3.4  c AP3.5  d AP3.6  d AP3.7  c AP3.8  a AP3.9  d AP3.10  c AP3.11  b AP3.12  c AP3.13  c AP3.14  d AP3.15  d AP3.16  e AP3.17  b AP3.18  b AP3.19  e AP3.20  c AP3.21  a AP3.22  d AP3.23  b AP3.24  e AP3.25  a AP3.26  b AP3.27  c AP3.28  d AP3.29  a AP3.30  b AP3.31  S: H0: md = 0 versus Ha: md < 0, where md is the true mean change in weight (after − before) in pounds for people like these who follow a five-week crash diet. P: Paired t test for md . Random: Random sample. 10%: nd = 15 is less than 10% of all dieters. Normal/Large Sample: There is no strong skewness or outliers.

– 25

– 20

– 15

– 10

–5

0

5

10

15

Differences (after – before)

D: x = −3.6 and sx = 11.53. t = −1.21. Using df = 14, the P-value is between 0.10 and 0.15 (0.1232). C: Because the P-value of 0.1232 is greater than a = 0.05, we fail to reject H0 . We do not have convincing evidence that the true mean change in weight (after − before) for people like these who follow a five-week crash diet is less than 0. AP3.32  (a) Observational study. No treatments were imposed on the individuals in the study. (b) H0: p1 − p2 = 0 versus Ha: p1 − p2 < 0, where p1 is the true proportion of VLBW babies who graduate from high school by age 20 and p2 is the true proportion of non-VLBW babies who graduate from high school by age 20. P: Two-sample z test for p1 − p2 . Random: Independent random samples. 10%: n1 = 242 is less than 10% of all VLBW babies and n2 = 233 is less than 10% of all non-VLBW babies. Large Counts: 179, 63, 193, 40 are all ≥ 10. Do: z = −2.34 and P-value = 0.0095. Conclude: Because the P-value of 0.0095 is less than a = 0.05, we reject H0 . We have convincing evidence that the true proportion of VLBW babies who graduate from high school by

11/20/13 5:23 PM

Solutions age 20 is less than the true proportion of non-VLBW babies who graduate from high school by age 20. AP3.33  (a) y^ = −73.64 + 5.7188x, where y^ = predicted distance and x = temperature (degrees Celsius) (b) For each increase of 1°C in the water discharge temperature, the predicted distance from the nearest fish to the outflow pipe increases by about 5.7188 meters. (c) Yes. The residual plot shows no leftover pattern. (d) residual = 78 − 92.21 = −14.21 meters. The actual distance on this afternoon was 14.21 meters closer than expected, based on the temperature of the water. AP3.34  (a) Define W = the weight of a randomly selected gift box. Then mW = 8(2) + 2(4) + 3 = 27 ounces and sW = "8(0.52) + 2(12) + 0.22 = 2.01 ounces. (b) We want to find 30 − 27 P(W > 30) using the N(27, 2.01) distribution. z = = 1.49 2.01 and P(W > 30) = 0.0681. Using technology: 0.0678. There is a 0.0678 probability of randomly selecting a box that weighs more than 30 ounces. (c) P(at least one box is greater than 30 ounces) = 1 − P(none of the boxes is greater than 30 ounces) = 1 − (1 − 0.0678)5 = 1 − (0.9322)5 = 0.2960. (d) Because the distribution of W is Normal, the distribution of W will also be Normal, with mean m — = 27 ounces and standard deviation s — = W W 30 − 27 2.01 = 3.34 = 0.899. We want to find P(W > 30). z = 0.899 "5

and P(Z > 3.34) = 0.0004. There is a 0.0004 probability of randomly selecting 5 boxes that have a mean weight of more than 30 ounces. AP3.35  (a) S: H0: mA − mB = 0 versus Ha: mA − mB ∙ 0, where mA is the true mean annualized return for stock A and mB is the true mean annualized return for stock B. P: Two-sample t test. Random: Independent random samples. 10%: nA = 50 is less than 10% of all days in the past 5 years and nB = 50 is less than 10% of all days in the past 5 years. Normal/Large Sample: nA = 50 ≥ 30 and nB = 50 ≥ 30. D: t = 2.07. Using df = 40, the P-value is between 0.04 and 0.05. Using df = 90.53, P-value = 0.0416. C: Because the P-value of 0.0416 is less than a = 0.05, we reject H0 . We have convincing evidence that the true mean annualized return for stock A is different than the true mean annualized return for stock B. (b) Ho: sA − sB = 0 vs. Ha: sA − sB > 0, where sA is the true standard deviation of returns for stock A and sB is the true standard deviation of returns for stock B. (c) When the standard deviation of stock A is greater than the standard deviation of stock B, the variance of stock A will be bigger than the variance of stock B. Thus, values of F that are significantly greater than 1 would indicate that the price volatility for stock A is higher than (12.9)2 that for stock B. (d)F = = 1.806. (e) In the simulation, a (9.6)2 test statistic of 1.806 or greater occurred in only 6 out of the 200 trials. Thus, the approximate P-value is 6∙200 = 0.03. Because the approximate P-value of 0.03 is less than a = 0.05, we reject H0. There is convincing evidence that the true standard deviation of returns for stock A is greater than the true standard deviation of returns for stock B.

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S-49

Chapter 11 Section 11.1 Answers to Check Your Understanding page 684:  1.  H0 : The company’s claimed color distribution for its Peanut M&M’S is correct versus Ha : The company’s claimed color distribution is not correct.  2.  The expected count of both blue and orange candies is 46(0.23) = 10.58, for green and yellow is 46(0.15) = 6.9, and for red and brown is 46(0.12) = 5.52.  (12 − 10.58)2 (7 − 10.58)2 (13 − 6.9)2 (4 − 6.9)2 + + + 10.58 10.58 6.9 6.9 2 2 (8 − 5.52) (2 − 5.52) + = 11.3724  + 5.52 5.52

3. c2 =

page 687:  1.  The expected counts are all at least 5. df = 6−1 = 5. 2.

P-value 0

5

10 15 11.3724

20

Chi-square distribution with 5 df

3.  The P-value is between 0.025 and 0.05 (0.0445).  4.  Because the P-value of 0.0445 < a = 0.05, we reject H0 . There is convincing evidence that the color distribution of M&M’S® Peanut Chocolate Candies is different from what the company claims.  page 691:  S: H0 : The distribution of eye color and wing shape is the same as what the biologists predict versus Ha : The distribution of eye color and wing shape is not what the biologists predict. P: Chi-square test for goodness of fit. Random: Random sample. 10%: n = 200 < 10% of all fruit flies. Large Counts: 112.5, 37.5, 37.5, 12.5 all ≥ 5. D: c2 = 6.1867, df = 3, the P-value is between 0.10 and 0.15 (0.1029). C: Because the P-value of 0.1029 > a = 0.01, we fail to reject H0 . We do not have convincing evidence that the distribution of eye color and wing shape is different from what the biologists predict.

Answers to Odd-Numbered Section 11.1 Exercises 11.1  (a) H0 : The company’s claimed distribution for its deluxe mixed nuts is correct versus Ha : The company’s claimed distribution is not correct. (b) Cashews: 150(0.52) = 78, almonds: 150(0.27) = 40.5, macadamia nuts: 150(0.13) = 19.5, brazil nuts: 150(0.08) = 12. 11.3  c2 =

(83 − 78)2 (29 − 40.5)2 (20 − 19.5)2 (18 − 12)2 + + + 78 40.5 19.5 12

= 6.599 11.5  (a) Expected counts are all at least 5 and df = 3. (b)

P-value 0

2

4 6.599 8

10 12 14

Chi-square distribution with 3 df

(c) The P-value is between 0.05 and 0.10 (0.0858). (d) Because the P-value of 0.0858 > a = 0.05, we fail to reject H0 . We do not have convincing evidence that the company’s claimed distribution for its deluxe mixed nuts is incorrect.

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Solutions

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11.23  The distribution of English grades for the heavy readers is skewed to the left, while the distribution of English grades for the light readers is roughly symmetric. The center of the distribution of English grades is greater for the heavy readers than for the light readers. The English grades are more variable for the light readers. There is one low outlier in the heavy reading group but no outliers in the light reading group. 11.25  (a) For each additional book read, the predicted English GPA increases by about 0.024. The predicted English grade for a student who has read 0 books is about 3.42. (b) residual 52.85 − 3.828 = −0.978. This student’s English GPA is 0.978 less than predicted, based on the number of books this student has read. (c) Not very strong. On the scatterplot, the points are quite spread out from the line. Also, the value of r2 is 0.083, which means that only 8.3% of the variation in English grades is accounted for by the linear model relating English GPA to number of books read.

Section 11.2 Answers to Check Your Understanding

Percent of campus

page 699:  1.  Main: 0.060 several times a month or less, 0.236 at least once a week, 0.703 at least once a day. Commonwealth: 0.121 several times a month or less, 0.250 at least once a week, 0.628 at least once a day.  2.  Because there was such a big difference in the sample size from the two different types of campuses.  3.  Students on the main campus are more likely to be everyday users of Facebook. Also, those on the commonwealth campuses are more likely to use Facebook several times a month or less.

Facebook use

70 60 50 40 30 20 10 0

er al tim A e tl ea s a m st on o nc A t tl ea h ea w st e on e ce k Se ad ve ra ay lt im A es tl ea am st on A onc t tl ea h ea w st on eek ce ad ay

11.7  S: H0 : Nuthatches do not prefer particular types of trees when searching for seeds and insects versus Ha : Nuthatches do prefer particular types of trees when searching for seeds and insects. P: Chi-square test for goodness of fit. Random: Random sample. 10%: n = 156 < 10% of all nuthatches. Large Counts: 84.24, 62.4, 9.36 all ≥ 5. D: c2 = 7.418. With df = 2, the P-value is between 0.02 and 0.025 (0.0245). C: Because the P-value of 0.0245 < a = 0.05, we reject H0 . There is convincing evidence that nuthatches prefer particular types of trees when they are searching for seeds and insects. 11.9  Time spent doing homework is quantitative. Chi-square tests for goodness of fit should be used only for distributions of categorical data. 11.11  (a) S: H0: The first digit of invoices from this company follow Benford’s law versus Ha: The first digit of invoices from this company do not follow Benford’s law. P: Chi-square test for goodness of fit. Random: Random sample. 10%: Assume n = 250 < 10% of all invoices from this company. Large Counts: 75.25, 44, 31.25, 24.25, 19.75, 16.75, 14.5, 12.75, 11.5 all ≥ 5. D: c2 = 21.563. With df = 8, the P-value is between 0.005 and 0.01 (0.0058). C: Because the P-value of 0.0058 < a = 0.05, we reject H0 . There is convincing evidence that the first digit of invoices from this company do not follow Benford’s law. Follow-up analysis: The largest contributors to the statistic are amounts with first digit 3, 4 and 7. There are more invoices that start with 3 or 4 than expected and fewer invoices that start with 7 than expected. (b) I: Finding convincing evidence that the company’s invoices do not follow Benford’s law (suggesting fraud), when in reality they are consistent with Benford’s law. A consequence is falsely accusing this company of fraud. II: Not finding convincing evidence that the invoices do not follow Benford’s law (suggesting fraud), when in reality they do not. A consequence is allowing this company to continue committing fraud. A Type I error would be more serious for the accountant. 11.13  (a) H0 : The true distribution of flavors for Skittles candies is the same as the company’s claim versus Ha : The true distribution of flavors for Skittles candies is not the same as the company’s claim. (b) Expected counts all = 12. (c) Using df = 4, c2 statistics greater than 9.49 would provide significant evidence at the a = 0.05 level and c2 values greater than 13.28 would provide significant evidence at the a = 0.01 level. (d) Answers will vary. 11.15  S: H0 : All 12 astrological signs are equally likely versus Ha : All 12 astrological signs are not equally likely. P: Chi-square test for goodness of fit. Random: Random sample. 10%: n = 4344 < 10% of all people in the United States. Large Counts: All expected counts 5 362, which are ≥ 5. D: c2 = 19.76. With df = 11, the P-value is between 0.025 and 0.05 (0.0487). C: Because the P-value of 0.0487 < a = 0.05, we reject H0 . There is convincing evidence that the 12 astrological signs are not equally likely. Follow-up analysis: The largest contributors to the statistic are Aries and Virgo. There are fewer Aries (321 − 362 = −41) and more Virgos (402 − 362 = 40) than we would expect. 11.17  S: H0 : Mendel’s 3:1 genetic model is correct versus Ha : Mendel’s 3:1 genetic model is not correct. P: Chi-square test for goodness of fit. Conditions are met. D: c2 = 0.3453. With df = 1, the P-value > 0.25 (0.5568). C: Because the P-value of 0.5568 > a = 0.05, we fail to reject H0 . We do not have convincing evidence that Mendel’s 3:1 genetic model is wrong. 11.19  d 11.21  c

Se v

S-50

Campus type

Main

Commonwealth

page 705: 1.  H0 : There is no difference in the distributions of Facebook use among students at the main campus and students at the commonwealth campuses versus Ha : There is a difference in the distributions of Facebook use among students at the main campus and students at the commonwealth campuses.  (131)(910) (131)(627) (372)(910) = 77.56, = 53.44, = 220.25, 2.  1537 1537 1537 (372)(627) (1034)(910) (1034)(627) = 151.75, = 612.19, = 421.81  1537 1537 1537 (55 − 77.56)2 c (394 − 421.81)2 3.  c2 = + + = 19.49  77.56 421.81 4.  With df = 2, the P-value < 0.0005 (0.000059).  5.  Assuming that no difference exists in the distributions of Facebook use between students on Penn State’s main campus and students at Penn State’s commonwealth campuses, there is a 0.000059 probability of observing samples that show a difference in the distributions of Facebook

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Solutions use among students at the main campus and the commonwealth campuses as large or larger than the one found in this study. 6.  Because the P-value of 0.000059 < a = 0.05, we reject H0. There is convincing evidence that the distribution of Facebook use is different among students at Penn State’s main campus and students at Penn State’s commonwealth campuses. page 711:  1. Percent of country

40 30 20 10

So Mu m ch A ewh be So bou at b tter m t th et ew e te h sa r M at w me uc o h rse w or se So Mu m ch e A wh be So bou at b tter m t th et ew e te h sa r M at w me uc o h rse w or se

0 Quality of life

Canada

Country

U.S.

2.  S: H0 : There is no difference in the distribution of quality of life for patients who have suffered a heart attack in Canada and the U.S. versus Ha : There is a difference. . . . P: Chi-square test for homogeneity. Random: Independent random samples. 10%: n1 = 311 < 10 % of all Canadian heart attack patients and n2 = 2165 < 10% of all U.S. heart attack patients. Large Counts: 77.37, 538.63, 71.47, 497.53, 109.91, 765.09, 41.70, 290.30, 10.55, 73.45 all ≥ 5. D: c2 = 11.725. With df = 4, the P-value is between 0.01 and 0.02 (0.0195). C: Because the P-value of 0.0195 > a = 0.01, we fail to reject H0 . There is not convincing evidence that a difference exists in the distribution of quality of life for heart attack patients in Canada and the United States.  page 717:  S: H0: There is no association between an exclusive territory clause and business survival versus Ha: There is an association. . . . P: Chi-square test for independence. Random: Random sample. 10%: We assume that n = 170 < 10% of all new franchise firms. Large Counts: 102.74, 20.26, 39.26, 7.74 all ≥ 5. D: c2 = 5.911. Using df = 1, the P-value is between 0.01 and 0.02 (0.0150). C: Because the P-value of 0.0150 > a = 0.01, we fail to reject H0. There is not convincing evidence of an association between exclusive territory clause and business survival.

Answers to Odd-Numbered Section 11.2 Exercises 11.27  (a) Female: 0.209, 0.104, 0.313, 0.373. Male: 0.463, 0.269, 0.075, 0.194. (b) Percent of gender

50 40 30 20 10

Gender

Female

SC H HM SC LS -LM CLS HM CLM

H

H

SC H HM SC LS -LM CLS HM CLM

0 Goal

Male

(c) In general, it appears that females were classified mostly as low social comparison, whereas males were classified mostly as high social comparison. However, about an equal percentage of males and females were classified as high mastery.

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S-51

11.29  (a) H0 : There is no difference in the distribution of sports goals for male and female undergraduates at this university versus Ha : There is a difference. . . . (b) 22.5, 12.5, 13, 19, 22.5, 12.5, 13, 19. (c) c2 = 24.898. 11.31  (a) Random: Independent random samples. 10%: n1 = 67 < 10% of all males and n2 = 67 < 10% of all females at the university. Large Counts: All expected counts ≥ 5. (b) With df = 3, the P-value < 0.0005 (0.000016). (c) Assuming that no difference exists in the distributions of goals for playing sports among males and females, there is a 0.000016 probability of observing independent random samples that show a difference in the distributions of goals for playing sports among males and females as large or larger than the one found in this study. (d) Because the P-value of 0.000016 < a = 0.05, we reject H0 . There is convincing evidence of a difference in the distribution of goals for playing sports among male and female undergraduates at this university. 11.33  (a) Cold: 0.593 hatched. Neutral: 0.679 hatched. Hot: 0.721 hatched. As the temperature warms up from cold to neutral to hot, the proportion of eggs that hatch appears to increase. (b) S: H0 : There is no difference in the true proportion of eggs that hatch in cold, neutral, or hot water versus Ha : There is a difference. . . . P: Chi-square test for homogeneity. Random: 3 groups in a randomized experiment. Large Counts: 18.63, 38.63, 71.74, 8.37, 17.37, 32.26 all ≥ 5. D: c2 = 1.703. With df = 2, the P-value > 0.25 (0.4267). C: Because the P-value of 0.4267 > a = 0.05, we fail to reject H0 . We do not have convincing evidence that there is a difference in the true proportions of eggs that hatch in cold, neutral, or hot water. 11.35  We do not have the actual counts of the travelers in each category. We also do not know if the sample was taken randomly or if the samples are independent. 11.37  (a) The data are given in the table below. The best success rate is for the patch plus the drug (0.355), followed by the drug alone (0.303). The patch alone (0.164) is just a little better than the placebo (0.156). Nicotine Patch

Drug

Patch plus drug

Placebo

Total

Success

 40

 74

 87

 25

226

Failure

204

170

158

135

667

Total

244

244

245

160

893

(b) Each of the four treatments has the same probability of success for smokers like these. (c) S: H0 : The true proportions of smokers like these who are able to quit for a year are the same for each of the four treatments versus Ha : The true proportions are not the same. . . . P: Chi-square test for homogeneity. Random: 4 groups in a randomized experiment. Large Counts: 61.75, 61.75, 62, 40.49, 182.25, 182.25, 183, 119.51 all ≥ 5. D:c2 = 34.937. With df = 3, the P-value < 0.0005. C: Because the P-value of approximately 0 < a = 0.05, we reject H0 . There is convincing evidence that the true proportions of smokers like these who are able to quit for a year are not the same for each of the four treatments. 11.39  The largest component comes from those who had success using both the patch and the drug (25 more than expected). The next largest component comes from those who had success using just the patch (21.75 less than expected). 11.41  Buyers are much more likely to think the quality of recycled coffee filters is higher, while nonbuyers are more likely to think the quality is the same or lower.

11/20/13 5:23 PM

Solutions 11.59  (a) Experiment, because a treatment (type of rating scale) was deliberately imposed on the students who took part in the study. (b) Several of the expected counts are less than 5.

60 50 40 30 20 10

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Stress management

Exercise

Usual care

Total

Suffered cardiac event

3

7

12

22

No cardiac event

30

27

28

85

Total

33

34

40

107

(b) The success rate was highest for stress management (0.909), followed by exercise (0.794) and usual care (0.70). (c) S: H0 : The true success rates for patients like these are the same for all three treatments versus Ha : The true success rates are not all the same. . . . P: Chi-square test for homogeneity. Random: 3 groups in a randomized experiment. Large Counts: 6.79, 6.99, 8.22, 26.21, 27.01, 31.78 all ≥ 5. D: c2 = 4.840. With df = 2, the P-value is between 0.05 and 0.10 (0.0889). C: Because the P-value > a = 0.05, we fail to reject H0 . We do not have convincing evidence that the true success rates for patients like these are not the same for all three treatments. R11.4  (a) The data could have been collected from 3 independent random samples—a random sample of ads from magazines aimed at young men, a random sample of ads from magazines aimed at young women, and a random sample of ads aimed at young adults in general. In each sample, the ads would be classified as sexual or not sexual. (b) The data could have been collected from a single random sample of ads from magazines aimed at young adults. Then each ad in the sample would be classified as sexual or not sexual, and the magazine that the ad was from would be classified as aimed at young men, young women, or young adults in general. (1113)(576) (351 − 424.8)2 351 (c) = 0.6094 = 60.94%. = 424.8. = 576 1509 424.8 12.82. (The difference is due to rounding error.) (d) The “sexual, Women” cell. There were 225 observed ads in this cell, which was 73.8 more than expected. R11.5  (a) 50 40 30 20 10 0 Goal Gender

Female

Gr ad es Po pu lar Sp ort s

11.43  (a) H0 : There is no association between beliefs about the quality of recycled products and whether or not a person buys recycled products in the population of adults versus Ha : There is an association. . . . (b)13.26, 35.74, 8.66, 23.34, 14.08, 37.92 (c) c2 = 7.64. With df = 2, the P-value is between 0.02 and 0.025 (0.022). (d) Because the P-value of 0.022 < a = 0.05, we reject H0. There is convincing evidence of an association between beliefs about the quality of recycled products and whether or not a person buys recycled products in the population of adults. 11.45  S: H0 : There is no association between education level and opinion about a handgun ban in the adult population versus Ha : There is an association. . . . P: Chi-square test for independence. Random: Random sample. 10%: n = 1201 < 10% of all adults. Large Counts: 46.94, 86.19, 187.36, 94.29, 71.22, 69.06, 126.81, 275.64, 138.71, 104.78 all ≥ 5. D: c2 = 8.525. With df = 4, the P-value is between 0.05 and 0.10 (0.0741). C: Because the P-value of 0.0741 > a = 0.05, we fail to reject H0 . We do not have convincing evidence that there is an association between educational level and opinion about a handgun ban in the adult population. 11.47  (a) Independence, because the data come from a single random sample. (b) H0 : There is no association between gender and where people live in the population of young adults versus Ha : There is an association. . . . (c) Random: Random sample. 10% : n = 4854 < 10% of all young adults. Large Counts: The expected counts are all at least 5. (d) P-value: If no association exists between gender and where people live in the population of young adults, there is a 0.012 probability of getting a random sample of 4854 young adults with an association as strong or even stronger than the one found in this study. Conclusion: Because the P-value of 0.012 < a = 0.05, we reject H0 . There is convincing evidence that an association exists between gender and where people live in the population of young adults. 11.49  (a) Hypotheses: H0 : There is no difference in the improvement rates for patients like these who receive gastric freezing and those who receive the placebo versus Ha : There is a difference. . . . P-value: Assuming that no difference exists in the improvement rates between those receiving gastric freezing and those receiving the placebo, there is a 0.570 probability of observing a difference in improvement rates as large or larger than the difference observed in the study by chance alone. Conclusion: Because the P-value of 0.570 is larger than a = 0.05, we fail to reject H0 . There is not convincing evidence that a difference exists in the improvement rates for patients like these who receive gastric freezing and those who receive the placebo. (b) The P-values are equal and z2 = (−0.57)2 = 0.3249 ≈ c2 = 0.322. 11.51  d 11.53  d 11.55  a 11.57  (a) One-sample t interval for a mean. (b) Two-sample z test for the difference between two proportions.

R11.1  S: H0 : The proposed 1:2:1 genetic model is correct versus Ha : The proposed 1:2:1 genetic model is not correct. P: Chi-square test for goodness of fit. Random: Random sample. 10% : n = 84 < 10% of all yellow-green parent plants. Large Counts: 21, 42, 21 all ≥ 5. D: c2 = 6.476. Using df = 2, the P-value is between 0.025 and 0.05 (0.0392). C: Because the P-value of 0.0392 > a = 0.01, we fail to reject H0 . We do not have convincing evidence that the proposed 1:2:1 genetic model is not correct. R11.2  Several of the expected counts are less than 5. R11.3  (a)

Gr ad es Po pu lar Sp ort s

Nonbuyers

Answers to Chapter 11 Review Exercises

Percent

Buyers

H

H Subject type

ig he r Sa m e Lo w er

0

Quality

ig he r Sa m e Lo w er

Percent of subject type

S-52

Male

11/26/13 3:08 PM

Solutions

S-53

Both groups of children have the largest percentage reporting grades as the goal. But after that, boys were more likely to pick sports, whereas girls were more likely to pick being popular. (b) S: H0 : There is no association between gender and goals for 4th, 5th, and 6th grade students versus Ha : There is an association. . . . P: Chi-square test for independence. Random: Random sample. 10% : n = 478 < 10% of all 4th, 5th, and 6th grade students. Large Counts: 129.70, 117.30, 74.04, 66.96, 47.26, 42.74 all ≥ 5. D: c2 = 21.455. With df = 2, the P-value < 0.0005 (0.00002). C: Because the P-value of 0.00002 < a = 0.05, we reject H0 . There is convincing evidence that an association exists between gender and goals for 4th, 5th, and 6th grade students.

T11.13  (a) S: H0: There is no association between smoking status and educational level among French men aged 20 to 60 years versus Ha: There is an association. . . . P: Chi-square test for independence. Random: Random sample. 10% : n = 459 < 10% of all French men aged 20 to 60 years. Large Counts: 59.48, 44.21, 42.31, 50.93, 37.85, 36.22, 42.37, 31.49, 30.14, 34.22, 25.44, 24.34 all ≥ 5. D: c2 = 13.305. With df = 6, the P-value is between 0.025 and 0.05 (0.0384). C: Because the P-value of 0.0384 < a = 0.05, we reject H0. There is convincing evidence of an association between smoking status and educational level among French men aged 20 to 60 years.

Answers to Chapter 11 AP® Statistics Practice Test

Chapter 12

Percent

T11.1  b T11.2  c T11.3  e T11.4  d T11.5  c T11.6  c T11.7  b T11.8  a T11.9  d T11.10  d T11.11  S: H0: The distribution of gas types is the same as the distributor’s claim versus Ha: The distribution of gas types is not the same as the distributor’s claim. P: Chi-square test for goodness of fit. Random: Random sample. 10% : n = 400 < 10% of all customers at this distributor’s service stations. Large Counts: 240, 80, 80 all ≥ 5. D: c2 = 13.15. With df = 2, the P-value is between 0.001 and 0.0025 (0.0014). C: Because the P-value of 0.0014 < a = 0.05, we reject H0. There is convincing evidence that the distribution of gas type is not the same as the distributor claims. T11.12  (a) Random assignment was used to create three roughly equivalent groups at the beginning of the study. (b) 90 80 70 60 50 40 30 20 10 0

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(c) H0 : The true proportion of spouse abusers like the ones in the study who will be arrested within 6 months is the same for all three police responses versus Ha : The true proportions are not all the same. (d) P-value: If the true proportion of spouse abusers like the ones in the study who will be arrested within 6 months is the same for all three police responses, there is a 0.0796 probability of getting differences between the three groups as large as or larger than the ones observed by chance alone. Conclusion: Because the P-value of 0.0796 is larger than a = 0.05, we fail to reject H0 . There is not convincing evidence that true proportion of spouse abusers like the ones in the study who will be arrested within 6 months is not the same for all three police responses.

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Section 12.1 Answers to Check Your Understanding page 752:  S: b = slope of the population regression line relating fat gain to change in NEA. P: t interval for the slope. Linear: There is no leftover pattern in the residual plot. Independent: The sample size (n = 16) is less than 10% of all healthy young adults. Normal: The histogram of the residuals shows no strong skewness or outliers. Equal SD: Other than one point with a large positive residual, the residual plot shows roughly equal scatter for all x values. Random: Random sample. D: With df = 14, (−0.005032, −0.001852). C: We are 95% confident that the interval from −0.005032 to −0.001852 captures the slope of the population regression line relating fat gain to change in NEA.  page 757:  S: H0: b = 0 versus Ha: b < 0, where b is the slope of the true regression line relating fat gain to NEA change. P: t test for the slope b. D: t = −4.64. P-value ≈ 0.0002 ≈ 0. C: Because the P-value of approximately 0 is less than a = 0.05, we reject H0 . There is convincing evidence that the slope of the true regression line relating fat gain to NEA change is negative.

Answers to Odd-Numbered Section 12.1 Exercises 12.1  The Equal SD condition is not met because the SD of the residuals clearly increases as the laboratory measurement (x) increases. 12.3  Linear: There is no leftover pattern in the residual plot. Independent: Knowing the BAC for one subject should not help us predict the BAC for another subject. Normal: The histogram of the residuals shows no strong skewness or outliers. Equal SD: The residual plot shows roughly equal scatter for all x values. Random: These data come from a randomized experiment. 12.5  a is the true y intercept, which measures the true mean BAC level if no beers had been drunk (a = −0.012701). b is the true slope, which measures how much the true mean BAC changes with the drinking of one additional beer (b = 0.018). Finally, s is the true standard deviation of the residuals, which measures how much the observed values of BAC typically vary from the population regression line (s = 0.0204). 12.7  (a) SEb = 0.0024. If we repeated the experiment many times, the slope of the sample regression line would typically vary by about 0.0024 from the slope of the true regression line for predicting BAC from the number of beers consumed. (b) With df = 14, 0.018 ± 2.977(0.0024) = (0.011,0.025). (c) We are 99% confident that the interval from 0.011 to 0.025 captures the slope of the true regression line for predicting BAC from the number of beers consumed. (d) If we repeated the experiment many times

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Solutions

Death rate (per 100,000)

and computed a confidence interval for the slope each time, about 99% of the resulting intervals would contain the slope of the true regression line for predicting BAC from the number of beers consumed. 12.9  S: b = the slope of the population regression line relating number of clusters of beetle larvae to number of stumps. P: t interval for b. D: With df = 21, (8.678, 15.11). C: We are 99% confident that the interval from 8.678 to 15.11 captures the slope of the population regression line relating number of clusters of beetle larvae to number of stumps. 12.11  (a) y^ = −1.286 + 11.894(5) = 58.184 clusters. (b) s = 6.419, so we would expect our prediction to be off from the actual number of clusters by about 6.419 clusters. 12.13  (a) y^ = 166.483 − 1.0987x, where y^ is the predicted corn yield and x is the number of weeds per meter. Slope: for each additional weed per meter, the predicted corn yield will decrease by about 1.0987 bushels/acre. y intercept: if there are no weeds per meter, we would predict a corn yield of 166.483 bushels/acre. (b) When using weeds per meter to predict corn yield, the actual yield will typically vary from the predicted yield by about 7.98 bushels/ acre. (c) S: H0: b = 0 versus Ha: b < 0, where b is the slope of the true regression line relating corn yield to weeds per meter. P: t test for b. D: t = −1.92. P-value = 0.0752 = 0.0375. C: Because the P-value of 0.0375 is less than a = 0.05, we reject H0 . There is convincing evidence that the slope of the true regression line relating corn yield to weeds per meter is negative. 12.15  S: H0: b = 0 versus Ha: b < 0, where b is the slope of the population regression line relating heart disease death rate to wine consumption in the population of countries. P: t test for b. Linear: There is no leftover pattern in the residual plot. Independent: The sample size (n = 19) is less than 10% of all countries. Normal: The histogram of residuals shows no strong skewness or outliers. Equal SD: The residual plot shows that the standard deviation of the death rates might be a little smaller for large values of wine consumption, x, but it is hard to tell with so few data values. Random: Random sample.

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D: t = −6.46, df = 17, and P-value ≈ 0. C: Because the P-value of approximately 0 is less than a = 0.05, we reject H0. There is convincing evidence of a negative linear relationship between wine consumption and heart disease death rate in the population of ­countries. 12.17  (a) With df = 19, 11,630.6 ± 2.093(1249) = (9016.4, 14,244.8). (b) Because the automotive group claims that people drive 15,000 miles per year, this says that for every increase of 1 year, the mileage would increase by 15,000 miles. (c) t = −2.70. With df = 19, the P-value is between 0.01 and 0.02 (0.0142). Because the P-value of 0.0142 is less than a = 0.05, we reject H0 . We have convincing evidence that the slope of the population regression line relating miles to years is not equal to 15,000. (d) Yes. Because the interval in part (a) does not include the value 15,000, the interval also provides convincing evidence that the slope of the population regression line relating miles to years is not equal to 15,000. 12.19  c 12.21  a 12.23  b 12.25  (a) The two treatments (say the color, read the word) were deliberately assigned to the students. (b) He used a randomized block design where each student was a block. He did this to help account for the different abilities of students to read the words or to say the color they were printed in. (c) To help average out the effects of the order in which people did the two treatments. If every subject said the color of the printed word first and were frustrated by this task, the times for the second treatment might be worse. Then we wouldn’t know the reason the times were longer for the second treatment—because of frustration or because the second method actually takes longer. 12.27  There is a small number of differences (nd = 16 < 30) and there is an outlier. 295 295 + 77 + 212 = 0.1933. (ii) = 0.3827. 12.29  (a) (i) 1526 1526 212 (iii) = 0.6951. (b) No. The probability that a person is a snow305 mobile owner (2951526 = 0.1933) is different from the probability that the person is a snowmobile owner given that he or she belongs to an environmental organization (16305 = 0.0525).

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(ii) P(at least one belongs to an environmental organization)

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= 1 − P(neither belong) = 1 − a

1221 1220 ba b = 0.3599 1526 1525

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Solutions

8

®

Option 3: ln(premium) = −0.063 + 0.0859(58) = 4.9192 S y^ = e4.9192 = $136.89 2.  Exponential (Option 3), because the scatterplot showing ln(premium) versus age was the most linear and this model had the most randomly scattered residual plot.

bounce number is more linear. (b) log y = 0.45374 − 0.11716x, where y = height in feet and x = bounce number. (c) log y = 0.45374 − 0.11716(7) = − 0.36638, so y^ = 10 −0.36638 = 0.43 feet. (d) The trend in the residual plot suggests that the residual for x = 7 would be positive, meaning that the predicted height will be less than the actual height. 12.45  (a) There is a strong, positive curved relationship between heart weight and length of left ventricle for mammals. 4000

Weight (g)

®

Option 2: ln(premium) = −12.98 + 4.416(ln 58) = 4.9509 S y^ = e4.9509 = $141.30

surviving bacteria and x is time in minutes. (d) ln y = 5.97316 − 0.218425(17) = 2.26, so y^ = e2.26 = 9.58 or 958 bacteria. 12.43  (a) Exponential, because the scatterplot of log(height) versus

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page 782:  1.  Option 1: premium = −343 + 8.63(58) = $157.54

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Section 12.2 Answers to Check Your Understanding

Answers to Odd-Numbered Section 12.2 Exercises

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Length (cm)

(b) Two scatterplots are given below. Because the relationship between ln(weight) and ln(length) is roughly linear, heart weight and length seem to follow a power model.

Length (cm)

(b) The class used the square root of x = length. (c) The class used the square of y = period. 12.33  (a) 1: y^ = −0.08594 + 0.21"x, where y is the period and x

ln (weight)

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12.37  log y = −0.73675 + 0.51701 log(80) = 0.24717 . Thus, y^ = 100.24717 = 1.77 seconds. h

12.39  log y = 1.01 + 0.72 log(127) = 2.525. Thus, y^ = 102.525 = 334.97 grams. 12.41  (a) The relationship between bacteria count and time is strong, negative, and curved with a possible outlier in the top lefthand corner.  400 300 200 100 0 0

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(b) Because the scatterplot of ln(count) versus time is fairly linear.  (c) ln y = 5.97316 − 0.218425x, where y is the count of

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(c) ln y = −0.314 + 3.1387 ln x, where y is the weight of the heart and x is the length of the cavity of the left ventricle. (d)  ln y = −0.314 + 3.1387 ln (6.8) = 5.703, so  y^ = e5.703 = 299.77 grams. 12.47  c 12.49  e 12.51  (a) For Marcela, X = the length of her shower on a randomly selected day follows a Normal distribution with mean 4.5 minutes and standard deviation 0.9 minutes. We want to find 3 − 4.5 6 − 4.5 P(3 < X < 6). z = = − 1.67 and z = = 1.67, so 0.9 0.9 P(3 < X < 6) = 0.9050. Using technology: 0.9044. There is a 0.9044 probability that Marcela’s shower lasts between 3 and 6 minQ1 − 4.5 gives Q1 = 3.897 minutes. utes. (b) Solving − 0.67 = 0.9 Q3 − 4.5 Solving 0.67 = gives Q3 = 5.103 minutes. Using technol0.9 ogy: Q1 = 3.893 minutes and Q3 = 5.107 minutes. Thus, an outlier is any value above 5.107 + 1.5(5.107 − 3.893) = 6.928. Because 7 > 6.928, a shower of 7 minutes would be considered an outlier for Marcela. (c) P(X > 7) = 0.0027. Let Y = the number of days that Marcela’s shower is 7 minutes or higher. Y is a binomial random variable with n = 10 and p = 0.0027. P(Y ≥ 2) = 1 − P(Y ≤ 1) = 1 − binomcdf(trials: 10, p: 0.0027, x value: 1) = 0.0003. (d) x follows a N(4.5, 0.285) distribution and we want to 5 − 4.5 = 1.75 and P(Z > 1.75) =. 0.0401 Using techfind P(x > 5) . z = 0.285 nology: 0.0397. There is a 0.0397 probability that the mean length of Marcela’s showers on these 10 days exceeds 5 minutes.

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terns. (b) log y = −0.73675 + 0.51701 log(x), where y is the period and x is the length.

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y2 = −0.15465 + 0.0428(80) = 3.269, so y^ = "3.269 = 1.808 seconds. 12.35  (a) The scatterplot of log(period) versus log(length) is roughly linear and the residual plot shows no obvious leftover pat-

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is the length. 2: y2 = −0.15465 + 0.0428x, where y is the period and x is the length. (b) 1: y^ = −0.08594 + 0.21"80 = 1.792 seconds. 2:

Count (hundreds of bacteria)

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Period (s)

12.31  (a) The scatterplot shows a fairly strong, positive, slightly curved association between length and period with one very unusual point (106.5, 2.115) in the top right corner. 2.2 2.0 1.8 1.6 1.4 1.2 1.0 0.8 0.6

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Solutions

–2.5 –0.0 –2.5 –5.0 –7.5 0

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(b)  There is a leftover pattern in the residual plot, so the relationship between practice time and percent of words recalled is not linear. (c)  Power, because the scatterplot showing ln(recall) versus ln(time) is more linear than the scatterplot showing ln(recall) versus time. (d) Power: ln y = 3.48 + 0.293 ln (25) = 4.423 and y^ = e4.423 = 83.35 percent of words recalled. Exponential: ln y = 3.69 + 0.0304(25) = 4.45 and y^ = e4.45 = 85.63 percent of words recalled. Based on my answer to part (c), I think the power model will give a better prediction.

Answers to Chapter 12 AP® Statistics Practice Test T12.1  c T12.2  b T12.3  d T12.4  a T12.5  d T12.6  d T12.7  e T12.8  d T12.9  d T12.10  c T12.11  (a) y^ = 4.546 + 4.832x, where y is the weight gain and x is the dose of growth hormone. (b) (i) For each 1-mg increase in growth hormone, the predicted weight gain increases by about 4.832 ounces. (ii) If a chicken is given no growth hormone (x = 0), the predicted weight gain is 4.546 ounces. (iii) When using the least-squares regression line with x = dose of growth hormone to predict y = weight gain, we will typically be off by about 3.135 ounces. (iv) If we repeated this experiment many times, the sample slope will typically vary by about 1.0164 from the true slope of the least-squares regression line with y = weight gain and x = dose of growth hormone. (v) About 38.4% of the variation in weight gain is accounted for by the linear model relating weight gain to the dose of growth hormone. (c) S: H0:b = 0 versus Ha:b ≠ 0, where b is the slope of the true regression line relating y = weight gain to x = dose of growth hormone. P: t test for b. D: t = 4.75, df = 13, and P-value = 0.0004. C: Because the P-value of 0.0004 is less than a = 0.05, we reject H0 . There is convincing evidence of a linear relationship between the dose of growth hormone and weight gain for chickens like these. (d) D: With df = 13, (2.6373, 7.0273). C: We are 95% confident that the interval from 2.6373 to 7.0273 captures the slope of the true regression line relating y = weight gain to x = dose of growth hormone for chickens like these. T12.12  (a) There is clear curvature evident in both the scatterplot and the residual plot. (b) 1: y^ = 2.078 + 0.0042597(30)3 = 117.09 board feet. 2: ln y = 1.2319 + 0.113417(30) = 4.63441 and y^ = e4.63441 = 102.967 board feet. (c) The residual plot for Option 1 is much more scattered, while the plot for Option 2 shows curvature, meaning that the model from Option 1 relating the amount of usable lumber to cube of the diameter is more appropriate.

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R12.1  (a) There is a moderately strong, positive linear relationship between the thickness and the velocity. (b) y^ = 70.44 + 274.78x, where y is the velocity and x is the thickness. (c) Residual = 104.8 − 180.352 = − 75.552, so the line overpredicts the velocity by 75.552 ft/sec. (d) The linear model is appropriate. The scatterplot shows a linear relationship and the residual plot has no leftover patterns. (e) Slope: For each increase of an inch in thickness, the predicted velocity increases by 274.78 feet/second. s: When using the least-squares regression line with x = thickness to predict y = velocity, we will typically be off by about 56.36 feet per second. r2: About 49.3% of the variation in velocity is accounted for by the linear relationship relating velocity to thickness. SEb : If we take many different random samples of 12 pistons and compute the least-squares regression line for each sample, the estimated slope will typically vary from the slope of the population regression line for predicting velocity from thickness by about 88.18. R12.2  S: H0:b = 0 versus Ha:b ≠ 0, where b is the slope of the population regression line relating thickness to velocity. P: t test for b. Linear: The residual plot shows no leftover patterns. Independent: Knowing the velocity for one piston should not help us predict the velocity for another piston. Also, the sample size (n = 12) is less than 10% of the pistons in the population. Normal: We are told that the Normal probability plot of the residuals is roughly linear. Equal SD: The residual plot shows roughly equal scatter for all x values. Random: The data come from a random sample. D: t = 3.116. With df = 10, the P-value is between 0.01 and 0.02 (0.0109). C: Because the P-value of 0.0109 is less than a = 0.05, we reject H0. There is convincing evidence of a linear relationship between thickness and gate velocity in the population of pistons formed from this alloy of metal. R12.3  D: With df = 12 − 2 = 10, (78.315, 471.245). C: We are 95% confident that the interval from 78.315 to 471.245 captures the slope of the population regression line for predicting velocity from thickness for the population of pistons formed from this alloy of metal. Because 0 is not in the interval, we reject 0 as a plausible value for the slope of the population regression line, as in R12.2. R12.4  The Linear condition is violated because there is clear curvature to the scatterplot and an obvious curved pattern in the residual plot. The Random condition may not be met because we weren’t told if the sample was selected at random. R12.5  (a) Yes, because there is no leftover pattern in the residual 1 plot. (b) y^ = − 0.000595 + 0.3a 2 b . Here, y = intensity and x = x 1 distance. (c) y^ = − 0.000595 + 0.3a b = 0.0674 candelas. (2.1)2

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Answers to Chapter 12 Review Exercises

R12.6  (a)

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12.53  (a) S: p = true proportion of all AP® teachers attending this workshop who have tattoos. P: One-sample z interval for p. Random: Random sample. 10%: The sample size (n = 98) is less than 10% of the population of teachers at this workshop (1100). Large Counts: 23 and 75 are both ≥ 10. D: (0.151, 0.319). C: We are 95% confident that the interval from 0.151 to 0.319 captures the true proportion of AP® teachers at this workshop who have tattoos. (b) Yes. Because the value 0.14 is not included in the interval, we have convincing evidence that the true proportion of teachers at the workshop who have a tattoo is not 0.14. (c) If we had two more failures, the interval will shift to lower values and might include the value 0.14. However, the new interval is (0.148, 0.312), which does not include the value 0.14. So the answer would not change if we got responses from the 2 nonresponders.

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Solutions

Answers to Cumulative AP® Practice Test 4

®

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not be generalizable to women in other racial and ethnic groups or who live in other states. (d) Two variables are confounded when their effects on the response variable cannot be distinguished from one another. For example, women who were physically active as teens might have also done other things differently as well, such as eating a healthier diet. We would be unable to determine if it was their physically active youth or their healthier diet that slowed their level of cognitive decline. AP4.43  (a) Because the first question called it a “fat tax,” people may have reacted negatively because they believe this is a tax on those who are overweight. The second question provides extra information that gets people thinking about the obesity problem in the U.S. and the increased health care that could be provided as a benefit with the tax money. Better: “Would you support or oppose a tax on non-diet sugared soda?” (b) This method samples only people at fast-food restaurants. They may go to these restaurants because they like the sugary drinks and wouldn’t want to pay a tax on their favorite beverages. Thus, it is likely that the proportion of those who would oppose such a tax will be overestimated with this method. Better: take a random sample of all New York State residents. (c) Use a stratified random sampling method in which each state is a stratum. AP4.44  (a) P(S) = (0.1)(0.3) + (0.4)(0.2) + (0.5)(0.1) = 0.16. (0.5)(0.1) (b) P(C 0 S) = = 0.3125. (0.1)(0.3) + (0.4)(0.2) + (0.5)(0.1) AP4.45  (a) No. The scatterplot exhibits a strong curved pattern. (b) B, because the scatterplot shows a much more linear pattern and its residual plot shows no leftover patterns. (c) ln(weight) = 15.491 − 1.5222 ln (3700) = 2.984, thus

h

AP4.1  e AP4.2  c AP4.3  e AP4.4  a AP4.5  b AP4.6  e AP4.7  d AP4.8  a AP4.9  e AP4.10  a AP4.11  b AP4.12  d AP4.13  e AP4.14  d AP4.15  b AP4.16  a AP4.17  d AP4.18  c AP4.19  a AP4.20  b AP4.21  e AP4.22  e AP4.23  b AP4.24  c AP4.25  d AP4.26  e AP4.27  c AP4.28  c AP4.29  b AP4.30  b AP4.31  d AP4.32  a AP4.33  e AP4.34  b AP4.35  a AP4.36  b AP4.37  d AP4.38  a AP4.39  d AP4.40  d AP4.41  S: H0:m1 − m2 = 0 vs. Ha:m1 − m2 ≠ 0, where m1 = true mean difference in electrical potential for diabetic mice and m2 = true mean difference in electrical potential for normal mice. P: Two-sample t test for m1 − m2. Random: Independent random samples. 10%: n1 = 24 is less than 10% of all diabetic mice and n2 = 18 is less than 10% of all normal mice. Normal/Large Sample Size: No outliers or strong skewness. D: t = 2.55. Using df = 23, the P-value is between 0.01 and 0.02. Using df = 38.46, P-value = 0.0149. C: Because the P-value of 0.0149 is less than a = 0.05, we reject H0. There is convincing evidence that the true mean difference in electric potential for diabetic mice is different than for normal mice. AP4.42  (a) H0:p1 − p2 = 0 vs. Ha:p1 − p2 < 0, where p1 = the true proportion of women like the ones in the study who were physically active as teens that would suffer a cognitive decline and p2 = the true proportion of women like the ones in the study who were not physically active as teens that would suffer a cognitive decline. (b) A two-sample z test for p1 − p2 . (c) No. Because the participants were mostly white women from only four states, the findings may

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weight = e2.984 = 19.77mg. (d) About 86.3% of the variation in ln(seed weight) is accounted for by the linear model relating ln(seed weight) to ln(seed count). AP4.46  (a) Let X = diameter of a randomly selected lid. Because X follows a Normal distribution, the sampling distribution of x also 0.02 follows a Normal distribution. mx = 4 inches and sx = = 0.004 "25 0.004 inches. (b) We want to find P(x < 3.99 or x > 4.01) using the 3.99 − 4 4.01 − 4 N(4, 0.004) distribution. z = = −2.50 and z = = 0.004 0.004 2.50. P(Z < −2.50 or Z > 2.50) = 0.0124. Assuming that the machine is working properly, there is a 0.0124 probability that the mean diameter of a sample of 25 lids is less than 3.99 inches or greater than 4.01 inches. (c) We want to find P(4 < x < 4.01) 4−4 using the N(4, 0.004) distribution. z = = 0 and 0.004 4.01 − 4 z= = 2.50. P(0 < Z < 2.50) = 0.4938. Assuming that 0.004 the machine is working properly, there is a 0.4938 probability that the mean diameter of a sample of 25 lids is between 4.00 and 4.01 inches. (d) Let Y = the number of samples (out of 5) in which the sample mean is between 4.00 and 4.01. The random variable Y has a binomial distribution with n = 5 and p = 0.4938. Using technology: P(X ≥ 4) = 1 − P(X ≤ 3) = 1 − binomcdf (trials:5, p:0.4938, x value:3) = 0.1798. (e) Because the probability found in part (b) is less than the probability found in part (d), getting a sample mean below 3.99 or above 4.01 is more convincing evidence that the machine needs to be shut down. This event is much less likely to happen by chance when the machine is working correctly. (f) Answers will vary.

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Appendix A  About the AP® Exam

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Appendix A  About the AP® Exam Chapter 1 If you learn to distinguish categorical from quantitative variables now, it will pay big rewards later. You will be expected to analyze categorical and quantitative variables correctly on the AP® exam. ●● 

When comparing distributions of quantitative data, it’s not enough just to list values for the center and spread of each distribution. You have to explicitly compare these values, using words like “greater than,” “less than,” or “about the same as.” ●● 

If you’re asked to make a graph on a free-response question, be sure to label and scale your axes. Unless your calculator shows labels and scaling, don’t just transfer a calculator screen shot to your paper. ●● 

You may be asked to determine whether a quantitative data set has any outliers. Be prepared to state and use the rule for identifying outliers. ●● 

Use statistical terms carefully and correctly on the AP® exam. Don’t say “mean” if you really mean “median.” Range is a single number; so are Q1, Q3, and IQR. Avoid colloquial use of language such as “the outlier skews the mean.” Skewed is a shape. If you misuse a term, expect to lose some credit.

●● 

Chapter 2 Normal probability plots are not included on the AP® Statistics topic outline. However, these graphs are very useful for assessing Normality. You may use them on the AP® exam if you wish—just be sure that you know what you’re looking for (a linear pattern).

●● 

Chapter 3 If you are asked to make a scatterplot for a free-response question, be sure to label and scale both axes. Don’t just copy an unlabeled calculator graph directly onto your paper. ●● 

If you’re asked to interpret a correlation, start by looking at a scatterplot of the data. Then be sure to address direction, form, strength, and outliers (sound familiar?) and put your answer in context. ●● 

When displaying the equation of a least-squares regression line, the calculator will report the slope and intercept with much more precision than is needed. However, there is no firm rule for how many decimal places to show for answers on the AP® exam. Our advice: Decide how much to round based on the context of the problem you are working on. ●● 

●● Students often have a hard time interpreting the value of r2 on AP® exam questions. They frequently leave out key words in the definition. Our advice: Treat this as a fill-in-the-blank exercise. Write “________% of the variation in [response variable name] is accounted for by the linear model relating [response variable name] to [explanatory variable name].”

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The formula sheet for the AP® exam uses different notation for sy these equations: b1 = r and b0 = y − b1x. That’s because the sx ­least-squares line is written as y^ = b0 + b1x. We prefer our simpler versions without the subscripts! ●● 

Chapter 4 If you’re asked to describe how the design of a study leads to bias, you’re expected to do two things: (1) identify a problem with the design, and (2) explain how this problem would lead to an underestimate or overestimate. Suppose you were asked, “Explain how using your statistics class as a sample to estimate the proportion of all high school students who own a graphing calculator could result in bias.” You might respond, “This is a convenience sample. It would probably include a much higher proportion of students with a graphing calculator than would the population at large because a graphing calculator is required for the statistics class. So this method would probably lead to an overestimate of the actual population proportion.” ●● 

If you are asked to identify a possible confounding variable in a given setting, you are expected to explain how the variable you choose (1) is associated with the explanatory variable and (2) affects the response variable. ●● 

If you are asked to describe the design of an experiment on the AP® exam, you won’t get full credit for a diagram like Figure 4.5 (page 246). You are expected to describe how the treatments are assigned to the experimental units and to clearly state what will be measured or compared. Some students prefer to start with a diagram and then add a few sentences. Others choose to skip the diagram and put their entire response in narrative form. ●● 

Don’t mix the language of experiments and the language of sample surveys or other observational studies. You will lose credit for saying things like “Use a randomized block design to select the sample for this survey” or “This experiment suffers from nonresponse since some subjects dropped out during the study.” ●● 

Chapter 5 On the AP® exam, you may be asked to describe how you will perform a simulation using rows of random digits. If so, provide a clear enough description of your simulation process for the reader to get the same results you did from only your written ­explanation. ●● 

Many probability problems involve simple computations that you can do on your calculator. It may be tempting to write down just your final answer without showing the supporting work. Don’t do it! A “naked answer,” even if it’s correct, will usually earn you no credit on a free-response question. ●● 

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Appendix A  About the AP® Exam

●● You can write statements like P(B | A) if events A and B are defined clearly, or you can use a verbal equivalent, such as P(reads New York Times | reads USA Today). Use the approach that makes the most sense to you.

paper to receive credit. You don’t have to draw multiple graphs— any appropriate graph will do.

Chapter 9 The conclusion to a significance test should always include three components: (1) an explicit comparison of the P-value to a stated significance level, (2) a decision about the null hypothesis: reject or fail to reject H0 , and (3) a statement in the context of the problem about whether or not there is convincing evidence for Ha . ●● 

Chapter 6 ●● If the mean of a random variable has a noninteger value but you report it as an integer, your answer will not get full credit.

When showing your work on a free-response question, you must include more than a calculator command. Writing ­normalcdf(68,70, 64, 2.7) will not earn you full credit for a Normal calculation. At a minimum, you must indicate what each of those calculator inputs represents. Better yet, sketch and label a Normal curve to show what you’re finding. ●● 

Don’t rely on “calculator speak” when showing your work on freeresponse questions. Writing binompdf(5, 0.25, 3)= 0.08789 will not earn you full credit for a binomial probability calculation. At the very least, you must indicate what each calculator input represents. For example, “I used binompdf(trials 5, p:0.25,  x value:3).” ●● 

Chapter 7 Terminology matters. Don’t say “sample distribution” when you mean sampling distribution. You will lose credit on free-response questions for misusing statistical terms.

●● When a significance test leads to a fail to reject H 0 decision, be sure to interpret the results as “We don’t have enough evidence to conclude Ha .” Saying anything that sounds like you believe H0 is (or might be) true will lead to a loss of credit. And don’t write textmessage-type responses, like “FTR the H0 .”

You can use your calculator to carry out the mechanics of a significance test on the AP® exam. But there’s a risk involved. If you give just the calculator answer with no work, and one or more of your values are incorrect, you will probably get no credit for the “Do” step. We recommend doing the calculation with the appropriate formula and then checking with your calculator. If you opt for the calculator-only method, be sure to name the procedure (one-proportion z test) and to report the test statistic (z = 1.15) and P-value (0.1243).

●● 

It is not enough just to make a graph of the data on your calculator when assessing Normality. You must sketch the graph on your paper to receive credit. You don’t have to draw multiple graphs— any appropriate graph will do.

●● 

●● 

●● Notation matters. The symbols p ^ , x, p, μ, s, μ p^ , s p^ , μ x , and s x all have specific and different meanings. Either use notation correctly—or don’t use it at all. You can expect to lose credit if you use incorrect notation.

Remember: If you give just calculator results with no work and one or more values are wrong, you probably won’t get any credit for the “Do” step. If you opt for the calculator-only method, name the procedure (t test) and report the test statistic (t = −0.94), degrees of freedom (df = 14), and P-value (0.1809).

Chapter 8 On a given problem, you may be asked to interpret the confidence interval, the confidence level, or both. Be sure you understand the difference: the confidence interval gives a set of plausible values for the parameter and the confidence level describes the long-run capture rate of the method. ●● 

●● 

Chapter 10 The formula for the two-sample z interval for p1 − p2 often leads to calculation errors by students. As a result, we recommend using the calculator’s 2-PropZInt feature to compute the confidence interval on the AP® exam. Be sure to name the procedure (twoproportion z interval) and to give the interval (0.076, 0.143) as part of the “Do” step.

●● 

If a free-response question asks you to construct and interpret a confidence interval, you are expected to do the entire four-step process. That includes clearly defining the parameter, identifying the procedure, and checking conditions.

●● 

You may use your calculator to compute a confidence interval on the AP® exam. But there’s a risk involved. If you give just the calculator answer with no work, you’ll get either full credit for the “Do” step (if the interval is correct) or no credit (if it’s wrong). We recommend showing the calculation with the appropriate formula and then checking with your calculator. If you opt for the calculatoronly method, be sure to name the procedure (e.g., one-proportion z interval) and to give the interval (e.g., 0.514 to 0.607).

●● 

●● 

●● 

If a question of the AP® exam asks you to calculate a confidence interval, all the conditions should be met. However, you are still required to state the conditions and show evidence that they are met.

●● 

It is not enough just to make a graph of the data on your calculator when assessing Normality. You must sketch the graph on your ●● 

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The formula for the two-sample z statistic for a test about p1 − p2 often leads to calculation errors by students. As a result, we recommend using the calculator’s 2-PropZTest feature to perform calculations on the AP® exam. Be sure to name the procedure (twoproportion z test) and to report the test statistic (z = 1.17) and Pvalue (0.2427) as part of the “Do” step. The formula for the two-sample t interval for m1 − m2 often leads to calculation errors by students. As a result, we recommend using the calculator’s 2-SampTInt feature to compute the confidence interval on the AP® exam. Be sure to name the procedure (twosample t interval) and to give the interval (3.9362, 17.724) and df (55.728) as part of the “Do” step.

When checking the Normal condition on an AP® exam question involving inference about means, be sure to include graphs. Don’t expect to receive credit for describing a graph that you made on your calculator but didn’t put on paper. ●● 

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Appendix A  About the AP® Exam

A-3

Chapter 12

The formula for the two-sample t statistic for m1 − m2 often leads to calculation errors by students. As a result, we recommend using the calculator’s 2-SampTTest feature to perform calculations on the AP® exam. Be sure to name the procedure (two-sample t test) and to report the test statistic (t = 1.60), P-value (0.0644), and df (15.59) as part of the “Do” step.

The AP® exam formula sheet gives y^ = b0 + b1x for the equation of the sample (estimated) regression line. We will stick with our simpler notation, y^ = a + bx, which is also used by TI calculators. Just remember: The coefficient of x is always the slope, no matter what symbol is used.

Chapter 11

The AP® exam formula sheet gives the formula for the standard error of the slope as

●● 

●● 

●● 

You can use your calculator to carry out the mechanics of a significance test on the AP® exam. But there’s a risk involved. If you give just the calculator answer with no work, and one or more of your values is incorrect, you will probably get no credit for the “Do” step. We recommend writing out the first few terms of the chi-square calculation followed by “. . .”. This approach might help you earn partial credit if you enter a number incorrectly. Be sure to name the procedure (c2 GOF-Test) and to report the test statistic (c2 = 11.2), degrees of freedom (df = 3), and P-value (0.011).

∙(y − y^ )

●● 

In the “Do” step, you aren’t required to show every term in the chi-square statistic. Writing the first few terms of the sum followed by “. . .” is considered as “showing work.” We suggest that you do this and then let your calculator tackle the computations. ●● 

You can use your calculator to carry out the mechanics of a significance test on the AP® exam. But there’s a risk involved. If you give just the calculator answer with no work and one or more of your values is incorrect, you will probably get no credit for the “Do” step. We recommend writing out the first few terms of the chi-square calculation followed by “. . .”. This approach might help you earn partial credit if you enter a number incorrectly. Be sure to name the procedure (c2 -Test for homogeneity) and to report the test statistic (c2 = 18.279), degrees of freedom (df = 4), and P-value (0.0011).

i

sb1 =

Ç

∙

i

2

n−2

" (xi − x)2

The numerator is just a fancy way of writing the standard deviation of the residuals s. Can you show that the denominator of this formula is the same as ours? ●● The formula for the t interval for the slope of a population (true) regression line often leads to calculation errors by students. As a result, we recommend using the calculator’s LinRegTInt feature to compute the confidence interval on the AP® exam. Be sure to name the procedure (t interval for slope) and to give the interval (−0.217, −0.108) and df (14) as part of the “Do” step. When you see a list of data values on an exam question, don’t just start typing the data into your calculator. Read the question first. Often, additional information is provided that makes it unnecessary for you to enter the data at all. This can save you valuable time on the AP® exam.

●● 

●● 

If you have trouble distinguishing the two types of chi-square tests for two-way tables, you’re better off just saying “chi-square test” than choosing the wrong type. Better yet, learn to tell the difference!

The formula for the test statistic in a t test for the slope of a population (true) regression line often leads to calculation errors by students. As a result, we recommend using the calculator’s LinRegTTest feature to perform calculations on the AP® exam. Be sure to name the procedure (t test for slope) and to report the test statistic (t = 3.065), P-value (0.002), and df (36) as part of the “Do” step.

●● 

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Appendix B

TI-NspireTM Technology Corners

Texas Instruments released the new TI-Nspire CX in March 2011. The new handheld no longer has an interchangeable TI-84 ­faceplate; however, the body of the Nspire CX is much slimmer and its display is in full color. When you click ctrl menu and arrow down to Color, there are several options available: Line Color, Fill Color, and Text Color. If you choose the Fill Color option, a color palette will appear. Then you can select the colors you want for your graphs. This feature is quite useful when displaying multiple graphs. When creating a bar graph, the CX even allows you to change the color of each bar!

The keystrokes used for the new CX are the same as for the TINspire Touchpad. The keystrokes for the older Nspire “clickpad” are still different in some ways; therefore they are still shown in parentheses when needed. Start by updating your device’s OS to ensure that your handheld has full capabilities. Go to education.ti.com and search under Downloads S Software, Apps, Operating Systems… to download the latest version of the OS. If you have the TI-Nspire computer link software, you should be asked automatically to update your handheld’s OS.

TOUCHPAD

CX

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A p p e n d i x B   T I - N s p i r e tm Te c h n o l o g y C o r n e r s

Chapter 1 TI-Nspire Technology Corners 2.  Histograms on the calculator 1. Insert a New Document by pressing ctrl N . 2. Insert a Lists & Spreadsheet page by arrowing down to Add Lists & Spreadsheet. • Name column A foreignbrn. • Type the data for the percent of state residents born outside the United States into the list. The data can be found on page 33.

3. Insert a Data & Statistics page: press ctrl I , arrow to Add Data & Statistics, and press enter . • Press tab and Click to Add Variable on the horizontal axis will show the variables available. Select foreignbrn.

• The data should now move into a dotplot. Notice the organization of the graph. Even though the data look “lopsided” in some places, you should consider the dots as being directly above each other in each column.

4. To make a better graphical display, let’s move the data into a histogram. Use the Navpad to position the pointer in an empty space within the graph. Press ctrl menu and select Histogram. You will now see the data move into a histogram.



You can now position the pointer ( ) over each bar to examine the classes. The pointer will become an open hand and the class size will be displayed along with the number of data values in the class.

5. Adjust the classes to match those in Figure 1.16 (page 34). • Arrow into an empty space inside the histogram. • Press ctrl menu and select Bin Settings, Equal Bin Width. Enter the values shown. tab to OK and press enter . The new histogram should be displayed.



Notice how the first bar is “off the page.” To adjust this, arrow over until the becomes .

• Press and hold until becomes . • Use the Navpad and, with the down arrow, “pull” the vertical axis down. Keep arrowing down until the top of the tallest histogram class is visible.

6. See if you can match the histogram in Figure 1.17 (page 35).

3.  Making calculator boxplots One of the added benefits of the TI-Nspire is its ability to plot more than three boxplots at a time in the viewing window. Let’s use the calculator to make parallel boxplots of the travel data for the samples from North Carolina and New York. 1. Insert a Lists & Spreadsheet page: press ctrl I , arrow to Add Lists & Spreadsheet, and press enter . • Name column A ncarolina and column B newyork.

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C h a p t e r 1   T I - N s p i r e TM Te c h n o l o g y C o r n e r s • Enter the travel time data from page 52.

2. Insert a Data & Statistics page: press ctrl I , arrow to Add Data & Statistics, and press enter . • Press tab and Click to Add Variable on the horizontal axis will show the variables available. Select ncarolina. • The data will move into a dotplot. Use the Navpad to position the pointer in an empty space within the graph. Press ctrl menu and select Box Plot. You will now see the data move into a boxplot. • Using the Navpad, arrow over the plot. You will see the values in the five-number summary display one by one as you move across the boxplot.

3. To add the boxplot of newyork travel times, arrow over “ncarolina” on the horizontal axis, press ctrl menu , and choose Add X Variable. Select newyork and the second boxplot will be added to the page.

A-7

• Press menu S Statistics S Stat Calculations S OneVariable Statistics. A dialogue box should appear asking for the number of lists. Press the up arrow ( ) to 2 or type “2.” tab to OK and press enter .

• Another dialogue box should appear. Select the lists in the drop-down boxes: X1 list: ncarolina and X2 list: newyork. tab between the entry boxes to enter the next list and the column where you want the one-variable stats listed: type “c”, tab to OK , and press enter . The numerical summaries for both states should now be displayed.

2. You can resize the columns to see which column contains values for which state: D has summary statistics for ncarolina, and E has summary statistics for newyork. • Use the Navpad to place the arrow between the columns. Press and hold until appears. Use the arrow keys to increase the column width. Press again to release the column.

4.  Computing numerical summaries with technology Let’s find numerical summaries for the travel times of North Carolina and New York workers from the previous Technology Corner. If you haven’t already done so, enter the North Carolina and New York data. I , arrow to • Insert a Lists & Spreadsheet page: press ctrl Add Lists & Spreadsheet, and press enter . Name column A ncarolina and column B newyork. • Arrow down to the first empty cell in column A and type in the data. Repeat the process for newyork in column B. 1. The Nspire can calculate one-variable statistics for several lists at the same time (unlike the TI-84 or TI-89).

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• Repeat the same process to resize the column with one-variable statistics for newyork.

Chapter 2 TI-Nspire Technology Corners 5. From z-scores to areas, and vice versa Finding areas: The normcdf command on the Nspire can be used to find areas under the Normal curve. The syntax is normalcdf (lower bound, upper bound, mean, standard deviation). Let’s use this command to confirm our answers to the examples on pages 116–118. 1. On the Home screen, select the Calculate scratchpad. This will

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A p p e n d i x B   T I - N s p i r e tm Te c h n o l o g y C o r n e r s

take you to a calculator page that is “outside” your current document. Therefore, you do not have to worry about losing/saving a document you are working on or about adding an unneeded page to that document.

value corresponding to a given percentile in a Normal distribution. For this command, the syntax is inv Norm(area to the left, m, s). 3. Let’s start with a “clean slate” by clearing the entries on our page. To do this, press menu S Actions S Clear History. Your scratchpad should now be blank. What is the 90th percentile of the standard Normal distribution? • Press menu S Statistics S Distributions S Inverse Normal. A dialogue box will appear. Type the numbers in the dialogue box as shown. To enter the numbers, tab between the entry boxes. When the last number is entered, tab to OK and press enter .

What proportion of observations from the standard Normal ­distribution are greater than –1.78? Recall that the standard Normal distribution has mean 0 and standard deviation 1. 2. On the Calculate scratchpad, press menu S Statistics S ­Distributions S Normal Cdf. In the dialogue box that appears, type the numbers shown. To move between the drop-down boxes, press tab after typing each number. When the last number is entered, tab to OK and press enter .

The proportion should now be displayed on the main screen. Note: We chose 10,000 as the upper bound because it is many standard deviations above the mean. These results agree with our previous answer using Table A: 0.9625.

What proportion of observations from the ­standard Normal distribution are between –1.25 and 0.81?

These results match our previous answer using Table A.

6. Normal probability plots We will use the state unemployment rates data from page 122 to demonstrate how to make a Normal probability plot for a set of quantitative data. 1. Insert a New Document by pressing ctrl N . 2. Insert a Lists & Spreadsheet page by arrowing down to Add Lists & Spreadsheet. • Name column A unemploy. • Arrow down to the first cell and type in the 50 data values. 3. Insert a Data & Statistics page by pressing ctrl I and use the Navpad to arrow to Add Data & Statistics. Press enter . 4. Press tab to select the “Click to add variable” for the horizontal axis. Arrow to unemploy and press enter to select it. The data will now move into a dotplot. 5. Arrow up into an empty region of the dotplot and press ctrl . menu . Select Normal Probability Plot and press

The following screen shot confirms our earlier result of 0.6854 using Table A.

Interpretation: The Normal probability plot is quite linear, so it is reasonable to believe that the data follow a Normal distribution. Working backward: The Nspire invNorm function calculates the

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C h a p t e r 3   T I - N s p i r e TM Te c h n o l o g y C o r n e r s Chapter 3 TI-Nspire Technology Corners 7. Scatterplots on the calculator 1. Insert a New Document by pressing ctrl N . 2. Insert a Lists & Spreadsheet page by arrowing down to Add Lists & Spreadsheet. • Name column A points and column B wins. • Type the corresponding values into each column. The data list follows. Points:

34.8

36.8

25.7

25.5

32.0

15.8

Wins:

12

11

8

7

10

5

Points:

35.7

16.1

25.3

30.1

20.3

26.7

Wins:

13

2

7

11

5

6

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Miles driven

70,583 129,484 29,932 29,953 24,495

75,678

Price (in dollars)

21,994

9500

28,986 31,891 37,991

Miles driven

34,077

58,023

44,447 68,474 144,162 140,776 29,397 131,385

Price (in ­dollars)

34,995

29,988

22,896 33,961 16,883

29,875 41,995 41,995

8359

4447

20,897 27,495 13,997

1. Insert a New Document by pressing ctrl N . 2. Insert a Lists & Spreadsheet page by arrowing down to Add Lists & Spreadsheet. • Name column A miles and column B price. • Type the corresponding values into each column. 3. Graph the data in a scatterplot putting miles on the horizontal axis and price on the vertical axis. Refer to the previous TI-Nspire Technology Corner. 4. To add a least-squares regression line, first ctrl back to the Lists & Spreadsheet page. 5. Press menu , and arrow to Statistics S Stat Calculations, Linear Regression (a + bx), enter . You should then see a dialogue box. In the drop-down boxes, arrow down to miles for the X List:, then press tab and arrow down to price for the Y List:. tab to OK and press enter .

I 3. Press ctrl and use the Navpad to arrow to Add Data & Statistics. Press enter . 4. Press tab to select the “Click to add variable” for the horizontal axis. Arrow to points and press enter to select it.

The linear regression information, a, b, r2, r, and resid will be displayed in another column within the Lists & Spreadsheet page.

5. Press tab again and the box will move to the vertical axis. ­Select wins. The data will now move into a scatterplot. TI-Nspire labels the x and y axes with the list names, making a well-labeled graph to insert into documents.

6. Press ctrl to return to the Data & Statistics page. Press menu ; arrow to ­Analyze S Regression S Show Linear (a + bx), and press enter . The least-squares regression line along with the equation will appear. If you arrow over the equation, the will appear. Click and hold . When the hand closes, , you can move the equation using the arrow keys.

8. Least-squares regression lines on the calculator Let’s use the Ford F-150 data to show how to find the equation of the least-squares regression line on the TI-Nspire. Here are the data.

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7. Save the document for later use. Press document Truck prices.

ctrl

S

. Name your

9.  Residual plots on the calculator Let’s continue the analysis of the Ford F-150 miles driven and price data from the previous Technology Corner. You should have already made a scatterplot, calculated the equation of the leastsquares regression line, and graphed the line on your plot. Now we want to make a residual plot. 1. Open the document Truck prices. Press ctrl N , arrow through My Documents S Truck prices and press enter . 2. Press ctrl to go to the Data & Statistics page. 3. Press menu ; arrow to Analyze S Residuals S Show Residual Plot. This will split the screen and the residual plot will be displayed below the graph of the least-squares regression line.

3. Name column A students. Arrow down to the formula cell and press enter . students:= should appear. Type seq(x,x,1,1750) and enter . This will put the digits 1 through 1750 in this list. 4. Name column B sampstudents. Arrow down to the formula cell and press enter . sampstudents:= should appear. Type randSamp(students,10,1) and press enter . This function will take a random sample of 10 students from the list. “1” lets the function know to do the sampling without replacement.

Note: Sampling with replacement is the default setting for this function. You can use 0 as the third input in the randSamp command or close the parentheses after the second input.

Chapter 6 TI-Nspire Technology Corners 11.  Analyzing random variables on the calculator

Chapter 4 TI-Nspire Technology Corner 10.  Choosing an SRS The TI-Nspire has a function called randSamp that will randomly select individuals for a sample with or without replacement from a population. 1. Check that your calculator’s random number generator is ­working properly. • Open the Calculator Scratchpad by pressing (or A on the keypad). • Type randint(1,1750) and press enter . • Compare results with your classmates. If several students have the same number, you’ll need to seed your calculator’s random number generator with different numbers before you proceed. Type randSeed, press (to insert a space), type < last four digits of your phone number> and press enter . Done should appear. Now your calculator is ready to generate numbers that are different from those of your classmates.

Let’s explore what the calculator can do using the random variable X = Apgar score of a randomly selected newborn from the example on page 349. 1. Insert a Lists & Spreadsheet page. Press ctrl I , arrow to Add Lists & Spreadsheet, and press enter . • Name column A apgar and column B apgrprob. • Enter the values of the random variable (0 − 10) in the apgar list and the corresponding probabilities in apgrprob.

2. Graph a histogram of the probability distribution. • Insert a Data & Statistics page. Press ctrl I , arrow to Add Data & Statistics, and press enter . • Press ctrl menu and select Add X Variable with Summary List. Press enter and a dialogue box should appear. a­ pgar should be in the X List and apgrprob should be in the ­Summary List. If they are not, use the drop-down boxes to select your variables. When your box looks like the one here, tab to OK and press enter .

2. Insert a Lists & Spreadsheet page. If you already have a document open, press ctrl I and select Add Lists & Spreadsheet. If you do not have a document already open, press ( on the clickpad), then 4 . Press menu and select Add Lists & Spreadsheet.

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13.  Binomial probability on the calculator There are two handy commands on the TI-Nspire for finding binomial probabilities: binomPdf(n,p,k) computes P(X = k) binomCdf(n,p,k) computes P(X ≤ k)

3. To calculate the mean and standard deviation of the random variable, use one-variable statistics with apgar as the Data List and apgrprob as the Frequency List. • Press ctrl to go back to the Lists & Spreadsheet page. • Press menu S Statistics S Stat Calculations S OneVariable Statistics. • Make sure your X1 List, Frequency List, and 1st Result Column have the variables/values shown (you can press the down arrow in the drop-down boxes to access the variable names and type C for 1st Result Column). tab to OK and press enter .

You will need to open the Calculator Scratchpad (press or A on the clickpad). These two commands can be found in the Distributions menu within the Statistics menu. You can access them by pressing menu S Statistics S Distributions. A dialogue box will appear. Input n (the number of observations), p (probability of success), and k (number of successes).

For the parents having n = 5 children, each with probability p = 0.25 of type O blood: P(X = 3) = binomPdf(5,0.25,3) = 0.08789 To find P(X > 3), we used the complement rule: P(X > 3) = 1 − P(X ≤ 3) = 1 − binomCdf(5,0.25,3)

• The statistics should now be displayed in your Lists & Spreadsheet page.

= 0.01563

Of course, we could also have done this as P(X > 3) = P(X = 4) + P(X = 5) = binomPdf(5,0.25,4) +   binomPdf(5,0.25,5) = 0.01465 + 0.00098 = 0.01563 On the TI-Nspire, you can also calculate using P(X > 3) = P(X = 4) + P(X = 5) = binomCdf(5,0.25,4,5) = 0.01563

12.  Binomial coefficients on the calculator

5 To calculate a binomial coefficient like a b on the TI-Nspire, 2 proceed as follows. Open the Calculator Scratchpad by pressing A on the clickpad). Press menu S Probability (or S Combinations, and then enter . nCr( will appear. Complete the command nCr(5,2) and press enter .

14.  Geometric probability on the calculator There are two handy commands on the TI-Nspire for finding geometric probabilities: geomPdf(p,k) computes P(Y = k) geomCdf(p,k) computes P(Y ≤ k)

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You will need to open the Calculator Scratchpad (press or A ). These two commands can be found in the Distributions menu within the Statistics menu. You can access them by pressing menu S Statistics S Distributions. A dialogue box will appear. Input p (probability of success) and k (number of trials to get the first success).

16. Inverse t on the calculator For the Lucky Day Game, with probability of success p = 17 on each trial, P(Y = 10) = geomPdf(1/7,10) = 0.0357 To find P(Y < 10), use geomCdf: P(Y < 10) = P(Y ≤ 9) = geomCdf(1/7,9) = 0.7503

The TI-Nspire allows you to find critical values t* using the inverse t command. As with the calculator’s inverse Normal command, you have to enter the area to the left of the desired critical value. Let’s use the inverse t command to find the critical values for parts (a) and (b) in the example on page 513. A ) to insert a Calculator Scratchpad. • Press (or • Press menu S Statistics S Distributions S Inverse t. • A dialogue box will appear. For part (a), enter .025 for the Area and 11 for the Deg of Freedom, df. tab to OK and press enter .

For part (b), enter .05 for the Area and 47 for the Deg of Freedom, df. tab to OK and press enter .

Chapter 8 TI-Nspire Technology Corners 15.  Confidence interval for a population ­proportion The TI-Nspire can be used to construct a confidence interval for an unknown population proportion. We’ll demonstrate using the example on page 500. Of n = 439 teens surveyed, x = 246 said they thought young people should wait to have sex until after marriage. To construct a confidence interval: A ) to insert a Calculator Scratchpad. • Press ( • Press menu S Statistics S Confidence Intervals S 1-Prop z Interval. • A dialogue box will appear: Enter the values as shown below. tab to OK and press enter .

• The critical values t* should now be displayed.

17. One-sample t intervals for m on the ­calculator

The lower and upper bounds of the confidence interval are reported, along with the sample proportion p^ , the margin of error (ME), and the sample size.

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Confidence intervals for a population mean using t procedures can be constructed on the TI-Nspire, thus avoiding the use of Table B. Here is a brief summary of the techniques when you have only numerical summaries and when you have the actual data values. 1. Using summary statistics: Auto pollution example, page 519 • Insert a Lists & Spreadsheet page: Press ctrl I and select Add Lists & Spreadsheet. • Press menu S Statistics S Confidence Intervals S t interval.

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• The first dialogue box that appears asks for Data or Stats in the drop-down box. Select Stats, tab to OK , and press enter . • In the next dialogue box, enter the values shown.

• tab to OK and press enter . • The results should now appear in the spreadsheet.

Chapter 9 TI-Nspire Technology Corners 18. One-proportion z test on the calculator The TI-Nspire can be used to test a claim about a population proportion. We’ll demonstrate using the example on page 559. In a random sample of size n = 500, the supervisor found x = 47 potatoes with blemishes. To perform a significance test: A ) to insert a Calculator Scratchpad. • Press (or • Press menu S Statistics S Stat Tests S 1-Prop z test. • A dialogue box will appear. Enter the values shown: p0 = 0.08, x = 47, and n = 500. Specify the alternative hypothesis as “Ha: prop > p0 .” tab to OK and press enter . Note: x is the number of successes and n is the number of trials. Both must be whole numbers!

2. Using raw data: Video screen tension example, page 520 Enter the 20 video screen tension readings data using the following procedure. • Insert a Lists & Spreadsheet page: Press ctrl I and select Add Lists & Spreadsheet. • Name the first column screen. • Arrow down to the first cell and enter the 20 values. You can see that the test statistic is z = 1.15392 and the P-value is 0.1243.

To construct the t interval: • Press menu S Statistics S Confidence Intervals S t interval. • The first dialogue box that appears asks for Data or Stats in the drop-down box. Select Data, tab to OK , and press enter . • In the next dialogue box, select the data list, screen, tab to OK , and press enter .

To display the P-value as a shaded area under the Normal curve: • Press and select the Lists & Spreadsheet icon . • Press menu S Statistics S Stat Tests S 1-Prop z test. • A dialogue box will appear: Enter the values shown below. Check the box to Shade P Value. tab to OK and press enter .

• The results should now appear in the spreadsheet. (You may have to scroll up to see them.)

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• The results should now appear in the spreadsheet.

19. Computing P-values from t distributions on the calculator You can use the tcdf command on the TI-Nspire to calculate areas under a t distribution curve. The syntax is tcdf(lower bound,upper bound,df). To use this command: A ) to insert a Calculator Scratchpad. • Press (or • Press menu S Statistics S Distributions S t Cdf. • In the dialogue box that appears, enter your lower and upper bound and degrees of freedom.

Use the t Cdf command to compute the P-values from the examples on pages 577 and 578. • Better batteries: To find P(t ≥ 1.54), use Lower Bound: 1.54, Upper Bound: 10000, and df:14. • Two-sided test: To find the P-value for the two-sided test with df = 36 and t = − 3.17, execute the command 2 # tCdf (−10000,−3.17,36).

The test statistic is t = − 0.94 and the P-value is 0.1809. If you check Shade P Value, you see a t-distribution curve (df = 14) with the lower tail shaded.

If you are given summary statistics instead of the original data, you would select the “Stats” option in the drop-down box.

Chapter 10 TI-Nspire Technology Corners 21.  Confidence interval for a difference in ­proportions

20. One-sample t test for a mean on the ­calculator You can perform a one-sample t test using either raw data or summary statistics on the TI-Nspire. Let’s use the calculator to carry out the test of H0: m = 5 versus Ha: m < 5 from the dissolved oxygen example on page 580. Start by entering the sample data into a column in a Lists & Spreadsheet page. Name the column oxygen. Then, to do the test: • Press menu S Statistics S Stats Tests S t Test. • The first dialogue box that appears asks for Data or Stats in the drop-down box. Make sure Data is selected. tab to OK and press enter . • In the next dialogue box, enter the values shown in the following box. To just “calculate,” leave the Shade PValue option unchecked. Then tab to OK and press enter .

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The TI-Nspire can be used to construct a confidence interval for p1 − p2 . We’ll demonstrate using the example on page 617. Of n1 = 799 teens surveyed, X = 639 said they used social networking sites. Of n2 = 2253 adults surveyed, X = 1555 said they engaged in social networking. To construct a confidence interval: A ) to insert a Calculator Scratchpad. ( • Press • Press menu S Statistics S Confidence Intervals S 2-Prop z Interval. • A dialogue box will appear. Enter the values shown below. tab to OK and press enter .

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22.  Significance test for a difference in ­proportions The TI-Nspire can be used to perform significance tests for comparing two proportions. Here, we use the data from the Hungry Children example on page 622. To perform a test of H0: p1 − p2 = 0: A ) to insert a Calculator Scratchpad. (or • Press • Press menu S Statistics S Stat Tests S 2-Prop z test. • A dialogue box will appear. Enter the values shown: x1 = 19, n1 = 80, x2 = 26, n2 = 150. Specify the alternative hypothesis Ha: p1 ≠ p2 as shown. • tab to OK and press enter .

23. Two-sample t intervals on the calculator You can use the two-sample t interval command on the TI-Nspire to construct a confidence interval for the difference between two means. We’ll show you the steps using the summary statistics from the pine trees example on page 641. A ) to insert a Calculator Scratchpad. • Press (or • Press menu S Statistics S Confidence Intervals S 2-Sample t interval. • In the first dialogue box, select Stats in the drop-down menu. tab to OK and press enter . Another dialogue box will appear. • Enter the summary statistics shown:

You will see that the z statistic is z = 1.168 and the P-value is 0.2427, as shown here. Do you see the combined proportion of students who didn’t eat breakfast? It’s the p^ value, 0.1957.

• Enter the confidence level: C level: .90. For pooled: choose “No.” (We’ll discuss pooling later.) tab to OK and press enter .

To display the P-value as a shaded area under the standard Normal curve: • Press and select the Lists & Spreadsheet icon . • Press menu S Statistics S Stat Tests S 2-Prop z test. A dialogue box will appear. • Enter the values shown in the following box. Check the box to Shade P Value. tab to OK and press enter .

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24. Two-sample t tests on the calculator Technology gives smaller P-values for two-sample t tests than the conservative method. That’s because calculators and software use the more complicated formula on page 640 to obtain a larger number of degrees of freedom. Start by entering the sample data into a column in a Lists & Spreadsheet page. Name column A calcium and enter the Group 1 data. Name column B placebo and enter the Group 2 data. Then do the test: • Press menu S Statistics S Stats Tests S 2-Sample t Test. • In the first dialogue box, select Data in the drop-down menu. tab to OK and press enter . • In the next dialogue box, enter the values shown, tab to OK , and press enter .

between c2 = 10.180 and a very large number (we’ll use 10,000) under the chi-square density curve with 5 degrees of freedom. • Press and the Calculator Scratchpad should appear. • Press menu S Statistics S Distributions S c2 Cdf. • In the dialogue box that appears, enter the values shown in the following box. tab to OK and press enter .

As the calculator screen shot shows, this method gives a more precise P-value than Table C. Note: To just “calculate,” leave the Shade P value option unchecked. • The results should now appear in the spreadsheet.

26.  Chi-square test for goodness of fit on the calculator You can use the TI-Nspire to perform the calculations for a chisquare test for goodness of fit. We’ll use the data from the hockey and birthdays example on page 688 to illustrate the steps. 1. Enter the observed counts and expected counts in two separate columns in a Lists & Spreadsheet page. Name the columns observed and expected. Birthday

If you check the Shade P value box, the appropriate t distribution will also be displayed, showing the same results and the shaded area corresponding to the P-value.

Observed Expected

Jan-Mar

32

20

Apr-June

20

20

July-Aug

16

20

Sept-Dec

12

20

2. Perform a chi-square test for goodness of fit. • Press menu S Statistics S Stat Tests S c2 GOF. • In the dialogue box that appears, enter the values shown in the following box. tab to OK and press enter . If you leave the Shade P value box unchecked, you’ll get the test results within the spreadsheet containing the test statistic, P-value, and df. If you check the Shade P value box, you’ll get a picture of the appropriate chi-square distribution with the test statistic marked and shaded area corresponding to the P-value.

Chapter 11 TI-Nspire Technology Corners 25. Finding P-values for chi-square tests on the calculator To find the P-value in the M&M’S® example on page 685 with your calculator, use the c2 cdf command. We ask for the area

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We’ll discuss the Comp List results later.

27.  Chi-square tests for two-way tables on the calculator You can use the TI-Nspire to perform calculations for a chi-square test for homogeneity. We’ll use the data from the restaurant study on page 704 to illustrate the process. A ) to insert a Calculator Scratchpad. 1. Press ( 2. Define a matrix by doing the following: • Name your matrix by typing musicinfluence ctrl .A box will appear with different math type options. Select and enter “3” for Number of rows and “3” for Number of columns.

• Type in the corresponding row data, pressing entries. Press enter when finished.

tab

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4. To see the expected counts and component matrix, press var and select stat.expmatrix for the expected matrix or stat.compmatrix for the component matrix.

between

Chapter 12 TI-Nspire Technology Corners 28.  Confidence interval for slope on the calculator

3. To perform the chi-square test, do the following steps: • Press menu S Statistics S Stat tests S c2 2-way Test. • Specify the observed matrix, tab to OK , and press enter . • The results will be displayed and the expected matrix and component matrix will be calculated.

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Let’s use the data from the Ford F-150 truck example on page A-9 to construct a confidence interval for the slope of a population (true) regression line on the TI-Nspire. 1. Insert a Lists & Spreadsheet page, and name column A miles and column B price. Type the corresponding values into each column. 2. To construct a confidence interval: • Press menu S Statistics S Confidence Intervals S Linear Reg t Intervals. • In the first dialogue box, select Slope. tab to OK and press enter .

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• In the next dialogue box, select miles for the X List and price for the Y List. Enter the rest of the values shown. tab to OK and press enter .

30.  Transforming to achieve linearity on the calculator We’ll use the planet data on page 779 to illustrate a general strategy for performing transformations with logarithms on the TI-Nspire. A similar approach could be used for transforming data with powers and roots. 1. Insert a Lists & Spreadsheet page, and name column A distance and column B period. Type the corresponding values into each column. 2. Make a scatterplot of y versus x and confirm that there is a curved pattern. • Insert a Data & Statistics page. Press ctrl I and select Add Data & Statistics. • Press tab and select distance for the horizontal axis. Press tab again and select period for the vertical axis.

29.  Significance test for slope on the calculator Let’s use the data from the crying and IQ study on page 754 to perform a significance test for the slope of the population regression line on the TI-Nspire. 1. Insert a Lists & Spreadsheet page, and name column A crycount and column B iqscore. Type the corresponding values into each column. 2. To do a significance test: • Press menu S Statistics S Stat Tests S Linear Reg t Test. • Select crycount for the X List and iqscore for the Y List. Enter the rest of the values as shown. tab to OK and press enter .

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3. To “straighten” the curve (that is, determine the relationship), we can use different models of the explanatory-response data to see which one provides a linear relationship. • Press ctrl to return your spreadsheet. Name column c lndistance and column d lnperiod. • In the formula cell for lndistance, press enter and enter ln(distance) to take the natural log of the distance values. • Repeat this step for lnperiod using the period data. 4. To see if an exponential model fits the data: • Insert another Data & Statistics page. • Put distance on the horizontal axis and lnperiod on the vertical axis. If the relationship looks linear, then an exponential model is appropriate.

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5. To see if a power model fits the data: • Using the same Data & Statistics page, change the horizontal axis to lndistance. • If this relationship looks linear, then a power model is appropriate.

6. If a linear pattern is present, calculate the equation of the leastsquares regression line: • In the spreadsheet, press menu S Statistics S Calculations S Linear Regression(a + bx). • In the dialogue box, select lndistance for X List, lnperiod for Y List, and enter the rest of the values as shown. tab to OK and press enter .

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7. Construct a residual plot to look for any departures from the linear pattern. • Insert another Data & Statistics page. • For the horizontal axis select lndistance. For Ylist, use the stat.resid list stored in the calculator.

8. To make a prediction for a specific value of the explanatory variable, compute log(x) or ln(x), if appropriate. Then do f1(k) to obtain the predicted value of log y or ln y. To get the predicted value of y, do 10^Ans or e^Ans to undo the logarithm transformation. Here’s our prediction of the period of revolution for Eris, which is at a distance of 102.15 AU from the sun.

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Glossary/Glosario

G-1

Formulas for AP® Statistics Exam Students are provided with the following formulas on both the multiple choice and free-response sections of the AP® Statistics exam.

I. Descriptive Statistics

∙x x=

m −x = m s −x =

i

n

s "n

III. Inferential Statistics

1 sx = ∙ (xi − x)2 Ån − 1

Standardized test statistic:

(n1 − 1)s21 + (n2 − 1)s22 sp = Ç (n1 − 1) + (n2 − 1) y^ = b0 + b1x

b1 =

If x is the mean of a random sample of size n from an infinite population with mean m and standard deviation s, then:

∙ (xi − x)(yi − y) ∙ (xi − x)2

statistic − parameter standard deviation of statistic

Confidence interval: statistic ± (critical value) ∙ (std. deviation of statistic)

Single-Sample Statistic Sample Mean

Standard Deviation of Statistic s "n

b0 = y − b1x xi − x yi − y 1 r= a ba b ∙ sx sy n−1 b1 = r

sy sx

sb1 =

s21 s22 + Å n1 n2

2

Special case when s1 = s2

II. Probability

s

P(A c B) = P(A) + P(B) − P(A d B) P(A 0 B) =

P(A d B) P(B)

Difference of sample proportions

E(X) = mx = ∙ xipi Var(X) =

s2x

Å

= ∙ (xi − mx) pi 2

If X has a binomial distribution with parameters n and p, then: n P(X = k) = a bpk(1 − p)n−k k

mx = np

p (1 − p) n

Standard Deviation of Statistic

Difference of sample means

n−2

$ ∙ (xi − x)

Å

Two-Sample Statistic

∙ (yi − y^ i)2 Ç

Sample Proportion

1 1 + Å n1 n2

p1(1 − p1) p2(1 − p2) + n1 n2 Special case when p1 = p2

"p (1 − p) Chi-square test statistic =

∙

1 1 + Å n1 n2

(observed − expected)2 expected

sx = "np(1 − p) mp^ = p sp^ =

p(1 − p) Å

n

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Probability

z

Table entry for z is the area under the standard Normal curve to the left of z.

Table A  Standard Normal probabilities z

.00

.01

.02

.03

.04

.05

.06

.07

.08

.09

-3.4 -3.3 -3.2 -3.1 -3.0 -2.9 -2.8 -2.7 -2.6 -2.5 -2.4 -2.3 -2.2 -2.1 -2.0 -1.9 -1.8 -1.7 -1.6 -1.5

.0003 .0005 .0007 .0010 .0013 .0019 .0026 .0035 .0047 .0062 .0082 .0107 .0139 .0179 .0228 .0287 .0359 .0446 .0548 .0668

.0003 .0005 .0007 .0009 .0013 .0018 .0025 .0034 .0045 .0060 .0080 .0104 .0136 .0174 .0222 .0281 .0351 .0436 .0537 .0655

.0003 .0005 .0006 .0009 .0013 .0018 .0024 .0033 .0044 .0059 .0078 .0102 .0132 .0170 .0217 .0274 .0344 .0427 .0526 .0643

.0003 .0004 .0006 .0009 .0012 .0017 .0023 .0032 .0043 .0057 .0075 .0099 .0129 .0166 .0212 .0268 .0336 .0418 .0516 .0630

.0003 .0004 .0006 .0008 .0012 .0016 .0023 .0031 .0041 .0055 .0073 .0096 .0125 .0162 .0207 .0262 .0329 .0409 .0505 .0618

.0003 .0004 .0006 .0008 .0011 .0016 .0022 .0030 .0040 .0054 .0071 .0094 .0122 .0158 .0202 .0256 .0322 .0401 .0495 .0606

.0003 .0004 .0006 .0008 .0011 .0015 .0021 .0029 .0039 .0052 .0069 .0091 .0119 .0154 .0197 .0250 .0314 .0392 .0485 .0594

.0003 .0004 .0005 .0008 .0011 .0015 .0021 .0028 .0038 .0051 .0068 .0089 .0116 .0150 .0192 .0244 .0307 .0384 .0475 .0582

.0003 .0004 .0005 .0007 .0010 .0014 .0020 .0027 .0037 .0049 .0066 .0087 .0113 .0146 .0188 .0239 .0301 .0375 .0465 .0571

.0002 .0003 .0005 .0007 .0010 .0014 .0019 .0026 .0036 .0048 .0064 .0084 .0110 .0143 .0183 .0233 .0294 .0367 .0455 .0559

-1.4 -1.3 -1.2 -1.1 -1.0 -0.9 -0.8 -0.7 -0.6 -0.5 -0.4 -0.3 -0.2

.0808 .0968 .1151 .1357 .1587 .1841 .2119 .2420 .2743 .3085 .3446 .3821 .4207

.0793 .0951 .1131 .1335 .1562 .1814 .2090 .2389 .2709 .3050 .3409 .3783 .4168

.0778 .0934 .1112 .1314 .1539 .1788 .2061 .2358 .2676 .3015 .3372 .3745 .4129

.0764 .0918 .1093 .1292 .1515 .1762 .2033 .2327 .2643 .2981 .3336 .3707 .4090

.0749 .0901 .1075 .1271 .1492 .1736 .2005 .2296 .2611 .2946 .3300 .3669 .4052

.0735 .0885 .1056 .1251 .1469 .1711 .1977 .2266 .2578 .2912 .3264 .3632 .4013

.0721 .0869 .1038 .1230 .1446 .1685 .1949 .2236 .2546 .2877 .3228 .3594 .3974

.0708 .0853 .1020 .1210 .1423 .1660 .1922 .2206 .2514 .2843 .3192 .3557 .3936

.0694 .0838 .1003 .1190 .1401 .1635 .1894 .2177 .2483 .2810 .3156 .3520 .3897

.0681 .0823 .0985 .1170 .1379 .1611 .1867 .2148 .2451 .2776 .3121 .3483 .3859

-0.1 -0.0

.4602 .5000

.4562 .4960

.4522 .4920

.4483 .4880

.4443 .4840

.4404 .4801

.4364 .4761

.4325 .4721

.4286 .4681

.4247 .4641

(Continued)

Starnes-Yates5e_Tables_001_005hr.indd 1

12/9/13 5:54 PM

T-2

Ta b l e s

Probability

z

Table entry for z is the area under the standard Normal curve to the left of z.

Table A  Standard Normal probabilities (continued) z 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 3.0 3.1 3.2 3.3 3.4

.00

.01

.02

.03

.04

.05

.06

.07

.08

.09

.5000 .5398 .5793 .6179 .6554 .6915 .7257 .7580 .7881 .8159 .8413 .8643 .8849 .9032 .9192 .9332 .9452 .9554 .9641 .9713 .9772 .9821 .9861 .9893 .9918 .9938 .9953 .9965 .9974 .9981 .9987 .9990 .9993 .9995 .9997

.5040 .5438 .5832 .6217 .6591 .6950 .7291 .7611 .7910 .8186 .8438 .8665 .8869 .9049 .9207 .9345 .9463 .9564 .9649 .9719 .9778 .9826 .9864 .9896 .9920 .9940 .9955 .9966 .9975 .9982 .9987 .9991 .9993 .9995 .9997

.5080 .5478 .5871 .6255 .6628 .6985 .7324 .7642 .7939 .8212 .8461 .8686 .8888 .9066 .9222 .9357 .9474 .9573 .9656 .9726 .9783 .9830 .9868 .9898 .9922 .9941 .9956 .9967 .9976 .9982 .9987 .9991 .9994 .9995 .9997

.5120 .5517 .5910 .6293 .6664 .7019 .7357 .7673 .7967 .8238 .8485 .8708 .8907 .9082 .9236 .9370 .9484 .9582 .9664 .9732 .9788 .9834 .9871 .9901 .9925 .9943 .9957 .9968 .9977 .9983 .9988 .9991 .9994 .9996 .9997

.5160 .5557 .5948 .6331 .6700 .7054 .7389 .7704 .7995 .8264 .8508 .8729 .8925 .9099 .9251 .9382 .9495 .9591 .9671 .9738 .9793 .9838 .9875 .9904 .9927 .9945 .9959 .9969 .9977 .9984 .9988 .9992 .9994 .9996 .9997

.5199 .5596 .5987 .6368 .6736 .7088 .7422 .7734 .8023 .8289 .8531 .8749 .8944 .9115 .9265 .9394 .9505 .9599 .9678 .9744 .9798 .9842 .9878 .9906 .9929 .9946 .9960 .9970 .9978 .9984 .9989 .9992 .9994 .9996 .9997

.5239 .5636 .6026 .6406 .6772 .7123 .7454 .7764 .8051 .8315 .8554 .8770 .8962 .9131 .9279 .9406 .9515 .9608 .9686 .9750 .9803 .9846 .9881 .9909 .9931 .9948 .9961 .9971 .9979 .9985 .9989 .9992 .9994 .9996 .9997

.5279 .5675 .6064 .6443 .6808 .7157 .7486 .7794 .8078 .8340 .8577 .8790 .8980 .9147 .9292 .9418 .9525 .9616 .9693 .9756 .9808 .9850 .9884 .9911 .9932 .9949 .9962 .9972 .9979 .9985 .9989 .9992 .9995 .9996 .9997

.5319 .5714 .6103 .6480 .6844 .7190 .7517 .7823 .8106 .8365 .8599 .8810 .8997 .9162 .9306 .9429 .9535 .9625 .9699 .9761 .9812 .9854 .9887 .9913 .9934 .9951 .9963 .9973 .9980 .9986 .9990 .9993 .9995 .9996 .9997

.5359 .5753 .6141 .6517 .6879 .7224 .7549 .7852 .8133 .8389 .8621 .8830 .9015 .9177 .9319 .9441 .9545 .9633 .9706 .9767 .9817 .9857 .9890 .9916 .9936 .9952 .9964 .9974 .9981 .9986 .9990 .9993 .9995 .9997 .9998

Starnes-Yates5e_Tables_001_005hr.indd 2

12/9/13 5:54 PM

Ta b l e s

T-3

Probability p

Table entry for p and C is the point t* with probability p lying to its right and probability C lying between −t* and t*.

t*

Table B  t distribution critical values df 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 40 50 60 80 100 1000

∞

.25 1.000 0.816 0.765 0.741 0.727 0.718 0.711 0.706 0.703 0.700 0.697 0.695 0.694 0.692 0.691 0.690 0.689 0.688 0.688 0.687 0.686 0.686 0.685 0.685 0.684 0.684 0.684 0.683 0.683 0.683 0.681 0.679 0.679 0.678 0.677 0.675 0.674

.20 1.376 1.061 0.978 0.941 0.920 0.906 0.896 0.889 0.883 0.879 0.876 0.873 0.870 0.868 0.866 0.865 0.863 0.862 0.861 0.860 0.859 0.858 0.858 0.857 0.856 0.856 0.855 0.855 0.854 0.854 0.851 0.849 0.848 0.846 0.845 0.842 0.841

.15 1.963 1.386 1.250 1.190 1.156 1.134 1.119 1.108 1.100 1.093 1.088 1.083 1.079 1.076 1.074 1.071 1.069 1.067 1.066 1.064 1.063 1.061 1.060 1.059 1.058 1.058 1.057 1.056 1.055 1.055 1.050 1.047 1.045 1.043 1.042 1.037 1.036

.10 3.078 1.886 1.638 1.533 1.476 1.440 1.415 1.397 1.383 1.372 1.363 1.356 1.350 1.345 1.341 1.337 1.333 1.330 1.328 1.325 1.323 1.321 1.319 1.318 1.316 1.315 1.314 1.313 1.311 1.310 1.303 1.299 1.296 1.292 1.290 1.282 1.282

.05 6.314 2.920 2.353 2.132 2.015 1.943 1.895 1.860 1.833 1.812 1.796 1.782 1.771 1.761 1.753 1.746 1.740 1.734 1.729 1.725 1.721 1.717 1.714 1.711 1.708 1.706 1.703 1.701 1.699 1.697 1.684 1.676 1.671 1.664 1.660 1.646 1.645

50%

60%

70%

80%

90%

Starnes-Yates5e_Tables_001_005hr.indd 3

Tail probability p .025 .02 12.71 15.89 4.303 4.849 3.182 3.482 2.776 2.999 2.571 2.757 2.447 2.612 2.365 2.517 2.306 2.449 2.262 2.398 2.228 2.359 2.201 2.328 2.179 2.303 2.160 2.282 2.145 2.264 2.131 2.249 2.120 2.235 2.110 2.224 2.101 2.214 2.093 2.205 2.086 2.197 2.080 2.189 2.074 2.183 2.069 2.177 2.064 2.172 2.060 2.167 2.056 2.162 2.052 2.158 2.048 2.154 2.045 2.150 2.042 2.147 2.021 2.123 2.009 2.109 2.000 2.099 1.990 2.088 1.984 2.081 1.962 2.056 1.960 2.054 95% 96% Confidence level C

.01 31.82 6.965 4.541 3.747 3.365 3.143 2.998 2.896 2.821 2.764 2.718 2.681 2.650 2.624 2.602 2.583 2.567 2.552 2.539 2.528 2.518 2.508 2.500 2.492 2.485 2.479 2.473 2.467 2.462 2.457 2.423 2.403 2.390 2.374 2.364 2.330 2.326

.005 63.66 9.925 5.841 4.604 4.032 3.707 3.499 3.355 3.250 3.169 3.106 3.055 3.012 2.977 2.947 2.921 2.898 2.878 2.861 2.845 2.831 2.819 2.807 2.797 2.787 2.779 2.771 2.763 2.756 2.750 2.704 2.678 2.660 2.639 2.626 2.581 2.576

.0025 127.3 14.09 7.453 5.598 4.773 4.317 4.029 3.833 3.690 3.581 3.497 3.428 3.372 3.326 3.286 3.252 3.222 3.197 3.174 3.153 3.135 3.119 3.104 3.091 3.078 3.067 3.057 3.047 3.038 3.030 2.971 2.937 2.915 2.887 2.871 2.813 2.807

.001 318.3 22.33 10.21 7.173 5.893 5.208 4.785 4.501 4.297 4.144 4.025 3.930 3.852 3.787 3.733 3.686 3.646 3.611 3.579 3.552 3.527 3.505 3.485 3.467 3.450 3.435 3.421 3.408 3.396 3.385 3.307 3.261 3.232 3.195 3.174 3.098 3.091

.0005 636.6 31.60 12.92 8.610 6.869 5.959 5.408 5.041 4.781 4.587 4.437 4.318 4.221 4.140 4.073 4.015 3.965 3.922 3.883 3.850 3.819 3.792 3.768 3.745 3.725 3.707 3.690 3.674 3.659 3.646 3.551 3.496 3.460 3.416 3.390 3.300 3.291

98%

99%

99.5%

99.8%

99.9%

12/9/13 5:54 PM

T-4

Ta b l e s

Probability p

2

Table entry for p is the point c2 with probability p lying to its right. Table C  Chi−square distribution critical values Tail probability p

df

.25

.20

.15

.10

.05

1

1.32

1.64

2.07

2.71

3.84

2

2.77

3.22

3.79

4.61

3

4.11

4.64

5.32

6.25

4

5.39

5.99

6.74

5

6.63

7.29

8.12

6

7.84

8.56

.025

.02

.01

.005

5.02

5.41

6.63

7.88

5.99

7.38

7.82

9.21

7.81

9.35

9.84

11.34

7.78

9.49

11.14

11.67

9.24

11.07

12.83

13.39

9.45

10.64

12.59

14.45

.0025

.001

.0005

9.14

10.83

12.12

10.60

11.98

13.82

15.20

12.84

14.32

16.27

17.73

13.28

14.86

16.42

18.47

20.00

15.09

16.75

18.39

20.51

22.11

15.03

16.81

18.55

20.25

22.46

24.10

7

9.04

9.80

10.75

12.02

14.07

16.01

16.62

18.48

20.28

22.04

24.32

26.02

8

10.22

11.03

12.03

13.36

15.51

17.53

18.17

20.09

21.95

23.77

26.12

27.87

9

11.39

12.24

13.29

14.68

16.92

19.02

19.68

21.67

23.59

25.46

27.88

29.67

10

12.55

13.44

14.53

15.99

18.31

20.48

21.16

23.21

25.19

27.11

29.59

31.42

11

13.70

14.63

15.77

17.28

19.68

21.92

22.62

24.72

26.76

28.73

31.26

33.14

12

14.85

15.81

16.99

18.55

21.03

23.34

24.05

26.22

28.30

30.32

32.91

34.82

13

15.98

16.98

18.20

19.81

22.36

24.74

25.47

27.69

29.82

31.88

34.53

36.48

14

17.12

18.15

19.41

21.06

23.68

26.12

26.87

29.14

31.32

33.43

36.12

38.11

15

18.25

19.31

20.60

22.31

25.00

27.49

28.26

30.58

32.80

34.95

37.70

39.72

16

19.37

20.47

21.79

23.54

26.30

28.85

29.63

32.00

34.27

36.46

39.25

41.31

17

20.49

21.61

22.98

24.77

27.59

30.19

31.00

33.41

35.72

37.95

40.79

42.88

18

21.60

22.76

24.16

25.99

28.87

31.53

32.35

34.81

37.16

39.42

42.31

44.43

19

22.72

23.90

25.33

27.20

30.14

32.85

33.69

36.19

38.58

40.88

43.82

45.97

20

23.83

25.04

26.50

28.41

31.41

34.17

35.02

37.57

40.00

42.34

45.31

47.50

21

24.93

26.17

27.66

29.62

32.67

35.48

36.34

38.93

41.40

43.78

46.80

49.01

22

26.04

27.30

28.82

30.81

33.92

36.78

37.66

40.29

42.80

45.20

48.27

50.51

23

27.14

28.43

29.98

32.01

35.17

38.08

38.97

41.64

44.18

46.62

49.73

52.00

24

28.24

29.55

31.13

33.20

36.42

39.36

40.27

42.98

45.56

48.03

51.18

53.48

25

29.34

30.68

32.28

34.38

37.65

40.65

41.57

44.31

46.93

49.44

52.62

54.95

26

30.43

31.79

33.43

35.56

38.89

41.92

42.86

45.64

48.29

50.83

54.05

56.41

27

31.53

32.91

34.57

36.74

40.11

43.19

44.14

46.96

49.64

52.22

55.48

57.86

28

32.62

34.03

35.71

37.92

41.34

44.46

45.42

48.28

50.99

53.59

56.89

59.30

29

33.71

35.14

36.85

39.09

42.56

45.72

46.69

49.59

52.34

54.97

58.30

60.73

30

34.80

36.25

37.99

40.26

43.77

46.98

47.96

50.89

53.67

56.33

59.70

62.16

40

45.62

47.27

49.24

51.81

55.76

59.34

60.44

63.69

66.77

69.70

73.40

76.09

50

56.33

58.16

60.35

63.17

67.50

71.42

72.61

76.15

79.49

82.66

86.66

60

66.98

68.97

71.34

74.40

79.08

83.30

84.58

88.38

91.95

95.34

99.61

80

88.13

90.41

93.11

96.58

100

109.1

111.7

Starnes-Yates5e_Tables_001_005hr.indd 4

114.7

118.5

89.56 102.7

101.9

106.6

108.1

112.3

116.3

120.1

124.8

128.3

124.3

129.6

131.1

135.8

140.2

144.3

149.4

153.2

12/9/13 5:54 PM

Ta b l e s

T-5

Table D  Random digits Line 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150

Starnes-Yates5e_Tables_001_005hr.indd 5

19223 73676 45467 52711 95592 68417 82739 60940 36009 38448 81486 59636 62568 45149 61041 14459 38167 73190 95857 35476 71487 13873 54580 71035 96746 96927 43909 15689 36759 69051 05007 68732 45740 27816 66925 08421 53645 66831 55588 12975 96767 72829 88565 62964 19687 37609 54973 00694 71546 07511

95034 47150 71709 38889 94007 35013 57890 72024 19365 48789 69487 88804 70206 32992 77684 26056 98532 32533 07118 55972 09984 81598 81507 09001 12149 19931 99477 14227 58984 64817 16632 55259 41807 78416 55658 44753 66812 68908 99404 13258 35964 50232 42628 88145 12633 59057 86278 05977 05233 88915

05756 99400 77558 93074 69971 15529 20807 17868 15412 18338 60513 04634 40325 75730 94322 31424 62183 04470 87664 39421 29077 95052 27102 43367 37823 36809 25330 06565 68288 87174 81194 84292 65561 18329 39100 77377 61421 40772 70708 13048 23822 97892 17797 83083 57857 66967 88737 19664 53946 41267

28713 01927 00095 60227 91481 72765 47511 24943 39638 24697 09297 71197 03699 66280 24709 80371 70632 29669 92099 65850 14863 90908 56027 49497 71868 74192 64359 14374 22913 09517 14873 08796 33302 21337 78458 28744 47836 21558 41098 45144 96012 63408 49376 69453 95806 83401 74351 65441 68743 16853

96409 27754 32863 40011 60779 85089 81676 61790 85453 39364 00412 19352 71080 03819 73698 65103 23417 84407 58806 04266 61683 73592 55892 72719 18442 77567 40085 13352 18638 84534 04197 43165 07051 35213 11206 75592 12609 47781 43563 72321 94591 77919 61762 46109 09931 60705 47500 20903 72460 84569

12531 42648 29485 85848 53791 57067 55300 90656 46816 42006 71238 73089 22553 56202 14526 62253 26185 90785 66979 35435 47052 75186 33063 96758 35119 88741 16925 49367 54303 06489 85576 93739 93623 37741 19876 08563 15373 33586 56934 81940 65194 44575 16953 59505 02150 02384 84552 62371 27601 79367

42544 82425 82226 48767 17297 50211 94383 87964 83485 76688 27649 84898 11486 02938 31893 50490 41448 65956 98624 43742 62224 87136 41842 27611 62103 48409 85117 81982 00795 87201 45195 31685 18132 04312 87151 79140 98481 79177 48394 00360 50842 24870 88604 69680 43163 90597 19909 22725 45403 32337

82853 36290 90056 52573 59335 47487 14893 18883 41979 08708 39950 45785 11776 70915 32592 61181 75532 86382 84826 11937 51025 95761 81868 91596 39244 41903 36071 87209 08727 97245 96565 97150 09547 68508 31260 92454 14592 06928 51719 02428 53372 04178 12724 00900 58636 93600 67181 53340 88692 03316

12/9/13 5:54 PM

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Glossary/Glosario English

Español

1.5 ∙ IQR rule for outliers  An observation is called an outlier if it falls more than 1.5 × IQR above the third quartile or below the first quartile. (p. 56)

regla 1.5 ∙ la gama entre cuartiles para valores atípicos Se le dice valor atípico a una observación si cae a más de 1.5 × la gama entre cuartiles por encima del tercer cuartil o por debajo del primer cuartil. (pág. 56)

10% condition  When taking an SRS of size n from a population 1 of size N, check that n ≤  N. (pp. 401, 494) 10

condición del 10%  Cuando se toma una muestra aleatoria sencilla de tamaño n de una población de tamaño N, se verifica que 1 n ≤  N. (págs. 401, 494) 10

68–95–99.7 rule (also known as the empirical rule)  In the Normal distribution with mean m and standard deviation s, (a) approximately 68% of the observations fall within s of the mean m, (b) approximately 95% of the observations fall within 2s of m, and (c) approximately 99.7% of the observations fall within 3s of m. (p. 110)

regla 68–95–99.7  A la que también se le dice la “regla empírica”. En la distribución normal con media m y desviación estándar s, (a) aproximadamente el 68% de las observaciones caen dentro de s de la media m, (b) aproximadamente el 95% de las observaciones caen dentro de 2s de m, y (c) aproximadamente el 99.7% de las observaciones caen dentro de 3s de m. (pág. 110)

A addition rule for mutually exclusive events  If A and B are mutually exclusive events, P(A or B) = P(A) + P(B). (p. 308)

regla de suma para eventos que se excluyen mutuamente  Si A y B son eventos que se excluyen entre sí, P(A o B) = P(A) + P(B). (pág. 308)

alternative hypothesis Ha  The claim that we are trying to find ­evidence for in a significance test. (p. 540)

hipótesis Ha alternativa  La proposición de que en una prueba de significancia estadística estamos tratando de hallar evidencia que esté a favor. (pág. 540)

anonymity  The names of individuals participating in a study are not known even to the director of the study. (p. 271)

anonimato  Cuando se desconocen los nombres de las personas que participan en un estudio; inclusive el director del estudio los ignora. (pág. 271)

association  Knowing the value of one variable helps predict the value of the other. If knowing the value of one variable does not help predict the value of the other, there is no association between the variables. (p. 18)

asociación  Saber el valor de una variable facilita la predicción del valor de la otra. Si saber el valor de una variable no facilita la predicción del valor de la otra, entonces no existe ninguna asociación entre las variables. (pág. 18)

B back-to-back stemplot (also called back-to-back stem-and-leaf plot)  Plot used to compare the distribution of a quantitative variable for two groups. Each observation in both groups is separated into a stem, consisting of all but the final digit, and a leaf, the final digit. The stems are arranged in a vertical column with the smallest at the top. The values from one group are plotted on the left side of the stem and the values from the other group are plotted on the right side of the stem. Each leaf is written in the row next to its stem, with the leaves arranged in increasing order out from the stem. (p. 32)

diagrama de tallos contiguos (también se le dice diagrama de tallos y hojas contiguos) Se utiliza para comparar la distribución de una variable cuantitativa en dos grupos. Cada observación efectuada en ambos grupos se separa en un tallo, que consiste de todos los dígitos salvo el último, y una hoja, que consta del último dígito. Los tallos se organizan en una columna vertical con las cifras más pequeñas arriba. Los valores de un grupo se diagraman al lado izquierdo del tallo y los valores del otro grupo se diagraman al lado derecho del tallo. Cada hoja se coloca en el renglón que está al lado de su tallo, y las hojas dispuestas en orden ascendiente extendiéndose hacia fuera a partir del tallo. (pág. 32)

G-1

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G-2

Glossary/Glosario

bar graph  Graph used to display the distribution of a categorical variable or to compare the sizes of different quantities. The horizontal axis of a bar graph identifies the categories or quantities being compared. The graph is drawn with blank spaces between the bars to separate the items being compared. (p. 8)

gráfico de barras  (pág. 8) Se usa para ilustrar la distribución de una variable categorizada o para comparar el tamaño de diferentes cantidades. El eje horizontal del gráfico de barras identifica las categorías o las cantidades que se han de comparar. Se puede dibujar con espacios en blanco entre las barras a fin de separar la s diversas categorías que se desea comparar. (pág. 8)

bias  The design of a statistical study shows bias if it would consistently underestimate or consistently overestimate the value you want to know. (p. 212)

sesgo  Al diseñar un estudio estadístico se demuestra un sesgo si de manera constante se subestima o sobrestima el valor que se desea saber. (pág. 212)

biased estimator  A statistic used to estimate a parameter is biased if the mean of its sampling distribution is not equal to the true value of the parameter being estimated. (p. 431)

calculador sesgado La estadística que se usa para computar un parámetro está sesgada si la media de la distribución de su muestreo no equivale al valor real del parámetro que se está computando. (pág. 431)

bimodal  A graph of quantitative data with two clear peaks. (p. 29)

bimodal  Gráfico de datos cuantitativos con dos picos bien definidos. (pág. 29)

binomial coefficient  The number of ways of arranging k successes among n observations is given by the binon! n mial coefficient a b = for k = 0, 1, 2, … n where k k!(n − k)! n! = n(n − 1)(n − 2) # ... # 3 # 2 # 1 and 0! = 1. (p. 392)

coeficiente binomial La cantidad de maneras de organizar k aciertos entre n observaciones se representa con el coeficin n! ente binomial a b = para k = 0, 1, 2, … n en el que k k!(n − k)! n! = n(n − 1)(n − 2) # ... # 3 # 2 # 1 y 0! = 1. (pág. 392)

binomial distribution In a binomial setting, suppose we let X = the number of successes. The probability distribution of X is a binomial distribution with parameters n and p, where n is the number of trials of the chance process and p is the probability of a success on any one trial. The possible values of X are the whole numbers from 0 to n. (p. 388)

distribución binomial En un entorno binomial, supongamos que se permite que X = la cantidad de aciertos. La distribución de la probabilidad de X es una distribución binomial con los parámetros n y p, en la que n es la cantidad de ensayos del proceso de probabilidad y p es la probabilidad de un acierto en cualquiera de los ensayos. Los posibles valores de X son los números enteros de 0 a n. (pág. 388)

binomial probability formula  If X has the binomial distribution with n trials and probability p of success on each trial, the possible values of X are 0, 1, 2, …, n. If k is any one of these values, n P(X = k) = a bpk(1 − p)n−k . (p. 409) k

fórmula de probabilidad binomial Si X tiene la distribución binomial con n ensayos y la probabilidad p de acierto en cada ensayo, los posibles valores de X son 0, 1, 2, …, n. Si k es cualquiera n de estos valores, P(X = k) = a bpk(1 − p)n−k (pág. 409) k

binomial setting  Arises when we perform several independent trials of the same chance process and record the number of times that a particular outcome occurs. The four conditions for a binomial setting: • Binary? The possible outcomes of each trial can be classified as “success” or “failure.” • Independent? Trials must be independent; that is, knowing the result of one trial must not tell us anything about the result of any other trial. • Number? The number of trials n of the chance process must be fixed in advance. • Success? There is the same probability p of success on each trial. (p. 388)

entorno binomial Surge cuando se realizan varios ensayos independientes del mismo proceso de probabilidad y se anota la cantidad de veces que se produce un resultado dado. Las cuatro condiciones que definen un entorno binomial son: • Binario? Los resultados posibles de cada ensayo se pueden clasificar como “acierto” o “fracaso”. • Independiente? Los ensayos han de ser independientes; es decir, saber el resultado de un ensayo no debe indicar nada acerca del resultado de otro ensayo. • Número? La cantidad de ensayos n del proceso de probabilidad se tiene que fijar con anticipación. • Acierto? Existe la misma probabilidad p de lograr un acierto en cada ensayo. (pág. 388)

block  Group of experimental units that are known before the experiment to be similar in some way that is expected to affect the response to the treatments. (p. 252)

bloque  Grupo de unidades experimentales que antes del experimento se sabe son similares de alguna manera previsible que afecte la respuesta a los tratamientos. (pág. 252)

binomial random variable  The count X of successes in a binomial setting. (p. 388)

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variable aleatoria binomial  La cuenta X de aciertos en un entorno binomial. (pág. 388)

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Glossary/Glosario

boxplot  Graph of the five-number summary. The box spans the quartiles and shows the spread of the central half of the distribution. The median is marked within the box. Lines extend from the box to the smallest and largest observations that are not outliers. Outliers are marked with a special symbol such as an asterisk (*). (p. 57)

G-3

diagrama de caja y bigotes  Un gráfico del resumen de cinco cifras. La caja abarca los cuartiles y muestra el alcance de la mitad central de la distribución. Dentro de la caja se marca la media. Las líneas se extienden a partir de la caja a las observaciones más pequeña y más grande que no son valores atípicos. Los valores atípicos se marcan con un símbolo especial tal como un asterisco (*). (pág. 57)

C categorical variable  Variable that places an individual into one of several groups or categories. (p. 3)

variable categorizada Coloca a un individuo en uno o varios grupos o categorías. (pág. 3)

census  Study that attempts to collect data from every individual in the population. (p. 210)

censo  Un estudio en el que se trata de recoger datos acerca de cada individuo en la población. (pág. 210)

central limit theorem (CLT) In an SRS of size n from any ­population with mean m and finite standard deviation s, when n is large, the sampling distribution of the sample mean x is approximately Normal. (p. 457)

teorema del límite central  Traza una muestra aleatoria sencilla de tamaño n a partir de una población con la media m y una desviación estándar finita de s. El teorema del límite central manifiesta que cuando n es grande, la distribución de muestreo de la media de la muestra x es aproximadamente normal. (pág. 457)

Chebyshev’s inequality In any distribution, the proportion of observations falling within k standard deviations of the mean is at 1 least 1 − 2 . (p. 112) k

desigualdad de Chebychov En cualquier distribución, la proporción de observaciones que yacen dentro de k desviaciones 1 estándar de la media es al menos 1 − 2 . (pág. 112) k

chi-square distribution Family of distributions that take only nonnegative values and are skewed to the right. A particular chisquare distribution is specified by giving its degrees of freedom. (p. 685)

distribución de ji cuadrado  Familia de distribuciones que acepta solo valores no negativos y que está sesgada hacia la derecha. Se especifica una distribución de ji cuadrado dada citando sus grados de libertad. (pág. 685)

chi-square statistic  Measure of how far the observed counts are from the expected counts. The formula is

estadística de ji cuadrado  Una medición de la distancia entre las cuentas observadas y las cuentas previstas. La fórmula es

c2 =

∙

(Observed − Expected)2 Expected

where the sum is over all possible values of the categorical v­ ariable or all cells in the two-way table. (p. 682)

chi-square test for goodness of fit  Suppose the Random, 10%, and Large Counts conditions are met. To determine whether a categorical variable has a specified distribution in the population of interest, expressed as the proportion of individuals falling into each possible category, perform a test of H0: The specified distribution of the categorical variable in the population of interest is correct. Ha: The specified distribution of the categorical variable in the population of interest is not correct. Start by finding the expected count for each category assuming that H0 is true. Then calculate the chi-square statistic c2 =

∙

(Observed − Expected)2 Expected

(Observadas − Previstas)2 Previstas en la que la suma está sobre todos los valores posibles de la variable categorizada o sobre todas las celdas en la tabla de doble vía. (pág. 682) c2 =

prueba de ji cuadrado para confirmar el cuadre Supongamos que se cumplen las condiciones de aleatorio, del 10% y de cuentas grandes. Para determinar si una variable categorizada tiene una distribución específica en la población de interés, expresada como la proporción de individuos que se encuentran dentro de cada categoría posible, se realiza una prueba de H0: La distribución especificada de la variable categorizada en la población de interés es correcta Ha: La distribución especificada de la variable categorizada en la población de interés no es correcta. Se comienza hallando la cuenta prevista para cada categoría, asumiendo que H0 es verdad. Luego, se calcula la estadística de ji cuadrado c2 =

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∙

∙

(Observadas − Previstas)2 Previstas

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G-4

Glossary/Glosario

where the sum is over the k different categories. The P-value is the area to the right of c2 under the density curve of the chi-square distribution with k − 1 degrees of freedom. (p. 680)

en la que la suma está sobre las k categorías diferentes. El valorP es el área a la derecha de c2 bajo la curva de densidad de la distribución ji cuadrado con k − 1 grados de libertad. (pág. 680)

chi-square test for homogeneity Suppose the Random, 10%, and Large Counts conditions are met. You can use the chi-square test for homogeneity to test H0 : There is no difference in the distribution of a categorical variable for several populations or treatments. Ha : There is a difference in the distribution of a categorical variable for several populations or treatments. Start by finding the expected counts. Then calculate the chisquare statistic

prueba de ji cuadrado de homogeneidad  Supongamos que se han cumplido las condiciones de aleatorio, del 10% y de cuentas grandes. Se puede usar la prueba de ji cuadrado de homogeneidad para verificar que H0 : No hay diferencia en la distribución de una variable categorizada entre varias poblaciones o tratamientos. Ha : Sí hay diferencia en la distribución de una variable categorizada entre varias poblaciones o tratamientos. Se comienza hallando las cuentas previstas. Luego se computa la estadística de ji cuadrado

c2 =

∙

(Observed − Expected)2 Expected

where the sum is over all cells (not including totals) in the two-way table. If H0 is true, the c2 statistic has approximately a chi-square distribution with degrees of freedom = (number of rows – 1)(number of columns – 1). The P-value is the area to the right of c2 under the corresponding chi-square density curve. (p. 708) chi-square test for independence  Suppose the Random, 10%, and Large Counts conditions are met. You can use the chi-square test for independence to test H0 : There is no association between two categorical variables in the population of interest. Ha : There is an association between two categorical variables in the population of interest. Or, alternatively, H0 : Two categorical variables are independent in the population of interest Ha : Two categorical variables are not independent in the population of interest. Start by finding the expected counts. Then calculate the chisquare statistic c2 =

∙

(Observed − Expected)2 Expected

where the sum is over all cells in the two-way table. If H0 is true, the c2 statistic has approximately a chi-square distribution with degrees of freedom = (number of rows – 1)(number of columns – 1). The P-value is the area to the right of c2 under the corresponding chi-square density curve. (p. 697)

cluster sample Sample obtained by classifying the population into groups of individuals that are located near each other, called clusters, and then choosing an SRS of the clusters. All individuals in the chosen clusters are included in the sample. (p. 221)

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c2 =

∙

(Observadas − Previstas)2 Previstas

en la que la suma está por sobre todas las celdas (sin incluir los totales) en la tabla de doble vía. Si H0 es verdad, la estadística c2 tiene una distribución de aproximadamente ji cuadrado con grados de libertad = (número de renglones – 1)(número de columnas – 1). El valor P es el área a la derecha de c2 bajo la curva de densidad de ji cuadrado correspondiente. (pág. 708) prueba de ji cuadrado de independencia  Supongamos que se han cumplido las c­ ondiciones de aleatorio, del 10% y de cuentas grandes. Se puede usar la ­prueba de ji cuadrado de independencia para verificar que H0 : No hay ninguna asociación entre dos variables categorizadas en la población de interés. Ha : Sí hay una asociación entre dos variables categorizadas en la población de interés. O alternativamente, H0 : Dos variables categorizadas son independientes en la población de interés. Ha : Dos variables categorizadas no son independientes en la población de interés. Se comienza encontrando las cuentas previstas. Luego se computa la estadística de ji cuadrado c2 =

∙

(Observadas − Previstas)2 Previstas

en la que la suma está sobre todas las celdas en la tabla de doble vía. Si H0 es verdad, la estadística c2 tiene una distribución de aproximadamente ji cuadrado con grados de libertad = (número de renglones – 1)(número de columnas – 1). El valor P es el área a la derecha de c2 bajo la curva de densidad de ji cuadrado correspondiente. (pág. 697) muestra de clúster  Muestra que se obtiene clasificando la población en grupos de individuos que están ubicados uno cerca del otro, llamados clústers, y luego escogiendo una muestra aleatoria sencilla de los clústers. Todos los individuos en los clústers escogidos se incluyen en la muestra. (pág. 221)

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Glossary/Glosario

coefficient of determination r2  Fraction of the variation in the values of y that is accounted for by the least-squares regression line of y on x. We can calculate r2 using the formula r2 = 1 −

∙ residuals2 ∙ (yi − y)2

G-5

coeficiente de determinación r2  La fracción de la variación en los valores y que se tiene en cuenta por la línea de regresión de mínimos cuadrados de y sobre x. Se puede computar r2 utilizando la fórmula r2 = 1 −

∙ residuales2 ∙ (yi − y)2

(p. 179)

(pág. 179)

comparison  Experimental design principle. Use a design that compares two or more treatments. (p. 240)

comparación  Principio de diseño experimental. Se usa un diseño que compara dos o más tratamientos. (pág. 240)

complement of an event AC  Event “not A”. (p. 307)

complemento de un evento AC  Se refiere al evento “que no es A”. (pág. 307)

complement rule  The probability that an event does not occur is 1 minus the probability that the event does occur. In symbols, P(AC) = 1 – P(A). (pp. 308, 314)

regla del complemento  La probabilidad de que no suceda un evento es 1 menos la probabilidad de que el evento sí suceda. En representación simbólica, P(AC) = 1 – P(A). (págs. 308, 314)

completely randomized design  Design in which the experimental units are assigned to the treatments completely by chance. (p. 245)

diseño completamente aleatorizado Cuando las unidades experimentales se les asignan a los tratamientos de manera completamente al azar. (pág. 245)

components  Individual terms

(Observed − Expected)2

Expected added together to produce the test statistic c2 . (p. 690)

that are

componentes  Los términos individuales

(Observades − Previstas)2 Previstas

que se suman para producir la estadística de prueba c2 . (pág. 690)

conditional distribution  Term that describes the values of one variable among individuals who have a specific value of another variable. There is a separate conditional distribution for each value of the other variable. (p. 15)

distribución condicional Describe los valores de una variable entre individuos que tienen un valor especifico de otra variable. Hay una distribución condicional separada para cada valor de la otra variable. (pág. 15)

conditional probability  Probability that one event happens given that another event is already known to have happened. Suppose we know that event A has happened. Then the probability that event B happens given that event A has happened is denoted by P(B | A). To find the conditional probability P(B | A), use the formula

probabilidad condicional La probabilidad de que un evento suceda a la luz de que se sabe que otro evento ya sucedió. Supongamos que nos consta que el evento A ya sucedió. Entonces la probabilidad de que el evento B suceda en vista de que el evento A ya sucedió, se denota con P(B | A). Para hallar la probabilidad condicional P(B | A), se usa la fórmula

(p. 320)

P(B 0 A) =

P(A d B) P(A)

conditions for regression inference  Suppose we have n observations on an explanatory variable x and a response variable y. Our goal is to study or predict the behavior of y for given values of x. • Linear: The actual relationship between x and y is linear. For any fixed value of x, the mean response y falls on the population (true) regression line my = a + bx. • Independent: Individual observations are independent. When sampling is done without replacement, check the 10% condition. • Normal: For any fixed value of x, the response y varies according to a Normal distribution.

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(pág. 320)

P(B 0 A) =

P(A d B) P(A)

condiciones para la inferencia de regresión  Supongamos que tenemos n observaciones en una variable explicativa x y una variable de respuesta y. Nuestra meta consiste en estudiar o predecir el comportamiento de y ante los valores dados de x. • Lineal (linear): La relación real entre x y y es lineal. Para todo valor fijo de x, la respuesta media y cae en la línea de regresión de la población (verdadera) my = a + bx. • Independiente (independent): Las observaciones individuales son independientes. Cuando el muestreo se hace sin reemplazo, se verifica la condición del 10%. • Normal (normal): Para cualquier valor fijo de x, la respuesta y varía según una distribución normal.

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G-6

Glossary/Glosario

• Equal SD: The standard deviation of y (call it s) is the same for all values of x.

• Desviación estándar equivalente: La desviación estándar de y es la misma para todos los valores de x.

• Random: The data are produced from a well-designed random sample or randomized experiment. (p. 743)

• Aleatorio (random): Los datos son producidos a partir de una muestra aleatoria o un experimento aleatorio, ambos bien diseñados. (pág. 743)

confidence interval  Gives an interval of plausible values for a parameter. The interval is calculated from the data and has the form

intervalo de confianza  Ofrece un intervalo de valores plausibles para un parámetro. El intervalo se computa a partir de los datos y tiene la forma

point estimate ± margin of error or, alternatively, statistic ± (critical value) # (standard deviation of statistic) (p. 480)

Estimado de punto ± margen de error o alternativamente, estadística ± (valor crítico) # (desviación estándar de la estadística) (pág. 480)

confidence level C  Success rate of the method for calculating the confidence interval. In C% of all possible samples, the method would yield an interval that captures the true parameter value. (p. 480)

nivel de confianza C  La tasa de aciertos del método con el que se computa el intervalo de confianza. En el C% de todas las muestras posibles, el método produciría un intervalo que capta el valor verdadero del parámetro. (pág. 480)

confidential  A basic principle of data ethics that requires that individual data be kept private. (p. 270)

confidencial  Principio básico de la ética de la gestión de datos. Requiere que los datos individuales se mantengan en reserva. (pág. 270)

confounding  When two variables are associated in such a way that their effects on a response variable cannot be distinguished from each other. (p. 236)

confuso  Cuando dos variables se asocian de tal manera que sus efectos en una variable de respuesta no se pueden distinguir el uno del otro. (pág. 236)

continuous random variable  Variable that takes all values in an interval of numbers. The probability distribution of a continuous random variable is described by a density curve. The probability of any event is the area under the density curve and above the values of the variable that make up the event. (p. 356)

variable aleatoria continua  Emplea todos los valores en un intervalo de cifras. La distribución de la probabilidad de una variable aleatoria continua se describe con una curva de densidad. La probabilidad de cualquier evento en el área debajo de la curva de densidad y encima de los valores de la variable que componen el evento. (pág. 356)

control  Experimental design principle that mandates keeping other variables that might affect the response the same for all groups. (p. 242)

control  Principio del diseño experimental. Se mantienen otras variables que podrían afectar la respuesta iguales para todos los grupos. (pág. 242)

control group  Experimental group whose primary purpose is to provide a baseline for comparing the effects of the other treatments. Depending on the purpose of the experiment, a control group may be given a placebo or an active treatment. (p. 246)

grupo de control  Grupo experimental cuyo fin primario es establecer una línea base mediante la cual se comparan los efectos de otros tratamientos. Según el objeto del experimento, a un grupo de control se le puede dar un placebo o un tratamiento activo. (pág. 246)

convenience sample  Sample selected by taking from the population individuals that are easy to reach. (p. 212)

muestra de conveniencia  Muestra escogida de individuos de la población con quienes es fácil hacer contacto. (pág. 212)

correlation  Measures the direction and strength of the linear relationship between two quantitative variables. Correlation is usually written as r. We can calculate r using the formula

correlación  Mide el sentido y la fuerza de la relación lineal entre dos variables cuantitativas. La correlación generalmente se denomina con una r. Calculamos la r con la fórmula

r=

1 n−1

∙a

x i − x yi − y ba b . (pp. 150, 154) sx sy

critical value Multiplier that makes the interval wide enough to have the stated capture rate. The critical value depends on both the confidence level C and the sampling distribution of the ­statistic. (pp. 486, 497)

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r=

1 n−1

∙a

xi − x yi − y ba b . (págs. 150, 154) sx sy

valor crítico  Multiplicador que amplía el intervalo lo suficiente para retener la tasa de captación indicada. El valor crítico depende de tanto el nivel de confianza C como de la distribución de muestreo de la estadística. (págs. 486, 497)

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Glossary/Glosario

cumulative relative frequency graph  Graph used to examine location within a distribution. Cumulative relative frequency graphs begin by grouping the observations into equal-width classes. The completed graph shows the accumulating percent of observations as you move through the classes in increasing order. (p. 87)

G-7

gráfico de la frecuencia relativa acumulada  Se usa para examinar la ubicación dentro de una distribución. Los gráficos de la frecuencia relativa acumulada se inician agrupando las observaciones en clases del mismo ancho. El gráfico completado muestra el porcentaje de observaciones que se van acumulando a medida que se desplaza por las clases en orden ascendiente. (pág. 87)

D data analysis Process of describing data using graphs and numerical summaries. (p. 2)

análisis de los datos  Proceso que describe los datos haciendo uso de gráficos y resúmenes numéricos. (pág. 2)

density curve  Curve that (a) is always on or above the horizontal axis and (b) has area exactly 1 underneath it. A density curve describes the overall pattern of a distribution. The area under the curve and above any interval of values on the horizontal axis is the proportion of all observations that fall in that interval. (p. 105)

curva de densidad  Curva que (a) siempre está sobre o por encima del eje horizontal y (b) tiene 1 área exactamente debajo. La curva de densidad describe el patrón general de una distribución. El área debajo de la curva y encima de todo intervalo de valores en el eje horizontal es la proporción de todas las observaciones que caen en dicho intervalo. (pág. 105)

describing a distribution  In any graph of data, look for the overall pattern and for striking departures from that pattern. Shape, center, and spread describe the overall pattern of the distribution of a quantitative variable. (p. 26)

descripción de una distribución  En un gráfico de datos, se observa cuál es el patrón general y se busca también valores atípicos que no se ajusten al patrón. Forma, centro y amplitud describen el patrón general de la distribución de una variable cuantitativa. (pág. 26)

describing a scatterplot  In any graph of data, look for the overall pattern and for striking departures from that pattern. Direction, form, and strength describe the overall pattern of a scatterplot. (p. 147)

descripción de un gráfico de dispersión  En todo gráfico de datos, se observa cuál es el patrón general y se busca también valores atípicos que no se ajusten al patrón. Dirección, forma y fuerza describen el patrón general de la distribución de un gráfico de dispersión. (pág. 147)

discrete random variable  Takes a fixed set of possible values with gaps between. The probability distribution of a discrete random variable gives its possible values and their probabilities. The probability of any event is the sum of the probabilities for the values of the variable that make up the event. (p. 348)

variable aleatoria discreta  Emplea un conjunto fijo de valores posibles entre los cuales hay brechas. La distribución de la probabilidad de una variable aleatoria discreta arroja valores posibles y sus probabilidades. La probabilidad de cualquier evento es la suma de las probabilidades de los valores de la variable que compone el evento. (pág. 348)

distribution  Tells what values a variable takes and how often it takes these values. (p. 4)

distribución  Indica qué valores adopta una variable y con qué frecuencia adopta dichos valores. (pág. 4)

distribution of sample data  Gives the values of the variable for all the individuals in the sample. (p. 428)

distribución de los datos de la muestra  Indica los valores de la variable que les corresponden a todos los individuos en la muestra. (pág. 428)

dotplot  Simple graph that shows each data value as a dot above its location on a number line. (p. 25)

gráfico de puntos  Un gráfico sencillo que muestra el valor de cada dato encima de su ubicación a lo largo de una línea de cifras. (pág. 25)

double-blind  An experiment in which neither the subjects nor those who interact with them and measure the response variable know which treatment a subject received. (p. 248)

doble ciego Experimento en el que ninguno de los sujetos ni aquellos que interactúan con los sujetos y que miden la variable de repuesta saben qué tratamiento recibió el sujeto. (pág. 248)

E event  Any collection of outcomes from some chance process. An event is a subset of the sample space. Events are usually designated by capital letters, like A, B, C, and so on. (p. 306)

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evento  Cualquier colección de los resultados de un proceso de probabilidad. Es decir, un evento es un subconjunto del espacio de muestras. Los eventos generalmente se designan con mayúsculas tales como A, B, C, y así sucesivamente. (pág. 306)

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G-8

Glossary/Glosario

expected counts  Expected numbers of individuals in the sample that would fall in each cell of the one-way or two-way table if H0 were true. (p. 681)

cuentas previstas Las cantidades previstas de individuos en la muestra que caerían en cada celda en la tabla, sea de una vía o de dos vías, si H0 fuera verdad. (pág. 681)

experiment  A study in which researchers deliberately impose treatments on individuals to measure their responses. (p. 235)

experimento  Estudio en el que los investigadores deliberadamente les imponen tratamientos a individuos, con el fin de medir sus ­respuestas. (pág. 235)

experimental units  Smallest collection of individuals to which treatments are applied. (p. 237)

unidades experimentales La colección más pequeña de individuos a quienes se les aplican los tratamientos. (pág. 237)

explanatory variable  Variable that may help explain or predict changes in a response variable. (pp. 143, 236)

variable explicativa  Variable que puede ayudar a explicar o predecir cambios en una variable de respuesta. (págs. 143, 236)

exponential model  Relationship of the form y = abx . If the relationship between two variables follows an exponential model and we plot the logarithm (base 10 or base e) of y against x, we should observe a straight-line pattern in the transformed data. (p. 775)

modelo exponencial  Relación de la forma y = abx . Si la relación entre dos variables se ajusta a un modelo exponencial, y trazamos el logaritmo (de base 10 o de base e) de y con respecto a x, se debe observar un patrón en línea recta en los datos transformados. (pág. 775)

extrapolation  Use of a regression line for prediction far outside the interval of values of the explanatory variable x used to obtain the line. Such predictions are often not accurate. (p. 168)

extrapolación  Uso de una línea de regresión para hacer predicciones muy por fuera del intervalo de valores de la variable explicativa x que se utiliza para obtener la línea. Tales predicciones a menudo carecen precisión. (pág. 168)

F factorial  For any positive whole number n, its factorial n! is n! = n ∙ (n − 1) # (n − 2) # … ∙ 3 # 2 # 1 In addition, we define 0! = 1. (p. 409)

factorial  Para cualquier número entero positivo n, su factorial n! es n! = n # (n − 1) # (n − 2) # … ∙ 3 # 2 # 1 Además, definimos 0! = 1. (pág. 409)

factors  Explanatory variables in an experiment. (p. 238)

factores  Las variables explicativas en un experimento. (pág. 238)

fail to reject H0  If the observed result is not very unlikely to occur when the null hypothesis is true, we should fail to reject H0 and say that we do not have convincing evidence for Ha. (p. 544)

no rechazar H0  Si no es muy improbable que el resultado observado suceda cuando es verdad la hipótesis nula, no se debe rechazar H0 y se ha de indicar que no contamos con evidencia convincente de Ha. (pág. 544)

first quartile Q1  If the observations in a data set are ordered from lowest to highest, the first quartile Q1 is the median of the observations whose position is to the left of the median. (p. 54)

primer cuartil Q1  Si las observaciones del conjunto de datos se organizan en orden ascendiente (del más bajo al más alto), el primer cuartil Q1 es la media de las observaciones cuya posición se encuentra a la izquierda de la media. (pág. 54)

five-number summary Smallest observation, first quartile, median, third quartile, and largest observation, written in order from smallest to largest. In symbols:

resumen de cinco cifras  Consta de la observación más pequeña, el primer cuartil, la media, el tercer cuartil y la observación más grande, enumeradas en orden ascendiente, desde la más pequeña hasta la más grande. Representado en forma simbólica, el resumen de cinco cifras es

Minimum  Q1 Median Q3 Maximum (p. 57)

Mínimo Q1 Media Q3 Máximo (pág. 57)

frequency table Table that displays the count (frequency) of ­observations in each category or class. (p. 8)

tabla de frecuencias  Muestra la cuenta (frecuencia) de observaciones en cada categoría o clase. (pág. 8)

G general addition rule If A and B are any two events resulting from some chance process, then the probability that event A or event B (or both) occur is P(A or B) = P(A c B) = P(A) + P(B) – P(A d B). (p. 310)

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regla general de adición Si A y B son dos eventos cualquiera que resulten de algún proceso de probabilidad, la probabilidad de que el evento A o el evento B (o ambos) suceda es P(A o B) = P(A c B) = P(A) + P(B) – P(A d B). (pág. 310)

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Glossary/Glosario

G-9

general multiplication rule  The probability that events A and B both occur can be found using the formula P(A ∩ B) = P(A) # P(B | A) where P(B | A) is the conditional probability that event B occurs given that event A has already occurred. (p. 321)

regla general de multiplicación  La probabilidad de que sucedan los eventos A y B se puede determinar utilizando la fórmula P(A ∩ B) = P(A) # P(B | A) en la que P(B | A) es la probabilidad condicional de que suceda el evento B a la luz de que el evento A ya sucedió. (pág. 321)

geometric distribution In a geometric setting, suppose we let Y = the number of trials it takes to get a success. The probability distribution of Y is a geometric distribution with parameter p, the probability of a success on any trial. The possible values of Y are 1, 2, 3, …. (p. 405)

distribución geométrica En un entorno geométrico, supongamos que se permite que Y = la cantidad de ensayos que se precisan para lograr un acierto. La distribución de la probabilidad de Y es una distribución geométrica con el parámetro p, la probabilidad de lograr n acierto en cualquier ensayo. Los valores posibles de Y son 1, 2, 3, …. (pág. 405)

geometric probability formula If Y has the geometric distribution with probability p of success on each trial, the possible values of Y are 1, 2, 3, …. If k is any one of these values, P(Y = k) = (1 − p)k−1p. (p. 406)

fórmula de probabilidad geométrica  Si Y tiene una distribución geometría con la probabilidad p de acierto en cada ensayo, los posibles valores de Y son 1, 2, 3, …. Si k es uno cualquiera de estos valores, P(Y = k) = (1 − p)k−1p. (pág. 406)

geometric random variable  The number of trials Y that it takes to get a success in a geometric setting. (p. 405)

variable aleatoria geométrica La cantidad de ensayos Y que se precisan para lograr un acierto en un entorno geométrico. (pág. 405)

geometric setting  Arises when we perform independent trials of the same chance process and record the number of trials it takes to get one success. On each trial, the probability p of success must be the same. (p. 404)

entorno geométrico Surge un entorno geométrico cuando se realizan ensayos independientes del mismo proceso de probabilidad y se graban la cantidad de ensayos que se precisan para lograr un acierto. En cada ensayo, la probabilidad p de lograr un acierto tiene que ser la misma. (pág. 404)

H histogram  Graph that displays the distribution of a quantitative variable. The horizontal axis is marked in the units of measurement for the variable. The vertical axis contains the scale of counts or percents. Each bar in the graph represents an equalwidth class. The base of the bar covers the class, and the bar height is the class frequency or relative frequency. (p. 33)

histograma  Muestra la distribución de una variable cuantitativa. En el eje horizontal se denotan las unidades de medición de la variable. El eje vertical contiene la escala de cuentas o porcentajes. Cada barra del gráfico representa una clase de ancho equivalente. La base de la barra abarca la clase, y la altura de la barra es la frecuencia o la frecuencia relativa de la clase. (pág. 33)

I independent events  Two events are independent if the occurrence of one event does not change the probability that the other event will happen. In other words, events A and B are independent if P(A | B) = P(A) and P(B | A) = P(B). (p. 327)

eventos independientes Dos eventos son independientes si el hecho de que uno suceda no cambia la probabilidad de que el otro suceda. Es decir, los eventos A y B son independientes si P(A | B) = P(A) y P(B | A) = P(B). (pág. 327)

independent random variables If knowing whether any event involving X alone has occurred tells us nothing about the occurrence of any event involving Y alone and vice versa, then X and Y are independent random variables. (p. 371)

variables aleatorias independientes  Saber que ha sucedido un evento que implique el valor X solo, no nos indica nada respecto al hecho de que suceda un evento que implique el valor Y solo, y viceversa. En tal caso, tanto X como Y son variables aleatorias independientes. (pág. 371)

individuals  Objects described by a set of data. Individuals may be people, animals, or things. (p. 2)

individuos  Objetos descritos por un conjunto de datos. Los individuos pueden ser personas, animales o cosas. (pág. 2)

inference  Drawing conclusions that go beyond the data at hand. (pp. 5, 223)

inferencia  Llegar a conclusiones que van más allá de los datos que están a la mano. (págs. 5, 223)

inference about cause and effect  Conclusion from the results of an experiment that the treatments caused the difference in responses. Requires a well-designed experiment in which the treatments are randomly assigned to the experimental units. (p. 266)

inferencia sobre causa y efecto  Uso de los resultados de un experimento para llegar a la conclusión de que son los tratamientos los que marcan la diferencia en las respuestas. Exige un experimento bien diseñado en el que los tratamientos se asignan de manera aleatoria a las unidades experimentales. (pág. 266)

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G-10

Glossary/Glosario

inference about a population Conclusion about the larger population based on sample data. Requires that the individuals taking part in a study be randomly selected from the population of interest. (p. 266)

inferencia sobre una población Conclusión sobre una población en general con base en datos muestrales. Se precisa que los participantes del estudio sean escogidos de manera aleatoria a partir de la población de interés. (pág. 266)

influential observation  An observation is influential for a statistical calculation if removing it would markedly change the result of the calculation. Points that are outliers in the x direction of a scatterplot are often influential for the least-squares regression line. (p. 189)

observación influyente  La observación es influyente en un cómputo estadístico si al retirarla se notaría un cambio sustancial en el resultado del cómputo. Los puntos que son valores atípicos en el sentido x en un gráfico de dispersión, a menudo son influyentes en al menos una línea de regresión de mínimos cuadrados. (pág. 189)

informed consent  Basic principle of data ethics that states that individuals must be informed in advance about the nature of a study and any risk of harm it may bring. Participating individuals must then consent in writing. (p. 270)

autorización informada  Principio básico de ética en la gestión de los datos. A los individuos se les ha de informar con antelación acerca de la naturaleza de un estudio y de los riesgos o perjuicios que podría conllevar. Los individuos que participen luego tendrán que dar su autorización por escrito. (pág. 270)

institutional review board Board charged with protecting the safety and well-being of the participants in advance of a planned study and with monitoring the study itself. (p. 270)

junta de revisión institucional  Principio básico de la ética de la gestión de datos. Todos los estudios planificados tienen que contar con aprobación anticipada y tienen que contar con un monitoreo por una junta de revisión institucional cuya función consiste en salvaguardar la seguridad y el bienestar de los participantes. (pág. 270)

interquartile range  IQR = Q3 – Q1. (p. 54)

gama entre cuartiles  IQR = Q3 – Q1. (pág. 54)

intersection  The intersection of events A and B, denoted by A d B, refers to the occurrence of both of two events at the same time. (p. 311)

intersección  El punto de cruce de los eventos A y B, designado con A d B, se refiere a la situación en la que ambos eventos suceden simultáneamente. (pág. 311)

L lack of realism  When the treatments, the subjects, or the environment of an experiment are not realistic. Lack of realism can limit researchers’ ability to apply the conclusions of an experiment to the settings of greatest interest. (p. 268)

falta de realismo  Cuando los tratamientos, los sujetos o el entorno de un experimento no son realistas. La carencia de realismo puede limitar la capacidad de los investigadores de aplicar las conclusiones de un experimento a los entornos de gran interés. (pág. 268)

Large Counts condition  It is safe to use Normal approximation for performing inference about a proportion p if np ≥ 10 and n(1 − p) ≥ 10. (p. 403)

condición de cuentas grandes  Se puede utilizar sin problemas la aproximación normal para realizar la inferencia de una proporción p si np ≥ 10 y n(1 − p) ≥ 10. (pág. 403)

Large Counts condition for a chi-square test  It is safe to use a chi-square distribution to perform calculations if all expected counts are at least 5. (p. 687)

condición de cuentas grandes en la prueba de ji cuadrado  Se puede utilizar sin problemas la distribución de ji cuadrado para realizar cómputos si todas las cuentas previstas son de al menos 5. (pág. 687)

law of large numbers  If we observe more and more repetitions of any chance process, the proportion of times that a specific outcome occurs approaches a single value, which we call the probability of that outcome. (p. 291)

ley de las cifras grandes  Si se observan más y más repeticiones en cualquier proceso de probabilidad, la proporción de veces que se da un resultado específico se aproxima a un valor sencillo, al cual se le denomina la probabilidad de dicho resultado. (pág. 291)

least-squares regression line  The line that makes the sum of the squared vertical distances of the data points from the line as small as possible. (p. 169)

línea de regresión de mínimos cuadrados  La línea que reduce al mínimo posible la suma de las distancias verticales cuadráticas de los puntos de datos a partir de la línea. (pág. 169)

level  Specific value of an explanatory variable (factor) in an ­experiment.

nivel  Valor específico de una variable explicativa (factor) en un experimento. (pág. 238)

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Glossary/Glosario

linear transformation A transformation of a random variable that involves adding a constant a, multiplying by a constant b, or both. We can write a linear transformation of the random variable X in the form Y = a + bX. The shape, center, and spread of the probability distribution of Y are as follows: Shape: Same as the probability distribution of X unless b is ­negative. Center: mY = a + bmX Spread: sY = 0 b 0 sX (p. 368)

G-11

transformación lineal  La transformación de una variable aleatoria que implica agregar una a constante, multiplicada por una b constante, o ambas. La transformación lineal de la variable aleatoria X se puede escribir en la forma Y = a + bX. La forma, el centro y la amplitud de la distribución de la probabilidad de Y son como sigue: Forma: Igual que la distribución de la probabilidad de X a ­menos que b sea negativo. Centro: mY = a + bmX Amplitud: sY = 0 b 0 sX (pág. 368)

M margin of error  The difference between the point estimate and the true parameter value will be less than the margin of error in C% of all samples, where C is the confidence level. (p. 480)

margen de error  La diferencia entre el estimado del punto y el valor real del parámetro será menor que el margen de error en C% de todas las muestras, en el que C es el nivel de confianza. (pág. 480)

marginal distribution The distribution of one of the categorical variables in a two-way table of counts among all individuals described by the table. (p. 12)

distribución marginal La distribución de una de las variables categorizadas en una tabla de doble vía de cuentas entre todos los individuos descritos por la tabla. (pág. 12)

matched pairs design  Common form of blocking for comparing just two treatments. In some matched pairs designs, each subject receives both treatments in a random order. In others, the subjects are matched in pairs as closely as possible, and each subject in a pair is randomly assigned to receive one of the treatments. (p. 255)

diseño de pares coincidentes Forma común de crear bloques para efectos de comparación de tan solo dos tratamientos. En algunos diseños de pares coincidentes, cada tema se somete a ambos tratamientos en un orden aleatorio. En otros, los temas se ponen en pares que coincidan lo más posible y cada tema en un par se asigna de manera aleatoria a fin de que reciba uno de los tratamientos. (pág. 255)

mean x  Arithmetic average. To find the mean of a set of observations, add their values and divide by the number of observations. xi In symbols, x– = (p. 49) n

media x  El promedio aritmético. Para hallar la media de un conjunto de observaciones, se suman todos los valores y se divide enxi tre el número de observaciones. En símbolos, x– = (pág. 49) n

mean of a density curve  Point at which a density curve would balance if made of solid material. (p. 107)

media de una curva de densidad  El punto en el cual la curva se equilibraría si estuviera elaborada de un material macizo. (pág. 107)

mean (expected value) of a discrete random variable  To find the mean (expected value) of X, multiply each possible value by its probability, then add all the products:

media (valor previsto) de una variable aleatoria discreta  Para hallar la media (un valor previsto) de X, se multiplica cada valor posible por su probabilidad y luego se suman todos los productos:

∙

mX = E(X) = x1 p1 + x2 p2 + x3 p3 + ... =

∙ x i pi

∙

mX = E(X) = x1p1 + x2p2 + x3p3 + ... =

∙ x i pi

(p. 351)

(pág. 351)

mean (expected value) of a geometric random variable  If Y is a geometric random variable with probability of success p on each 1 trial, then its mean (expected value) is mY = E(Y) = . That is, p the expected number of trials required to get the first success is 1/p. (p. 408)

media (valor previsto) de una variable aleatoria geométrica  Si Y es una variable aleatoria geométrica con probabilidad de acierto p en cada ensayo, entonces su media (un valor previsto) es 1 mY = E(Y) = . Es decir, la cifra de ensayos prevista para lograr el p primer acierto es 1/p. (pág. 408)

mean and standard deviation of a binomial random variable  If a count X of successes has the binomial distribution with number of trials n and probability of success p, the mean and standard

media y desviación estándar de una variable binomial aleatoria  Si una cuenta X de aciertos tiene una distribución binomial con la cantidad de ensayos n y la probabilidad de aciertos p, la media y la desviación estándar de X son mX = np and

deviation of X are mX = np and sX = "np(1 − p). (p. 398)

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sX = "np(1 − p) (pág. 398)

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G-12

Glossary/Glosario

mean and standard deviation of the sampling distribution of a sample mean x  Suppose that x is the mean of an SRS of size n from a large population with mean m and standard deviation s. Then • The mean of the sampling distribution of x is mx = m. • The standard deviation of the sampling distribution of x is s 1 sx = as long as the 10% condition is satisfied: n ≤  N. 10 "n (p. 452)

media y desviación estándar de la distribución de muestreo de la media de una muestra Supongamos que x es la media de una muestra aleatoria sencilla de tamaño n a partir de una población grande con media y una desviación estándar s. Así • La media de la distribución de muestreo de x es mx = m. • La desviación estándar de la distribución de muestreo de x es s sx = siempre y cuando se cumpla con la condición del 10%: "n 1 n ≤ N. (pág. 452) 10

median  The midpoint of a distribution; the number such that about half the observations are smaller and about half are larger. To find the median of a distribution: (1) Arrange all observations in order of size, from smallest to largest. (2) If the number of observations n is odd, the median is the center observation in the ordered list. (3) If the number of observations n is even, the median is the average of the two center observations in the ordered list. (p. 51)

media  El punto intermedio de una distribución, con una cifra tal que aproximadamente la mitad de las observaciones son más pequeñas y la mitad son más grandes. Para hallar la media de una distribución: (1) Se organizan todas las observaciones en orden de su tamaño, de las más pequeñas a las más grandes. (2) Si la cantidad de observaciones n es impar, la media es el la observación central en la lista organizada. (3) Si la cantidad de observaciones n es par, la media es el promedio de las dos observaciones centrales en la lista organizada. (pág. 51)

median of a density curve  The point with half the area under the curve to its left and the remaining half of the area to its right. (p. 106)

media de una curva de densidad  El punto en el que la mitad del área que está debajo de la curva está a la izquierda y la otra mitad del área está a la derecha. (pág. 106)

mode  Value or class in a statistical distribution having the greatest frequency. (p. 26)

modo  En una distribución estadística, el valor o clase que tiene la mayor frecuencia. (pág. 26)

multimodal  A graph of quantitative data with more than two clear peaks. (p. 29)

multimodal  Gráfico de datos cuantitativos que tiene más de dos picos claros. (pág. 29)

multiple comparisons  Problem of how to do many comparisons at once with an overall measure of confidence in all our conclusions. (p. 700)

comparaciones múltiples  El problema de cómo hacer muchas comparaciones a la vez con una medida de confianza general en todas las conclusiones a las que se llega. (pág. 700)

multiplication rule for independent events  If A and B are independent events, then the probability that A and B both occur is P(A ∙ B) = P(A) ∙ P(B). (p. 328)

regla de multiplicación de eventos independientes  Si A y B son eventos independientes, la probabilidad de que sucedan ambos, tanto A como B es P(A ∙ B) = P(A) ∙ P(B). (pág. 328)

mutually exclusive (disjoint)  Two events that have no outcomes in common and so can never occur together. (p. 307)

exclusivos mutuamente (desencajamiento)  Dos eventos que no tienen resultados en común y por lo tanto nunca pueden suceder a la vez. (pág. 307)

N negative association  When above-average values of one variable tend to accompany below-average values of the other. (p. 148)

asociación negativa  Cuando los valores por encima del promedio de una variable tienden a acompañar a los valores por debajo del promedio de la otra. (pág. 148)

nonresponse  Occurs when an individual chosen for the sample can’t be contacted or refuses to participate. (p. 225)

no respondió  Sucede cuando a un individuo escogido para la muestra no se le puede contactar o el sujeto se niega a participar. (pág. 225)

Normal approximation to a binomial distribution Suppose that a count X of successes has the binomial distribution with n trials and success probability p. When n is large, the distribution of X is approximately Normal with mean np and standard devia-

aproximación normal hacia una distribución binomial  Supongamos que una cuenta X de aciertos tiene la distribución binomial con n ensayos y una probabilidad de acierto p. Cuando n es grande, la distribución de X es aproximadamente normal con

tion "np(1 − p). We use this approximation when np ≥ 10 and n(1 – p) ≥ 10. (p. 403)

media np y desviación estándar "np(1 − p). Se hace uso de esta aproximación cuando np ≥ 10 and n(1 – p) ≥ 10. (pág. 403)

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Glossary/Glosario

G-13

Normal curves  Important class of density curves that are symmetric, single-peaked, and bell-shaped. (p. 109)

curvas normales Clase importante de curvas de densidad que son simétricas, de un solo pico y con la forma de curva de campana. (pág. 109)

Normal distribution  Distribution described by a Normal density curve. Any particular Normal distribution is completely specified by two numbers, its mean m and standard deviation s. The mean of a Normal distribution is at the center of the symmetric Normal curve. The standard deviation is the distance from the center to the change-of-curvature points on either side. We abbreviate the Normal distribution with mean m and standard deviation s as N(m, s). (p. 109)

distribución normal  Según la describe una curva de densidad normal. Cualquier distribución normal dada se especifica completamente con dos cifras, su media m y la desviación estándar s. La media de una distribución normal yace en el centro de la curva normal simétrica. La desviación estándar es la distancia del centro a los puntos a ambos lados en los que cambia la curva. La distribución normal con la media m y la desviación estándar s se abrevia N(m, s). (pág. 109)

Normal/Large Sample condition for inference about a mean  The population has a Normal distribution or the sample size is large (n ≥ 30). If the population distribution has unknown shape and n < 30, use a graph of the sample data to assess the Normality of the population. Do not use t procedures if the graph shows strong skewness or outliers. (p. 515)

condición de muestra normal/grande para inferir sobre una media  La población tiene una distribución normal o la muestra es de tamaño grande (n ≥ 30). Si la distribución de la población tiene una forma desconocida y n < 30, se usa un gráfico de los datos de la muestra para evaluar la normalidad de la población. No se han de usar los procedimientos t si en el gráfico se aprecian valores atípicos o un sesgo marcado. (pág. 515)

Normal probability plot Plot used to assess whether a data set follows a Normal distribution. To make a Normal probability plot, (1) arrange the data values from smallest to largest and record the percentile of each observation, (2) use the standard Normal distribution to find the z-scores at these same percentiles, and (3) plot each observation x against the corresponding z. If the points on a Normal probability plot lie close to a straight line, the plot indicates that the data are approximately Normal. (p. 122)

gráfico de probabilidad normal  Se usa para evaluar si un conjunto de datos se ciñe a una distribución normal. Para trazar un gráfico de probabilidad normal, (1) se dispone de los valores de los datos del más pequeño al más grande y se anota el percentil de cada observación, (2) se usa la distribución normal estándar para hallar los puntos z en esos mismos percentiles, y (3) se traza cada observación x con la z correspondiente. Si los puntos en un gráfico de probabilidad normal yacen cerca de una línea recta, el gráfico indica que los datos son aproximadamente normales. (pág. 122)

null hypothesis H0  Claim we weight evidence against in a ­significance test. Often the null hypothesis is a statement of “no difference.” (p. 540)

hipótesis nula H0  Contrapeso de la evidencia en una prueba de significancia. A menudo la hipótesis nula es una declaración de “no hay diferencia.” (pág. 540)

O observational study Study that observes individuals and measures variables of interest but does not attempt to influence the responses. (p. 235)

estudio de observación  Se observan los individuos y se miden las variables de interés pero no se trata de influir en las respuestas. (pág. 235)

observed counts Actual numbers of individuals in the sample that fall in each cell of the one-way or two-way table. (p. 681)

cuentas observadas  Las cifras reales que corresponden a individuos en la muestra que caen en cada celda de la tabla de una vía o en la de dos vías. (pág. 681)

one-sample t interval for a mean  When the Random, 10%, and Normal/Large Sample conditions are met, a C% confidence interval for m is

intervalo t de una sola muestra para una media Cuando se cumplen las condiciones de aleatorio, del 10% y de cuentas normales/grandes, un intervalo de confianza C% para es

x ± t*

sx "n

where t* is the critical value for the t distribution with df = n − 1, with C% of the area between −t* and t*. (p. 518)

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x ± t*

sx "n

en el que t* es el valor crítico para la distribución t con df = n − 1, con C% del área entre −t* y t*. (pág. 518)

12/2/13 9:37 AM

G-14

Glossary/Glosario

one-sample t test for a mean  Suppose that the Random, 10%, and Normal/Large Sample conditions are met. To test the hypothesis H0: m = m 0, compute the one-sample t statistic t=

x − m0 sx "n

Find the P-value by calculating the probability of getting a t statistic this large or larger in the direction specified by the alternative hypothesis Ha in a t distribution with df = n – 1. (p. 580)

one-sample z interval for a proportion When the Random, 10%, and Large Counts conditions are met, a C% confidence interval for the unknown proportion p is p^ ± z*

Ç

p^ (1 − p^ ) n

where z* is the critical value for the standard Normal curve with C% of its area between −z* and z*. (p. 498) one-sample z test for a proportion  Suppose that the Random, 10%, and Large Counts conditions are met. To test the hypothesis H0: p = p0, compute the z statistic p^ − p0

z= Å

p0(1 − p0) n

Find the P-value by calculating the probability of getting a z statistic this large or larger in the direction specified by the alternative hypothesis Ha . (p. 559)

intervalo t de una sola muestra para una media  Supongamos que se han cumplido las condiciones de aleatorio, del 10% y de muestra normal/grande. Para probar la hipótesis H0: m = m0, se computa la estadística t de una sola muestra t=

x − m0 sx "n

Se halla el valor P computando la probabilidad de obtener una estadística t de este tamaño o más grande en el sentido especificado por la hipótesis alternativa en una distribución t con df = n – 1. (pág. 580) intervalo z de una sola muestra para una proporción  Cuando se cumplen las condiciones de aleatorio, del 10% y de cuentas grandes, un intervalo de confianza C% para la proporción desconocida p es p^ ± z*

Ç

p^ (1 − p^ ) n

en el que z* es el valor crítico para la curva normal estándar con C% de su área entre − z* y z*. (pág. 498) prueba z de una sola muestra para una proporción  Supongamos que se han cumplido las condiciones de aleatorio, del 10% y de muestras grandes. Para probar la hipótesis H0: p = p0, se computa la estadística p^ − p0

z= Å

p0(1 − p0) n

Se halla el valor P computando la probabilidad de obtener una estadística z de este tamaño o más grande en el sentido especificado por la hipótesis alternativa Ha . (pág. 559)

one-sided alternative hypothesis  An alternative hypothesis that states that a parameter is larger than the null hypothesis value or that states that the parameter is smaller than the null value. (p. 541)

hipótesis alternativa unilateral   Hipótesis alternativa que indica que un parámetro es más grande que el valor de la hipótesis nula o que indica que el parámetro es más pequeño que el valor nulo. (pág. 541)

one-way table  Table used to display the distribution of a single categorical variable. (p. 680)

tabla de una vía  Se usa para mostrar la distribución de una sola variable categorizada. (pág. 680)

outlier  Individual value that falls outside the overall pattern of a distribution. (p. 26)

valor atípico  Un valor individual que cae por fuera del patrón general de la distribución. (pág. 26)

outlier in regression Observation that lies outside the overall pattern of the other observations. Points that are outliers in the y direction but not the x direction of a scatterplot have large residuals. Other outliers may not have large residuals. (p. 189)

valor atípico en regresión Observación que yace por fuera del patrón general de las otras observaciones. Los puntos que son valores atípicos en el sentido y pero no en el sentido x de un gráfico de dispersión tienen residuales grandes. Es posible que otros valores atípicos no tengan residuales grandes. (pág. 189)

P P-value  The probability, computed assuming H0 is true, that the statistic would take a value as extreme as or more extreme than the one actually observed, in the direction specified by Ha. The smaller the P-value, the stronger the evidence against H0 and in favor of Ha provided by the data. (p. 543)

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valor P  La probabilidad, computada suponiendo que H0 es verdad, de que la estadística tomaría un valor tan extremo o más extremo que el que de hecho se observa, en el sentido especificado por Ha. Cuanto menor sea el valor P, más fuerte será la evidencia contra H0 y en favor de Ha que proporcionan los datos. (pág. 543)

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Glossary/Glosario

G-15

paired data  Study designs that involve making two observations on the same individual or one observation on each of two similar individuals result in paired data. (p. 586)

datos apareados Se estudian diseños que implican hacer dos observaciones del mismo individuo, o una observación de cada uno de dos individuos parecidos, resultando en datos apareados. (pág. 586)

paired t procedures When paired data result from measuring the same quantitative variable twice, we can make comparisons by analyzing the differences in each pair. If the conditions for inference are met, we can use one-sample t procedures to perform inference about the mean difference md . These methods are sometimes called paired t procedures. (p. 586)

procedimientos t apareados  Cuando la misma variable cuantitativa se mide dos veces pueden resultar datos apareados, con los cuales se pueden hacer comparaciones que analizan las diferencias en cada par. Si se cumplen las condiciones para la inferencia, se pueden usar procedimientos t de una sola muestra para realizar una inferencia acerca de la diferencia media md . A estos métodos a veces se les denomina procedimientos t apareados. (pág. 586)

parameter  A number that describes some characteristic of the population. (p. 424)

parámetro  Número que describe algunas de las características de la población. (pág. 424)

percentile  The pth percentile of a distribution is the value with p percent of the observations less than it. (p. 85)

percentil  El percentil pavo de una distribución es el valor cuyo porcentaje de las observaciones es menor que la cifra. (pág. 85)

pie chart  Chart that shows the distribution of a categorical variable as a “pie” whose slices are sized by the counts or percents for the categories. A pie chart must include all the categories that make up a whole. (p. 8)

gráfico circular  Muestra la distribución de una variable categorizada n la forma de un círculo subdividido según las cuentas o los porcentajes de las categorías. El gráfico circular tiene que incluir todas las categorías que componen la totalidad. (pág. 8)

placebo  Inactive (fake) treatment. (p. 244)

placebo  Tratamiento inactivo (falso). (pág. 244)

placebo effect  Describes the fact that some subjects respond favorably to any treatment, even an inactive one (placebo). (p. 247)

efecto placebo  Describe el hecho de que algunos sujetos responden de manera favorable a cualquier tratamiento, incluso uno inactivo (con placebo). (pág. 247)

point estimate  Specific value of a point estimator from a sample. (p. 477)

estimado de punto  El valor especifico de un estimador de punto tomado de una muestra. (pág. 477)

point estimator  Statistic that provides an estimate of a population parameter. (p. 477)

estimador de punto  Estadística que nos da un estimado de un parámetro de la población. (pág. 477)

pooled or combined sample proportion  The overall proportion of successes in the two samples is

proporción combinada de la muestra  La proporción total de aciertos en las dos muestras es

p^ C =

count of successes in both samples combined count of individuals in both samples combined

=

X1 + X2 n1 + n2

(p. 621)

cuenta de aciertos en X + X2 ^pC = ambas muestras combinadas = 1 cuenta de individuos en n1 + n2 ambas muestras combinadas (pág. 621)

population  In a statistical study, the entire group of individuals we want information about. (p. 210)

población  En un estudio estadístico, la población es el grupo completo de individuos sobre el cual deseamos contar con información. (pág. 210)

population distribution Gives the values of the variable for all the individuals in the population. (p. 428)

distribución de la población  Presenta los valores de la variable para todos los individuos en la población. (pág. 428)

population regression line  Regression line my = a + bx based on the entire population of data. (p. 739)

línea de regresión de la población  La línea de regresión my = a + bx basada en la totalidad de la población de datos. (pág. 739)

positive association  When above-average values of one variable to accompany above-average values of the other and also of below-average values to occur together. (p. 148)

asociación positiva  Cuando los valores por encima del promedio de una variable tienden a acompañar a los valores por encima del promedio de la otra, y los valores por debajo del promedio también tienden a suceder juntos. (pág. 148)

power  The probability that a test will reject H0 at a chosen significance level a when a specified alternative value of the parameter is true. The power of a test against any alternative is 1 minus the probability of a Type II error for that alternative; that is, power = 1 − b. (p. 565)

poder  La probabilidad de que una prueba rechace H0 en un nivel de significancia dado cuando un valor alternativo especificado del parámetro es verdad. El poder de una prueba con respecto a cualquier alternativa es 1 menos la probabilidad de un error Tipo II para dicha alternativa; es decir, poder = 1 − b. (pág. 565)

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G-16

Glossary/Glosario

power model Relationship of the form y = axp . When experience or theory suggests that the relationship between two variables is described by a power model, you can transform the data to achieve linearity in two ways: (1) raise the values of the explanatory variable x to the p power and plot the points (xp, y), or (2) take the pth root of the values of the response variable y and plot p the points (x, !y). If you don’t know what power to use, taking the logarithms of both variables should produce a linear pattern. (p. 767)

modelo de poder  Relación de la forma y = axp . Cuando la experiencia o una teoría sugiere que la relación entre dos variables la describe un modelo de poder, se pueden transformar los datos para que logren la linealidad de dos maneras: (1) elevar los valores de la variable explicativa x a la potencia p y trazar los puntos (xp, y), o (2) tomar la raíz p de los valores de la variable de respup esta y y trazar los puntos (x, !y). Si no se sabe qué potencia se ha de usar, se toman los logaritmos de ambas variables para producir un patrón lineal. (pág. 767)

predicted value  y^ (read “y hat”) is the predicted value of the response variable y for a given value of the explanatory variable x. (p. 166)

valor proyectado  y^ es el valor proyectado de la variable de respuesta y para un valor dado de la viariable explicativa x. (pág. 166)

probability  A number between 0 and 1 that describes the proportion of times an outcome of a chance process would occur in a very long series of repetitions. (p. 291)

probabilidad  Cifra entre 0 y 1 que describe la proporción de veces que un resultado de un proceso aleatorio sucedería en una serie muy prolongada de repeticiones. (pág. 291)

probability distribution  Gives the possible values of a random variable and their probabilities. (p. 348)

distribución de la probabilidad  Presenta los valores posibles de una variable aleatoria y sus posibles probabilidades. (pág. 348)

probability model Description of some chance process that consists of two parts: a sample space S and a probability for each outcome. (p. 305)

modelo de probabilidad  Descripción de un proceso de probabilidad que consta de dos partes: un espacio de muestra s y una probabilidad para cada resultado. (pág. 305)

Q quantitative variable Variable that takes numerical values for which it makes sense to find an average. (p. 3)

variable cuantitativa Toma valores numéricos para los cuales tiene sentido hallar un promedio. (pág. 3)

R random assignment  Experimental design principle. Use chance to assign experimental units to treatments. Doing so helps create roughly equivalent groups of experimental units by b ­ alancing the effects of other variables among the treatment groups. (p. 241)

asignación aleatoria  Principio de diseño experimental. Se usa el azar para asignar unidades experimentales a los tratamientos a fin de ayudar a formar grupos de unidades experimentales más o menos equivalentes al equilibrar los efectos de otras variables entre los grupos de tratamiento. (pág. 241)

random condition  The data come from a well-designed random sample or randomized experiment. (p. 493)

condición aleatoria  Los datos provienen de una muestra aleatoria bien diseñada o de un experimento aleatorizado. (pág. 493)

random sampling  Using a chance process to determine which members of a population are included in the sample. (p. 214)

muestreo aleatorio  Uso de un proceso de probabilidad para determinar cuáles miembros de la población se han de incluir en la muestra. (pág. 214)

random variable Variable that takes numerical values that describe the outcomes of some chance process. (p. 348)

variable aleatoria  Toma valores numéricos que describen los resultados de algún proceso de azar y ­probabilidad. (pág. 348)

randomization distribution Distribution of a statistic (like p^ 1 − p^ 2 or x1 − x2 ) in repeated random assignments of experimental units to treatment groups assuming that the specific treatment received doesn’t affect individual responses. When the conditions are met, usual inference procedures based on the sampling distribution of the statistic will be approximately correct. (p. 627)

distribución de la aleatoriedad  La distribución de una estadística (como p^ 1 − p^ 2 o x1 − x2) en designaciones aleatorizadas reiteradas de unidades experimentales a grupos de tratamiento, asumiendo que el tratamiento específico no afecte las respuestas individuales. Cuando se cumplan las condiciones, los procedimientos de inferencia corrientes que se basan en la distribución del muestreo de la estadística serán aproximadamente correctos. (pág. 627)

randomized block design  Experimental design begun by forming blocks consisting of individuals that are similar in some way that is important to the response. Random assignment of treatments is then carried out separately within each block. (p. 252)

diseño de bloques aleatorios  Se comienza con la formación de bloques compuestos por individuos que son similares de alguna manera que sea importante para la respuesta. La asignación aleatoria de tratamientos luego se realiza separadamente dentro de cada bloque. (pág. 252)

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Glossary/Glosario

G-17

range  The maximum value minus the minimum value for a set of quantitative data. (p. 54)

gama  El valor máximo menos el valor mínimo de un conjunto de datos cuantitativos. (pág. 54)

regression line Line that describes how a response variable y changes as an explanatory variable x changes. We often use a regression line to predict the value of y for a given value of x. (p. 164)

línea de regresión  Una línea que describe cómo una variable de respuesta y cambia a medida que cambia una variable explicativa x. A menudo se usa una línea de regresión para predecir el valor de y para un valor dado de x. (pág. 164)

reject H0  If the observed result is too unlikely to occur just by chance when the null hypothesis is true, we can reject H0 and say that there is convincing evidence for Ha. (p. 544)

rechazar H0  Si es demasiado improbable que el resultado observado suceda por simple azar cuando la hipótesis nula es verdad, se puede rechazar H0 y decir que existe evidencia convincente a favor de Ha. (pág. 544)

relative frequency table  Table that shows the percents (relative frequencies) of observations in each category or class. (p. 8)

tabla de frecuencia relativa Permite apreciar los porcentajes (frecuencias relativas) de las observaciones en cada categoría o clase. (pág. 8)

replication  Experimental design principle. Use enough experimental units in each group so that any differences in the effects of the treatments can be distinguished from chance differences between the groups. (p. 242)

replicación  Principio de diseño experimental. Se usan suficientes unidades experimentales en cada grupo a fin de que todas las diferencias en los efectos de los tratamientos se puedan distinguir de las diferencias de azar entre los grupos. (pág. 242)

residual  Difference between an observed value of the response variable and the value predicted by the regression line:

residual  La diferencia entre un valor observado de la variable de respuesta y el valor proyectado por la línea de regresión. Es decir,

residual = observed y – predicted y = y − y^

residual = observada y – proyectada y = y − y^ .

(p. 169)

(pág. 169)

residual plot  Scatterplot of the residuals against the explanatory variable. Residual plots help us assess whether a linear model is appropriate. (p. 173)

gráfico residual  Gráfico de dispersión de los residuales en comparación con la variable explicativa. Los trazados residuales nos ayudan a evaluar si el modelo lineal es apropiado. (pág. 173)

resistant measure  Statistic that is not affected very much by extreme observations. (p. 50)

medida resistente  Estadística que no se ve muy afectada por observaciones extremas. (pág. 50)

response bias  Systemic pattern of inaccurate answers. (p. 227)

sesgo de la respuesta  Patrón sistémico de respuestas imprecisas. (pág. 227)

response variable  Variable that measures an outcome of a study. (pp. 143, 236)

variable de respuesta  Variable que mide un resultado de un estudio. (págs. 143, 236)

roundoff error Difference between the calculated approximation of a number and its exact mathematical value. (p. 8)

error de redondeo La diferencia entre la aproximación computada y su valor matemático exacto. (pág. 8)

S sample  Subset of individuals in the population from which we actually collect data. (p. 210)

muestra  Subconjunto de individuos en la población a partir de la cual de hecho se recogen datos. (pág. 210)

sample regression line (estimated regression line)  Least-squares regression line y^ = a + bx computed from the sample data. (p. 739)

línea de regresión de la muestra (línea de regresión estimada)  La línea de regresión de mínimos cuadrados y^ = a + bx computada a partir de los datos de la muestra. (pág. 739)

sample space S  Set of all possible outcomes of a chance process. (p. 305)

espacios S de la muestra  El conjunto de todos los resultados posibles de un proceso de probabilidad. (pág. 305)

sample survey Study that uses an organized plan to choose a sample that represents some specific population. We base conclusions about the population on data from the sample. (p. 210)

valoración de la muestra  Estudio que usa un plan organizado para escoger una muestra que represente una población específica. Basamos las conclusiones sobre la población en datos tomados de la muestra. (pág. 210)

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G-18

Glossary/Glosario

sampling distribution  The distribution of values taken by a statistic in all possible samples of the same size from the same population. (p. 427)

distribución del muestreo La distribución de valores tomados por la estadística en todas las muestras posibles del mismo tamaño tomadas de la misma población. (pág. 427)

sampling distribution of a sample mean x  Suppose that x is the mean of an SRS of size n drawn from a large population with mean m and standard deviation s. Then • The mean of the sampling distribution of x is mx = m. • The standard deviation of the sampling distribution of x is

distribución de muestreo de la media de una muestra x  Supongamos que x es la media de una muestra aleatoria sencilla de tamaño n tomada de una población grande con media m y una desviación estándar s. ­Entonces • La media de la distribución del muestreo de x es mx = m. • La desviación estándar de la distribución del muestreo de x es

sx =

s !n

1  N. 10 • If the population has a Normal distribution, then the sampling distribution of x also has a Normal distribution. Otherwise, the central limit theorem tells us that the sampling distribution of x will be approximately Normal in most cases when n ≥ 30. (p. 452) as long as the 10% condition is satisfied: n ≤

sampling distribution of a sample proportion p^   Choose an SRS of size n from a population of size N with proportion p of successes. Let p^ be the sample proportion of successes. Then • The mean of the sampling distribution of p^ is mp^ = p. • The standard deviation of the sampling distribution of p^ is sp^ =

p(1 − p) Å

n

1  N. 10 • As n increases, the sampling distribution of p^ becomes approximately Normal. Before you perform Normal calculations, check that the Large Counts condition is satisfied: np ≥ 10 and n(1 – p) ≥ 10. (p. 444) as long as the 10% condition is satisfied: n ≤

sampling distribution of a slope  Choose an SRS of n observations (x, y) from a population of size N with least-squares regression line my = a + bx. Let b be the slope of the sample regression line. Then • The mean of the sampling distribution of b is mb = b. • The standard deviation of the sampling distribution of b is s 1 sb = as long as the 10% condition is satisfied: n ≤  N. 10 sx"n

• The sampling distribution of b will be approximately Normal if the values of the response variable y follow a Normal distribution for each value of the explanatory variable x (the Normal ­condition). (p. 741)

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sx =

s !n

1  N. 10 • Si la población tiene una distribución normal, entonces la distribución del muestreo x también tiene una distribución normal. De no ser así, el teorema del límite central nos indica que en la mayoría de los casos, la distribución del muestreo x sera aproximadamente normal en la mayoría de casos cuando n ≥ 30. siempre y cuando se cumpla con la condición del 10%: n ≤

distribución del muestreo de una proporción de la muestra p^   Se escoge una muestra aleatoria sencilla de tamaño n a partir de una población de tamaño N con proporción p de aciertos. Permita que p^ sea la proporción de aciertos de la muestra. Entonces • La media de la distribución del muestreo de p^ es mp^ = p. • La desviación estándar de la distribución del muestreo de p^ es sp^ =

p(1 − p) Å

n

siempre y cuando que se cumpla con la condición del 10%: 1 n ≤  N. 10 • A medida que aumenta n, la distribución del muestreo de p^ se torna aproximadamente normal. Antes de realizar los cómputos normales, se verifica que se haya cumplido con la condición de cuentas grandes: np ≥ 10 y n(1 – p) ≥ 10. (pág. 444) distribución del muestreo de una pendiente Se escoge una muestra aleatoria sencilla de n observaciones (x, y) a partir de una población tamaño N con línea de regresión de mínimos cuadrados my = a + bx. Se permite que b sea la pendiente de una línea de regresión de la muestra. ­Entonces • La media de la distribución del muestreo de b es mb = b. • La desviación estándar de la distribución del muestreo de b es s sb = siempre y cuando se cumpla con la condición del sx"n 1 10%: n ≤  N. 10 • La distribución del muestreo de b será aproximadamente normal si los valores de la variable de respuesta y siguen una distribución normal para cada valor de la variable explicativa x (la condición normal). (pág. 741)

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Glossary/Glosario

sampling distribution of p^ 1 ∙ p^ 2   Choose an SRS of size n1 from population 1 with proportion of successes p1 and an independent SRS of size n2 from population 2 with proportion of successes p2 . • Shape: When n1 p1, n1(1– p1), n2 p2, and n2(1– p2) are all at least 10, the sampling distribution of p^ 1 − p^ 2 is approximately Normal. • Center: The mean of the sampling distribution is p1 − p2 . • Spread: The standard deviation of the sampling distribution of p^ 1 − p^ 2 is

Å

p1(1 − p1) n1

+

p2(1 − p2)

distribución del muestreo p^ 1 ∙ p^ 2   Se escoge una muestra aleatoria sencilla de tamaño n1 a partir de una población 1 con una proporción de aciertos p1 y una muestra aleatoria sencilla independiente de tamaño n2 a partir de la población 2 con una proporción de aciertos p2 . • Forma: Cuando n1 p1, n1(1– p1), n2 p2, and n2(1– p2) son por lo menos 10, la distribución del muestreo de p^ 1 − p^ 2 es aproximadamente normal. • Centro: La media de la distribución del muestreo es p1 − p2 . •A  mplitud: La desviación estándar de la distribución del muestreo de p^ 1 − p^ 2 es

n2

as long as each sample is no more than 10% of its population. (p. 612) sampling distribution of x1 ∙ x2   Choose an SRS of size n1 from population 1 with mean m1 and standard deviation s1 and an independent SRS of size n2 from population 2 with mean m2 and standard deviation s2 . • Shape: When the population distributions are Normal, the sampling distribution of x1 − x2 is Normal. In other cases, the sampling distribution of x1 − x2 will be approximately Normal if the sample sizes are large enough (n1 ≥ 30 and n2 ≥ 30). • Center: The mean of the sampling distribution is m1 − m2 . • Spread: The standard deviation of the sampling distribution of x1 − x2 is s21 s22 + Å n1 n2

as long as each sample is no more than 10% of its population. (p. 638)

G-19

Å

p1(1 − p1) n1

+

p2(1 − p2) n2

siempre y cuando cada muestra es de no más del 10% de su población. (pág. 612) distribución de muestreo de x1 ∙ x2   Se escoge una muestra aleatoria sencilla de tamaño a partir de una población 1 con una media m1 y una desviación estándar s1 y una muestra aleatoria sencilla independiente de tamaño n2 a partir de una población 2 con una media m2 y una desviación estándar s2 . • Forma: Cuando las distribuciones de la población son normales, la distribución del muestreo de x1 − x2 es normal. En otros casos, la distribución del muestreo de x1 − x2 será aproximadamente normal si los tamaños de las muestras son suficientemente grandes (n1 ≥ 30 y n2 ≥ 30). • Centro: La media de la distribución del muestreo es m1 − m2 . • Amplitud: La desviación estándar de la distribución del muestreo de x1 − x2 es s21 s22 + Å n1 n2

siempre y cuando cada muestra es de no más del 10% de su población. (pág. 638)

sampling variability  The value of a statistic varies in repeated random sampling. (p. 425)

variabilidad del muestreo  El valor de una estadística varía en muestras aleatorias reiteradas. (pág. 425)

scatterplot  Plot that shows the relationship between two quantitative variables measured on the same individuals. The values of one variable appear on the horizontal axis, and the values of the other variable appear on the vertical axis. Each individual in the data appears as a point in the graph. (p. 145)

gráfico de dispersión  Permite apreciar la relación entre dos variables cuantitativas midiendo los mismos individuos. Los valores de una variable figuran en el eje horizontal, y los valores de la otra variable figuran en el eje vertical. Cada individuo en los datos figura como un punto en el gráfico. (pág. 145)

segmented bar graph  Graph used to compare the distribution of a categorical variable in each of several groups. For each group, there is a single bar with “segments” that correspond to the different values of the categorical variable. The height of each segment is determined by the percent of individuals in the group with that value. Each bar has a total height of 100%. (p. 17)

gráfico de barras segmentado  Se usa para comparar la distribución de una variable categorizada en cada uno de varios grupos. Para cada grupo, hay una sola barra que tiene “segmentos” que corresponden a los diferentes valores de la variable categorizada. La altura de cada segmento la determina el porcentaje de individuos en el grupo que tengan ese valor. Cada barra tiene una altura total del 100%. (pág. 17)

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G-20

Glossary/Glosario

side-by-side bar graph  Graph used to compare the distribution of a categorical variable in each of several groups. For each value of the categorical variable, there is a bar corresponding to each group. The height of each bar is determined by the count or percent of individuals in the group with that value. (p. 17)

gráfico de barras contiguas  Se usa para comparar la distribución de una variable categorizada en cada uno de varios grupos. Para cada valor de la variable categorizada, hay una barra que corresponde a cada grupo. La altura de la barra la determina el cuenteo o el porcentaje de individuos en el grupo que tengan ese valor. (pág. 17)

significance level  Fixed value a that we use as a cutoff for deciding whether an observed result is too unlikely to happen by chance alone when the null hypothesis is true. The significance level gives the probability of a Type I error. (p. 545)

nivel de significancia  Valor fijo a que se usa como punto de corte para decidir si un resultado observado es demasiado improbable para suceder solo al azar cuando la hipótesis nula es verdad. El nivel de significancia genera la probabilidad de un error Tipo 1. (pág. 545)

significance test  Procedure for using observed data to decide between two competing claims (also called hypotheses). The claims are often statements about a parameter. (p. 539)

prueba de significancia  Procedimiento en el que se usan datos observados para decidir entre dos opciones que compiten entre sí (también se les dice hipótesis). Las opciones a menudo son enunciados acerca de un parámetro. (pág. 539)

simple random sample (SRS) Sample chosen in such a way that every group of n individuals in the population has an equal chance to be selected as the sample. (p. 214)

muestra aleatoria sencilla  Muestra tomada de tal manera que cada grupo de n individuos en la población tenga la misma oportunidad de ser escogido como la muestra. (pág. 214)

simulation  Imitation of chance behavior, based on a model that accurately reflects the situation. (p. 295)

simulación  Imitación de conducta de azar, basada en un modelo que refleja la situación con precisión. (pág. 295)

single-blind  An experiment in which either the subjects or those who interact with them and measure the response variable, but not both, know which treatment a subject received. (p. 248)

ciego sencillo  Experimento en el que ya sea los sujetos o bien aquellos que interactúan con los sujetos y miden la variable de respuesta, pero no ambos, saben cuál fue el tratamiento que recibió un sujeto. (pág. 248)

skewness  A distribution is skewed to the right if the right side of the graph (containing the half of the observations with larger values) is much longer than the left side. It is skewed to the left if the left side of the graph is much longer than the right side. (p. 27)

asimetría  Distribución que está sesgada hacia la derecha si la derecha del gráfico (que contiene la mitad de las observaciones con valores más grandes) es mucho más larga que el lado izquierdo. Está sesgada hacia la izquierda si el lado izquierdo del gráfico es mucho más largo que el lado derecho. (pág. 27)

slope  Suppose that y is a response variable (plotted on the vertical axis) and x is an explanatory variable (plotted on the horizontal axis). A regression line relating y to x has an equation of the form y^ = a + bx. In this equation, b is the slope, the amount by which y is predicted to change when x increases by one unit. (p. 166)

pendiente  Supongamos que y es una variable de respuesta (trazada en el eje vertical) y x es una variable explicativa (trazada en el eje horizontal). Una línea de regresión que relacione y con x tiene una ecuación en la forma de y^ = a + bx. En esta ecuación, b es la pendiente, la cantidad mediante la cual se predice que y cambiará cuando x tenga un aumento de una unidad. (pág. 166)

splitting stems  Method for spreading out a stemplot that has too few stems. (p. 32)

tallos separados Método para separar un gráfico de tallos que tiene una carencia de tallos. (pág. 32)

standard deviation sx  Statistic that measures the typical distance of the values in a distribution from the mean. It is calculated by finding an “average” of the squared distances and then taking the square root. In symbols,

desviación estándar sx  Estadística que mide la distancia típica de los valores en una distribución a partir de la media. Se computa hallando un “promedio” de las distancias al cuadrado a las que luego se les computa la raíz cuadrada. En símbolos se representa

sx = (p. 61)

1 ∙ (xi − x)2 Ån − 1

sx = (pág. 61)

1 ∙ (xi − x)2 Ån − 1

standard deviation of a random variable Square root of the variance of a random variable s2X . The standard deviation measures the typical distance of the values in a distribution from their mean. In symbols,

desviación estándar de una variable aleatoria   La raíz cuadrada de la variación de una variable aleatoria s2X . La desviación estándar mide la distancia típica de los valores en una distribución a partir de su media. En símbolos se represent

sX = " (xi − mX)2pi

sX = " (xi − mX)2pi

∙

(p. 353)

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∙

(pág. 353)

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Glossary/Glosario

standard deviation of the residuals (s)  If we use a least-squares line to predict the values of a response variable y from an explanatory variable x, the standard deviation of the residuals (s) is given by s=

∙ residuals2 Ç

n−2

=

Ç

G-21

desviación estándar de las residuales (s)  Si se hace uso de la línea de cuadrados mínimos para predecir los valores de una variable de respuesta y a partir de una variable explicativa x, la desviación estándar de las residuales (s) la da

∙ (yi − y^ )2

s=

n−2

∙ residuales2 Ç

n−2

=

Ç

∙ (yi − y^ )2 n−2

This value gives the approximate size of a “typical” prediction error (residual). (p. 177)

Este valor ofrece un tamaño aproximado de un error de predicción “típico” (residual). (pág. 177)

standard error  When the standard deviation of a statistic is estimated from data, the result is the standard error of the statistic. (p. 497)

error estándar  Cuando se computa la desviación estándar de una estadística a partir de datos, el resultado es el error estándar de la estadística. (pág. 497)

standard error of p^ 1 − p^ 2   Estimated standard deviation of the statistic p^ 1 − p^ 2 , given by

error estándar de p^ 1 − p^ 2   La desviación estándar coputada de la estadística p^ 1 − p^ 2 , dada por

(p. 616)

Ç

p^ 1(1 − p^ 1) n1

+

p^ 2(1 − p^ 2) n2

(pág. 616)

standard error of the sample mean 

sx

where sx is the sample

"n standard deviation. It describes how far x will typically be from m in repeated SRSs of size n. (p. 518)

standard error of the sample proportion 

p^ (1 − p^ )

where p^

Ç

p^ 1(1 − p^ 1) n1

+

p^ 2(1 − p^ 2) n2

error estándar de la media de la muestra 

sx

en el que sx es la "n desviación estándar de la muestra. Describe la distancia típica en la que x se separa de m en muestras aleatorias sencillas reiteradas de tamaño n. (pág. 518) error estándar de la proporción de la muestra 

p^ (1 − p^ )

n Ç is the sample proportion. It describes how far p^ will typically be from p in repeated SRSs of size n. (p. 496)

n Ç en la que p^ es la proporción de la muestra. Se describe la distancia típica en la que p^ se separa de p en muestras aleatorias sencillas reiteradas de tamaño n. (pág. 496)

standard error of the slope  Formula used to estimate the spread of the sampling distribution of b: s SEb = sx"n − 1 (p. 747)

error estándar de la pendiente  Se usa para computar la amplitud de la distribución del muestreo de b: s SEb = sx"n − 1 (pág. 747)

standard error of x1 ∙ x2   Estimated standard deviation of the statistic x1 − x2 , given by

error estándar de x1 ∙ x2   La desviación estándar computada de la estadística x1 − x2 , dada por

s21 s22 + Å n1 n2

s21 s22 + Å n1 n2

(p. 640)

(pág. 640)

standard Normal distribution  Normal distribution with mean 0 and standard deviation 1. (p. 113)

distribución normal estándar  La distribución normal con media de 0 y desviación estándar 1. (pág. 113)

standard Normal table (Table A)  Table of areas under the standard Normal curve. The table entry for each value z is the area under the curve to the left of z. (p. 113)

tabla normal estándar (Tabla A) Tabla de áreas debajo de la curva normal estándar. La ­entrada a la tabla de cada valor z es el área debajo de la curva a la ­izquierda de z. (pág. 113)

standardized score (z-score)  If x is an observation from a distribution that has known mean and standard deviation, the stanx − mean dardized value of x is z = . A standardized standard deviation

puntuación estandarizada Si x es una observación a partir de una distribución que tiene una media y desviación estándar conox − media cidas, el valor estandarizado de x es z = . Al desviación estándar valor estandarizado a menudo se le dice puntuación z. (pág. 90)

value is often called a z-score. (p. 90)

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G-22

Glossary/Glosario

statistic  Number that describes some characteristic of a sample. (p. 424)

estadística  Número que describe alguna característica de una muestra. (pág. 424)

statistically significant  (1) Observed effect so large that it would rarely occur by chance. (p. 249) (2) If the P-value is smaller than alpha, we say that the results of a statistical study are significant at level a. In that case, we reject the null hypothesis H0 and conclude that there is convincing evidence in favor of the alternative hypothesis Ha. (p. 545)

estadísticamente significativo  (1) Efecto observado que es tan grande que raramente sucedería producto del azar. (pág. 249) (2) Si el valor P es menor que alfa, se dice que los resultados de un estudio estadístico son significativos al nivel a. En tal caso, se rechaza la hipótesis nula H0 y se concluye que hay evidencia convincente a favor de la hipótesis Ha alternativa. (pág. 545)

stemplot (also called stem-and-leaf plot)  Simple graphical display for fairly small data sets that gives a quick picture of the shape of a distribution while including the actual numerical values in the graph. Each observation is separated into a stem, consisting of all but the final digit, and a leaf, the final digit. The stems are arranged in a vertical column with the smallest at the top. Each leaf is written in the row to the right of its stem, with the leaves arranged in increasing order out from the stem. (p. 31)

gráfico de tallos al que también se le dice gráfico de tallos y hojas  Representación gráfica sencilla de conjuntos de datos relativamente pequeños que dan una imagen rápida de la forma de una distribución, al tiempo que incluyen los valores numéricos mismos en el gráfico. Cada observación se separa en un tallo compuesto de todos menos el último dígito, y una hoja, que es ese último dígito. Los tallos se disponen en una columna vertical en la cual el valor más pequeño está arriba. Cada hoja se escribe en el renglón a la derecha del tallo, con las hojas dispuestas en orden ascendiente comenzando a partir del tallo. (pág. 31)

stratified random sample Sample obtained by classifying the population into groups of similar individuals, called strata, then choosing a separate SRS in each stratum and combining these SRSs to form the sample. (p. 219)

muestra aleatoria estratificada Muestra que se obtiene clasificando la población en grupos de individuos parecidos, llamados estratos. Luego, se escoge una muestra aleatoria sencilla separada en cada estrato y se combinan estas muestras aleatorias sencillas para conformar la muestra. (pág. 219)

subjects  Experimental units that are human beings. (p. 237)

sujetos  Unidades experimentales que son seres humanos. (pág. 237)

symmetric  A graph in which the right and left sides are approximately mirror images of each other. (p. 27)

simétrico  Si los lados derecho e izquierdo de un gráfico son reflejos aproximados uno del otro, se dice que son simétricos. (pág. 27)

T t distribution  Draw an SRS of size n from a large population that has a Normal distribution with mean m and standard deviation s. The statistic t=

x−m sx !n

has the t distribution with degrees of freedom df = n − 1. This statistic will have approximately a tn−1 distribution if the sample size is large enough. (p. 512) t interval for the slope b  When the conditions for regression inference are met, a C% confidence interval for the slope b of the population (true) regression line is b ± t*SEb . In this formula, the standard error of the slope is SEb =

s sx"n − 1

and t* is the critical value for the t distribution with df = n − 2 having C% of its area between −t* and t*. (p. 548)

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distribución t  Se grafica una muestra aleatoria sencilla de tamaño n a partir de una población grande que tiene una distribución normal con media de m y desviación estándar s. La estadística t=

x−m sx !n

tiene la distribución t con grados de libertad df = n − 1. Esta estadística tendrá una distribución aproximada tn−1 si el tamaño de la muestra es suficientemente grande. (pág. 512) intervalo t para la pendiente b  (pág. 748) Cuando se cumple con las condiciones para lograr una inferencia de regresión, el intervalo de confianza C% para la pendiente b de la línea de regresión de la población (verdadera) es b ± t*SEb . En esta fórmula, el error estándar de la pendiente es SEb =

s sx"n − 1

y t* es el valor crítico para la distribución t y df = n − 2 tiene el C% de su área entre −t* y t*. (pág. 548)

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Glossary/Glosario

t test for the slope  Suppose the conditions for inference are met. To test the hypothesis H0: b = b0 , compute the test statistic t=

b − b0 SEb

Find the P-value by calculating the probability of getting a t statistic this large or larger in the direction specified by the alternative hypothesis Ha . Use the t distribution with df = n − 2. (p. 753)

test statistic  Calculation that measures how far a sample statistic diverges from what we would expect if the null hypothesis H0 were true, in standardized units. That is, test statistic =

statistic − parameter standard deviation of statistic

G-23

prueba t para la pendiente Supongamos que se cumple con todas las condiciones para la inferencia. Para poner a prueba la hipótesis H0: b = b0 , se computa la estadística de prueba t=

b − b0 SEb

Se halla el valor P computando la probabilidad de obtener una ­estadística t de este tamaño o más grande en el sentido especificado por la hipótesis alternativa Ha . Se usa la distribución t con df = n − 2. (pág. 753) estadística de prueba  Mide la divergencia entre la estadística de muestra y lo que esperaríamos si la hipótesis nula H0 fuera verdad, expresado en unidades estandarizadas. Es decir, estadística de prueba =

estadística − parametro desviación estándar de la estadística

(p. 556)

(pág. 556)

third quartile Q3  In a data set in which the observations are ordered from lowest to highest, the median of the observations whose position is to the right of the median. (p. 54)

tercer cuartil Q3  Si las observaciones en un conjunto de datos se organizan de la más baja a la más alta, el tercer cuartil Q3 es la media de las observaciones cuya posición está a la derecha de la media. (pág. 54)

transforming  Applying a function such as the logarithm or square root to a quantitative variable. (p. 767)

transformación  La aplicación de una función tal como el logaritmo o la raíz cuadrada a una variable cuantitativa se denomina transformación del dato. (pág. 767)

treatment  Specific condition applied to the individuals in an experiment. If an experiment has several explanatory variables, a treatment is a combination of specific values of these variables. (p. 237)

tratamiento  Una condición específica que se les aplica a los individuos en un experimento. Si un experimento tiene varias variables explicativas, el tratamiento es una combinación de los valores específicos de estas variables. (pág. 237)

tree diagram Diagram used to display the sample space for a chance process that involves a sequence of outcomes. (p. 322)

diagrama de árbol Se usa para mostrar el espacio de muestra para un proceso de probabilidad que produce una secuencia de resultados. (pág. 322)

two-sample t interval for a difference between two means  When the Random, 10%, and Normal/Large Sample conditions are met, an approximate C% confidence interval for m1 − m2 is

intervalo t de dos muestras para obtener la diferencia entre dos medias  Cuando se cumple con las condiciones aleatoria, del 10% y de muestra normal/grande, un intervalo de confianza C% aproximado para m1 − m2 es

s21 s22 (x1 − x2) ± t* + Å n1 n2

Here t* is the critical value with area C% between − t* and t* for the t distribution with degrees of freedom from either Option 1 (technology) or Option 2 (the smaller of n1 − 1 and n2 − 1). (p. 641) two-sample t statistic  When we standardize the estimate x1 − x2 , the result is the two-sample t statistic t=

(x1 − x2) − (m1 − m2) s21 Å n1

+

s22 n2

The statistic t has the same interpretation as any z or t statistic: it says how far x1 − x2 is from its mean in standard deviation units. (p. 640)

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s21 s22 (x1 − x2) ± t* + Å n1 n2

en el que t* es el valor crítico con área C% entre − t* y t* para la distribución t con grados de libertad ya sea la Opción 1 (tecnología) o la Opción 2 (el valor menor de n1 − 1 y n2 − 1). (pág. 641) estadística t de dos muestras  Cuando se estandariza el estimado x1 − x2 , el resultado es una estadística t de dos muestras t=

(x1 − x2) − (m1 − m2) s21 s22 + Å n1 n2

La estadística t tiene la misma interpretación de cualquier estadística z o t: indica la distancia de x1 − x2 a partir de su media en unidades de desviación estándar. (pág. 640)

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G-24

Glossary/Glosario

two-sample t test for the difference between two means  Suppose the Random, 10%, and Normal/Large Sample conditions are met. To test the hypothesis H0: m1 − m2 = hypothesized value, compute the two-sample t statistic t=

(x1 − x2) − (m1 − m2) s21 s22 + Å n1 n2

t=

Find the P-value by calculating the probability of getting a t statistic this large or larger in the direction specified by the alternative hypothesis Ha . Use the t distribution with degrees of freedom approximated by technology or the smaller of n1 − 1 and n2 − 1. (p. 645) two-sample z interval for a difference between two proportions  When the Random, 10%, and Large Counts conditions are met, an approximate C% confidence interval for p1 − p2 is (p^ 1 − p^ 2) ± z* Ç

p^ 1(1 − p^ 1) n1

prueba t de dos muestras para obtener la diferencia entre dos medias  Supongamos que se ha cumplido con las condiciones aleatoria, del 10% y de muestra normal/grande. Para someter a prueba la hipótesis H0: m1 − m2 = valor hipotético, se computa la estadística t de dos muestras

+

p^ 2(1 − p^ 2) n2

where z* is the critical value for the standard Normal curve with C% of its area between −z* and z*. (p. 617)

two-sample z test for the difference between two proportions  Suppose the Random, 10%, and Large Counts conditions are met. To test the hypothesis H0: p1 − p2 = 0, first find the pooled proportion p^ C of successes in both samples combined. Then compute the z statistic

Ç

p^ C(1 − p^ C) n1

+

p^ C(1 − p^ C) n2

Find the P-value by calculating the probability of getting a z statistic this large or larger in the direction specified by the alternative hypothesis Ha . (p. 622)

s21 s22 + Å n1 n2

Halla el valor P computando la probabilidad de obtener una estadística t de este tamaño o más grande en el sentido especificado por la hipótesis Ha alternativa. Se usa la distribución t con grados de libertad aproximados por la tecnología o el valor menor de n1 − 1 y n2 − 1. (pág. 645) intervalo z de dos muestras para obtener la diferencia entre dos proporciones  Cuando se cumplen las condiciones de aleatorio, del 10% y de muestras grandes, el intervalo de confianza C% aproximado para p1 − p2 es (p^ 1 − p^ 2) ± z* Ç

p^ 1(1 − p^ 1) n1

+

p^ 2(1 − p^ 2) n2

en el que z* es el valor crítico para la curva estándar normal con C% de su área entre −z* y z*. (pág. 617) prueba z de dos muestras para obtener la diferencia entre dos proporciones  Supongamos que se cumplen las condiciones aleatoria, del 10% y de muestras grandes. Para someter a prueba la hipótesis H0: p1 − p2 = 0, primero se halla la proporción combinada p^ C de aciertos en ambas muestras combinadas. Luego se computa la estadística z

(p^ 1 − p^ 2) − 0

z=

(x1 − x2) − (m1 − m2)

(p^ 1 − p^ 2) − 0

z= Ç

p^ C(1 − p^ C) n1

+

p^ C(1 − p^ C) n2

Se halla el valor P computando la probabilidad de obtener una estadística z de este tamaño o más grande en el sentido especificado por la hipótesis Ha alternativa. (pág. 622)

two-sided alternative hypothesis The alternative hypothesis is two-sided if it states that the parameter is different from the null value (it could be either smaller or larger). (p. 541)

hipótesis alternativa bilateral  La hipótesis alternativa es bilateral si indica que el parámetro es diferente del valor nulo (podría ser más pequeño o más grande). (pág. 541)

two-way table  Table of counts that organizes data about two categorical variables. (p. 12)

tabla de doble vía  Una tabla de doble vía de cuentas organiza los datos acerca de dos variables categorizadas. (pág. 12)

Type I error  Occurs if we reject H0 when H0 is true. (p. 547)

error Tipo I Sucede si rechazamos H0 cuando H0 es verdad. (pág. 547)

Type II error Occurs if we fail to reject H0 when Ha is true. (p. 547)

error Tipo II  Sucede si no rechazamos H0 cuando Ha es verdad. (pág. 547)

U unbiased estimator  A statistic used for estimating a parameter is unbiased if the mean of its sampling distribution is equal to the true value of the parameter being estimated. (p. 429)

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estimador sin sesgo  La estadística que se usa para computar un parámetro es un estimador sin sesgo si la media de distribución de su muestreo equivale al valor verdadero del parámetro que se está computando. (pág. 429)

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Glossary/Glosario

G-25

undercoverage  Occurs when some members of the population cannot be chosen in a sample. (p. 225)

subcobertura  Sucede cuando algunos miembros de la población no pueden ser escogidos para la muestra. (pág. 225)

unimodal  A graph of quantitative data with a single peak. (p. 28)

unimodal  Gráfico de datos cuantitativos con un solo pico. (pág. 28)

union  The union of events A and B, denoted by A c B, consists of all outcomes in A or B or both. (p. 311)

unión  La unión de los eventos A y B, denotados por A c B, consiste en todos los resultados en A o B, o ambos. (pág. 311)

V variability of a statistic  Spread of a statistic’s sampling distribution. Statistics from larger samples have less variability. (p. 433)

variabilidad de una estadística  La variabilidad de una estadística se describe por la amplitud de la distribución de su muestreo. Las estadísticas de muestras más grandes tienen menos variabilidad. (pág. 433)

variable  Any characteristic of an individual. A variable can take different values for different individuals. (p. 2)

variable  Toda característica de un individuo. Una variable puede tener diferentes valores para diferentes individuos. (pág. 2)

variance of a random variable s2X   Weighted average of the squared deviations of the values of the variable from their mean. In symbols,

variación de una variable aleatoria s2X   Promedio sopesado de las desviaciones cuadráticas de los valores de la variable a partir de su media. En símbolos se expresa

s2X =

∙ (xi − mX)2pi

s2X =

∙ (xi − mX)2pi

(p. 352)

(pág. 352)

variance s2x   “Average” squared deviation of the observations in a data set from their mean. In symbols,

variación s2x   La desviación cuadrática “promedio” de las observaciones en el conjunto de datos según su separación de la media. En símbolos se expresa

s2x =

(x1 − x)2 + (x2 − x)2 + ... + (xn − x)2 1 = ∙ (xi − x)2 n−1 n−1

(p. 61)

s2x =

(x1 − x)2 + (x2 − x)2 + ... + (xn − x)2 1 = ∙ (xi − x)2 n−1 n−1

(pág. 61)

Venn diagrams  Used to display the sample space for a chance process. Venn diagrams can also be used to find probabilities involving events A and B. (p. 314)

diagramas Venn  Se usan para mostrar el espacio de muestras para un proceso de probabilidad. Los diagramas Venn también se pueden usar para hallar las probabilidades que existen para los eventos A y B. (pág. 314)

voluntary response sample  People decide whether to join a sample by responding to a general invitation. (p. 212)

muestra de respuesta voluntaria  La gente que decide vincularse a una muestra mediante una respuesta a una invitación generalizada. (pág. 212)

W wording of questions  Most important influence on the answers given in a survey. Confusing or leading questions can introduce strong bias, and changes in wording can greatly change a survey’s outcome. Even the order in which questions are asked matters. (p. 227)

terminología de las preguntas  La influencia más importante sobre las respuestas que se dan en un sondeo. Toda pregunta confusa, capciosa o que sugiera una respuesta puede introducirle al proceso un sesgo marcado, y los cambios en la terminología pueden modificar de manera marcada los resultados de tal sondeo. Incluso los resultados se pueden ver afectados por el orden en que se hacen las preguntas. (pág. 227)

Y y intercept  Suppose that y is a response variable (plotted on the vertical axis of a graph) and x is an explanatory variable (plotted on the horizontal axis). A regression line relating y to x has an equation of the form y^ = a + bx. In this equation, the number a is the y intercept, the predicted value of y when x = 0. (p. 166)

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interceptación y  Supongamos que y es una variable de respuesta (graficada en el eje vertical) y x es una variable explicativa (graficada en el eje horizontal). La línea de regresión que relaciona y con x tiene una ecuación y^ = a + bx. Según esta ecuación, el número a es la interceptación y, el valor proyectado de y cuando x = 0. (pág. 166)

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Index

Index A Addition of constant, 93–94, 95–97, 364, 366–368 of Normal random variables, 378–380 of random variables, 370–376, 378–380 Addition rule general, 310–311, 312 for mutually exclusive events, 308, 312, 330 Alternative hypothesis, 539–540 for chi-square test, 681, 710 American Community Survey (ACS), 226 Anonymity, vs. confidentiality, 271 Applets Correlation and Regression, 152, 170 Mean and Median, 53 Normal Approximation to Binomial, 402 Normal Density Curve, 110, 116 One-Variable Statistical Calculator, 36 Probability, 290 Reese’s Pieces®, 441 Rice University Sampling Distributions, 453–454, 456 Statistical Power, 590–591 Test of Significance, 538 Area under standard Normal curve, 115 Association, 18–19 confounding and, 236 independent events and, 328 negative, 146, 147–148, 151 positive, 148, 151 statistically significant, 249 vs. causation, 156 Average, vs. single measurement, 451, 456 B Back-to-back stemplots, 32 Bar graphs, 8–11, 16 segmented, 17 vs. histograms, 38 Bayes’s theorem, 323 Bell curve. See Normal distributions Bias, 212, 223, 225–227, 429–432 in experiments, 240–241 variability and, 434–435 Biased estimator, 431–432 Bimodal distributions, 29

Binomial coefficients, 398–400 Binomial counts, 443 Binomial distributions, 388–404 definition of, 388 mean of, 397–400 Normal approximation for, 402–404 sampling and, 401–404 standard deviation of, 397–400 Binomial probability, 390–396 calculation of, 394–396 formula for, 393 Binomial random variables, 382–404 definition of, 388 mean of, 398–400 sampling and, 401–404 standard deviation of, 398–400 variance of, 398–400 Binomial settings, 387–389, 401 BINS mnemonic, 388 Blocking, 251–257 matched pairs and, 255 Boxplots, 57–59 C Calculator for binomial coefficients, 392–393 for binomial probabilities, 394–395 for boxplots, 59 for chi-square test for goodness of fit, 689–690, 691 for chi-square test for homogeneity, 706–707 for confidence intervals, 501, 521–522, 618–619 for discrete random variables, 354 for geometric probabilities, 406–407 for histograms, 36–37 for inverse t, 513–514 for least-squares regression lines, 171–172 for Normal distributions, 116–117, 125 for Normal probability plots, 125 for numerical summaries, 63–64 for one-proportion z test, 561 for one-sample t intervals, 521–522 for one-sample t tests, 582–583 for P-values, 579, 686

for residual plots, 175–176 for scatterplots, 150, 155 for significance test for difference in proportions, 624 for significance test for slope of regression line, 756–757 for simple random samples, 215–216 for standard deviations, 62 for transformation, 781–782 for two-sample t interval, 643–644 for two-sample t test, 647–648 for two-way tables, 706–707 Call-in polls, 212 Cases, 4 Categorical variables, 3–4 association between, 17–19, 711–713 definition of, 3 distributions of, 8, 11–18, 697–701 graphs for, 8–11, 16–18 in two-way tables, 11–12, 697–704, 711–713 values of, 7 vs. quantitative variables, 508, 520, 720 Causation evidence for, 268–270 experiments and, 238, 249, 268–270 explanatory variables and, 144 inference about, 266–268. See also Inference response variables and, 144 vs. association, 156 vs. correlation, 190–191 Center of distribution, 26, 49–53, 64–65, 429–432, 613. See also Mean; Median for difference between means, 637 for difference between proportions, 613 measures of, 64–65 for sample mean, 451–453, 459, 478 for slope of regression line, 740, 741 Central limit theorem (CLT), 456–459 Chance behavior. See Randomness Chebyshev’s inequality, 112 Chi-square distributions, 684–687

I-1

P-values from, 704–705 Chi-square statistic, 682–684 components of, 690–691 for multiple comparisons, 700–701 Chi-square test, 680–692 alternative hypothesis in, 681 calculator for, 689–690, 691 carrying out, 687–691 conditions for, 687–688 expected counts and, 680–684, 701–704 follow-up analysis for, 690, 710 hypothesis statement for, 680–681 Large Counts condition for, 687 for multiple comparisons, 697–704 null hypothesis in, 681 observed counts and, 681–684 Random condition for, 687 for two-way tables, 697–704, 711–713 Chi-square test for homogeneity, 697, 703, 707–710 calculator for, 706–707 for comparing proportions, 710, 719–720 conditions for, 703 for difference between proportions, 710 guidelines for, 717–721 vs. test for independence, 717 Chi-square test for independence, 697, 713–717 guidelines for, 717–721 vs. test for homogeneity, 717 Cluster sample, 220–222 Coefficient binomial, 392–393 of determination, 178–180 Column variables, 12 Combined sample proportion, 621 Comparative experiments, 240 Complement rule, 307–308, 311, 314 Completely randomized design, 244–247 Components, of chi-square statistic, 690–691 Conditional distributions, 14–18, 320 comparing, 697–701

I-1

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I-2

Index

Conditional probability, 318–328 calculation of, 319–320 conditional distributions and, 320 definition of, 319 formula for, 319–320 independence and, 326–328 notation for, 319 tree diagrams and, 322–326 Confidence intervals, 477–489, 493–494 calculation of, 487 conditions for, 493–496 construction of, 485–487, 496–499 critical values for, 486 for difference between means, 641–643, 648 for difference between proportions, 616–619, 623 formula for, 486–487 four-step process for, 499–500 guidelines for, 487–488 interpretation of, 481–485 margin of error in, 487, 500, 501–503, 523–524 Normal/Large Sample condition for, 640 one-sided tests and, 585–586 for population mean, 518–521 for population proportion, 496–499, 563–564 Random condition for, 640 sample size and, 501–503, 522–524 significance tests and, 563–564, 583–586 for slope of regression line, 747–753 standard errors and, 496–497 two-sided tests and, 564, 583–586 Confidence levels, 480–485 critical values for, 497 interpretation of, 481–485 Confidentiality, 271 Confounding, 236, 251 Consent, informed, 271 Continuous random variables definition of, 356 from measuring, 356 probability distributions of, 355–358 vs. discrete random variables, 356 Control groups, 246–247 Convenience sample, 211–212. See also Sample/sampling Correlation, 150–157 calculation of, 154–155

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definition of, 150 explanatory variables and, 156, 187 formula for, 154–155 interpreting, 155–157 limitations of, 186–190 linear association and, 150–157, 187–188 nonsense, 190 positive/negative association and, 151 properties of, 156 response variables and, 156, 187 scatterplots and, 145–155 strength of, 146, 151 units of measure and, 156 vs. causation, 190–191 Correlation and Regression (applet), 152, 170 Correlation coefficient, 150 Counts binomial, 443 in chi-square tests, 681–684, 701–704 expected, 680–684, 701–704, 720–721 in frequency tables, 7, 38–39 in histograms, 38–39 observed, 681–684 in two-way tables, 701–704, 720–721 Critical values for confidence intervals, 486, 640 for t-distributions, 512–513, 640 Cube root transformation, 769 Cumulative relative frequency graphs, 86–89 Curves bell. See Normal distributions density, 104–107, 355–358, 685. See also Density curves Normal, 108, 109, 115, 116–117, 356–358 standard Normal, area under, 115, 116–117 D Data analysis, 2–6 categorical, 7–20. See also Categorical variables definition of, 2 Data ethics, 270–271 Data tables. See Tables Data transformation. See Transformation Degrees of freedom, 511–512 for chi-square distributions, 685 Density curves, 84, 104–107

for chi-square distributions, 685 definition of, 105 mean of, 107 median of, 106–107 probability distributions and, 355–358 skewed, 107 standard deviation of, 107 symmetric, 106 for t-distributions, 512 Dependent variables, 144 Difference between means, 634–651 confidence intervals for, 641–643, 648 sampling distribution of, 636–639 significance tests for, 644–649, 651 standard error of, 640 two-sample t procedures for, 644–651 two-sample t statistic and, 639–641, 650–651 Difference between proportions, 612–629 calculator for, 624 chi-square test for homogeneity for, 710 confidence intervals for, 616–619, 623 inference for experiments and, 625–627 randomization distribution of, 627 sampling distribution of, 612–615 significance tests for, 619–624 standard error of, 616 two-sample z interval for, 616–617 two-sample z test for, 621–623, 710 Difference of random variables mean of, 376–377 variance of, 376–377 Discrete random variables from counting, 356 definition of, 348 mean (expected value) of, 350–352 probability distributions of, 348–350, 354 standard deviation of, 352–353 variance of, 352–353, 373 vs. continuous random variables, 356 Disjoint events, 307–308, 311 addition rule for, 308, 312, 330 vs. independent events, 331 Distributions, 4, 83–128 bimodal, 29

binomial, 382–404 of categorical variables, 7–8, 11–18, 697–701 center of. See Center of distribution chi-square, 684–687, 704–705, 717–721 comparing, 29–31 conditional, 14–18, 320, 697–701 cumulative relative frequency graphs and, 86–89 density curves and, 104–107 five-number summary for, 57–59, 63–64 geometric, 404–408 graphing of. See Graph(s) marginal, 12–14 mean of, 49–51. See also Mean median of, 51–53. See also Median mode of, 26, 28–29, 35 multimodal, 29 Normal, 108–109. See also Normal distributions outliers in, 26, 56–57. See also Outliers overall pattern of, 26, 27–29 percentiles of, 84, 85–86 pie charts for, 8–9 population, 428 probability. See Probability distributions of quantitative variables, 25–40 quartiles of, 54–55, 57–58 randomization, 627 range of, 53–55 of sample data, 428 sampling. See Sampling distribution shape of, 26, 27–29, 35, 108, 435, 440–445, 451–456, 478, 613 skewed, 27–29, 52–53, 107, 125, 516–517 spread of, 26, 53–55, 432–433, 440–445, 478–479, 613, 637, 740, 741 symmetric, 27–29 t, 510–514 transforming data and, 92–98. See also Transformation uniform, 355–356 unimodal, 28–29 z-scores and, 89–91, 97 Division, by constant, 94, 95–97, 364–366, 367–368 Dotplots, 25–27 shape of, 29 Double-blind experiments, 248

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Index E Effect size, 568 Empirical (68-95-99.7) rule, 109–112 Equal standard deviation, for regression inference, 743–746 Error root mean squared, 746 roundoff, 8 standard. See Standard error Type I, 547–550 Type II, 547–548, 565–569 Estimators biased, 431–432 point, 477 unbiased, 429–431, 432 Ethical issues, 270–271 Events complements of, 307–308, 311, 314 definition of, 306 independent, 326–331 mutually exclusive, 307–308, 311, 330–331 in Venn diagrams, 311 Expected counts in chi-square test, 680–684, 701–704 computation of, 701–704 for multiple comparisons, 701–704 in two-way tables, 701–704, 720–721 Expected value. See also Mean of difference of random variables, 376–377 of discrete random variables, 350–352 of sum of random variables, 370–376 Experiment(s), 208, 234–257 bias in, 240–241 blocking in, 251–257 causation and, 238, 249, 268–270 comparative, 240 confidentiality in, 271 confounding in, 236 control groups in, 242, 246–247 ethical aspects of, 270–271 experimental units in, 237 inference for, 249–251, 266–268, 625–627 informed consent for, 271 multifactor, 245–247 placebos in, 244 problems in, 239–240, 247–248 random assignment in, 244–247 statistically significant results in, 249 subjects in, 237 terminology for, 237–239

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types of. See Experimental design vs. observational studies, 208, 235–236, 268–269 Experimental design, 240–244 analytic methods and, 650 block, 251–257 completely randomized, 244–247 control and, 242, 246–247 double-blind, 248 lack of realism and, 268 matched pairs, 255 placebo-controlled, 243–244 principles of, 242 problems with, 239–240, 247–248 random assignment and, 241–242, 249–251 randomized block, 251–257 randomized comparative, 240–244 replication and, 242 scope of inference and, 266–268. See also Inference single-blind, 248 Experimental units, 237 Explanatory variables, 143–144, 236 causation and, 144 correlation and, 156, 187 in experiments, 238 as factors, 238 multiple, 244–247 regression and, 164, 187 Exponential models, 774–778 Extrapolation, 168 F Factorial notation, 392 Factors, in experiments, 238 False negative, 287 False positive, 287 Finite population correction (FPC), 445 First quartile, 54–55 in five-number summary, 57–58 Fisher, Ronald A., 546–547 Five-number summary, 57–58, 63–64 Four-step process, for statistical problems, 65 Frequency histograms, 34, 35 Frequency tables, 8, 86–87 G Galton, Francis, 184 Gauss, Carl Friedrich, 109 General addition rule, 310–311, 312 General Social Survey (GSS), 226 Geometric distributions, 404–408

Geometric probability, 404–408 Geometric random variables, 387, 404–408 definition of, 405 distributions of, 405 mean of, 408 standard deviation of, 408 Geometric settings, 404–406 German tank problem, 422–423, 435 Gosset, William S., 512 Graphing calculator. See Calculator Graph(s), 25–40 bar, 8–11, 16, 17–18, 38 boxplots, 57–59 for categorical variables, 8–11, 16, 17–18 cumulative relative frequency, 86–89 density curves and, 104–107 dotplots, 25–27 histograms, 33–40 interpreting, 25–27 Normal probability plots, 122–125 outliers in, 26, 56–57, 147, 188–190. See also Outliers pie charts, 8–9 for quantitative variables, 25–40 scatterplots, 145–155 stemplots, 31–33 H Histograms, 33–40 cautions for, 38–39 choosing classes for, 35–36 constructing, 33–38 counts and percents in, 38–39 definition of, 33 density curves and, 104–107 frequency, 34, 35 frequency tables and, 8 interpreting, 35 relative frequency, 34, 35 relative frequency tables and, 8 shape of distributions in, 35 vs. bar graphs, 38 Homogeneity, chi-square test for, 703, 707–710 Hypothesis alternative, 539–540, 618, 710 null, 539–541, 543. See also Null hypothesis one-sided, 540 populations vs. samples and, 541 two-sided, 540 Hypothesis tests. See Significance tests

I-3

I Independent condition, for regression inference, 743–746 Independent events, 326–328 association and, 328 multiplication rule for, 328–331 vs. mutually exclusive events, 330–331 Independent variables, 144, 371–376 adding, 370–376 definition of, 371 subtracting, 376–378 Individuals, 2, 4 Inequality, Chebyshev’s, 112 Inference, 5–6 cause and effect and, 266–268 for difference between means, 640, 648–649, 650 for difference between proportions, 625–627 for experiments, 249–251, 266–268, 625–627, 648–649 for linear regression, 738–758 for mean, 520, 586–589 populations and, 266–268 probability and, 289 for proportion, 520 sample size and, 565–569 for sampling, 223–225, 266–268 scope of, 266–268 for slope of regression line, 746–747 Inference problems, four-step process for, 65 Inflection points, 108 Influential observations, 189 Informed consent, 271 Institutional review boards, 270–271 Internet polls, 212–213 Interquartile range (IQR), 54–55, 64–65 Intersections, in Venn diagrams, 311 Interval estimate, 480. See also Confidence intervals J JMP (software), regression output for, 181–182 L Large Counts condition, 403 for chi-square tests, 687 for confidence interval for proportion, 494–496

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I-4

Index

Large Counts condition (cont.) for significance tests, 555, 621–622 Large-sample test for proportion, 558 Law of averages, 294–295 Law of large numbers, 291, 294 Laws of probability, 224 Least-squares regression, 164–192, 736–758. See also Regression explanatory variables and, 156, 187 inference for, 738–758 influential observations in, 189 limitations of, 186–190 response variables and, 164, 187 software output for, 181–182 transformation for, 765–785 Least-squares regression line, 168–172, 739–758. See also Regression line calculation of, 183–185 coefficient of determination and, 178–181 influential points and, 189 population (true), 739 residuals and, 169–176 sample (estimated), 739 slope of. See Slope of regression line y intercept for, 166, 746, 749 Least-squares residuals, 169–176 Left-skewed distributions, 27–28 Levels, in experiments, 238 Linear condition, for regression inference, 743–746 Linear regression. See Regression Linear relationships, 150–157. See also Correlation direction of, 146 form of, 146 regression lines and, 164–172, 187–188. See also Regression line strength of, 147–148, 150–151 Linear transformation, 364–369. See also Transformation Logarithm transformation, 771–782 exponential models and, 774–778 power models and, 771–774, 778–780 Logarithmic models, 771 Long-run regularity, 293

Starnes-Yates5e_index_I01-I08hr.indd 4

M Margin of error in confidence intervals, 487, 500, 501–503 definition of, 480 sample size and, 501–503, 523–524 Marginal distributions, 12–14 Matched pairs design, 255 Mean, 49–51 as arithmetic average, 49 as balance point, 51 of binomial distribution, 397–400 of binomial random variable, 398–400 comparison of, 634–651. See also Difference between means definition of, 49 of density curve, 107 of difference of random variables, 376–377 of discrete random variable, 350–352 of geometric random variable, 408 inference for, 520, 586–589 linear transformations and, 368 as nonresistant measure, 50 of Normal distribution, 108 notation for, 49, 107 outliers and, 50 population. See Population mean sample, 49, 450–461. See also Sample mean of sampling distribution, 429–432, 443–446, 451–453, 459, 478, 613, 637, 740, 741 of sum of random variables, 371–376 vs. median, 52–53, 65 Mean and Median (applet), 53 Measurement units, conversion of, 92–98 Median, 51–53 calculation of, 51–52 definition of, 51 of density curve, 106–107 in five-number summary, 57 as midpoint, 49 as resistant measure, 52–53 vs. mean, 52–53, 65 Minitab (software) quartiles and, 64 regression output for, 181–182 Mnemonics BINS, 388 SOCS, 26 Mode, 26, 28–29 Multimodal distributions, 29 Multiple comparisons, 700–701

Multiplication, by constant, 94–97, 364–368 Multiplication rule, for independent events, 328–331 Multistage sample, 222 Mutually exclusive events, 307–308, 311, 331 addition rule for, 308, 312, 330 vs. independent events, 331 N Negative association, 146, 147–148, 151. See also Association 95% confidence interval, 480 Nonlinear relationships. See also Curves transformation of, 765–786 Nonresistant measures correlation as, 188–189 standard deviation as, 62 Nonresponse, in sample surveys, 225 Nonsense correlation, 190 Normal/Large Sample condition for confidence intervals, 640 for population mean, 514–515 for sampling distributions, 444, 458 for significance tests, 575–576 for two-sample t procedures, 649 Normal approximation of binomial distribution, 402–404 of sample proportion, 443, 445–446 Normal Approximation to Binomial (applet), 402 Normal condition for regression inference, 743–746 Normal curve, 109, 356–358 Normal Curve (applet), 110, 116–117 Normal distributions, 84, 108–109 assessing normality, 121–125 calculating probabilities involving, 118–121 central limit theorem and, 456–459 Chebyshev’s inequality and, 112 distribution of sample proportion and, 444 mean of, 108 as probability distributions, 356–358 of sample proportion, 444–446

shape of, 108 significance tests and, 556 68–95–99.7 rule for, 109–112, 122 standard, 112–117 standard deviation of, 108–109 usefulness of, 109 vs. non-Normal distributions, 109, 121–125. See also Skewed distributions Normal population, sampling from, 453–456 Normal probability plots, 122–125 Normal random variables adding, 378–380 subtracting, 379–381 Notation for binomial coefficient, 392 for conditional probability, 319 factorial, 392 for mean, 49, 107 for standard deviation, 61 for variance, 61 for Venn diagrams, 311–312 Null hypothesis, 539–541, 543 for chi-square test, 681, 710 definition of, 539 failure to reject, 547–550 P-values and, 543–544 rejection of, 547–550 significance tests and, 619–620 Numerical summaries, 57–58, 63–64 O Observational studies, 207–229. See also Sample/sampling definition of, 235–237 ethical aspects of, 270–271 vs. experiments, 208, 235–236, 268–269 Observed counts, in chi-square tests, 681–684 One-proportion z test, calculator for, 561 One-sample t interval, for population mean, 518–520 One-sample t test, 579–583 calculator for, 582–583 paired data and, 586–589 One-sample z interval for proportion, 498 One-sample z test for proportion, 557–561 One-sided alternative hypothesis, 540 One-sided significance tests, 562, 564 confidence intervals and, 564, 585–586 One-Variable Statistical Calculator (applet), 36

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Index One-way table, 680 1.5 × IQR rule, 56, 58 Opinion polls. See Polls; Sample/sampling Outliers, 26, 56–57 definition of, 26 identifying, 56–57 1.5 × IQR rule for, 56, 58 in regression, 188–190 in scatterplots, 147 t procedures and, 516 vs. influential observations, 189 P P-values, 543–544 calculation of, 576–579 from chi-square distributions, 684–687, 704–705 definition of, 543 interpreting, 543–544 for negative t-values, 577–578 statistical significance and, 545–547 for t test for slope, 753–754 test statistic and, 556–557, 576–578 for two-sample t test, 644–645 for two-sided tests, 562–563 Paired data, 586–589 Parameters population, 424 regression, 746–747 vs. statistics, 424 Percentiles, 84, 85–86 in cumulative relative frequency graphs, 86–89 quartiles and, 88 Percents, in relative frequency tables, 8 Physicians’ Health Study, 243–244 Pictographs, 10–11 Pie charts, 8–9 Placebos, 244 Plus four estimate, 500 Point estimate, 477, 480 Polls, 210–213. See also Sample/sampling call-in, 212 Internet, 212–213 write-in, 212 Pooled sample proportion, 621 Pooled two-sample t procedures, 650–651 Population(s), 5, 208 definition of, 210 mean of. See Population mean in sample survey, 210 vs. sample, 210 Population distribution, 428

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Population mean, 49 confidence intervals and, 518–521 estimation of, 507–526 Normal/Large Sample condition for, 514–515 one-sample t interval for, 518–520 Random condition for, 514–515 sampling distribution and, 510–511 t-distributions and, 510–514 Population parameters, 424, 541 Population proportion, 424 confidence interval for, 496–499, 563–564 estimation of, 492–504 one-sample z interval for, 498 significance tests about, 554–570 Population regression line, 739. See also Regression line Population size, sampling variability and, 432–433 Population variance, 431 Positive association, 148, 151. See also Association Power, 565–569, 590–591 definition of, 565 sample size and, 647 Type I error and, 568 Type II error and, 565–569 vs. significance level, 566 Power models, logarithm transformation and, 771–774, 778–780 Probability, 286–333 addition rule for mutually exclusive events and, 308, 312 basic rules of, 307–311 binomial, 390–396 conditional, 318–328 definition of, 291 general addition rule for, 310–311, 312 geometric, 404–408 inference and, 289 law of averages and, 294–295 law of large numbers and, 291, 294 long-run regularity and, 293 mutually exclusive events and, 307–308, 311 overview of, 289–292 randomness and, 292–295 simulation and, 295–299 tree diagrams and, 322–326 two-way tables and, 309–311 Probability (applet), 290 Probability distributions, 348–349. See also Random variables

adding/subtracting constant and, 364, 366–368 of continuous random variables, 355–358 definition of, 348 density curves and, 355–358 of discrete random variables, 348–350 multiplying/dividing by constant and, 364–368 Normal distributions as, 356–358 uniform, 355–356 Probability laws, 224, 291, 294–295 Probability models, 305–307 Problem solving, four-step process for, 65 Proportions difference between, 616–629. See also Difference between proportions one-sample z test for, 557–561 two-sample z test for, 616–617 PROVE-IT Study, 610 pth percentile, 85–86 Q Quantitative variables, 3–4. See also Variables correlation between, 156. See also Correlation definition of, 3 distribution of, 25–40. See also Distributions relationships between, 140–203 vs. categorical variables, 508, 520, 720 Quartiles first, 54–55 in five-number summary, 57–58 in interquartile range, 54–55 percentiles and, 88 third, 54–55 R r2 (R-sq; coefficient of determination), 178–181 Random assignment, 241–242, 249–251 inference and, 266–268. See also Inference Random condition for chi-square tests, 687, 703 for confidence intervals, 493, 494–496, 640 for population mean, 514–515 for regression inference, 743–746

I-5

for significance tests, 555, 575–576 Random digits, 216 Random digits table for simple random samples, 216–217 for simulations, 296–297 Random samples, 213–225. See also Sample/sampling choosing with calculator, 215–216 choosing with Table D, 216 choosing with technology, 215 confidence intervals and, 493 inference and, 266–268. See also Inference population mean and, 514–515 significance tests and, 555–556 simple, 213–218 stratified, 219–220 Random variables, 344–410 adding, 364, 366–368, 370–376, 378–380 adding/subtracting constant and, 364, 366–368 binomial, 387–404. See also under Binomial combining, 370–376 continuous, 355–358 definition of, 348 discrete, 348–354, 356, 373 geometric, 387, 404–408 independent, 371–376 linear transformations of, 364–369 mean of, 350–352, 371, 376–377 multiplying/dividing by constant and, 364–368 Normal, 378–381 subtracting, 376–381 Randomization distribution, for difference between proportions, 627 Randomized block design, 251–257 matched pairs, 255 Randomized comparative experiments, 240–244 Randomness, 292–295 law of averages and, 294–295 long-run regularity and, 293 model for, 305–307 myths about, 293–294 predictability of, 293–294 Range, 54 definition of, 54 interquartile, 54–55 Reese’s Pieces® (applet), 441 Regression coefficient of determination and, 178–181

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I-6

Index

Regression (cont.) explanatory variables and, 156, 187 least-squares, 164–192. See also Least-squares regression limitations of, 186–190 outliers in, 188–190 response variables and, 164, 187 Regression inference, 738–758 conditions for, 742–746 estimating parameters and, 746–747 time-series data for, 744 Regression line, 164–172 definition of, 164 equation of, 166 extrapolation and, 168 interpreting, 165–167 least-squares, 168–172. See also Least-squares regression line linear relationships and, 164–172, 187–188 predicted value and, 166, 167–168 slope of. See Slope of regression line y intercept for, 166, 746, 749 Regression output, interpreting, 181–182 Regression standard error, 746 Regression to the mean, 182–185 Relative frequency graphs, cumulative, 86–89 Relative frequency tables, 8, 34, 35, 86–87 Replication, in experimental design, 242 Residual(s), 169–176 calculating, 169–172 definition of, 169 interpreting, 172–176 squaring, 170. See also Leastsquares regression line standard deviation of, 177–178, 180–181 Residual plots, 172–176 on calculator, 175–176 Resistant measures, 50, 52–53, 55 Response variables, 143–144, 236 correlation and, 156, 187 regression and, 164, 187 Rice University Sampling Distributions (applet), 453–454, 456 Right-skewed distributions, 27–28 Root mean squared error, 746 Roundoff error, 8 Row variables, 12

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S Sample data, distribution of, 428 Sample mean, 49, 450–461. See also Mean Normal/Large Sample condition for, 458 sampling distribution of. See Sampling distribution of sample mean standard error of, 518 Sample proportion, 424, 440–446 Large Counts condition for, 444 Normal approximation for, 445–446 Normal distribution of, 444–446 pooled (combined), 621 sampling distribution of. See Sampling distribution of sample proportion 10% condition for, 444 as unbiased estimator, 443 Sample regression line, 739. See also Regression line Sample/sampling, 5, 207–229. See also Observational studies bias in, 212, 223, 429–431. See also Bias binomial distributions and, 401–404 cluster, 220–222 convenience, 211–212 definition of, 210 ethical aspects of, 270–271 inference for, 223–225, 266–268 laws of probability and, 224 margin of error in, 480. See also Margin of error mean of. See Sample mean multistage, 222 nonresponse in, 225 populations and, 208, 210. See also Population(s) problems in, 225–227 question order in, 227 question wording in, 227 random, 213–222, 266–268. See also Random samples sample size and, 223–225, 432–433 sampling errors in, 225–227 self-selected, 212 unbiased estimates and, 432–433 voluntary response, 212, 225 vs. population, 210 without replacement, 402 Sample size, 224 confidence intervals and, 501–503, 522–524

margin of error and, 501–503, 523–524 power and, 565–569, 647 for significance tests, 565–569, 589–590, 647 variability and, 432–433 Sample spaces, 305 Venn diagrams and, 311–314 Sample surveys, 210–211. See also Sample/sampling multistage, 222 as observational studies, 208, 235–236 sampling errors in, 225–227 Sample variance, estimating, 431–432 Sampling distribution, 420–461 bias and, 429–432. See also Bias binomial count and, 443 center (mean) of, 429–432, 443–444, 451–453, 459, 478, 613, 637–638, 740, 741 characteristics of, 429–435 definition of, 427 of difference between means, 636–639 of difference between proportions, 612–615 distribution of sample data and, 428 Normal/Large Sample condition for, 458 overview of, 422–424 population distribution and, 428 shape of, 435, 440–445, 453–456, 478, 613, 636, 740, 741 of slope of regression line, 740–742 spread of, 432–433, 440–445, 478–479, 613, 637, 740, 741 standard deviation of, 444, 451–453, 459, 478–479, 613, 614–615 vs. sample data distribution, 428 Sampling distribution of sample mean, 451–456 center (mean) of, 451–453, 459, 478 central limit theorem and, 456–459 for Normal population, 453–456 shape of, 453–456, 459, 478 standard deviation of, 451–453, 459, 478–479 Sampling distribution of sample proportion, 440–445 center (mean) of, 440–445

Normal approximation and, 444–446 shape of, 440–445 spread of, 440–445 standard deviation of, 444 10% condition and, 444 Sampling variability, 425–428 bias and, 434–435 definition of, 425 Satterthwaite, F. E., 634 Scatterplots, 145–155 calculators for, 150, 155 constructing, 145–146 correlation and, 150–157. See also Correlation definition of, 145 describing, 146–149 direction of relationship and, 146–149 form in, 146 linearizing curved patterns in. See Transformation outliers and, 147 patterns in, 146–147 positive/negative associations and, 146, 148, 151 regression lines and, 164–172, 738. See also Regression line strength of relationship and, 146, 151 Segmented bar graphs, 17 Self-selected samples, 212 Shape of distributions, 26, 27–29, 35, 108–109, 435, 441–443, 453–456, 478, 613 Significance level, 545–546, 589 definition of, 566 Type I and II errors and, 565–569 vs. power, 566 Significance tests, 536–607 about population mean, 575–579 about population proportion, 554–570 alternative hypothesis and, 539–540 carrying out, 554–557, 575–579 chi-square. See Chi-square test confidence intervals and, 563–564, 583–586 definition of, 539 for difference between means, 644–649 for difference between proportions, 619–624 four-step process for, 558 guidelines for, 589–593 Large Counts condition for, 555, 621–622

12/9/13 5:52 PM

Index Normal/Large Sample condition for, 575–576 null hypothesis and, 539–541 one-sample, 579–583 one-sided, 562, 585–586 P-value for, 543–547, 562–563, 576–578 pitfalls in, 589–593 power of, 565–569, 647 purpose of, 560 Random condition for, 555, 575–576 reasoning of, 538–539, 541–543 sample size for, 565–569, 589–590, 647 for slope of regression line, 753–757 statistical significance and, 544–547. See also Significance level test statistic in, 556–557, 576 two-sided, 562–564, 583–586 Type I and II errors and, 547–550 Simple random sample (SRS), 213–218. See also Sample/ sampling Simulations, 295–299 Single-blind experiments, 248 68–95–99.7 rule, 109–112, 122 Size. See Sample size Skewed distributions, 27–29, 516–517 density curves and, 107 left- vs. right-skewed, 125 mean vs. median for, 52–53 Normal probability plot for, 125 Slope of regression line, 166, 739–758 calculator for, 756–757 center of distribution for, 740, 741 confidence interval for, 747–753 inference about, 746–747 P-value for, 753–754 sampling distribution of, 740–742 significance test for, 753–757 standard error of, 746 t interval for, 747–749 t test for, 753–754 SOCS mnemonic, 26 Solving problems, four-step process for, 65 Splitting stems, 32 Spread, 26, 53–55 in five-number summary, 57 interquartile range and, 53–55, 64–65 measures of, 53–55

Starnes-Yates5e_index_I01-I08hr.indd 7

of sampling distribution, 432–433, 441–443, 478–479, 613, 637, 740, 741 standard deviation and, 60–62, 64–65 Standard deviation, 60–62, 372–373 of binomial distributions, 397–400 of binomial random variables, 398–400 calculation of, 62 definition of, 61 of density curve, 107 of difference of random variables, 377–378 of discrete random variable, 352–353 estimated. See Standard error of geometric random variable, 408 linear transformations and, 368 as nonresistant measure, 62 of Normal distribution, 108–109 notation for, 61 regression line and, 746 of residuals, 177–178, 180–181 of sampling distribution, 444, 451–453, 459, 478–479, 613, 614–615, 637–638 spread and, 60–62, 64–65 of sum of random variables, 371–376 usefulness of, 62, 64 variance and, 60–61. See also Variance z-score and, 90 Standard error confidence interval and, 496–497 definition of, 496–497 of difference between means, 640 of difference between proportions, 616 regression, 746 of sample mean, 518 of slope, 746 Standard Normal curve, area under, 115, 116–117 Standard Normal distributions, 112–117 Standard Normal tables, 113–115 Standardized values, 89–91 Standardizing, 89–91 State-plan-do-conclude process, 65 Statistic definition of, 424 variability of, 432–433 vs. parameter, 424

Statistical Power (applet), 590–591 Statistical problems, four-step process for, 65 Statistical significance, 544–547. See also Significance level multiple analyses and, 593 practical importance of, 592 sample size and, 589–591 Statistically significant association, 249 Statistically significant at level a, 545 Statistics, definition of, 2 Stemplots, 31–33 Strata, 219–220 vs. clusters, 221 Stratified random sample, 219–220 Student’s t-distribution, 512 Study design, 207–273. See also Experimental design; Sample/sampling Subjects, experimental, 237 confidentiality for, 271 informed consent of, 271 Subtraction of constant, 93–97, 364, 366–368 of random variables, 376–381 Sum of random variables, mean of, 371–376 Survey, sample, 210–211. See also Sample/sampling Symmetry, 27–29 T t distribution, 510–514 critical values for, 512–513, 640 degrees of freedom and, 511–512 Student’s, 512 t interval one-sample, 518–522 for population mean, 518–522 for slope of regression line, 747–749 two-sample, 641–643 t procedures one-sample, 510–514, 579–583, 586–589 outliers and, 516 paired, 586–589 sample size and, 522–524 t statistic, two-sample, 639–641, 644, 650–651 t test one-sample, 579–583, 586–589 paired data and, 586–589 for slope, 753–754 t values, negative, 577–578 Tables, 3–4

I-7

frequency, 8, 86–87 one-way, 680 random digit, 216–217, 296–297 relative frequency, 8, 34, 35, 86–87 standard Normal (Table A), 113–115 Table B, 513, 577–578, 641 Table C, 685–686 Table D, 216 two-way. See Two-way tables Technology. See Applets; Calculator 10% condition, 401–402, 444 for regression inference, 743 Test of Significance (applet), 538 Test statistic, 556–557, 576 calculation of, 576–578, 588 calculator for, 561 P-value and, 556–557 z, distribution of, 627 Theorem, Bayes’s, 323 Third quartile, 54–55 in five-number summary, 57–58 Time-series data, for regression inference, 744 Transformation, 92–98, 765–785 to achieve linearity, 765–785 adding/subtracting constant and, 93–97 calculator for, 781–782 definition of, 767 exponential models and, 774–778 linear, 364–369 with logarithms, 771–782 model selection for, 778–780 multiplying/dividing by constant and, 94–97 power models and, 767–774, 778–780 z-scores and, 97 Treatment, 237 Tree diagrams, 322–326 True regression line, 739 Truncated data, 32 Two-sample t interval, 641–643 Two-sample t procedures calculator for, 647–648 for difference between means, 644–649 guidelines for, 650–651 Normal condition for, 649 pooled, 650–651 sample size for, 647 Two-sample t statistic, 639–641, 644 pooled, 650–651 Two-sample z interval, for difference between proportions, 616–617

12/9/13 5:52 PM

I-8

Index

Two-sample z statistic, 639–641 chi-square statistic and, 710 Two-sample z test for comparing proportions, 710, 719–720 for difference between proportions, 621–623 Two-sided alternative hypothesis, 539–540 Two-sided significance tests, 562–564 confidence intervals and, 564, 583–586 Two-way tables, 11–12, 15–16, 309–311 calculator for, 706–707 chi-square tests for, 697–704, 711–713 collapsing, 721 expected counts for, 701–704, 720–721 Type I errors, 547–550 Type II errors, 547–548, 565–569 U Unbiased estimator, 429–431, 432. See also Bias Uniform distributions, 355–356 Unimodal distributions, 28–29

Starnes-Yates5e_index_I01-I08hr.indd 8

Unions, in Venn diagrams, 311 Units of measure, conversion of, 92–98 V Variability sampling, 425–428, 434–435 of statistic, 432–433 Variables, 2–4 association between, 17–19. See also Association categorical, 3–4, 508, 520, 697–701, 711–713, 720. See also Categorical variables column, 12 confounding, 236, 251 definition of, 2 dependent, 144 distribution of. See Distributions explanatory, 143–144, 156, 187, 236, 238 independent, 144 negatively associated, 146, 148, 151 positively associated, 148, 151 quantitative, 3–4, 508. See also Quantitative variables

random, 344–410. See also Random variables response, 143–144, 156, 164, 187, 236 row, 12 standardizing, 184–185 Variance. See also Standard deviation adding, 373 of binomial random variables, 398–400 definition of, 61 of difference of independent random variables, 376–377 of discrete random variables, 352–353 equal vs. unequal, 650 of Normal random variables, 378–380 notation for, 61 population, 431 sample, 431–432 of sum of independent random variables, 373 Venn diagrams, 311–314 intersections in, 311 notation for, 311–312 unions in, 311 Voluntary response sample, 212, 225

W Welch, B.L., 634 Wilson, Edwin Bidwell, 500 Write-in polls, 212 X x-bar. See Mean; Sample mean Y y intercept for regression line, 166, 746 confidence interval for, 749 Z z interval for difference between proportions, 616–617 z scores, 89–98 for standard Normal curve, 116–117 transforming data and, 97. See also Transformation z statistic distribution of, 627 two-sample, 639–641 z test one-sample, 557–561 two-sample, 621–623

12/9/13 5:52 PM

Technology Corners Reference TI-Nspire instructions in Appendix B; HP Prime instructions at www.whfreeman.com/tps5e

  1. Analyzing two-way tables

page   16

  2. Histograms on the calculator

page   36

  3. Making calculator boxplots

page   59

  4. Computing numerical summaries with technology

page   63

  5. From z-scores to areas, and vice versa

page 116

  6. Normal probability plots

page 125

  7. Scatterplots on the calculator

page 150

  8. Least-squares regression lines on the calculator

page 171

  9. Residual plots on the calculator

page 175

10. Choosing an SRS

page 215

11. Analyzing random variables on the calculator

page 354

12. Binomial coefficients on the calculator

page 392

13. Binomial probability on the calculator

page 394

14. Geometric probability on the calculator

page 406

15. Confidence interval for a population proportion

page 501

16. Inverse t  on the calculator

page 513

17. One-sample t  intervals for m on the calculator

page 521

18. One-proportion z  test on the calculator

page 561

19. Computing P-values from t  distributions on the calculator

page 578

20. One-sample t  test for a mean on the calculator

page 582

21. Confidence interval for a difference in proportions

page 618

22. Significance test for a difference in proportions

page 624

23. Two-sample t  intervals on the calculator

page 643

24. Two-sample t  tests on the calculator

page  647

25. Finding P-values for chi-square tests on the calculator

page 686

26. Chi-square tests for goodness of fit on the calculator

page 689

27. Chi-square tests for two-way tables on the calculator

page 706

28. Confidence interval for slope on the calculator

page 751

29. Significance test for slope on the calculator

page 756

30. Transforming to achieve linearity on the calculator

page 781

Starnes-Yates5e_BM_endpp.hr.indd 1

11/22/13 9:41 AM

Inference Summary STEP

4 How to Organize a Statistical Problem: A Four-Step Process Confidence intervals (CIs)

Significance tests

STATE:

What parameter do you want to estimate, and at what confidence level ?

What hypotheses do you want to test, and at what ­significance level? Define any parameters you use.

PLAN:

Choose the appropriate inference method. Check conditions.

Choose the appropriate inference method. Check conditions.

DO:

If the conditions are met, perform calculations.

If the conditions are met, perform calculations.   • Compute the test statistic.   • Find the P -value.

CONCLUDE:

Interpret your interval in the context of the problem.

Make a decision about the hypotheses in the context of the problem.

CI: statistic ± (critical value).(standard deviation of statistic)

Inference about

Number of samples (groups)

Name of procedure (TI Calculator function) Formula

Interval or test

Interval

One-sample z interval for p (1-PropZInt) p^ ± z *

1

Interval

Two-sample z interval for p1 − p2 (2-PropZInt) (p^1 − p^2) ± z *

2

p^(1 − p^) Å n

One-sample z test for p (1-PropZTest) p^ − p0 z= p0(1 − p0) Å n

Test

Proportions

p^1(1 − p^1) p^2(1 − p^2) + Å n1 n2

Two-sample z test for p1 − p2 (2-PropZTest) z=

Test

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Standardized test statistic =

(p^1 − p^2) − 0

p^c(1 − p^c) p^c(1 − p^c) + Å n1 n2 where X1 + X2 total successes p^c = = total sample size n1 + n2

statistic − parameter standard deviation of statistic

Conditions Random Data from a random sample or ­randomized experiment n ≤ 0.10N if sampling without replacement

~  10%: 

Large Counts At least 10 successes and failures; that is, np^ ≥ 10 and n(1 − p^ ) ≥ 10

Random Data from a random sample or randomized experiment ~ 

10%:  n ≤ 0.10N if sampling without replacement

Large Counts np0 ≥ 10 and n(1 − p0) ≥ 10 Random Data from independent random samples or randomized experiment n1 ≤ 0.10N1 and n2 ≤ 0.10N2 if sampling without replacement

~  10%: 

Large Counts At least 10 successes and failures in both samples/groups; that is, n1p^ 1 ≥ 10, n1(1 − p^ 1) ≥ 10 , n2p^ 2 ≥ 10, n2(1 − p^ 2) ≥ 10

Random Data from independent random samples or randomized experiment n1 ≤ 0.10N1 and n2 ≤ 0.10N2 if sampling without replacement

~  10%: 

Large Counts At least 10 successes and failures in both samples/groups; that is, n1p^ 1 ≥ 10, n1(1 − p^ 1) ≥ 10 , n2p^ 2 ≥ 10, n2(1 − p^ 2) ≥ 10

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Inference about

Number of samples (groups)

Estimate or test

Interval

Name of procedure (TI Calculator function) Formula

(or paired data) Test

Random Data from a random sample or ­randomized experiment

One-sample t interval for m (TInterval) x ± t*

1

Conditions

sx "n

~ 

with df = n − 1

One-sample t test for m (T-Test)

Random Data from a random sample or randomized experiment

x − m0 t= with df = n − 1 sx

Normal/Large Sample Population distribution Normal or large sample (n ≥ 30); no strong skewness or outliers if n < 30 and population distribution has unknown shape

~ 

"n

Two-sample t interval for m1 − m2 (2-SampTInt)

Means Interval

2

Test

1

Test

Distribution of categorical variables

10%:  n ≤ 0.10N if sampling without replacement

Normal/Large Sample Population distribution Normal or large sample (n ≥ 30); no strong skewness or outliers if n < 30 and population distribution has unknown shape

(x 1 − x 2) ± t *

s 21 s 22 + Å n1 n2

10%:  n ≤ 0.10N if sampling without replacement

Random Data from independent random samples or randomized experiment n1 ≤ 0.10N1 and n2 ≤ 0.10N2 if sampling without replacement

~  10%: 

df from technology or min(n1 − 1, n2 − 1)

Normal/Large Sample Population distributions Normal or large samples (n1 ≥ 30 and n2 ≥ 30); no strong skewness or outliers if sample size < 30 and population distribution has unknown shape

Two-sample t test for m1 − m2 (2-SampTTest)

Random Data from independent random samples or randomized experiment

t=

(x 1 − x 2) − (m1 − m2) s 21

n1 ≤ 0.10N1 and n2 ≤ 0.10N2 if sampling without ­replacement

~  10%: 

s 22

+ Å n1 n2 df from technology or min(n1 − 1, n2 − 1)

Normal/Large Sample Population distributions Normal or large samples (n1 ≥ 30 and n2 ≥ 30); no strong skewness or outliers if sample size < 30 and population distribution has unknown shape

Chi-square test for goodness of fit (x2GOF-Test)

Random Data from a random sample or randomized experiment

(observed − expected)2 c2 = ∙ expected

Large Counts All expected counts at least 5

~ 

10%:  n ≤ 0.10N if sampling without replacement

with df = number of ­categories − 1 Chi-square test for homogeneity (x2-Test) 2 or more

Test

c2 =

∙

(observed − expected)2 expected

df = (# of rows - 1)(# of columns - 1) Chi-square test for ­independence Relationship between 2 categorical variables

2

(x -Test) 1

Test

c2 =

∙

Random Data from independent random samples or randomized experiment n1 ≤ 0.10N1 , n2 ≤ 0.10N2 , and so on if sampling without replacement

~  10%: 

Large Counts All expected counts at least 5 Random Data from a random sample or randomized experiment ~ 

2

(observed − expected) expected

10%:  n ≤ 0.10N if sampling without replacement

Large Counts All expected counts at least 5

df = (# of rows - 1)(# of columns - 1)

One-sample t interval for b Relationship between 2 quantitative variables (slope)

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Interval

(LinRegTInt) b ± t *(SEb)

with df = n - 2

1

One-sample t test for b Test

(LinRegTTest) b − b0 t= with df = n - 2 SEb

Linear Relationship between the variables is linear Independent observations; check the 10% condition if sampling without replacement. Normal Responses vary Normally around regression line for all x-values Equal SD around regression line for all x-values Random Data from a random sample or randomized experiment

11/22/13 9:41 AM