SM McClave Stat10 Wm

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INSTRUCTOR'S SOLUTIONS MANUAL to Accompany James T. McClave P. George Benson and Terry Sincich's

STATISTICS FOR BUSINESS AND ECONOMICS Tenth Edition

Nancy S. Boudreau

Bowling Green State University

Upper Saddle River, New Jersey Columbus, Ohio

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Contents Preface

v

Chapter 1

Statistics, Data, and Statistical Thinking

1

Chapter 2

Methods for Describing Sets of Data The Kentucky Milk Case

5 46

Chapter 3

Probability

55

Chapter 4

Random Variables and Probability Distributions The Furniture Fire Case

82 136

Chapter 5

Inferences Based on a Single Sample: Estimation with Confidence Intervals

137

Chapter 6

Inferences Based on a Single Sample: Tests of Hypothesis

161

Chapter 7

Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses The Kentucky Milk Case – Part II

201 243

Chapter 8

Design of Experiments and Analysis of Variance

256

Chapter 9

Categorical Data Analysis Discrimination in the Work Place

300 328

Chapter 10

Simple Linear Regression

332

Chapter 11

Multiple Regression and Model Building The Condo Sales Case

379 444

Chapter 12

Methods for Quality Improvement

448

Chapter 13

Time Series: Descriptive Analyses, Models, and Forecasting The Gasket Manufacturing Case

476 522

Chapter 14

Nonparametric Statistics

529

iii

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iv

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Preface This solutions manual is designed to accompany the text, Statistics for Business and Economics, Tenth Edition, by James T. McClave, P. George Benson, and Terry Sincich. It provides answers to most evennumbered exercises for each chapter in the text. Other methods of solution may also be appropriate; however, the author has presented one that she believes to be most instructive to the beginning Statistics student. This manual is provided to help instructors save time in preparing presentations of the solutions and to possibly provide another point of view regarding their meaning. Some of the exercises are subjective in nature. Subjective decisions regarding these exercises have been made and are explained by the author. Solutions based on these decisions are presented; the solution to this type of exercise is often most instructive. When an alternative interpretation of an exercise may occur, the author has often addressed it and given justification for the approach taken. I would like to thank Kelly Barber for creating the art work and for typing this work.

Nancy S. Boudreau Bowling Green State University Bowling Green, Ohio

v

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Statistics, Data, and Statistical Thinking

Chapter 1

1.2

Descriptive statistics utilizes numerical and graphical methods to look for patterns, to summarize, and to present the information in a set of data. Inferential statistics utilizes sample data to make estimates, decisions, predictions, or other generalizations about a larger set of data.

1.4

The first element of inferential statistics is the population of interest. The population is a set of existing units. The second element is one or more variables that are to be investigated. A variable is a characteristic or property of an individual population unit. The third element is the sample. A sample is a subset of the units of a population. The fourth element is the inference about the population based on information contained in the sample. A statistical inference is an estimate, prediction, or generalization about a population based on information contained in a sample. The fifth and final element of inferential statistics is the measure of reliability for the inference. The reliability of an inference is how confident one is that the inference is correct.

1.6

Quantitative data are measurements that are recorded on a meaningful numerical scale. Qualitative data are measurements that are not numerical in nature; they can only be classified into one of a group of categories.

1.8

A population is a set of existing units such as people, objects, transactions, or events. A sample is a subset of the units of a population.

1.10

An inference without a measure of reliability is nothing more than a guess. A measure of reliability separates statistical inference from fortune telling or guessing. Reliability gives a measure of how confident one is that the inference is correct.

1.12

Statistical thinking involves applying rational thought processes to critically assess data and inferences made from the data. It involves not taking all data and inferences presented at face value, but rather making sure the inferences and data are valid.

1.14

a.

The two variables measured are ‘type of credit card used’ and ‘amount of purchase.’ ‘Type of credit card used’ is qualitative. It has no meaningful number associated with it, only the name of the card used. ‘Amount of purchase’ is quantitative. It has a meaningful number associated with it.

b.

In Study 1, it says that all purchases were tracked. Thus, the data represent a population.

a.

High school GPA is a number usually between 0.0 and 4.0. Therefore, it is quantitative.

b.

Honors/awards would have responses that name things. Therefore, it would be qualitative.

1.16

Statistics, Data, and Statistical Thinking

1

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1.18

1.20

1.22.

2

c.

The scores on the SAT's are numbers between 200 and 800. Therefore, it is quantitative.

d.

Gender is either male or female. Therefore, it is qualitative.

e.

Parent's income is a number: $25,000, $45,000, etc. Therefore, it is quantitative.

f.

Age is a number: 17, 18, etc. Therefore, it is quantitative.

a.

1.

The variable of interest is the status of a company’s e-commerce strategy. Since a company either has an e-commerce strategy or not, the variable is qualitative.

2.

The variable of interest is when the company will implement an e-commerce plan. Since the time of implementation will be a date, this variable will be qualitative.

3.

The variable of interest is whether the company is delivering products over the internet or not. Since the company is either delivering products or not, the variable is qualitative.

4.

The variable of interest is the company’s total revenue in the last fiscal year. Since this is a meaningful number, this variable is quantitative.

b.

Since there are many more that 154 companies in the U.S., this represents a sample rather than a population.

a.

The population of interest is the collection of computer security personnel at all U.S. corporations and government agencies.

b.

Surveys were sent to computer security personnel at all U. S. corporations and government agencies. However, in 2006, only 616 organizations responded to the survey. There could be nonresponse bias. Often, only those subjects with strong opinions will respond to a survey. Thus, the responses may not reflect what the population as a whole thinks.

c.

The variable measured in the survey is whether or not there was unauthorized use of computer systems at the firms during the year. Since the responses will be either ‘Yes’ or “No’, the variable is qualitative.

d.

If we assume that the responses were a random sample from the population, we could infer that about 52% of all computer security personnel will admit to unauthorized use of computer systems at their firms during the year.

a.

The data collection method used is a designed experiment.

b.

The experimental units in the study are the 50,000 smokers.

c.

The variable of interest is the age at which the scanning method first detects a tumor. Since this is a meaningful number, this variable is quantitative.

Chapter 1

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1.24

1.26

1.28

1.30

d.

The population of interest is the set of all smokers in the U.S. The sample of interest is the set of 50,000 smokers surveyed.

e.

The researchers want to compare the age at first detection for the 2 methods to see if one is more sensitive than the other.

a.

The variable of interest to the researchers is the rating of highway bridges.

b.

Since the rating of a bridge can be categorized as one of three possible values, it is qualitative.

c.

The data set analyzed is a population since all highway bridges in the U.S. were categorized.

d.

The data were collected observationally. Each bridge was observed in its natural setting.

a.

The population of interest is the set of all New York accounting firms employing two or more professionals. There are two variables of interest: Whether or not the firm uses audit sampling methods, and if so, whether or not it uses random sampling. The sample is the set of 163 firms whose responses were useable. The inference of interest to the New York Society of CPAs is the proportion of all New York accounting firms employing two or more professionals that use sampling methods in auditing their clients.

b.

The four responses that were unusable could have been returned blank or could have been filled out incorrectly.

c.

Any time a survey is mailed it is questionable whether the returned questionnaires represent a random sample. Often times, only those with very strong opinions return the surveys. In such a case, the returned surveys would not be representative of the entire population.

a.

The experimental units in this study are the 24 projects.

b.

The population from which the sample was selected is the set of all new software development projects.

c.

The variable of interest in this project is the outcome of reusing previously developed software for the new software development projects.

d.

In the sample, 9 of the 24 projects were judged failures. This is (9 / 24)*100% = 37.5%. We could infer that approximately 37.5% of all projects would be judged failures.

a.

The process being studied is the process of filling beverage cans with softdrink at CCSB's Wakefield plant.

b.

The variable of interest is the amount of carbon dioxide added to each can of beverage.

c.

The sampling plan was to monitor five filled cans every 15 minutes. The sample is the total number of cans selected.

Statistics, Data, and Statistical Thinking

3

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4

d.

The company's immediate interest is learning about the process of filling beverage cans with softdrink at CCSB's Wakefield plant. To do this, they are measuring the amount of carbon dioxide added to a can of beverage to make an inference about the process of filling beverage cans. In particular, they might use the mean amount of carbon dioxide added to the sampled cans of beverage to estimate the mean amount of carbon dioxide added to all the cans on the process line.

e.

The technician would then be dealing with a population. The cans of beverage have already been processed. He/she is now interested in the outputs.

Chapter 1

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Methods for Describing Sets of Data 2.2

a.

To find the frequency for each class, count the number of times each letter occurs. The frequencies for the three classes are: Class X Y Z Total

b.

Chapter 2

Frequency 8 9 3 20

The relative frequency for each class is found by dividing the frequency by the total sample size. The relative frequency for the class X is 8/20 = .40. The relative frequency for the class Y is 9/20 = .45. The relative frequency for the class Z is 3/20 = .15. Class X Y Z Total

Frequency 8 9 3 20

Relative Frequency .40 .45 .15 1.00

c.

The frequency bar chart is:

d.

The pie chart for the frequency distribution is:

Methods for Describing Sets of Data

5

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2.4

a.

The variable summarized in the table is ‘Reason for requesting the installation of the passenger-side on-off switch.’ The values this variable could assume are: Infant, Child, Medical, Infant & Medical, Child & Medical, Infant & Child, and Infant & Child & Medical. Since the responses name something, the variable is qualitative.

b.

The relative frequencies are found by dividing the number of requests for each category by the total number of requests. For the category ‘Infant’, the relative frequency is 1,852/30,337 = .061. The rest of the relative frequencies are found in the table below: Reason Infant

Number of Requests 1,852

1,852/30,337

Relative frequencies .061

Child

17,148

17,148/30,337

.565

Medical

8,377

8,377/30,337

.276

Infant & Medical

44

44/30,337

.0014

Child & Medical

903

903/30,337

.030

1,878

1,878/30,337

.062

135

135/30,337

.0045

Infant & Child Infant & Child & Medical TOTAL c.

30,337

.9999

Using MINITAB, a pie chart of the data is:

Pie Chart of Reason Child

(17148, 56.5%)

Child&Medica ( 903, 3.0%) Inf &Chd&Med ( 135, 0.4%) Inf ant

( 1852, 6.1%)

Medical

( 8377, 27.6%)

Inf ant&Child ( 1878, 6.2%) Inf ant&Medic (

d.

6

44, 0.1%)

There are 4 categories where Medical is mentioned as a reason: Medical, Infant & Medical, Child & Medical, and Infant & Child & Medical. The sum of the frequencies for these 4 categories is 8,377 + 44 + 903 + 135 = 9,459. The proportion listing Medical as one of the reasons is 9,459/30,337 = .312.

Chapter 2

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2.6

a.

To find relative frequencies, we divide the frequencies of each category by the total number of incidents. The relative frequencies of the number of incidents for each of the cause categories are: Management System Cause Category Engineering & Design Procedures & Practices Management & Oversight Training & Communication TOTAL

b.

Number of Incidents

Relative Frequencies

27 24 22 10 83

27 / 83 = .325 24 / 83 = .289 22 / 83 = .265 10 / 83 = .120 1

The Pareto diagram is: Management Systen Cause Category 35 30

P er cent

25 20 15 10 5 0

2.8

E ng&D es

P roc&P ract M gmt&O v er C ategor y

Trn&C omm

c.

The category with the highest relative frequency of incidents is Engineering and Design. The category with the lowest relative frequency of incidents is Training and Communication.

a.

The data collection method was a survey.

b.

Since the data were numbers (percentage of US labor and materials), the variable is quantitative. Once the data were collected, they were grouped into 4 categories.

Methods for Describing Sets of Data

7

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c.

Using MINITAB, a pie chart of the data is: Pie Chart of Made in USA

100% (64, 60.4%)

30 and σ 15 σx = μ x = μ = 840 = = 2.1213 n 50

b. c.

134

830 − 840 ⎞ ⎛ P( x ≤ 830) = P ⎜ z ≤ ⎟ = P(z ≤ −4.71) ≈ .5 − .5 = 0 2.1213 ⎠ ⎝ Since the probability of observing a mean of 830 or less is extremely small (≈0) if the true mean is 840, we would tend to believe that the mean is not 840, but something less.

Chapter 4

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d.

By the Central Limit Theorem, the sampling distribution of is approximately normal since n > 30 and σ 45 σx = μ x = μ = 840 = = 6.3640 n 50 830 − 840 ⎞ ⎛ P( x ≤ 830) = P ⎜ z ≤ ⎟ = P(z ≤ −1.57) ≈ .5 − .4418 = .0582 6.3640 ⎠ ⎝

4.198

Let x = length of time a bus is late. Then x is a uniform random variable with probability distribution: ⎧1 (0 ≤ x ≤ 20) ⎪ f(x) = ⎨ 20 ⎪⎩ 0 otherwise 0 + 20 = 10 2

a.

μ=

b.

⎛ 1 ⎞ 1 P(x ≥ 19) = (20 − 19) ⋅ ⎜ ⎟ = = .05 ⎝ 20 ⎠ 20

c.

It would be doubtful that the director's claim is true, since the probability of the bus being more than 19 minutes late is so small.

Random Variables and Probability Distributions

135

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The Furniture Fire Case (To accompany Chapters 3–4)

Using the entire data set of 3,005 invoices as the population, the mean profit margin is 48.9% and the standard deviation is 13.8291%. If a random sample is selected from this population, the sampling distribution of the sample mean ( x ) is approximately normal with a mean of 48.901% and a standard deviation of 13.8291%/ n by the Central Limit Theorem. If a random sample of 253 invoices is selected, then the probability of obtaining a sample mean of 50.8% or higher is:

50.8 − 48.901 ⎞ ⎛ P(x ≥ 50.8) = P ⎜ z ≥ ⎟ = P(z ≥ 2.18) = .5 − .4854 = .0146 13.8291/ 253 ⎠ ⎝ Since the probability of obtaining a sample mean of 50.8% or higher from this population is extremely small (.0146), we would conclude that there is evidence of fraud. If we look at the two samples separately, the evidence becomes even more damning. For the sample of 134 invoices, the probability of obtaining a sample mean of 50.6% or higher is: 50.6 − 48.901 ⎞ ⎛ P( x1 ≥ 50.6) = P ⎜ z ≥ ⎟ = P(z ≥ 1.42) = .5 − .4222 = .0778 13.8291/ 134 ⎠ ⎝ For the sample of 119 invoices, the probability of obtaining a sample mean of 51.0% or higher is: 51.0 − 48.901 ⎞ ⎛ P( x2 ≥ 51.0) = P ⎜ z ≥ ⎟ = P(z ≥ 1.66) = .5 − .4515 = .0485 13.8291/ 119 ⎠ ⎝ The probability of observing one sample mean of 50.6% or higher AND a second sample mean of 51.0% or higher is:

P( x1 ≥ 50.6, x2 ≥ 51.0) = .0778(.0485) = .0038 Again, since the probability of obtaining two sample means of 50.8% or higher and 51.0% or higher from this population is extremely small (.0038), we would conclude that there is evidence of fraud.

136

The Furniture Fire Case

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Inferences Based on a Single Sample: Estimation with Confidence Intervals

5.2

5.4

a.

zα/2 = 1.96, using Table IV, Appendix B, P(0 ≤ z ≤ 1.96) = .4750. Thus, α/2 = .5000 − .4750 = .025, α = 2(.025) = .05, and 1 - α = 1 - .05 = .95. The confidence level is 100% × .95 = 95%.

b.

zα/2 = 1.645, using Table IV, Appendix B, P(0 ≤ z ≤ 1.645) = .45. Thus, α/2 = .50 − .45 = .05, α = 2(.05) = .1, and 1 − α = 1 − .1 = .90. The confidence level is 100% × .90 = 90%.

c.

zα/2 = 2.575, using Table IV, Appendix B, P(0 ≤ z ≤ 2.575) = .495. Thus, α/2 = .500 − .495 = .005, α = 2(.005) = .01, and 1 − α = 1 − .01 = .99. The confidence level is 100% × .99 = 99%.

d.

zα/2 = 1.282, using Table IV, Appendix B, P(0 ≤ z ≤ 1.282) = .4. Thus, α/2 = .5 − .4 = .1, α = 2(.1) = .2, and 1 − α = 1 − .2 = .80. The confidence level is 100% × .80 = 80%.

e.

zα/2 = .99, using Table IV, Appendix B, P(0 ≤ z ≤ .99) = .3389. Thus, α/2 = .5000 − .3389 = .1611, α = 2(.1611) = .3222, and 1 − α = 1 − .3222 = .6778. The confidence level is 100% × .6778 = 67.78%.

a.

For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B, z.025 = 1.96. The confidence interval is:

x ± z.025 b.

c.

s 2.7 ⇒ 25.9 ± 1.96 ⇒ 25.9 ± .56 ⇒ (25.34, 26.46) 90 n

For confidence coefficient .90, α = .10 and α/2 = .10/2 = .05. From Table IV, Appendix B, z.05 = 1.645. The confidence interval is:

x ± z.05

s n

⇒ 25.9 ± 1.645

2.7 90

⇒ 25.9 ± .47 ⇒ (25.43, 26.37)

For confidence coefficient .99, α = .01 and α/2 = .01/2 = .005. From Table IV, Appendix B, z.005 = 2.58. The confidence interval is: x ± z.005

5.6

Chapter 5

s 2.7 ⇒ 25.9 ± 2.58 ⇒ 25.9 ± .73 ⇒ (25.17, 26.63) 90 n

If we were to repeatedly draw samples from the population and form the interval x ± 1.96 σ x each time, approximately 95% of the intervals would contain μ. We have no way of knowing whether our interval estimate is one of the 95% that contain μ or one of the 5% that do not.

Inferences Based on a Single Sample: Estimation with Confidence Intervals

137

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5.8

a.

For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B, z.025 = 1.96. The confidence interval is: x ± z.025

5.10

s 3.3 ⇒ 33.9 ± 1.96 ⇒ 33.9 ± .323 ⇒ (33.577, 34.223) n 400

b.

x ± z.025

c.

For part a, the width of the interval is 2(.647) = 1.294. For part b, the width of the interval is 2(.323) = .646. When the sample size is quadrupled, the width of the confidence interval is halved.

a.

A point estimate for the average number of latex gloves used per week by all healthcare workers with latex allergy is x = 19.3 .

b.

For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B, z.025 = 1.96. The confidence interval is: x ± zα / 2

138

s n

⇒ 19.3 ± 1.96

11.9 46

⇒ 19.3 ± 3.44 ⇒ (15.86, 22.74)

c.

We are 95% confident that the true average number of latex gloves used per week by all healthcare workers with a latex allergy is between 15.86 and 22.74.

d.

The conditions required for the interval to be valid are: a. b.

5.12

s 3.3 ⇒ 33.9 ± 1.96 ⇒ 33.9 ± .647 ⇒ (33.253, 34.547) 100 n

The sample selected was randomly selected from the target population. The sample size is sufficiently large, i.e. n > 30.

a.

The point estimate for the mean charitable commitment of tax-exempt organizations is x = 74.9667.

b.

From the printout, the 95% confidence interval is (68.2371, 81.6962).

c.

The probability of estimating the true mean charitable commitment with a single number is 0. By estimating the true mean charitable commitment with an interval, we can be pretty confident that the true mean is in the interval.

Chapter 5

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5.14

Using MINITAB, the descriptive statistics are: Descriptive Statistics: r Variable r

N 34

Mean 0.4224

Median 0.4300

TrMean 0.4310

Variable r

Minimum -0.0800

Maximum 0.7400

Q1 0.2925

Q3 0.6000

StDev 0.1998

SE Mean 0.0343

For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B, z.025 = 1.96. The confidence interval is: x ± zα / 2

s

⇒ .4224 ± 1.96

n ⇒ (.3552, .4895)

.1998 34

⇒ .4224 ± .0672

We are 95% confident that the mean value of r is between .3552 and .4895. 5.16

a.

Using MINITAB, the descriptive statistics are:

Descriptive Statistics: Rate Variable Rate

N 30

Mean 79.73

Median 80.00

TrMean 80.15

Variable Rate

Minimum 60.00

Maximum 90.00

Q1 76.75

Q3 84.00

StDev 5.96

SE Mean 1.09

For confidence coefficient .90, α = .10 and α/2 = .10/2 = .05. From Table IV, Appendix B, z.05 = 1.645. The confidence interval is: x ± zα / 2

s 5.96 ⇒ 79.73 ± 1.645 ⇒ 79.73 ± 1.79 n 30 ⇒ (77.94, 81.52)

b.

We are 90% confident that the mean participation rate for all companies that have 401(k) plans is between 77.94% and 81.52%.

c.

We must assume that the sample size (n = 30) is sufficiently large so that the Central Limit Theorem applies.

d.

Yes. Since 71% is not included in the 90% confidence interval, it can be concluded that this company's participation rate is lower than the population mean.

e.

The center of the confidence interval is . If 60% is changed to 80%, the value of will increase, thus indicating that the center point will be larger. The value of s2 will decrease if 60% is replaced by 80%, thus causing the width of the interval to decrease.

Inferences Based on a Single Sample: Estimation with Confidence Intervals

139

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5.18

a.

Using MINITAB, I generated 30 random numbers using the uniform distribution from 1 to 308. The random numbers were: 9, 15, 19, 36, 46, 47, 63, 73, 90, 92, 108, 112, 117, 127, 144, 145, 150, 151, 172, 178, 218, 229, 230, 241, 242, 246, 252, 267, 274, 282 I numbered the 308 observations in the order that they appear in the file. Using the random numbers generated above, I selected the 9th, 15th, 19th, etc. observations for the sample. The selected sample is: .31, .34, .34, .50, .52, .53, .64, .72, .70, .70, .75, .78, 1.00, 1.00, 1.03, 1.04, 1.07, 1.10, .21, .24, .58, 1.01, .50, .57, .58, .61, .70, .81, .85, 1.00

b.

Using MINITAB, the descriptive statistics for the sample of 30 observations are:

Descriptive Statistics: carats-samp Variable carats-s

N 30

Mean 0.6910

Median 0.7000

TrMean 0.6965

Variable carats-s

Minimum 0.2100

Maximum 1.1000

Q1 0.5150

Q3 1.0000

StDev 0.2620

SE Mean 0.0478

From above, x =.6910 and s = .2620. c.

For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B, z.025 = 1.96. The confidence interval is: x ± zα / 2

5.20

s n

⇒ .691 ± 1.96

.262 30

⇒ .691 ± .094 ⇒ (.597, .785)

d.

We are 95% confident that the mean number of carats is between .597 and .785.

e.

From Exercise 2.47, we computed the “population” mean to be .631. This mean does fall in the 95% confidence interval we computed in part d.

x=

11,298 = 2.26 5,000

For confidence coefficient, .95, α = .05 and α/2 = .025. From Table IV, Appendix B, z.025 = 1.96. The confidence interval is: 1 .5 s ⇒ 2.26 ± 1.96 ⇒ 2.26 ± .04 ⇒ (2.22, 2.30) x ± zα/2 5000 n We are 95% confident the mean number of roaches produced per roach per week is between 2.22 and 2.30.

140

Chapter 5

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5.22

5.24

a.

If x is normally distributed, the sampling distribution of x is normal, regardless of the sample size.

b.

If nothing is known about the distribution of x, the sampling distribution of x is approximately normal if n is sufficiently large. If n is not large, the distribution of x is unknown if the distribution of x is not known.

a.

P(t ≥ t0) = .025 where df = 11 t0 = 2.201

b.

P(t ≥ t0) = .01 where df = 9 t0 = 2.821

c.

P(t ≤ t0) = .005 where df = 6 Because of symmetry, the statement can be rewritten P(t ≥ −t0) = .005 where df = 6 t0 = −3.707

d.

5.26

P(t ≤ t0) = .05 where df = 18 t0 = −1.734

For this sample, ∑ x = 1567 = 97.9375 x= n 16 s2 = s=

∑x

2

(∑ x) −

n −1

2

n

=

1567 2 16 = 159.9292 16 − 1

155,867 −

s 2 = 12.6463

a.

For confidence coefficient, .80, α = 1 − .80 = .20 and α/2 = .20/2 = .10. From Table VI, Appendix B, with df = n − 1 = 16 − 1 = 15, t.10 = 1.341. The 80% confidence interval for μ is: s 12.6463 x ± t.10 ⇒ 97.94 ± 1.341 ⇒ 97.94 ± 4.240 ⇒ (93.700, 102.180) n 16

b.

For confidence coefficient, .95, α = 1 − .95 = .05 and α/2 = .05/2 = .025. From Table VI, Appendix B, with df = n − 1 = 24 − 1 = 23, t.025 = 2.131. The 95% confidence interval for μ is: x ± t.025

s n

⇒ 97.94 ± 2.131

12.6463 16

⇒ 97.94 ± 6.737 ⇒ (91.203, 104.677)

The 95% confidence interval for μ is wider than the 80% confidence interval for μ found in part a.

Inferences Based on a Single Sample: Estimation with Confidence Intervals

141

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c.

For part a: We are 80% confident that the true population mean lies in the interval 93.700 to 102.180. For part b: We are 95% confident that the true population mean lies in the interval 91.203 to 104.677. The 95% confidence interval is wider than the 80% confidence interval because the more confident you want to be that μ lies in an interval, the wider the range of possible values.

5.28

a.

Using MINITAB, the descriptive statistics are:

Descriptive Statistics: MTBE Variable MTBE

N 12

N* 0

Mean 97.2

SE Mean 32.8

StDev 113.8

Minimum 8.00

Q1 12.0

Median 50.5

Q3 146.0

Maximum 367.0

A point estimate for the true mean MTBE level for all well sites located near the New Jersey gasoline service station is x = 97.2 . b.

For confidence coefficient .99, α = .01 and α/2 = .01/2 = .005. From Table VI, Appendix B, with df = n – 1 = 12 – 1 = 11, t.005 = 3.106. The 99% confidence interval is: s

x ± t.005

n

⇒ 97.2 ± 3.106

113.8 12

⇒ 97.2 ± 102.04 ⇒ (−4.84, 199.24)

We are 99% confident that the true mean MTBE level for all well sites located near the New Jersey gasoline service station is between −4.84 and 199.24. c.

We must assume that the data were sampled from a normal distribution. We will use the four methods to check for normality. First, we will look at a histogram of the data. Using MINITAB, the histogram of the data is: Histogram of MTBE 5

Fr equency

4

3

2

1

0

142

0

50

100

150 200 M T BE

250

300

350

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From the histogram, the data do not appear to be mound-shaped. This indicates that the data may not be normal. Next, we look at the intervals x ± s, x ± 2 s, x ± 3s . If the proportions of observations falling in each interval are approximately .68, .95, and 1.00, then the data are approximately normal. Using MINITAB, the summary statistics are: x ± s ⇒ 97.2 ± 113.8 ⇒ (−16.6, 211.0) 10 of the 12 values fall in this interval. The proportion is .83. This is not very close to the .68 we would expect if the data were normal. x ± 2 s ⇒ 97.2 ± 2(113.8) ⇒ 97.2 ± 227.6 ⇒ (−130.4, 324.8) 11 of the 12 values fall in this interval. The proportion is .92. This is a somewhat smaller than the .95 we would expect if the data were normal. x ± 2 s ⇒ 97.2 ± 3(113.8) ⇒ 97.2 ± 341.4 ⇒ (−244.2, 438.6) 12 of the 12 values fall in this interval. The proportion is 1.00. This is exactly the 1.00 we would expect if the data were normal. From this method, it appears that the data may not be normal. Next, we look at the ratio of the IQR to s. IQR = QU – QL = 146.0 – 12.0 = 134.0. IQR 134.0 = = 1.18 This is somewhat smaller than the 1.3 we would expect if the data s 113.8 were normal. This method indicates the data may not be normal.

Finally, using MINITAB, the normal probability plot is: Probability Plot of MTBE N ormal - 95% C I 99

95 90

Mean StDev

97.17 113.8

N AD P-Value

12 0.929 0.012

P er cent

80 70 60 50 40 30 20 10 5

1

-300

-200

-100

0

100 200 M T BE

300

400

500

Since the data do not form a fairly straight line, the data may not be normal. From above, the all methods indicate the data may not be normal. It appears that the data probably are not normal.

Inferences Based on a Single Sample: Estimation with Confidence Intervals

143

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5.30

We must assume that the distribution of the LOS's for all patients is normal. a.

For confidence coefficient .90, α = 1 − .90 = .10 and α/2 = .10/2 = .05. From Table VI, Appendix B, with df = n − 1 = 20 − 1 = 19, t.05 = 1.729. The 90% confidence interval is:

x ± t.05

5.32

5.34

s n

⇒ 3.8 ± 1.729

1.2 20

⇒ 3.8 ± .464 ⇒ (3.336, 4.264)

b.

We are 90% confident that the mean LOS is between 3.336 and 4.264 days.

c.

“90% confidence” means that if repeated samples of size n are selected from a population and 90% confidence intervals are constructed, 90% of all intervals thus constructed will contain the population mean.

a.

The 95% confidence interval for the mean surface roughness of coated interior pipe is (1.63580, 2.12620).

b.

No. Since 2.5 does not fall in the 95% confidence interval, it would be very unlikely that the average surface roughness would be as high as 2.5 micrometers.

a.

The population is the set of all DOT permanent count stations in the state of Florida.

b.

Yes. There are several types of routes included in the sample. There are 3 recreational areas, 7 rural areas, 5 small cities, and 5 urban areas.

c.

Using MINITAB, the descriptive statistics are:

Descriptive Statistics: 30th hour, 100th hour Variable 30th hou 100th ho

N 20 20

Mean 2206 2096

Median 2064 1999

TrMean 2165 2048

Variable 30th hou 100th ho

Minimum 252 229

Maximum 4905 4815

Q1 1429 1318

Q3 3068 2877

StDev 1224 1203

SE Mean 274 269

For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table VI, Appendix B, with df = n – 1 = 20 – 1 = 19, t.025 = 2.093. The 95% confidence interval is: x ± t.025

s n

⇒ 2, 206 ± 2.093

1, 224 20

⇒ 2, 206 ± 572.84 ⇒ (1,633.16, 2,778.84)

We are 95% confident that the mean traffic count at the 30th highest hour is between 1,633.16 and 2,778.84. d.

144

We must assume that the distribution of the traffic counts at the 30th highest hour is normal. From the stem-and-leaf display, the data look fairly mound-shaped. Thus, the assumption of normality is probably met.

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e.

For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table VI, Appendix B, with df = n – 1 = 20 – 1 = 19, t.025 = 2.093. The 95% confidence interval is: x ± t.025

s n

⇒ 2,096 ± 2.093

1, 203 20

⇒ 2,096 ± 563.01 ⇒ (1,532.99, 2,659.01)

We are 95% confident that the mean traffic count at the 100th highest hour is between 1,532.99 and 2,659.01. We must assume that the distribution of the traffic counts at the 100th highest hour is normal. From the stem-and-leaf display, the data look fairly mound-shaped. Thus, the assumption of normality is probably met. f.

If μ = 2,700, it is very possible that it is the mean count for the 30th highest hour. It falls in the 95% confidence interval for the mean count for the 30th highest hour. It is not very likely that the mean count for the 100th highest hour is 2,700. It does not fall in the 95% confidence interval for the mean count for the 100th highest hour. (See parts c and e above.)

5.36

By the Central Limit Theorem, the sampling distribution of is approximately normal with pq mean μ pˆ = p and standard deviation σ pˆ = . n

5.38

a.

The sample size is large enough if the interval pˆ ± 3σ pˆ does not include 0 or 1.

pˆ ± 3σ pˆ ⇒ pˆ ± 3

ˆˆ pq pq .88(1 − .88) ⇒ .88 ± .089 ⇒ pˆ ± 3 ⇒ .88 ± n n 121 ⇒ (.791, .969)

Since the interval lies within the interval (0, 1), the normal approximation will be adequate. b.

For confidence coefficient .90, α = .10 and α/2 = .05. From Table IV, Appendix B, z.05 = 1.645. The 90% confidence interval is: pˆ ± z .05

c.

pq ⇒ pˆ ± 1.645 n

ˆˆ pq .88(.12) ⇒ .88 ± .049 ⇒ .88 ± 1.645 1.645 n 121 ⇒ (.831, .929)

We must assume that the sample is a random sample from the population of interest.

Inferences Based on a Single Sample: Estimation with Confidence Intervals

145

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5.40

a.

Of the 50 observations, 15 like the product ⇒ pˆ =

15 = .30. 30

To see if the sample size is sufficiently large:

pˆ ± 3 σ pˆ ≈ pˆ ± 3

ˆˆ pq .3(.7) ⇒ .3 ± 3 ⇒ .3 ± .194 ⇒ (.106, .494) n 50

Since this interval is wholly contained in the interval (0, 1), we may conclude that the normal approximation is reasonable. For the confidence coefficient .80, α = .20 and α/2 = .10. From Table IV, Appendix B, z.10 = 1.28. The confidence interval is: pˆ ± z.10

5.42

ˆˆ pq .3(.7) ⇒ .3 ± 1.28 ⇒ .3 ± .083 ⇒ (.217, .383) n 50

b.

We are 80% confident the proportion of all consumers who like the new snack food is between .217 and .383.

a.

The point estimate of p is pˆ = .11 .

b.

To see if the sample size is sufficiently large: ˆˆ pq .11(.89) ⇒ .11 ± 3 ⇒ .11 ± .077 ⇒ (.033, .187) n 150 Since the interval is wholly contained in the interval (0, 1), we may conclude that the normal approximation is reasonable. pˆ ± 3σ pˆ ≈ pˆ ± 3

For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B, z.025 = 1.96. The confidence interval is: pˆ ± z.025

5.44

ˆˆ pq .11(.89) ⇒ .11 ± 1.645 ⇒ .11 ± .05 ⇒ (.06, .16) n 150

c.

We are 95% confident that the true proportion of MSDS that are satisfactorily completed is between .06 and .16.

a.

The point estimate of p is pˆ =

x 16 = = .052 . n 308

To see if the sample size is sufficiently large: ˆˆ pq .052(.948) pˆ ± 3σ pˆ ≈ pˆ ± 3 ⇒ .052 ± 3 ⇒ .052 ± .038 ⇒ (.014, .090) n 308 Since the interval is wholly contained in the interval (0, 1), we may conclude that the normal approximation is reasonable.

146

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For confidence coefficient .99, α = .01 and α/2 = .01/2 = .005. From Table IV, Appendix B, z.005 = 2.58. The confidence interval is: pˆ ± z.05

b.

ˆˆ pq .052(.948) ⇒ .052 ± 2.58 ⇒ .052 ± .033 ⇒ (.019, .085) n 308

We are 99% confident that the true proportion of diamonds for sale that are classified as “D” color is between .019 and .085. x 81 = .263 . The point estimate of p is pˆ = = n 308 To see if the sample size is sufficiently large: pˆ ± 3σ pˆ ≈ pˆ ± 3

ˆˆ pq .263(.737) ⇒ .263 ± 3 ⇒ .263 ± .075 ⇒ (.188, .338) n 308

Since the interval is wholly contained in the interval (0, 1), we may conclude that the normal approximation is reasonable. For confidence coefficient .99, α = .01 and α/2 = .01/2 = .005. From Table IV, Appendix B, z.005 = 2.58. The confidence interval is: pˆ ± z.05

ˆˆ pq .263(.737) ⇒ .263 ± 2.58 ⇒ .263 ± .065 ⇒ (.198, .328) n 308

We are 99% confident that the true proportion of diamonds for sale that are classified as “VS1” clarity, is between .198 and .328. 5.46

a.

The population is all senior human resource executives at U.S. companies.

b.

The population parameter of interest is p, the proportion of all senior human resource executives at U.S. companies who believe that their hiring managers are interviewing too many people to find qualified candidates for the job.

c.

The point estimate of p is pˆ =

x 211 = = .42 . To see if the sample size is sufficiently n 502

large: pˆ ± 3σ pˆ ≈ pˆ ± 3

ˆˆ pq .42(.58) ⇒ .42 ± 3 ⇒ .42 ± .066 ⇒ (.354, .486) n 502

Since the interval is wholly contained in the interval (0, 1), we may conclude that the normal approximation is reasonable.

Inferences Based on a Single Sample: Estimation with Confidence Intervals

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d.

For confidence coefficient .98, α = .02 and α/2 = .02/2 = .01. From Table IV, Appendix B, z.01 = 2.33. The confidence interval is: pˆ ± z.01

ˆˆ pq .42(.58) ⇒ .42 ± 2.33 ⇒ .42 ± .051 ⇒ (.369, .471) n 502

We are 98% confident that the true proportion of all senior human resource executives at U.S. companies who believe that their hiring managers are interviewing too many people to find qualified candidates for the job is between .369 and .471.

5.48

e.

A 90% confidence interval would be narrower. If the interval was narrower, it would contain fewer values, thus, we would be less confident.

a.

The point estimate of p is

b.

We must check to see if the sample size is sufficiently large:

pˆ ± 3σ pˆ ≈ pˆ ± 3

pˆ = x/n = 35/55 = .636.

ˆˆ pq .636(.364) ⇒ .636 ± 3 ⇒ .636 ± .195 ⇒ (.441, .831) n 55

Since the interval is wholly contained in the interval (0, 1) we may assume that the normal approximation is reasonable. For confidence coefficient, .99, α = .01 and α/2 = .01/2 = .005. From Table IV, Appendix B, z.005 = 2.575. The confidence interval is: pˆ ± z.005 c. d.

5.50

ˆˆ pq .636(.364) ⇒ .636 ± 2.575 ⇒ .636 ± .167 ⇒ (.469, .803) n 55

We are 99% confident that the true proportion of fatal accidents involving children is between .469 and .803. The sample proportion of children killed by air bags who were not wearing seat belts or were improperly restrained is 24/35 = .686. This is rather large proportion. Whether a child is killed by an airbag could be related to whether or not he/she was properly restrained. Thus, the number of children killed by air bags could possibly be reduced if the child were properly restrained.

The point estimate of p is pˆ =

x 36 = = .434 . n 83

To see if the sample size is sufficiently large: pˆ ± 3σ pˆ ≈ pˆ ± 3

ˆˆ pq .434(.566) ⇒ .434 ± 3 ⇒ .434 ± .163 ⇒ (.271, .597) n 83

Since the interval is wholly contained in the interval (0, 1), we may conclude that the normal approximation is reasonable.

148

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For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B, z.025 = 1.96. The confidence interval is: ˆˆ pq .434(.566) ⇒ .434 ± 1.96 ⇒ .434 ± .107 ⇒ (.327, .541) n 83

pˆ ± z.025

We are 95% confident that the true proportion of healthcare workers with latex allergies actually suspects the he or she actually has the allergy is between .327 and .541. 5.52

To compute the necessary sample size, use

n=

2 ( zα / 2 ) σ 2

where α = 1 − .95 = .05 and α/2 = .05/2 = .025.

SE 2

From Table IV, Appendix B, z.025 = 1.96. Thus, n=

(1.96) 2 (7.2) = 307.328 ≈ 308 .32

You would need to take 308 samples. 5.54

a.

To compute the needed sample size, use:

n=

Thus, n =

( zα / 2 ) SE

2

pq

2

where z.025 = 1.96 from Table IV, Appendix B.

(1.96) 2 (.2)(.8) = 96.04 ≈ 97 .08 2

You would need to take a sample of size 97. b.

To compute the needed sample size, use:

n=

( zα / 2 )

2

SE

2

pq

=

(1.96) 2(.5)(.5) = 150.0625 ≈ 151 .08 2

You would need to take a sample of size 151. 5.56

a.

For a width of 5 units, SE = 5/2 = 2.5. To compute the needed sample size, use

( zα / 2 ) σ 2 2

n=

SE

2

where α = 1 − .95 = .05 and α/2 = .025.

Inferences Based on a Single Sample: Estimation with Confidence Intervals

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From Table IV, Appendix B, z.025 = 1.96. Thus,

n=

(1.96) 2 (14) 2 = 120.47 ≈ 121 2.52

You would need to take 121 samples at a cost of 121($10) = $1210. Yes, you do have sufficient funds. b.

For confidence coefficient .90, α = 1 − .90 = .10 and α/2 = .10/2 = .05. From Table IV, Appendix B, z.05 = 1.645.

n=

(1.645) 2 (14) 2 = 84.86 ≈ 85 2.52

You would need to take 85 samples at a cost of 85($10) = $850. You still have sufficient funds but have an increased risk of error. 5.58

The sample size will be larger than necessary for any p other than .5.

5.60

a.

The confidence level desired by the researchers is 90%.

b.

The sampling error desired by the researchers is SE = .05.

c.

For confidence coefficient .90, α = .10 and α/2 = .10/2 = .05. From Table IV, x 64 Appendix B, z.05 = 1.645. From the problem, we will use pˆ = = = .604 n 106 to estimate p. Thus,

n=

( zα / 2 ) 2 pq 1.6452.604(.396) = = 258.9 ≈ 259 ( SE ) 2 .052

Thus, we would need a sample of size 259. 5.62

For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B, z.025 = 1.96. For this study, n=

( zα / 2 ) 2 σ 2 1.962 (5) 2 ≈ = 96.04 ≈ 97 SE 2 12

The sample size needed is 97.

150

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5.64

For confidence coefficient .90, α = .10 and α/2 = .05. From Table IV, Appendix B, z.05 = 1.645. For a width of .06, SE = .06/2 = .03 ( zα / 2 ) 2 pq (.1645) 2 (.17)(.83) = 424.2 ≈ 425 = The sample size is n = 2 .032 SE You would need to take n = 425 samples.

5.66

To compute the necessary sample size, use n=

( zα / 2 ) 2 σ 2 where α = 1 − .90 = .10 and α/2 = .05. SE 2

From Table IV, Appendix B, z.05 = 1.645. Thus, n= 5.68

a.

(1.645) 2 (10) 2 = 270.6 ≈ 271 12

To compute the needed sample size, use n=

( zα / 2 ) 2 σ 2 where α = 1 − .90 = .10 and α/2 = .05. SE 2

From Table IV, Appendix B, z.10 = 1.645. Thus, n=

(1.645) 2 (2) 2 = 1,082.41 ≈ 1,083 .12

b.

As the sample size decreases, the width of the confidence interval increases. Therefore, if we sample 100 parts instead of 1,083, the confidence interval would be wider.

c.

To compute the maximum confidence level that could be attained meeting the management's specifications, n=

( zα / 2 ) 2 σ 2 ( zα / 2 )(2) 2 100(.01) ⇒ 100 = ⇒ ( zα / 2 ) 2 = = .25 ⇒ zα/2 = .5 2 2 4 SE .1

Using Table IV, Appendix B, P(0 ≤ z ≤ .5) = .1915. Thus, α/2 = .5000 − .1915 = .3085,

α = 2(.3085) = .617, and 1 − α = 1 − .617 = .383. The maximum confidence level would be 38.3%.

Inferences Based on a Single Sample: Estimation with Confidence Intervals

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5.70

5.72

σx =

σ n

N −n N 2500 − 1000 = 4.90 2500

a.

σx=

200 1000

b.

σx =

200 5000 − 1000 = 5.66 5000 1000

c.

σx =

10,000 − 1000 = 6.00 10,000 1000

d.

σx =

200 100,000 − 1000 = 6.293 100,000 1000

a.

For n = 36, with the finite population correction factor: ⎛ N − n ⎞ 24 ⎛ 5000 − 64 ⎞ σˆ x = s / n ⎜⎜ ⎟= ⎜ ⎟ = 3 .9872 = 2.9807 N ⎟⎠ 5000 ⎟⎠ 64 ⎜⎝ ⎝

200

without the finite population correction factor: 24 σˆ x = s / n = =3 64

σˆ x without the finite population correction factor is slightly larger. b.

For n = 400, with the finite population correction factor: ⎛ N −n ⎞ 24 ⎛ 5000 − 400 ⎞ σˆ x = s / n ⎜⎜ ⎟⎟ = ⎜ ⎟ = 1.2 .92 = 1.1510 N ⎠ 5000 ⎟⎠ 400 ⎜⎝ ⎝ without the finite population correction factor: 24 σˆ x = s / n = = 1.2 400

c.

5.74

In part a, n is smaller relative to N than in part b. Therefore, the finite population correction factor did not make as much difference in the answer in part a as in part b.

An approximate 95% confidence interval for μ is: s N −n 14 375 − 40 x ± 2σˆ x ⇒ x ± 2 ⇒ 422 ± 2 375 N 40 n ⇒ 422 ± 4.184 ⇒ (417.816, 426.184)

152

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5.76

a.

For N = 2,193, n = 223, x =116,754, and s = 39,185, the 95% confidence interval is:

s N −n 39,185 2,193 − 223 ⇒ 116,754 ± 2 N 2,193 n 223 ⇒ 116,754 ± 4,974.06 ⇒ (111,779.94, 121,728.06)

x ± 2σˆ x ⇒ x ± 2

5.78

b.

We are 95% confident that the mean salary of all vice presidents who subscribe to Quality Progress is between $111,777.94 and $121,728.06.

a.

The population of interest is the set of all households headed by women that have incomes of $25,000 or more in the database.

b.

Yes. Since n/N = 1,333/25,000 = .053 exceeds .05, we need to apply the finite population correction.

c.

The standard error for pˆ should be:

σˆ pˆ = d.

.708(1 − .708) ⎛ 25,000 − 1,333 ⎞ pˆ (1 − pˆ ) ⎛ N − n ⎞ ⎜ ⎟= ⎜ ⎟ = .012 1333 25,000 n ⎝ N ⎠ ⎝ ⎠

For confidence coefficient .90, α = 1 − .90 = .10 and α/2 = .10/2 = .05. From Table IV, Appendix B, z.05 = 1.645. The approximate 90% confidence interval is:

pˆ ± 1.645σˆ pˆ ⇒ .708 ± 1.645(.012) ⇒ (.688, .728) 5.80

For N = 1,500, n = 35, x = 1, and s = 124, the 95% confidence interval is:

⎛ s ⎞ N −n ⎛ 124 ⎞ 1,500 − 35 x ± 2σˆ x ⇒ x ± 2 ⎜ ⇒ 1 ± 2⎜ ⇒ 1 ± 41.43 ⎟ ⎟ 1,500 N ⎝ n⎠ ⎝ 35 ⎠ ⇒ (−40.43, 42.43) We are 95% confident that the mean error of the new system is between -$40.43 and $42.43.

5.82

a.

For a small sample from a normal distribution with unknown standard deviation, we use the t statistic. For confidence coefficient .95, α = 1 − .95 = .05 and α/2 = .05/2 = .025. From Table VI, Appendix B, with df = n − 1 = 23 − 1 = 22, t.025 = 2.074.

b.

For a large sample from a distribution with an unknown standard deviation, we can estimate the population standard deviation with s and use the z statistic. For confidence coefficient .95, α = 1 − .95 = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B, z.025 = 1.96.

c.

For a small sample from a normal distribution with known standard deviation, we use the z statistic. For confidence coefficient .95, α = 1 − .95 = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B, z.025 = 1.96.

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5.84

d.

For a large sample from a distribution about which nothing is known, we can estimate the population standard deviation with s and use the z statistic. For confidence coefficient .95, α = 1 − .95 = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B, z.025 = 1.96.

e.

For a small sample from a distribution about which nothing is known, we can use neither z nor t.

a.

Of the 400 observations, 227 had the characteristic ⇒ pˆ = 227/400 = .5675. To see if the sample size is sufficiently large: pˆ ± 3σ pˆ ⇒ pˆ ± 3

ˆˆ pq pq .5675(.4325) ⇒ pˆ ± 3 ⇒ .5675 ± 3 ⇒ .5675 ± .0743 n n 400 ⇒ (.4932, .6418)

Since the interval lies within the interval (0, 1), the normal approximation will be adequate. For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B, z.025 = 1.96. The confidence interval is: pˆ ± z.025

b.

pq ⇒ ± 1.96 n

ˆˆ pq .5675(.4325) ⇒ .5675 ± 1.96 ⇒ .5675 ± .0486 n 400 ⇒ (.5189, .6161)

For this problem, SE = .02. For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B, z.025 = 1.96. Thus, n=

( zα / 2 ) 2 pq (1.96) 2 (.5675)(.4325) = = 2,357.2 ≈ 2,358 SE 2 .022

Thus, the sample size was 2,358. 5.86

a.

The finite population correction factor is: ( N − n) = N

b.

The finite population correction factor is: ( N − n) = N

c.

(100 − 20) = .8944 100

The finite population correction factor is: ( N − n) = N

154

(2,000 − 50) = .9874 2,000

(1,500 − 300) = .8944 1,500

Chapter 5

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5.88

5.90

a.

From the printout, the 90% confidence interval is (4.277, 6.184). We are 90% confident that the mean number of offices operated by all Florida law firms is between 4.277 and 6.184.

b.

From the histogram, it appears that the data probably are not from a normal distribution. The data appear to be skewed to the right.

c.

The interval constructed in part a depends on the assumption that the data came from a normal distribution. From part b, it appears that this assumption is not valid. Thus, the confidence interval is probably not valid.

a.

The point estimate of p is pˆ =

b.

To see if the sample size is sufficiently large:

x 67 = = .638 . n 105

ˆˆ pq .638(.362) ⇒ .638 ± 3 ⇒ .638 ± .141 ⇒ (.497, .779) n 105 Since the interval is wholly contained in the interval (0, 1), we may conclude that the normal approximation is reasonable. pˆ ± 3σ pˆ ≈ pˆ ± 3

For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B, z.025 = 1.96. The confidence interval is: pˆ ± z.025

5.92

ˆˆ pq .638(.362) ⇒ .638 ± 1.96 ⇒ .638 ± .092 ⇒ (.546, .730) n 105

c.

We are 95% confident that the true proportion of on-the-job homicide cases that occurred at night is between .546 and .730.

a.

Using MINITAB, the descriptive statistics are:

Descriptive Statistics: NJValues Variable NJValues

N 20

N* 0

Mean 440.4

SE Mean 67.8

StDev 303.0

Minimum 159.0

Q1 212.3

Median 297.5

Q3 660.5

Maximum 1190.0

For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table VI, Appendix B, with df = n – 1 = 20 – 1 = 19, t.025 = 2.093. The 95% confidence interval is: x ± t.025 b.

s n

⇒ 440.4 ± 2.093

303.0 20

⇒ 440.4 ± 141.81 ⇒ (298.59, 582.21)

We are 95% confident that the true mean sales price is between $298,590 and $582,210.

Inferences Based on a Single Sample: Estimation with Confidence Intervals

155

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c.

"95% confidence" means that in repeated sampling, 95% of all confidence intervals constructed will contain the true mean sales price and 5% will not.

d.

Using MINITAB, a histogram of the data is: Histogram of NJValues 9 8

Fr equency

7 6 5 4 3 2 1 0

200

400

600 800 NJValues

1000

1200

Since the sample size is small (n = 20), we must assume that the distribution of sales prices is normal. From the histogram, it does not appear that the data come from a normal distribution. Thus, this confidence interval is probably not valid. 5.94

a.

For confidence coefficient .90, α = .10 and α/2 = .05. From Table IV, Appendix B, z.05 = 1.645. The 90% confidence interval is: x ± z.05

σ n

⇒ x ± 1.645

s n

⇒ 12.2 ± 1.645

10 100

⇒ 12.2 ± 1.645 ⇒ (10.555, 13.845)

We are 90% confident that the mean number of days of sick leave taken by all its employees is between 10.555 and 13.845. b.

For confidence coefficient .99, α = .01 and α/2 = .005. From Table IV, Appendix B, z.005 = 2.58. The sample size is n =

2 ( zα / 2 ) σ 2

SE 2

=

(2.58) 2 (10) 2 = 166.4 ≈ 167 22

You would need to take n = 167 samples.

156

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5.96

a.

For confidence coefficient .99, α = .01 and α/2 = .01/2 = .005. From Table IV, Appendix B, z.005 = 2.58. The confidence interval is: x ± zα / 2

2.21 s ⇒ 1.13 ± 2.58 ⇒ 1.13 ± .67 72 n ⇒ (.46, 1.80)

We are 99% confident that the mean number of pecks at the blue string is between .46 and 1.80.

5.98

b.

Yes. The mean number of pecks at the white string is 7.5. This value does not fall in the 99% confident interval for the blue string found in part a. Thus, the chickens are more apt to peck at white string.

a.

First we must compute pˆ : pˆ =

x 124 = .78 = n 159

To see if the sample size is sufficiently large: ˆˆ pq .78(22) ⇒ .78 ± 3 ⇒ .78 ± .099 ⇒ (.681, .879) n 159 Since this interval is wholly contained in the interval (0, 1), we may conclude that the normal approximation is reasonable.

pˆ ± 3σ pˆ ≈ pˆ ± 3

For confidence coefficient .90, α = .10 and α/2 = .10/2 = .05. From Table IV, Appendix B, z.05 = 1.645. The confidence interval is: pˆ ± z.05

pq ≈ pˆ ± 1.645 n

ˆˆ pq .78(.22) ⇒ .78 ± 1.645 ⇒ .78 ± .054 n 159 ⇒ (.726, .834)

We are 90% confident that the true proportion of all truck drivers who suffer from sleep apnea is between .726 and .834.

5.100

b.

Sleep researchers believe that 25% of the population suffer from obstructive sleep apnea. Since the 90% confidence interval for the proportion of truck drivers who suffer from sleep apnea does not contain .25, it appears that the true proportion of truck drivers who suffer from sleep apnea is larger than the proportion of the general population.

a.

The population of interest is the set of all debit cardholders in the U.S.

c.

Of the 1252 observations, 180 had used the debit card to purchase a product or service on the Internet ⇒ pˆ =

180 = .144 1252

Inferences Based on a Single Sample: Estimation with Confidence Intervals

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To see if the sample size is sufficiently large: pˆ ± 3σ pˆ ≈ pˆ ± 3

ˆˆ pq .144(.856) ⇒ .144 ± 3 ⇒ .144 ± .030 ⇒ (.114, .174) n 1252

Since this interval is wholly contained in the interval (0, 1), we may conclude that the normal approximation is reasonable. d.

For confidence coefficient .98, α = 1 − .98 = .02 and α/2 = .02/2 = .01. From Table IV, Appendix B, z.01 = 2.33. The confidence interval is: pˆ ± z.01

ˆˆ pq .144(.856) ⇒ .144 ± .023 ⇒ (.121, .167) ⇒ .144 ± 2.33 n 1252

We are 98% confident that the proportion of debit cardholders who have used their card in making purchases over the Internet is between .121 and .167.

5.102

e.

Since we would have less confidence with a 90% confidence interval than with a 98% confidence interval, the 90% interval would be narrower.

a.

Of the 100 cancer patients, 7 were fired or laid off ⇒ = 7/100 = .07. To see if the sample size is sufficiently large: pˆ ± 3σ pˆ ⇒ pˆ ± 3

ˆˆ pq pq .07(.93) ⇒ pˆ ± 3 ⇒ .07 ± 3 ⇒ .07 ± .077 n n 100 ⇒ (−.007, .145)

Since the interval does not lie within the interval (0, 1), the normal approximation will not be adequate. We will go ahead and construct the interval anyway. For confidence coefficient .90, α = .10 and α/2 = .10/2 = .05. From Table IV, Appendix B, z.05 = 1.645. The confidence interval is: pˆ ± z.05

pq ⇒ pˆ ± 1.645 n

ˆˆ pq .07(.93) ⇒ .07 ± 1.645 ⇒ .07 ± .042 n 100 ⇒ (.028, .112)

Converting these to percentages, we get (2.8%, 11.2%).

158

b.

We are 90% confident that the percentage of all cancer patients who are fired or laid off due to their illness is between 2.8% and 11.2%.

c.

Since the rate of being fired or laid off for all Americans is 1.3% and this value falls outside the confidence interval in part b, there is evidence to indicate that employees with cancer are fired or laid off at a rate that is greater than that of all Americans.

Chapter 5

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5.104

a.

x 9296 = = .9296 n 10,000

pˆ =

The approximate 95% confidence interval is: pˆ (1 − pˆ ) N − n .9296(.0704) 500,000 − 10,000 ⇒ .9296 ± 2 10,000 500,000 n N

pˆ ± 2

⇒ .9296 ± 2 .000006413 ⇒ .9296 ± .0051 ⇒ (.9245, .9347)

5.106

10,000 × 100% = 2% of the subscribers returned the questionnaire. Often in mail 500,000 surveys, those that respond are those with strong views. Thus, the 10,000 that responded may not be representative. I would question the estimate in part a.

b.

Only

a.

The point estimate for the fraction of the entire market who refuse to purchase bars is:

pˆ = b.

x 23 = = .094 n 244

To see if the sample size is sufficient:

pˆ ± 3

ˆˆ pq (.094)(.906) ⇒ .094 ± 3 ⇒ .094 ± .056 ⇒ (.038, .150) 244 n

Since the interval above is contained in the interval (0, 1), the sample size is sufficiently large. c.

For confidence coefficient .95, α = 1 − .95 = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B, z.025 = 1.96. The confidence interval is:

pˆ ± z.025 d.

ˆˆ pq (.094)(.906) ⇒ .094 ± 1.96 ⇒ .094 ± .037 ⇒ (.057, .131) 244 n

The best estimate of the true fraction of the entire market who refuse to purchase bars six months after the poisoning is .094. We are 95% confident the true fraction of the entire market who refuse to purchase bars six months after the poisoning is between .057 and .131.

Inferences Based on a Single Sample: Estimation with Confidence Intervals

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5.108

The bound is SE = .1. For confidence coefficient .99, α = 1 − .99 = .01 and α/2 = .01/2 = .005. From Table IV, Appendix B, z.005 = 2.575. We estimate p with from Exercise 7.48 which is = .636. Thus,

n=

( zα / 2 ) 2 pq 2.5752 (.636)(.364) ≈ = 153.5 ⇒ 154 .12 SE 2

The necessary sample size would be 154. 5.110

Since the manufacturer wants to be reasonably certain the process is really out of control before shutting down the process, we would want to use a high level of confidence for our inference. We will form a 99% confidence interval for the mean breaking strength. For confidence coefficient .99, α = .01 and α/2 = .01/2 = .005. From Table VI, Appendix B, with df = n – 1 = 9 – 1 = 8, t.005 = 3.355. The 99% confidence interval is:

x ± t.005

s 22.9 ⇒ 985.6 ± 3.355 ⇒ 985.6 ± 25.61 ⇒ (959.99, 1,011.21) 9 n

We are 99% confident that the true mean breaking strength is between 959.99 and 1,011.21. Since 1,000 is contained in this interval, it is not an unusual value for the true mean breaking strength. Thus, we would recommend that the process is not out of control.

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Inferences Based on a Single Sample: Tests of Hypothesis

Chapter 6

6.2

The test statistic is used to decide whether or not to reject the null hypothesis in favor of the alternative hypothesis.

6.4

A Type I error is rejecting the null hypothesis when it is true. A Type II error is accepting the null hypothesis when it is false.

α = the probability of committing a Type I error. β = the probability of committing a Type II error. 6.6

We can compute a measure of reliability for rejecting the null hypothesis when it is true. This measure of reliability is the probability of rejecting the null hypothesis when it is true which is α. However, it is generally not possible to compute a measure of reliability for accepting the null hypothesis when it is false. We would have to compute the probability of accepting the null hypothesis when it is false, β, for every value of the parameter in the alternative hypothesis.

6.8

Let p = proportion of U.S. companies that have formal, written travel and entertainment policies for their employees. The null hypothesis would be: H0: p = .80

6.10

Let μ = average Libor rate for 3-month loans. Since many Western banks think that the reported average Libor rate (.054) is too high, they want to show that the average is less than .054. The appropriate hypotheses would be: H0: μ = .054 Ha: μ < .054

6.12

Let p = proportion of time the camera correctly detects liars. The null hypothesis would be: H0: p = .75

6.14

a.

A Type I error would be concluding the proposed user is unauthorized when, in fact, the proposed user is authorized. A Type II error would be concluding the proposed user is authorized when, in fact, the proposed user is unauthorized. In this case, a more serious error would be a Type II error. One would not want to conclude that the proposed user is authorized when he/she is not.

b.

The Type I error rate is 1%. This means that the probability of concluding the proposed user is unauthorized when, in fact, the proposed user is authorized is .01.

Inferences Based on a Single Sample: Tests of Hypothesis

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The Type II error rate is .00025%. This means that the probability of concluding the proposed user is authorized when, in fact, the proposed user is unauthorized is .0000025. c.

The Type I error rate is .01%. This means that the probability of concluding the proposed user is unauthorized when, in fact, the proposed user is authorized is .0001. The Type II error rate is .005%. This means that the probability of concluding the proposed user is authorized when, in fact, the proposed user is unauthorized is .00005.

6.16

6.18

a.

The null hypothesis is: Ho: There is no intrusion.

b.

The alternative hypothesis is: Ha: There is an intrusion.

c.

α = P(warning | no intrusion) =

1 = .001 . 1000

β = P(no warning | intrusion) =

500 = .5 . 1000

a.

The decision rule is to reject H0 if x > 270. Recall that z=

x − μ0

σx

Therefore, reject H0 if x > 270 can be written reject H0 if z >

x − μ0

σx 270 − 255 z> 63/ 81 z > 2.14

The decision rule in terms of z is to reject H0 if z > 2.14. b.

6.20

a.

P(z > 2.14) = .5 − P(0 < z < 2.14) = .5 − .4838 = .0162 H0: μ = .36 Ha: μ < .36

The test statistic is z =

x − μ0

σx

=

.323 − .36 .034 / 64

= −1.61

The rejection region requires α = .10 in the lower tail of the z-distribution. From Table IV, Appendix B, z.10 = 1.28. The rejection region is z < −1.28.

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Since the observed value of the test statistic falls in the rejection region (z = −1.61 < −1.28), H0 is rejected. There is sufficient evidence to indicate the mean is less than .36 at α = .10. b.

H0: μ = .36 Ha: μ ≠ .36

The test statistic is z = −1.61 (see part a). The rejection region requires α/2 = .10/2 = .05 in the each tail of the z-distribution. From Table IV, Appendix B, z.05 = 1.645. The rejection region is z < −1.645 or z > 1.645. Since the observed value of the test statistic does not fall in the rejection region (z = −1.61 1.96.

e.

Since the observed value of the test statistic falls in the rejection region (z = 2.48 > 1.96), Ho is rejected. There is sufficient evidence to indicate the mean July, 2006 dealer price of the Toyota Prius differs from $25,000 at α = .05.

σx

=

25, 476.69 − 25,000 = 2.48 2, 429.8267 160

c.

Inferences Based on a Single Sample: Tests of Hypothesis

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6.24

a.

A Type I error is rejecting H0 when H0 is true. In this case, we would conclude that the mean number of carats per diamond is different from .6 when, in fact, it is equal to .6. A Type II error is accepting H0 when H0 is false. In this case, we would conclude that the mean number of carats per diamond is equal to .6 when, in fact, it is different from .6.

b.

From Exercise 5.18, the random sample of 30 diamonds yielded x = .691 and s = .262. Let μ = mean number of carats per diamond. To determine if the mean number of carats per diamond is different from .6, we test: H0: μ = .6 Ha: μ ≠ .6 The test statistic is z =

x − μ0

σx

=

.691 − .6 .262

30

= 1.90

The rejection region requires α/2 = .05/2 = .025 in each tail of the z-distribution. From Table IV, Appendix B, z.025 = 1.96. The rejection region is z > 1.96 or z < −1.96. Since the observed value of the test statistic does not fall in the rejection region (z = 1.90 >/ 1.96), H0 is not rejected. There is insufficient evidence to indicate the mean number of carats per diamond is different from .6 carats at α = .05. c.

When α is changed, H0, Ha, and the test statistic remain the same. The rejection region requires α/2 = .10/2 = .05 in each tail of the z-distribution. From Table IV, Appendix B, z.05 = 1.645. The rejection region is z > 1.645 or z < −1.645. Since the observed value of the test statistic falls in the rejection region (z = 1.90 > 1.645), H0 is rejected. There is sufficient evidence to indicate the mean number of carats per diamond is different from .6 carats at α = .10.

d.

6.26

When the value of α changes, the decision can also change. Thus, it is very important to include the level of α used in all decisions.

Using MINITAB, the descriptive statistics are: Descriptive Statistics: GASTURBINE Variable GASTURBINE

N 67

N* 0

Variable GASTURBINE

Maximum 16243

Mean 11066

SE Mean 195

StDev 1595

Minimum 8714

Q1 9918

Median 10656

Q3 11842

To determine if the mean heat rate of gas turbines augmented with high pressure inlet fogging exceeds 10,000 kJ/kWh, we test: H0: μ = 10,000 H0: μ > 10,000

164

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x − μo

The test statistic is z =

σx

=

11,066 − 10,000 = 5.47 1,595 67

The rejection region requires α = .05 in the upper tail of the z-distribution. From Table IV, Appendix B, z.05 = 1.645. The rejection region is z > 1.645. Since the observed value of the test statistics falls in the rejection region (z = 5.47 > 1.645), H0 is rejected. There is sufficient evidence to indicate the true mean heat rate of gas turbines augmented with high pressure inlet fogging exceeds 10,000 kJ/kWh at α = .05. 6.28

a.

Let μ = average full-service fee (in thousands of dollars) of U.S. funeral homes in 2006. To determine if the average full-service fee exceeds $6,500, we test: H0: μ = 6.50 Ha: μ > 6.50

b.

Using MINTAB, the output is: Descriptive Statistics: FUNERAL Variable Fee Variable Fee

N 36

Mean 6.819 Minimum 5.200

Median 6.600 Maximum 11.600

StDev 1.265 Q1 6.025

SE Mean 0.211 Q3 7.400

H0: μ = 6.50 Ha: μ > 6.50 The test statistic is z =

x − μ0

σx

=

6.819 − 6.50 = 1.51 1.265 36

The rejection region requires α = .05 in the upper tail of the z-distribution. From Table IV, Appendix B, z.05 = 1.645. The rejection region is z > 1.645. Since the observed value of the test statistic does not fall in the rejection region (z = 1.51 >/ 1.645), H0 is not rejected. There is insufficient evidence to indicate the true mean full-service fee of U.S. funeral homes in 2006 exceeds $6,500 at α = .05. c.

No. Since the sample size (n = 36) is greater than 30, the Central Limit Theorem applies. The distribution of x is approximately normal regardless of the population distribution.

Inferences Based on a Single Sample: Tests of Hypothesis

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6.30

a.

To determine if the sample data refute the manufacturer's claim, we test:

H0: μ = 10 Ha: μ < 10 b.

A Type I error is concluding the mean number of solder joints inspected per second is less than 10 when, in fact, it is 10 or more. A Type II error is concluding the mean number of solder joints inspected per second is at least 10 when, in fact, it is less than 10.

c.

Using MINITAB, the descriptive statistics are:

Descriptive Statistics: PCB Variable PCB

N 48

Mean 9.292

Median 9.000

TrMean 9.432

Variable PCB

Minimum 0.000

Maximum 13.000

Q1 9.000

Q3 10.000

StDev 2.103

SE Mean 0.304

H0: μ = 10 Ha: μ < 10 The test statistic is z =

x − μ0

σx

=

9.292 − 10 2.103 / 48

= −2.33

The rejection region requires α = .05 in the lower tail of the z-distribution. From Table IV, Appendix B, z.05 = 1.645. The rejection region is z < −1.645. Since the observed value of the test statistic falls in the rejection region (z = −2.33 < −1.645), H0 is rejected. There is sufficient evidence to indicate the mean number of inspections per second is less than 10 at α = .05. 6.32

166

We will reject H0 if the p-value < α. a.

.06 75 at α = .10.

d.

For this two-tailed test, the p-value = .1032. Since the p-value = .1032 > α = .01, H0 is not rejected. There is insufficient evidence to indicate μ ≠ 75 at α = .01.

6.40

The p-value is p = 0.014. The probability of observing a test statistic of t = 2.48 or anything more unusual if μ = 25,000 is 0.014. Since p = 0.014 is so small, we would reject H0. There is sufficient evidence to indicate the mean prices for hybrid Toyota Prius cars is different than $25,000 for any value of α > .014.

6.42

From the printout, the p-value = .000. Since the p-value = .000 < α = .01, H0 is rejected. There is sufficient evidence to indicate that the true population mean weight of plastic golf tees is different from .250 at α = .01.

Inferences Based on a Single Sample: Tests of Hypothesis

167

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6.44

a.

z=

x − μo

σx

=

52.3 − 51 7.1

= 1.29

50

The p-value is p = P ( z ≥ 1.29)+P ( z ≤ −1.29) = (.5 − .4015) + (.5 − .4015) = .1970 . (Using Table IV, Appendix B.)

b.

The p-value is p = P ( z ≥ 1.29)= (.5 − .4015) = .0985 . (Using Table IV, Appendix B.)

c.

z=

x − μo

σx

=

52.3 − 51 10.4

50

= 0.88

The p-value is p = P ( z ≥ 0.88)+P ( z ≤ −0.88) = (.5 − .3106) + (.5 − .3106) = .3788 . (Using Table IV, Appendix B.) d.

In part a, in order to reject H0, α would have to be greater than .1970. In part b, in order to reject H0, α would have to be greater than .0985. In part c, in order to reject H0, α would have to be greater than .3788.

e.

For a two-tailed test, α/2 = .01/2 = .005. From Table IV, Appendix B, z.005 = 2.58. z=

x − μo

σx

⇒ 2.58 =

52.3 − 51 s

50

⇒ 2.58

s 50

= 52.3 − 51 ⇒ .3649s = 1.3 ⇒ s = 3.56

For a one-tailed test, α = .01. From Table IV, Appendix B, z.01 = 2.33. z=

6.46

a.

z=

x − μo

σx

x − μ0

σx

⇒ 2.33 =

=

52.3 − 51 s

10.2 − 0

50

⇒ 2.33

s 50

= 52.3 − 51 ⇒ .3295s = 1.3 ⇒ s = 3.95

= 2.30

31.3 / 50

b.

For this two-sided test, the p-value = P(z ≥ 2.30) + P(z ≤ −2.30) = (.5 − .4893) + (.5 − .4893) = .0214. Since this value is so small, there is evidence to reject H0. There is sufficient evidence to indicate the mean level of feminization is different from 0% for any value of α > .0214.

c.

z=

x - μ0

σx

=

15.0 − 0

= 4.23

25.1/ 50

For this two-sided test, the p-value = P(z ≥ 4.23) + P(z ≤ −4.23) ≈ (.5 − .5) + (.5 − .5) = 0. Since this value is so small, there is evidence to reject H0. There is sufficient evidence to indicate the mean level of feminization is different from 0% for any value of α > 0.0.

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6.48

6.50

a.

P(t > 1.440) = .10 (Using Table VI, Appendix B, with df = 6)

b.

P(t < −1.782) = .05 (Using Table VI, Appendix B, with df = 12)

c.

P(t < −2.060) + P(t > 2.060) = .025 + .025 = .05 (Using Table VI, Appendix B, with df = 25)

d.

The probability of a Type I error is computed above for each of the parts.

a.

H0: μ = 6 Ha: μ < 6 The test statistic is t =

x − μ0 s/ n

=

4.8 − 6 1.3/ 5

= −2.064

The necessary assumption is that the population is normal. The rejection region requires α = .05 in the lower tail of the t-distribution with df = n − 1 = 5 − 1 = 4. From Table VI, Appendix B, t.05 = 2.132. The rejection region is t < −2.132. Since the observed value of the test statistic does not fall in the rejection region (t = −2.064 2.776. Since the observed value of the test statistic does not fall in the rejection region (t = −2.064 2.447 or t < −2.447. Since the value of the test statistic does not fall in the rejection region (t = .92 >/ 2.447), H0 is not rejected. There is insufficient evidence to indicate the mean consumption rate of salad dressings in the Southeastern U.S. is different than the mean national consumption rate at α = .05.

6.56

d.

The observed significance level is p-value = P(t ≥ .92) + P(t ≤ −.92). Since we did not reject H0 in part c, we know that the p-value must be greater than .05. Using Table VI, Appendix B, with df = n − 1 = 7 − 1 = 6, p-value = P(t ≥ .92) + P(t ≤ −.92) > .1 + .1 = .2 Thus, with this table, we only know that the p-value is greater than .2.

a.

To determine if the mean repellency percentage of the new mosquito repellent is less than 95, we test:

H0: μ = 95 Ha: μ < 95 The test statistic is t =

x − μ0 s/ n

=

83 − 95 15 / 5

= −1.79

The rejection region requires α = .10 in the lower tail of the t distribution. From Table VI, Appendix B, with df = n − 1 = 5 − 1 = 4, t.10 = 1.533. The rejection region is t < −1.533. Since the observed value of the test statistic falls in the rejection region (t = −1.79 < −1.533), H0 is rejected. There is sufficient evidence to indicate that the true mean repellency percentage of the new mosquito repellent is less than 95 at α = .10.

6.58

b.

We must assume that the population of percent repellencies is normally distributed.

a.

Using MINITAB, the descriptive statistics are:

Descriptive Statistics: Plants Variable Plants

N 20

Mean 4.000

Median 3.500

TrMean 3.667

Variable Plants

Minimum 1.000

Maximum 13.000

Q1 1.250

Q3 5.000

StDev 3.061

SE Mean 0.684

Let μ = mean number of active nuclear power plants operating in all states. To determine if the mean number of active nuclear power plants operating in all states exceeds 3, we test:

H0: μ = 3 Ha: μ > 3

Inferences Based on a Single Sample: Tests of Hypothesis

171

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The test statistic is t =

x − μo s

n

=

4−3 3.061

20

= 1.46

The rejection region requires α = .10 in the upper tail of the t-distribution with df = n – 1 = 20 – 1 = 19. From Table VI, Appendix B, t.10 = 1.328. The rejection region is t > 1.328. Since the observed value of the test statistic falls in the rejection region (t = 1.46 > 1.328), H0 is rejected. There is sufficient evidence to indicate the mean number of active nuclear power plants operating in all states exceeds 3 at α = .10. b.

We will look at the 4 methods for determining if the data are normal. First, we will look at a histogram of the data. Using MINITAB, the histogram of the number of power plants is:

7 6

Frequency

5 4 3 2 1 0 2

4

6

8

10

12

14

Plants

From the histogram, the data appear to be skewed to the right. This indicates that the data may not be normal. Next, we look at the intervals x ± s, x ± 2 s, x ± 3s . If the proportions of observations falling in each interval are approximately .68, .95, and 1.00, then the data are approximately normal.

x ± s ⇒ 4 ± 3.061 ⇒ (.939, 7.061) 18 of the 20 values fall in this interval. The proportion is .90. This is much greater than the .68 we would expect if the data were normal. x ± 2s ⇒ 4 ± 2(3.061) ⇒ 4 ± 6.122 ⇒ (−2.122, 10.122) 19 of the 20 values fall in this interval. The proportion is .95. This is the same as the .95 we would expect if the data were normal. x ± 3s ⇒ 4 ± 3(3.061) ⇒ 4 ± 9.183 ⇒ (−5.183, 13.183) 20 of the 20 values fall in this interval. The proportion is 1.000. This is equal to the 1.00 we would expect if the data were normal.

172

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From this method, it appears that the data are not normal. Next, we look at the ratio of the IQR to s. IQR = QU – QL = 5.00 – 1.25 = 3.75.

IQR 3.75 = = 1.22 This is close to the 1.3 we would expect if the data were normal. s 3.061 This method indicates the data may be normal. Finally, using MINITAB, the normal probability plot is: Normal Probability Plot for Plants ML Estimates - 95% CI

99

ML Estimates

95

Mean

4

StDev

2.98329

90

Goodness of Fit

Percent

80

AD*

70 60 50

1.298

40 30 20 10 5

1 -5

0

5

10

Data

Since the data do not form a straight line, the data are not normal. From 3 of the 4 different methods, the indications are that the number of power plants data are not normal. c.

The two largest values are 9 and 13. The two lowest values are 1 and 1. Using MINITAB with the data deleted yields the descriptive statistics:

Descriptive Statistics: Plants2 Variable Plants2

N 16

Mean 3.500

Median 3.500

TrMean 3.429

Variable Plants2

Minimum 1.000

Maximum 7.000

Q1 2.000

Q3 5.000

StDev 1.826

SE Mean 0.456

To determine if the mean number of active nuclear power plants operating in all states exceeds 3 (using the reduced data set), we test: H0: μ = 3 Ha: μ > 3

Inferences Based on a Single Sample: Tests of Hypothesis

173

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The test statistic is t =

x − μo s

n

=

3.5 − 3 1.826

16

= 1.10

The rejection region requires α = .10 in the upper tail of the t-distribution with df = n – 1 = 16 – 1 = 15. From Table VI, Appendix B, t.10 = 1.341. The rejection region is t > 1.341. Since the observed value of the test statistic does not fall in the rejection region (t = 1.10 >/ 1.341), H0 is not rejected. There is insufficient evidence to indicate the mean number of active nuclear power plants operating in all states exceeds 3 at α = .10. By eliminating the top two and bottom two observations, we have changed the decision from rejecting H0 to not rejecting H0. d.

6.60

It is very dangerous to eliminate data points to satisfy assumptions. The data may, in fact, not be normal. By eliminating data points, one has changed the kind of data that come from the parent population. Thus, incorrect decisions could be made.

Using MINITAB, the descriptive statistics for the 2 plants are: Descriptive Statistics: AL1, AL2 Variable aximum AL1 AL2

N

N*

Mean

SE Mean

StDev

Minimum

Q1

Median

Q3

2 2

0 0

0.00750 0.0700

0.00250 0.0200

0.00354 0.0283

0.00500 0.0500

* *

0.00750 0.0700

* *

M 0.01000 0.0900

To determine if plant 1 is violating the OSHA standard, we test: H0: μ = .004 Ha: μ > .004 The test statistic is t =

x − μo s

n

=

.0075 − .004 .00354

2

= 1.40

Since no α level was given, we will use α = .05. The rejection region requires α = .05 in the upper tail of the t-distribution with df = n – 1 = 2 – 1 = 1. From Table VI, Appendix B, t.05 = 6.314. The rejection region is t > 6.314. Since the observed value of the test statistic does not fall in the rejection region (t = 1.40 >/ 6.314), H0 is not rejected. There is insufficient evidence to indicate the OSHA standard is violated by plant 1 at α = .05. To determine if plant 2 is violating the OSHA standard, we test: H0: μ = .004 Ha: μ > .004 The test statistic is t =

174

x − μo s

n

=

.07 − .004 .0283

2

= 3.30

Chapter 6

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Since no α level was given, we will use α = .05. The rejection region requires α = .05 in the upper tail of the t-distribution with df = n – 1 = 2 – 1 = 1. From Table VI, Appendix B, t.05 = 6.314. The rejection region is t > 6.314. Since the observed value of the test statistic does not fall in the rejection region (t = 3.30 >/ 6.314), H0 is not rejected. There is insufficient evidence to indicate the OSHA standard is violated by plant 2 at α = .05. 6.62

b.

First, check to see if n is large enough. p0 ± 3σ pˆ ⇒ p0 ± 3

p0 q0 (.70)(.30) ⇒ .70 ± 3 ⇒ .70 ± .14 ⇒ (.56, .84) 100 n

Since the interval lies within the interval (0, 1), the normal approximation will be adequate. H0: p = .70 Ha: p < .70 The test statistic is z =

pˆ − p0

σ pˆ

=

pˆ − p0 p0 q0 n

=

.63 − .70 = −1.53 .70(.30) 100

The rejection region requires α = .05 in the lower tail of the z-distribution. From Table IV, Appendix B, z.05 = 1.645. The rejection region is z < −1.645. Since the observed value of the test statistic does not fall in the rejection region (−1.53 2.58 or z < −2.58.

e.

Since the observed value of the test statistic does not fall in the rejection region (z = −2.16 .02 c.

The test statistic is z =

pˆ − po po qo n

=

.083 − .02 .02(.98) 1000

= 14.23

The rejection region requires α = .05 in the upper tail of the z-distribution. From Table IV, Appendix B, z.05 = 1.645. The rejection region is z > 1.645. d.

Since the observed value of the test statistic falls in the rejection region (z = 14.23 > 1.645), H0 is rejected. There is sufficient evidence to indicate that the true proportion of items scanned at California Wal-Mart stores with the wrong price exceeds the 2% NIST standard at α = .05. This means that the proportion of items with wrong prices at California Wal-Mart stores is much higher than what is allowed.

e.

In order for the inference to be valid, the sampling distribution of pˆ must be approximately normal. We check this assumption: po ± 3σ pˆ ⇒ po ± 3

po qo .02(.98) ⇒ .02 ± 3 ⇒ .02 ± .013 ⇒ (.007, .033) n 1000

Since the above interval falls completely in the interval (0, 1), the normal distribution will be adequate.

176

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6.70

a.

Let p = proportion of vacation-home owners who are minorities in 2003. pˆ =

x 46 = = .111 n 416

To determine if the percentage of vacation-home owners in 2006 who are minorities is larger than 6%, we test: H0: p = .06 Ha: p > .06 The test statistic is z =

pˆ − po po qo n

=

.111 − .06 = 4.38 .06(.94) 416

The rejection region requires α = .01 in the upper tail of the z-distribution. From Table IV, Appendix B, z.01 = 2.33. The rejection region is z > 2.33. Since the observed value of the test statistic falls in the rejection region (z = 4.38 > 2.33), H0 is rejected. There is sufficient evidence to indicate that the true percentage of vacation-home owners in 2006 who are minorities is larger than 6% at α = .01. b.

6.72

Since the return rate of the questionnaire was so small compared to the number sent out, one should be very skeptical of the results. It would be fairly unusual that the sample of returned questionnaires would be representative of the entire population.

Let p = proportion of firms in violation of the new 4-day rule for reporting material changes. pˆ =

x 23 = = .050 n 462

To determine if the percentage of firms in violation of the new 4-day rule for reporting material changes is less than 10%, we test: H0: p = .10 Ha: p < .10 The test statistic is z =

pˆ − po po qo n

=

.050 − .10 = −3.58 .10(.90) 462

The rejection region requires α = .01 in the lower tail of the z-distribution. From Table IV, Appendix B, z.01 = 2.33. The rejection region is z < −2.33. Since the observed value of the test statistic falls in the rejection region (z = −3.58 < −2.33), Ho is rejected. There is sufficient evidence to indicate that the true percentage of firms in violation of the new 4-day rule for reporting material changes is less than 10% at α = .01.

Inferences Based on a Single Sample: Tests of Hypothesis

177

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6.74

Let p = proportion of patients taking the pill who reported an improved condition. First we check to see if the normal approximation is adequate: p0 ± 3σ pˆ ⇒ p0 ± 3

p0 q0 .5(.5) ⇒± 3 ⇒ .5 ± .018 ⇒ (.482, .518) 7000 n

Since the interval falls completely in the interval (0, 1), the normal distribution will be adequate. To determine if there really is a placebo effect at the clinic, we test: H0: p = .5 Ha: p > .5 The test statistic is z =

pˆ − p0 p0 q0 n

=

.7 − .5 = 33.47 .5(.5) 7000

The rejection region requires α = .05 in the upper tail of the z distribution. From Table IV, Appendix B, z.05 = 1.645. The rejection region is z > 1.645. Since the observed value of the test statistic falls in the rejection region (z = 33.47 > 1.645), H0 is rejected. There is sufficient evidence to indicate that there really is a placebo effect at the clinic at α = .05. 6.76

a.

The power of a test increases when: 1. 2. 3.

b.

178

The distance between the null and alternative values of μ increases. The value of α increases. The sample size increases.

The power of a test is equal to 1 − β. As β increases, the power decreases.

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6.78

6.80

From Exercise 6.77 we want to test H0: μ = 500 against Ha: μ > 500 using α = .05, σ = 100, n = 25, and x = 532.9. ⎛

532.9 − 575 ⎞ ⎟ = P(z < −2.11) 100 / 25 ⎠ = .5 − .4826 = .0174

a.

β = P( x0 < 532.9 when μ = 575) = P ⎜ z <

b.

Power = 1 − β = 1 − .0174 = .9826

c.

In Exercise 6.77, β = .1949 and the power is .8051. The value of β has decreased in this exercise since μ = 575 is further from the hypothesized value than μ = 550. As a result, the power of the test in this exercise has increased (when β decreases, the power of the test increases).

a.

From Exercise 6.79, we want to test H0: μ = 75 against Ha: μ < 75 using α = .10, σ = 15, n = 49, and x = 72.257.

⎝

If μ = 74,

⎛

β = P( x0 > 72.257 when μ = 74) = P ⎜ z > ⎝

If μ = 72,

⎛

μ = P( x0 > 72.257 when μ = 72) = P ⎜ z > ⎝

If μ = 70,

72.257 − 74 ⎞ ⎟ = P(z > −.81) 15 / 49 ⎠ = .5 + .2910 = .7910 72.257 − 72 ⎞ ⎟ = P(z > .12) 15 / 49 ⎠ = .5 − .0478 = .4522

β = P( x0 > 72.257 when μ = 70) = .1469 (Refer to Exercise 6.69, part c.) If μ = 68,

⎛

β = P( x0 > 72.257 when μ = 68) = P ⎜ z > ⎝

If μ = 66,

⎛

β = P( x0 > 72.257 when μ = 66) = P ⎜ z > ⎝

In summary,

μ β

74 .7910

72 .4522

70 .1469

Inferences Based on a Single Sample: Tests of Hypothesis

72.257 − 68 ⎞ ⎟ = P(z > 1.99) 15 / 49 ⎠ = .5 − .4767 = .0233 72.257 − 66 ⎞ ⎟ = P(z > 2.92) 15 / 49 ⎠ = .5 − .4982 = .0018

68 .0233

66 .0018

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b.

c.

Looking at the graph, β is approximately .62 when μ = .73.

d.

Power = 1 − β Therefore, 74 μ .7910 β Power .2090

72 .4522 .5478

70 .1469 .8531

68 .0233 .9767

66 .0018 .9982

The power curve starts out close to 1 when μ = 66 and decreases as μ increases, while the β curve is close to 0 when μ = 66 and increases as μ increases.

6.82

e.

As the distance between the true mean μ and the null hypothesized mean μ0 increases, β decreases and the power increases. We can also see that as β increases, the power decreases.

a.

To determine if the mean size of California homes exceeds the national average, we test: H0: μ = 2230 Ha: μ > 2230

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The test statistic is z =

x − μ0

σx

=

2347 − 2230 = 4.55 257 / 100

The rejection region requires α = .01 in the upper tail of the z-distribution. From Table IV, Appendix B, z.05 = 2.33. The rejection region is z > 2.33. Since the observed value of the test statistic falls in the rejection region (z = 4.55 > 2.33), H0 is rejected. There is sufficient evidence to indicate the mean size of California homes exceeds the national average at α = .01. b.

To compute the power, we must first set up the rejection regions in terms of . ⎛ s ⎞ ⎛ 257 ⎞ x0 = μ0 + zα σ x ≈ μ0 + 2.33 ⎜ ⎟ = 2, 230 + 2.33 ⎜ ⎟ = 2,289.88 ⎝ n⎠ ⎝ 100 ⎠

We would reject H0 if x > 2,289.88 The power of the test when μ = 2,330 would be: ⎛ x − μa ⎞ 2, 289.88 − 2,330 ⎞ ⎛ Power = P( x > 2289.88⏐μ = 2,330) = P ⎜ z > 0 ⎟ = P⎜ z > ⎟ σx ⎠ 257 / 100 ⎝ ⎠ ⎝ = P(z > −1.56) = .5 + .4406 = .9406

c.

The power of the test when μ = 2,280 would be: ⎛ x − μa Power = P( > 2289.88⏐μ = 2,280) = P ⎜ z > 0 σx ⎝ = P(z > 0.38) = .5 − .1480 = .3520

6.84

a.

⎞ 2, 289.88 − 2, 280 ⎞ ⎛ ⎟ = P⎜ z > ⎟ 257 / 100 ⎝ ⎠ ⎠

To determine if the mean mpg for 2006 Honda Civic autos is greater than 38 mpg, we test: H0: μ = 38 Ha: μ > 38

b.

The test statistic is z =

x − μ0

σx

=

40.3 − 38 = 2.16 6.4 / 36

The rejection region requires α = .05 in the upper tail of the z-distribution. From Table IV, Appendix B, z.05 = 1.645. The rejection region is z > 1.645. Since the observed value of the test statistic falls in the rejection region (z = 2.16 > 1.645), H0 is rejected. There is sufficient evidence to indicate that the mean mpg for 2006 Honda Civic autos is greater than 38 mpg at α = .05. We must assume that the sample was a random sample.

Inferences Based on a Single Sample: Tests of Hypothesis

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c.

First find: x0 = μ0 + zα σ x = μ0 + zα

Thus, x0 = 38 + 1.645

σ n

where zα = 1.645 from Table IV, Appendix B.

6.4 = 39.75 36

For μ = 38.5: 39.75 − 38.5 ⎞ ⎛ Power = P( x > 39.75│μ = 38.5) = P ⎜ z > ⎟ = P(z > 1.17) 6.4 / 36 ⎠ ⎝ For μ = 39:

= .5 − .3790 = .1210

39.75 − 39 ⎞ ⎛ Power = P( x > 39.75│μ = 39) = P ⎜ z > ⎟ = P(z > .70) 6.4 / 36 ⎠ ⎝ For μ = 39.5:

= .5 − .2580 = .2420

39.75 − 39.5 ⎞ ⎛ Power = P( x > 39.75│μ = 39.5) = P ⎜ z > ⎟ = P(z > .23 ) 6.4 / 36 ⎠ ⎝ For μ = 40:

= .5 − .0910 = .4090

39.75 − 40 ⎞ ⎛ Power = P( x > 39.75│μ = 40) = P ⎜ z > ⎟ = P(z > −.23) 6.4 / 36 ⎠ ⎝ For μ = 40.5:

= .5 + .0910 = .5910

39.75 − 40.5 ⎞ ⎛ Power = P( x > 39.75│μ = 40.5) = P ⎜ z > ⎟ = P(z > −.70) 6.4 / 36 ⎠ ⎝ = .5 + .2580 = .7580 d.

182

The plot is:

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e.

From the plot, the power is approximately .5. For μ = 39.75 : ⎛ 39.75 − 39.75 ⎞ Power = P( x > 39.75 | μ = 39.75) = P ⎜ z > ⎟ = P( z > 0) = .5 ⎜ 6.4 36 ⎟⎠ ⎝

f.

From the plot, the power is approximately 1. For μ = 43 : ⎛ 39.75 − 43 ⎞ Power = P( x > 39.75 | μ = 43) = P ⎜ z > ⎟ = P( z > −3.05) ⎜ 6.4 36 ⎟⎠ ⎝ = .5 + .4989 = .9989 If the true value of μ is 40, the approximate probability that the test will fail to reject H0 is 1 − .9989 = .0011.

6.86

Using Table VII, Appendix B: a.

For n = 12, df = n − 1 = 12 − 1 = 11 P(χ2 > χ 02 ) = .10 ⇒ χ 02 = 17.2750

b.

For n = 9, df = n − 1 = 9 − 1 = 8 P(χ2 > χ 02 ) = .05 ⇒ χ 02 = 15.5073

c.

For n = 5, df = n − 1 = 5 − 1 = 4 P(χ2 > χ 02 ) = .025 ⇒ χ 02 = 11.1433

6.88

a.

It would be necessary to assume that the population has a normal distribution.

b.

H0: σ2 = 1 Ha: σ2 > 1 The test statistic is χ2 =

(n − 1) s 2

σ

2 0

=

6(4.84) = 29.04 1

The rejection region requires α = .05 in the upper tail of the χ2 distribution with 2 = 12.5916. The rejection df = n − 1 = 7 − 1 = 6. From Table VII, Appendix B, χ.05 region is χ2 > 12.5916. Since the observed value of the test statistic falls in the rejection region (29.04 > 12.5916), H0 is rejected. There is sufficient evidence to indicate that the variance is greater than 1 at α = .05.

Inferences Based on a Single Sample: Tests of Hypothesis

183

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c.

H0: σ2 = 1 Ha: σ2 ≠ 1 (n − 1) s 2

The test statistic is χ2 =

σ

2 0

=

6(4.84) = 29.04 1

The rejection region requires α/2 = .025 in the upper tail of the χ2 distribution with 2 = 1.237347 and df = n − 1 = 7 − 1 = 6. From Table VII, Appendix B, χ.975 2 χ.025 = 14.4494. The rejection region is χ2 < 1.237347 or χ2 > 14.4494.

Since the observed value of the test statistic falls in the rejection region (29.04 > 14.4494), H0 is rejected. There is sufficient evidence to indicate that the variance is not equal to 1 at α = .05. 6.90

Some preliminary calculations are:

s2 =

∑x

(∑ x) −

2

n −1

n

2

=

302 7 = 7.9048 7 −1

176 −

To determine if σ2 < 1, we test: H0: σ2 = 1 Ha: σ2 < 1 The test statistic is χ2 =

(n − 1) s 2

σ

2 0

=

(7 − 1)7.9048 = 47.43 1

The rejection region requires α = .05 in the lower tail of the χ2 distribution with df = n − 1 = 7 2 = 1.63539. The rejection region is χ2 < 1.63539. − 1 = 6. From Table VII, Appendix B, χ.95 Since the observed value of the test statistic does not fall in the rejection region (χ2 = 47.43 1,5002

Inferences Based on a Single Sample: Tests of Hypothesis

185

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The test statistic is χ 2 =

( n − 1) s 2

σ o2

=

(67 − 1)1,5952 = 74.625 . 1,5002

The rejection region requires α = .05 in the upper tail of the χ2 distribution with 2 ≈ 85.95148. The rejection df = n – 1 = 67 – 1 = 66. From Table VII, Appendix B, χ.05 2 region is χ > 85.95148. Since the observed value of the test statistic does not fall in the rejection region (χ2 = 74.625 >/ 85.95148), H0 is not rejected. There is insufficient evidence to indicate the heat rates of the augmented gas turbine engine are more variable than the heat rates of the standard gas turbine engine at α = .05. 6.98

For a large sample test of hypothesis about a population mean, no assumptions are necessary because the Central Limit Theorem assures that the test statistic will be approximately normally distributed. For a small sample test of hypothesis about a population mean, we must assume that the population being sampled from is normal. The test statistic for the large sample test is the z statistic, and the test statistic for the small sample test is the t statistic.

6.100

The elements of the test of hypothesis that should be specified prior to analyzing the data are: null hypothesis, alternative hypothesis, and rejection region based on α.

6.102

α = P(Type I error) = P(rejecting H0 when it is true). Thus, if rejection of H0 would cause your firm to go out of business, you would want this probability or α to be small.

6.104

a.

H0: μ = 8.3 Ha: μ ≠ 8.3 The test statistic is z =

x − μ0

σx

=

8.2 − 8.3 .79 / 175

= −1.67

The rejection region requires α/2 = .05/2 = .025 in each tail of the z-distribution. From Table IV, Appendix B, z.025 = 1.96. The rejection region is z < −1.96 or z > 1.96. Since the observed value of the test statistic does not fall in the rejection region (−1.67 1.96.

186

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Since the observed value of the test statistic falls in the rejection region (−3.35 < −1.96), H0 is rejected. There is sufficient evidence to indicate that the mean is different from 8.4 at α = .05. c.

H0: σ = 1 Ha: σ ≠ 1

H0: σ2 = 1 or

Ha: σ2 ≠ 1

The test statistic is χ 2 =

(n − 1) s 2

σ 02

=

(175 − 1)(.79) 2 = 108.59 1

The rejection region requires α/2 = .05/2 = .025 in each tail of the χ 2 distribution with df 2 2 ≈ 129.561 and χ.975 ≈ = n – 1 = 175 – 1 = 174. From Table VII, Appendix B, χ.025

74.2219. The rejection region is χ 2 > 129.561 or χ 2 < 74.2219. Since the observed value of the test statistic does not fall in the rejection region ( χ 2 = 108.59 >/ 129.561 and χ 2 = 108.59 1.96. In terms of x , the rejection region would be:

z=

x − μ0

z=

x − μ0

σx

σx

⇒ 1.96 =

xU − 8.3 .79

⇒ −1.96 =

175

⇒ .117 = xU − 8.3 ⇒ xU = 8.417

xL − 8.3 .79

175

⇒ −.117 = xL − 8.3 ⇒ xL = 8.183

Based on x , the rejection region would be: Reject H0 if x < 8.183 or x > 8.417 The power of the test is the probability the test statistic falls in the rejection region, given the alternative hypothesis is true. In this case, we will let μa = 8.5. Power = P( x < 8.183 | μa = 8.5) + P( x > 8.417 | μa = 8.5) ⎛ ⎛ 8.183 − 8.5 ⎞ 8.417 − 8.5 ⎞ = P ⎜⎜ z < ⎟ + P ⎜⎜ z > ⎟ ⎟ .79 175 ⎠ .79 175 ⎟⎠ ⎝ ⎝ = P( z < −5.31) + P ( z > −1.39) = (.5 − .5) + (.5 + .4177) = .9177 (Using Table IV, Appendix B)

Inferences Based on a Single Sample: Tests of Hypothesis

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6.106

6.108

a.

The p-value = .1288 = P(t ≥ 1.174). Since the p-value is not very small, there is no evidence to reject H0 for α ≤ .10. There is no evidence to indicate the mean is greater than 10.

b.

We must assume that a random sample was selected from a population that is normally distributed.

c.

For the alternative hypothesis Ha: μ ≠ 10, the p-value is 2 times the p-value for the onetailed test. The p-value = 2(.1288) = .2576. There is no evidence to reject H0 for α ≤ .10. There is no evidence to indicate the mean is different from 10.

a.

If we wish to test the research hypothesis that the mean GHQ score for all unemployed men exceeds 10, we test: H0: μ = 10 Ha: μ > 10 This is a one-tailed test. We are only interested in rejecting H0 if the mean GHQ score for all unemployed men is greater than 10.

b.

The rejection region requires α = .05 in the upper tail of the z-distribution. From Table IV, Appendix B, z.05 = 1.645. The rejection region is z > 1.645.

c.

The test statistic is z =

x − μ0

σx

=

10.94 − 10.0 = 1.29 5.10 / 49

Since the observed value of the test statistic does not fall in the rejection region (z = 1.29 >/ 1.645), H0 is not rejected. There is insufficient evidence to indicate the mean GHQ score for all unemployed men is greater than 10 at α = .05. d.

The p-value is P(z ≥ 1.29) = .5 − .4015 = .0985. (Using Table IV, Appendix B) The probability of observing our test statistic or anything more unusual, given H0 is true, is .0985. Since this value is not less than α = .05, we do not reject H0. There is insufficient evidence to indicate the mean GHO score is greater than 10.

6.110

a.

The population parameter of interest is p = proportion of all television viewers with access to cable-TV who agree with the statement “Overall, I find the quality of news on cable networks to be better than news on the ABC, CBS, and NBC networks.

b.

pˆ =

c.

To determine if the true proportion of TV-viewers who find cable news to be better quality than network news differs from .50, we test:

x 248 = = .496 n 500

H0: p = .50 Ha: p ≠ .50

188

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d.

The test statistic is z =

pˆ − p0 p0 q0 n

=

.496 − .50 = −0.18 .50(.50) 500

The rejection region requires α/2 = .10/2 = .05 in each tail of the z-distribution. From Table IV, Appendix B, z.05 = 1.645. The rejection region is z > 1.645 or z < −1.645. Since the observed value of the test statistic does not fall in the rejection region (z = −0.18 1.645.

Inferences Based on a Single Sample: Tests of Hypothesis

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Since the observed value of the test statistic falls in the rejection region (z = 15.61 > 1.645), H0 is rejected. There is sufficient evidence to indicate that the percentage of truckers who suffer from sleep apnea differs from 25% at α = .05. b.

The observed significance level is the p-value and is: p-value = P(z ≥ 15.61) + P(z ≤ −15.61) ≈ (.5 − .5) + (.5 − .5) = 0 Since the p-value is so small, we would reject H0 for any reasonable value of α. There is sufficient evidence to indicate that the percentage of truckers who suffer from sleep apnea differs from 25%.

6.114

c.

The inference from a confidence interval and a test of hypothesis must agree because the same numbers are used in both if the same level of significance is used.

a.

Let p = proportion of shoppers using cents-off coupons. To determine if the proportion of shoppers using cents-off coupons exceeds .65, we test: H0: p = .65 Ha: p > .65 The test statistic is z =

pˆ − p0 p0 q0 n

=

.77 − .65 .65(.35) 1, 000

= 7.96

The rejection region requires α = .05 in the upper tail of the z-distribution. From Table IV, Appendix B, z.05 = 1.645. The rejection region is z > 1.645. Since the observed value of the test statistic falls in the rejection region (z = 7.96 > 1.645), H0 is rejected. There is sufficient evidence to indicate the proportion of shoppers using cents-off coupons exceeds .65 at α = .05. b.

The sample size is large enough if the interval does not include 0 or 1. p0 q0 .65(.35) ⇒ .65 ± 3 ⇒ .65 ± .045 ⇒ (.605, .695) n 1, 000 Since the interval falls completely in the interval (0, 1), the normal distribution will be adequate. p0 ± 3σ pˆ ⇒ p0 ± 3

c.

190

The p-value is p = P ( z ≥ 7.96) = (.5 − .5) ≈ .0 . (Using Table IV, Appendix B.) Since the p-value is smaller than α = .05, H0 is rejected. There is sufficient evidence to indicate the proportion of shoppers using cents-off coupons exceeds .65 at α = .05.

Chapter 6

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6.116

Using MINITAB, the descriptive statistics are: Descriptive Statistics: Tunnel Variable Tunnel

N 10

Mean 989.8

Median 970.5

TrMean 987.9

Variable Tunnel

Minimum 735.0

Maximum 1260.0

Q1 862.5

Q3 1096.8

StDev 160.7

SE Mean 50.8

To determine whether peak hour pricing succeeded in reducing the average number of vehicles attempting to use the Lincoln Tunnel during the peak rush hour, we test: H0: μ = 1,220 Ha: μ < 1,220 The test statistic is t =

x − μ0 s/ n

=

989.8 − 1, 220 160.7 / 10

= −4.53

Since no α is given, we will use α = .05. The rejection region requires α = .05 in the lower tail of the t-distribution with df = n − 1 = 10 − 1 = 9. From Table VI, Appendix B, t.05 = 1.833. The rejection region is t < −1.833. Since the observed value of the test statistic falls in the rejection region (t = −4.53 < −1.833), H0 is rejected. There is sufficient evidence to indicate that peak hour pricing succeeded in reducing the average number of vehicles attempting to use the Lincoln Tunnel during the peak rush hour at α = .05. 6.118

a.

To determine if the true mean number of pecks at the blue string is less than 7.5, we test: H0: μ = 7.5 Ha: μ < 7.5 The test statistic is z =

x − μ0

σx

=

1.13 − 7.5 2.21

72

= −24.46

The rejection region requires α = .01 in the lower tail of the z-distribution. From Table IV, Appendix B, z.01 = 2.33. The rejection region is z < −2.33. Since the observed value of the test statistic falls in the rejection region (z = −24.46 < −2.33), H0 is rejected. There is sufficient evidence to indicate the true mean number of pecks at the blue string is less than 7.5 at α = .01.

b.

From Exercise 5.96, the 99% confidence interval is (.46, 1.80). Since the hypothesized value of the mean (μ = 7.5) does not fall in the confidence interval, it is not a likely candidate for the true value of the mean. Thus, you would reject it. This agrees with the conclusion in part a.

Inferences Based on a Single Sample: Tests of Hypothesis

191

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6.120

a.

pˆ = 24/40 = .6 To determine if the proportion of shoplifters turned over to police is greater than .5, we test: H0: p = .5 Ha: p > .5 The test statistic is z =

pˆ − p0 p0 q0 n

=

.6 − .5 .5(.5) 40

= 1.26

The rejection region requires α = .05 in the upper tail of the z-distribution. From Table IV, Appendix B, z.05 = 1.645. The rejection region is z > 1.645. Since the observed value of the test statistic does not fall in the rejection region (z = 1.26 >/ 1.645), H0 is not rejected. There is insufficient evidence to indicate the proportion of shoplifters turned over to police is greater than .5 at α = .05. b.

To determine if the normal approximation is appropriate, we check: p0 ± 3σ pˆ ± 3

p0 q0 (.5)(.5) ≈ .5 ± 3 ⇒ .5 ± .237 ⇒ (.263, .737) n 40

Since the interval falls completely in the interval (0, 1), the normal distribution will be adequate. c.

The observed significance level of the test is p-value = P(z ≥ 1.26) = .5 − .3962 = .1038. The probability of observing the value of our test statistic or anything more unusual if the true value of p is .5 is .4038. Since this p-value is so large, there is no evidence to reject H0. There is no evidence to indicate the true proportion of shoplifters turned over to police is greater than .5.

6.122

d.

Any value of α that is greater than the p-value would lead one to reject H0. Thus, for this problem, we would reject H0 for any value of α > .1038.

a.

To determine whether the mean profit change for restaurants with frequency programs is greater than $1047.34, we test: H0: μ = 1047.34 Ha: μ > 1047.34

b.

Some preliminary calculations are: x =

192

∑ x = 30,113.17 n

12

= 2,509.43

Chapter 6

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(∑ x) −

2

30,113.17 2 n 12 s2 = = 4,619,331.955 = n −1 12 − 1 s = 4,619,331.955 = 2149.2631

∑x

2

The test statistic is t =

126,379,568.8 −

x − μ0 s/ n

=

2509.43 − 1047.34 2149.2631/ 12

= 2.36

The rejection region requires α = .05 in the upper tail of the t-distribution with df = n − 1 = 12 − 1 = 11. From Table VI, Appendix B, t.05 = 1.796. The rejection region is t > 1.796. Since the observed value of the test statistic falls in the rejection region (t = 2.36 > 1.796), H0 is rejected. There is sufficient evidence to indicate the mean profit change for restaurants with frequency programs is greater than $1047.34 for α = .05. It appears that the frequency program would be profitable for the company if adopted nationwide. 6.124

a.

A Type II error would be concluding the mean amount of PCB in the air is less than or equal to 3 parts per million when, in fact, it is more than 3 parts per million.

b.

From Exercise 6.123, z =

x0 − μ

σ/ n

⇒ x0 = z

σ n

.5 +3 50 ⇒ x0 = 3.165

+ μ0 ⇒ x0 = 2.33

⎛ ⎞ ⎜ 3.165 − 3.1 ⎟ ⎟ = P(z ≤ .92) = .5 + .3212 = .8212 For μ = 3.1, β = P( x ≤ 3.165) = P ⎜ z ≤ .5 ⎜ ⎟ ⎜ ⎟ 50 ⎝ ⎠ (from Table IV, Appendix B) c.

Power = 1 − β = 1 − .8212 = .1788

d.

⎛ ⎞ ⎜ 3.165 − 3.2 ⎟ ⎟ = P(z ≤ −.49) = .5 − .1879 = .3121 For μ = 3.2, β = P( x ≤ 3.165) = P ⎜ z ≤ .5 ⎜ ⎟ ⎜ ⎟ 50 ⎝ ⎠ Power = 1 − β = 1 − .3121 = .6879 As the plant's mean PCB departs further from 3, the power increases.

Inferences Based on a Single Sample: Tests of Hypothesis

193

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6.126

a.

Some preliminary calculations: x =

s2 = s=

∑ x = 79.93 n

∑x

5

2

(∑ x) −

= 15.986 2

n = n −1 .00043 = .0207

1, 277.7627 − 5 −1

79.932 5 = .00043

To determine if the mean measurement differs from 16.01, we test: H0: μ = 16.01 Ha: μ ≠ 16.01 The test statistic is t =

x − μ0 s/ n

=

15,986 − 16.01 .0207 / 5

= −2.59

The rejection region requires α/2 = .05/2 = .025 in each tail of the t-distribution with df = n − 1 = 5 − 1 = 4. From Table VI, Appendix B, t.025 = 2.776. The rejection region is t < −2.776 or t > 2.776. Since the observed value of the test statistic does not fall in the rejection region (t = −2.59 .012 The test statistic is χ 2 =

( n − 1) s 2

σ o2

=

(5 − 1).0207 2 = 16.0684 . .012

The rejection region requires α = .05 in the upper tail of the χ2 distribution with df = n – 1 = 5 – 1 = 4. From Table VII, Appendix B, χ .205 = 9.48773. The rejection region is χ2 > 9.48773. Since the observed value of the test statistic falls in the rejection region (χ2 = 16.0684 > 9.48773), H0 is rejected. There is sufficient evidence to indicate the standard deviation of the weight measurements is greater than .01 at α = .05.

194

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6.128

a.

Let pi = proportion of first round games won by the ith seed. To determine if the higher seed has a better than 50-50 chance of winning a first-round game, we test: H0: pi = .5 Ha: pi > .5 for i = 1, 2, 3, …, 8 The test statistic is zi =

pˆ i − p0 po qo n

.

No value of α was given. We will use α = .05. The rejection region requires α = .05 in the upper tail of the z-distribution. From Table IV, Appendix B, z.05 = 1.645. The rejection region is z > 1.645. xi x x 52 x 49 41 = 1 , pˆ 2 = 2 = = .942 , pˆ 3 = 3 = = .788 , . Thus, pˆ1 = 1 = n 52 n 52 n n 52 x 37 x x x 42 36 35 pˆ 4 = 4 = = .808 , pˆ 5 = 5 = = .712 , pˆ 6 = 6 = = .692 , pˆ 7 = 7 = = .673 , n 52 n 52 n 52 n 52 x 22 pˆ 8 = 8 = = .423 n 52 pˆ i =

The corresponding test statistics are: z1 =

z3 =

z5 =

z7 =

pˆ1 − p0 po qo n pˆ 3 − p0 po qo n pˆ 5 − p0 po qo n pˆ 7 − p0 po qo n

=

1.00 − .5

=

=

=

.5(.5) 52 .788 − .5 .5(.5) 52 .712 − .5 .5(.5) 52 .673 − .5 .5(.5) 52

= 7.21 , z2 =

= 4.15 , z4 =

= 3.06 , z6 =

= 2.50 , z8 =

pˆ 2 − p0 po qo n pˆ 4 − p0 po qo n pˆ 6 − p0 po qo n pˆ 8 − p0 po qo n

=

.942 − .5

=

=

=

.5(.5) 52 .808 − .5 .5(.5) 52 .692 − .5 .5(.5) 52 .423 − .5 .5(.5) 52

= 6.37 ,

= 4.44 ,

= 2.77 ,

= −1.11

For games matching 1 and 16, since the observed value of the test statistic falls in the rejection region (z1 = 7.21 > 1.645), H0 is rejected. There is sufficient evidence to indicate the #1 seed has a better than 50-50 chance of winning a first-round game at α = .05.

Inferences Based on a Single Sample: Tests of Hypothesis

195

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For games matching 2 and 15, since the observed value of the test statistic falls in the rejection region (z2 = 6.37 > 1.645), H0 is rejected. There is sufficient evidence to indicate the #2 seed has a better than 50-50 chance of winning a first-round game at α = .05. For games matching 3 and 14, since the observed value of the test statistic falls in the rejection region (z3 = 4.15 > 1.645), H0 is rejected. There is sufficient evidence to indicate the #3 seed has a better than 50-50 chance of winning a first-round game at α = .05. For games matching 4 and 13, since the observed value of the test statistic falls in the rejection region (z4 = 4.44 > 1.645), H0 is rejected. There is sufficient evidence to indicate the #4 seed has a better than 50-50 chance of winning a first-round game at α = .05. For games matching 5 and 12, since the observed value of the test statistic falls in the rejection region (z5 = 3.06 > 1.645), H0 is rejected. There is sufficient evidence to indicate the #5 seed has a better than 50-50 chance of winning a first-round game at α = .05. For games matching 6 and 11, since the observed value of the test statistic falls in the rejection region (z6 = 2.77 > 1.645), H0 is rejected. There is sufficient evidence to indicate the #6 seed has a better than 50-50 chance of winning a first-round game at α = .05. For games matching 7 and 10, since the observed value of the test statistic falls in the rejection region (z7 = 2.50 > 1.645), H0 is rejected. There is sufficient evidence to indicate the #7 seed has a better than 50-50 chance of winning a first-round game at α = .05. For games matching 8 and 9, since the observed value of the test statistic does not fall in the rejection region (z8 = −1.11 >/ 1.645), H0 is not rejected. There is insufficient evidence to indicate the #8 seed has a better than 50-50 chance of winning a first-round game at α = .05. b.

Let μi = mean margin of victory. To determine if the mean margin of victory is greater than 10 points, we test: H0: μi = 10 Ha: μi > 10 i = 1, 2, 3, and 4 The test statistic is zi =

xi − μ0

σx

No value of α was given. We will use α = .05. The rejection region requires α = .05 in the upper tail of the z-distribution. From Table IV, Appendix B, z.05 = 1.645. The rejection region is z > 1.645.

196

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The test statistics are: z1 =

z3 =

x1 − μ0

σx

x3 − μ0

σx

=

22.9 − 10 12.4

=

52

= 7.50 , z2 =

10.6 − 10 12.0

52

= 0.36 , z4 =

x2 − μ0

σx

x4 − μ0

σx

17.2 − 10

=

11.4

=

52

= 4.55 ,

10.0 − 10 12.5

52

=0

For games matching 1 and 16, since the observed value of the test statistic falls in the rejection region (z1 = 7.50 > 1.645), H0 is rejected. There is sufficient evidence to indicate the #1 seed wins by more than 10 points in first-round games at α = .05. For games matching 2 and 15, since the observed value of the test statistic falls in the rejection region (z2 = 4.55 > 1.645), H0 is rejected. There is sufficient evidence to indicate the #2 seed wins by more than 10 points in first-round games at α = .05. For games matching 3 and 14, since the observed value of the test statistic does not fall in the rejection region (z3 = 0.36 >/ 1.645), H0 is not rejected. There is insufficient evidence to indicate the #3 seed wins by more than 10 points in first-round games at α = .05.

c.

For games matching 4 and 13, since the observed value of the test statistic does not fall in the rejection region (z4 = 0 >/ 1.645), H0 is not rejected. There is insufficient evidence to indicate the #4 seed wins by more than 10 points in first-round games at α = .05. Let μi = mean margin of victory. To determine if the mean margin of victory is less than 5 points, we test: H0: μi = 5 Ha: μi < 5 i = 5, 6, 7, and 8 The test statistic is zi =

xi − μ0

σx

No value of α was given. We will use α = .05. The rejection region requires α = .05 in the upper tail of the z-distribution. From Table IV, Appendix B, z.05 = 1.645. The rejection region is z < −1.645. The test statistics are: z5 =

z7 =

x5 − μ0

σx

x7 − μ0

σx

=

=

5.3 − 5 10.4

52

3.2 − 5 10.5

52

= 0.21 , z6 =

x6 − μ0

= −1.24 , z8 =

σx

=

x8 − μ0

Inferences Based on a Single Sample: Tests of Hypothesis

σx

4.3 − 5 10.7 =

52

= −.47 ,

−2.1 − 5 11.0

52

= −4.65

197

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For games matching 5 and 12, since the observed value of the test statistic does not fall in the rejection region (z5 = 0.21 / 1.96), H0 is not rejected. There is insufficient evidence to indicate there is a difference in the game outcome and point spread at α = .05. There is no evidence to indicate the point spread is not a good predictor of the victory margin. 6.130

Using MINITAB, the descriptive statistics are: Descriptive Statistics: Candy Variable Candy

N 5

N* 0

Mean 24.00

SE Mean 1.67

StDev 3.74

Minimum 21.00

Q1 21.00

Median 23.00

Q3 27.50

Maximum 30.00

To give the benefit of the doubt to the students we will use a small value of α. (We do not want to reject H0 when it is true to favor the students.) Thus, we will use α = .001.

We must also assume that the sample comes from a normal distribution. To determine if the mean number of candies exceeds 15, we test: H0: μ = 15 Ha: μ > 15 The test statistic is z =

x − μo

σ

n

=

22 − 15 3

5

= 5.22

The rejection region requires α = .001 in the upper tail of the z-distribution. From Table IV, Appendix B, z.001 = 3.08. The rejection region is z > 3.08. Since the observed value of the test statistic falls in the rejection region (z = 5.22 > 3.08), H0 is rejected. There is sufficient evidence to indicate the mean number of candies exceeds 15 at α = .001.

200

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Inferences Based on Two Samples: Confidence Intervals and Tests of Hypothesis 7.2

a.

μ x = μ1 = 12

σx =

b.

μ x = μ2 = 10

σx =

c.

μ x − x = μ1 − μ2 = 12 − 10 = 2 1

1

7.4

2

2

σ1 n1

σ2 n2

=

4 = .5 64

=

3 64

= .375

2

σ x −x = d.

1

1

Chapter 7

2

σ 12 n1

+

σ 22 n2

=

42 32 25 + = = .625 64 64 64

Since n1 ≥ 30 and n2 ≥ 30, the sampling distribution of x1 − x2 is approximately normal by the Central Limit Theorem.

Assumptions about the two populations: 1. 2.

Both sampled populations have relative frequency distributions that are approximately normal. The population variances are equal.

Assumptions about the two samples: The samples are randomly and independently selected from the population. 7.6

a.

sp2 =

(n1 − 1) s12 + (n2 − 1) s22 (25 − 1)120 + (25 − 1)100 5280 = 110 = = n1 + n2 − 2 25 + 25 + 2 48

b.

sp2 =

(20 − 1)12 + (10 − 1)20 408 = = 14.5714 20 + 10 − 2 28

c.

sp2 =

(6 − 1).15 + (10 − 1).2 2.55 = = .1821 6 + 10 − 2 14

d.

sp2 =

(16 − 1)3000 + (17 − 1)2500 85,000 = = 2741.9355 16 + 17 − 2 31

e.

sp2 falls near the variance with the larger sample size.

Inferences Based on Two Samples: Confidence Intervals and Tests of Hypothesis

201

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7.8

σ 12

σ 22

9 16 + = .25 = .5 100 100

a.

σ x −x =

b.

The sampling distribution of x1 − x2 is approximately normal by the Central Limit Theorem since n1 ≥ 30 and n2 ≥ 30.

1

2

n1

+

n2

=

μ x − x = μ1 − μ2 = 10 1

c.

2

x1 − x2 = 15.5 − 26.6 = −11.1 Yes, it appears that x1 − x2 = −11.1 contradicts the null hypothesis H0: μ1 − μ2 = 10.

d.

The rejection region requires α/2 = .025 = .05/2 in each tail of the z-distribution. From Table IV, Appendix B, z.025 = 1.96. The rejection region is z < −1.96 or z > 1.96.

e.

H0: μ1 − μ2 = 10 Ha: μ1 − μ2 ≠ 10 The test statistic is z =

( x1 − x2 ) − 10

σ 12 n1

+

σ 22

=

(15.5 − 26.6) − 10 = −42.2 .5

n2

The rejection region is z < −1.96 or z > 1.96. (Refer to part d.) Since the observed value of the test statistic falls in the rejection region (z = −42.2 < −1.96), H0 is rejected. There is sufficient evidence to indicate the difference in the population means is not equal to 10 at α = .05. f.

The form of the confidence interval is: ( x1 − x2 ) ± zα / 2

σ 12 n1

+

σ 22 n2

For confidence coefficient .95, α = 1 − .95 = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B, z.025 = 1.96. The confidence interval is: 9 16 + ⇒ −11.1 ± .98 ⇒ (−12.08, −10.12) (15.5 − 26.6) ± 1.96 100 100

We are 95% confident that the difference in the two means is between −12.08 and −10.12. g.

202

The confidence interval gives more information.

Chapter 7

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7.10

Some preliminary calculations: x1 =

∑x

1

n1

∑x

2 1

s12 = x1 =

sp2 = a.

654 15

(∑ x ) −

2

∑x

2

n2

∑x

6542 15 = 419.6 = 29.3167 15 − 1 14

1

n1

=

28934 −

=

n1 − 1

2 2

s22 =

=

858 = 53.625 16

(∑ x ) −

2

2

n2 n2 − 1

=

8582 16 = 439.75 = 29.3167 16 − 1 15

46450 −

(n1 − 1) s12 + (n2 − 1) s22 (15 − 1)29.9714 + (16 − 1)29.3167 859.3501 = 29.6328 = = 29 n1 + n2 − 2 15 + 16 − 2

H0: μ2 − μ1 = 10 Ha: μ2 − μ1 > 10 The test statistic is t =

( x1 − x2 ) − D0 ⎛1 1⎞ sp2 ⎜ + ⎟ ⎝ n1 n2 ⎠

=

(53.625 − 43.6) − 10 ⎛1 1⎞ 29.6328 ⎜ + ⎟ ⎝ 15 16 ⎠

=

.025 = .013 1.9564

The rejection region requires α = .01 in the upper tail of the t-distribution with df = n1 + n2 − 2 = 15 + 16 − 2 = 29. From Table VI, Appendix B, t.01 = 2.462. The rejection region is t > 2.462. Since the test statistic does not fall in the rejection region (t = .013 >/ 2.462), H0 is not rejected. There is insufficient evidence to conclude μ2 − μ1 > 10 at α = .01. b.

For confidence coefficient .98, α = .02 and α/2 = .01. From Table VI, Appendix B, with df = n1 + n2 − 2 = 15 + 16 − 2 = 29, t.01 = 2.462. The 98% confidence interval for (μ2 − μ1) is:

⎛1 1⎞ ⎛1 1⎞ ( x1 − x2 ) ± tα / 2 sp2 ⎜ + ⎟ ⇒ (53.625 − 43.6) ± 2.462 29.6328 ⎜ + ⎟ ⎝ 15 16 ⎠ ⎝ n1 n2 ⎠ ⇒ 10.025 ± 4.817 ⇒ (5.208, 14.842) We are 98% confident that the difference between the mean of population 2 and the mean of population 1 is between 5.208 and 14.842.

Inferences Based on Two Samples: Confidence Intervals and Tests of Hypothesis

203

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7.12

a.

Let μ1 = mean carat size of diamonds certified by GIA and μ2 = mean carat size of diamonds certified by HRD. For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B, z.025 = 1.96. The 95% confidence interval is:

σ 12

( x1 − x2 ) ± zα / 2

n1

+

σ 22 n2

⇒ (.6723 − .8129) ± 1.96

.24562 .18312 + 151 79

⇒ −.1406 ± .0563 ⇒ (−.1969, − .0843)

b.

We are 95% confident that the difference in mean carat size between diamonds certified by GIA and those certified by HRD is between -.1969 and -.0843.

c.

Let μ3 = mean carat size of diamonds certified by IGI. ( x1 − x3 ) ± zα / 2

σ 12 n1

+

σ 32 n3

⇒ (.6723 − .3665) ± 1.96

.24562 .21632 + 151 78

⇒ .3058 ± .0620 ⇒ (.2438, .3678)

7.14

d.

We are 95% confident that the difference in mean carat size between diamonds certified by GIA and those certified by IGI is between .2438 and .3678.

e.

( x2 − x3 ) ± zα / 2

f.

We are 95% confident that the difference in mean carat size between diamonds certified by HRD and those certified by IGI is between .3837 and .5091.

a.

Let μ1 = mean score for males and μ2 = mean score for females. For confidence coefficient .90, α = .10 and α/2 = .10/2 = .05. From Table IV, Appendix B, z.025 = 1.645. The 90% confidence interval is:

σ 22 n2

( x1 − x2 ) ± zα / 2

+

σ 32 n3

σ 12 n1

.18312 .21632 + 79 78 ⇒ .4464 ± .0627 ⇒ (.3837, .5091)

⇒ (.8129 − .3665) ± 1.96

+

σ 22 n2

⇒ (39.08 − 38.79) ± 1.645

6.732 6.942 + 127 114

⇒ 0.29 ± 1.452 ⇒ (−1.162, 1.742)

We are 90% confident that the difference in mean service-rating scores between males and females. b.

204

Because 0 falls in the 90% confidence interval, we are 90% confident that there is no difference in the mean service-rating scores between males and females.

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7.16

a.

The descriptive statistics are:

Descriptive Statistics: US, Japan Variable US Japan

N 5 5

Mean 6.562 3.118

Median 6.870 3.220

TrMean 6.562 3.118

Variable US Japan

Minimum 4.770 1.920

Maximum 8.000 4.910

Q1 5.415 1.970

Q3 7.555 4.215

s 2p =

StDev 1.217 1.227

SE Mean 0.544 0.549

(n1 − 1) s12 + (n2 − 1) s22 (5 − 1)1.217 2 + (5 − 1)1.227 2 = 1.4933 = 5+5−2 n1 + n2 − 2

To determine if the mean annual percentage turnover for U.S. plants exceeds that for Japanese plants, we test:

H0: μ1 − μ2 = 0 Ha: μ1 − μ2 > 0 The test statistic is t =

( x1 − x2 ) − D0 ⎛1 1 ⎞ sp2 ⎜ + ⎟ ⎝ n1 n2 ⎠

=

(6.562 − 3.118) − 0 ⎛1 1⎞ 1.4933 ⎜ + ⎟ ⎝5 5⎠

= 4.456

The rejection region requires α = .05 in the upper tail of the t-distribution with df = n1 + n2 − 2 = 5 + 5 − 2 = 8. From Table VI, Appendix B, t.05 = 1.860. The rejection region is t > 1.860. Since the observed value of the test statistic falls in the rejection region (t = 4.46 > 1.860), H0 is rejected. There is sufficient evidence to indicate the mean annual percentage turnover for U.S. plants exceeds that for Japanese plants at α = .05. b.

The p-value = P(t ≥ 4.456). Using Table VI, Appendix B, with df = n1 + n2 − 2 = 5 + 5 – 2 = 8, .005 < P(t ≥ 4.456) < .001. Since the p-value is so small, there is evidence to reject H0 for α > .005.

c.

The necessary assumptions are: 1. 2. 3.

Both sampled populations are approximately normal. The population variances are equal. The samples are randomly and independently sampled.

There is no indication that the populations are not normal. Both sample variances are similar, so there is no evidence the population variances are unequal. There is no indication the assumptions are not valid.

Inferences Based on Two Samples: Confidence Intervals and Tests of Hypothesis

205

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7.18

Let μ1 = the mean relational intimacy score for participants in the CMC group and μ2 = the mean relational intimacy score for participants in the FTF group. Using MINITAB, the descriptive statistics are: Descriptive Statistics: CMC, FTF Variable CMC FTF

N 24 24

N* 0 0

Mean 3.500 3.542

SE Mean 0.159 0.134

StDev 0.780 0.658

Minimum 2.000 2.000

Q1 3.000 3.000

Median 3.500 4.000

Q3 4.000 4.000

Maximum 5.000 5.000

Some preliminary calculations are: s 2p =

( n1 − 1) s12 + ( n2 − 1) s22 = ( 24 − 1) .7802 + ( 24 − 1) .6582 n1 + n2 − 2

24 + 24 − 2

= 0.5207

To determine if the mean relational intimacy score for participants in the CMC group is lower than the mean relational intimacy score for participants in the FTF group, we test: H0: μ1 − μ2 = 0 Ha: μ1 − μ2 < 0 The test statistic is t =

( x1 − x2 ) − Do ⎛1 1 ⎞ s 2p ⎜ + ⎟ ⎝ n1 n2 ⎠

=

( 3.500 − 3.542 ) − 0 = −0.042 = −.20 1 ⎞ ⎛ 1 .5207 ⎜ + ⎟ ⎝ 24 24 ⎠

.20831

The rejection region requires α= .10 in the lower tail of the t-distribution with df = n1 + n2 – 2 = 24 + 24 – 2 = 46. From Table VI, Appendix B, t.10 ≈1.303. The rejection region is t < −1.303. Since the observed value of the test statistic does not fall in the rejection region (t = −.20 ≤/ −1.303), H0 is not rejected. There is insufficient evidence to indicate that the mean relational intimacy score for participants in the CMC group is lower than the mean relational intimacy score for participants in the FTF group at α = .10. 7.20

206

a.

The first population is the set of responses for all business students who have access to lecture notes and the second population is the set of responses for all business students not having access to lecture notes.

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b.

To determine if there is a difference in the mean response of the two groups, we test: H0: μ1 − μ2 = 0 Ha: μ1 − μ2 ≠ 0 The test statistic is z =

( x1 − x2 ) − 0 s12 s22 + n1 n2

=

(8.48 − 7.80) − 0 = 2.19 .94 2.99 + 86 35

The rejection region requires α/2 = .01/2 = .005 in each tail of the z-distribution. From Table IV, Appendix B, z.005 = 2.58. The rejection region is z < −2.58 or z > 2.58. Since the observed value of the test statistic does not fall in the rejection region (z = 2.19 >/ 2.58), H0 is not rejected. There is insufficient evidence to indicate a difference in the mean response of the two groups at α = .01. c.

For confidence coefficient .99, α = .01 and α/2 = .01/2 = .005. From Table IV, Appendix B, z.005 = 2.58. The confidence interval is: s12 s22 .94 2.99 + ⇒ (8.48 − 7.80) ± 2.58 + n1 n2 86 35

( x1 − x2 ) ± z.005

⇒ .68 ± .801 ⇒ (−.121, 1.481) We are 99% confident that the difference in the mean response between the two groups is between −.121 and 1.481.

7.22

d.

A 95% confidence interval would be smaller than the 99% confidence interval. The z value used in the 95% confidence interval is z.025 = 1.96 compared with the z value used in the 99% confidence interval of z.005 = 2.58.

a.

The bacteria counts are probably normally distributed because each count is the median of five measurements from the same specimen.

b.

Let μ1 = mean of the bacteria count for the discharge and μ2 = mean of the bacteria count upstream. Since we want to test if the mean of the bacteria count for the discharge exceeds the mean of the count upstream, we test: H0: μ1 − μ2 = 0 Ha: μ1 − μ2 > 0

c.

Using MINITAB, the descriptive statistics are: Descriptive Statistics: Plant, Upstream

Variable Plant Upstream

N 6 6

Mean 32.10 29.617

Median 31.75 30.000

TrMean 32.10 29.617

Variable Plant Upstream

Minimum 28.20 26.400

Maximum 36.20 32.300

Q1 29.40 27.075

Q3 35.23 31.850

StDev 3.19 2.355

Inferences Based on Two Samples: Confidence Intervals and Tests of Hypothesis

SE Mean 1.30 0.961

207

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(n1 − 1) s12 + (n2 − 1) s22 (6 − 1)3.192 + (6 − 1)2.3552 = 7.861 s = = n1 + n2 − 2 6+6−2 2 p

The test statistic is t =

( x1 − x2 ) − 0 ⎛1 1 ⎞ s ⎜ + ⎟ ⎝ n1 n2 ⎠

=

(32.10 − 29.617) − 0

2 p

⎛1 1⎞ 7.861 ⎜ + ⎟ ⎝6 6⎠

= 1.53

No α level was given, so we will use α = .05. The rejection region requires α = .05 in the upper tail of the t-distribution with df = n1 + n2 − 2 = 6 + 6 – 2 = 10. From Table VI, Appendix B, t.05 = 1.812. The rejection region is t > 1.812. Since the observed value of the test statistic does not fall in the rejection region (t = 1.53 >/ 1.812), H0 is not rejected. There is insufficient evidence to indicate the mean bacteria count for the discharge exceeds the mean of the count upstream at α = .05. d.

We must assume: 1. 2. 3.

7.24

The mean counts per specimen for each location is normally distributed. The variances of the 2 distributions are equal. Independent and random samples were selected from each population.

a.

We cannot make inferences about the difference between the mean salaries of male and female accounting/finance/banking professionals because no standard deviations are provided.

b.

To determine if the mean salary for males is significantly greater than that for females, we test: H0: μ1 − μ2 = 0 Ha: μ1 − μ2 > 0 The rejection region requires α = .05 in the upper tail of the z-distribution. From Table IV, Appendix B, z.05 = 1.645. To make things easier, we will assume that the standard deviations for the 2 groups are the same. The test statistic is z=

208

( x1 − x2 ) − Do = ( 69, 484 − 52,012 ) − 0 = ⎛ σ 12 σ 22 ⎞ + ⎜ ⎟ ⎝ n1 n2 ⎠

1 ⎞ ⎛ 1 + ⎟ 1400 1400 ⎝ ⎠

σ2⎜

17,836 471,896.2038 = σ (.037796) σ

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In order to reject H0 this test statistic must fall in the rejection region, or be greater than 1.645. Solving for σ we get: z=

471,896.2038

σ

> 1.645 ⇒ σ <

471,896.2038 = 286,866.99 1.645

Thus, to reject H0 the average of the two standard deviations has to be less than $286,866.99.

7.26

c.

Yes. In fact, reasonable values for the standard deviation will be around $5,000. which is much smaller than the required $286,866.99.

d.

These data were collected from voluntary subjects who responded to a Web-based survey. Thus, this is not a random sample, but a self-selected sample. Generally, subjects who respond to surveys tend to have very strong opinions, which may not be the same as the population in general. Thus, the results from this self-selected sample may not reflect the results from the population in general.

a. Pair

Difference

1 2 3 4 5 6

3 2 2 4 0 1

nd

d=

∑d i =1

nd

i

=

12 =2 6

⎛ nd ⎞ ⎜ ∑ di ⎟ nd i =1 2 ⎠ di − ⎝ ∑ n d sd2 = i =1 nd − 1

2

⎛ (12) 2 ⎞ ⎜ 34 − ⎟ 6 ⎠ ⎝ =2 = 5

b.

μd = μ1 − μ2

c.

For confidence coefficient .95, α = .05 and α/2 = .025. From Table VI, Appendix B, with df = nD − 1 = 6 − 1 = 5, t.025 = 2.571. The confidence interval is:

d ± tα / 2

sd nd

= 2.571

2 6

⇒ 2 ± 1.484 ⇒ (.516, 3.484)

Inferences Based on Two Samples: Confidence Intervals and Tests of Hypothesis

209

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d.

H0: μd = 0 Ha: μd ≠ 0 The test statistic is t = t =

d sd

nd

=

2 = 3.46 2/ 6

The rejection region requires α/2 = .05/2 = .025 in each tail of the t-distribution with df = nD − 1 = 6 − 1 = 5. From Table VI, Appendix B, t.025 = 2.571. The rejection region is t < −2.571 or t > 2.571. Since the observed value of the test statistic falls in the rejection region (3.46 > 2.571), H0 is rejected. There is sufficient evidence to indicate that the mean difference is different from 0 at α = .05. 7.28

a.

H0: μ1 − μ2 = 0 Ha: μ1 − μ2 < 0 The rejection region requires α = .10 in the lower tail of the z-distribution. From Table IV, Appendix B, z.10 = 1.28. The rejection region is z < −1.28.

b.

H0: μ1 − μ2 = 0 Ha: μ1 − μ2 < 0 The test statistic is z =

d − 0 −3.5 − 0 = = −4.71 . sd 21 nd 38

The rejection region is z < −1.28 (Refer to part a.) Since the observed value of the test statistic falls in the rejection region (z = −4.71 < −1.28), H0 is rejected. There is sufficient evidence to indicate μ1 − μ2 < 0 at α = .10. c.

Since the sample size of the number of pairs is greater than 30, we do not need to assume that the population of differences is normal. The sampling distribution of d is approximately normal by the Central Limit Theorem. We must assume that the differences are randomly selected.

d.

For confidence coefficient .90, α = .10 and α/2 = .10/2 = .05. From Table IV, Appendix B, z.05 = 1.645. The 90% confidence interval is:

d ± z.05

e.

210

sd nd

⇒ −3.5 ± 1.645

21 38

⇒ −3.5 ± 1.223 ⇒ (−4.723, − 2.277)

The confidence interval provides more information since it gives an interval of possible values for the difference between the population means.

Chapter 7

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7.30

a.

Let μ1 = the mean salary of technology professionals in 2003 and μ2 = the mean salary of technology professionals in 2005. Let μd = μ1 - μ2. To determine if the mean salary of technology professionals at all U.S. metropolitan areas has increased between 2003 and 2005, we test:

H0: μ1 − μ2 = 0

H0: μd = 0 OR

Ha: μ1 − μ2 < 0

Ha: μd < 0

b. Metro Area

2003 Salary ($ thousands) 87.7 78.6 71.4 70.8 73.0 76.3 73.6 71.1 69.5 69.0 71.0 73.0 62.3

Silicon Valley New York Washington, D.C. Los Angeles Denver Boston Atlanta Chicago Philadelphia San Diego Seattle Dallas-Ft. Worth Detroit

2005 Salary ($ thousands) 85.9 80.3 77.4 77.1 77.1 80.1 73.2 73.0 69.8 77.1 66.9 71.0 64.1

Difference (2003 – 2005) 1.8 −1.7 −6.0 −6.3 −4.1 −3.8 0.4 −1.9 −0.3 −8.1 4.1 2.0 −1.8

nd

c.

d=

∑ di 1

nd

=

−25.7 = −1.977 13 2

⎛ nd ⎞ ⎜⎜ ∑ di ⎟⎟ nd 1 ⎠ 2 (−25.7) 2 di − ⎝ ∑ 206.59 − nd 13 = = 12.9819 sd2 = 1 nd − 1 13 − 1 sd = sd2 = 12.9819 = 3.603 d − μo

The test statistic is t =

e.

The rejection region requires α = .10 in the lower tail of the t-distribution with df = nd – 1 = 13 – 1 = 14. From Table VI, Appendix B, t.10 = 1.345. The rejection region is t < −1.345.

sd

nd

=

−1.977 − 0 = −1.978 3.603 13

d.

Inferences Based on Two Samples: Confidence Intervals and Tests of Hypothesis

211

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f.

Since the observed value of the test statistic falls in the rejection region (t = −1.978 < −1.345), H0 is rejected. There is sufficient evidence to indicate the mean salary of technology professionals at all U.S. metropolitan areas has increased between 2003 and 2005 at α = .10.

g.

In order for the inference to be valid, we must assume that the population of differences is normal and that we have a random sample. Using MINITAB, the histogram of the differences is: Histogram of Diff 3.0

Fr equency

2.5 2.0

1.5 1.0

0.5 0.0

-7.5

-5.0

-2.5

0.0

2.5

5.0

Diff

The graph is fairly mound-shaped although it is somewhat skewed to the right. Since there are only 13 observations, this graph is close enough to being mound-shaped to indicate the normal assumption is reasonable. 7.32

212

a.

The data should be analyzed as a paired difference experiment because each actor who won an Academy Award was paired with another actor with similar characteristics who did not win the award.

b.

Let μ1 = mean life expectancy of Academy Award winners and μ2 = mean life expectancy of non-Academy Award winners. To compare the mean life expectancies of Academy Award winners and non-winners, we test: H0: μ1 − μ2 = μd = 0 Ha: μd ≠ 0

c.

Since the p-value was so small, there is sufficient evidence to indicate the mean life expectancies of the Academy Award winners and non-winners are different for any value of α > .003. Since the sample mean life expectancy of Academy Award winners is greater than that for non-winners, we can conclude that Academy Award winners have a longer mean life expectancy than non-winners.

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7.34

a.

Let μ1 = mean driver chest injury rating and μ2 = mean passenger chest injury rating. Because the data are paired, we are interested in μ1 − μ2 = μd, the difference in mean chest injury ratings between drivers and passengers.

b.

The data were collected as matched pairs and thus, must be analyzed as matched pairs. Two ratings are obtained for each car – the driver’s chest injury rating and the passenger’s chest injury rating.

c.

Using MINITAB, the descriptive statistics are:

Descriptive Statistics: DrivChst, PassChst, diff Variable DrivChst PassChst diff

N 98 98 98

Mean 49.663 50.224 -0.561

Median 50.000 50.500 0.000

TrMean 49.682 50.148 -0.420

Variable DrivChst PassChst diff

Minimum 34.000 35.000 -15.000

Maximum 68.000 69.000 13.000

Q1 45.000 45.000 -4.000

Q3 54.000 55.000 3.000

StDev 6.670 7.107 5.517

SE Mean 0.674 0.718 0.557

For confidence coefficient .99, α = .01 and α/2 = .01/2 = .005. From Table IV, Appendix B, z.005 = 2.58. The 99% confidence interval is: d ± z.005

7.36

sd nd

⇒ −0.561 ± 2.58

5.517 98

⇒ −0.561 ± 1.438 ⇒ (−1.999, 0.877)

d.

We are 99% confidence that the difference between the mean chest injury ratings of drivers and front-seat passengers is between −1.999 and 0.877. Since 0 is in the confidence interval, there is no evidence that the true mean driver chest injury rating exceeds the true mean passenger chest injury rating.

e.

Since the sample size is large, the sampling distribution of d is approximately normal by the Central Limit Theorem. We must assume that the differences are randomly selected.

a.

Let μC1 = mean relational intimacy score for the CMC group on the first meeting and μC3 = mean relational intimacy score for the CMC group on the third meeting. Let μCd = difference in mean relational intimacy score between the first and third meetings for the CMC group. To determine if the mean relational intimacy score will increase between the first and third meetings, we test: Ho: μCd = 0 Ha: μCd < 0

b.

The researchers used the paired t-test because the same individuals participated in each of the three meeting sessions. Thus, the samples would not be independent.

c.

Since the p-value is so small (p = .003), H0 would be rejected. There is sufficient evidence to indicate that the mean relational intimacy score for participants in the CMC group increased from the first to the third meeting for any value of α > .003.

Inferences Based on Two Samples: Confidence Intervals and Tests of Hypothesis

213

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d.

Let μF1 = mean relational intimacy score for the FTF group on the first meeting and μF3 = mean relational intimacy score for the FTF group on the third meeting. Let μFd = difference in mean relational intimacy score between the first and third meetings for the FTF group. To determine if the mean relational intimacy score will change between the first and third meetings, we test: H0: μFd = 0 Ha: μFd ≠ 0

e.

7.38

Since the p-value is not small (p = .39), H0 would be not be rejected. There is insufficient evidence to indicate that the mean relational intimacy score for participants in the FTF group changed from the first to the third meeting for any value of α < .39.

Using MINITAB, the descriptive statistics are: Descriptive Statistics: Method1, Method2, Diff Variable Method1 Method2 Diff

N 10 10 10

N* 0 0 0

Mean 13.39 13.10 0.290

SE Mean 4.18 3.96 0.553

StDev 13.22 12.51 1.750

Minimum 1.00 1.40 -2.200

Q1 1.30 1.78 -0.875

Median 10.35 9.50 -0.150

Q3 24.63 25.05 1.575

Maximum 34.40 30.70 3.700

To determine if the mean transition error for method 1 differs from the mean transition error for method 2, we test: H0: μ1 − μ2 = 0

H0: μd = 0 OR

Ha: μ1 − μ2 ≠ 0 The test statistic is t =

d − μo sd

nd

=

Ha: μd ≠ 0

0.290 − 0 = 0.52 1.750 10

The rejection region requires α/2 = .10/2 = .05 in each tail of the t-distribution with df = nd – 1 = 10 – 1 = 9. From Table VI, Appendix B, t.05 = 1.833. The rejection region is t < −1.833 or t > 1.833. Since the observed value of the test statistic does not fall in the rejection region (t = 0.52 >/ 1.833), H0 is not rejected. There is insufficient evidence to indicate the mean transition error for method 1 differs from the mean transition error for method 2 at α = .10. 7.40

Using MINITAB, the descriptive statistics are: Descriptive Statistics: HMETER, HSTATIC, Diff

214

Variable HMETER HSTATIC Diff

N 40 40 40

N* 0 0 0

Mean 1.0405 1.0410 -0.000523

Variable HMETER HSTATIC Diff

Median 1.0232 1.0237 -0.000165

SE Mean 0.00638 0.00649 0.000204

Q3 1.0883 1.0908 0.000317

StDev 0.0403 0.0410 0.001291

Minimum 0.9936 0.9930 -0.004480

Q1 1.0047 1.0043 -0.001078

Maximum 1.1026 1.1052 0.001580

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For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table VI, Appendix B, with df = nd – 1 = 40 – 1 = 39, t.025 ≈ 2.021. The 95% confidence interval is: d ± t.025

sd

⇒ −0.000523 ± 2.021

n ⇒ (−0.000936,

0.001291 ⇒ −0.000523 ± 0.000413 40

− 0.000110)

We are 95% confident that the true difference in mean density measurements between the two methods is between -0.000936 and -0.000110. Since the absolute value of this interval is completely less than the desired maximum difference of .002, the winery should choose the alternative method of measuring wine density. 7.42

a.

The rejection region requires α = .01 in the lower tail of the z-distribution. From Table IV, Appendix B, z.01 = 2.33. The rejection region is z < −2.33.

b.

The rejection region requires α = .025 in the lower tail of the z-distribution. From Table IV, Appendix B, z.025 = 1.96. The rejection region is z < −1.96.

c.

The rejection region requires α = .05 in the lower tail of the z-distribution. From Table IV, Appendix B, z.05 = 1.645. The rejection region is z < −1.645. The rejection region requires α = .10 in the lower tail of the z-distribution. From Table IV, Appendix B, z.10 = 1.28. The rejection region is z < −1.28.

d.

7.44

For confidence coefficient .95, α = 1 − .95 = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B, z.025 = 1.96. The 95% confidence interval for p1 − p2 is approximately: a.

( pˆ1 − pˆ 2 ) ± zα / 2

pˆ1qˆ1 pˆ 2 qˆ2 .65(1 − .65) .58(1 − .58) + ⇒ (.65 − .58) ± 1.96 + n1 n2 400 400 ⇒ .07 − .067 ⇒ (.003, .137)

b.

( pˆ1 − pˆ 2 ) ± zα / 2

pˆ1qˆ1 pˆ 2 qˆ2 + ⇒ (.31 − .25) − 1.96 n1 n2

.31(1 − .31) .25(1 − .25) + 180 250

⇒ .06 ± .086 ⇒ (−.026, .146) c.

( pˆ1 − pˆ 2 ) ± zα / 2

pˆ1qˆ1 pˆ 2 qˆ2 .46(1 − .46) .61(1 − .61) + ⇒ (.46 − .61) ±1.96 + 100 120 n1 n2 ⇒ −.15 ± .131 ⇒ (−.281, −.019)

7.46

pˆ =

n1 pˆ1 + n2 pˆ 2 55(.7) + 65(.6) 78 = = = .65 55 + 65 120 n1 + n2

qˆ = 1 − pˆ = 1 − .65 = .35

H0: p1 − p2 = 0 Ha: p1 − p2 > 0

Inferences Based on Two Samples: Confidence Intervals and Tests of Hypothesis

215

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The test statistic is z =

( pˆ1 − pˆ 2 ) − 0 ⎛1 1⎞ ˆ ˆ⎜ + ⎟ pq ⎝ n1 n2 ⎠

=

(.7 − .6) − 0 1 ⎞ ⎛ 1 .65(.35) ⎜ + ⎟ 55 65 ⎝ ⎠

=

.1 = 1.14 .08739

The rejection region requires α = .05 in the upper tail of the z-distribution. From Table IV, Appendix B, z.05 = 1.645. The rejection region is z > 1.645. Since the observed value of the test statistic does not fall in the rejection region (z = 1.14 >/ 1.645), H0 is not rejected. There is insufficient evidence to indicate the proportion from population 1 is greater than that for population 2 at α = .05. 7.48

a.

Let p1 = proportion of men who prefer to keep track of appointments in their head and p2 = proportion of women who prefer to keep track of appointments in their head. To determine if the proportion of men who prefer to keep track of appointments in their head is greater than that of women, we test:

H0: p1 − p2 = 0 Ha: p1 − p2 > 0 b.

pˆ =

n1 pˆ1 + n2 pˆ 2 500(.56) + 500(.46) = .51 and qˆ = 1 − pˆ = 1 − .51 = .49 = n1 + n2 500 + 500

The test statistic is z =

7.50

( pˆ1 − pˆ 2 ) − 0 ⎛1 1⎞ ˆ ˆ⎜ + ⎟ pq ⎝ n1 n2 ⎠

=

(.56 − .46) − 0 1 ⎞ ⎛ 1 + .51(.49) ⎜ ⎟ ⎝ 500 500 ⎠

= 3.16

c.

The rejection region requires α = .01 in the upper tail of the z distribution. From Table IV, Appendix B, z.01 = 2.33. The rejection region is z > 2.33.

d.

The p-value is p = P(z ≥ 3.16) ≈ .5 − .5 = 0.

e.

Since the observed value of the test statistic falls in the rejection region (z = 3.16 > 2.33), H0 is rejected. There is sufficient evidence to indicate the proportion of men who prefer to keep track of appointments in their head is greater than that of women at α = .01.

a.

Let p1 = proportion of customers returning the printed survey and p2 = proportion of customers returning the electronic survey. Some preliminary calculations are: pˆ1 =

x1 261 = = .414 n1 631

pˆ 2 =

x2 155 = = .374 n2 414

For confidence coefficient .90, α = .10 and α/2 = .10/2 = .05. From Table IV, Appendix B, z.05 = 1.645. The 90% confidence interval is:

216

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( pˆ1 − pˆ 2 ) ± z.05

pˆ1qˆ1 pˆ 2 qˆ2 .414(.586) .374(.626) + ⇒ (.414 − .374) ± 1.645 + n1 n2 631 414 ⇒ .04 ± .051 ⇒ (−.011, .091)

We are 90% confidence that the difference in the response rates for the two types of surveys is between −.011 and .091.

7.52

b.

Since the value .05 falls in the 90% confidence interval, it is not an unusual value. Thus, there is no evidence that the difference in response rates is different from .05. The researchers would be able to make this inference.

a.

Let p1 = proportion of managers and professionals who are male and p2 = proportion of part-time MBA students who are male. To see if the samples are sufficiently large: pˆ1 ± 3σ pˆ1 ⇒ pˆ1 ± 3

p1q1 pˆ qˆ (.95)(0.5) ⇒ pˆ1 ± 3 1 1 ⇒ .95 ± 3 n1 n1 162

⇒ .95 ± .05 ⇒ (.90, 1.00) pˆ 2 ± 3σ pˆ 2 ⇒ pˆ 2 ± 3

p2 q2 pˆ qˆ (.689)(.311) ⇒ pˆ 2 ± 3 2 2 ⇒ .95 ± 3 n2 n2 109

⇒ .689 ± .133 ⇒ (.556, .822) Since both intervals are contained within the interval (0, 1), the normal approximation will be adequate. First, we calculate the overall estimate of the common proportion under H0. pˆ =

n1 pˆ1 + n2 pˆ 2 162(.95) + 109(.689) = .845 = n1 + n2 162 + 109

To determine if the population of managers and professionals consists of more males than the part-time MBA population, we test: H0: p1 = p2 Ha: p1 > p2 The test statistic is z =

( pˆ1 − pˆ 2 ) − 0 ⎛1 1⎞ ˆ ˆ⎜ + ⎟ pq ⎝ n1 n2 ⎠

=

(.95 − .689) − 0 1 ⎞ ⎛ 1 + .845(.155) ⎜ ⎟ ⎝ 162 109 ⎠

= 5.82

The rejection region requires α = .05 in the upper tail of the z-distribution. From Table IV, Appendix B, z.05 = 1.645. The rejection region is z > 1.645. Since the observed value of the test statistic falls in the rejection (z = 5.82 > 1.645), H0 is rejected. There is sufficient evidence to indicate that population of managers and professionals consists of more males than the part-time MBA population at α = .05.

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b.

We had to assume: 1. Both samples were randomly selected 2. Both sample sizes are sufficiently large.

c.

First, we calculate the overall estimate of the common proportion under H0. pˆ =

n1 pˆ1 + n2 pˆ 2 162(.912) + 109(.534) = = .760 n1 + n2 162 + 109

To determine if the population of managers and professionals consists of more married individuals than the part-time MBA population, we test: H0: p1 = p2 Ha: p1 > p2 The test statistic is z =

( pˆ1 − pˆ 2 ) − 0 ⎛1 1⎞ ˆ ˆ⎜ + ⎟ pq ⎝ n1 n2 ⎠

(.912 − .534) − 0

=

1 ⎞ ⎛ 1 + .760(.240) ⎜ ⎟ ⎝ 162 109 ⎠

= 7.14

The rejection region requires α = .01 in the upper tail of the z-distribution. From Table IV, Appendix B, z.01 = 2.33. The rejection region is z > 2.33. Since the observed value of the test statistic falls in the rejection (z = 7.14 > 2.33), H0 is rejected. There is sufficient evidence to indicate that population of managers and professionals consists of more married individuals than the part-time MBA population at α = .01. d.

We had to assume: 1. Both samples were randomly selected 2. Both sample sizes are sufficiently large.

7.54

Let p1 = accuracy rate for modules with correct code and p2 = accuracy rate for modules with defective code. Some preliminary calculations are:

pˆ 1 =

218

x1 400 = = .891 n1 449

pˆ 2 =

x 2 20 = = .408 n2 49

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For confidence coefficient .99, α = .01 and α/2 = .01/2 = .005. From Table IV, Appendix B, z.005 = 2.58. The 99% confidence interval is: ( pˆ1 − pˆ 2 ) ± z.005

pˆ1qˆ1 pˆ 2 qˆ2 .891(.109) .408(.592) + ⇒ (.891 − .408) ± 2.58 + n1 n2 449 49 ⇒ .483 ± .185 ⇒ (.298, .668)

We are 99% confident that the difference in accuracy rates between modules with correct code and modules with defective code is between .298 and .668. 7.56

a.

Let p = proportion of all children who recognize Joe Camel. pˆ =

x 15 + 46 = = .735 n 28 + 55

qˆ = 1 − pˆ = 1 − .735 = .265

To see if the sample is sufficiently large: pˆ ± 3σ pˆ ⇒ pˆ ± 3

ˆˆ pq pq .735(.265) ⇒ pˆ ± 3 ⇒ .735 ± 3 ⇒ .735 ± .145 n n 83 ⇒ (.590, .880)

Since the interval lies within the interval (0, 1), the normal approximation will be adequate. For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B, z.025 = 1.96. The 95% confidence interval is: pˆ ± z.025

ˆˆ pq .735(.265) ⇒ .735 ± 1.96 ⇒ .735 ± .095 ⇒ (.640, .830) n 83

We are 95% confident that the proportion of all children who recognize Joe Camel is between .640 and .830. b.

Let p1 = proportion of children under the age of 6 who recognize Joe Camel and p2 = proportion of children age 6 and over who recognize Joe Camel. x1 15 = = .536 n1 28 x 46 pˆ 2 = 2 = = .836 n2 55 pˆ1 =

qˆ1 = 1 − pˆ1 = 1 − .536 = .464 qˆ2 = 1 − pˆ 2 = 1 − .836 = .164

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To see if the samples are sufficiently large:

pˆ1 ± 3σ pˆ1 ⇒ pˆ1 ± 3

pˆ 2 ± 3σ pˆ 2 ⇒ pˆ 2 ± 3

p1q1 pˆ qˆ .536(.464) ⇒ pˆ1 ± 3 1 1 ⇒ .536 \± 3 28 n1 n1 p2 q2 n2

⇒ .536 ± .283 ⇒ (.253, .819) pˆ qˆ .836(.164) ⇒ pˆ 2 ± 3 2 2 ⇒ .836 ± 3 n2 55 ⇒ .836 ± .150 ⇒ (.686, .986)

Since both intervals lie within the interval (0, 1), the normal approximation will be adequate. To determine if the recognition of Joe Camel increases with age, we test: H0: p1 − p2 = 0 Ha: p1 − p2 < 0 The test statistic is z =

( pˆ1 − pˆ 2 ) − 0 ⎛1 1⎞ ˆ ˆ⎜ + ⎟ pq ⎝ n1 n2 ⎠

=

(.536 − .836) − 0 1 ⎞ ⎛ 1 .735(.265) ⎜ + ⎟ ⎝ 28 55 ⎠

= −2.93

The rejection region requires α = .05 in the lower tail of the z-distribution. From Table IV, Appendix B, z.05 = 1.645. The rejection region is z < −1.645. Since the observed value of the test statistic falls in the rejection region (z = −2.93 < −1.645), H0 is rejected. There is sufficient evidence to indicate that the recognition of Joe Camel increases with age at α = .05. 7.58

a.

For confidence coefficient .95, α = 1 − .95 = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B, z.025 = 1.96. n1 = n2 =

b.

2 ( zα / 2 ) (σ 12 + σ 22 )

ME 2

=

(1.96) 2 (152 + 17 2 ) = 192.83 ≈ 193 3.22

If the range of each population is 40, we would estimate σ by:

σ ≈ 60/4 = 15 For confidence coefficient .99, α = 1 − .99 = .01 and α/2 = .01/2 = .005. From Table IV, Appendix B, z.005 = 2.58. n1 = n2 =

220

2 ( zα / 2 ) (σ 12 + σ 22 )

ME 2

=

(2.58) 2 (152 + 152 ) = 46.80 ≈ 47 82

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c.

For confidence coefficient .9, α = 1 − .9 = .1 and α/2 = .1/2 = .05. From Table IV, Appendix B, z.05 = 1.645. For a width of 1, the bound is .5. n1 = n2 =

7.60

2 ( zα / 2 ) (σ 12 + σ 22 )

ME

2

=

(1.645) 2 (5.82 + 7.52 ) = 143.96 ≈ 144 .52

First, find the sample sizes needed for width 5, or margin of error 2.5. For confidence coefficient .9, α = 1 − .9 = .1 and α/2 = .1/2 = .05. From Table IV, Appendix B, z.05 = 1.645. n1 = n2 =

2 ( zα / 2 ) (σ 12 + σ 22 )

ME 2

=

(1.645) 2 (102 + 102 ) = 86.59 ≈ 87 2.52

Thus, the necessary sample size from each population is 87. Therefore, sufficient funds have not been allocated to meet the specifications since n1 = n2 = 100 are large enough samples. 7.62

For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B, z.025 = 1.96.

n1 = n2 =

2 ( zα / 2 ) (σ 12 + σ 22 )

( ME ) 2

=

1.962 (3.1892 + 2.3552 ) = 26.8 ≈ 27 1.52

We would need to sample 27 specimens from each location. 7.64

For confidence coefficient .90, α = 1 − .90 = .10 and α/2 = .10/2 = .05. From Table IV, Appendix B, z.05 = 1.645. Since no information is given about the values of p1 and p2, we will be conservative and use .5 for both. A width of .04 means the bound is .04/2 = .02. n1 = n2 =

7.66

a.

( zα / 2 )

( p1 q1 + p2 q2 ) ( ME ) 2

=

1.6452 (.5(.5) + .5(.5) ) .022

= 3,382.5 ≈ 3,383

For confidence coefficient .80, α = 1 − .80 = .20 and α/2 = .20/2 = .10. From Table IV, Appendix B, z.10 = 1.28. Since we have no prior information about the proportions, we use p1 = p2 = .5 to get a conservative estimate. For a width of .06, the margin of error is .03. n1 = n2 =

b.

2

( zα / 2 )

2

( p1q1 + p2 q2 ) ME 2

=

(1.28) 2 (.5(1 − .5) + .5(1 − .5) ) .032

= 910.22 ≈ 911

For confidence coefficient .90, α = 1 − .90 = .10 and α/2 = .10/2 = .05. From Table IV, Appendix B, z.05 = 1.645. Using the formula for the sample size needed to estimate a proportion from Chapter 7, n=

( zα / 2 )

2

ME

2

pq

=

1.6452 (.5(1 − .5) ) .02

2

=

.6765 = 1691.27 ≈ 1692 .0004

No, the sample size from part a is not large enough.

Inferences Based on Two Samples: Confidence Intervals and Tests of Hypothesis

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7.68

For confidence coefficient .95, α = 1 − .95 = .05 and α/2 = .025. From Table IV, Appendix B, z.025 = 1.96.

n1 = n2 = 7.70

2 ( zα / 2 ) (σ 12 + σ 22 )

( ME ) 2

=

1.962 (352 + 802 ) = 292.9 ≈ 293 102

a.

With ν1 = 2 and ν2 = 30, P(F ≥ 5.39) = .01 (Table XI, Appendix B)

b.

With ν1 = 24 and ν2 = 10, P(F ≥ 2.74) = .05 (Table IX, Appendix B) Thus, P(F < 2.74) = 1 − P(F ≥ 2.74) = 1 − .05 = .95.

c.

With ν1 = 7 and ν2 = 1, P(F ≥ 236.8) = .05 (Table VIII, Appendix B) Thus, P(F < 236.8) = 1 − P(F ≥ 236.8) = 1 − .05 = .95.

d.

7.72

With ν1 = 40 and ν2 = 40, P(F > 2.11) = .01 (Table XI, Appendix B)

To test H0: σ 12 = σ 22 against Ha: σ 12 ≠ σ 22 , the rejection region is F > Fα/2 with ν1 = 10 and ν2 = 12. a.

α = .20, α/2 = .10 Reject H0 if F > F.10 = 2.19 (Table VIII, Appendix B)

b.

α = .10, α/2 = .05 Reject H0 if F > F.05 = 2.75 (Table IX, Appendix B)

c.

α = .05, α/2 = .025 Reject H0 if F > F.025 = 3.37 (Table X, Appendix B)

d.

α = .02, α/2 = .01 Reject H0 if F > F.01 = 4.30 (Table XI, Appendix B)

7.74

a.

To determine if a difference exists between the population variances, we test: H0: σ 12 = σ 22 Ha: σ 12 ≠ σ 22

The test statistic is F =

222

s22 8.75 = = 2.26 s12 3.87

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The rejection region requires α/2 = .10/2 = .05 in the upper tail of the F-distribution with ν1 = n2 − 1 = 27 − 1 = 26 and ν2 = n1 − 1 = 12 − 1 = 11. From Table IX, Appendix B, F.05 ≈ 2.60. The rejection region is F > 2.60. Since the observed value of the test statistic does not fall in the rejection region (F = 2.26 >/ 2.60), H0 is not rejected. There is insufficient evidence to indicate a difference between the population variances. b.

The p-value is 2P(F ≥ 2.26). From Tables VIII and IX, with ν1 = 26 and ν2 = 11, 2(.05) < 2P(F ≥ 2.26) < 2(.10) ⇒ .10 < 2P(F ≥ 2.26) < .20 There is no evidence to reject H0 for α ≤ .10.

7.76

Let σ 12 = variance of carat size for diamonds certified by GIA, σ 22 = variance of carat size for diamonds certified by HRD, and σ 32 = variance of carat size for diamonds certified by IGI. a.

To determine if the variation in carat size differs for diamonds certified by GIA and diamonds certified by HRD, we test: H0: σ 12 = σ 22

Ha: σ 12 ≠ σ 22 The test statistic is F =

Larger sample variance s12 .24562 = = = 1.799 Smaller sample variance s22 .18312

The rejection region requires α/2 = .05/2 = .025 in the upper tail of the F-distribution with ν1 = n1 – 1 = 151 – 1 = 150 and ν2 = n2 – 1 = 79 – 1 = 78. From Table X, Appendix B, F.025 ≈ 1.43. The rejection region is F > 1.43.

Since the observed value of the test statistic falls in the rejection region (F = 1.799 > 1.43), H0 is rejected. There is sufficient evidence to indicate the variation in carat size differs for diamonds certified by GIA and those certified by HRD at α = .05. b.

To determine if the variation in carat size differs for diamonds certified by GIA and diamonds certified by IGI, we test: H0: σ 12 = σ 32 Ha: σ 12 ≠ σ 32

The test statistic is F =

Larger sample variance s12 .24562 = = = 1.289 Smaller sample variance s32 .21632

The rejection region requires α/2 = .05/2 = .025 in the upper tail of the F-distribution with ν1 = n1 – 1 = 151 – 1 = 150 and ν2 = n3 – 1 = 78 – 1 = 77. From Table X, Appendix B, F.025 ≈ 1.43. The rejection region is F > 1.43.

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Since the observed value of the test statistic does not fall in the rejection region (F = 1.289 >/ 1.43), H0 is not rejected. There is insufficient evidence to indicate the variation in carat size differs for diamonds certified by GIA and those certified by IGI at α = .05. c.

To determine if the variation in carat size differs for diamonds certified by HRD and diamonds certified by IGI, we test: H0: σ 22 = σ 32 Ha: σ 22 ≠ σ 32

Larger sample variance s32 .21632 = = = 1.396 Smaller sample variance s22 .18312 The rejection region requires α/2 = .05/2 = .025 in the upper tail of the F-distribution with ν1 = n3 – 1 = 78 – 1 = 77 and ν2 = n2 – 1 = 79 – 1 = 78. From Table X, Appendix B, F.025 ≈ 1.67. The rejection region is F > 1.67. The test statistic is F =

Since the observed value of the test statistic does not fall in the rejection region (F = 1.396 >/ 1.67), H0 is not rejected. There is insufficient evidence to indicate the variation in carat size differs for diamonds certified by HRD and those certified by IGI at α = .05. d.

We will look at the 4 methods for determining if the data are normal. First, we will look at histograms of the data. Using MINITAB, the histograms of the carat sizes for the 3 certification bodies are:

40 40

30

Percent

Percent

30

20

20

10 10

0

0

0.0

0.5

1.0

0.0

GIA

0.5

1.0

HRD

40

Percent

30

20

10

0 0.0

0.5

1.0

IGI

From the histograms, none of the data appear to be mound-shaped. It appears that none of the data sets are normal.

224

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Next, we look at the intervals x ± s, x ± 2 s, x ± 3s . If the proportions of observations falling in each interval are approximately .68, .95, and 1.00, then the data are approximately normal. Using MINITAB, the summary statistics are: Descriptive Statistics: GIA, IGI, HRD Variable GIA IGI HRD

N 151 78 79

Mean 0.6723 0.3665 0.8129

Median 0.7000 0.2900 0.8100

TrMean 0.6713 0.3406 0.8169

Variable GIA IGI HRD

Minimum 0.3000 0.1800 0.5000

Maximum 1.1000 1.0100 1.0900

Q1 0.5000 0.2100 0.6500

Q3 0.9000 0.4850 1.0000

StDev 0.2456 0.2163 0.1831

SE Mean 0.0200 0.0245 0.0206

For GIA:

x ± s ⇒ .6723 ± .2456 ⇒ (.4267, .9179) 84 of the 151 values fall in this interval. The proportion is .56. This is much smaller than the .68 we would expect if the data were normal. x ± 2 s ⇒ .6723 ± 2(.2456) ⇒ .6723 ± .4912 ⇒ (.1811, 1.1635) 151 of the 151 values fall in this interval. The proportion is 1.00. This is much larger than the .95 we would expect if the data were normal. x ± 3s ⇒ .6723 ± 3(.2456) ⇒ .6723 ± .7368 ⇒ (−.0645, 1.4091) 151 of the 151 values fall in this interval. The proportion is 1.00. This is the same as the 1.00 we would expect if the data were normal. From this method, it appears that the data are not normal. For IGI:

x ± s ⇒ .3665 ± .2163 ⇒ (.1502, .5828) 69 of the 78 values fall in this interval. The proportion is .88. This is much larger than the .68 we would expect if the data were normal. x ± 2s ⇒ .3665 ± 2(.2163) ⇒ .3665 ± .4326 ⇒ (−.0661, .7991) 74 of the 78 values fall in this interval. The proportion is .95. This is the same as the .95 we would expect if the data were normal. x ± 3s ⇒ .3665 ± 3(.2163) ⇒ .3665 ± .6489 ⇒ (−.2824, 1.0154) 78 of the 78 values fall in this interval. The proportion is 1.00. This is the same as the 1.00 we would expect if the data were normal. From this method, it appears that the data are not normal.

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For HRD:

x ± s ⇒ .8129 ± .1831 ⇒ (.6298, .9960) 30 of the 79 values fall in this interval. The proportion is .38. This is much smaller than the .68 we would expect if the data were normal. x ± 2 s ⇒ .8129 ± 2(.1831) ⇒ .8129 ± .3662 ⇒ (.4467, 1.1791) 79 of the 79 values fall in this interval. The proportion is 1.00. This is much larger than the .95 we would expect if the data were normal. x ± 3s ⇒ .8129 ± 3(.1831) ⇒ .8129 ± .5493 ⇒ (.2636, 1.3622) 79 of the 79 values fall in this interval. The proportion is 1.00. This is the same as the 1.00 we would expect if the data were normal. From this method, it appears that the data are not normal. Next, we look at the ratio of the IQR to s. For GIA:

IQR = QU – QL = 1.1 − .3 = .8. IQR .8 = = 3.26 This is much larger than the 1.3 we would expect if the data were s .2456 normal. This method indicates the data are not normal. For IGI:

IQR = QU – QL = 1.01 - .18 = .83. IQR .83 = = 3.84 This is much larger than the 1.3 we would expect if the data were s .2163 normal. This method indicates the data are not normal. For HRD:

IQR = QU – QL = 1.09 - .5 = .59. IQR .59 = = 3.22 This is much larger than the 1.3 we would expect if the data were s .1831 normal. This method indicates the data are not normal.

226

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Finally, using MINITAB, the normal probability plot for GIA is: Normal Probability Plot for GIA ML Estimates - 95% CI

ML Estimates 99

Percent

95 90

Mean

0.672252

StDev

0.244757

Goodness of Fit

80 70 60 50 40 30 20

AD*

3.332

10 5 1

0.0

0.5

1.0

1.5

Data

Since the data do not form a straight line, the data are not normal. Using MINITAB, the normal probability plot for IGI is: Normal Probability Plot for IGI ML Estimates - 95% CI

ML Estimates 99

Mean

0.366538

StDev

0.214863

Percent

95 90

Goodness of Fit

80 70 60 50 40 30 20

AD*

5.622

10 5 1

0.0

0.5

1.0

Data

Since the data do not form a straight line, the data are not normal.

Inferences Based on Two Samples: Confidence Intervals and Tests of Hypothesis

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Using MINITAB, the normal probability plot for HRD is: Normal Probability Plot for HRD ML Estimates - 95% CI

ML Estimates 99

Percent

95 90

Mean

0.812911

StDev

0.181890

Goodness of Fit

80 70 60 50 40 30 20

AD*

3.539

10 5 1

0.5

1.0

1.5

Data

Since the data do not form a straight line, the data are not normal. From the 4 different methods, all indications are that the carat size data are not normal for any of the certification bodies. 7.78

a.

The amount of variability of GHQ scores tells us how similar or different the members of the group are on GHQ scores. The larger the variability, the larger the differences are among the members on the GHQ scores. The smaller the variability, the smaller the differences are among the members on the GHQ scores.

b.

Let σ 12 = variance of the mental health scores of the employed and σ 22 = variance of the mental health scores of the unemployed. To determine if the variability in mental health scores differs for employed and unemployed workers, we test: H0: σ 12 = σ 22 Ha: σ 12 ≠ σ 22

c.

The test statistic is F =

Larger sample variance s12 5.102 = 2.45 = = Smaller sample variance s22 3.262

The rejection region requires α/2 = .05/2 = .025 in the upper tail of the F-distribution with ν1 = n2 − 1 = 49 − 1 = 48 and ν2 = n1 − 1 = 142 − 1 = 141. From Table XI, Appendix B, F.025 ≈ 1.61. The rejection region is F > 1.61. Since the observed value of the test statistic falls in the rejection region (F = 2.45 > 1.61), H0 is rejected. There is sufficient evidence to indicate that the variability in mental health scores differs for employed and unemployed workers for α = 05.

228

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d.

7.80

We must assume that the 2 populations of mental health scores are normally distributed. We must also assume that we selected 2 independent random samples.

Let σ 12 = variance zinc measurements from the text-line, σ 22 = variance zinc measurements from the witness-line, and σ 32 = variance zinc measurements from the intersection. Using MINITAB, the descriptive statistics are: Descriptive Statistics: Text-line, Witness-line, Intersection Variable Text-lin WitnessIntersec

N 3 6 5

Mean 0.3830 0.3042 0.3290

Median 0.3740 0.2955 0.3190

TrMean 0.3830 0.3042 0.3290

Variable Text-lin WitnessIntersec

Minimum 0.3350 0.1880 0.2850

Maximum 0.4400 0.4390 0.3930

Q1 0.3350 0.2045 0.2900

Q3 0.4400 0.4075 0.3730

a.

StDev 0.0531 0.1015 0.0443

SE Mean 0.0306 0.0415 0.0198

To determine if the variation in the zinc measurements for the text-line and the intersection differ, we test: H0: σ 12 = σ 32 Ha: σ 12 ≠ σ 32 The test statistic is F =

Larger sample variance s12 .05312 = 1.437 = = Smaller sample variance s32 .04432

The rejection region requires α/2 = .05/2 = .025 in the upper tail of the F-distribution with ν1 = n1 – 1 = 3 – 1 = 2 and ν2 = n3 – 1 = 5 – 1 = 4. From Table X, Appendix B, F.025 = 10.65. The rejection region is F > 10.65. Since the observed value of the test statistic does not fall in the rejection region (F = 1.437 >/ 10.65), H0 is not rejected. There is insufficient evidence to indicate the variation in the zinc measurements for the text-line and the intersection differ at α = .05. b.

To determine if the variation in the zinc measurements for the witness-line and the intersection differ, we test: H0: σ 22 = σ 32 Ha: σ 22 ≠ σ 32 The test statistic is F =

Larger sample variance s22 .10152 = 5.250 = = Smaller sample variance s32 .04432

The rejection region requires α/2 = .05/2 = .025 in the upper tail of the F-distribution with ν1 = n2 – 1 = 6 – 1 = 5 and ν2 = n3 – 1 = 5 – 1 = 4. From Table X, Appendix B, F.025 = 9.36. The rejection region is F > 9.36. Since the observed value of the test statistic does not fall in the rejection region (F = 5.250 >/ 9.36), H0 is not rejected. There is insufficient evidence to indicate the variation in the zinc measurements for the witness-line and the intersection differ at α = .05.

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7.82

c.

There is no indication that the variances of the zinc measurements for three locations differ.

d.

With only 3, 6, and 5 measurements, it is very difficult to check the assumptions.

Using MINITAB, some preliminary calculations are: Descriptive Statistics: HEATRATE Variable HEATRATE

ENGINE Advanced Aeroderiv Traditional

N 21 7 39

Variable HEATRATE

ENGINE Advanced Aeroderiv Traditional

Q3 10060 14628 11964

a.

N* 0 0 0

Mean 9764 12312 11544

SE Mean 139 1002 205

StDev 639 2652 1279

Minimum 9105 8714 10086

Q1 9252 9469 10592

Median 9669 12414 11183

Maximum 11588 16243 14796

To determine if the heat rate variances for traditional and aeroderivative augmented gas turbines differ, we test: H0: σ 22 = σ 32 Ha: σ 22 ≠ σ 32 89) The test statistic is

F=

Larger sample variance s22 26522 = 4.299 = = Smaller sample variance s32 12792

The rejection region requires α/2 = .05/2 = .025 in the upper tail of the F distribution with numerator df = ν2 = n2 – 1 = 7 – 1 = 6 and denominator df = ν3 = n3 – 1 = 39 – 1 = 38. From Table X, Appendix B, F.025 ≈ 2.74. The rejection region is F > 2.74. Since the observed value of the test statistic falls in the rejection region (F = 4.299 > 2.74), H0 is rejected. There is sufficient evidence to indicate the heat rate variances for traditional and aeroderivative augmented gas turbines differ at α = .05. Since the test in Exercise 7.23 a assumes that the population variances are the same, the validity of the test is suspect since we just found the variances are different. b.

To determine if the heat rate variances for advanced and aeroderivative augmented gas turbines differ, we test: H0: σ 12 = σ 22 Ha: σ 12 ≠ σ 22

230

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Larger sample variance s 212 26522 = 17.224 = = The test statistic is F = Smaller sample variance s12 6392 The rejection region requires α/2 = .05/2 = .025 in the upper tail of the F distribution with numerator df = ν1 = n1 – 1 = 7 – 1 = 6 and denominator df = ν2 = n2 – 1 = 21 – 1 = 20. From Table X, Appendix B, F.025 = 3.13. The rejection region is F > 3.13. Since the observed value of the test statistic falls in the rejection region (F = 17.224 > 3.13), H0 is rejected. There is sufficient evidence to indicate the heat rate variances for advanced and aeroderivative augmented gas turbines differ at α = .05. Since the test in Exercise 7.23 b assumes that the population variances are the same, the validity of the test is suspect since we just found the variances are different. 7.84

a.

The 2 samples are randomly selected in an independent manner from the two populations. The sample sizes, n1 and n2, are large enough so that x1 and x2 each have approximately normal sampling distributions and so that s12 and s22 provide good approximations to σ 12 and σ 22 . This will be true if n1 ≥ 30 and n2 ≥ 30.

b.

7.86

1. 2. 3.

Both sampled populations have relative frequency distributions that are approximately normal. The population variances are equal. The samples are randomly and independently selected from the populations.

c.

1. 2.

The relative frequency distribution of the population of differences is normal. The sample of differences are randomly selected from the population of differences.

d.

The two samples are independent random samples from binomial distributions. Both samples should be large enough so that the normal distribution provides an adequate approximation to the sampling distributions of pˆ1 and pˆ 2 .

e.

The two samples are independent random samples from populations which are normally distributed.

a.

H0: σ 12 = σ 22 Ha: σ 12 ≠ σ 22 The test statistic is F =

s2 Larger sample variance 120.1 = 22 = = 3.84 Smaller sample variance s1 31.3

The rejection region requires α/2 = .05/2 = .025 in the upper tail of the F-distribution with numerator df ν1 = n2 − 1 = 15 − 1 = 14 and denominator df ν2 = n1 − 1 = 20 − 1 = 19. From Table XI, Appendix B, F.025 ≈ 2.66. The rejection region is F > 2.66. Since the observed value of the test statistic falls in the rejection region (F = 3.84 > 2.66), H0 is rejected. There is sufficient evidence to conclude σ 12 ≠ σ 22 at α = .05.

Inferences Based on Two Samples: Confidence Intervals and Tests of Hypothesis

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b.

7.88

No, we should not use a small sample t test to test H0: (μ1 − μ2) = 0 against Ha: (μ1 − μ2) ≠ 0 because the assumption of equal variances does not seem to hold since we concluded σ 12 ≠ σ 22 in part b.

Some preliminary calculations are:

pˆ1 = a.

x1 110 x 130 x +x 110 + 130 240 = = = .55; pˆ 2 = 2 = = .65; pˆ = 1 2 = n1 200 n2 200 n1 + n2 200 + 200 400

H0: (p1 − p2) = 0 Ha: (p1 − p2) < 0 The test statistic is z =

( pˆ1 − pˆ 2 ) − 0 ⎛1 1⎞ ˆ ˆ⎜ + ⎟ pq ⎝ n1 n2 ⎠

=

(.55 − .65) − 0 1 ⎞ ⎛ 1 .6(1 − .6) ⎜ + ⎟ ⎝ 200 200 ⎠

=

−.10 = −2.04 .049

The rejection region requires α = .10 in the lower tail of the z-distribution. From Table IV, Appendix B, z.10 = 1.28. The rejection region is z < −1.28. Since the observed value of the test statistic falls in the rejection region (z = −2.04 < −1.28), H0 is rejected. There is sufficient evidence to conclude (p1 − p2 < 0) at α = .10. b.

For confidence coefficient .95, α = 1 − .95 = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B, z.025 = 1.96. The 95% confidence interval for (p1 − p2) is approximately: pˆ qˆ pˆ qˆ ( pˆ1 − pˆ 2 ) ± zα / 2 1 2 + 2 2 n1 n2 .55(1 − .55) .65(1 − .65) + 200 200 ⇒ −.10 ± .096 ⇒ (−.196, −.004) ⇒ (.55 − .65) ± 1.96

c.

From part b, z.025 = 1.96. Using the information from our samples, we can use p1 = .55 and p2 = .65. For a width of .01, the margin of error is .005. n1 = n2 =

232

( zα / 2 )

2

( p1q1 + p2 q2 ) ME

2

=

(1.96) 2 (.55(1 − .55) + .65(1 − .65) )

.005 = 72990.4 ≈ 72,991

2

=

1.82476 .000025

Chapter 7

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7.90

a.

Let p1 = proportion of Opening Doors students enrolled full time and p2 = proportion of traditional students enrolled full time. The target parameter for this comparison is p1 – p2.

b.

Let μ1 = mean GPA of Opening Doors students and μ2 = mean GPA of traditional students. The target parameter for this comparison is μ1 – μ2.

7.92

Using MINITAB, some preliminary calculations are: Descriptive Statistics: Spillage Variable Spillage

Cause Collision Fire Grounding HullFail Unknown

Variable Spillage

Cause Q3 Maximum Collision 102.0 257.0 Fire 80.5 239.0 Grounding 59.00 124.00 HullFail 46.0 221.0 Unknown * 27.00

a.

N 10 12 14 12 2

N* 0 0 0 0 0

Mean 76.6 70.9 47.79 54.4 26.00

SE Mean 22.3 17.5 7.61 16.3 1.00

StDev 70.4 60.7 28.47 56.4 1.41

Minimum 31.0 26.0 21.00 24.0 25.00

Q1 35.0 32.3 30.25 29.3 *

Median 41.5 49.0 37.50 31.5 26.00

Let μ1 = mean spillage for accidents caused by collision and μ2 = mean spillage for accidents caused by fire/explosion. s 2p =

( n1 − 1) s12 + ( n2 − 1) s22 = (10 − 1) 70.42 + (12 − 1) 60.72 n1 + n2 − 2

10 + 12 − 2

= 4, 256.7415

For confidence coefficient .90, α = 1 − .90 = .10 and α/2 = .10/2 = .05. From Table VI, Appendix B, with df = n1 + n2 − 2 = 10 + 12 – 2 = 20, t.05 = 1.725. The confidence interval is:

⎛1 1 ⎞ ⎛ 1 1 ⎞ ( x1 − x2 ) ± t.05 s 2p ⎜ + ⎟ ⇒ (76.6 − 70.9) ± 1.725 4256.7415 ⎜ + ⎟ ⎝ 10 12 ⎠ ⎝ n1 n2 ⎠ ⇒ 5.7 ± 48.19 ⇒ ( −42.59, 53.89) b.

Let μ3 = mean spillage for accidents caused by grounding and μ4 = mean spillage for accidents caused by hull failure. s 2p =

( n1 − 1) s12 + ( n2 − 1) s22 = (14 − 1) 28.47 2 + (12 − 1) 56.42 n1 + n2 − 2

14 + 12 − 2

= 1,896.9830

Inferences Based on Two Samples: Confidence Intervals and Tests of Hypothesis

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To determine if the mean spillage amount for accidents caused by grounding is different from the mean spillage amount caused by hull failure, we test: H0: μ3 − μ4 = 0 Ha: μ3 − μ4 ≠ 0 The test statistic is t =

( x1 − x2 ) − Do ⎛1 1 ⎞ s 2p ⎜ + ⎟ ⎝ n1 n2 ⎠

=

( 47.79 − 54.4 ) − 0 ⎛ 1 1⎞ 1896.983 ⎜ + ⎟ ⎝ 14 12 ⎠

=

−6.61 = −.39 17.1342

The rejection region requires α/2 = .05/2 = .025 in each tail of the t-distribution with df = n1 + n2 – 2 = 14 +12 – 2 = 24. From Table VI, Appendix B, t.025 = 2.064. The rejection region is t < −2.064 or t > 2.064. Since the observed value of the test statistic does not fall in the rejection region (t = −.39 4.19. Since the observed value of the test statistic falls in the rejection region (F = 6.11 > 4.19), H0 is rejected. There is sufficient evidence to indicate the variances of the amounts of spillage due to collision and grounding differ at α = .02. 7.94

a.

Let μ1 = mean rating of concern about product tampering for males and μ2 = mean rating of concern about product tampering for females. To determine whether a difference exists in the mean level of concern about product tampering between males and females, we test: H0: μ1 − μ2 = 0 Ha: μ1 − μ2 ≠ 0

7.96

b.

The p-value = .008. Since the p-value is so small, there is evidence to reject H0. There is sufficient evidence to indicate a difference exists in the mean level of concern about product tampering between males and females for α > .008.

c.

We must assume the sample sizes were sufficiently large so that the Central Limit Theorem applies. We must also assume that we selected two random and independent samples from the two populations.

For confidence coefficient .95, α = .05 and α/2 = .025. From Table IV, Appendix B, z.025 = 1.96. n1 − n 2 =

7.98

236

a.

( zα / 2 )

2

( p1q1 + p2 q2 ) ( ME ) 2

=

1.962 (.395(.605) + .293(.707) ) .032

= 1904.26 ≈ 1905

Let p1999 = proportion of adult Americans who would vote for a woman president in 1999 and p1975 = proportion of adult Americans who would vote for a woman president in 1975.

Chapter 7

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b.

To see if the samples are sufficiently large: pˆ1999 ± 3σ pˆ1999 ⇒ pˆ1999 ± 3

p1999 q1999 pˆ qˆ .92(.08) ⇒ pˆ1999 ± 3 1999 1999 ⇒ .92 ± 3 n1999 n1999 2000

⇒ .92 ± .02 ⇒ (.90, .94) pˆ1975 ± 3σ pˆ1975 ⇒ pˆ1975 ± 3

p1975 q1975 pˆ qˆ .73(.27) ⇒ pˆ1975 ± 3 1975 1975 ⇒ .73 ± 3 n1975 n1975 2000

⇒ .73 ± .03 ⇒ (.70, .76) Since both intervals are contained within the interval (0, 1), the normal approximation will be adequate. c.

For confidence coefficient .90, α = .10 and α/2 = .10/2 = .05. From Table IV, Appendix B, z.05 = 1.645. The 90% confidence interval is: ( pˆ1 − pˆ 2 ) ± z.05

pˆ1 qˆ1 pˆ 2 qˆ2 .92(.08) .73(.27) + ⇒ (.92 − .73) ± 1.645 + n1 n2 2000 1500

⇒ .19 ± .02 ⇒ (.17, .21) We are 90% confident that the difference in the proportions of adult Americans who would vote for a woman president between 1999 and 1975 is between .17 and .21. d.

To see if the samples are sufficiently large: pˆ1999 ± 3σ pˆ1999 ⇒ pˆ1999 ± 3

p1999 q1999 pˆ qˆ .92(.08) ⇒ pˆ1999 ± 3 1999 1999 ⇒ .92 ± 3 n1999 n1999 20

⇒ .92 ± .18 ⇒ (.74, 1.10) pˆ1975 ± 3σ pˆ1975 ⇒ pˆ1975 ± 3

p1975 q1975 pˆ qˆ .73(.27) ⇒ pˆ1975 ± 3 1975 1975 ⇒ .73 ± 3 n1975 n1975 50

⇒ .73 ± .19 ⇒ (.54, .92) Since the first interval is not contained within the interval (0, 1), the normal approximation will not be adequate. 7.100

a.

For each measure, let μ1 = mean job satisfaction for day-shift nurses and μ2 = mean job satisfaction for night-shift nurses. To determine whether a difference in job satisfaction exists between day-shift and night-shift nurses, we test: H0: μ1 − μ2 = 0 Ha: μ1 − μ2 ≠ 0

Inferences Based on Two Samples: Confidence Intervals and Tests of Hypothesis

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b.

Hours of work: The p-value = .813. Since the p-value is so large, there is no evidence to reject H0. There is insufficient evidence to indicate a difference in mean job satisfaction exists between day-shift and night-shift nurses on hours of work for α ≤ .10. Free time: The p-value = .047. Since the p-value is so small, there is evidence to reject H0. There is sufficient evidence to indicate a difference in mean job satisfaction exists between day-shift and night-shift nurses on free time for α > .047. Breaks: The p-value = .0073. Since the p-value is so small, there is evidence to reject H0. There is sufficient evidence to indicate a difference in mean job satisfaction exists between day-shift and night-shift nurses on breaks for α > .0073.

c.

We must make the following assumptions for each measure: 1. 2. 3.

7.102

The job satisfaction scores for both day-shift and night-shift nurses are normally distributed. The variances of job satisfaction scores for both day-shift and night-shift nurses are equal. Random and independent samples were selected from both populations of job satisfaction scores.

For confidence coefficient .90, α = 1 − .90 = .10 and α/2 = .10/2 = .05. From Table IV, Appendix B, z.05 = 1.645. We estimate p1 = p2 = .5. n1 − n 2 =

7.104

( zα / 2 )

2

( p1q1 + p2 q2 ) ( ME ) 2

=

(1.645)2 (.5(.5) + .5(.5) ) .052

= 541.205 ≈ 542

Let p1 = proportion of larvae that died in containers containing high carbon dioxide levels and p2 = proportion of larvae that died in containers containing normal carbon dioxide levels. The parameter of interest for this problem is p1 − p2, or the difference in the death rates for the two groups. Some preliminary calculations are: pˆ =

x1 + x2 .10(80) + .05(80) = = .075 n1 + n2 80 + 80

qˆ = 1 − pˆ = 1 − .075 = .925

To determine if an increased level of carbon dioxide is effective in killing a higher percentage of leaf-eating larvae, we test: H0: p1 − p2 = 0 Ha: p1 − p2 > 0 The test statistic is z =

238

( pˆ1 − pˆ 2 ) − 0 1 1 ˆ ˆ ⎛⎜ + ⎞⎟ pq ⎝ 80 80 ⎠

=

(.10 − .05) − 0 1 ⎞ ⎛ 1 .075(.925) ⎜ + ⎟ ⎝ 80 80 ⎠

= 1.201

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The rejection region requires α = .01 in the upper tail of the z distribution. From Table IV, Appendix B, z.01 = 2.33. The rejection region is z > 2.33. Since the observed value of the test statistic does not fall in the rejection region (z = 1.201 >/ 2.33), H0 is not rejected. There is insufficient evidence to indicate that an increased level of carbon dioxide is effective in killing a higher percentage of leaf-eating larvae at α = .01. 7.106

a.

Let p1 = proportion of female students who switched due to loss of interest in SME and p2 = proportion of male students who switched due to lack of interest in SME. Some preliminary calculations are: pˆ1 =

x1 74 x x +x 72 74 + 72 = = .43; pˆ 2 = 2 = = .44; pˆ = 1 2 = = .436 n1 172 n2 163 n1 + n2 172 + 163

To determine if the proportion of female students who switch due to lack of interest in SME differs from the proportion of males who switch due to a lack of interest, we test: H0: p1 − p2 = 0 Ha: p1 − p2 ≠ 0 The test statistic is z =

( pˆ1 − pˆ 2 ) − 0 ⎛1 1⎞ ˆ ˆ⎜ + ⎟ pq ⎝ n1 n2 ⎠

=

(.43 − .44) − 0 1 ⎞ ⎛ 1 + .436(.564) ⎜ ⎟ ⎝ 172 163 ⎠

= −0.18

The rejection region requires α/2 = .10/2 = .05 in each tail of the z-distribution. From Table IV, Appendix B, z.05 = 1.645. The rejection region is z < −1.645 or z > 1.645. Since the observed value of the test statistic does not fall in the rejection region (z = −0.18 6.39. Since the observed value of the test statistic does not fall in the rejection region (F = 2.79 >/ 6.39), H0 is not rejected. There is insufficient evidence of a difference in the precision of the two instruments at α = .10.

240

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7.112

a.

Let μ1 = mean change in bond prices handled by underwriter 1 and μ2 = mean change in bond prices handled by underwriter 2. sp2 =

(n1 − 1) s12 + ( n2 − 1) s22 (27 − 1).0098 + (23 − 1).002465 .30903 = = .006438 = 48 n1 + n2 − 2 27 + 23 − 2

To determine if there is a difference in the mean change in bond prices handled by the 2 underwriters, we test: H0: μ1 − μ2 = 0 Ha: μ1 − μ2 ≠ 0 The test statistic is t =

( x1 − x2 ) − D0 ⎛1 1⎞ s ⎜ + ⎟ ⎝ n1 n2 ⎠

=

2 p

−.0491 − (−.0307) − 0 1 ⎞ ⎛ 1 .006438 ⎜ + ⎟ ⎝ 27 23 ⎠

= −.81

The rejection region requires α/2 = .05/2 = .025 in each tail of the t-distribution with df = n1 + n2 − 2 = 27 + 23 − 2 = 48. From Table VI, Appendix B, t.025 ≈ 1.96. The rejection region is t < −1.96 or t > 1.96. Since the observed value of the test statistic does not fall in the rejection region (t = −.81 μF

b.

The test statistic is z =

( xM − xF ) − 0 s

2 xM

+s

2 xF

=

(61, 340 − 32, 227) 2,1852 + 9322

= 12.26

Inferences Based on Two Samples: Confidence Intervals and Tests of Hypothesis

241

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242

c.

The rejection region requires α = .01 in the upper tail of the z-distribution. From Table IV, Appendix B, z.01 = 2.33. The rejection region is z > 2.33.

d.

Since the observed value of the test statistic falls in the rejection region (z = 12.26 > 2.33), H0 is rejected. There is sufficient evidence to indicate the mean salary of all males with post-graduate degrees exceeds the mean salary of all females with post-graduate degrees at α = .01.

Chapter 7

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The Kentucky Milk Case—Part II (To accompany Chapters 5–7)

(1)Incumbency Rates I have repeated the incumbency rates for the Tri-county market. If the "normal" incumbency rate is .7 in competitive markets, then we would like to test to see if the incumbency rate in the Tri-county market is larger than .7. We will run a test for each of the years from 1985 through 1988, and also for the four years combined.

Year 1984 1985 1986 1987 1988 1989 1990 1991

Tri-County Market Number of Same Incumbency Districts Vendors Rate 10 8 .800 12 12 1.000 13 13 1.000 13 12 .923 13 13 1.000 13 9 .692 13 10 .769 13 11 .846

1985 One of the assumptions necessary for this test is that the sample size is sufficiently large. In order for the sample size to be sufficiently large, the interval p0 ± 3σ pˆ must not contain 0 or 1. For this problem, the interval is: p0 ± 3σ pˆ ⇒ .7 ± 3

.7(.3) ⇒ .7 ± .397 ⇒ (.303, 1.097) 12

Since 1 is included in the interval, the sample size is not sufficiently large. The following test may not be valid. To see if the incumbency rate in 1985 exceeds .7, we test: H0: p = .7 Ha: p > .7

The test statistic is z =

pˆ − p0 p0 q0 n

The Kentucky Milk Case—Part II

=

1 − .7 = 2.27 .7(.3) 12

243

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The rejection region requires α = .05 in the upper tail of the z-distribution. From Table IV, Appendix B, z.05 = 1.645. The rejection region is z > 1.645. Since the observed value of the test statistic falls in the rejection region (z = 2.27 > 1.645), H0 is rejected. There is sufficient evidence to indicate that the incumbency rate exceeds .7 in the Tricounty market at α = .05. 1986

In order for the sample size to be sufficiently large, the interval p0 ± 3σ pˆ must not contain 0 or 1. For this problem, the interval is: p0 ± 3σ pˆ ⇒ .7 ± 3

.7(.3) ⇒ .7 ± .381 ⇒ (.319, 1.081) 13

Since 1 is included in the interval, the sample size is not sufficiently large. The following test may not be valid. To see if the incumbency rate in 1986 exceeds .7, we test: H0: p = .7 Ha: p > .7

The test statistic is z =

pˆ − p0 p0 q0 n

=

1 − .7 .7(.3) 13

= 2.36

The rejection region requires α = .05 in the upper tail of the z-distribution. From Table IV, Appendix B, z.05 = 1.645. The rejection region is z > 1.645. Since the observed value of the test statistic falls in the rejection region (z = 2.36 > 1.645), H0 is rejected. There is sufficient evidence to indicate that the incumbency rate exceeds .7 in the Tricounty market at α = .05. 1987

In order for the sample size to be sufficiently large, the interval p0 ± 3σ pˆ must not contain 0 or 1. For this problem, the interval is: p0 ± 3σ pˆ ⇒ .7 ± 3

244

.7(.3) ⇒ .7 ± .381 ⇒ (.319, 1.081) 13

The Kentucky Milk Case—Part II

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Since 1 is included in the interval, the sample size is not sufficiently large. The following test may not be valid. To see if the incumbency rate in 1987 exceeds .7, we test: H0: p = .7 Ha: p > .7 The test statistic is z =

pˆ − p0 p0 q0 n

=

.923 − .7 = 1.75 .7(.3) 13

The rejection region requires α = .05 in the upper tail of the z-distribution. From Table IV, Appendix B, z.05 = 1.645. The rejection region is z > 1.645. Since the observed value of the test statistic falls in the rejection region (z = 1.75 > 1.645), H0 is rejected. There is sufficient evidence to indicate that the incumbency rate exceeds .7 in the Tricounty market at α = .05. 1988

This test is the same as the test for 1986. Combined 1985-1988

To see if the sample size is sufficiently large, the interval p0 ± 3σ pˆ must not contain 0 or 1. For this problem, the interval is: p0 ± 3σ pˆ ⇒ .7 ± 3

.7(.3) ⇒ .7 ± .193 ⇒ (.507, .893) 51

Since neither 0 nor 1 is included in the interval, the sample size is sufficiently large. pˆ =

50 = .980 51

To see if the incumbency rate in 1985–1988 exceeds .7, we test: H0: p = .7 Ha: p > .7

The test statistic is z =

pˆ − p0 p0 q0 n

=

980 − .7 = 4.36 .7(.3) 51

The rejection region requires α = .05 in the upper tail of the z-distribution. From Table IV, Appendix B, z.05 = 1.645. The rejection region is z > 1.645.

The Kentucky Milk Case—Part II

245

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Since the observed value of the test statistic falls in the rejection region (z = 4.36 > 1.645), H0 is rejected. There is sufficient evidence to indicate that the incumbency rate exceeds .7 in the Tricounty market at α = .05. Thus, there is evidence, based on the incumbency rates, that bid collusion is present in the Tricounty market.

(2)Bid Price Dispersion

Again, we can use only the data provided which are the winning bids in each of the school districts in both markets. The sample sizes and the variances for each of the milk products for each year and each market are provided in the table. Whole White Milk

YR 83 84 85 86 87 88 89 90 91

N 22 22 26 33 36 36 37 35 5

Surround Market VAR 0.000212 0.000188 0.000174 0.000120 0.000105 0.000128 0.000056 0.000063 0.000042

N 8 9 10 10 12 12 12 12 13

Tri-County Market VAR 0.000213 0.000022 0.000028 0.000019 0.000027 0.000024 0.000089 0.000010 0.000020

N 10 12 13 13 13 13 13 13 12

Tri-County Market VAR 0.000155 0.000040 0.000028 0.000028 0.000049 0.000038 0.000068 0.000025 0.000034

Lowfat White Milk

YR 83 84 85 86 87 88 89 90 91

246

N 24 26 29 33 35 35 35 34 5

Surround Market VAR 0.000279 0.000216 0.000210 0.000139 0.000152 0.000165 0.000043 0.000091 0.000051

The Kentucky Milk Case—Part II

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Lowfat Chocolate Milk

YR 83 84 85 86 87 88 89 90 91

N 24 25 28 34 36 36 36 33 5

Surround Market VAR 0.000287 0.000234 0.000248 0.000163 0.000163 0.000184 0.000060 0.000098 0.000098

N 5 6 6 6 7 9 9 10 11

Tri-County Market VAR 0.000015 0.000060 0.000038 0.000027 0.000040 0.000087 0.000087 0.000014 0.000042

I will write out the first test and then summarize the others in a table. The first test will be for the year 1983 and will compare the variances of the whole white milk. To determine if the variances in the winning bid prices differ for the two markets, we test:

σ 12 =1 σ 22 σ2 Ha: 12 ≠ 1 σ2 H0:

The test statistic is F =

s2 larger sample variance .000213 = 1.005 = 12 = s2 smaller sample variance .000212

The rejection region requires α/2 = .05/2 = .025 in the upper tail of the F-distribution with ν1 = n2 − 1 = 8 − 1 = 7 and ν2 = n1 − 1 = 22 − 1 = 21. From Table IX, Appendix B, F.025 = 2.97. The rejection region is F > 2.97. Since the observed value of the test statistic does not fall in the rejection region (F = 1.005 >/ 2.97), H0 is not rejected. There is insufficient evidence to indicate that the variances of the winning bids are different for the two markets. Whole White Milk Year 1983 1984 1985 1986 1987 1988 1989 1990 1991

ν 1, ν 2 7,21 21,8 25,9 32,9 35,11 35,11 11,36 34,11 4,12

F.025 2.97 4.00 3.61 3.56 2.96 2.96 2.51 3.12 4.12

The Kentucky Milk Case—Part II

F 1.005 8.545 6.214 6.316 3.889 5.333 1.589 6.300 2.100

Decision Do not reject Reject Reject Reject Reject Reject Do not reject Reject Do not reject

247

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In all cases where there was a significant difference in the variances of the winning bids between the two markets, the variance in the Surrounding market was larger than the variance in the Tri-county market. This implies that collusion might be present in the Tri-county market. Lowfat White Milk Year 1983 1984 1985 1986 1987 1988 1989 1990 1991

ν 1, ν 2 23,9 25,11 28,12 32,12 34,12 34,12 12,34 33,12 4,11

F.025 3.62 3.17 3.02 2.96 2.96 2.96 2.41 2.96 4.28

F 1.800 5.400 7.500 4.964 3.102 4.342 1.581 3.640 1.500

Decision Do not reject Reject Reject Reject Reject Reject Do not reject Reject Do not reject

Again, in all cases where there was a significant difference in the variances of the winning bids between the two markets, the variance in the Surrounding market was larger than the variance in the Tri-county market. This implies that collusion might be present in the Tri-county market. Lowfat Chocolate Milk Year 1983 1984 1985 1986 1987 1988 1989 1990 1991

v1,v2 23,4 24,5 27,5 33,5 35,6 35,8 8,35 32,9 4,10

F.025 8.56 6.28 6.28 6.23 5.07 3.89 2.65 3.56 4.47

F 19.133 3.900 6.526 6.037 4.075 10.222 1.450 7.000 2.333

Decision Reject Do not reject Reject Do not reject Do not reject Reject Do not reject Reject Do not reject

Again, in all cases where there was a significant difference in the variances of the winning bids between the two markets, the variance in the Surrounding market was larger than the variance in the Tri-county market. This implies that collusion might be present in the Tri-county market. Based on the analysis of the three milk products, there appears to be collusion in the Tri-county market.

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(3)Average Winning Bid Price

I have provided the SAS output for computing the t-tests to compare the mean winning bid prices between the two markets for each of the years and each of the milk products. I will discuss the findings for each milk product separately. For t-tests, we must assume that the two population variances are the same. If the population variances are not the same, there is an approximate test that takes into consideration the different variances. The SAS printout provided allows for the test of equal variances first. I used a p-value of .25 as the cutoff point. If the p-value was less than or equal to .25 for the F-test, I assumed that the variances were different and used the approximate test designated as UNEQUAL. If the p-value for the F-test was greater than .25, I assumed that the population variances were the same and used the test designated as EQUAL. Whole White Milk: Variable: Whole White Milk - 1983 MARKET

N

Mean

Std Dev

Std Error

Variances

T

DF

Prob>|T|

-----------------------------------------------------------------------------SUR

22

0.1318

0.01458844

0.00311027

Unequal

2.4045

12.4

0.0326

TRI

8

0.1173

0.01462038

0.00516909

Equal

2.4071

28.0

0.0229*

For H0: Variances are equal, F' = 1.00

DF = (7,21)

Prob>F' = 0.9116 ************************************************************************ Variable: Whole White Milk MARKET

N

Mean

- 1984

Std Dev

Std Error

Variances

T

DF Prob>|T|

-----------------------------------------------------------------------------SUR

22

0.1309

0.01374189

0.00292978

Unequal

-2.3904

28.6

0.0236*

TRI

9

0.1389

0.00474871

0.00158290

Equal

-1.6825

29.0

0.1032

For H0: Variances are equal, F' = 8.37

DF = (21,8)

Prob>F' = 0.0044 ************************************************************************ Variable: Whole White Milk MARKET

N

Mean

- 1985

Std Dev

Std Error

Variances

T

DF Prob>|T|

-----------------------------------------------------------------------------SUR

26

0.1279

0.01321810

0.00259228

Unequal

-4.3968

33.8

0.0001*

TRI

10

0.1415

0.00534266

0.00168950

Equal

-3.1348

34.0

0.0035

For H0: Variances are equal, F' = 6.12

DF = (25,9)

Prob>F' = 0.0077 ************************************************************************

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Variable: Whole White Milk MARKET

N

Mean

- 1986

Std Dev

Std Error

Variances

T

DF Prob>|T|

----------------------------------------------------------------------------SUR

33

0.1253

0.01098665

0.00191253

Unequal

-8.1534

37.3

0.0001*

TRI

10

0.1446

0.00442846

0.00140040

Equal

-5.3943

41.0

0.0000

For H0: Variances are equal, F' = 6.15

DF = (32,9)

Prob>F' = 0.0070 ************************************************************************ Variable: Whole White Milk MARKET

N

Mean

- 1987

Std Dev

Std Error

Variances

T

DF

Prob>|T|

-----------------------------------------------------------------------------SUR

36

0.1264

0.01026078

0.00171013

Unequal

TRI

12

0.1495

0.00527196

0.00152188

Equal

For H0: Variances are equal, F' = 3.79

-10.0785

37.5

0.0001*

-7.4313

46.0

0.0000

DF = (35,11)

Prob>F' = 0.0224 ************************************************************************ Variable: Whole White Milk MARKET

N

Mean

- 1988

Std Dev

Std Error

Variances

T

DF

Prob>|T|

-----------------------------------------------------------------------------SUR

36

0.1277

0.01135449

0.00189242

Unequal

-9.9271

42.2

0.0001*

TRI

12

0.1513

0.00499090

0.00144075

Equal

-6.9441

46.0

0.0000

For H0: Variances are equal, F' = 5.18

DF = (35,11)

Prob>F' = 0.0060 ************************************************************************ Variable: Whole White Milk MARKET

N

Mean

- 1989

Std Dev

Std Error

Variances

T

DF

Prob>|T|

-----------------------------------------------------------------------------SUR

37

0.1299

0.00752173

0.00123657

Unequal

-0.4890

15.8

0.6316

TRI

12

0.1314

0.00944991

0.00272795

Equal

-0.5501

47.0

0.5849NS

For H0: Variances are equal, F' = 1.58

DF = (11,36)

Prob>F' = 0.2947 ************************************************************************ Variable: Whole White Milk MARKET

N

Mean

- 1990

Std Dev

Std Error Variances

T

DF

Prob>|T|

--------------------------------------------------------------------------SUR

35

0.1609

0.00794659

0.00134322 Unequal

-1.1177

43.7

0.2698NS

TRI

12

0.1628

0.00317904

0.00091771 Equal

-0.7673

45.0

0.4469

For H0: Variances are equal, F' = 6.25

DF = (34,11)

Prob>F' = 0.0026 ************************************************************************

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Variable: Whole White Milk MARKET

N

Mean

- 1991

Std Dev

Std Error

Variances

T

DF

Prob>|T|

-----------------------------------------------------------------------------SUR

5

0.1452

0.00652012

0.00291589

Unequal

1.2585

5.6

TRI

13

0.1412

0.00458169

0.00127073

Equal

1.4813

16.0

For H0: Variances are equal, F' = 2.03

0.2585 0.1580NS

DF = (4,12)

Prob>F' = 0.3095 ************************************************************************

The mean winning bid prices were significantly different between the markets for all years except 1989, 1990, and 1991. In 1983, the mean winning bid for the Surrounding market was significantly larger than that for the Tri-county market. For the years 1984–1988, the mean winning bid price for the Tri-county market was significantly larger than that for the Surrounding market. This implies evidence of collusion for the years 1984–1988. Lowfat White Milk: Variable: Lowfat White Milk - 1983 MARKET

N

Mean

Std Dev

Std Error

Variances

T

DF

Prob>|T|

-----------------------------------------------------------------------------SUR

24

0.1243

0.01672220

0.00341341

Unequal

2.5085

22.6

0.0198

TRI

10

0.1112

0.01246237

0.00394095

Equal

2.2214

32.0

0.0335*

For H0: Variances are equal, F' = 1.80

DF = (23,9)

Prob>F' = 0.3627 ************************************************************************ Variable: Lowfat White Milk - 1984 MARKET

N

Mean

Std Dev

Std Error

Variances

T

DF

Prob>|T|

----------------------------------------------------------------------------SUR

26

0.1236

0.01469859

0.00288263

Unequal

-3.0061

36.0

0.0048*

TRI

12

0.1338

0.00635717

0.00183516

Equal

-2.3099

36.0

0.0267

For H0: Variances are equal, F' = 5.35

DF = (25,11)

Prob>F' = 0.0059 ************************************************************************ Variable: Lowfat White Milk - 1985 MARKET

N

Mean

Std Dev

Std Error

Variances

T

DF

Prob>|T|

----------------------------------------------------------------------------SUR

29

0.1200

0.01452245

0.00269675

Unequal

-5.3857

39.2

0.0001*

TRI

13

0.1366

0.00537445

0.00149061

Equal

-3.9769

40.0

0.0003

For H0: Variances are equal, F' = 7.30

DF = (28,12)

Prob>F' = 0.0008 ************************************************************************

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Variable: Lowfat White Milk - 1986 MARKET

N

Mean

Std Dev

Std Error

Variances

T

DF

Prob>|T|

-----------------------------------------------------------------------------SUR

33

0.1178

0.01180640

0.00205523

Unequal

-8.4010

43.0

0.0001*

TRI

13

0.1391

0.00533205

0.00147884

Equal

-6.2183

44.0

0.0000

For H0: Variances are equal, F' = 4.90

DF = (32,12)

Prob>F' = 0.0055 ************************************************************************ Variable: Lowfat White Milk - 1987 MARKET

N

Mean

Std Dev

Std Error

Variances

T

DF

Prob>|T|

------------------------------------------------------------------------------SUR

35

0.1173

0.01235100

0.00208770

Unequal

-8.7991

37.8

0.0001*

TRI

13

0.1424

0.00701738

0.00194627

Equal

-6.8995

46.0

0.0000

For H0: Variances are equal, F' = 3.10

DF = (34,12)

Prob>F' = 0.0404 ************************************************************************ Variable: Lowfat White Milk - 1988 MARKET

N

Mean

Std Dev

Std Error

Variances

T

DF

Prob>|T|

----------------------------------------------------------------------------SUR

35

0.1182

0.01285522

0.00217293

Unequal

-9.6219

42.7

0.0001*

TRI

13

0.1448

0.00618019

0.00171408

Equal

-7.1332

46.0

0.0000

For H0: Variances are equal, F' = 4.33

DF = (34,12)

Prob>F' = 0.0095 ************************************************************************ Variable: Lowfat White Milk - 1989 MARKET

N

Mean

Std Dev

Std Error

Variances

T

DF

Prob>|T|

----------------------------------------------------------------------------SUR

35

0.1187

0.00655938

0.00110874

Unequal

-2.1005

17.9

0.0501

TRI

13

0.1240

0.00828350

0.00229743

Equal

-2.3400

46.0

0.0237*

For H0: Variances are equal, F' = 1.59

DF = (12,34)

Prob>F' = 0.2798 ************************************************************************ Variable: Lowfat White Milk - 1990 MARKET

N

Mean

Std Dev

Std Error

Variances

T

DF

Prob>|T|

-----------------------------------------------------------------------------SUR

34

0.1519

0.00954524

0.00163700

Unequal

-2.3772

39.8

0.0223*

TRI

13

0.1570

0.00508486

0.00141029

Equal

-1.8347

45.0

0.0732

For H0: Variances are equal, F' = 3.52

DF = (33,12)

Prob>F' = 0.0238 ************************************************************************

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Variable: Lowfat White Milk - 1991 MARKET

N

Mean

Std Dev

Std Error

Variances

T

DF

Prob>|T|

----------------------------------------------------------------------------SUR

5

0.1364

0.00718485

0.00321316

Unequal

0.2745

6.3

TRI

12

0.1354

0.00585768

0.00169097

Equal

0.3001

15.0

For H0: Variances are equal, F' = 1.50

0.7925 0.7682NS

DF = (4,11)

Prob>F' = 0.5343 ************************************************************************

The mean winning bid prices were significantly different between the markets for all years except 1991. In 1983, the mean winning bid for the Surrounding market was significantly larger than that for the Tri-county market. For the years 1984–1990, the mean winning bid price for the Tricounty market was significantly larger than that for the Surrounding market. This implies evidence of collusion for the years 1984–1990. Lowfat Chocolate Milk: Variable: Lowfat Chocolate Milk - 1983 MARKET

N

Mean

Std Dev

Std Error

Variances

T

DF

Prob>|T|

-----------------------------------------------------------------------------SUR

24

0.1267

0.01696642

0.00346326

Unequal

5.3313

26.3

0.0001*

TRI

5

0.1060

0.00394740

0.00176533

Equal

2.6795

27.0

0.0124

For H0: Variances are equal, F' =

18.47

DF = (23,4)

Prob>F' = 0.0117 ************************************************************************ Variable: Lowfat Chocolate Milk - 1984 MARKET

N

Mean

Std Dev

Std Error

Variances

T

DF

Prob>|T|

----------------------------------------------------------------------------SUR

25

0.1251

0.01530156

0.00306031

Unequal

-2.1693

15.7

0.0457*

TRI

6

0.1347

0.00778522

0.00317830

Equal

-1.4733

29.0

0.1514

For H0: Variances are equal, F' = 3.86

DF = (24,5)

Prob>F' = 0.1379 ************************************************************************ Variable: Lowfat Chocolate Milk - 1985 MARKET

N

Mean

Std Dev

Std Error

Variances

T

DF

Prob>|T|

----------------------------------------------------------------------------SUR

28

0.1206

0.01575587

0.00297758

Unequal

-4.6215

20.9

0.0001*

TRI

6

0.1387

0.00621914

0.00253895

Equal

-2.7384

32.0

0.0100

For H0: Variances are equal, F' = 6.42

DF = (27,5)

Prob>F' = 0.0472 ************************************************************************

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Variable: Lowfat Chocolate Milk - 1986 MARKET

N

Mean

Std Dev

Std Error

Variances

T

DF

Prob>|T|

----------------------------------------------------------------------------SUR

34

0.1169

0.01279357

0.00219408

Unequal

-8.0140

18.2

0.0001*

TRI

6

0.1414

0.00521130

0.00212751

Equal

-4.5821

38.0

0.0000

For H0: Variances are equal, F' = 6.03

DF = (33,5)

Prob>F' = 0.0533 ************************************************************************ Variable: Lowfat Chocolate Milk - 1987 MARKET

N

Mean

Std Dev

Std Error

Variances

T

DF

Prob>|T|

----------------------------------------------------------------------------SUR

36

0.1184

0.01280507

0.00213418

Unequal

-7.8853

17.5

0.0001*

TRI

7

0.1436

0.00632926

0.00239224

Equal

-5.0675

41.0

0.0000

For H0: Variances are equal, F' = 4.09

DF = (35,6)

Prob>F' = 0.0832 ************************************************************************ Variable: Lowfat Chocolate Milk - 1988 MARKET

N

Mean

Std Dev

Std Error

Variances

T

DF

Prob>|T|

----------------------------------------------------------------------------SUR

36

0.1192

0.01359999

0.00226666

Unequal

10.3636

40.6

0.0001*

TRI

9

0.1470

0.00425532

0.00141844

Equal

-5.9934

43.0

0.0000

For H0: Variances are equal, F' =

10.21

DF = (35,8)

Prob>F' = 0.0019 ************************************************************************ Variable: Lowfat Chocolate Milk - 1989 MARKET

N

Mean

Std Dev

Std Error

Variances

T

DF

Prob>|T|

----------------------------------------------------------------------------SUR

36

0.1200

0.00776605

0.00129434

Unequal

-1.7178

10.9

0.1140

TRI

9

0.1258

0.00932923

0.00310974

Equal

-1.9216

43.0

0.0613NS

For H0: Variances are equal, F' = 1.44

DF = (8,35)

Prob>F' = 0.4274 ************************************************************************ Variable: Lowfat Chocolate Milk - 1990 MARKET

N

Mean

Std De

Std Error

Variances

T

DF

Prob>|T|

-----------------------------------------------------------------------------SUR

33

0.1531

0.00993298

0.00172911

Unequal

-3.9472

38.3

0.0003*

TRI

10

0.1614

0.00383030

0.00121125

Equal

-2.5773

41.0

0.0137

For H0: Variances are equal, F' = 6.73

DF = (32,9)

Prob>F' = 0.0050 ************************************************************************

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Variable: Lowfat Chocolate Milk - 1991 MARKET

N

Mean

Std Dev

Std Error

Variances

T

DF

Prob>|T|

----------------------------------------------------------------------------SUR

5

0.1402

0.00991020

0.00443197

Unequal

-0.4431

5.6

TRI

11

0.1423

0.00650294

0.00196071

Equal

-0.5216

14.0

For H0: Variances are equal, F' = 2.32

0.6743 0.6101NS

DF = (4,10)

Prob>F' = 0.2552

The mean winning bid prices were significantly different between the markets for all years except 1989 and 1991. In 1983, the mean winning bid for the Surrounding market was significantly larger than that for the Tri-county market. For the years 1984–1988 and 1990, the mean winning bid price for the Tri-county market was significantly larger than that for the Surrounding market. This implies evidence of collusion for the years 1984–1988.

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Design of Experiments and Analysis of Variance 8.2

Chapter 8

The treatments are the combinations of levels of each of the two factors. There are 2 × 5 = 10 treatments. They are: (A, 50), (A, 60), (A, 70), (A, 80), (A, 90) (B, 50), (B, 60), (B, 70), (B, 80), (B, 90)

8.4

8.6

a.

College GPA's are measured on college students. The experimental units are college students.

b.

Household income is measured on households. The experimental units are households.

c.

Gasoline mileage is measured on automobiles. The experimental units are the automobiles of a particular model.

d.

The experimental units are the sectors on a computer diskette.

e.

The experimental units are the states.

a.

The response variable is the amount of the purchase.

b. There is one factor in this problem: type of credit card. c. There are 4 treatments, corresponding to the 4 levels of the factor. The treatments are VISA, MasterCard, American Express, and Discover. d. The experimental units are the credit card holders. 8.8

8.10

256

a.

The response variable in this problem is the consumer’s opinion on the value of the discount offer.

b.

There are two treatments in this problem: Within-store price promotion and betweenstore price promotion.

c.

The experimental units are the consumers.

a.

There are 2 factors in the problem: Type of yeast and Temperature. Type of yeast has 2 levels – Brewer’s yeast and baker’s yeast. Temperature has 4 levels – 45o, 48o, 51o and 54oC.

b.

The response variable is the autolysis yield.

c.

There are a total of 2 × 4 = 8 treatments in this experiment. The treatments are all the type of yeast-temperature combinations.

d.

This is a designed experiment.

Chapter 8

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8.12

8.14

8.16

a.

The response is the evaluation by the undergraduate student of the ethical behavior of the salesperson.

b.

There are two factors—type of sales job at two levels (high tech. vs. low tech.) and sales task at two levels (new account development vs. account maintenance).

c.

The treatments are the 2 × 2 = 4 combinations type of sales job and sales task.

d.

The experimental units are the college students.

a.

From Table IX with ν1 = 4 and ν2 = 4, F.05 = 6.39.

b.

From Table XI with ν1 = 4 and ν2 = 4, F.01 = 15.98.

c.

From Table VIII with ν1 = 30 and ν2 = 40, F.10 = 1.54.

d.

From Table X with ν1 = 15, and ν2 = 12, F.025 = 3.18.

a.

In the second dot diagram #2, the difference between the sample means is small relative to the variability within the sample observations. In the first dot diagram #1, the values in each of the samples are grouped together with a range of 4, while in the second diagram #2, the range of values is 8.

b. For diagram #1,

∑x

7 + 8 + 9 + 10 + 11 54 = =9 n 6 6 ∑ x2 = 12 + 13 + 14 + 14 + 15 + 16 = 84 = 14 x2 = n 6 6 x1 =

1

=

For diagram #2,

∑x

5 + 5 + 7 + 11 + 13 + 13 54 = =9 n 6 6 ∑ x2 = 10 + 10 + 12 + 16 + 18 + 18 = 84 = 14 x2 = n 6 6

x1 =

c.

1

=

For diagram #1, 2

SST =

∑ n (x i =1

i

i

− x ) 2 1 = 6(9 − 11.5)2 + 6(14 − 11.5)2 = 75

⎛ ∑ x = 54 + 84 = 11.5 ⎞⎟ ⎜⎜ x = ⎟ 12 n ⎝ ⎠ For diagram #2, 2

SST =

∑ n (x i =1

i

i

− x ) 2 = 6(9 - 11.5)2 + 6(14 - 11.5)2 = 75

Design of Experiments and Analysis of Variance

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d.

For diagram #1,

∑x

2 1

s12 =

(∑ x ) −

2

1

n1

=

n1 − 1

∑x

2 2

s22 =

(∑ x ) −

2 542 54 496 − 6 = 6 =2 6 −1 6 −1

496 −

2

2

n2 n2 − 1

=

842 6 =2 6 −1

1186 −

SSE = (n1 − 1) s12 + (n2 − 1) s22 = (6 − 1)2 + (6 − 1)2 = 20 For diagram #2,

∑x

2 1

s12 =

(∑ x ) −

2

1

n1

=

n1 − 1

∑x

2 2

s22 =

(∑ x ) −

2

2

n2 n2 − 1

542 6 = 14.4 6 −1

558 −

=

842 6 = 14.4 6 −1

1248 −

SSE = (n1 − 1) s12 + (n2 − 1) s22 = (6 − 1)14.4 + (6 − 1)14.4 = 144 e.

For diagram #1, SS(Total) = SST + SSE = 75 + 20 = 95 SST is

SST 75 × 100% = × 100% = 78.95% of SS(Total) SS(Total) 95

For diagram #2, SS(Total) = SST + SSE = 75 + 144 = 219 SST is

f.

SST 75 × 100% = × 100% = 34.25% of SS(Total) SS(Total) 219 SST 75 = = 75 k −1 2 −1 SSE 20 = MSE = =2 n − k 12 − 2

For diagram #1, MST =

SST 75 = = 75 k −1 2 −1 SSE 144 = = 14.4 MSE = n − k 12 − 2

F=

MST 75 = = 37.5 MSE 2

F=

MST 75 = = 5.21 MSE 14.4

For diagram #2, MST =

258

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g.

The rejection region for both diagrams requires α = .05 in the upper tail of the Fdistribution with ν1 = p − 1 = 2 − 1 = 1 and ν2 = n − p = 12 − 2 = 10. From Table IX, Appendix B, F.05 = 4.96. The rejection region is F > 4.96. For diagram #1, the observed value of the test statistic falls in the rejection region (F = 37.5 > 4.96). Thus, H0 is rejected. There is sufficient evidence to indicate the samples were drawn from populations with different means at α = .05. For diagram #2, the observed value of the test statistic falls in the rejection region (F = 5.21 > 4.96). Thus, H0 is rejected. There is sufficient evidence to indicate the samples were drawn from populations with different means at α = .05.

h. 8.18

We must assume both populations are normally distributed with common variances.

Refer to Exercise 8.16, the ANOVA table is: For diagram #1: Source Treatment Error Total

Df 1 10 11

SS 75 20 95

MS 75 2

F 37.5

SS 75 144 219

MS 75 14.4

F 5.21

For diagram #2: Source Treatment Error Total

8.20

a.

Df 1 10 11

df for Error is 41 − 6 = 35 SSE = SS(Total) − SST = 46.5 − 17.5 = 29.0 SST 17.5 = = 2.9167 k −1 6 MST 2.9167 = F= = 3.52 MSE .8286

MST =

MSE =

SSE 29.0 = = .8286 n−k 35

The ANOVA table is: Source Treatment Error Total

df 6 35 41

SS 17.5 29.0 46.5

MS 2.9167 .8286

Design of Experiments and Analysis of Variance

F 3.52

259

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b.

The number of treatments is k. We know k − 1 = 6 ⇒ k = 7.

c.

To determine if there is a difference among the population means, we test: H0: μ1 = μ2 = ⋅⋅⋅ = μ7 Ha: At least one of the population means differs from the rest The test statistic is F = 3.52. The rejection region requires α = .10 in the upper tail of the F-distribution with numerator df = k − 1 = 6 and denominator df = n − k = 35. From Table VIII, Appendix B, F.10 ≈ 1.98. The rejection region is F > 1.98. Since the observed value of the test statistic falls in the rejection region (F = 3.52 > 1.98), H0 is rejected. There is sufficient evidence to indicate a difference among the population means at α = .10.

d.

The observed significance level is P(F ≥ 3.52). With numerator df = 6 and denominator df = 35, and Table XI, P(F ≥ 3.52) < .01.

e.

H0: μ1 = μ2 Ha: μ1 ≠ μ2 The test statistic is t =

x1 − x2 ⎛1 1 ⎞ MSE ⎜ + ⎟ ⎝ n1 n2 ⎠

=

3.7 − 4.1 ⎛1 1⎞ .8286 ⎜ + ⎟ ⎝6 6⎠

= −.76

The rejection region requires α/2 = .10/2 = .05 in each tail of the t-distribution with df = n − p = 35. From Table VI, Appendix B, t.05 ≈ 1.697. The rejection region is t < −1.697 or t > 1.697. Since the observed value of the test statistic does not fall in the rejection region (t = −.76 4.98), H0 is rejected. There is sufficient evidence to indicate differences exist in the mean rates of return among the three types of fund groups at α = .01.

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8.28

a.

The response variable for this study is the safety rating of nuclear power plants.

b.

There are three treatments in this study. The treatment groups are the scientists, the journalists, and the federal government policymakers.

c.

To determine whether there are differences in the attitudes of scientists, journalists, and government officials regarding the safety of nuclear power plants, we test:

H0: μ1 = μ2 = μ3 Ha: At least two means differ d.

The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = k − 1 = 3 − 1 = 2 and ν2 = n − k = 300 − 3 = 297. From Table IX, Appendix B, F.05 ≈ 3.00. The rejection region is F > 3.00. In order to reject H0, the test statistic F must be greater than 3.00.

F=

MST > 3.00 MSE

⇒ MST > 3.00(MSE) ⇒ 3.00 (2.355) = 7.065. Thus, MST must be greater than 7.065.

8.30

MST 11.28 = = 4.79 MSE 2.355

e.

For MST = 11.280, F =

f.

With ν1 = k − 1 = 3 − 1 = 2 and ν2 = n − k = 300 − 3 = 297, P(F > 4.79) ≈ .01, using Table XI, Appendix B. The approximate p-value is .01.

a.

We will select size as the quantitative variable and color as the qualitative variable. To determine if the mean size of diamonds differ among the 6 colors, we test:

H0: μ1 = μ2 = μ3 = μ4 = μ5 = μ6 Ha: At least two means differ b.

Using MINITAB, the ANOVA table is:

One-way ANOVA: Carats versus Color Analysis of Variance for Carats Source DF SS MS Color 5 0.7963 0.1593 Error 302 22.7907 0.0755 Total 307 23.5869 Level D E F G H I

N 16 44 82 65 61 40

Pooled StDev =

262

Mean 0.6381 0.6232 0.5929 0.5808 0.6734 0.7310 0.2747

StDev 0.3195 0.2677 0.2648 0.2792 0.2643 0.2918

F 2.11

P 0.064

Individual 95% CIs For Mean Based on Pooled StDev ----------+---------+---------+-----(-------------*------------) (-------*-------) (-----*-----) (------*------) (------*------) (-------*--------) ----------+---------+---------+-----0.60 0.70 0.80

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The test statistic is F = 2.11 and the p-value is p = 0.064. Since the p-value (0.064) is less than α = .10, H0 is rejected. There is sufficient evidence to indicate the mean size of diamonds differ among the 6 colors at α = .10. c.

We will check the assumptions of normality and equal variances. Using MINITAB, the stem-and-leaf plots are: Stem-and-Leaf Display: Carats Stem-and-leaf of Carats Leaf Unit = 0.010 1 3 5 5 7 7 (4) 5 5 5

1 2 3 4 5 6 7 8 9 10

1 2 3 4 5 6 7 8 9 10

1 2 3 4 5 6 7 8 9 10

1156 00011

1 2 3 4 5 6 7 8 9 10

Color = 2

N

= 44

Color = 3

N

= 82

N

= 65

9 123 0011345 6 00012245668 23 000123 113 0000011113

88999 1356667 01124445567 0178 000111122333345566678 0 00001112367 0012555 0 00000011112224

Stem-and-leaf of Carats Leaf Unit = 0.010 5 12 21 23 (12) 30 26 16 12 12

= 16

23

Stem-and-leaf of Carats Leaf Unit = 0.010 5 12 23 27 (21) 34 33 22 15 14

N

9 01 01

Stem-and-leaf of Carats Leaf Unit = 0.010 1 4 11 12 (11) 21 19 13 10 10

Color= 1

Color = 4

88899 0001359 000124455 08 000013556789 0034 0000001348 0125 000000011126

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Stem-and-leaf of Carats Leaf Unit = 0.010 5 14 16 21 27 (13) 21 14 14 1

2 3 4 5 6 7 8 9 10 11

Color = 5

2 457 012344567 03 25778 001466 0001112233448 0014669

2 3 4 5 6 7 8 9 10

= 61

1 89

0000011111266 0

Stem-and-leaf of Carats Leaf Unit = 0.010 4 8 11 13 15 20 20 17 16

N

Color = 6

N

= 40

5689 0113 115 26 25 03355 002 0 0000001111114579

The data for the 6 colors do not look particularly mound-shaped, so the assumption of normality is probably not valid. However, departures from this assumption often do not invalidate the ANOVA results. Using MINITAB, the box plots are:

1.1 1.0 0.9

Carats

0.8 0.7 0.6 0.5 0.4 0.3 0.2 D

E

F

G

H

I

Color

The spreads of all the colors appear to be about the same, so the assumption of constant variance is probably valid.

264

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8.32

a.

The df for Groups = ν1 = k – 1 = 3 – 1 = 2. The df for Error = ν2 = n – k = 71 – 3 = 68. The completed ANOVA table is: Source Groups Error

b.

df

2 68

SS 128.70 27,124.52

MS 64.35 398.89

F 0.16

To determine if the total number of activities undertaken differed among the three groups of entrepreneurs, we test:

H0: μ1 = μ2 = μ3 Ha: At least one mean differs The test statistic is F = 0.16. The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = k − 1 = 3 − 1 = 2 and ν2 = n − k = 71 − 3 = 68. From Table IX, Appendix B, F.05 ≈ 3.15. The rejection region is F > 3.15. Since the observed value of the test statistic does not fall in the rejection region (F = 0.16 >/ 3.15), H0 is not rejected. There is insufficient evidence to indicate that the total number of activities differed among the groups of entrepreneurs at α = .05. c.

The p-value of the test is P(F > 0.16). From Table VIII, Appendix B, with ν1 = 2 and

ν2 = 68, P(F > 0.16) > .10.

d.

No. Since our conclusion was that there was no evidence of a difference in the total number of activities among the groups, there would be no evidence to indicate a difference between two specific groups.

e.

This study would be observational. The group that each entrepreneur fell into was observed, not controlled. Since no differences were found, the type of study does not have an impact on the conclusions.

8.34

The experimentwise error rate is the probability of making a Type I error for at least one of all of the comparisons made. If the experimentwise error rate is α = .05, then each individual comparison is made at a value of α which is less than .05.

8.36

a.

From the diagram, the following pairs of treatments are significantly different because they are not connected by a line: A and E, A and B, A and D, C and E, C and B, C and D, and E and D. All other pairs of means are not significantly different because they are connected by lines.

b.

From the diagram, the following pairs of treatments are significantly different because they are not connected by a line: A and B, A and D, C and B, C and D, E and B, E and D, and B and D. All other pairs of means are not significantly different because they are connected by lines.

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8.38

8.40

c.

From the diagram, the following pairs of treatments are significantly different because they are not connected by a line: A and E, A and B, and A and D. All other pairs of means are not significantly different because they are connected by lines.

d.

From the diagram, the following pairs of treatments are significantly different because they are not connected by a line: A and E, A and B, A and D, C and E, C and B, C and D, E and D, and B and D. All other pairs of means are not significantly different because they are connected by lines.

a.

The total number of comparisons conducted is k(k – 1)/2 = 4(4 – 1)/2 = 6.

b.

The mean energy expended by robots in the 12 robot colony is significantly smaller than the mean energy expended by robots in any of the other size colonies. There is no difference in the mean energy expended by robots in the 3 robot colony, the 6 robot colony, and the 9 robot colony.

a.

There will be c =

b.

Comparing the mean safety scores for government officials and journalists, the difference in mean safety scores is 4.2 − 3.7 = .5, The critical value for the Tukey comparison is .23. Since .5 > .23, we conclude that the mean safety score for government officials is higher than the mean safety score for journalists.

k (k − 1) 3(3 − 1) = 3 pairwise comparisons. = 2 2

Comparing the mean safety scores for government officials and scientists, the difference in mean safety scores is 4.2 − 4.1 = .1. Since .1 < .23, we conclude that there is no difference in mean safety scores between government officials and scientists. Comparing the mean safety scores for scientists and journalists, the difference in mean safety scores is 4.1 − 3.7 = .4, The critical value for the Tukey comparison is .23. Since .4 > .23, we conclude that the mean safety score for scientists is higher than the mean safety score for journalists. A display of these conclusions is: Journalists 3.7 8.42

Scientists 4.1

Gov. Officials 4.2

a.

The probability of declaring at least one pair of means different when they are not is .01.

b.

There are a total of

k (k − 1) 3(3 − 1) = = 3 pair-wise comparisons. They are: 2 2

‘Under $30 thousand’ to ‘Between $30 and $60 thousand’ ‘Under $30 thousand’ to ‘Over $60 thousand’ ‘Between $30 and $60 thousand’ to ‘Over $60 thousand’

266

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c.

Means for groups in homogeneous subsets are displayed in the table: Income Group

Subsets

Under $30,000 $30,000-$60,000 Over $60,000 d.

N 379 392 267

1 4.60

2 5.08 5.15

Two of the comparisons in part b will yield confidence intervals that do not contain 0. They are: ‘Under $30 thousand’ to ‘Between $30 and $60 thousand’ ‘Under $30 thousand’ to ‘Over $60 thousand’

8.44

From Exercise 8.30, we found that there were differences in the mean carats among the 6 levels of color From Exercise 8.30, the mean carats for the 6 colors are: G F E D H I

0.5808 0.5929 0.6232 0.6381 0.6734 0.7310

Using MINITAB, the Tukey confidence intervals are: Tukey's pairwise comparisons Family error rate = 0.100 Individual error rate = 0.0101 Critical value = 3.66 Intervals for (column level mean) - (row level mean) D

E

F

G

E

-0.1926 0.2225

F

-0.1491 0.2395

-0.1026 0.1631

G

-0.1411 0.2558

-0.0964 0.1812

-0.1059 0.1302

H

-0.2350 0.1644

-0.1909 0.0904

-0.2007 0.0397

-0.2194 0.0341

I

-0.3032 0.1174

-0.2631 0.0475

-0.2752 -0.0010

-0.2931 -0.0074

Design of Experiments and Analysis of Variance

H

-0.2022 0.0871

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There are only 2 intervals that do not contain 0: The confidence interval for the difference in mean carats between colors G and I is (−0.2931, −0.0074). The confidence interval for the difference in mean carats between colors F and I is (−0.2752, −0.0010). Since 0 is not contained in these confidence intervals, there is sufficient evidence of a difference in the mean number of carats between colors G and I and between colors F and I. No other differences exist. 8.46

a.

There are 3 blocks used since Block df = b − 1 = 2 and 5 treatments since the treatment df = k − 1 = 4.

b.

There were 15 observations since the Total df = n − 1 = 14.

c.

H0: μ1 = μ2 = μ3 = μ4 = μ5 Ha: At least two treatment means differ

d.

The test statistic is F =

e.

The rejection region requires α = .01 in the upper tail of the F distribution with ν1 = k − 1 = 5 − 1 = 4 and ν2 = n − k − b + 1 = 15 − 5 − 3 + 1 = 8. From Table XI, Appendix B, F.01 = 7.01. The rejection region is F > 7.01.

f.

Since the observed value of the test statistic falls in the rejection region (F = 9.109 > 7.01), H0 is rejected. There is sufficient evidence to indicate that at least two treatment means differ at α = .01.

g.

The assumptions necessary to assure the validity of the test are as follows: 1. 2.

8.48

a.

The probability distributions of observations corresponding to all the blocktreatment combinations are normal. The variances of all the probability distributions are equal.

The ANOVA Table is as follows: Source Treatment Block Error Total

268

MST = 9.109 MSE

df 2 3 6 11

SS 12.032 71.749 .708 84.489

MS 6.016 23.916 .118

F 50.958 202.586

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b.

To determine if the treatment means differ, we test:

H0: μA = μB = μC Ha: At least two treatment means differ B

The test statistic is F =

MST = 50.958 MSE

The rejection region requires α = .05 in the upper tail of the F distribution with ν1 = k − 1 = 3 − 1 = 2 and ν2 = n − k − b + 1 = 12 − 3 − 4 + 1 = 6. From Table IX, Appendix B, F.05 = 5.14. The rejection region is F > 5.14. Since the observed value of the test statistic falls in the rejection region (F = 50.958 > 5.14), H0 is rejected. There is sufficient evidence to indicate that the treatment means differ at α = .05. c.

To see if the blocking was effective, we test:

H0: μ1 = μ2 = μ3 = μ4 Ha: At least two block means differ The test statistic is F =

MSB = 202.586 MSE

The rejection region requires α = .05 in the upper tail of the F distribution with ν1 = k − 1 = 4 − 1 = 3 and ν2 = n − k − b + 1 = 12 − 3 − 4 + 1 = 6. From Table IX, Appendix B, F.05 = 4.76. The rejection region is F > 4.76. Since the observed value of the test statistic falls in the rejection region (F = 202.586 > 4.76), H0 is rejected. There is sufficient evidence to indicate that blocking was effective in reducing the experimental error at α = .05. d.

From the printouts, we are given the differences in the sample means. The difference between Treatment B and both Treatments A and C are positive (1.125 and 2.450), so Treatment B has the largest sample mean. The difference between Treatment A and C is positive (1.325), so Treatment A has a larger sample mean than Treatment C. So Treatment B has the largest sample mean, Treatment A has the next largest sample mean and Treatment C has the smallest sample mean. From the printout, all the means are significantly different from each other.

e.

The assumptions necessary to assure the validity of the inferences above are: 1. 2.

The probability distributions of observations corresponding to all the blocktreatment combinations are normal. The variances of all the probability distributions are equal.

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8.50

a.

This is a randomized block design. The blocks are the 12 plots of land. The treatments are the three methods used on the shrubs: fire, clipping, and control. The response variable is the mean number of flowers produced. The experimental units are the 36 shrubs.

b.

Plot

c.

To determine if there is a difference in the mean number of flowers produced among the three treatments, we test:

H0: μ1 = μ2 = μ3 Ha: The mean number of flowers produced differ for at least two of the methods. The test statistic is F = 5.42 and p = .009. We can reject the null hypothesis at the α > .009 level of significance. At least two of the methods differ with respect to mean number of flowers produced by pawpaws. d.

8.52

270

The means of Control and Clipping do not differ significantly. The means of Clipping and Burning do not differ significantly. The mean of treatment Burning exceeds that of the Control.

From the printout, the p-value for treatments or Decoy is p = .589. Since the p-value is not small, we cannot reject H0. There is insufficient evidence to indicate a difference in mean percentage of a goose flock to approach to within 46 meters of the pit blind among the three decoy types. This conclusion is valid for any reasonable value of α.

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8.54

Using SAS, the ANOVA Table is: The ANOVA Procedure Dependent Variable: temp Source

DF

Sum of Squares

Mean Square

F Value

Pr > F

Model

11

18.53700000

1.68518182

0.52

0.8634

Error

18

58.03800000

3.22433333

Corrected Total

29

76.57500000

R-Square

Coeff Var

Root MSE

temp Mean

0.242076

1.885189

1.795643

95.25000

Source STUDENT PLANT

DF

Anova SS

Mean Square

F Value

Pr > F

9 2

18.41500000 0.12200000

2.04611111 0.06100000

0.63 0.02

0.7537 0.9813

To determine if there are differences among the mean temperatures among the three treatments, we test:

H0: μ1 = μ2 = μ3 Ha: At least two treatment means differ The test statistic is F = 0.02. The associated p-value is p = .9813. Since the p-value is very large, there is no evidence of a difference in mean temperature among the three treatments. Since there is no difference, we do not need to compare the means. It appears that the presence of plants or pictures of plants does not reduce stress. 8.56

a.

Some preliminary calculations are:

(∑ y ) CM =

2

n SS(Total) =

2.952 = .435125 10 y 2 − CM = .4705 − .435125 = .035375

=

∑

1.622 1.332 T12 T22 + − CM = + − .435125 = .004205 10 10 b b SST .004205 = = .004205, df = k − 1 = 1 MST = 2 −1 k −1 B2 B2 B2 SSB = SS(DOG) = 1 + 2 + ⋅⋅⋅ + 10 − CM k k k 2 2 2 2 2 .32 + .38 + .27 + .36 + .42 + .312 + .19 2 + .192 + .32 + .212 = 2 − .435125 = .028925 SSB .028925 = MSB = = .003214, df = b − 1 = 9 b −1 10 − 1

SST = SS(DRUG) =

SSE = SS(Total) − SST − SSB = .035375 − .004205 − .028925 = .002245

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MSE =

F=

SSE .002245 = = .0002494 n − k − b + 1 20 − 2 − 10 + 1

MST .004205 = = 16.86 MSE .0002494

F=

MSB .003214 = = 12.89 MSE .0002494

To determine if there is a difference in mean pressure readings for the two treatments, we test:

H0: μA = μB Ha: μA ≠ μB B

B

The test statistic is F =

MST = 16.86 MSE

The rejection region requires α = .05 in the upper tail of the F distribution with ν1 = k − 1 = 2 − 1 = 1 and ν2 = n − k − b + 1 = 20 − 2 − 10 + 1 = 9. From Table IX, Appendix B, F.05 = 5.12. The rejection region is F > 5.12. Since the observed value of the test statistic falls in the rejection region (F = 16.86 > 5.12), H0 is rejected. There is sufficient evidence to indicate a difference in mean pressure readings for the two drugs at α = .05. b.

Since there is expected to be much variation between the dogs, we use the dogs as blocks to eliminate this identified source of variation.

c.

272

Dog

Drug A

Drug B

1 2 3 4 5 6 7 8 9 10

.17 .20 .14 .18 .23 .19 .12 .10 .16 .13

.15 .18 .13 .18 .19 .12 .07 .09 .14 .08

(A − B) Differences .02 .02 .01 .00 .04 .07 .05 .01 .02 .05

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Some preliminary calculations are: d=

sd2 = sd =

∑d

i

nd

∑d

.29 = .029 10

= 2 i

(∑ d ) −

2

i

nd nd − 1

=

(.29) 2 10 = .00449 = .0004989 10 − 1 9

.0129 −

sd2 = .0004989 = .02234

To determine if there is a difference in mean pressure readings for the two treatments, we test: H0: μA = μB Ha: μA ≠ μB B

B

The test statistic is t =

d −0 sd / nd

=

.029 − 0

= 4.105

.02234 / 10

The rejection region requires α/2 = .05/2 = .025 in each tail of the t distribution with df = n − 1 = 10 − 1 = 9. From Table VI, Appendix B, t.025 = 2.262. The rejection region is t < −2.262 or t > 2.262. Since the observed value of the test statistic falls in the rejection region (t = 4.105 > 2.262), H0 is rejected. There is sufficient evidence to indicate a difference in the treatment means at α = .05. d.

In part a, F = 16.86; and in part c, t = 4.105. Note that t2 = 4.1052 = 16.85 = F. In part a, F.05 = 5.12; and in part c, t.025 = 2.262. Note that t.2025 = 2.2622 = 5.12 = F.05.

e.

p-value = P(F ≥ 16.86) with ν1 = 1 and ν2 = 9. Using Table XI, Appendix B, P(F ≥ 10.56) < .01. Thus, the p-value is < .01. The probability of a test statistic this extreme if the treatment means are the same is less than .01. This is very significant. We would reject H0 in favor of Ha if α is larger than the p-value.

8.58

a.

There are two factors.

b.

No, we cannot tell whether the factors are qualitative or quantitative.

c.

Yes. There are four levels of factor A and three levels of factor B.

d.

A treatment would consist of a combination of one level of factor A and one level of factor B. There are a total of 4 × 3 = 12 treatments.

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8.60

e.

One problem with only one replicate is there are no degrees of freedom for error. This is overcome by having at least two replicates.

a.

Factor A has 3 + 1 = 4 levels and factor B has 1 + 1 = 2 levels.

b.

There are a total of 23 + 1 = 24 observations and 4 × 2 = 8 treatments. Therefore, there were 24/8 = 3 observations for each treatment.

c.

AB df = (a − 1)(b − 1) = (4 − 1)(2 − 1) = 3 Error df = n − ab = 24 − 4(2) = 16 SS A ⇒ SSA = (a − 1)MSA = (4 − 1)(.75) = 2.25 a −1 SSB .95 = MSB = = .95 b −1 2 −1 SS AB MSAB = ⇒ SSAB = (a − 1)(b − 1)MSAB = (4 − 1)(2 − 1)(.30) = .9 (a − 1)(b − 1) SSE = SS(Total) − SSA − SSB − SSAB = 6.5 − 2.25 − .95 − .9 = 2.4 SSE 2.4 = MSE = = .15 n − ab 24 - 4(2)

MSA =

SST = SSA + SSB + SSAB = 2.25 + .95 + .90 = 4.1 Treatment df = ab − 1 = 4(2) − 1 = 7 SST 4.1 MST .5857 MST = = .5857 FT = = 3.90 = = ab − 1 7 MSE .15 MSA .75 = = 5.00 MSE .15 MSAB .30 = = 2.00 FAB = MSE .15

FA =

FB = B

MSB .95 = = 6.33 MSE .15

The ANOVA table is: Source Treatments A B AB Error Total

274

df 7 3 1 3 16 23

SS 4.1 2.25 .95 .90 2.40 6.50

MS .59 .75 .95 .30 .15

F 3.90 5.00 6.33 2.00

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d.

To determine whether the treatment means differ, we test: H0: μ1 = μ2 = ⋅⋅⋅ = μ8 Ha: At least two treatment means differ The test statistic is F =

MST = 3.90 MSE

The rejection region requires α = .10 in the upper tail of the F-distribution with ν1 = ab − 1 = 4(2) − 1 = 7 and ν2 = n − ab = 24 − 4(2) = 16. From Table VIII, Appendix B, F.10 = 2.13. The rejection region is F > 2.13. Since the observed value of the test statistic falls in the rejection region (F = 3.90 > 2.13), H0 is rejected. There is sufficient evidence to indicate the treatment means differ at α = .10. e.

To determine if the factors interact, we test: H0: Factors A and B do not interact to affect the response mean Ha: Factors A and B do interact to affect the response mean The test statistic is F = 2.00. The rejection region requires α = .10 in the upper tail of the F-distribution with ν1 = (a − 1)(b − 1) = (4 − 1)(2 − 1) = 3 and ν2 = n − ab = 24 − 4(2) = 16. From Table VIII, Appendix B, F.10 = 2.46. The rejection region is F > 2.46. Since the observed value of the test statistic does not fall in the rejection region (F = 2.00 >/ 2.46), H0 is not rejected. There is insufficient evidence to indicate factors A and B interact at α = .10. To determine if the four means of factor A differ, we test: H0: There is no difference in the four means of factor A Ha: At least two of the factor A means differ The test statistic is F = 5.00. The rejection region requires α = .10 in the upper tail of the F-distribution with ν1 = a − 1 = 4 − 1 = 3 and ν2 = n − ab = 24 - 4(2) = 16. From Table VIII, Appendix B, F.10 = 2.46. The rejection region is F > 2.46. Since the observed value of the test statistic falls in the rejection region (F = 5.00 > 2.46), H0 is rejected. There is sufficient evidence to indicate at least two of the four means of factor A differ at α = .10. To determine if the 2 means of factor B differ, we test: H0: There is no difference in the two means of factor B Ha: At least two of the factor B means differ

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The test statistic is F = 6.33. The rejection region requires α = .10 in the upper tail of the F-distribution with ν1 = b − 1 = 2 − 1 = 1 and ν2 = n − ab = 24 − 4(2) = 16. From Table VIII, Appendix B, F.10 = 3.05. The rejection region is F > 3.05. Since the observed value of the test statistic falls in the rejection region (F = 6.33 > 3.05), H0 is rejected. There is sufficient evidence to indicate the two means of factor B differ at α = .10. All of the tests performed are warranted because interaction was not significant. 8.62

a.

The treatments are the combinations of the levels of factor A and the levels of factor B. There are 2 × 2 = 4 treatments. The treatment means are: x11 = x21 =

∑x

11

2 ∑ x21 2

=

29.6 + 35.2 = 32.4 2

x12 =

=

12.9 + 17.6 = 15.25 2

x22 =

∑x

12

2 ∑ x22 2

=

47.3 + 42.1 2

=

28.4 + 22.7 2

The factors do not appear to interact—the lines are almost parallel. The treatment means do appear to differ because the sample means range from 15.25 to 44.7.

b.

276

(∑ x )

2

235.82 8 n 2 SS(Total) = ∑ x − CM = 7922.92 − 6950.205 = 972.715

CM =

i

SSA =

∑A

SSB =

∑B

2 i

br 2 i

ar

=

− CM=

154.22 81.62 + = 7609.05 − 6950.205 = 658.845 2(2) 2(2)

− CM=

95.32 140.52 + = 7205.585 − 6950.205 = 255.38 2(2) 2(2)

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∑∑ AB

2 ij

64.82 89.42 30.52 51.12 + + + r 2 4 2 2 − 658.845 − 255.38 − 6950.205 = 7866.43 − 7864.43 = 2 SSE = SS(Total) − SSA − SSB − SSAB = 972.715 − 658.845 − 255.38 − 2 = 56.49

SSAB =

− SSA − SSB − CM =

df = a − 1 = 2 − 1 = 1 df = b − 1 = 2 − 1 = 1 df = (a − 1)(b − 1) = (2 − 1)(2 − 1) = 1 df = n − ab = 8 − 2(2) = 4 df = n − 1 = 8 − 1 = 7

A B AB Error Total

SSA 658.845 = = 658.845 a −1 1 SSAB 2 = =2 MSAB = (a − 1)(b − 1) 1

MSA =

MSB =

SSB 255.38 = = 255.38 b −1 1

MSE =

SSE 56.49 = 14.1225 = n - ab 4

MSA 658.845 = = 46.65 MSE 14.1225

FA = FAB =

FB = B

MSB 255.38 = = 18.08 MSE 14.1225

MSAB 2 = = .14 MSE 14.1225

The ANOVA table is: Source A B AB Error Total

c.

df

1 1 1 4 7

SS 658.845 255.380 2.000 56.490 972.715

MS 658.845 255.380 2.000 14.1225

F 46.65 18.08 .14

SST = SSA + SSB + SSAB = 658.845 + 255.380 + 2.000 = 916.225 df = ab − 1 = 2(2) − 1 = 3 SST 916.225 MST 305.408 = 21.63 MST = = = 305.408 FT = = ab − 1 3 MSE 14.1225 To determine whether the treatment means differ, we test: H0: μ1 = μ2 = μ3 = μ4 Ha: At least two of the treatment means differ The test statistic is F = 21.63. The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = ab − 1 = 2(2) − 1 = 3 and ν2 = n − ab = 8 − 2(2) = 4. From Table IX, Appendix B, F.05 = 6.59. The rejection region is F > 6.59.

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Since the observed value of the test statistic falls in the rejection region (F = 21.63 > 6.59), H0 is rejected. There is sufficient evidence to indicate the treatment means differ at α = .05. This agrees with the conclusion in part a. d.

Since there are differences among the treatment means, we test for the presence of interaction: H0: Factors A and B do not interact to affect the response means Ha: Factors A and B do interact to affect the response means The test statistic is F = .14. The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = (a − 1)(b − 1) = (2 − 1)(2 − 1) = 1 and ν2 = n − ab = 8 − 2(2) = 4. From Table IX, Appendix B, F.05 = 7.71. The rejection region is F > 7.71. Since the observed value of the test statistic does not fall in the rejection region (F = .14 >/ 7.71), H0 is not rejected. There is insufficient evidence to indicate the factors interact at α = .05.

e.

Since the interaction was not significant, we test for main effects. To determine whether the two means of factor A differ, we test: H0: μ1 = μ2 Ha: μ1 ≠ μ2 The test statistic is F = 46.65. The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = a − 1 = 2 − 1 = 1 and ν2 = n − ab = 8 − 2(2) = 4. From Table IX, Appendix B, F.05 = 7.71. The rejection region is F > 7.71. Since the observed value of the test statistic falls in the rejection region (F = 46.65 > 7.71), H0 is rejected. There is sufficient evidence to indicate the two means of factor A differ at α = .05. To determine whether the two means of factor B differ, we test: H0: μ1 = μ2 Ha: μ1 ≠ μ2 The test statistic is F = 18.08. The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = b − 1 = 2 − 1 = 1 and ν2 = n − ab = 8 − 2(2) = 4. From Table IX, Appendix B, F.05 = 7.71. The rejection region is F > 7.71.

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Since the observed value of the test statistic falls in the rejection region (F = 18.08 > 7.71), H0 is rejected. There is sufficient evidence to indicate the two means of factor B differ at α = .05. f.

The results of all the tests agree with those in part a.

g.

Since no interaction is present, but the means of both factors A and B differ, we compare the two means of factor A and compare the two means of factor B. Since there are only two means to compare for each factor, the higher population mean corresponds to the higher sample mean. Factor A: x1 = x2 =

∑x

1

br

∑x

2

br

=

29.6 + 35.2 + 47.3 + 42.1 = 38.55 2(2)

=

12.9 + 17.6 + 28.4 + 22.7 = 20.4 2(2)

The mean for level 1 of factor A is significantly higher than the mean for level 2. Factor B: x1 = x2 =

∑x

1

ar

∑x

2

ar

=

29.6 + 35.2 + 12.9 + 17.6 = 23.825 2(2)

=

47.3 + 42.1 + 28.4 + 22.7 = 35.125 2(2)

The mean for level 2 of factor B is significantly higher than the mean for level 1. 8.64

a.

There are a total of 2 × 4 = 8 treatments.

b.

The interaction between temperature and type was significant. This means that the effect of type of yeast on the mean autolysis yield depends on the level of temperature.

c.

To determine if the main effect of type of yeast is significant, we test: H0: μBa = μBr Ha: μBa ≠ μBr To determine if the main effect of temperature is significant, we test: H0: μ1 = μ2 = μ3 = μ4 Ha: At least one mean differs

d.

The tests for the main effects should not be run until after the test for interaction is conducted. If interaction is significant, then these interaction effects could cover up the main effects. Thus, the main effect tests would not be informative. If the test for interaction is not significant, then the main effect tests could be run.

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e.

Baker’s yeast:

The mean yield for temperature 54o is significantly lower than the mean yields for the other 3 temperatures. There is no difference in the mean yields for the temperatures 45o, 48o and 51o. Brewer’s yeast: The mean yield for temperature 54o is significantly lower than the mean yields for the other 3 temperatures. There is no difference in the mean yields for the temperatures 45o, 48o and 51o.

8.66

a.

This is an observational experiment. The researcher recorded the number of users per hour for each of 24 hours per day, 7 days per week, for 7 weeks. The researcher did not manipulate the weeks or days or hours.

b.

The two factors are (1) the day of the week with 7 levels and (2) the hour of the day with 24 levels.

c.

In a factorial experiment, a is the number of levels of factor A and b is the number of levels of factor B. If we let factor A be the day of the week and factor B be the hour of the day, then a = 7 and b = 24.

d.

To determine if the a × b = 7 × 24 = 168 treatment means differ, we test: H0: μ1 = μ2 = μ3 = . . . = μ168 Ha: At least two means differ The test statistic is F =

MST 1143.99 = = 25.06 MSE 45.65

The rejection region requires α = .01 in the upper tail of the F distribution with v1 = p − 1 = 168 − 1 = 167 and v2 = n − p = 1172 – 168 = 1004. From Table XI, Appendix B, F.01 ≈ 1.00. The rejection region is F > 1.00. Since the observed value of the test statistic falls in the rejection region (F = 25.06 > 1.00), H0 is rejected. There is sufficient evidence to indicate a difference in mean usage among the day-hour combinations at α = .01. e.

The hypotheses used to test if an interaction effect exists are: H0: Days and hours do not interact to affect the mean usage Ha: Days and hours interact do affect the mean usage

f.

The test statistic is F =

MSAB 55.69 = 1.22 = MSE 45.65

The p-value is p = .0527. Since the p-value is not less than α = .01, H0 is not rejected. There is insufficient evidence to indicate days and hours interact to affect usage at α = .01.

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g.

To determine if the mean usage differs among the days of the week, we test: H0: μ1 = μ2 = μ3 = μ4 = μ5 = μ6 = μ7 Ha: At least two means differ The test statistic is F =

MSA 3122.02 = 68.39 = MSE 45.65

The p-value is p = .0001. Since the p-value is less than α = .01, H0 is rejected. There is sufficient evidence to indicate the mean usage differs among the days of the week at α = .01. To determine if the mean usage differs among the hours of the day, we test: H0: μ1 = μ2 = μ3 = . . . = μ24 Ha: At least two means differ The test statistic is F =

MSB 7157.82 = 156.80 = MSE 45.65

The p-value is p = .0001. Since the p-value is less than α = .01, H0 is rejected. There is sufficient evidence to indicate the mean usage differs among the hours of the day at α = .01. 8.68

a.

The degrees of freedom for “Type of message retrieval system” is a − 1 = 2 − 1 = 1. The degrees of freedom for “Pricing option” is b − 1 = 2 − 1 = 1. The degrees of freedom for the interaction of Type of message retrieval system and Pricing option is (a − 1)(b – 1) = (2 − 1)(2 − 1) = 1. The degrees of freedom for error is n − ab = 120 − 2(2) = 116. Source Type of message retrieval system Pricing Option Type of system × pricing option Error Total

b.

Df 1 1 1 116 119

SS -

MS -

F 2.001 5.019 4.986

To determine if “Type of system” and “Pricing option” interact to affect the mean willingness to buy, we test: H0: “Type of system” and “Pricing option” do not interact Ha: “Type of system” and “Pricing option” interact

c.

The test statistic is F =

MSAB = 4.986 MSE

The rejection region requires α = .05 in the upper tail of the F distribution with ν1 = (a − 1)(b − 1) = (2 − 1)(2 − 1) = 1 and ν2 = n − ab = 120 − 2(2) = 116. From Table IX, Appendix B, F.05 ≈ 3.92. The rejection region is F > 3.92.

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Since the observed value of the test statistic falls in the rejection region (F = 4.986 > 3.92), H0 is rejected. There is sufficient evidence to indicate “Type of system” and “Pricing option” interact to affect the mean willingness to buy at α = .05.

8.70

d.

No. Since the test in part c indicated that interaction between “Type of system” and “Pricing option” is present, we should not test for the main effects. Instead, we should proceed directly to a multiple comparison procedure to compare selected treatment means. If interaction is present, it can cover up the main effects.

a.

The treatments are the 3 × 3 = 9 combinations of PES and Trust. The nine treatments are: (BC, Low), (PC, Low), (NA, Low), (BC, Med), (PC, Med), (NA, Med), (BC, High), (PC, High), and (NA, High).

b.

df(Trust) = 3 − 1 = 2; SSE = SSTot − SS(PES) − SS(Trust) − SSPT = 161.1162 − 2.1774 − 7.6367 − 1.7380 = 149.5641 SS(PES) 2.1774 = = 1.0887 MS(PES) = 2 df(PES) SS(Trust) 7.6367 = = 3.81835 MS(Trust) = 2 df(Trust) SS(PT) 1.7380 = = 0.4345 MS(PT) = df(PT) 4 SSE 149.5641 MSE = = 0.7260 = df(Error) 206 MS(PES) MS(Trust) 1.0887 3.81835 FPES = = 1.50 FTrust = = 5.26 = = MSE MSE 0.7260 0.7260 MS(PT) 0.4345 FPT = = = 0.60 0.7260 MSE The ANOVA table is: Source PES Trust PES × Trust Error Total

c.

df 2 2 4 206 214

SS 2.1774 7.6367 1.7380 149.5641 161.1162

MS 1.0887 3.81835 0.4345 0.7260

F 1.50 5.26 0.60

To determine if PES and Trust interact, we test: H0: PES and Trust do not interact to affect the mean tension Ha: PES and Trust do interact to affect the mean tension The test statistic is F = 0.60.

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The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = (a − 1)(b − 1) = (3 − 1)(3 − 1) = 4 and ν2 = n − ab = 215 − 3(3) = 206. From Table IX, Appendix B, F.05 ≈ 2.37. The rejection region is F > 2.37. Since the observed value of the test statistic does not fall in the rejection region (F = 0.60 >/ 2.37), H0 is not rejected. There is insufficient evidence to indicate that PES and Trust interact at α = .05. d.

The plot of the treatment means is: The mean tension scores for Low Trust are relatively the same for each level of PES. Similarly, the mean tension scores for Medium Trust are relatively the same for each level of PES. However, the mean tension scores for High Trust are not the same for each level of PES. For both PES levels BC and PC, as the level of trust increases, the mean tension scores decrease. However, for PES level NA, as trust goes from low to medium, the mean tension decreases. As the trust goes from medium to high, the mean tension increases. This indicates that interaction is present which was also found in part d.

e.

8.72

Because the interaction of PES and Trust was found to be significant, the tests for the main effects are irrelevant. If the factors interact, the interaction effect can cover up any main effect differences. In addition, interaction implies that the effects of one factor on the dependent variable are different at different levels of the second factor. Thus, there is no one "main" effect of the factor.

Using MINITAB, the ANOVA results are: General Linear Model: Deviation versus Group, Trail Factor Group Trail

Type Levels Values fixed 4 F G M N fixed 2 C E

Analysis of Variance for Deviatio, using Adjusted SS for Tests Source Group Trail Group*Trail Error Total

DF 3 1 3 112 119

Seq SS 16271.2 46445.5 2245.2 82131.7 147093.6

Adj SS 13000.6 46445.5 2245.2 82131.7

Adj MS 4333.5 46445.5 748.4 733.3

F 5.91 63.34 1.02

P 0.001 0.000 0.386

First, we must test for treatment effects. SST = SS(Group) + SS(Trail) + SS(GxT) = 16,271.2 + 46,445.5 + 2,245.2 = 64,961.9. The df = 3 + 1 + 3 = 7.

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MST =

SST 64, 961.9 = = 9, 280.2714 ab − 1 4(2) − 1

F=

MST 9, 280.2714 = = 12.66 MSE 733.3

To determine if there are differences in mean ratings among the 8 treatments, we test: H0: All treatment means are the same Ha: At least two treatment means differ The test statistic is F = 12.66. Since no α was given, we will use α = .05. The rejection region requires α = .05 in the upper tail of the F distribution with ν1 = ab – 1 = 4(2) – 1 = 7 and ν2 = n – ab = 120 – 4(2) = 112. From Table IX, Appendix B, F.05 ≈ 2.09. The rejection region is F > 2.09. Since the observed value of the test statistic falls in the rejection region (F = 12.66 > 2.09), H0 is rejected. There is sufficient evidence that differences exist among the treatment means at α = .05. Since differences exist, we now test for the interaction effect between Trail and Group. To determine if Trail and Group interact, we test: H0: Trail and Group do not interact Ha: Trail and Group do interact The test statistic is F = 1.02 and p = .386 Since the p-value is greater than α (p = .386 > .05), H0 is not rejected. There is insufficient evidence that Trail and Group interact at α = .05. Since the interaction does not exist, we test for the main effects of Trail and Group. To determine if there are differences in the mean rating between the two levels of Trail, we test: H0: μ1 = μ2 Ha: μ1 ≠ μ2 The test statistics is F = 63.34 and p = 0.000. Since the p-value is greater than α (p = .000 < .05), H0 is rejected. There is sufficient evidence that the mean trail deviations differ between the fecal extract trail and the control trail α = .05. To determine if there are differences in the mean rating between the four levels of Group, we test: H0: μ1 = μ2 = μ3 = μ4 Ha: At least 2 means differ The test statistics is F = 5.91 and p = 0.001. Since the p-value is less than α (p = 0.001 < .05), Ho is rejected. There is sufficient evidence that the mean trail deviations differ among the four groups at α = .05.

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8.74

There are 3 × 2 = 6 treatments. They are A1B1, A1B2, A2B1, A2B2, A3B1, and A3B2.

8.76

a.

B

B

B

B

B

B

SSE = SSTot − SST = 62.55 − 36.95 = 25.60 df Treatment = p − 1 = 4 − 1 = 3 df Error = n − p = 20 − 4 = 16 df Total = n − 1 = 20 − 1 = 19 36.95 = 12.32 MST = SST/df = 3 25.60 = 1.60 MSE = SSE/df = 16 MST 12.32 F= = = 7.70 MSE 1.60 The ANOVA table: Source Treatment Error Total

b.

df

3 16 19

SS 36.95 25.60 62.55

MS 12.32 1.60

F 7.70

To determine if there is a difference in the treatment means, we test: H0: μ1 = μ2 = μ3 = μ4 Ha: At least two of the means differ where the μi represents the mean for the ith treatment. The test statistic is F =

MST = 7.70 MSE

The rejection region requires α = .10 in the upper tail of the F-distribution with ν1 = (p − 1) = (4 − 1) = 3 and ν2 = (n − p) = (20 − 4) = 16. From Table VIII, Appendix B, F.10 = 2.46. The rejection region is F > 2.46. Since the observed value of the test statistic falls in the rejection region (F = 7.70 > 2.46), H0 is rejected. There is sufficient evidence to conclude that at least two of the means differ at α = .10. c.

x4 =

∑x

4

n4

=

57 = 11.4 5

For confidence level .90, α = .10 and α/2 = .10/2 = .05. From Table VI, Appendix B, with df = 16, t.05 = 1.746. The confidence interval is: x4 ± t.05 MSE/n4 ⇒ 11.4 ± 1.746⋅ 1.6 / 5 ⇒ 11.4 ± .99 ⇒ (10.41, 12.39)

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8.78

a.

df(AB) = (a − 1)(b - 1) = 3(5) = 15 df(Error) = n − ab = 48 − 4(6) = 24 SSAB = MSAB(df) = 3.1(15) = 46.5 SS(Total) = SSA + SSB + SSAB + SSE = 2.6 + 9.2 + 46.5 + 18.7 = 77 SS A 2.6 SSB 9.2 = = .8667 = = 1.84 MSA = MSB = a −1 3 b −1 5 SSE 18.7 = = .7792 MSE = n − ab 24 MSA .8667 MSB 1.84 = = 1.11 = = 2.36 FB = FA = MSE .7792 MSE .7792 MS AB 3.1 = = 3.98 FAB = MSE .7792 B

Source A B AB Error Total

df 3 5 15 24 47

SS 2.6 9.2 46.5 18.7 77.0

MS .8667 1.84 3.1 .7792

F 1.11 2.36 3.98

b.

Factor A has a = 3 + 1 = 4 levels and factor B has b = 5 + 1 = 6 levels. The number of treatments is ab = 4(6) = 24. The total number of observations is n = 47 + 1 = 48. Thus, two replicates were performed.

c.

SST = SSA + SSB + SSAB = 2.6 + 9.2 + 46.5 = 58.3 SST 58.3 = = 2.5347 MST = ab − 1 4(6) − 1

F=

MST 2.5347 = = 3.25 MSE .7792

To determine whether the treatment means differ, we test: H0: μ1 = μ2 = ⋅⋅⋅ = μ24 Ha: At least one treatment mean is different The test statistic is F =

MST = 3.25 MSE

The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = ab − 1 = 4(6) − 1 = 23 and ν2 = n − ab = 48 - 4(6) = 24. From Table IX, Appendix B, F.05 ≈ 2.03. The rejection region is F > 2.03. Since the observed value of the test statistic falls in the rejection region (F = 3.25 > 2.03), H0 is rejected. There is sufficient evidence to indicate the treatment means differ at α = .05.

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d.

Since there are differences among the treatment means, we test for the presence of interaction: H0: Factor A and factor B do not interact to affect the response mean Ha: Factor A and factor B do interact to affect the response mean The test statistic is F =

MS AB = 3.98 MSE

The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = (a − 1)(b − 1) = (4 − 1)(6 − 1) = 15 and ν2 = n − ab = 48 − 4(6) = 24. From Table IX, Appendix B, F.05 = 2.11. The rejection region is F > 2.11. Since the observed value of the test statistic falls in the rejection region (F = 3.98 > 2.11), H0 is rejected. There is sufficient evidence to indicate factors A and B interact to affect the response means at α = .05. Since the interaction is significant, no further tests are warranted. Multiple comparisons need to be performed. 8.80

a.

This is a two-factor factorial design. It is also a completely randomized design.

b.

The two factors are "involvement in topic" and "question wording." Both are qualitative variables because neither are measured on numerical scales.

c.

There are two levels of "involvement in topic": high and low. There are two levels of "question wording": positive and negative.

d.

There are 2 × 2 = 4 treatments. The are: (high, positive), (high, negative), (low, positive), and (low, negative)

8.82

e.

The experiment's dependent variable is the level of agreement.

a.

To determine if the mean vacancy rates of the eight office-property submarkets in Atlanta differ, we test: H0: μ1 = μ2 = μ3 = μ4 = μ5 = μ6 = μ7 = μ8 Ha: At least two means differ

b.

If quarterly data were used for nine years, there are 4 × 9 = 36 observations per submarket. Since there are 8 submarkets, the total sample size is 8 × 36 = 288. Since no value of α is given, we will use α = .05. The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = k − 1 = 8 − 1 = 7 and ν2 = n − k = 288 – 8 = 280. From Table X, Appendix B, F.05 ≈ 2.01. The rejection region is F > 2.01.

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Since the observed value of the test statistic falls in the rejection region (F = 17.54 > 2.01), H0 is rejected. There is sufficient evidence to indicate the mean vacancy rates of the eight office-property submarkets in Atlanta differ at α = .05.

8.84

c.

With ν1 = k − 1 = 8 − 1 = 7 and ν2 = n − k = 288 – 8 = 280, P(F > 17.54) < .01, using Table XI, Appendix B. Thus, the p-value is less than .01.

d.

We must assume that all eight samples are randomly drawn from normal populations, the eight populations variances are the same, and the samples are independent.

e.

The mean vacancy rate for the South submarket is significantly larger than the mean vacancy rates for all other submarkets. The mean vacancy rate of the Downtown submarket is significantly larger than the mean vacancy rates for all other submarkets except the South. The mean vacancy rate of the North Lake submarket is significantly larger than the mean vacancy rates for all other submarkets except the South and Downtown. The mean vacancy rate of the Midtown submarket is significantly larger than the mean vacancy rates for all other submarkets except the South, Downtown, and North Lake. There are no other significant differences.

a.

The response is the weight of a brochure. There is one factor and it is carton. The treatments are the five different cartons, while the experimental units are the brochures.

b.

(∑ y)

2

.750052 = .01406437506 n 40 SS(Total) = ∑ y 2 − CM = .014066537 − .01406437506 = .00000216264

CM =

SST =

=

2 Ti 2 . .15028 2 .14962 2 .15217 2 .150312 + + + + − .01406437506 ∑ n − CM = 14767 8 8 8 8 8 i

= .01406568209 - .01406437506 = .00000130703 SSE = SS(Total) − SST = .00000216264 - .00000130703 = .00000085561 SST .00000130703 = MST = = .000000326756 k −1 5 −1 SSE .00000085561 = = .000000024446 MSE = n−k 40 − 5 MST .000000326756 F= = = 13.37 MSE .000000024446 Source Treatments Error Total

df 4 35 39

SS .00000130703 .00000085561 .00000216264

MS F .000000326756 13.37 .000000024446

To determine whether there are differences in mean weight per brochure among the five cartons, we test:

H0: μ1 = μ2 = μ3 = μ4 = μ5 Ha: At least two treatment means differ

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The test statistic is F = 13.37. The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = k − 1 = 5 − 1 = 4 and ν2 = n − k = 40 − 5 = 35. From Table IX, Appendix B, F.05 ≈ 2.53. The rejection region is F > 2.53. Since the observed value of the test statistic falls in the rejection region (F = 13.37 > 2.53), H0 is rejected. There is sufficient evidence to indicate a difference in mean weight per brochure among the five cartons at α = .05. c.

We must assume that the distributions of weights for the brochures in the five cartons are normal, that the variances of the weights for the brochures in the five cartons are equal, and that random and independent samples were selected from each of the cartons.

d.

Using MINITAB, the results of Tukey’s multiple comparison procedure are:

Level Carton1 Carton2 Carton3 Carton4 Carton5

N 8 8 8 8 8

Mean 0.018459 0.018785 0.018703 0.019021 0.018789

Individual 95% CIs For Mean Based on Pooled StDev ---+---------+---------+---------+----(-----*-----) (----*-----) (----*-----) (-----*-----) (----*-----) ---+---------+---------+---------+-----0.01840 0.01860 0.01880 0.01900

StDev 0.000105 0.000101 0.000109 0.000232 0.000188

Pooled StDev = 0.000156 Tukey 95% Simultaneous Confidence Intervals All Pairwise Comparisons Individual confidence level = 99.32% Carton1 subtracted from: Carton2 Carton3 Carton4 Carton5

Lower 0.0001013 0.0000188 0.0003375 0.0001050

Center 0.0003262 0.0002437 0.0005625 0.0003300

Upper 0.0005512 0.0004687 0.0007875 0.0005550

Carton2 Carton3 Carton4 Carton5

------+---------+---------+---------+--(-----*------) (-----*-----) (-----*-----) (-----*------) ------+---------+---------+---------+---0.00035 0.00000 0.00035 0.00070

Carton2 subtracted from: Carton3 Carton4 Carton5

Lower -0.0003075 0.0000113 -0.0002212

Center -0.0000825 0.0002363 0.0000037

Carton3 Carton4 Carton5

------+---------+---------+---------+--(------*-----) (------*-----) (-----*------) ------+---------+---------+---------+---0.00035 0.00000 0.00035 0.00070

Design of Experiments and Analysis of Variance

Upper 0.0001425 0.0004612 0.0002287

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Carton3 subtracted from: Carton4 Carton5

Lower 0.0000938 -0.0001387

Center 0.0003187 0.0000862

Upper 0.0005437 0.0003112

Carton4 Carton5

------+---------+---------+---------+--(-----*------) (-----*------) ------+---------+---------+---------+---0.00035 0.00000 0.00035 0.00070

Carton4 subtracted from: Carton5

Lower -0.0004575

Center -0.0002325

Upper -0.0000075

Carton5

------+---------+---------+---------+--(-----*------) ------+---------+---------+---------+---0.00035 0.00000 0.00035 0.00070

The means arranged in order are: Carton 1 Carton 3 Carton 2 .018459.018703.018785.018789.019021

Carton 5

Carton 4

The interpretation of the Tukey results are: The mean weight for carton 4 is significantly higher than the mean weights of all the other cartons. The mean weights of cartons 5, 4, and 3 are significantly higher than the mean weight of carton 1.

8.86

e.

Since there are differences among the cartons, management should sample from many cartons.

a.

This is a randomized block design. Response: Factor: Factor type: Treatments: Experimental units:

290

the length of time required for a cut to stop bleeding drug qualitative drugs A, B, and C subjects

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b.

Using MINITAB, the results are: General Linear Model: Y versus Drug, Person Factor Drug Person

Type Levels Values fixed 3 A B C fixed 5 1 2 3 4 5

Analysis of Variance for Y, using Adjusted SS for Tests Source Drug Person Error Total

DF 2 4 8 14

Seq SS 156.4 7645.8 160.1 7962.3

Adj SS 156.4 7645.8 160.1

Adj MS 78.2 1911.5 20.0

F 3.91 95.51

P 0.066 0.000

Tukey 90.0% Simultaneous Confidence Intervals Response Variable Y All Pairwise Comparisons among Levels of Drug Drug = A subtracted from: Drug B C

Lower -11.56 -3.72

Center -4.820 3.020

Upper 1.922 9.762

-----+---------+---------+---------+(-------*-------) (--------*-------) -----+---------+---------+---------+-8.0 0.0 8.0 16.0

Upper 14.58

-----+---------+---------+---------+(--------*-------) -----+---------+---------+---------+-8.0 0.0 8.0 16.0

Drug = B subtracted from: Drug C

Lower 1.098

Center 7.840

Let μ1, μ2, and μ3 represent the mean clotting time for the three drugs.

H0: μ1 = μ2 = μ3 Ha: At least two means differ The test statistic is F =

MS(Drug) = 3.91 MSE

The p-value is p = 0.066. Since the observed level of significance is less than α = .10, H0 is rejected. There is sufficient evidence to indicate differences in the mean clotting times among the three drugs at α = .10. c.

The observed level of significance is given as 0.066.

d.

To determine if there is a significant difference in the mean response over blocks, we test:

H0: μ1 = μ2 = μ3 = μ4 = μ5 Ha: At least two block means differ The test statistic is F =

MS(Person) = 95.51 MSE

Design of Experiments and Analysis of Variance

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The p-value is p = 0.000. Since the observed level of significance is less than α = .10, H0 is rejected. There is sufficient evidence to indicate differences in the mean clotting times among the five people at α = .10. e.

The confidence interval to compare drugs A and B is (-11.56, 1.922). Since 0 is in the interval, there is no evidence of a difference in mean clotting times between drugs A and B. The confidence interval to compare drugs A and C is (-3.72, 9.762). Since 0 is in the interval, there is no evidence of a difference in mean clotting times between drugs A and C. The confidence interval to compare drugs B and C is (1.098, 14.58). Since 0 is not in the interval, there is evidence of a difference in mean clotting times between drugs B and C. Since the numbers are positive, the mean clotting time for drug C is greater than that for drug B. In summary, the mean clotting time for drug C is greater than that for drug B. No other differences exist.

8.88

a.

243.2 57.8 SS A SSB = = 243.2 MSB = = = 57.8 1 1 df B df A SSAB = SSTot- SSA - SSB - SSE = 976.3 - 243.2 - 57.8 - 670.8 = 4.5 SS AB SSE 4.5 670.8 = 4.5 MSE = = 8.712 = = MSAB = 1 77 df AB df E

MSA =

MS A 243.2 = 27.92 = MSE 8.712 MSAB 4.5 = 0.52 FAB = = 8.712 MSE

FA =

FB = B

MSB 57.8 = 6.63 = MSE 8.712

The ANOVA table is: Source Recent Performance (A) Risk attitude(B) AB Error Total

b.

df

1 1 1 77 80

SS 243.2 57.8 4.5 670.8 976.3

MS 243.2 57.8 4.5 8.712

F 27.92 6.63 0.52

To determine if factors A and B interact, we test:

H0: Factors A and B do not interact to affect the mean decision Ha: Factors A and B do interact to affect the mean decision The test statistic is F = 0.52.

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The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = (a − 1)(b − 1) = (2 − 1)(2 − 1) = 1 and ν2 = n − ab = 81 − 2(2) = 77. From Table IX, Appendix B, F.05 ≈ 4.00. The rejection region is F > 4.00. Since the observed value of the test statistic does not fall in the rejection region (F = .52 >/ 4.00), H0 is not rejected. There is insufficient evidence to indicate that factors A and B interact at α = .05. c.

Since the interaction is not significant, the main effect tests are meaningful. To determine if an individual's risk attitude affects his or her budgetary decisions, we test:

H0: No difference exists between the risk attitude means Ha: The risk attitude means differ The test statistic is F = 6.63. The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = b − 1 = 2 − 1 = 1 and ν2 = n − ab = 81 − 2(2) = 77. From Table IX, Appendix B, F.05 ≈ 4.00. The rejection region is F > 4.00. Since the observed value of the test statistic falls in the rejection region (F = 6.63 > 4.00), H0 is rejected. There is sufficient evidence to indicate an individual's risk attitude affects his or her budgetary decisions at α = .05. d.

To determine if recent performance affects budgeting decisions, we test:

H0: No difference exists between the recent performance means Ha: The recent performance means differ The test statistic is F = 27.92. The rejection region requires α = .01 in the upper tail of the F-distribution with ν1 = a − 1 = 2 − 1 = 1 and ν2 = n − ab = 81 − 2(2) = 77. From Table XI, Appendix B, F.01 ≈ 7.08. The rejection region is F > 7.08. Since the observed value of the test statistic falls in the rejection region (F = 27.92 > 7.08), H0 is rejected. There is sufficient evidence to indicate that recent performance affects his or her budgetary decisions at α = .01.

Design of Experiments and Analysis of Variance

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8.90

Let factor A be second plastic and factor B be metal density. Some preliminary calculations are:

(∑ y)

2

5.562 = 3.8642 n 8 SS(Total) = ∑ y 2 − CM = 9.1646 − 3.8642 = 5.3004

CM =

SSA = SSB =

=

Ai2 .922 4.642 ∑ br − CM = 2(2) + 2(2) − 3.8642 = 5.594 − 3.8642 = 1.7298 B 2j

∑ ar

SSAB =

∑

− CM =

ABij2 ar

.57 2 4.992 + − 3.8642 = 6.30625 − 3.8642 = 2.44205 2(2) 2(2)

− SSA − SSB − CM

.062 .862 .512 4.132 + + + − 1.7298 − 2.44205 − 3.8642 2 2 2 2 = 9.0301 − 8.03605 = .99405 SSE = SS(Total) − SSA − SSB − SSAB = 5.3004 − 1.7298 − 2.44205 − .99405 = .1345 SSA 1.7298 MSA = = = 1.7298 a −1 2 −1 SSB 2.44205 = = 2.44205 MSB = b −1 2 −1 SS AB .99405 = MSAB = = .99405 (a − 1)(b − 1) (1)(1) SSE .1345 = MSE = = .033625 n − ab 8 − (2)(2) MSA 1.7298 F(A) = = = 51.44 MSE .033625 MSB 2.44205 = F(B) = = 72.63 MSE .033625 MS AB .99405 F(AB) = = = 29.56 MSE .033625 =

Source A B AB Error Total

df 1 1 1 4

SS 1.72980 2.44205 .99405 .13450 7 5.30040

MS 1.72980 2.44205 .99405 .033625

F 51.44 72.63 29.56

SST = SSA + SSB + SSAB = 1.7298 + 2.44205 + .99405 = 5.1659 SST 5.1659 = = 1.7220 MST = ab − 1 2(2) − 1

294

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F(T) =

MST 1.7220 = 51.21 = MSE .033625

To determine whether differences exist among the treatment means, we test:

H0: μ1 = μ2 = μ3 = μ4 Ha: At least two treatment means differ The test statistic is F = 51.21. The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = ab − 1 = 2(2) − 1 = 3 and ν2 = n − ab = 8 − 2(2) = 4. From Table IX, Appendix B, F.05 = 6.59. The rejection region is F > 6.59. Since the observed value of the test statistic falls in the rejection region (F = 51.21 > 6.59), H0 is rejected. There is sufficient evidence to indicate differences in mean radiation among the four treatments at α = .05. Since there are differences among the treatment means, we next test to see if the two factors interact.

H0: Second plastic and metal density do not interact Ha: Second plastic and metal density do interact The test statistic is F =

MS AB = 29.56 MSE

The rejection requires α= .05 in the upper tail of the F-distribution with ν1 = (a − 1)(b − 1) = 1 and ν2 = n − ab = 8 − 2(2) = 4. From Table IX, Appendix B, F.05 = 7.71. The rejection region is F > 7.71. Since the observed value of the test statistic falls in the rejection region (F = 29.56 > 7.71), H0 is rejected. There is sufficient evidence to indicate second plastic and metal density interact at α = .05. Since interaction is present, no tests for main effects are necessary. Since we want to find the preferred method to protect patients, we will compare all four treatment means. There are four p ( p − 1) 4(4 − 1) treatments, so c = = = 6. For α* = α/c = .05/6 = .0083 and α*/2 = .0083/2 = 2 2 .0042 ≈ .005 and df = n - ab = 4, t.005 = 4.604 from Table VI, Appendix B.

Design of Experiments and Analysis of Variance

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We now form confidence intervals for the differences between each pair of means using the formula: ( xi − x j ) ± t.005 s

1 1 + where s = ni n j

MSE = .033625 = .1834

Pair

11 – 12 11 – 21 11 – 22 12 – 21 12 – 22 21 – 22

1 1 + ⇒ −.40 ± .844 ⇒ (−1.244, .444) 2 2 (.03 − .255) ± .844 ⇒ −.255 ± .844 ⇒ (−1.069, .619) (.03 − 2.065) ± .844 ⇒ −2.035 ± .844 ⇒ (−2.879, −1.191) (.43 − .255) ± .844 ⇒ .175 ± .844 ⇒ (−.669, 1.019) (.43 − 2.065) ± .844 ⇒ −1.635 ± .844 ⇒ (−2.479, −.791) (.255 - 2.065) ± .844 ⇒ −1.81 ± .844 ⇒ (−2.654, −.966) (.03 − .43) ± 4.604(.1834)

The means that differ are 11 and 22, 12 and 22, and 21 and 22. No other means are significantly different. Since we are looking for the treatment that gives the best protection (allows the smallest amount of radiation), we would pick any treatment except 22. Thus, use second plastic present and heavy alloy, second plastic present and light alloy, or second plastic not present and heavy alloy. Pick the one of these three which is the cheapest or the most convenient. 8.92

a.

There are a total of a × b = 3 × 3 = 9 treatments in this study.

b.

Using MINITAB, the ANOVA results are: General Linear Model: Y versus Display, Price Factor Display Price

Type Levels Values fixed 3 1 2 3 fixed 3 1 2 3

Analysis of Variance for Y, using Adjusted SS for Tests Source Display Price Display*Price Error Total

DF 2 2 4 18 26

Seq SS 1691393 3089054 510705 8905 5300057

Adj SS 1691393 3089054 510705 8905

Adj MS F 845696 1709.37 1544527 3121.89 127676 258.07 495

P 0.000 0.000 0.000

To get the SS for Treatments, we must add the SS for Display, SS for Price, and the SS for Interaction. Thus, SST = 1,691,393 + 3,089,054 + 510,705 = 5,291,152. The df = 2 + 2 + 4 = 8. SST 5, 291,152 MST 661,394 = = 661,394 MST = F= = = 1336.15 3(3) − 1 ab − 1 MSE 495

296

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To determine whether the treatment means differ, we test:

H0: μ1 = μ2 = ⋅⋅⋅ = μ9 Ha: At least two treatment means differ The test statistic is F =

MST = 1336.15 MSE

The rejection region requires α = .10 in the upper tail of the F-distribution with ν1 = ab − 1 = 3(3) − 1 = 8 and ν2 = n − ab = 27 − 3(3) = 18. From Table VIII, Appendix B, F.10 = 2.04. The rejection region is F > 2.04. Since the observed value of the test statistic falls in the rejection region (F = 1336.15 > 2.04), H0 is rejected. There is sufficient evidence to indicate the treatment means differ at α = .10. c.

Since there are differences among the treatment means, we next test for the presence of interaction.

H0: Factors A and B do not interact to affect the response means Ha: Factors A and B do interact to affect the response means The test statistic is F =

MSAB = 258.07 MSE

The rejection region requires α = .10 in the upper tail of the F-distribution with ν1 = (a − 1)(b − 1) = (3 − 1)(3 − 1) = 4 and ν2 = n − ab = 17 − 3(3) = 18. From Table VIII, Appendix B, F.10 = 2.29. The rejection region is F > 2.29. Since the observed value of the test statistic falls in the rejection region (F = 258.07 > 2.29), H0 is rejected. There is sufficient evidence to indicate the two factors interact at α = .10. d.

The main effect tests are not warranted since interaction is present in part c.

e.

The nine treatment means need to be compared.

f.

From the graph, if the like letters are connected, the lines are not parallel. This implies interaction is present. This agrees with the results of part c.

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8.94

a.

This is a completely randomized design with a complete four-factor factorial design.

b.

There are a total of 2 × 2 × 2 × 2 = 16 treatments.

c.

Using SAS, the output is: Analysis of Variance Procedure Dependent Variable: Y Sum of

Mean

Source

DF

Squares

Square

F Value

Pr > F

Model

15

546745.50

36449.70

5.11

0.0012

Error

16

114062.00

7128.88

Corrected Total

31

660807.50

R-Square

C.V.

Root MSE

Y Mean

0.827390

41.46478

84.433

203.63

DF

Anova SS

Mean Square

F Value

Pr > F

SPEED

1

56784.50

56784.50

7.97

0.0123

FEED

1

21218.00

21218.00

2.98

0.1037

SPEED*FEED

1

55444.50

55444.50

7.78

0.0131

COLLET

1

165025.13

165025.13

23.15

0.0002

SPEED*COLLET

1

44253.13

44253.13

6.21

0.0241

FEED*COLLET

1

142311.13

142311.13

19.96

0.0004

SPEED*FEED*COLLET

1

54946.13

54946.13

7.71

0.0135

WEAR

1

378.13

378.13

0.05

0.8208

SPEED*WEAR

1

1540.13

1540.13

0.22

0.6483

FEED*WEAR

1

946.13

946.13

0.13

0.7204

SPEED*FEED*WEAR

1

528.13

528.13

0.07

0.7890

COLLET*WEAR

1

1682.00

1682.00

0.24

0.6337

SPEED*COLLET*WEAR

1

512.00

512.00

0.07

0.7921

FEED*COLLET*WEAR

1

72.00

72.00

0.01

0.9212

SPEE*FEED*COLLE*WEAR

1

1104.50

1104.50

0.15

0.6991

Source

d.

To determine if the interaction terms are significant, we must add together the sum of squares for all interaction terms as well as the degrees of freedom. SS(Interaction) = 55,444.50 + 44,253.13 + 142,311.13 + 54,946.13 + 1,540.13 + 946.13 + 528.13 + 1,682.00 + 512.00 + 72.00 + 1,104.50 = 303,339.78 df(Interaction) = 11 SS(Interacton) 303, 339.78 = = 27,576.34364 MS(Interaction) = 11 df(Interaction) MS(Interaction) 27, 576.34364 = 3.87 F(Interaction) = = MSE 7128.88

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To determine if interaction effects are present, we test:

H0: No interaction effects exist Ha: Interaction effects exist The test statistic is F = 3.87. The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = 11 and ν2 = 16. From Table IX, Appendix B, F.05 ≈ 2.49. The rejection region is F > 2.49. Since the observed value of the test statistic falls in the rejection region (F = 3.87 > 2.49), H0 is rejected. There is sufficient evidence to indicate that interaction effects exist at α = .05. Since the sums of squares for a balanced factorial design are independent of each other, we can look at the SAS output to determine which of the interaction effects are significant. The three-way interaction between speed, feed, and collet is significant (p = .0135). There are three two-way interactions with p-values less than .05. However, all of these two-way interaction terms are imbedded in the significant three-way interaction term. e.

Yes. Since the significant interaction terms do not include wear, it would be necessary to perform the main effect test for wear. All other main effects are contained in a significant interaction term. To determine if the mean finish measurements differ for the different levels of wear, we test:

H0: The mean finish measurements for the two levels of wear are the same Ha: The mean finish measurements for the two levels of wear are different The test statistic is t = 0.05. The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = 1 and ν2 = 16. From Table IX, Appendix B, F.05 = 4.49. The rejection region is F > 4.49. Since the observed value of the test statistic does not fall in the rejection region (F = .05 >/ 4.49), H0 is not rejected. There is insufficient evidence to indicate that the mean finish measurements differ for the different levels of wear at α = .05. f.

We must assume that: i. ii. iii.

The populations sampled from are normal. The population variances are the same. The samples are random and independent.

Design of Experiments and Analysis of Variance

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Categorical Data Analysis

9.2

Chapter 9

The characteristics of the multinomial experiment are: 1. 2. 3. 4. 5.

The experiment consists of n identical trials. There are k possible outcomes to each trial. The probabilities of the k outcomes, denoted p1, p2, ... , pk, remain the same from trial to trial, where p1 + p2 + ⋅⋅⋅ + pk = 1. The trials are independent. The random variables of interest are the counts n1, n2, ... , nk in each of the k cells.

The characteristics of the binomial are the same as those for the multinomial with k = 2. 9.4

The hypotheses of interest are: H0: p1 = .25, p2 = .25, p3 = .50 Ha: At least one of the probabilities differs from the hypothesized value E(n1) = np1,0 = 320(.25) = 80 E(n2) = np2,0 = 320(.25) = 80 E(n3) = np3,0 = 320(.50) = 160 The test statistic is χ = 2

∑

[ ni − E (ni )] E (ni )

2

=

(78 − 80) 2 (60 − 80) 2 (182 − 160)2 = 8.075 + + 80 80 160

The rejection region requires α = .05 in the upper tail of the χ2 distribution with df = k − 1 2 = 5.99147. The rejection region is χ2 > = 3 − 1 = 2. From Table VII, Appendix B, χ.05 5.99147. Since the observed value of the test statistic falls in the rejection region (χ2 = 8.075 > 5.99147), H0 is rejected. There is sufficient evidence to indicate that at least one of the probabilities differs from its hypothesized value at α = .05. 9.6

300

a.

The qualitative variable of interest is the location of professional sports stadiums and ballparks. There are 3 levels or categories of this variable – downtown, central city, and suburban.

b.

Let p1 = proportion of major sports facilities located in downtown areas, p2 = proportion of major sports facilities located in central city areas, and p3 = proportion of major sports facilities located in suburban areas in 1997.

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To determine if the proportions of major sports facilities in downtown, central city, and suburban areas in 1997 are the different than in 1985, we test: H0: p1 = .40, p2 = .30, p3 = .30 Ha: At least one of the proportions differs from their hypothesized values c.

E(n1) = np1,0 = 113(.40) = 45.2; E(n2) = np2,0 = 113(.30) = 33.9; E(n3) = np3,0 = 113(.30) = 33.9

d.

The test statistic is [n − E (ni )]2 (58 − 45.2) 2 (26 − 33.9) 2 (29 − 33.9) 2 = + + = 6.174 χ2 = ∑ i 45.2 33.9 33.9 E ( ni )

e.

The degrees of freedom for the test statistic is k – 1 = 3 – 1 = 2. The p-value is p = P ( χ 2 ≥ 6.174) .

Using Table VII, Appendix B, with df = 2, .025 > P ( χ 2 ≥ 6.174) > .01 . Thus, .01 < p < .025. Since the p-value is smaller than α = .05, H0 is rejected. There is sufficient evidence to indicate the proportions of major sports facilities in downtown, central city, and suburban areas in 1997 are the different than in 1985. 9.8

a.

The categorical variable is the rating of the student exposure to social and environmental issues. It has 5 levels: 1-star, 2-stars, 3-stars, 4-stars, and 5-stars.

b.

If there were no difference in the category proportions, then each proportion should be pi = 1/5 = .20. There were a total of n = 30 business schools sampled. The expected number would be: E(n1) = E(n2) = E(n3) = E(n4) = E(n5) = n(pi,0) = 30(.20) = 6

c.

To determine if there are differences in the star rating category proportions of all MBA programs, we test: H0: p1 = p2 = p3 = p4 = p5 = .20 Ha: At least one pi differs from its hypothesized value

d.

The test statistic is

⎡ ni − E ( ni ) ⎦⎤ ( 2 − 6 )2 ( 9 − 6 )2 (14 − 6 )2 ( 5 − 6 )2 ( 0 − 6 )2 = + + + + = 21 χ =∑⎣ E ( ni ) 6 6 6 6 6 2

2

e.

The rejection region requires α = .05 in the upper tail of the χ2 distribution with 2 = 9.48773. The rejection df = k – 1 = 5 – 1 = 4. From Table VII, Appendix B, χ.05 2 region is χ > 9.48773.

Categorical Data Analysis

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f.

Since the observed value of the test statistic falls in the rejection region (χ2 = 21 > 9.48773), H0 is rejected. There is sufficient evidence to indicate differences in the star rating category proportions of all MBA programs at α = .05.

g.

Some preliminary calculations are: pˆ 3 =

x3 14 = = .467 n 30

For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B, z.025 = 1.96. The 95% confidence interval is:

pˆ 3 ± z.025

pˆ 3qˆ3 .467(.533) ⇒ .467 ± 1.96 ⇒ .467 ± .179 ⇒ (.288, .646) n 30

We are 95% confident that the proportion of all MBA programs that are ranked in the 3-star category is between .288 and .646. 9.10

a.

Some preliminary calculations are: E(n1) = np1,0 = 1000(.50) = 500

E(n2) = np2,0 = 1000(.22) = 220

E(n3) = np3,0 = 1000(.11) = 110

E(n4) = np4,0 = 1000(.17) = 170

To determine if the percentages disagree with the percentages reported by Nielson/NetRatings, we test: H0: p1 = .50, p2 = .22, p3 = .11, and p4 = .17 Ha: At least one pi differs from its hypothesized value The test statistic is 2 2 2 2 ⎡⎣ ni − E ( ni ) ⎤⎦ 487 − 500 ) 245 − 220 ) 121 − 110 ) 147 − 170 ) ( ( ( ( = + + + χ =∑ 500 220 110 170 E ( ni ) 2

2

= 7.391 The rejection region requires α = .05 in the upper tail of the χ2 distribution with 2 df = k – 1 = 4 – 1 = 3. From Table VII, Appendix B, χ.05 = 7.81473. The rejection 2 region is χ > 7.81473. Since the observed value of the test statistic does not fall in the rejection region (χ2 = 7.391 >/ 7.81473), H0 is not rejected. There is insufficient evidence to indicate the percentages disagree with the percentages reported by Nielson/NetRatings at α = .05.

302

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b.

Some preliminary calculations are: pˆ1 =

x1 487 = = .487 n 1000

For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B, z.025 = 1.96. The 95% confidence interval is: pˆ1 ± z.025

pˆ1qˆ1 .487(.513) ⇒ .487 ± 1.96 ⇒ .487 ± .031 ⇒ (.456, .518) n 1000

We are 95% confident that the percentage of all Internet searches that use the Google Search Engine is between 45.6% and 51.8%. 9.12

Some preliminary calculations are: E(n1) = np1,0 = 2,023(.45) = 910.35

E(n2) = np2,0 = 2,023 (.35) = 708.05

E(n3) = np3,0 = 2,023 (.15) = 303.45

E(n4) = np4,0 = 2,023 (.05) = 101.15

To determine if the percentages of all adults falling into the four response categories changed after the Enron scandal, we test: H0: p1 = .45, p2 = .35, p3 = .15, and p4 = .05 Ha: At least one pi differs from its hypothesized value The test statistic is 2 2 2 2 ⎡⎣ ni − E ( ni ) ⎤⎦ 1,173 − 910.35 ) 587 − 708.05 ) 182 − 303.45 ) 81 − 101.15 ) ( ( ( ( χ =∑ = + + + 910.35 708.05 303.45 101.15 E ( ni ) 2

2

= 149.096 The rejection region requires α = .01 in the upper tail of the χ2 distribution with 2 = 11.3449. The rejection region is df = k – 1 = 4 – 1 = 3. From Table VII, Appendix B, χ.01 2 χ > 11.3449. Since the observed value of the test statistic falls in the rejection region (χ2 = 149.096 > 11.3449), H0 is rejected. There is sufficient evidence to indicate the percentages of all adults falling into the four response categories changed after the Enron scandal at α = .01.

Categorical Data Analysis

303

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9.14

a.

Some preliminary calculations are: E(n1) = np1,0 = 700(.09) = 63 E(n3) = np3,0 = 700(.02) = 14 E(n5) = np5,0 = 700(.12) = 84 E(n7) = np7,0 = 700(.03) = 21 E(n9) = np9,0 = 700(.09) = 63 E(n11) = np11,0 = 700(.01) = 7 E(n13) = np13,0 = 700(.02) = 14 E(n15) = np15,0 = 700(.08) = 56 E(n17) = np17,0 = 700(.01) = 7 E(n19) = np19,0 = 700(.04) = 28 E(n21) = np21,0 = 700(.04) = 28 E(n23) = np23,0 = 700(.02) = 14 E(n25) = np25,0 = 700(.02) = 14 E(n27) = np27,0 = 700(.02) = 14

χ2 = ∑

E(n2) = np2,0 = 700(.02) = 14 E(n4) = np4,0 = 700(.04) = 28 E(n6) = np6,0 = 700(.02) = 14 E(n8) = np8,0 = 700(.02) = 14 E(n10) = np10,0 = 700(.01) = 7 E(n12) = np12,0 = 700(.04) = 28 E(n14) = np14,0 = 700(.06) = 42 E(n16) = np16,0 = 700(.02) = 14 E(n18) = np18,0 = 700(.06) = 42 E(n20) = np20,0 = 700(.06) = 42 E(n22) = np22,0 = 700(.02) = 14 E(n24) = np24,0 = 700(.01) = 7 E(n26) = np26,0 = 700(.01) = 7

[ ni − E (ni )]2 (39 − 63) 2 (18 − 14) 2 (30 − 14) 2 (34 − 14) 2 = + + + ... + = 360.48 E (ni ) 63 14 14 14

To determine if ScrabbleExpress “ presents the player with unfair word selection opportunities” that are different from the Scrabble board game, we test: H0: Proportions in ScrabbleExpress are the same as in the Scrabble board game Ha: Proportions in ScrabbleExpress are different from those in the Scrabble board game The test statistic is χ 2 = 360.47 The rejection region requires α = .05 in the upper tail of the χ 2 distribution with df = k – 1 = 27 – 1 = 26. From Table VII, Appendix B, χ 2 = 38.8852. The rejection region is χ 2 > 38.8852. Since the observed value of the test statistic falls in the rejection region ( χ 2 = 360.47 > 38.8852), H0 is rejected. There is sufficient evidence to indicate the ScrabbleExpress “presents the player with unfair word selection opportunities” that are different from the Scrabble board game at α = .05. b.

The relative frequency of vowels for the board game is P(A) + P(E) + P(I) + P(O) + P(U) = .09 + .12 + .09 +.08 + .04 = .42 pˆ v =

304

39 + 31 + 25 + 20 + 21 136 = = .194 700 700

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For confidence level .95, α = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B, z.025 = 1.96. The 95% confidence interval is: pˆ v (1 − pˆ v ) .194(.806) ⇒ .194 ± 1.96 ⇒ .194 ± .029 ⇒ (.165, .223) n 700

pˆ v ± z.025

We are 95% confident that the true proportion of vowels in the ScrabbleExpress game is between .165 and .223. The true proportion from the board game is .42 which is much greater than the values in the interval. 9.16

2 df = (r − 1)(c − 1) = (5 − 1)(5 − 1) = 16. From Table VII, Appendix B, χ.05 = 26.2962. 2 The rejection region is χ > 26.2962.

a.

b.

9.18

2 df = (r − 1)(c − 1) = (3 − 1)(6 − 1) = 10. From Table VII, Appendix B, χ.10 = 15.9871. 2 The rejection region is χ > 15.9871.

c.

df = (r − 1)(c − 1) = (2 − 1)(3 − 1) = 2. From Table VII, Appendix B, χ2 = 9.21034. The rejection region is χ2 > 9.21034.

a.

To convert the frequencies to percentages, divide the numbers in each column by the column total and multiply by 100. Also, divide the row totals by the overall total and multiply by 100. The column totals are 25, 64, and 78, while the row totals are 96 and 71. The overall sample size is 165. The table of percentages are: Column 2

1 Row 1

b.

3

9 ⋅ 100 = 36% 25

34 ⋅ 100 = 53.1% 64

53 ⋅ 100 = 67.9% 78

96 ⋅ 100 = 57.5% 167

2 16 ⋅ 100 = 64% 25

30 ⋅ 100 = 46.9% 64

25 ⋅ 100 = 32.1% 78

71 ⋅ 100 = 42.5% 167

Using MINITAB, the graph is:

70 60

57.5%

50

Percent

40 30 20 10 0 1

2

3

Column

Categorical Data Analysis

305

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c.

9.20

If the rows and columns are independent, the row percentages in each column would be close to the row total percentages. This pattern is not evident in the plot, implying the rows and columns are not independent.

a-b. To convert the frequencies to percentages, divide the numbers in each column by the column total and multiply by 100. Also, divide the row totals by the overall total and multiply by 100. B B2

B1 B

c.

B3

B

B

Totals

A1 40 ⋅ 100 = 29.9% 134

72 ⋅ 100 = 44.2% 163

42 ⋅ 100 = 29.6% 142

154 ⋅ 100 = 35.1% 439

A2 63 ⋅ 100 = 47.0% Row 134

53 ⋅ 100 = 32.5% 163

70 ⋅ 100 = 49.3% 142

186 ⋅ 100 = 42.4% 439

A3 31 ⋅ 100 = 23.1% 134

38 ⋅ 100 = 23.3% 163

30 ⋅ 100 = 21.1% 142

99 ⋅ 100 = 22.6% 439

Using MINITAB, the graph is:

45 40 35

35.1%

30

Percent

25 20 15 10 5 0 1

2

3

B

The graph supports the conclusion that the rows and columns are not independent. If they were, then the height of all the bars would be essentially the same. 9.22

a.

The contingency table would be: Taxmotivation Yes No Total

306

Itemize Deductions Yes No 691 381 794 899 1,482 1,280

Total 1,072 1,693 2,765

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b.

c.

E11 =

R1C1 1,072(1, 485) = = 575.7 n 2,765

E21 =

R2C1 1,693(1, 485) = = 909.3 n 2,765

E12 =

R1C2 1,072(1, 280) = = 496.3 n 2,765

E22 =

R2C2 1,693(1, 280) = = 783.7 n 2,765

The test statistic is:

χ 2 = ∑∑

[nij − Eij ]2 Eij

[691 − 575.7]2 [381 − 496.3]2 [794 − 909.3]2 [899 − 783.7]2 + + + 575.7 496.3 909.3 783.7 = 81.46 =

d.

To determine if tax-motivation and itemize-deduction are related for charitable givers, we test: H0: Tax-motivation and itemize-deduction are independent Ha: Tax-motivation and itemize-deduction are dependent The test statistic is χ 2 = 81.46. The rejection region requires α = .05 in the upper tail of the χ 2 distribution with df = 2 = 3.84146. The (r – 1)(c – 1) = (2 – 1)(2 – 1) = 1. From Table VII, Appendix B, χ.05

rejection region is χ 2 > 3.84146. Since the observed value of the test statistic falls in the rejection region ( χ 2 = 81.46 > 3.84146), H0 is rejected. There is sufficient evidence to indicate that tax-motivation and itemize-deduction are related for charitable givers at α = .05. e.

To compute the bar graph, we first convert frequencies to percentages by dividing the numbers in each column by the column total and multiplying by 100%. Also, divide the row totals by the overall total and multiply by 100%.

Taxmotivation Yes No Total

Itemize Deductions Yes 691 ⋅ 100% = 46.5% 1485 794 ⋅ 100% = 53.5% 1485 1,485

Categorical Data Analysis

No 381 ⋅ 100% = 29.8% 1280 899 ⋅ 100% = 70.2% 1280 1,280

Total 1072 ⋅ 100% = 38.8% 2765 1693 ⋅ 100% = 61.28% 2765 2,765

307

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Using MINITAB, the bar graph is:

50

40

38.8%

Percent

30

20

10

0 Yes

No

Itemize

9.24.

a.

Some preliminary calculations are: pˆ C1 =

xC1 175 = = .028 n1 6, 222

pˆ C 2 =

xC 2 236 = = .050 4,692 n2

pˆ C 3 =

xC 3 319 = = .045 7,140 n3

pˆ C 4 =

xC 4 231 = = .038 6,120 n4

pˆ C 5 =

xC 5 480 = = .046 n5 10,353

pˆ C 6 =

xC 6 187 = = .039 4794 n6

The proportions range from .028 to .050. Since .050 is about twice as big as .028, there may be evidence to conclude some of the proportions are different. b.

308

Some preliminary calculations are: E11 =

R1C1 6, 222(37,693) = = 5,964.39 n 39,321

E12 =

R1C2 6, 222(1628) = = 257.61 n 39,321

E21 =

R2C1 4,692(37,693) = = 4497.74 n 39,321

E22 =

R2C2 4,692(1,628) = = 194.26 n 39,321

E31 =

R3C1 7,140(37,693) = = 6,844.38 n 39,321

E32 =

R3C2 7,140(1,628) = = 295.62 n 39,321

E41 =

R4C1 6,120(37,693) = = 5,866.61 n 39,321

E42 =

R4C2 6,120(1,628) = = 253.39 n 39,321

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E51 =

R5C1 10,353(37,693) = = 9,924.36 n 39,321

E52 =

R5C2 10,353(1,628) = = 428.64 n 39,321

E61 =

R6C1 4,794(37,693) = = 4,595.51 39,321 n

E62 =

R6C2 4,794(1,628) = = 198.49 39,321 n

To determine if the proportions of censored measurements differ for the six tractor lines, we test: H0: Tractor lines and Censored measurements are independent Ha: Tractor lines and Censored measurements are dependent The test statistic is 2

2 2 2 ⎡ nij − Eij ⎤ 6047 − 5964.39 ) 175 − 257.61) 4456 − 4497.74 ) ( ( ( ⎣ ⎦ χ = ∑∑ = + + 5964.39 257.61 4497.74 Eij 2

2 187 − 198.49 ) ( + ⋅⋅⋅ +

198.49

= 48.0978

The rejection region requires α = .01 in the upper tail of the χ2 distribution with 2 = 15.0863. df = (r – 1)(c – 1) = (6 – 1)(2 − 1) = 5. From Table VII, Appendix B, χ.01 2 The rejection region is χ > 15.0863. Since the observed value of the test statistic falls in the rejection region (χ2 = 48.0978 > 15.0863), H0 is rejected. There is sufficient evidence to indicate that the proportions of censored measurements differ for the six tractor lines at α = .01. c.

9.26

Even though there are differences in the proportions of censured data among the 6 tractor lines, these proportions range from .028 to .050. In practice, there is very little difference between .028 and .050.

Some preliminary calculations are: E11 =

R1C1 95(118) = = 42.8 262 n

E21 =

R2 C1 69(118) = = 31.1 n 262

E31 =

R3 C1 42(118) = = 18.9 n 262

E32 =

R3 C2 42(144) = = 23.1 n 262

E41 =

R4 C1 56(118) = = 25.2 n 262

E42 =

R4 C2 56(144) = = 30.8 n 262

Categorical Data Analysis

E12 =

R1C2 95(144) = = 52.2 262 n

E22 =

R2 C2 69(144) = = 37.9 n 262

309

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To determine whether a pig farmer’s education level has an impact on the size of the pig farm, we test: H0: Pig farmer’s education level and size of pig farm are independent Ha: Pig farmer’s education level and size of pig farm are dependent The test statistic is

χ 2 = ∑∑ +

[nij − Eij ]2 Eij

=

(42 − 42.8) 2 (53 − 52.2) 2 (27 − 31.1) 2 (42 − 37.9) 2 (22 − 18.9) 2 + + + + 42.8 52.2 31.1 37.9 18.9

(20 − 23.1) 2 (27 − 25.2)2 (29 − 30.8) 2 + + = 2.17 23.1 25.2 30.8

The rejection region requires α = .05 in the upper tail of the χ 2 distribution with df 2 = (r – 1)(c – 1) = (4 – 1)(2 – 1) = 3. From Table VII, Appendix B, χ.05 = 7.81473. The

rejection region is χ 2 > 7.81473. Since the observed value of the test statistic does not fall in the rejection region ( χ 2 = 2.17 >/ 7.81473), H0 is not rejected. There is insufficient evidence to indicate that a pig farmer’s education level has an impact on the size of the pig farm at α = .05. To compute the bar graph, we first convert frequencies to percentages by dividing the numbers in each row by the row total and multiplying by 100%. Also, divide the column totals by the overall total and multiply by 100%. Farm Size 5,000 pigs Total

310

Education Level No college College 42 53 ⋅ 100% = 44.2% ⋅ 100% = 55.8% 95 95 27 42 ⋅ 100% = 39.1% ⋅ 100% = 60.9% 69 69 22 20 ⋅ 100% = 52.4% ⋅ 100% = 47.6% 42 42 27 29 ⋅ 100% = 48.2% ⋅ 100% = 51.8% 56 56 118 144 ⋅ 100% = 45.0% ⋅ 100% = 55.0% 262 262

Total 95

69 42 56 262

Chapter 9

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Using MINITAB, the bar graph is:

50 45.0%

Percent

40

30 20

10 0 5,000

Farm Size

9.28

a.

Some preliminary calculations are: R1C1 53(35) = = 26.5 n 70 R C 17(35) = 8.5 E21 = 2 1 = n 70

R1C2 53(35) = = 26.5 n 70 R C 17(35) E22 = 2 2 = = 8.5 n 70

E11 =

E12 =

To determine if the severity of the ethical issue influenced whether the issue was identified or not by the auditors, we test: H0: Severity of ethical issue and identification are independent Ha: Severity of ethical issue and identification are dependent ⎡ nij − Eij ⎤⎦ The test statistic is χ = ∑∑ ⎣ Eij

2

2

(27 − 26.5) (26 − 26.5) (8 − 8.5) (9 − 8.5) + + + = = .078 26.5 26.5 8.5 8.5 2

2

2

2

The rejection region requires α = .05 in the upper tail of the χ2 distribution with df = 2 = 3.84146. The (r − 1)(c − 1) = (2 − 1)(2 − 1) = 1. From Table VII, Appendix B, χ.05 2 rejection region is χ > 3.84146. Since the observed value of the test statistic does not fall in the rejection region (χ2 = .078 >/ 3.84146), H0 is not rejected. There is insufficient evidence to indicate that the severity of the ethical issue influenced whether the issue was identified or not by the auditors at α = .05. b.

No. If there were 0 in the bottom cell of the column, then the expected count for that cell will be less than 5. One of the assumptions necessary for the test statistic to have a χ2 distribution will not hold.

Categorical Data Analysis

311

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c.

Suppose we change the numbers in the table to be as follows: Severity of Ethical Issue Moderate Severe 32 21 3 14

Ethical Issue Identified Ethical Issue Not Identified

Since the row and column totals are the same, the expected cell counts are the same as above.

⎡⎣ nij − Eij ⎤⎦ The test statistic is χ = ∑∑ Eij

2

2

=

(32 − 26.5) 2 (21 − 26.5) 2 (3 − 8.5) 2 (14 − 8.5) 2 + + + = 9.401 26.5 26.5 8.5 8.5

Now the test statistic would fall in the rejection region. 9.30

a.

The contingency table is:

Altitude < 300 300-600 ≥ 600 Totals

b.

Flight Response Low High 85 105 77 121 17 59 179 285

Totals 190 198 76 464

Some preliminary calculations are:

E11 =

R1C1 190(179) = = 73.297 n 464

E12 =

R1C2 190(285) = = 116.703 n 464

E21 =

R2C1 198(179) = = 76.384 n 464

E22 =

R2C2 198(285) = = 121.616 n 464

E31 =

R3C1 76(179) = = 29.319 464 n

E32 =

R3C2 76(285) = = 46.681 464 n

To determine if flight response of the geese depends on the altitude of the helicopter, we test:

H0: Flight response and Altitude of helicopter are independent Ha: Flight response and Altitude of helicopter are dependent

312

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The test statistic is

⎡ nij − Eij ⎤ ⎦ χ = ∑∑ ⎣ Eij

2

2

=

(85 − 73.297 )2 (105 − 116.703)2 ( 77 − 76.384 )2 (121 − 121.616 )2 73.297

+

+

+

116.703

(17 − 29.319 )

2

29.319

+

( 59 − 46.681)

+

76.384

121.616

2

46.681

= 11.477 The rejection region requires α = .01 in the upper tail of the χ2 distribution with 2 df = (r – 1)(c – 1) = (3 – 1)(2 − 1) = 2. From Table VII, Appendix B, χ.01 = 9.21034. 2 The rejection region is χ > 9.21034. Since the observed value of the test statistic falls in the rejection region (χ2 = 11.477 > 9.21034), H0 is rejected. There is sufficient evidence to indicate that the flight response of the geese depends on the altitude of the helicopter at α = .01. c.

The contingency table is: Flight Response Lateral Distance < 1000 1000-2000 2000-3000 ≥ 3000 Totals

d.

Low 37 68 44 30 179

High 243 37 4 1 285

Totals 280 105 48 31 464

Some preliminary calculations are:

E11 =

R1C1 280(179) = = 108.017 n 464

E12 =

R1C2 280(285) = = 171.983 n 464

E21 =

R2C1 105(179) = = 40.506 n 464

E22 =

R2C2 105(285) = = 64.494 n 464

E31 =

R3C1 48(179) = = 18.517 464 n

E32 =

R3C2 48(285) = = 29.483 464 n

E41 =

R 4 C1 31(179) = = 11.959 n 464

E42 =

R4C2 31(285) = = 19.041 n 464

Categorical Data Analysis

313

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To determine if flight response of the geese depends on the lateral distance of the helicopter, we test:

H0: Flight response and Lateral distance of the helicopter are independent Ha: Flight response and Lateral distance of the helicopter are dependent The test statistic is ⎡ nij − Eij ⎤ ⎦ χ 2 = ∑∑ ⎣ Eij =

2

( 37 − 108.017 )2 ( 243 − 171.983)2 ( 68 − 40.506 )2 ( 37 − 64.494 )2 108.017 +

+

171.983

( 44 − 18.517 ) 18.517

2

+

+

( 4 − 29.494 )

40.506 2

29.494

+

+

( 30 − 11.959 )

64.494 2

11.959

+

(1 − 19.041)2 19.041

= 207.814 The rejection region requires α = .01 in the upper tail of the χ2 distribution with 2 df = (r – 1)(c – 1) = (4 – 1)(2 − 1) = 3. From Table VII, Appendix B, χ.01 = 11.3449. 2 The rejection region is χ > 11.3449. Since the observed value of the test statistic falls in the rejection region (χ2 = 207.814 > 11.3449), H0 is rejected. There is sufficient evidence to indicate that the flight response of the geese depends on the lateral distance of the helicopter at α = .01. e.

Using SAS, the contingency table for altitude by response with the column percents is: Table of ALTGRP by RESPONSE ALTGRP

RESPONSE

Frequency| Percent | Row Pct | Col Pct |LOW |HIGH | Total ---------+--------+--------+

9 (1 − p 0 ) p0

=

9(1 − .01) = 891 .01

The minimum sample size is 892. b.

The sample size is determined by the following: n>

9 (1 − p 0 ) p0

=

9(1 − .05) = 171 .05

The minimum sample size is 172.

462

Chapter 12

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c.

The sample size is determined by the following: n>

9 (1 − p 0 ) p0

=

9(1 − .10) = 81 .10

The minimum sample size is 82. d.

The sample size is determined by the following: n>

9 (1 − p 0 ) p0

=

9(1 − .20) = 36 .20

The minimum sample size is 37. 12.36

a.

The sample size is determined by the following: n>

9 (1 − p 0 ) p0

=

9(1 − .07) = 119.6 ≈ 120 .07

The minimum sample size is 120. b.

To compute the proportion of defectives in each sample, divide the number of defectives by the number in the sample, 120: pˆ =

No. defectives No. in sample

The sample proportions are listed in the table: Sample No.

1 2 3 4 5 6 7 8 9 10

pˆ .092 .042 .033 .067 .083 .108 .075 .067 .083 .092

Sample No. pˆ 11 .083 12 .100 13 .067 14 .050 15 .083 16 .042 17 .083 18 .083 19 .025 20 .067

To get the total number of defectives, sum the number of defectives for all 20 samples. The sum is 171. To get the total number of units sampled, multiply the sample size by the number of samples: 120(20) = 2400.

Methods for Quality Improvement

463

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p =

Total defective in all samples 171 = .071 = 2400 Total units sampled

Centerline = p = .071 Upper control limit = p + 3 Lower control limit = p − 3

p (1 − p ) .071(.929) = .071 + 3 = .141 n 120 p (1 − p ) .071(.929) = .071 − 3 = .001 n 120

p (1 − p ) .071(.929) = .071 + 2 = .118 n 120 p (1 − p ) .071(.929) = .071 − 2 = .024 Lower A–B boundary = p − 2 n 120 p (1 − p ) .071(.929) Upper B–C boundary = p + = .071 + = .094 n 120 p (1 − p ) .071(.929) Lower B–C boundary = p − = .071 − = .048 n 120 Upper A–B boundary = p + 2

The p-chart is:

c.

To determine if the process is in or out of control, we check the four rules: Rule 1: Rule 2: Rule 3: Rule 4:

One point beyond Zone A: No points are beyond Zone A. Nine points in a row in Zone C or beyond: No sequence of nine points are in Zone C (on one side of the centerline) or beyond. Six points in a row steadily increasing or decreasing: No sequence of six points steadily increase or decrease. Fourteen points in a row alternating up and down: This pattern does not exist.

The process appears to be in control.

464

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12.38

d.

Since the process is in control, it is appropriate to use the control limits to monitor future process output.

e.

No. The number of defectives recorded was per day, not per hour. Therefore, the p-chart is not capable of signaling hour-to-hour changes in p.

a.

To compute the proportion of defectives in each sample, divide the number of defectives by the number in the sample, 200: pˆ =

No. defectives No. in sample

The sample proportions are listed in the table: Sample No.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

pˆ .065 .025 .010 .015 .010 .015 .005 .010 .005 .005 .055 .030 .010 .015 .005

Sample No. pˆ 16 .015 17 .005 18 .010 19 .015 20 .005 21 .045 22 .025 23 .010 24 .005 25 .015 26 .010 27 .020 28 .010 29 .005 30 .005

To get the total number of defectives, sum the number of defectives for all 30 samples. The sum is 96. To get the total number of units sampled, multiply the sample size by the number of samples: 200(30) = 6000. p =

Total defective in all samples 96 = = .016 Total units sampled 6000

The centerline is p = .016 p (1 − p ) .016(1 − .016) = .016 + 3 = .0426 n 200 p (1 − p ) .016(1 − .016) Lower control limit = p − 3 = .016 − 3 = -.0106 n 200 p (1 − p ) .016(1 − .016) Upper A–B boundary = p + 2 = .016 + 2 = .0337 n 200 Upper control limit = p + 3

Methods for Quality Improvement

465

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Lower A–B boundary = p − 2 Upper B–C boundary = p + Lower B–C boundary = p −

p (1 − p ) .016(1 − .016) = .016 − 2 = -.0017 n 200 p (1 − p ) .016(1 − .016) = .016 + = .0249 n 200 p (1 − p ) .016(1 − .016) = .016 − = .0071 n 200

The p-chart is:

b.

To determine if the process is in or out of control, we check the four rules: Rule 1: Rule 2: Rule 3: Rule 4:

One point beyond Zone A: There are 3 points beyond Zone A—Points 1, 11, and 21. Nine points in a row in Zone C or beyond: No sequence of nine points are in Zone C (on one side of the centerline) or beyond. Six points in a row steadily increasing or decreasing: This pattern is not present. Fourteen points in a row alternating up and down: This pattern does not exist.

The process does not appear to be in control. Rule 1 indicates that the process is out of control. 12.40

Specification spread is the difference between the upper specification limit and the lower specification spread. The specification spread is determined by customers, management, and product designers. Process spread is the spread of the actual output and is a function of the standard deviation of the data.

12.42

There are two reasons why CP should not be used in isolation. First, CP is a statistic and is subject to sampling error. The sample standard deviation is used to estimate the population standard deviation which is used to calculate the process spread. Thus, the estimate of the process spread can vary from sample to sample. Second, CP does not reflect the shape of the output distribution. Distributions with different shapes can have the same CP value.

466

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12.44

12.46

12.48

The specification spread is the difference between the upper specification limit and the lower specification limit. a.

Specification spread = USL − LSL = 19.65 − 12.45 = 7.20

b.

Specification spread = USL − LSL = .0010 − .0008 = .0002

c.

Specification spread = USL − LSL = 1.43 − 1.27 = 0.16

d.

Specification spread = USL − LSL = 490 − 486 = 4

CP =

Specification spread USL − LSL = 6σ Process spread

a.

CP ≈

1.0065 − 1.0035 USL − LSL .003 =1 = = 6s .003 6(.0005)

b.

CP ≈

22 − 21 USL − LSL 1 = = = .8333 6s 1.2 6(.2)

c.

CP ≈

875 − 870 USL − LSL 5 = 1.111 = = 6s 4.5 6(.75)

a.

If the output distribution is normal with a mean of 1000 and a standard deviation of 100, then the proportion of the output that is unacceptable is: P(x < 980) + P(x > 1,020) 980 − 1, 000 ⎞ 1, 020 − 1, 000 ⎞ ⎛ ⎛ = P⎜ z < ⎟ + P⎜ z > ⎟ 100 100 ⎝ ⎠ ⎝ ⎠ = P(z < −.2) + P(z > .2) = (.5 − .0793) + (.5 − .0793) = .8414 (using Table IV, Appendix B) The percentage of unacceptable output is 84.14%.

b.

CP =

USL − LSL 1, 020 − 980 40 = .067 ≈ = 6σ 600 6(100)

Since the value of CP is less than 1, the process is not capable.

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12.50

a.

A capability diagram is: LSL = 35 is off the chart.

b.

Fifty-two of the observations are above the upper specification limit. Thus, the percentage is (52/100) × 100% = 52%.

c.

From the sample, x = 37.007 and s = .083. CP =

d.

37 − 35 USL − LSL 2 ≈ = 4.016 = 6s .498 6(.083)

Since the CP value is greater than 1, the process is capable.

12.52

The quality of a good or service is indicated by the extent to which it satisfies the needs and preferences of its users. Its eight dimensions are: performance, features, reliability, conformance, durability, serviceability, aesthetics, and other perceptions that influence judgments of quality.

12.54

A process is a series of actions or operations that transform inputs to outputs. A process produces output over time. Organizational process: Manufacturing a product. Personnel Process: Balancing a checkbook.

12.56

The six major sources of process variation are: people, machines, materials, methods, measurements, and environment.

12.62

Common causes of variation are the methods, materials, equipment, personnel, and environment that make up a process and the inputs required by the process. That is, common causes are attributable to the design of the process. Special causes of variation are events or actions that are not part of the process design. Typically, they are transient, fleeting events that affect only local areas or operations within the process for a brief period of time. Occasionally, however, such events may have a persistent or recurrent effect on the process.

12.64

If a process is capable, then it is necessarily in control. If a process is in control, then the control chart should be used to monitor the process.

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12.66

The probability of observing a value of more than 3 standard deviations from its mean is: P( x > μ + 3 σ x ) + P( x < μ − 3 σ x ) = P(z > 3) + P(z < 3) = .5000 − .4987 + .5000 − .4987 = .0026 If we want to find the number of standard deviations from the mean the control limits should be set so the probability of the chart falsely indicating the presence of a special cause of variation is .10, we must find the z score such that: P(z > z0) + P(z < −z0) = .1000 or P(z > z0) = .0500 Using Table IV, Appendix B, z0 = 1.645. Thus the control limits should be set 1.645 standard deviations above and below the mean.

12.68

a.

The centerline = x =

∑ x = 150.58 n

20

= 7.529

The time series plot is:

12.70

b.

The variation pattern that best describes the pattern in this time series is the level shift. Points 1 through 10 all have fairly low values, while points 11 through 20 all have fairly high values.

a.

Yes. The minimum sample size necessary so the lower control limit is not negative is: n>

9 (1 − p 0 ) p0

From the data, p0 ≈ .06 Thus, n >

9(1 − .06) = 141. Our sample size was 200. .06

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b.

To compute the proportion of defectives in each sample, divide the number of defectives by the number in the sample, 200: p =

No. of defectives No. in sample

The sample proportions are listed in the table: Sample No. 1 2 3 4 5 6 7 8 9 10 11

p .02 .03 .055 .06 .025 .05 .04 .08 .085 .10 .14

Sample No. 12 13 14 15 16 17 18 10 20 21

p .10 .10 .085 .065 .05 .055 .035 .03 .04 .045

To get the total number of defectives, sum the number of defectives for all 21 samples. The sum is 258. To get the total number of units sampled, multiply the sample size by the number of samples: 200(21) = 4200. p =

No. of defectives 258 = = .0614 No. in sample 4200

Centerline = p = .0614 Upper control limit = p + 3 Lower control limit = p − 3 Upper A-B boundary = p + Lower A-B boundary = p − Upper B-C boundary = p + Lower B-C boundary = p −

470

p (1 − p ) = .0614 + 3 n p (1 − p ) = .0614 − 3 n p (1 − p ) 2 = .0614 + n p (1 − p ) 2 = .0614 − n p (1 − p ) = .0614 + n p (1 − p ) = .0614 − n

.0614(.9386) = .1123 200 .0614(.9386) = .0105 200 .0614(.9386) 2 = .0953 200 .0614(.9386) 2 = .0275 200 .0614(.9386) = .0784 200 .0614(.9386) = .0444 200

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The p-chart is:

c.

To determine if the control limits should be used to monitor future process output, we need to check the four rules. Rule 1: Rule 2: Rule 3: Rule 4:

One point beyond Zone A: The 11th point is beyond Zone A. This indicates the process is out of control. Nine points in a row in Zone C or beyond: There are not nine points in a row in Zone C (on one side of the centerline) or beyond. Six points in a row steadily increasing or decreasing: No sequence of six points steadily increase or decrease. Fourteen points in a row alternating up and down: This pattern does not exist.

Rule 1 indicates the process is out of control. These control limits should not be used to monitor future process output. 12.72

a.

In order for the x -chart to be meaningful, we must assume the variation in the process is constant (i.e., stable). ∑ x and R = range = largest measurement - smallest For each sample, we compute x = n measurement. The results are listed in the table: Sample No. 1 2 3 4 5 6 7 8 9 10 11 12

x 32.325 30.825 30.450 34.525 31.725 33.850 32.100 28.250 32.375 30.125 32.200 29.150

Methods for Quality Improvement

R 11.6 12.4 7.8 10.2 9.1 10.4 10.1 6.8 8.7 6.3 7.1 9.3

Sample No. 13 14 15 16 17 18 19 20 21 22 23 24

x 31.050 34.400 31.350 28.150 30.950 32.225 29.050 31.400 30.350 34.175 33.275 30.950

R 13.3 9.6 7.3 8.6 7.6 5.6 10.0 8.7 8.9 10.5 13.0 8.9

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x1 + x2 + " + x24 755.225 = = 31.4677 k 24 R + R2 + " + R24 221.8 = R = 1 = 9.242 k 24 x =

Centerline = x = 31.468 From Table XII, Appendix B, with n = 4, A2 = .729. Upper control limit = x + A2 R = 31.468 + .729(9.242) = 38.205 Lower control limit = x − A2 R = 31.468 - .729(9.242) = 24.731 2 2 ( A2 R ) = 31.468 + (.729)(9.242) = 35.960 3 3 2 2 Lower A-B boundary = x − ( A2 R) = 31.468 − (.729)(9.242) = 26.976 3 3 1 1 Upper B-C boundary = x + ( A2 R ) = 31.468 + (.729)(9.242) = 33.714 3 3 1 1 Lower B-C boundary = x − ( A2 R) = 31.468 − (.729)(9.242) = 29.222 3 3

Upper A-B boundary = x +

The x -chart is:

b.

To determine if the process is in or out of control, we check the six rules. Rule 1: Rule 2: Rule 3: Rule 4:

472

One point beyond Zone A: No points are beyond Zone A. Nine points in a row in Zone C or beyond: No sequence of nine points are in Zone C (on one side of the centerline) or beyond. Six points in a row steadily increasing or decreasing: No sequence of six points steadily increase or decrease. Fourteen points in a row alternating up and down: This pattern does not exist.

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Rule 5: Rule 6:

Two out of three points in Zone A or beyond: There are no groups of three consecutive points that have two or more in Zone A or beyond. Four out of five points in a row in Zone B or beyond: No sequence of five points has four or more in Zone B or beyond.

The process appears to be in control. There are no indications that special causes of variation are affecting the process.

12.74

c.

Since the process appears to be in control, these limits should be used to monitor future process output.

a.

A capability analysis diagram is:

b.

For an upper specification limit of 5, there are 27 observations above this limit. Thus, (27/100) × 100% = 27% of the observations are unacceptable. It does not appear that the process is capable.

c.

From Exercise 14.73, the process appears to be in control. Thus, it is appropriate to estimate CP. From the sample, x = 3.867 and s = 2.190 CP =

5−0 USL − LSL 5 = .381 = ≈ 6s 6(2.19) 13.14

Since the CP value is less than 1, the process is not capable.

12.76

d.

There is no lower specification limit because management has no time limit below which is unacceptable. The variable being measured is time customers wait in line. The actual lower limit would be 0.

a.

To get the total number of defectives, sum the number of defectives for all 36 samples. The sum is 279. To get the total number of units sampled, multiply the sample size by the number of samples: 160(36) = 5760. p =

Total defective in all samples 279 = .048 = 5760 Total units sampled

The centerline is p = .048

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p (1 − p ) .048(1 − .048) = .048 + 3 = .099 N 160 p (1 − p ) .048(1 − .048) = .048 − 3 = -.003 −3 N 160 p (1 − p ) .048(1 − .048) p +2 = .048 + 2 = .082 N 160 p (1 − p ) .048(1 − .048) = .048 − 2 = .014 p −2 N 160 p (1 − p ) .048(1 − .048) = .048 + = .065 p + N 160 p (1 − p ) .048(1 − .048) = .048 − = .031 p − N 160

Upper control limit = p + 3 Lower control limit = p Upper A–B boundary = Lower A–B boundary = Upper B–C boundary = Lower B–C boundary = The p-chart is:

b

To determine if the process is in or out of control, we check the four rules of the R-chart: Rule 1: Rule 2: Rule 3: Rule 4:

One point beyond Zone A: No points are beyond Zone A. Nine points in a row in Zone C or beyond: No sequence of nine points are in Zone C (on one side of the centerline) or beyond. Six points in a row steadily increasing or decreasing: This pattern is not present. Fourteen points in a row alternating up and down: This pattern does not exist.

The process appears to be in control. Thus, there is no indication that special causes of variation are present.

474

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c.

The Pareto diagram is:

Most of the defects are due to microcracks. Thus, "microcracks" are the "vital few." The other types of defectives are broken stands, gaps between layers, and internal voids. These are the "trivial many."

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Time Series: Descriptive Analyses, Models, and Forecasting

13.2

a.

Chapter 13

The simple composite index is calculated as follows: First, sum the observations for all the series of interest at each time period. Select the base time period. Divide each sum by the sum in the base time period and multiply by 100.

b.

To calculate a weighted composite index, we follow the following steps: First, multiply the observations in each time series by its appropriate weight. Then sum the weighted observations across all times series for each time period. Select the base time period. Divide each weighted sum by the weighted sum in the base time period and multiply by 100.

c.

The steps necessary to compute a Laspeyres Index are: 1. 2. 3. 4. 5.

d.

The steps necessary to compute a Paasche index are: 1. 2. 3. 4.

5.

13.4

476

a.

Collect data for each of k price series. Select a base time period and collect purchase quantity information for each of the k series at the base time period. Using the purchase quantity values at the base period as weights, multiply each value in the kth series by its corresponding weight. Sum the products for each time period. Divide each sum by the sum corresponding to the base period and multiply by 100.

Collect data for each of k price series. Select a base period. Collect purchase quantity information for each series at each time period. For each time period, multiply the value in each price series by its corresponding purchase quantity for that time period. Sum the products for each time period. To find the value of the Paasche index at a particular time period, multiply the purchase quantity values (weights) for that time period by the corresponding price values of the base time period. Sum the results for the base period. The Paasche Index is then found by dividing the sum found in (4) by the sum found in (5).

The simple index for the quarter 4 price of product A, using quarter 1 as the base period is (4.25 / 3.25) × 100 = 130.77.

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13.6

b.

The simple index for the quarter 2 price of product B, using quarter 1 as the base period is (1.25 / 1.75) × 100 = 71.43.

c.

To find the simple composite index, we must first sum the prices for all three products over the base period and the quarter for which we want to compute the simple composite index. The sum for quarter 1 is 3.25 + 1.75 + 8.00 = 13.00. The sum for quarter 4 is 4.25 + 1.00 + 10.50 = 15.75. The simple composite index for quarter 4 using quarter 1 as the base period is (15.75 / 13.00) × 100 = 121.15.

d.

The sum of all the products for quarter 2 is 3.50 + 1.25 + 9.35 = 14.10. The simple composite index for quarter 4 using quarter 2 as the base period is (15.75 / 14.10) × 100 = 111.70.

a.

To find the simple index, divide each value by the value for the base year and multiply by 100. The index numbers are:

Year 1975 1980 1985 1990 1995 2000 b.

Simple Index (Base Year = 1975) (13,719/13,719) × 100 = 100.00 (21,023/13,719) × 100 = 153.24 (27,735/13,719) × 100 = 202.16 (35,353/13,719) × 100 = 257.69 (40,611/13,719) × 100 = 296.02 (50,890/13,719) × 100 = 370.95

Simple Index (Base Year = 1980) (13,719/21,023) × 100 = 65.26 (21,023/21,023) × 100 = 100.00 (27,735/21,023) × 100 = 131.93 (35,353/21,023) × 100 = 168.16 (40,611/21,023) × 100 = 193.17 (50,890/21,023) × 100 = 242.07

The index value for 1990 is 257.69 when the base is 1975. Thus, the median annual family income for 1990 increased by 257.69 – 100 = 157.69% over the median annual family income in 1975. The index value for 1990 is 168.16 when the base is 1980. Thus, the median annual family income for 1990 increased by 168.16 – 100 = 68.16% over the median annual family income in 1980.

13.8

a.

To compute the simple index, divide each housing start value by the 2001, Quarter 1 value, 274 and then multiply by 100. Year 2001

2002

2003

Quarter 1 2 3 4 1 2 3 4 1 2

Simple Index (274/274) × 100 = (374/274) × 100 = (341/274) × 100 = (285/274) × 100 = (293/274) × 100 = (386/274) × 100 = (361/274) × 100 = (319/274) × 100 = (304/274) × 100 = (406/274) × 100 =

100.00 136.50 124.45 104.01 106.93 140.88 131.75 116.42 110.95 148.18

Year 2003 2004

2005

Time Series: Descriptive Analyses, Models, and Forecasting

Quarter 3 4 1 2 3 4 1 2 3 4

Simple Index (412/274) x 100 = (377/274) x 100 = (345/274) x 100 = (456/274) x 100 = (440/274) x 100 = (370/274) x 100 = (369/274) x 100 = (485/274) x 100 = (471/274) x 100 = (392/274) x 100 =

150.36 137.59 125.91 166.42 160.58 135.04 134.67 177.01 171.90 143.07

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13.10

b.

The value of the index for Quarter 2, 2004 is 166.42. Thus, the housing starts in Quarter 2, 2004 increased by 166.42 – 100 = 66.42% over the housing starts in the base quarter, Quarter 1, 2001.

c.

The value of the index for Quarter 4, 2005 is 143.07. Thus, the housing starts in Quarter 4, 2005 increased by 143.07 – 100 = 43.07% over the housing starts in the base quarter, Quarter 1, 2001.

d.

The number of housing starts for Quarter 1, 2003 is 304 thousand. The number of housing starts for Quarter 4, 2005 is 392 thousand. Using Quarter 1, 2003 as the base, the index for Quarter 4, 2005 is (392/304) × 100 = 128.95. Thus, the number of housing starts in Quarter 4, 2005 increased by 128.95 – 100 = 28.95% over the housing starts in Quarter 1, 2003.

a.

To compute the simple index for the agricultural data, divide each farm value by the 1980 value 3,364 and then multiply by 100. To compute the simple index for the nonagricultural data, divide each nonfarm value by the 1980 value 95,938 and then multiply by 100. The two indices are:

Year 1980 1985 1990 1995 2000 2003

Farm Index (3,364/3,364) × 100 = (3,179/3,364) × 100 = (3,223/3,364) × 100 = (3,440/3,364) × 100 = (2,464/3,364) × 100 = (2,275/3,364) × 100 =

Nonfarm Index (95,938/95,938) × 100 = (10,3971/95,938) × 100 = (115,570/95,938) × 100 = (121,460/95,938) × 100 = (134,427/95,938) × 100 = (135,461/95,938) × 100 =

100.00 108.37 120.46 126.60 140.12 141.20

b.

The nonfarm segment has shown the greater percentage change in employment over the time period. The nonfarm employment in 2003 was 41.20% greater than in 1980. The farm employment in 2003 was 32.37% lower than in 1980.

c.

To compute the simple composite index, first sum the two values (farm and nonfarm) for every time period. Then divide the sum by the sum in 1980, 99,302, and then multiply by 100. The simple composite index is: Year 1980 1985 1990 1995 2000 2003

d.

478

100.00 94.50 95.81 102.26 73.25 67.63

Sum 99,302 107,150 118,793 124,900 136,891 137,736

Simple Composite Index (99,302/99,302) × 100 = (107,150/99,302) × 100 = (118,793/99,302) × 100 = (124,900/99,302) × 100 = (136,891/99,302) × 100 = (137,736/99,302) × 100 =

100.00 107.90 119.63 125.78 137.85 138.70

The simple composite index value for 2003 is 138.70. The composite employment is 38.70% higher in 2003 than in 1980.

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13.12

a.

The find Laspeyres index, we multiply the durable goods by 10.9, the nondurable goods by 14.02, and the services by 42.6. The three products are then summed. The index is found by dividing the weighted sum at each time period by the weighted sum of 1970, 17,108.86, and then multiplying by 100. The Laspeyres index and the simple composite index for 1970 (computed in Exercise 13.11) are: Year

Simple Composite Index-1970 51.43 68.77 100.00 158.52 270.39 412.59 581.78 768.60 1,033.83 1,272.99

1960 1965 1970 1975 1980 1985 1990 1995 2000 2004 b.

Weighted Sum 8,409.95 11,442.51 17,108.86 27,509.89 48,215.53 76,167.86 110,254.64 150,193.08 202,856.51 251,152.45

Laspeyres Index 49.16 66.88 100.00 160.79 281.82 445.20 644.43 877.87 1,185.68 1,467.97

The plot of the two indices is: 1600

Variable I-1970 Laspeyres

1400 1200

Index

1000 800 600 400 200 0 1960

1965

1970

1975 1980

1985

1990

1995

2000

Y ear

The two indices are very similar from 1960 to approximately 1980. After 1980, the difference between the two indices becomes larger, with the Laspeyres index increasing faster than the simple composite index.

Time Series: Descriptive Analyses, Models, and Forecasting

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13.14

a.

To get the simple composite price index, sum the prices for the three metals for each month, divide by 2,090.35 (the sum of the prices for the base period January), and multiply by 100. To get the simple composite quantity index, sum the quantities for the three metals for each month, divide by 8,793.40 (the sum of the quantities for the base period January), and multiply by 100. The indices are:

Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec b.

Price Total 2,090.35 2,495.72 2,536.85 2,409.55 2,550.70 2,603.20 2,719.30 2,998.52 2,978.98 2,997.82 3,038.80 3,018.57

Price Index 100.00 119.39 121.36 115.27 122.02 124.53 130.09 143.45 142.51 143.41 145.37 144.41

Quantity Index 100.00 97.02 106.97 102.89 105.80 104.08 105.78 107.56 106.70 110.29 103.79 100.16

To compute the Laspeyres index, multiply the price for each month by the quantity for each of the metals for January, sum the products for the three metals, divide by 1,768,700.64 (the sum for the base period January), and multiply by 100. The Laspeyres index is: Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

480

Quantity Total 8,793.40 8,531.70 9,406.50 9,047.10 9,303.20 9,152.10 9,301.80 9,457.90 9,382.90 9,698.20 9,127.00 8,807.90

Total 1,768,700.64 2,077,067.24 2,345,138.00 2,114,563.64 1,760,956.32 1,746,326.88 2,117,568.80 2,377,017.20 2,100,958.72 2,276,109.40 2,366,980.72 2,155,654.92

Laspeyres Index 100.00 117.43 132.59 119.55 99.56 98.74 119.72 134.39 118.79 128.69 133.83 121.88

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c.

The plots of the simple composite price index, the simple composite quantity index, and Laspeyres index are: 150

Variable Price Quantity Laspeyres

140

Index

130

120

110

100

90 Jan

Feb Mar

Apr

May

Jun

Jul

Aug

Sep Oct

Nov Dec

M onth

The quantity index appears to be fairly stable while the price index steadily increases. The Laspeyres index is rather unstable, as it varies much more than the other two indices. d.

The following steps are used to compute the Paasche index: 1. 2.

3.

First, multiply the price × production for copper, steel, and lead for each month. The numerator of the index is the sum of these three quantities at each month. Next, multiply the production values of copper by 1,133, the production of steel by 187.75, and the production of lead by 769.6. The denominator is the sum of these three quantities at each month. The values of the Paasche index are the ratios of these two values at each month times 100.

The Paasche index is: Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

Paasche Numerator 1,768,700.64 2,013,192.24 2,500,128.80 2,180,640.81 1,858,912.26 1,822,735.92 2,230,984.40 2,549,791.96 2,244,369.96 2,504,067.86 2,450,159.20 2,175,046.70

Paasche Denominator 1,768,700.64 1,714,396.58 1,884,813.60 1,823,938.71 1,867,861.77 1,844,379.26 1,864,385.48 1,898,332.74 1,888,977.74 1,946,822.77 1,831,683.15 1,781,166.44

Time Series: Descriptive Analyses, Models, and Forecasting

Paasche Index 100.00 117.43 132.65 119.56 99.52 98.83 119.66 134.32 118.81 128.62 133.77 122.11

481

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e.

The plot of the Laspeyres index and the Paasche index is: The two indices are almost identical. Time Series Plot of Laspeyres, Paasche 135

Variable Laspeyres Paasche

130 125

Data

120 115 110 105 100 Jan

Feb Mar

Apr

May

Jun

Jul

Aug

Sep Oct

Nov Dec

M onth

13.16

f.

The values of Laspeyres index for September and December are 118.79 and 121.88 The values of the Paasche index for September and December are 118.81 and 122.11. These values are almost identical. Both the Laspeyres and Paasche indices are so close to being the same, neither is superior to the other.

a.

The exponentially smoothed employment for the first period is equal to the employment for that period. For the rest of the time periods, the exponentially smoothed employment values are found by multiplying .5 times the employment value of that time period and adding to that (1 − .5) times the value of the exponentially smoothed employment figure of the previous time period. The exponentially smoothed employment value for the time period 2 is .5(281) + (1 − .5)(280) = 280.5. The rest of the values are shown in the table.

Month Jan. Feb. Mar. Apr. May June July Aug. Sept. Oct. Nov. Dec.

482

t 1 2 3 4 5 6 7 8 9 10 11 12

Yt 280 281 250 246 239 218 218 210 205 206 200 200

Exponentially Smoothed Series w = .5 280.0 280.5 265.3 255.6 247.3 232.7 225.3 217.7 211.3 208.7 204.3 202.2

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b.

The graph of the time series and the exponentially smoothed series is:

280 270

Exponentially Smoothed Series

260

Yt

250 240

Series

230 220 210 200 2

4

6

8

10

12

Time Period

13.18

a.

The exponentially smoothed fish catch for Chile for the first period is equal to the fish catch for that period. For the rest of the time periods, the exponentially smoothed fish catch values are found by multiplying .5 times the fish catch of that time period and adding to that (1 − .5) times the value of the exponentially smoothed fish catch figure of the previous time period. The exponentially smoothed fish catch for Chile for the time period 1995 is .5(7,590.5) + (1 − .5)(5,195.4) = 6,392.95. The rest of the values are shown in the table. Similarly, the exponentially smoothed fish catch for Brazil for the first period is equal to the fish catch for that period. For the rest of the time periods, the exponentially smoothed fish catch values are found by multiplying .5 times the fish catch of that time period and adding to that (1 − .5) times the value of the exponentially smoothed fish catch figure of the previous time period. The exponentially smoothed fish catch for Brazil for time period 1995 is .5(800.0) + (1 − .5)(802.9) = 801.45. The rest of the values are shown in the table.

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Year 1990 1995 1998 1999 2000 2001 2002

b.

Chile Catch 5,195.4 7,590.5 3,265.3 5,050.2 4,300.0 3,797.1 4,271.5

Chile w=.5 Exponentially Smoothed Catch 5,195.40 6,392.95 4,829.13 4,939.66 4,619.83 4,208.47 4,239.98

Brazil Catch 802.9 800.0 706.8 703.9 766.8 806.7 822.1

Brazil w=.5 Exponentially Smoothed Catch 802.90 801.45 754.13 729.01 747.91 777.30 799.70

The plot of the two time series and the two exponentially smoothed series is: 8000

Variable Chile Brazil Chile-Exp Brazil-Exp

7000

Fish C atch

6000 5000 4000 3000 2000 1000 0 1990

1992

1994

1996 Y ear

1998

2000

2002

Both the time series and the exponentially smoothed series for the fish catch in Brazil are fairly stable over time. There is a decrease and then increase for both series in Brazil. Both the time series and exponentially smoothed series for the fish catch in Chile show a decrease over time. The exponentially smoothed series is more stable than the actual time series.

484

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13.20

a.

The exponentially smoothed expenditure for the first time period is equal to the expenditure for that period. For the rest of the time periods, the exponentially smoothed expenditures are found by multiplying the expenditures for the time period by w = .2 and adding to that (1 − .2) times the exponentially smoothed value above it. The exponentially smoothed value for the year 1991 is .2(548.9) + (1 − .2)(590.1) = 581.86. The rest of the values appear in the table. The process is repeated with w = .8.

Year 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000 2001 2002 2003 2004 2005

b.

Expenditure s 590.1 548.9 581.1 607.6 643.2 654.6 687.1 727.4 779.3 831.6 853.4 872.0 890.9 912.3 925.6 931.5

w = .2 Exponentially Smoothed Value 590.10 581.86 581.71 586.89 598.15 609.44 624.97 645.46 672.23 704.10 733.96 761.57 787.43 812.41 835.05 854.34

w = .8 Exponentially Smoothed Value 590.10 557.14 576.31 601.34 634.83 650.65 679.81 717.88 767.02 818.68 846.46 866.89 886.10 907.06 921.89 929.58

The plot of the two series is:

Variable Expend Exp-.2 Exp-.8

900

Expenditur es

800

700

600

500 1991

1993

1995

1997 1999 Y ear

2001

2003

There trend in personal consumption expenditure on transportation increased at a faster rate in the 1990s than in the 2000s. In the 2000s, the consumption expenditure is increasing but at a slower rate.

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13.22

a.

The exponentially smoothed Stock Index for the first time period is equal to the Stock Index for that time period. For the rest of the time periods, the exponentially smoothed stock price is found by multiplying w = .3 times the stock prices for that time period and adding to that (1 − .3) times the value of the exponentially smoothed stock price for the previous time period. The exponentially smoothed stock prices for the second time period is .3(1372.7) + (1 − .3)(1286.4) = 1312.29. The rest of the values are shown in the table.

Year 1999

2000

2001

2002

2003

2004

2005

2006

486

Quarter 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3

S&P 500 1286.4 1372.7 1282.7 1469.2 1498.6 1454.6 1436.5 1320.3 1160.3 1224.4 1040.9 1148.1 1147.4 989.8 815.3 879.8 848.2 974.5 996 1111.9 1126.2 1140.8 1114.6 1211.9 1180.6 1191.3 1228.8 1248.3 1294.9 1270.2 1335.8

Exponentially Smoothed Series w = .3 1286.4 1312.3 1303.4 1353.1 1396.8 1414.1 1420.8 1390.7 1321.6 1292.4 1217.0 1196.3 1181.6 1124.1 1031.4 986.0 944.6 953.6 966.3 1010.0 1044.9 1073.6 1085.9 1123.7 1140.8 1155.9 1177.8 1198.9 1227.7 1240.5 1269.1

Exponentially Smoothed Series w = .7 1286.4 1346.8 1301.9 1419.0 1474.7 1460.6 1443.7 1357.3 1219.4 1222.9 1095.5 1132.3 1142.9 1035.7 881.4 880.3 857.8 939.5 979.0 1072.0 1110.0 1131.5 1119.7 1184.2 1181.7 1188.4 1216.7 1238.8 1278.1 1272.6 1316.8

Chapter 13

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The plot of the original series and the exponentially smoothed series with w = .3 is:

Variable S&P 500 Exp-.3

1500 1400

S & P 500

1300 1200 1100 1000 900 800 Q uarter Year

b.

Q1 1999

Q1 Q1 2000 2001

Q1 Q1 2002 2003

Q1 2004

Q1 Q1 2005 2006

The same procedure is followed for w = .7. The exponentially smoothed Stock Index for the first time period is equal to the Stock Index for that time period. For the rest of the time periods, the exponentially smoothed stock price is found by multiplying w = .7 times the stock prices for that time period and adding to that (1 − .7) times the value of the exponentially smoothed stock price for the previous time period. The exponentially smoothed stock prices for the second time period is .7(1372.7) + (1 − .7)(1286.4) = 1346.8. The rest of the values are shown in the table in part a. The plot of the original series and the exponentially smoothed series with w = .7 is:

Variable S&P 500 Exp-.7

1500 1400

S & P 500

1300 1200 1100 1000 900 800 Q uarter Year

c.

Q1 1999

Q1 Q1 2000 2001

Q1 Q1 2002 2003

Q1 2004

Q1 Q1 2005 2006

The exponentially smoothed series with w = .3 better describes the trends in the series. The exponentially smoothed series with w = .7 is almost exactly like the original series.

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13.24

13.26

a.

The missing trend value for quarter 3 is: T3 = v(E3 – E2) + (1 – v)T2 = .6(3.78 – 3.50) + (1 − .6)(.25) = .27

b.

The missing smoothed value for quarter 4 is: E4 = wY4 + (1 – w)(E3 + T3) = .2(4.25) + (1 − .2)(3.78 + .27) = 4.09.

c.

The forecast for quarter 5 is: FQ5 = Ft+1 = Et + Tt = 4.09 + .29 = 4.38.

a.

To compute the exponentially smoothed values, we follow these steps: E1 = Y1 = 345 E2 = wY2 + (1 – w)E1 = .6(456) + (1 − .6)(345) = 411.60 E3 = wY3 + (1 – w)E2 = .6(440) + (1 − .6)(411.60) = 428.64 The rest of the values are computed in a similar manner and are listed in the table: Year 2004

Quarter 1 2 3 4 1 2 3 4

2005

b.

Exponentially Smoothed w = .6 345.00 411.60 428.64 393.46 378.78 442.51 459.61 419.04

Housing Starts 345 456 440 370 369 485 471 392

Using MINITAB, the plot is: 500

Variable Housing Exp-.6

475

Star ts

450

425

400

375

350 Q uarter Year

Q1 2004

Q2

Q3

Q4

Q1 2005

Q2

Q3

Q4

c. To forecast using exponentially smoothed values, we use the following: F2006,1 = Ft+1 = Et = 419.04 F2006,2 = Ft+2 = Ft+1 = 419.04 F2006,3 = Ft+3 = Ft+1 = 419.04 F2006,4 = Ft+4 = Ft+1 = 419.04

488

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13.28

a.

Using the information from Exercise 13.21, the forecast using the exponentially smoothed values with w = .9 is: F2006 = Ft+2 = Ft+ 1 = Et = 1815.3

b.

We first compute the Holt-Winters values for years 1974-2004. With w = .3 and v = .8, E2 = Y2 = 1171 E3 = wY3 + (1 – w)(E2 + T2) = .3(1663) + (1 − .3)(1171 + 245) = 1490.1 T2 = Y2 – Y1 = 1171 – 926 = 245 T3 = v(E3 – E2) + (1 – v)T2 = .8(1490.1 – 1171) + (1 − .8)(245) = 304.28 The rest of the Et’s and Tt’s appear in the table:

Year 1974 1975 1976 1977 1978 1979 1980 1981 1982 1983 1984 1985 1986 1987 1988 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000 2001 2002 2003 2004

t 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 31

Imports 926 1,171 1,663 2,058 1,892 1,866 1,414 1,067 633 540 553 479 771 876 987 1,232 1,282 1,233 1,247 1,339 1,307 1,303 1,258 1,378 1,522 1,543 1,664 1,770 1,490 1,671 1,833

Et w = .3 v = .8

Tt w = .3 v = .8

1171.00 1490.10 1873.47 2136.31 2253.87 2107.47 1734.46 1182.96 637.02 235.48 8.40 50.00 283.65 622.67 1020.93 1365.36 1571.76 1639.14 1619.79 1529.25 1411.34 1289.30 1232.36 1270.65 1364.08 1508.72 1679.04 1736.09 1771.26 1820.42

245.00 304.28 367.55 283.79 150.80 −86.96 −315.80 −504.36 −537.62 −428.76 −267.41 −20.21 182.88 307.79 380.16 351.58 235.43 100.99 4.72 −71.48 −108.63 −119.36 −69.42 16.75 78.09 131.33 162.52 78.14 43.77 48.08

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To forecast using the Holt-Winters Model: For w = .3 and v = .8, F2006 = Ft+2 = Ft+1 = Et + 2Tt = 1,820.42 + 2(48.08) = 1,916.58 c.

The error forecast for the exponentially smoothed series is Yt+2 – Ft+2 = 2,100 – 1815.3 = 284.7 The error forecast for the Holt-Winters series is Yt+2 – Ft+2 = 2,100 – 1,916.58 = 183.42 The error for the Holt-Winters forecast is smaller than the error for the exponentially smoothed forecast.

13.30

a.

We first compute the Holt-Winters values for the years 2003-2005. With w = .3 and v = .5, E2 = Y2 = 974.5 E3 = wY3 + (1 – w)(E2 + T2) =.3(996.0) + (1 − .3)(974.5 + 126.3) = 1,069.36. T2 = Y2 – Y1 = 974.5 – 848.2 = 126.3 T3 = v(E3 – E2) + (1 – v)T2 = .5(1,069.36 – 974.5) + (1 − .5)(126.3) = 110.58 The rest of the Et’s and Tt’s appear in the table that follows.

Year 2003

2004

2005

2006

490

Quarter 1 2 3 4 1 2 3 4 1 2 3 4 1 2 3

S&P 500 848.2 974.5 996.0 1111.9 1126.2 1140.8 1114.6 1211.9 1180.6 1191.3 1228.8 1248.3 1294.9 1270.2 1335.8

Et w = .3 v = .5

Tt w = .3 v = .5

Et w = .7 v = .5

Tt w = .7 v = .5

974.5 1069.36 1159.53 1219.79 1252.32 1250.50 1258.03 1246.99 1232.52 1227.45 1229.96

126.30 110.58 100.37 80.32 56.42 27.30 17.42 3.19 -5.64 -5.35 -1.42

974.5 1027.44 1113.45 1148.72 1161.64 1139.88 1192.62 1193.28 1196.53 1221.92 1245.60

126.30 89.62 87.81 61.54 37.23 7.74 30.24 15.45 9.35 17.37 20.52

Chapter 13

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To forecast using the Holt-Winters Model with w = .3 and v = .5: F2006,1 = Ft+1 = Et + Tt = 1,229.96 – 1.42 = 1,228.54 F2006,2 = Ft+2 = Et + 2Tt = 1,229.96 + 2(–1.42) = 1,227.12 F2006,3 = Ft+3 = Et + 3Tt = 1,229.96 + 3(–1.42) = 1,225.70 With w = .7 and v = .5, E2 = Y2 = 974.5 E3 = wY3 + (1 – w)(E2 + T2) =.7(996.0) + (1 − .7)(974.5 + 126.3) = 1,027.44. T2 = Y2 – Y1 = 974.5 – 848.2 = 126.3 T3 = v(E3 – E2) + (1 – v)T2 = .5(1,027.44 – 974.5) + (1 − .5)(126.3) = 89.62 The rest of the Et’s and Tt’s appear in the table above. To forecast using the Holt-Winters Model with w = .7 and v = .5: F2006,1 = Ft+1 = Et + Tt = 1,245.60 + 20.52 = 1,266.12 F2006,2 = Ft+2 = Et + 2Tt = 1,245.60 + 2(20.52) = 1,286.64 F2006,3 = Ft+3 = Et + 3Tt = 1,245.60 + 3(20.52) = 1,307.16 13.32

a.

From Exercise 13.25a, the forecasts for 2003-2005 using w = .3 are: F2003 = 199.48 F2004 = 199.48 F2005 = 199.48 The errors are the differences between the actual values and the predicted values. Thus, the errors are: Y2003 − F2003 = 195 − 199.48 = −4.48 Y2004 − F2004 = 197 − 199.48 = −2.48 Y2005 − F2005 = 195 − 199.48 = −4.48

b.

From Exercise 13.25a, the forecasts for 2003-2005 using w = .7 are: F2003 = 199.74 F2004 = 199.74 F2005 = 199.74 The errors are: Y2003 − F2003 = 195 − 199.74 = −4.74 Y2004 − F2004 = 197 − 199.74 = −2.74 Y2005 − F2005 = 195 − 199.74 = −4.74

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c.

For the exponentially smoothed forecasts with w = .3, m

∑ | Yt − Ft |

|195 − 199.48 | + |197 − 199.48 | + |195 − 199.48 | 11.44 = = 3.81 m 3 3 ⎡ m (Yt − Ft ) ⎤ ⎡ 195 − 199.48 197 − 199.48 195 − 199.48 ⎤ ⎢∑ ⎥ + + ⎢ ⎥ Y 195 197 195 i =1 t ⎢ ⎥ ⎢ ⎥ 100 = MAPE = ⎢ 100 ⎥ m 3 ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎢⎣ ⎥⎦ ⎡ .0585 ⎤ =⎢ ⎥ 100 = 1.9512 ⎣ 3 ⎦ MAD =

i =1

=

m

∑ (Yt − Ft )

2

i =1

RMSE =

=

m

= d.

(195 − 199.48)2 + (197 − 199.48)2 + (195 − 199.48)2 3 46.2912 = 3.928 3

For the exponentially smoothed forecasts with w = .7, m

MAD =

∑ | Yt − Ft | i =1

m

=

|195 − 199.74 | + |197 − 199.74 | + |195 − 199.74 | 12.22 = = 4.07 3 3

⎡ m (Yt − Ft ) ⎢∑ Yt i =1 MAPE = ⎢⎢ m ⎢ ⎢⎣

⎤ ⎡ 195 − 199.74 197 − 199.74 195 − 199.74 ⎥ + + ⎢ 195 197 195 ⎥ 100 = ⎢ ⎥ 3 ⎢ ⎥ ⎢ ⎣ ⎥⎦

⎤ ⎥ ⎥ 100 ⎥ ⎥ ⎦

⎡ .0625 ⎤ =⎢ ⎥ 100 = 2.0841 ⎣ 3 ⎦ m

RMSE =

∑ (Yt − Ft ) i =1

m

2

= =

492

(195 − 199.74 )2 + (197 − 199.74 )2 + (195 − 199.74 )2 3 52.4428 = 4.181 3

Chapter 13

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13.34

a.

From Exercise 13.29a, the forecasts for the 3 quarters of 2006 using w = .7 are: F2006,1 = 1,238.8 F2006,2 = 1,238.8 F2006,3 = 1,238.8 For the exponentially smoothed forecasts with w = .7: m

MAD =

∑ | Yt − Ft | i =1

m

=

|1294.9 − 1238.8 | + |1270.2 − 1238.8 | + |1335.8 − 1238.8 | 184.5 = = 61.5 3 3

⎡ m (Yt − Ft ) ⎢∑ Yt i =1 ⎢ MAPE = ⎢ m ⎢ ⎢⎣

⎤ ⎡ 1294.9 − 1238.8 1270.2 − 1238.8 1335.8 − 1238.8 ⎥ + + ⎢ 1294.9 1270.2 1335.8 ⎥ 100 = ⎢ ⎥ 3 ⎢ ⎥ ⎢ ⎣ ⎥⎦

⎤ ⎥ ⎥ 100 ⎥ ⎥ ⎦

⎡ .1407 ⎤ =⎢ ⎥ 100 = 4.689 ⎣ 3 ⎦ m

RMSE =

∑ (Yt − Ft ) i =1

m

2

= =

b.

(1294.9 − 1238.8)2 + (1270.2 − 1238.8)2 + (1335.8 − 1238.8)2 3 13,542.17 = 67.187 3

From Exercise 13.29b, the forecasts for the 3 quarters of 2006 using w = .3 are: F2006,1 = 1,198.9 F2006,2 = 1,198.9 F2006,3 = 1,198.9 For the exponentially smoothed forecasts with w = .3: m

MAD =

∑ | Yt − Ft | i =1

=

m 304.2 = = 101.4 3

|1294.9 − 1198.9 | + |1270.2 − 1198.9 | + |1335.8 − 1198.9 | 3

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⎡ m (Yt − Ft ) ⎢∑ Yt i =1 MAPE = ⎢⎢ m ⎢ ⎢⎣

⎤ ⎡ 1294.9 − 1198.9 1270.2 − 1198.9 1335.8 − 1198.9 ⎥ + + ⎢ 1294.9 1270.2 1335.8 ⎥ 100 = ⎢ ⎥ 3 ⎢ ⎥ ⎢ ⎣ ⎥⎦

⎤ ⎥ ⎥ 100 ⎥ ⎥ ⎦

⎡ .2328 ⎤ =⎢ ⎥ 100 = 7.759 ⎣ 3 ⎦ m

∑ (Yt − Ft )

2

i =1

RMSE =

m

= =

13.36

(1294.9 − 1198.9 )2 + (1270.2 − 1198.9 )2 + (1335.8 − 1198.9 )2 3 33,041.3 = 104.946 3

c.

For all three measures of error, the exponentially smoothed series with w = .7 is smaller than the exponentially smoothed series with w = .3. Thus, the more accurate series would be the exponentially smoothed series with w = .7.

a.

From Exercise 13.31, the actual data and the forecasts using the exponential smoothing and the Holt-Winters forecasts are:

Year 2005

Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

Gold Price 424.2 423.4 434.2 428.9 421.9 430.7 424.5 437.9 456.0 469.9 476.7 509.8

Exponential Forecast w =.5 433.47 433.47 433.47 433.47 433.47 433.47 433.47 433.47 433.47 433.47 433.47 433.47

Holt-Winters Forecast w =.5, v =.5 454.09 466.55 479.01 491.47 503.93 516.39 528.85 541.31 553.77 566.23 578.69 591.15

For the exponential smoothing forecasts with w = .5: m

MAD =

∑ | Yt − Ft | i =1

=

| 424.2 − 433.47 | + | 423.4 − 433.47 | + ⋅⋅⋅ + | 509.8 − 433.47 | 12

m 230.9 = = 19.242 12

494

Chapter 13

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⎡ m (Yt − Ft ) ⎢∑ Yt i =1 MAPE = ⎢⎢ m ⎢ ⎢⎣

⎤ ⎡ 424.2 − 433.47 423.4 − 433.47 509.8 − 433.47 ⎥ + + ⋅⋅⋅ + ⎢ 424.2 423.4 509.8 ⎥ 100 = ⎢ ⎥ 12 ⎢ ⎥ ⎢ ⎣ ⎥⎦

⎤ ⎥ ⎥ 100 ⎥ ⎥ ⎦

⎡ .4904 ⎤ =⎢ ⎥ 100 = 4.087 ⎣ 12 ⎦ m

∑ (Yt − Ft )

2

i =1

RMSE =

=

m

=

( 424.2 − 433.47 )2 + ( 423.4 − 433.47 )2 + ⋅⋅⋅ + ( 509.8 − 433.47 )2 12 9,980.2268 = 28.839 12

For the Holt-Winters forecasts with w = .5 and v = .5: m

MAD =

∑ | Yt − Ft | i =1

=

| 424.2 − 454.09 | + | 423.4 − 466.55 | + ⋅⋅⋅ + | 509.8 − 591.15 | 12

m 933.34 = = 77.778 12

⎡ m (Yt − Ft ) ⎢∑ Yt i =1 MAPE = ⎢⎢ m ⎢ ⎢⎣

⎤ ⎡ 424.2 − 454.09 423.4 − 466.55 509.8 − 591.15 ⎥ + + ⋅⋅⋅ + ⎢ 424.2 423.4 509.8 ⎥ 100 = ⎢ ⎥ 12 ⎢ ⎥ ⎢ ⎣ ⎥⎦

⎤ ⎥ ⎥ 100 ⎥ ⎥ ⎦

⎡ 2.0897 ⎤ =⎢ ⎥ 100 = 17.415 ⎣ 12 ⎦ m

RMSE =

∑ (Yt − Ft ) i =1

m

2

= =

( 424.2 − 454.09 )2 + ( 423.4 − 466.55)2 + ⋅⋅⋅ + ( 509.8 − 591.15)2 12 80,190.7476 = 81.747 12

For all three measures of forecast errors, the exponential smoothing forecasts had smaller errors. Thus, the exponential smoothing forecasts are better.

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b.

From Exercise 13.31, the actual data and the forecasts using the exponential smoothing one-step-ahead and the Holt-Winters one-step-ahead forecasts are:

Year 2005

Month Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

Gold Price 424.2 423.4 434.2 428.9 421.9 430.7 424.5 437.9 456.0 469.9 476.7 509.8

Exponential Forecast w =.5 433.47 428.83 426.12 430.16 429.53 425.71 428.21 426.35 432.13 444.06 456.98 466.84

Holt-Winters Forecast w =.5, v =.5 454.09 444.12 433.57 433.84 430.10 422.67 425.37 423.40 432.74 452.27 473.40 488.19

For the exponential smoothing one-step-ahead forecasts with w = .5: m

MAD =

∑ | Yt − Ft | i =1

=

| 424.2 − 433.47 | + | 423.4 − 428.83 | + ⋅⋅⋅ + | 509.8 − 466.84 | 12

m 164.32 = = 13.693 12

⎡ m (Yt − Ft ) ⎢∑ Yt i =1 MAPE = ⎢⎢ m ⎢ ⎢⎣

⎤ ⎡ 424.2 − 433.47 423.4 − 428.83 509.8 − 466.84 ⎥ + + ⋅⋅⋅ + ⎢ 424.2 423.4 509.8 ⎥ 100 = ⎢ ⎥ 12 ⎢ ⎥ ⎢ ⎣ ⎥⎦

⎤ ⎥ ⎥ 100 ⎥ ⎥ ⎦

⎡ .3540 ⎤ =⎢ ⎥ 100 = 2.950 ⎣ 12 ⎦ m

RMSE =

∑ (Yt − Ft ) i =1

m

2

= =

496

( 424.2 − 433.47 )2 + ( 423.4 − 428.83)2 + ⋅⋅⋅ + ( 509.8 − 466.84 )2 12 3,884.9754 = 17.993 12

Chapter 13

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For the Holt-Winters one-step-ahead forecasts with w = .5 and v = .5: m

MAD =

∑ | Yt − Ft | i =1

| 424.2 − 454.09 | + | 423.4 − 444.12 | + ⋅⋅⋅ + | 509.8 − 488.19 | 12

=

m 153.58 = = 12.798 12

⎡ m (Yt − Ft ) ⎢∑ Yt i =1 MAPE = ⎢⎢ m ⎢ ⎢⎣

⎤ ⎡ 424.2 − 454.09 423.4 − 444.12 509.8 − 488.19 ⎥ + + ⋅⋅⋅ + ⎢ 424.2 423.4 509.8 ⎥ 100 = ⎢ ⎥ 12 ⎢ ⎥ ⎢ ⎣ ⎥⎦

⎤ ⎥ ⎥ 100 ⎥ ⎥ ⎦

⎡ .3434 ⎤ =⎢ ⎥ 100 = 2.862 ⎣ 12 ⎦ m

RMSE =

∑ (Yt − Ft )

2

i =1

m

= =

( 424.2 − 454.09 )2 + ( 423.4 − 444.12 )2 + ⋅⋅⋅ + ( 509.8 − 488.19 )2 12 3,019.9854 = 15.864 12

For all three measures of forecast errors, the Holt-Winters forecasts have smaller errors. Thus, the Holt-Winters forecasts are better. 13.38

a.

Using MINITAB, the output is: Regression Analysis: Price versus t The regression equation is Price = 24.7 + 0.0910 t Predictor Constant t

Coef 24.6975 0.09103

S = 1.497

SE Coef 0.7851 0.08119

R-Sq = 8.2%

T 31.46 1.12

P 0.000 0.281

R-Sq(adj) = 1.7%

Analysis of Variance Source Regression Residual Error Total

DF 1 14 15

SS 2.817 31.379 34.197

MS 2.817 2.241

F 1.26

P 0.281

Predicted Values for New Observations New Obs 1

Fit 26.245

SE Fit 0.785

(

95.0% CI 24.561, 27.929)

(

95.0% PI 22.619, 29.871)

Values of Predictors for New Observations New Obs 1

t 17.0

Time Series: Descriptive Analyses, Models, and Forecasting

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Predicted Values for New Observations New Obs 2

Fit 26.336

SE Fit 0.857

(

95.0% CI 24.497, 28.175)

(

95.0% PI 22.636, 30.036)

Values of Predictors for New Observations New Obs 2

b.

t 18.0

The estimates of the parameters in the model, E(Yt) = β0 + β1t, are

βˆ0 = 24.6975 The price is estimated to be 24.6975 cents/pound for t = 0 or for 1991. βˆ1 = .09103

c.

The price is estimated to increase by .091 cents/pound for each additional year.

The forecast for 2007 is: Using t = 17, Yˆ 2003 = 24.6975 + .09103(17) = 26.2450 The forecast for 2008 is: Using t = 18, Yˆ 2004 = 24.6975 + .09103(18) = 26.3360 Yes, these agree with the predicted values on the printout.

d.

From the printout, the 95% forecast intervals are: 2007 (22.619, 29.871) 2008 (22.636, 30.036) We are 95% confident that the actual price in 2007 will be between 22.619 and 29.871. We are 95% confident that the actual price in 2008 will be between 22.636 and 30.036.

e.

13.40

498

No, we would not recommend that this model be used to forecast annual price. If we were to test if there is a significant linear relationship between time and annual price (H0: β1 = 0 vs Ha: β1 ≠ 0), the test statistic would be t = 1.12 and the p-value would be p = .281. Thus, we would conclude there is insufficient evidence to indicate a linear relationship exists between time and annual price. (Do not reject H0.)

The major advantage of regression forecasts over the exponentially smoothed forecasts is that prediction intervals can be formed using the regression forecasts and not using the exponentially smoothed forecasts.

Chapter 13

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13.42

a.

Using MINITAB, the results are: Regression Analysis: Price versus Time The regression equation is Price = 4.76 + 0.309 Time Predictor Constant Time

Coef 4.7608 0.30857

S = 0.769971

SE Coef 0.4184 0.04601

R-Sq = 77.6%

T 11.38 6.71

P 0.000 0.000

R-Sq(adj) = 75.8%

Analysis of Variance Source Regression Residual Error Total

DF 1 13 14

SS 26.661 7.707 34.368

MS 26.661 0.593

F 44.97

P 0.000

Unusual Observations Obs 15

Time 15.0

Price 10.740

Fit 9.389

SE Fit 0.379

Residual 1.351

St Resid 2.01R

R denotes an observation with a large standardized residual. Predicted Values for New Observations New Obs 1

Fit 9.698

SE Fit 0.418

95% CI (8.794, 10.602)

95% PI (7.805, 11.591)

Values of Predictors for New Observations New Obs 1

Time 16.0

Predicted Values for New Observations New Obs 1

Fit 10.006

SE Fit 0.459

95% CI (9.014, 10.999)

95% PI (8.069, 11.943)

Values of Predictors for New Observations New Obs 1

Time 17.0

From the printout:

βˆo = 4.7608 . The price of gas is estimated to be 4.7608 dollars per 1,000 cubic feet in 1989.

βˆ1 = .30857 . For each additional year, the price of gas is estimated to increase by .30857 dollars per 1,000 cubic feet.

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b.

To determine the model fit, we test: H0: β = 0 Ha: β ≠ 0 The test statistic is t = 6.71 (from the printout). The p-value is p = 0.000. Since the p-value is so small, H0 is rejected for any reasonable value of α. There is sufficient evidence that the model has an adequate fit.

c.

The 95% prediction interval for 2005 is (7.805, 11.591). We are 95% confident that the actual annual price of natural gas in 2005 is between 7.805 and 11.591 dollars per 1,000 cubic feet. The 95% prediction interval for 2006 is (8.069, 11.943). We are 95% confident that the actual annual price of natural gas in 2006 is between 8.069 and 11.943 dollars per 1,000 cubic feet.

13.44

d.

There are basically two problems with using simple linear regression for predicting time series data. First, we must predict values of the time series for values of time outside the observed range. We observe data for time periods 1, 2, … , t and use the regression model to predict values of the time series for t + 1, t + 2, … . The second problem is that simple linear regression does not allow for any cyclical effects such as seasonal trends.

a.

The regression model is: E (Yt ) = β o + β1t + β 2 Q1 + β3 Q2 + β 3 Q3

b.

Using MINITAB, the output is: Regression Analysis: Sales versus t, Q1, Q2, Q3 The regression equation is Sales = 120 + 16.5 t + 262 Q1 + 223 Q2 + 106 Q3 Predictor Constant t Q1 Q2 Q3

Coef 119.85 16.512 262.34 222.83 105.51

S = 26.00

SE Coef 16.95 1.028 16.73 16.57 16.48

R-Sq = 96.9%

T 7.07 16.07 15.68 13.45 6.40

P 0.000 0.000 0.000 0.000 0.000

R-Sq(adj) = 96.1%

Analysis of Variance Source Regression Residual Error Total Source t Q1 Q2 Q3

500

DF 1 1 1 1

DF 4 15 19

SS 318560 10139 328700

MS 79640 676

F 117.82

P 0.000

Seq SS 114343 81883 94610 27724

Chapter 13

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Predicted Values for New Observations New Obs 1

Fit 728.95

SE Fit 16.95

(

95.0% CI 692.82, 765.08)

(

95.0% PI 662.80, 795.10)

(

95.0% PI 639.80, 772.10)

(

95.0% PI 539.00, 671.30)

(

95.0% PI 450.00, 582.30)

Values of Predictors for New Observations New Obs 1

t 21.0

Q1 1.00

Q2 0.000000

Q3 0.000000

Predicted Values for New Observations New Obs 1

Fit 705.95

SE Fit 16.95

(

95.0% CI 669.82, 742.08)

Values of Predictors for New Observations New Obs 1

t 22.0

Q1 0.000000

Q2 1.00

Q3 0.000000

Predicted Values for New Observations New Obs 1

Fit 605.15

SE Fit 16.95

(

95.0% CI 569.02, 641.28)

Values of Predictors for New Observations New Obs 1

t 23.0

Q1 0.000000

Q2 0.000000

Q3 1.00

Predicted Values for New Observations New Obs 1

Fit 516.15

SE Fit 16.95

(

95.0% CI 480.02, 552.28)

Values of Predictors for New Observations New Obs 1

t 24.0

Q1 0.000000

Q2 0.000000

Q3 0.000000

The least squares equation is: Yˆt = 119.85 + 16.512t + 262.34Q1 + 222.83Q2 + 105.51Q3

βˆ1 = 16.512 βˆ2 = 262.34 βˆ3 = 222.83 βˆ4 = 105.51

For every increase in time period (1 quarter), the mean sales index increases by an estimated 16.512. The difference in mean sales index between the first and fourth quarters is estimated to be 262.34. The difference in the mean sales index between the second and fourth quarters is estimated to be 222.83. The difference in the mean sales index between the third and fourth quarters is estimated to be 105.51.

To determine if the model is useful, we test: H0: β1 = β2 = β3 = β4 = 0 Ha: At least one βi ≠ 0, i = 1, 2, 3, 4 The test statistic is F = 117.82

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Since no α is given, we will use α = .05. The rejection region requires α = .05 in the upper tail of the F-distribution with numerator df = k = 4 and denominator df = n − (k + 1) = 20 − (4 + 1) = 15. From Table IX, Appendix B, F = 3.06. The rejection region is F > 3.06. Since the observed value of the test statistic falls in the rejection region (F = 117.82 > 3.06), H0 is rejected. There is sufficient evidence to indicate the model is useful at α = .05. c.

The assumption of independent error terms is in doubt.

d.

The forecasts and the 95% prediction intervals are found at the bottom of the printout and are:

2007

13.46

13.48

I II III IV

Forecast 728.95 705.95 605.15 516.115

95% Lower Limit 95% Upper Limit 662.8 795.1 639.8 772.1 539.0 671.3 450.0 582.3

a.

d = 3.9 indicates the residuals are very strongly negatively autocorrelated.

b.

d = .2 indicates the residuals are very strongly positively autocorrelated.

c.

d = 1.99 indicates the residuals are probably uncorrelated.

a.

To determine if the overall model contributes information for the prediction of monthly passenger car and light truck sales, we test: H0: β1 = β2 = β3 = β4 = β5 = 0 Ha: At least 1 βi ≠ 0 The test statistic is F =

R2 / k .856 / 5 = = 164.067 2 (1 − R ) /[n − (k + 1)] (1 − .856) /[144 − (5 + 1)]

The rejection region requires α = .05 in the upper tail of the F distribution with ν1 = k = 5 and ν2 = n – (k + 1) = 144 – (5 + 1) = 138. From Table IX, Appendix B, F.05 ≈ 2.29. The rejection region is F > 2.29. Since the observed value of the test statistic falls in the rejection region (F = 164.067 > 2.29), H0 is rejected. There is sufficient evidence to indicate the overall model contributes information for the prediction of monthly passenger car and light truck sales at α = .05. b.

To determine if positive autocorrelation is present, we test: H0: No first-order autocorrelation Ha: Positive first-order autocorrelation of residuals

502

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The test statistics is d = 1.01. For α = .05, the rejection region is d < dL, α = dL,.05 ≈ 1.57. The value dL,.05 is found in Table XIII, Appendix B, with k = 5, n = 144, and α = .05. Since the observed value of the test statistic falls in the rejection region (d = 1.01 < 1.57, H0 is rejected. There is sufficient evidence to indicate the time series residuals are positively autocorrelated at α = .05.

13.50

c.

One of the requirements for the validity of the test in part b is that the error terms are independent. Since H0 was rejected in part a, there is evidence that positive autocorrelation exists. Since the error terms are not independent, the test in part b may not be valid.

a.

There is a tendency for the residuals to have long positive runs and negative runs. Residuals 1 through 6 are positive, while residuals 7 through 25 are negative. Residuals 26 through 35 are positive. This indicates the error terms are correlated.

b.

From the printout, the Durbin-Watson d is d = .0627. To determine if the time series residuals are autocorrelated, we test: H0: No first-order autocorrelation of residuals Ha: Positive or negative first-order autocorrelation of residuals The test statistic is d = .0627. For α = .10, the rejection region is d < dL,α/2 = dL,.05 = 1.40 or (4 − d) < dL,.05 = 1.40. The value dL,.05 is found in Table XIII, Appendix B, with k = 1, n = 35, and α = .10. Since the observed value of the test statistic falls in the rejection region (d = .0627 < 1.40), H0 is rejected. There is sufficient evidence to indicate the time series residuals are autocorrelated at α = .10.

c.

We must assume the residuals are normally distributed.

Time Series: Descriptive Analyses, Models, and Forecasting

503

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13.52

a.

Using MINITAB, the plot of the residuals against t is: Scatterplot of RESI1 vs Time 1.5

1.0

RESI1

0.5

0

0.0

-0.5

-1.0 0

2

4

6

8 T ime

10

12

14

16

There is not a random scattering of the residuals. The first 5 residuals are positive, the next 6 are negative, the next one is positive, the next one is negative and the last 2 are positive. This does not appear to be a random scattering. The plot suggests the possibility of autocorrelation. b.

Using MINITAB, the output is: Regression Analysis: Price versus Time The regression equation is Price = 4.76 + 0.309 Time Predictor Constant Time

Coef 4.7608 0.30857

S = 0.769971

SE Coef 0.4184 0.04601

R-Sq = 77.6%

T 11.38 6.71

P 0.000 0.000

R-Sq(adj) = 75.8%

Analysis of Variance Source Regression Residual Error Total

DF 1 13 14

SS 26.661 7.707 34.368

MS 26.661 0.593

F 44.97

P 0.000

Unusual Observations Obs 15

Time 15.0

Price 10.740

Fit 9.389

SE Fit 0.379

Residual 1.351

St Resid 2.01R

R denotes an observation with a large standardized residual. Durbin-Watson statistic = 1.39909

504

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To determine if positive autocorrelation is present, we test: H0: No first-order autocorrelation Ha: Positive first-order autocorrelation of residuals The test statistics is d = 1.399. For α = .05, the rejection region is d < dL, α = dL,.05 = 1.08. The value dL,.05 is found in Table XIII, Appendix B, with k = 1, n = 15, and α = .05. Since the observed value of the test statistic does not fall in the rejection region (d = 1.399 1.36), there is insufficient evidence to indicate the time series residuals are positively autocorrelated at α = .05.

13.54

c.

Since the error terms do not appear to be dependent, the validity of the test for the model adequacy appears to be fine.

a.

Using MINITAB, the plot of the residuals against t is: Scatterplot of RESI1 vs t 30 20

RESI1

10 0

0

-10 -20 -30 0

5

10

15

20

25

30

35

t

Since there appear to be groups of consecutive positive and groups of consecutive negative residuals, the data appear to be autocorrelated.

Time Series: Descriptive Analyses, Models, and Forecasting

505

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b.

Using MINITAB, the output is: Regression Analysis: Policies versus t The regression equation is Policies = 385 - 0.363 t Predictor Constant t

Coef 385.326 -0.3632

S = 15.0555

SE Coef 5.280 0.2632

R-Sq = 5.6%

T 72.98 -1.38

P 0.000 0.177

R-Sq(adj) = 2.7%

Analysis of Variance Source Regression Residual Error Total

DF 1 32 33

SS 431.6 7253.3 7685.0

MS 431.6 226.7

F 1.90

P 0.177

Unusual Observations Obs 1

t 1.0

Policies 355.00

Fit 384.96

SE Fit 5.05

Residual -29.96

St Resid -2.11R

R denotes an observation with a large standardized residual. Durbin-Watson statistic = 0.424942

To determine if positive autocorrelation is present, we test: H0: No first-order autocorrelation Ha: Positive first-order autocorrelation of residuals The test statistics is d = 0.42. For α = .05, the rejection region is d < dL, α = dL,.05 = 1.39. The value dL,.05 is found in Table XIII, Appendix B, with k = 1, n = 34, and α = .05. Since the observed value of the test statistic falls in the rejection region (d = .42 < 1.39), H0 is rejected. There is sufficient evidence to indicate the time series residuals are positively autocorrelated at α = .05. c.

506

Since the error terms do not appear to be independent, the validity of the test for model adequacy is in question.

Chapter 13

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13.56

a.

Year 1995 2000 2001 2002 2003 2004 b.

The exponentially smoothed price for the first time period is equal to the price for that period. For the rest of the time periods, the exponentially smoothed prices are found by multiplying the price for that time period by w = .5 and adding to that (1 − .5) times the exponentially smoothed price for the time period preceeding it. The exponentially smoothed values for each of the price series appear in the table:

Cold Finished Price 25.70 23.08 22.76 23.26 25.15 38.67

Exponentially Smoothed Value w = .5 25.70 24.39 23.58 23.42 24.28 31.48

Exponentially Smoothed Value w = .5 25.32 20.50 16.10 16.28 15.54 23.19

Hot Rolled Price 25.32 15.67 11.71 16.46 14.80 30.84

Galvanized Price 34.47 21.38 16.41 22.00 20.08 36.69

Exponentially Smoothed Value w = .5 34.47 27.93 22.17 22.08 21.08 28.89

The plot of the three price series and the exponentially smoothed series are: Cold Finished 40

Variable CF CF-Exp-.5

P r ice

35

30

25

1995 1996 1997

1998 1999

2000 2001

2002 2003

2004

Y ear

Time Series: Descriptive Analyses, Models, and Forecasting

507

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Hot Rolled Variable HR HR-Exp-.5

30

P r ice

25

20

15

10 1995 1996 1997 1998

1999 2000 2001

2002 2003 2004

Y ear

Galvanized Variable Gal Gal-Exp-.5

35

P r ice

30

25

20

15 1995

1996 1997 1998 1999 2000

2001 2002 2003 2004

Y ear

c.

The exponential smoothing forecasts for 2005 are: Cold Finished: F2005 = E2004 = 31.48 Hot Rolled: F2005 = E2004 = 23.19 Galvanized: F2005 = E2004 = 28.89 One of the main drawbacks of this kind of forecast is the inability to forecast future values using prediction intervals.

508

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13.58

a.

To compute the Laspeyres index, multiply the price for each year by the quantity for each of the items for 1990, sum the products for the four items, divide by 14.05 (the sum for the base period 1990), and multiply by 100. The Laspeyres index is:

Year 1990 1995 2000 2004

13.60

Spaghetti 0.85 0.88 0.88 0.95

Ground Beef 1.63 1.40 1.63 2.14

Eggs 1.00 1.16 0.96 0.98

Potatoes 0.32 0.38 0.35 0.51

Total 14.05 13.72 14.37 18.68

Laspeyres 100.00 97.65 102.28 132.95

b.

From 1990 to 2004, the “basket” of foods increased by 132.95 – 100 = 32.95%.

a.

We first calculate the exponentially smoothed values for 1980–1999. E1 = Y1 = 56.50 E2 = .8Y2 + (1 − .8)E1 = .8(27.0) + .2(56.50) = 32.90 E3 = .8Y3 + (1 − .8)E2 = .8(38.75) + .2(32.90) = 37.58 The rest of the values appear in the table. Year

1980 1981 1982 1983 1984 1985 1986 1987 1988 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000 2001 2002 2003 2004 2005 2006

Closing Exponentially Smoothed Value Price (w = .8) 56.50 56.50 27.00 32.90 38.75 37.58 45.25 43.72 41.75 42.14 68.37 63.12 45.62 49.12 48.02 48.24 48.01 48.06 64.03 60.84 45.00 48.17 68.07 64.09 30.03 36.84 29.05 30.61 32.05 31.76 41.05 39.19 50.75 48.44 65.50 62.09 49.00 51.62 36.31 39.37 48.44 46.63 55.75 53.93 40.00 42.79 46.60 45.84 46.65 46.49 39.43 40.84 43.80 43.21

Time Series: Descriptive Analyses, Models, and Forecasting

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The forecasts for 2007 and 2008 are: F2007 = Ft+1 = Et = 43.21 F2008 = Ft+2 = Et = 43.21 The expected gain is F2008 – Y2006 = 43.21 – 43.80 = −.59. Since this number is negative, it is actually a loss. b.

We first calculate the Holt-Winters values for 1980-2006. For w = .8 and v = .5, E2 = Y2 = 27.00 E3 = .8Y3 + (1 − .8)(E2 + T2) = .8(38.75) + .2(27 − 29.50) = 30.50 T2 = Y2 − Y1 = 27.00 − 56.50 = −29.50 T3 = .5(E3 − E2) + (1 − .5)(T2) = .5(30.50 − 27.00) + .5(−29.50) = -13.00 The rest of the values appear in the table. Year

1980 1981 1982 1983 1984 1985 1986 1987 1988 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000 2001 2002 2003 2004 2005 2006

510

Closing Price

56.50 27.00 38.75 45.25 41.75 68.37 45.62 48.02 48.01 64.03 45.00 68.07 30.03 29.05 32.05 41.05 50.75 65.50 49.00 36.31 48.44 55.75 40.00 46.60 46.65 39.43 43.80

Holt-Winters w = .8 v = .5 Et Tt

27.00 −29.5 30.50 −13.00 39.70 −1.90 40.96 −0.32 62.82 10.77 51.22 −0.42 48.58 −1.53 47.82 −1.14 60.56 5.80 49.27 −2.74 63.76 5.87 37.95 −9.97 28.84 −9.54 29.50 −4.44 37.85 1.96 48.56 6.33 63.38 10.58 53.99 0.59 39.96 −6.72 45.40 −0.64 53.55 3.76 43.46 −3.17 45.34 −0.65 46.26 0.14 40.82 −2.65 42.67 −0.40

Chapter 13

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The forecasts for 2007 and 2008 are: F2007 = Ft+1 = Et + Tt = 42.67 + (−.40) = 42.27 F2008 = Ft+2 = Et + 2Tt = 42.67 + 2(−.40) = 41.87 The expected gain is F2008 – Y2006 = 41.87 – 43.80 = −1.93. Since this number is negative, it is actually a loss. 13.62

a.

To compute the simple index for the IRA series, divide each IRA value by the 1990 value, 140, and then multiply by 100. To compute the simple index for the 401(k) series, divide each 401(k) value by the 1990 value, 35, and then multiply by 100. The values for the indices are in the table:

Year 1990 1994 1995 1996 1997 1998 1999 2000 2001 2002 2003 2004

b.

IRA 140 350 476 598 767 960 1234 1232 1161 1034 1307 1490

IRA Simple Index 100.00 250.00 340.00 427.14 547.86 685.71 881.43 880.00 829.29 738.57 933.57 1064.29

401(k) 35 184 266 346 466 616 810 815 794 706 919 1086

401(k) Simple Index 100.00 525.71 760.00 988.57 1331.43 1760.00 2314.29 2328.57 2268.57 2017.14 2625.71 3102.86

The time series plot is: 3500

Variable IRAindex 401(K)index

3000

Index

2500 2000 1500 1000 500 0 1990 1992

1994

1996 1998 Y ear

2000

2002

2004

Time Series: Descriptive Analyses, Models, and Forecasting

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13.64

c.

Both the IRA and 401(K) finds have increased since 1990. However, the 401(K) fund has increased at a higher rate than has the IRA fund.

a.

Using MINITAB, the results from fitting the model E(Yt) = βo + β1t are: Regression Analysis: GDP versus t The regression equation is GDP = 9595 + 79.5 t Predictor Constant t

Coef 9594.96 79.537

S = 97.4825

SE Coef 45.28 3.780

R-Sq = 96.1%

T 211.89 21.04

P 0.000 0.000

R-Sq(adj) = 95.9%

Analysis of Variance Source Regression Residual Error Total

DF 1 18 19

SS 4206863 171051 4377914

MS 4206863 9503

F 442.70

P 0.000

Unusual Observations Obs 1

t 1.0

GDP 9876.0

Fit 9674.5

SE Fit 42.0

Residual 201.5

St Resid 2.29R

R denotes an observation with a large standardized residual. Durbin-Watson statistic = 0.236602 Predicted Values for New Observations New Obs 1

Fit 11265.2

SE Fit 45.3

95% CI (11170.1, 11360.4)

95% PI (11039.4, 11491.1)

Values of Predictors for New Observations New Obs 1

t 21.0

Predicted Values for New Observations New Obs 1

Fit 11344.8

SE Fit 48.6

95% CI (11242.6, 11446.9)

95% PI (11115.9, 11573.6)

Values of Predictors for New Observations New Obs 1

512

t 22.0

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Predicted Values for New Observations New Obs 1

Fit 11424.3

SE Fit 52.0

95% CI (11315.0, 11533.6)

95% PI (11192.2, 11656.5)

Values of Predictors for New Observations New Obs 1

t 23.0

Predicted Values for New Observations New Obs 1

Fit 11503.8

SE Fit 55.5

95% CI (11387.3, 11620.4)

95% PI (11268.2, 11739.5)X

X denotes a point that is an outlier in the predictors. Values of Predictors for New Observations New Obs 1

t 24.0

The fitted regression line is: Yˆt = 9,594.96 + 79.537t From the printout, the 2006 quarterly GDP forecasts are:

Year 2006

b.

Quarter Q1 Q2 Q3 Q4

Forecast 11,265.2 11,344.8 11,424.3 11,503.8

95% Lower Limit 11,039.4 11,115.9 11,192.2 11,268.2

95% Upper Limit 11,491.1 11,573.6 11,656.5 11,739.5

The following model is fit: E(Yt) = βo + β1t + β1t + β2Q1 + β3Q2 + β4Q3 ⎧1 if quarter 1 where Q1 = ⎨ ⎩0 otherwise

⎧1 if quarter 2 Q2 = ⎨ ⎩0 otherwise

⎧1 if quarter 3 Q3 = ⎨ ⎩0 otherwise

The MINITAB printout is: Regression Analysis: GDP versus t, Q1, Q2, Q3 The regression equation is GDP = 9573 + 79.8 t + 29.4 Q1 + 21.1 Q2 + 25.8 Q3 Predictor Constant t Q1 Q2 Q3

Coef 9572.60 79.850 29.35 21.10 25.85

S = 105.993

SE Coef 69.10 4.190 68.20 67.56 67.17

R-Sq = 96.2%

T 138.53 19.06 0.43 0.31 0.38

P 0.000 0.000 0.673 0.759 0.706

R-Sq(adj) = 95.1%

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Analysis of Variance Source Regression Residual Error Total Source t Q1 Q2 Q3

DF 1 1 1 1

DF 4 15 19

SS 4209395 168519 4377914

MS 1052349 11235

F 93.67

P 0.000

Seq SS 4206863 656 212 1664

Unusual Observations Obs 1

t 1.0

GDP 9876.0

Fit 9681.8

SE Fit 58.1

Residual 194.2

St Resid 2.19R

R denotes an observation with a large standardized residual. Durbin-Watson statistic = 0.238059 Predicted Values for New Observations New Obs 1

Fit 11278.8

SE Fit 69.1

95% CI (11131.5, 11426.1)

95% PI (11009.1, 11548.5)

Values of Predictors for New Observations New Obs 1

t 21.0

Q1 1.00

Q2 0.000000

Q3 0.000000

Predicted Values for New Observations New Obs 1

Fit 11350.4

SE Fit 69.1

95% CI (11203.1, 11497.7)

95% PI (11080.7, 11620.1)

Values of Predictors for New Observations New Obs 1

t 22.0

Q1 0.000000

Q2 1.00

Q3 0.000000

Predicted Values for New Observations New Obs 1

Fit 11435.0

SE Fit 69.1

95% CI (11287.7, 11582.3)

95% PI (11165.3, 11704.7)

Values of Predictors for New Observations New Obs 1

t 23.0

Q1 0.000000

Q2 0.000000

Q3 1.00

Predicted Values for New Observations New Obs Fit SE Fit 95% CI 95% PI 1 11489.0 69.1 (11341.7, 11636.3) (11219.3, 11758.7) Values of Predictors for New Observations New Obs 1

514

t 24.0

Q1 0.000000

Q2 0.000000

Q3 0.000000

Chapter 13

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The fitted regression line is: Yˆt = 9,572.6 + 79.85t + 29.35Q1 + 21.10Q2 + 25.85Q3 To determine whether the data indicate a significant seasonal component, we test:

H0: β2 = β3 = β4 = 0 Ha: At least one βi ≠ 0

i = 2, 3, 4

The test statistic is F=

(SSE R − SSE C ) /(k − g ) (171,051 − 168,519) /(4 − 1) 844 = = = 0.075 SSE C [ n − ( k + 1)] 168,519 /[20 − (4 + 1)] 11, 234.6

Since no α is given, we will use α = .05. The rejection region requires α = .05 in the upper tail of the F-distribution with ν1 = k – g = 4 – 1 = 3 and ν2 = n – (k + 1) = 20 – (4 + 1) = 15. From Table IX, Appendix B, F.05 = 3.29. The rejection region is F > 3.29. Since the observed value of the test statistic does not fall in the rejection region (F = .075 >/ 3.29), H0 is not rejected. There is insufficient evidence to indicate a seasonal component at α = .05. This supports the assertion that the data have been seasonally adjusted. c.

From the printout, the 2006 quarterly forecasts are:

Year 2006

d.

Quarter Q1 Q2 Q3 Q4

Forecast 11,278.8 11,350.4 11,435.0 11,489.0

95% Lower Limit 11,009.1 11,080.7 11,165.3 11,219.3

95% Upper Limit 11,548.5 11,620.1 11,704.7 11,758.7

To determine if the time series residuals are autocorrelated, we test: H0: No first-order autocorrelation of residuals Ha: Positive or negative first-order autocorrelation of residuals The test statistic is d = 0.24. For α = .10, the rejection region is d < dL,α/2 = dL,.05 = .90 or (4 – d) < dL,.01 = .90. The value of dL,.05 is found in Table XIII, Appendix B, with k = 4 and n = 20. Since the observed value of the test statistic falls in the rejection region (d = 0.24 < .90), H0 is rejected. There is sufficient evidence to indicate the time series residuals are autocorrelated at α = .10.

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13.66

a.

Using MINITAB, the results from fitting the model E(Yt) = β0 + β1t are: Regression Analysis: Revolving versus t The regression equation is Revolving = - 84.5 + 33.8 t Predictor Constant t

Coef -84.54 33.768

S = 56.7803

SE Coef 23.41 1.575

R-Sq = 95.2%

T -3.61 21.44

P 0.001 0.000

R-Sq(adj) = 95.0%

Analysis of Variance Source Regression Residual Error Total

DF 1 23 24

SS 1482334 74152 1556486

MS 1482334 3224

F 459.78

P 0.000

Unusual Observations Obs 1

t 1.0

Revolving 55.0

Fit -50.8

SE Fit 22.0

Residual 105.8

St Resid 2.02R

R denotes an observation with a large standardized residual. Predicted Values for New Observations New Obs 1

Fit 827.2

SE Fit 24.8

95% CI (775.9, 878.5)

95% PI (699.0, 955.4)

Values of Predictors for New Observations New Obs 1

t 27.0

Predicted Values for New Observations New Obs 1

Fit 861.0

SE Fit 26.2

95% CI (806.7, 915.2)

95% PI (731.6, 990.3)

Values of Predictors for New Observations New Obs 1

t 28.0

The fitted regression line is: Yˆt = −84.54 + 33.768t

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For the years 2006 and 2007, t = 27 and 28. From the printout, the predicted values and 95% prediction intervals for 2006 and 2007 are:

Year 2006 2007

b.

Forecast 827.2 861.0

95% Lower Limit 699.0 731.6

95% Upper Limit 955.4 990.3

To compute the Holt-Winters values for the years 1980-2004: With w = .7 and v = .7, E2 = Y2 = 61 E3 = wY3 + (1 – w)(E2 + T2) =.7(66) + (1 − .7)(61 + 6) = 66.3. T2 = Y2 – Y1 = 61 – 55 = 6 T3 = v(E3 – E2) + (1 – v)T2 = .7(66.3 – 61) + (1 − .7)(6) = 5.51 The rest of the values appear in the table:

Year 1980 1981 1982 1983 1984 1985 1986 1987 1988 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000 2001 2002 2003 2004

Revolving 55 61 66 79 100 122 136 153 174 198 239 245 257 288 338 443 499 530 579 608 678 722 738 759 794

Holt-Winters w = .7 v = .7 Et Tt

61.00 66.30 76.84 95.76 118.91 137.17 153.98 173.24 196.19 232.66 250.91 261.89 284.49 327.99 419.44 497.62 543.45 584.91 614.75 669.40 720.81 748.01 765.97 792.44

Time Series: Descriptive Analyses, Models, and Forecasting

6.00 5.51 9.03 15.95 20.99 19.08 17.49 18.73 21.69 32.04 22.38 14.40 20.14 36.49 74.97 77.22 55.24 45.59 34.57 48.62 50.57 34.22 22.83 25.38

517

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Using the Holt-Winters series, the forecasts for 2006 and 2007 are: F2006 = Ft+2 = Et + 2Tt = 792.44 + 2(25.38) = 843.20 F2007 = Ft+3 = Et + 3Tt = 792.44 + 3(25.38) = 868.58 These values are very similar to forecasts found using regression. 13.68

a.

From Example 13.4, the exponentially smoothed value for September 2005 is 80.333. The forecasts for October through December 2005 are: F2005,Oct = Ft+1 = Et = 80.333 F2005,Nov = Ft+2 = Ft+1 = 80.333 F2005,Dec = Ft+3 = Ft+1 = 80.333 The forecast errors are the differences between the actual values and the forecasted values. The forecast errors are: Year 2005,Oct 2005,No v 2005,Dec

b.

Yt+i 81.88

Ft+i 80.333

Difference 1.55

88.90 82.20

80.333 80.333

8.57 1.87

Using MINITAB, the results of fitting the model are: Regression Analysis: IBM versus Time The regression equation is IBM = 95.8 - 0.740 Time Predictor Constant Time

Coef 95.777 -0.7401

S = 5.79351

SE Coef 2.622 0.2088

R-Sq = 39.8%

T 36.53 -3.54

P 0.000 0.002

R-Sq(adj) = 36.6%

Analysis of Variance Source Regression Residual Error Total

DF 1 19 20

SS 421.71 637.73 1059.44

MS 421.71 33.56

F 12.56

P 0.002

Unusual Observations Obs 12

Time 12.0

IBM 98.58

Fit 86.90

SE Fit 1.28

Residual 11.68

St Resid 2.07R

R denotes an observation with a large standardized residual. Durbin-Watson statistic = 0.688518

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Predicted Values for New Observations New Obs 1

Fit 79.50

SE Fit 2.62

95% CI (74.01, 84.98)

95% PI (66.19, 92.81)

Values of Predictors for New Observations New Obs 1

Time 22.0

Predicted Values for New Observations New Obs 1

Fit 78.76

SE Fit 2.81

95% CI (72.88, 84.63)

95% PI (65.28, 92.23)

Values of Predictors for New Observations New Obs 1

Time 23.0

Predicted Values for New Observations New Obs 1

Fit 78.02

SE Fit 2.99

95% CI (71.75, 84.28)

95% PI (64.37, 91.67)

Values of Predictors for New Observations New Obs 1

Time 24.0

The least squares fitted model is: Yˆt = 95.777 − .7401t

βˆo = 95.777

The estimated stock price for IBM in December 2003 is 95.777.

βˆ1 = −.7401

The estimated decrease in the value of the stock for IBM for each additional month is .7401.

c.

The approximate precision is ±2s or ±2(5.79) or ±11.58 .

d.

The forecasts and prediction intervals are found at the bottom of the printout in part b.

Year 2005, Oct 2005, Nov 2005, Dec

Forecast 79.50 78.76 78.02

95% Lower Limit 66.19 65.28 64.37

The precision for October is approximately

95% Upper Limit 92.81 92.23 91.67

92.81 − 66.19 = 13.31 . 2

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The precision for November is approximately

92.23 − 65.28 = 13.48 . 2

The precision for December is approximately

91.67 − 64.37 = 13.65 . 2

All of these are close to the 11.58 from part c. e.

The MAD, MAPE, and RMSE for the smoothed series are: m

MAD =

∑ | Yt − Ft | i =1

m

=

| 81.88 − 80.33 | + | 88.90 − 80.33 | + | 82.20 − 80.33 | 11.98 = = 3.994 3 3

⎡ m (Yt − Ft ) ⎢∑ Yt ⎢ i =1 MAPE = ⎢ m ⎢ ⎢ ⎣

m

∑ (Yt − Ft )

2

i =1

RMSE =

⎤ ⎡ 81.88 − 80.33 88.90 − 80.33 82.20 − 80.33 ⎥ + + ⎢ ⎥ 81.88 88.90 88.90 ⎥ 100 = ⎢⎢ 3 ⎥ ⎢ ⎥ ⎣ ⎦ ⎡ .1380 ⎤ =⎢ ⎥ 100 = 4.599 ⎣ 3 ⎦

m

= =

⎤ ⎥ ⎥ 100 ⎥ ⎥ ⎦

(81.88 − 80.33)2 + (88.90 − 80.33)2 + (82.20 − 80.33)2 3 79.2724 = 5.140 3

The MAD, MAPE, and RMSE for the regression model are: m

MAD =

∑ | Yt − Ft | i =1

=

m 16.70 = = 5.567 3

| 81.88 − 79.50 | + | 88.90 − 78.76 | + | 82.20 − 78.02 | 3

⎡ m (Yt − Ft ) ⎢∑ Yt ⎢ i =1 MAPE = ⎢ m ⎢ ⎢ ⎣

520

⎤ ⎡ 81.88 − 79.50 88.90 − 78.76 82.20 − 78.02 ⎥ + + ⎢ ⎥ 81.88 88.90 88.90 ⎥ 100 = ⎢⎢ 3 ⎥ ⎢ ⎥ ⎣ ⎦ ⎡ .1940 ⎤ =⎢ ⎥ 100 = 6.466 ⎣ 3 ⎦

⎤ ⎥ ⎥ 100 ⎥ ⎥ ⎦

Chapter 13

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m

RMSE =

∑ (Yt − Ft ) i =1

m

2

= =

(81.88 − 79.50 )2 + (88.90 − 78.76 )2 + (82.20 − 78.02 )2 3 125.9564 = 6.480 3

The values of MAD, MAPE, and RMSE for the exponentially smoothed model are all smaller than their corresponding values for the regression model. f.

We have to assume that the error terms are independent.

g.

To determine if positive autocorrelation is present, we test: H0: No first-order autocorrelation of residuals Ha: Positive first-order autocorrelation of residuals The test statistic is d = 0.69. The rejection region is d < dL,α = dL,.05 = 1.22. The value of dL,.05 is found in Table XIII, Appendix B, with k = 1 and n = 21 . Since the observed value of the test statistic falls in the rejection region (d = .69 < 1.22), H0 is rejected. There is sufficient evidence to indicate the time series residuals are positively autocorrelated at α = .05. Since there is evidence of positive autocorrelation, the validity of the regression model is questioned.

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The Gasket Manufacturing Case (To accompany Chapters 12–13)

For this study, I constructed an R chart and an x -chart for both the original data (5.1) and for the new data (5.2). First, we will analyze the data set, 5.1 (that collected under the discretion of the operator). We must compute the mean and range for each sample. The range = R = largest measurement smallest measure. The results are listed in the table:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

0.0440 0.0438 0.0453 0.0451 0.0459 0.0449 0.0472 0.0457 0.0464 0.0451 0.0456 0.0448 0.0459 0.0456 0.0472 0.0462 0.0427 0.0431 0.0425 0.0429 0.0443 0.0443 0.0429 0.0448

Samples 0.0446 0.0425 0.0428 0.0441 0.0466 0.0471 0.0477 0.0459 0.0457 0.0447 0.0455 0.0423 0.0468 0.0471 0.0465 0.0463 0.0437 0.0448 0.0442 0.0447 0.0441 0.0423 0.0427 0.0451

0.0437 0.0443 0.0433 0.0434 0.0476 0.0451 0.0452 0.0472 0.0447 0.0457 0.0445 0.0442 0.0452 0.0450 0.0461 0.0471 0.0445 0.0429 0.0432 0.0450 0.0450 0.0447 0.0464 0.0428

x 0.0441 0.0435 0.0438 0.0442 0.0467 0.0457 0.0467 0.0463 0.0456 0.0452 0.0452 0.0438 0.0460 0.0459 0.0466 0.0465 0.0436 0.0436 0.0433 0.0442 0.0445 0.0438 0.0440 0.0442

Range 0.0009 0.0018 0.0025 0.0017 0.0017 0.0022 0.0025 0.0015 0.0017 0.0010 0.0011 0.0025 0.0016 0.0021 0.0011 0.0009 0.0018 0.0019 0.0017 0.0021 0.0009 0.0024 0.0037 0.0023

x1 + x2 + " + x24 1.0770 = = .0449 n 24 R + R2 + " + R24 .0436 = R = 1 = .0018 n 24 x =

We now construct an R chart. From Table XVII, Appendix B, with n = 3, D3 = .000 and D4 = 2.574.

522

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R = .0018 Upper control limit = RD4 = .0018(2.574) = .0046 Since D3 = 0, the lower control limit is negative and is not included on the chart. From Table XVII, Appendix B, with n = 3, d2 = 1.693 and d3 = .888.

Upper A–B boundary = R + 2d3

.0018 R = .0018 + 2(.888) = .0037 1.693 d2

Lower A−B boundary = R − 2d3

.0018 R = .0018 − 2(.888) = −.0001 = 0 1.693 d2

Upper B–C boundary = R + d3

.0018 R = .0018 + (.888) = .0027 d2 1.693

Lower B–C boundary = R − d3

.0018 R = .0018 − (.888) = .0009 1.693 d2

The R-chart is:

To determine if the process is in control, we check the four rules. Rule 1: One point beyond Zone A: There are no points beyond Zone A. Rule 2: Nine points in a row in Zone C or beyond: No sequence of nine points are in Zone C (on one side of the centerline) or beyond. Rule 3: Six points in a row steadily increasing or decreasing: This pattern is not present. Rule 4: Fourteen points in a row alternating up and down: This pattern does not exit. The process appears to be in control. No rule is violated. Next, we construct the x -chart.

The Gasket Manufacturing Case

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Centerline = x = .0449 From Table XVII, Appendix B, with n = 3, A2 = 1.023

Upper control limit = x + A2 R = .0449 + 1.023(.0018) = .0467 Lower control limit = x − A2 R = .0449 − 1.023(.0018) = .0431

Upper A-B boundary = x =

2 2 ( A2 R ) = .0449 + (1.023(.0018) ) = .0461 3 3

Lower A–B boundary = x −

2 2 ( A2 R ) = .0449 − (1.023(.0018) ) = .0437 3 3

Upper B–C boundary = x +

1 1 ( A2 R ) = .0449 + (1.023(.0018) ) = .0455 3 3

Lower B–C boundary = x −

1 1 ( A2 R ) = .0449 − (1.023(.0018) ) = .0443 3 3

The x -chart is:

To determine if the process is in or out of control, we check the six rules: Rule 1: One point beyond Zone A: No points are beyond Zone A. Rule 2: Nine points in a row in Zone C or beyond: No sequence of nine points are in Zone C (on one side of the centerline) or beyond. Rule 3: Six points in a row steadily increasing or decreasing: This pattern is not present. Rule 4: Fourteen points in a row alternating up and down: This pattern does not exit. Rule 5: Two out of three points in Zone A or beyond: There are six groups of at least three points in Zone A or beyond—points 5–7, points 6–8, points 7–9, points 14–16, points 17–19, and points 18–20. Rule 6: Four out of five points in a row in Zone B or beyond: There are six groups of points that satisfy this rule—points 5–9, points 6–10, points 17–21, points 18–22, points 19–23, and points 20–24.

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The process appears to be out of control. Rules 5 and 6 indicate that the process is out of control. Since the process is out of control, a capability analysis is not appropriate. However, I will include a dot diagram which indicates that many of the actual observations are outside of the specification limits. The dot plot is: . : : ::: ....

.

:. .:

. .

..

:. .::: :.::.:::. .:: : ...:.. .

::

..

-------+---------+---------+---------+---------+--------0.0430

0.0440

0.0450

0.0460

0.0470

0.0480

The specification limits are .043 to .047. There are 11 points below .043 and 8 above .047. Thus, 19 out of the 72 points or .264 of the points are outside of the specification limits. This indicates that the present system, when the operator is allowed to adjust the system at his/her discretion, is not capable of reaching the needs of the customers. Next, we analyze the second set of data, 5.2. First, we must compute the mean and range for each sample. The range = R = largest measurement smallest measure. The results are listed in the table:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

0.0445 0.0435 0.0438 0.0449 0.0433 0.0455 0.0455 0.0445 0.0443 0.0449 0.0465 0.0461 0.0443 0.0456 0.0447 0.0454 0.0445 0.0438 0.0453 0.0455 0.0440 0.0444 0.0445 0.0450

Samples 0.0455 0.0453 0.0459 0.0449 0.0461 0.0454 0.0458 0.0451 0.0450 0.0448 0.0449 0.0439 0.0434 0.0459 0.0442 0.0445 0.0471 0.0445 0.0444 0.0435 0.0438 0.0450 0.0447 0.0463

The Gasket Manufacturing Case

0.0457 0.0450 0.0428 0.0467 0.0451 0.0461 0.0445 0.0436 0.0441 0.0467 0.0448 0.0452 0.0454 0.0452 0.0457 0.0451 0.0465 0.0472 0.0451 0.0443 0.0444 0.0467 0.0461 0.0456

x 0.0452 0.0446 0.0442 0.0455 0.0448 0.0457 0.0453 0.0444 0.0445 0.0455 0.0454 0.0451 0.0444 0.0456 0.0449 0.0450 0.0460 0.0452 0.0449 0.0444 0.0441 0.0454 0.0451 0.0456

Range 0.0012 0.0018 0.0031 0.0018 0.0028 0.0007 0.0013 0.0015 0.0009 0.0019 0.0017 0.0022 0.0020 0.0007 0.0015 0.0009 0.0026 0.0034 0.0009 0.0020 0.0006 0.0023 0.0016 0.0013

525

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x1 + x2 + " + x24 1.0808 = = .0450 n 24 R + R2 + " + R24 .0407 = R = 1 = .0017 24 n x =

First, we construct an R chart. From Table XVII, Appendix B, with n = 3, D3 = .000 and D4 = 2.574.

R = .0017 Upper control limit = RD4 = .0017(2.574) = .0044 Since D3 = 0, the lower control limit is negative and is not included on the chart. From Table XVII, Appendix B, with n = 3, d2 = 1.693 and d3 = .888.

.0017 Upper A–B boundary = R + 2d3 R = .0017 + 2(.888) = .0035 1.693 d2 .0017 Lower A–B boundary = R − 2d3 R = .0017 − 2(.888) = -.0001 = 0 1.693 d2 .0017 Upper B–C boundary = R + d3 R = .0017 + (.888) = .0026 1.693 d2 .0017 Lower B–C boundary = R − d3 R = .0017 − (.888) = .0008 1.693 d2 The R-chart is:

To determine if the process is in control, we check the four rules. Rule 1: One point beyond Zone A: There are no points beyond Zone A. Rule 2: Nine points in a row in Zone C or beyond: No sequence of nine points are in Zone C (on one side of the centerline) or beyond.

526

The Gasket Manufacturing Case

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Rule 3: Six points in a row steadily increasing or decreasing: This pattern is not present. Rule 4: Fourteen points in a row alternating up and down: This pattern does not exit. The process appears to be in control. No rule is violated. Next, we construct the x -chart. Centerline = x = .0450 From Table XVII, Appendix B, with n = 3, A2 = 1.023

Upper control limit = x + A2 R = .0450 + 1.023(.0017) = .0467 Lower control limit = x − A2 R = .0450 − 1.023(.0017) = .0433 Upper A-B boundary = x +

2 2 ( A2 R ) = .0450 + (1.023(.0017) ) = .0462 3 3

Lower A–B boundary = x −

2 2 ( A2 R ) = .0450 − (1.023(.0017) ) = .0438 3 3

Upper B–C boundary = x +

1 1 ( A2 R ) = .0450 + (1.023(.0017) ) = .0456 3 3

Lower B–C boundary = x −

1 1 ( A2 R ) = .0450 − (1.023(.0017) ) = .0444 3 3

The x -chart is:

To determine if the process is in or out of control, we check the six rules: Rule 1: One point beyond Zone A: No points are beyond Zone A. Rule 2: Nine points in a row in Zone C or beyond: No sequence of nine points are in Zone C (on one side of the centerline) or beyond.

The Gasket Manufacturing Case

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Rule 3: Rule 4: Rule 5: Rule 6:

Six points in a row steadily increasing or decreasing: This pattern is not present. Fourteen points in a row alternating up and down: This pattern does not exit. Two out of three points in Zone A or beyond: This pattern does not exist. Four out of five points in a row in Zone B or beyond: This pattern does not exist.

The process appears to be in control. No rules are violated. Since the process is in control, we will perform a capability analysis to see if the process can meet the customer's demand. I will include a dot diagram which indicates that many of the actual observations are outside of the specification limits. The dot plot is: . : . ..: : :: . : : . . .. :. : .... ::: ::: ::::: :::. : : . : : .. -----+---------+---------+---------+---------+---------+0.04320 0.04400 0.04480 0.04560 0.04640 0.04720

The specification limits are .043 to .047. There is one point below .043 and two points above .047. Thus, 3 out of the 72 points or .042 of the points are outside of the specification limits. This indicates that the present system, when the operator does not adjust the system at his/her discretion, might be able to meet the needs of the customers. We will also compute the capability index. The capability index is defined as the ratio of the specification limits to 6 standard deviations or:

Cp =

upper specification limit − lower specification limit 6σ

Since σ is not known, we will estimate it with s. In this case, s = .00095. The capability index is:

Cp =

.047 - .043 = .702 6(.00095)

Since the capability index is less than 1, it indicates that the process is not capable of meeting the customer's needs. Even though this process (operator does not make adjustments) is in control, it is not capable of meeting the needs of the customers. In conclusion, it appears that the engineers are correct—the present equipment is not capable of producing gasket material within the necessary limits.

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Nonparametric Statistics

14.2

14.4

Chapter 14

a.

Since the normal distribution is symmetric, the probability that a randomly selected observation exceeds the mean of a normal distribution is .5.

b.

By the definition of "median," the probability that a randomly selected observation exceeds the median of a normal distribution is .5.

c.

If the distribution is not normal, the probability that a randomly selected observation exceeds the mean depends on the distribution. With the information given, the probability cannot be determined.

d.

By definition of "median," the probability that a randomly selected observation exceeds the median of a non-normal distribution is .5.

a.

H0: η = 9 Ha: η > 9 The test statistic is S = {Number of observations greater than 9} = 7. The p-value = P(x ≥ 7) where x is a binomial random variable with n = 10 and p = .5. From Table II, p-value = P(x ≥ 7) = 1 − P(x ≤ 6) = 1 − .828 = .172 Since the p-value = .172 > α = .05, H0 is not rejected. There is insufficient evidence to indicate the median is greater than 9 at α = .05.

b.

H0 : η = 9 Ha: η ≠ 9 S1 = {Number of observations less than 9} = 3 and S2 = {Number of observations greater than 9} = 7 The test statistic is S = larger of S1 and S2 = 7. The p-value = 2P(x ≥ 7) where x is a binomial random variable with n = 10 and p = .5. From Table II, p-value = 2P(x ≥ 7) = 2(1 − P(x ≤ 6)) = 2(1 - .828) = .344 Since the p-value = .344 > α = .05, H0 is not rejected. There is insufficient evidence to indicate the median is different than 9 at α = .05.

Nonparametric Statistics

529

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c.

H0: η = 20 Ha: η < 20 The test statistic is S = {Number of observations less than 20} = 9. The p-value = P(x ≥ 9) where x is a binomial random variable with n = 10 and p = .5. From Table II, p-value = P(x ≥ 9) = 1 − P(x ≤ 8) = 1 − .989 = .011 Since the p-value = .011 < α= .05, H0 is rejected. There is sufficient evidence to indicate the median is less than 20 at α = .05.

d.

H0: η = 20 Ha: η ≠ 20 S1 = {Number of observations less than 20} = 9 and S2 = {Number of observations greater than 20} = 1 The test statistic is S = larger of S1 and S2 = 9. The p-value = 2P(x ≥ 9) where x is a binomial random variable with n = 10 and p = .5. From Table II, p-value = 2P(x ≥ 9) = 2(1 − P(x ≤ 8)) = 2(1 − .989) = .022 Since the p-value = .022 < α = .05, H0 is rejected. There is sufficient evidence to indicate the median is different than 20 at α = .05.

e.

For all parts, μ = np = 10(.5) = 5 and σ =

npq = 10(.5)(.5) = 1.581.

(7 − .5) − 5 ⎞ ⎛ For part a, P(x ≥ 7) ≈ P ⎜ z ≥ = P(z ≥ .95) = .5 − .3289 = .1911 1.581 ⎟⎠ ⎝

This is close to the probability .172 in part a. The conclusion is the same. (7 − .5) − 5 ⎞ ⎛ For part b, 2P(x ≥ 7) ≈ 2 P ⎜ z ≥ = 2P(z ≥ .95) = 2(.5 − .3289) 1.581 ⎟⎠ ⎝ = .3422 This is close to the probability .344 in part b. The conclusion is the same. (9 − .5) − 5 ⎞ ⎛ = P(z ≥ 2.21) = .5 − .4864 For part c, P(x ≥ 9) ≈ P ⎜ z ≥ 1.581 ⎟⎠ ⎝ = .0136

This is close to the probability .011 in part c. The conclusion is the same.

530

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(9 − .5) − 5 ⎞ ⎛ For part d, 2P(x ≥ 9) ≈ 2 P ⎜ z ≥ = 2P(z ≥ 2.21) = 2(.5 − .4864) 1.581 ⎟⎠ ⎝ = .0272 This is close to the probability .022 in part d. The conclusion is the same.

14.6

f.

We must assume only that the sample is selected randomly from a continuous probability distribution.

a.

To determine if the median amount of caffeine in Breakfast Blend coffee exceeds 300 milligrams, we test: H0: η = 300 Ha: η > 300

b.

S=4

c.

Using Table II, Appendix B, with n = 6 and p = .5,

P ( x ≥ 4) = 1 − P ( x ≤ 3) = 1 − .656 = .344 d.

14.8

a.

Since the probability in part c is greater than α = .05, H0 is not rejected. There is insufficient evidence to indicate the median amount of caffeine in Breakfast Blend coffee exceeds 300 milligrams at α = .05. To determine if cohesiveness will deteriorate after storage, we test: H0: η = 0 Ha: η > 0

b.

The test statistic is S = {number of measurements greater than 0} = 13. The p-value = P(x ≥ 13) where x is a binomial random variable with n = 20 and p = .5. From Table II, p-value = P(x ≥ 13) = 1 – P(x ≤ 12) = 1 − .868 = .132

14.10

c.

Since the p-value = .132 > α = .05, H0 is not rejected. There is insufficient evidence to indicate cohesiveness will deteriorate after storage at α = .05.

a.

I would recommend the sign test because five of the sample measurements are of similar magnitude, but the 6th is about three times as large as the others. It would be very unlikely to observe this sample if the population were normal.

b.

To determine if the airline is meeting the requirement, we test: H0: η = 30 Ha: η < 30

Nonparametric Statistics

531

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c.

The test statistic is S = number of measurements less than 30 = 5. H0 will be rejected if the p-value < α = .01.

d.

The test statistic is S = 5. The p-value = P(x ≥ 5) where x is a binomial random variable with n = 6 and p = .5. From Table II, p-value = P(x ≥ 5) = 1 − P(x ≤ 4) = 1 − .891 = .109 Since the p-value = .109 is not less than α = .01, H0 is not rejected. There is insufficient evidence to indicate the airline is meeting the maintenance requirement at α = .01.

14.12

To determine if the median surface roughness of coated interior pipe differs from 2 micrometers, we test: H0: η = 2 Ha: η ≠ 2 S1 = {Number of measurements < 2} = 9. S2 = {Number of measurements > 2} = 11. The test statistic is S = Larger of S1 and S2 = 11. The p-value = 2 P(x ≥ 11) where x is a binomial random variable with n = 20 and p = .5 From Table II, Appendix B, p-value = 2 P(x ≥ 11) = 2(1 − P( x ≤ 10)) = 2(1 − .588) = .824 Since the p-value = .824 1.96. Since the observed value of the test statistic does not fall in the rejection region (z = −1.929 1,96. Since the observed value of the test statistic does not fall in the rejection region (z = 1.767 >/ 1.96), H0 is not rejected. There is insufficient evidence to indicate the distribution of quality scores for the successfully implemented systems differs from that for the unsuccessfully implemented systems at α = .05. b.

We could use the two-sample t-test if: 1. 2.

14.26

a.

Both populations are normal. The variances of the two populations are the same.

The test statistic is T− or T+, the smaller of the two. The rejection region is T ≤ 152, from Table XVI, Appendix B, with n = 30, α = .10, and two-tailed.

b.

The test statistic is T−. The rejection region is T− ≤ 60, from Table XVI, Appendix B, with n = 20, α = .05, and one-tailed.

c.

The test statistic is T+. The rejection region is T+ ≤ 0, from Table XVI, Appendix B, with n = 8, α = .005, and one-tailed.

14.28

a.

The rejection region requires α = .05 in the upper tail of the z-distribution. From Table IV, Appendix B, z.05 = 1.645. The rejection region is z > 1.645. n(n + 1) 25(26) 273 − 4 4 = 2.97 = n(n + 1)(2n + 1) 25(26)(51) 24 24 T+−

b.

The large sample test statistic is z =

Since the observed value of the test statistic falls in the rejection region (z = 2.97 > 1.645), H0 is rejected. There is sufficient evidence to indicate that the responses for A tend to be larger than those for B at α = .05.

538

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c.

p-value = P(z ≥ 2.97) = .5 − P(0 < z < 2.97) = .5 − .4985 = .0015 (from Table IV, Appendix B) Thus, we can reject H0 for any preselected α greater than .0015.

14.30

a.

To determine if the chest injury ratings of drivers and front-seat passengers differ, we test: H0: The two sampled populations have identical probability distributions Ha: The probability distribution of drivers is shifted to the right or left of that for front-seat passengers

b.

Using MINITAB, the results are: Wilcoxon Signed Rank Test: Diff Test of median = 0.000000 versus median not = 0.000000

Diff

N 18

N for Test 16

Wilcoxon Statistic 23.0

P 0.021

Estimated Median -4.000

From the printout, the test statistic is T+ = 23.

c.

The rejection region is T+ ≤ To where To corresponds to α = .01 (two-tailed) and n = 16. From Table XVI, Appendix B, To = 19. The rejection region is T+ ≤ 19.

d.

Since the observed value of the test statistic does not fall in the rejection region (T+ = 23 ≤/ 19), H0 is not rejected. There is insufficient evidence to indicate the chest injury ratings of drivers and front-seat passengers differ at α = .01. From the printout, the p-value is p = .021.

14.32

Some preliminary calculations: Theme

Tourism Physical Transportation People History Climate Forestry Agriculture Fishing Energy Mining Manufacturing

Nonparametric Statistics

High School Teachers 10 2 7 1 2 6 5 7 9 2 10 12

Geography Alumni 2 1 3 6 5 4 8 10 7 8 11 12

Difference Rank of Absolute T-A Differences 8 11 1 1.5 4 8 9 −5 6 −3 2 3.5 6 −3 6 −3 2 3.5 10 −6 1.5 −1 0 (eliminated) Positive rank sum T+ = 27.5

539

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To determine if the distributions of theme rankings for the two groups differ, we test: H0: The probability distributions for the two populations are identical Ha: The probability distribution of the high school teachers is shifted to the right or left of the probability distribution of the geography alumni The test statistic is T+ = 27.5. Reject H0 if T+ ≤ T0 where T0 is based on α = .05 and n = 11 (two-tailed): Reject H0 if T+ ≤ 11 (from Table XVI, Appendix B) Since the observed value of the test statistic does not fall in the rejection region (T+ = 27.5 ≤/ 11), H0 is not rejected. There is insufficient evidence to indicate that the distributions of these rankings for the two groups differ at α = .05. Practically, this means that the thematic content of a new atlas could be based on the views of either educators or geography alumni. 14.34

Some preliminary calculations are:

Employee 1 2 3 4 5 6 7 8 9 10

Before Flextime 54 25 80 76 63 82 94 72 33 90

After Flextime 68 42 80 91 70 88 90 81 39 93

Difference (B − A) −4 −17 0 −15 −7 −6 4 −9 −6 −3

Difference 7 9 (Eliminated) 8 5 3.5 2 6 3.5 1 T+ = 2

To determine if the pilot flextime program is a success, we test: H0: The two probability distributions are identical Ha: The probability distribution before is shifted to the left of that after The test statistic is T+ = 2. The rejection region is T+ ≤ 8, from Table XVI, Appendix B, with n = 9 and α = .05. Since the observed value of the test statistic falls in the rejection region (T+ = 2 ≤ 8), H0 is rejected. There is sufficient evidence to indicate the pilot flextime program has been a success at α = .05.

540

Chapter 14

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14.36

Some preliminary calculations are:

Science 0 4 3 1 3 2 4 2 3 4

Math 2 3 0 1 1 3 0 1 1 1

Rank of Difference Absolute ScienceDifference Math −2 5 1 2 3 7.5 0 eliminate 2 5 −1 2 4 9 1 2 2 5 3 7.5 Negative rank sum T_ = 7 Positive rank sum T+ = 38

To determine if there are differences in the levels of family involvement between math and science homework, we test; H0: The distributions of the science and math levels of family involvement are the same Ha: The distributions of the science and math levels of family involvement differ The test statistic is T_ = 7. The rejection region is T_ ≤ To where To corresponds to α = .05 (two-tailed) and n = 9. From Table XVI, Appendix B, To = 6. The rejection region is T_ ≤ 6. Since the observed value of the test statistic does not fall in the rejection region (T_ = 7 ≤/ 6), H0 is not rejected. There is insufficient evidence to indicate there are differences in the levels of family involvement between math and science homework at α = .05. 14.38

a.

The hypotheses are: H0: The three probability distributions are identical Ha: At least two of the three probability distributions differ in location

b.

The test statistic is: H=

2 12 12 ⎡ 230 2 440 2 365 2 ⎤ Rj + + − 3(n + 1) = ∑ ⎢ ⎥ − 3(46) n( n + 1) 45(46) ⎣ 15 15 15 ⎦ nj

= 146.754 − 138 = 8.754

Nonparametric Statistics

541

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The rejection region requires α = .05 in the upper tail of the χ2 distribution with 2 = 5.99147. The rejection df = p − 1 = 3 − 1 = 2. From Table VII, Appendix B, χ.05 region is H > 5.99147. Since the observed value of the test statistic falls in the rejection region (H = 8.754 > 5.99147), H0 is rejected. There is sufficient evidence to indicate that the probability distributions of at least two of the populations A, B, and C, differ in location at α = .05. c.

d.

14.40

a.

The approximate p-value is P(χ2 ≥ 8.754). From Table VII, Appendix B, with df = 2, .01 ≤ P(χ2 ≤ 8.754) ≤ .025. RB 440 R A = 230 = = 29.333 = 15.333 RB = 15 15 15 15 RC 365 n + 1 45 + 1 = = 24.333 = = 23 R = RC = 15 15 2 2 12 H= ∑ n j ( R j − R )2 n(n + 1) 12 ⎡ = 15(15.333 − 23) 2 + 15(29.333 − 23) 2 + 15(24.333 − 23) 2 ⎤⎦ = 8.754 ⎣ 45(46) In order to compare the three population means using parametric techniques, we must assume that all populations being sampled from are normal and all population variances are the same. It is quite possible that these two conditions are not met with this data. RA =

b.

Since we want to compare 3 groups, we will use the Kruskal-Wallis test.

c.

The test statistic is H=

R 2j ⎛ 53352 3937 2 37692 12 12 − + = + + 3( 1) n ⎜ ∑n n(n + 1) 161(161 + 1) ⎝ 67 57 37 j

⎞ ⎟ − 3(161 + 1) ⎠

= 11.201

14.42

d.

Since the p-value is so small (p = .0037), H0 will be rejected. There is sufficient evidence to indicate DEF distributions differ for the 3 tax litigation forums for α > .0037.

a.

To determine if the distributions of office rental growth rates differ among the four market cycle phases, we test: H0: The four probability distributions are identical Ha: At least two of the growth rate distributions differ

542

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b. Phase I 2.7 −1.0 1.1 3.4 4.2 3.5

14.44

The ranks of the measurements are: Rank 9 4.5 6 10 12 11 R1 = 52.5

Phase II 10.5 11.5 9.4 12.2 8.6 10.9

Rank 20 23 19 24 18 21 R2 = 125

Phase III 6.1 1.2 11.4 4.4 6.2 7.6

Rank 14 7 22 13 15.5 17 R3 = 88.5

Phase IV −1.0 6.2 −10.8 2.0 −1.1 −2.3

Rank 4.5 15.5 1 8 3 2 R4 = 34

c.

The rank sums appear in the table above. The test statistic is: R 2j ⎛ 52.52 1252 88.52 342 ⎞ 12 12 − + = + + + 3( 1) H= n ⎜ ⎟ − 3(24 + 1) ∑n n( n + 1) 24(24 + 1) ⎝ 6 6 6 6 ⎠ j = 16.23

d.

The rejection region requires α = .05 in the upper tail of the χ2 distribution with df = 2 p – 1 = 4 – 1 = 3. From Table VII, Appendix B, χ.05 = 7.81473. The rejection region is H > 7.81473.

e.

Since the observed value of the test statistic falls in the rejection region (H = 16.23 > 7.81473), H0 is rejected. There is sufficient evidence to indicate the distributions of office rental growth rates differ among the four market cycle phases at α = .05.

Some preliminary calculations are: Aromatics 1.06 0.79 0.82 0.89 1.05 0.95 0.65 1.15 1.12

Ranks 26 19 20 22 25 24 18 29 27.5

R1 = 210.5

Nonparametric Statistics

Chloroalkanes 1.58 1.45 0.57 1.16 1.12 0.91 0.83 0.43

Ranks 32 31 15 30 27.5 23 21 9.5

R2 = 189

Esters 0.29 0.06 0.44 0.61 0.55 0.43 0.51 0.10 0.34 0.53 0.06 0.09 0.17 0.60 0.17

Ranks 7 1.5 11 17 14 9.5 12 4 8 13 1.5 3 5.5 16 5.5 R3 = 128.5

543

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To determine if the sorption rate distributions differ among the three solvents, we test: H0: The three probability distributions are identical Ha: At least two of the three probability distributions differ in location

The test statistic is R 2j ⎛ 210.52 1892 128.52 ⎞ 12 12 − 3(n + 1) = + + H= ⎜ ⎟ − 3(32 + 1) ∑ n( n + 1) nj 32(32 + 1) ⎝ 9 8 15 ⎠ = 20.197

The rejection region requires α = .01 in the upper tail of the χ2 distribution with df = p – 1 = 3 2 – 1 = 2. From Table VII, Appendix B, χ.01 = 9.21034. The rejection region is H > 9.21034. Since the observed value of the test statistic falls in the rejection region (H = 20.197 > 9.21034), H0 is rejected. There is sufficient evidence to indicate the sorption rate distributions differ among the three solvents at α = .01. 14.46

a.

The F-test would be appropriate if: 1. 2. 3.

b. c.

All p populations sampled from are normal. The variances of the p populations are equal. The p samples are independent.

The variances for the three populations are probably not the same and the populations are probably not normal. To determine whether the salary distributions differ among the three cities, we test: H0: The three probability distributions are identical Ha: At least two of the three probability distributions differ in location

Some preliminary calculations are: 1 Atlanta 34,600 84,900 61,700 38,900 77,200 83,600 59,800

544

Rank 1 19 11 3 17 18 10 R1 = 79

2 Los Angeles 42,400 135,000 63,000 43,700 69,400 97,000 49,500

Rank 4 21 12 5 13 20 7 R2 = 82

3 Washington, D.C. 38,000 76,900 48,000 72,600 73,200 51,800 55,000

Rank 2 16 6 14 15 8 9 R3 = 70

Chapter 14

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The test statistic is H = =

2 12 Rj − 3(n + 1) ∑ n( n + 1) nj

12 ⎛ 79 2 82 2 70 2 ⎞ + + ⎜ ⎟ − 3(22) = 66.2894 − 66 = .2894 21(22) ⎝ 7 7 7 ⎠

The rejection region requires α = .05 in the upper tail of the χ2 distribution with df = p − 2 1 = 3 − 1 = 2. From Table VII, Appendix B, χ.05 = 5.99147. The rejection region is H > 5.99147. Since the observed value of the test statistic does not fall in the rejection region (H = .2894 >/ 5.99147), H0 is not rejected. There is insufficient evidence to indicate the salary distributions differ among the three cities at α = .05. We must assume we have independent random samples, sample sizes greater than or equal to 5 from each population, and that all populations are continuous. 14.48

a.

The hypotheses are: H0: The probability distributions for three treatments are identical Ha: At least two of the probability distributions differ in location

b.

The rejection region requires α = .10 in the upper tail of the χ2 distribution with df = 2 p − 1 = 3 − 1 = 2. From Table VII, Appendix B, χ.10 = 4.60517. The rejection region is Fr > 4.60517.

c.

Some preliminary calculations are: Block 1 2 3 4 5 6 7

A

9 13 11 10 9 14 10

Rank

1 2 1 1 2 2 1 RA = 10

B 11 13 12 15 8 12 12

Rank 2 2 2.5 2 1 1 2 RB = 12.5

C 18 13 12 16 10 16 15

Rank 3 2 2.5 3 3 3 3 RC = 19.5

12 R 2j − 3b( p + 1) ∑ bp ( p + 1) 12 ⎡102 + 12.52 + 19.52 ⎤ − 3(7)(4) = 90.9286 − 84 = 6.9286 = ⎦ 7(3)(4) ⎣

The test statistic is Fr =

Since the observed value of the test statistic falls in the rejection region (Fr = 6.9286 > 4.60517), H0 is rejected. There is sufficient evidence to indicate the effectiveness of the three different treatments differ at α = .10.

Nonparametric Statistics

545

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14.50

a.

The Friedman test statistic is Fr = =

14.52

12 ∑ R 2j − 3b( p + 1) bp ( p + 1)

12 (27 2 + 252 + 182 + 112 + 92 ) − 3(6)(5 + 1) = 17.333 6(5)(5 + 1)

b.

The rejection region requires α = .05 in the upper tail of the χ2 distribution with df = 2 p – 1 = 5 – 1 = 4. From Table VII, Appendix B, χ.05 = 9.48773. The rejection region is Fr > 9.48773.

c.

Since the observed value of the test statistic falls in the rejection region (Fr = 17.333 > 9.48773), H0 is rejected. There is sufficient evidence to indicate there is a difference in the levels of farm production among the five conditions at α = .05.

a.

To determine if the distributions of rotary oil rigs differ among the three states, we test: H0: The probability distributions of the rotary oil rigs for the 3 states are the same Ha: At least two of the probability distributions of rotary oil rigs differ in location

b.

The ranked data are: Month/Year Nov. 2000 Oct. 2001 Nov. 2001

c.

Utah 2 2 2 R2 = 6

Alaska 1 1 1 R3 = 3

The test statistic is Fr =

546

California 3 3 3 R1 = 9

(

)

12 12 92 + 62 + 32 − 3(3)(3 + 1) = 6 R 2j − 3b( p + 1) = ∑ 3(3)(3 + 1) bp ( p + 1)

d.

The rejection region requires α = .05 in the upper tail of the χ2 distribution with df = 2 p – 1 = 3 – 1 = 2. From Table VII, Appendix B, χ.05 = 5.99147. The rejection region is H > 5.99147.

e.

Since the observed value of the test statistic falls in the rejection region (H = 6 > 5.99147), H0 is rejected. There is sufficient evidence to indicate the distributions of rotary oil rigs differ among the three states at α = .05.

Chapter 14

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14.54

Some preliminary calculations are:

Location Anguilla Antigua Dominica Guyana Jamaica St. Lucia Suriname

Temephos Rank 4.6 5 9.2 5 7.8 5 1.7 2 3.4 3 6.7 4 1.4 1 R1 = 13

Malsathion Rank 1.2 1 2.9 3 1.4 1 1.9 4 3.7 4 2.7 1.5 1.9 3 R2 = 15

Fenitrothion Rank 1.5 2.5 2.0 1.5 2.4 2 2.2 5 2.0 2 2.7 1.5 2.0 4 R3 = 18.5

Fenthion Rank 1.8 4 7.0 4 4.2 4 1.5 1 1.5 1 4.8 3 2.1 5 R4 = 22

Chlorpyrifos Rank 1.5 2.5 2.0 1.5 4.1 3 1.8 3 7.1 5 8.7 5 1.7 2 R5 = 22

To determine if the resistance ratio distributions of the 5 insecticides differ, we test: H0: The distributions of the 5 insecticide ratios are the same Ha: At least two of the distributions of insecticide ratios differ 12 R 2j − 3b( p + 1) ∑ bp ( p + 1) 12 (252 + 17.52 + 18.52 + 222 + 222 ) − 3(7)(5 + 1) = 2.086 = 7(5)(5 + 1)

The test statistic is Fr =

Since no α was given, we will use α = .05. The rejection region requires α = .05 in the upper 2 tail of the χ2 distribution with df = p – 1 = 5 – 1 = 4. From Table VII, Appendix B, χ.05 = 9.48773. The rejection region is Fr > 9.48773. Since the observed value of the test statistic does not fall in the rejection region (Fr = 2.086 >/ 9.48773), H0 is not rejected. There is insufficient evidence to indicate that the resistance ratio distributions of the 5 insecticides differ at α = .05. 14.56

Some preliminary calculations are:

Week 1 2 3 4 5 6 7 8 9

Monday 5 5 2.5 2 5 4 5 4 1 R1 = 33.5

Tuesday 1 4 2.5 1 1 2 3.5 2 2 R2 = 19

Wednesday 4 3 5 3.5 2 3 1.5 1 5 R3 = 28

Thursday 2 1 1 5 3 1 3.5 3 3 R1 = 22.5

Friday 3 2 4 3.5 4 5 1.5 5 4 R2 = 32

To determine if the distributions of days of the weeks differ, we test: H0: The probability distributions of the 5 days of the week are the same Ha: At least two of the probability distributions of the 5 days of the week differ in location

Nonparametric Statistics

547

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The test statistic is 12 R 2j − 3b( p + 1) ∑ bp ( p + 1) 12 33.52 + 192 + 282 + 22.52 + 322 − 3(9)(5 + 1) = 6.778 = 9(5)(5 + 1)

Fr =

(

)

Since no α was given we will use α = .05. The rejection region requires α = .05 in the upper 2 tail of the χ2 distribution with df = p – 1 = 5 – 1 = 4. From Table VII, Appendix B, χ.05 = 9.48773. The rejection region is H > 9.48773. Since the observed value of the test statistic does not fall in the rejection region (H = 6.778 >/ 9.48773), H0 is not rejected. There is insufficient evidence to indicate the distributions of the absentee rate for the days of the weeks differ at α = .05. 14.58

14.60

a.

From Table XVII with n = 10, rs,α/2 = rs,.025 = .648. The rejection region is rs > .648 or rs < −.648.

b.

From Table XVII with n = 20, rs,α = rs,.025 = .450. The rejection region is rs > .450.

c.

From Table XVII with n = 30, rs,α = rs,.01 = .432. The rejection region is rs < −.432.

a.

H0: ρs = 0 Ha: ρs ≠ 0

b.

The test statistic is rs =

x 0 3 0 −4 3 0 4

548

Rank, u 3 5.5 3 1 5.5 3 7 ∑ u = 28

SSuv =

∑ uv −

SSuu =

∑u

SSvv =

∑v

2

2

SSuv SSuuSSvv y 0 2 2 0 3 1 2

Rank, v 1.5 5 5 1.5 7 3 5 ∑ v = 28

( ∑ u )( ∑ v ) = 131 − 28(28) n

(∑u ) −

2

n

(∑ v) − n

7

= 137.5 −

(20) 2 7

= 137.5 −

(20) 2 7

2

u2 9 30.25 9 1 30.25 9 49 ∑ u 2 = 137.5

v2 2.25 25 25 2.25 49 9 25 ∑ v 2 = 137.5

uv 45 27.5 15 1.5 38.5 9 35 ∑ uv = 131

= 19

Chapter 14

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rs =

19 = .745 25.5(25.5) Reject H0 if rs < −rs,α/2 or rs > rs,α/2 where α/2 = .025 and n = 7: Reject H0 if rs < −.786 or rs > .786 (from Table XVII, Appendix B).

Since the observed value of the test statistic does not fall in the rejection region, (rs = .745 >/ .786), do not reject H0. There is insufficient evidence to indicate x and y are correlated at α = .05.

14.62

c.

The p-value is P(rs ≥ .745) + P(rs ≤ −.745). For n = 7, rs = .745 is above rs,.025 where α/2 = .025 and below rs,.05 where α/2 = .05. Therefore, 2(.025) = .05 < p-value < 2(.05) = .10.

d.

The assumptions of the test are that the samples are randomly selected and the probability distributions of the two variables are continuous.

a.

Some preliminary calculations are: Expert 1 6 5 1 3 2 4

Brand A B C D E F

rs = 1 − b.

6∑ di2 n(n − 1) 2

= 1−

Expert 2 5 6 2 1 4 3

Difference di 1 −1 −1 2 −2 1

di2 1 1 1 4 4 1 ∑ di2 = 12

6(12) = 1 − .343 = .657 6(62 − 1)

To determine if there is a positive correlation in the rankings of the two experts, we test: H0: ρs = 0 Ha: ρs > 0 The test statistic is rs = .657. Reject H0 if rs > rs,α where α = .05 and n = 6. From Table XVII, Appendix B, rs,.01 = .829. Reject H0 if rs > .829. Since the observed value of the test statistic does not fall in the rejection region (rs = .657 >/ .829), H0 is not rejected. There is insufficient evidence to indicate a positive correlation in the rankings of the two experts at α = .05.

Nonparametric Statistics

549

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14.64

a.

Some preliminary calculations are: x u y v 5.2 1 220 4.5 5.5 7 227 7.5 6.0 23.5 259 15.5 5.9 20.5 210 1 5.8 16 224 6 6.0 23.5 215 3 5.8 16 231 9 5.6 10 268 19 5.6 10 239 11 5.9 20.5 212 2 5.4 5 410 24 5.6 10 256 14 5.8 16 306 22 5.5 7 259 15.5 5.3 3 284 21 5.3 3 383 23 5.7 12.5 271 20 5.5 7 264 18 5.7 12.5 227 7.5 5.3 3 263 17 5.9 20.5 232 10 5.8 16 220 4.5 5.8 16 246 13 5.9 20.5 241 12 ∑ u =300 ∑ v = 300 SSuv =

∑ uv −

SSuu =

∑u

SSvv =

∑v

rs =

2

2

u-sq 1 49 552.25 420.25 256 552.25 256 100 100 420.25 25 100 256 49 9 9 156.25 49 156.25 9 420.25 256 256 420.25 2 ∑ u =4878

( ∑ u )( ∑ v ) = 3197.5 − 300(300) n

(∑u ) −

2

n

(∑v) −

SSuv SSuuSSvv

n

=

24

= 4878 −

v-sq 20.25 56.25 240.25 1 36 9 81 361 121 4 576 196 484 240.25 441 529 400 324 56.25 289 100 20.25 169 144 2 ∑ v =4898.5

uv 4.5 52.5 364.25 20.5 96 70.5 144 190 110 41 120 140 352 108.5 63 69 250 126 93.75 51 205 72 208 246 ∑ uv =3197.5

= −552.5

3002 = 1128 24

2

= 4898.5 − −552.5

1128(1148.5)

3002 = 1148.5 24 = −.4854

Since the magnitude of the correlation coefficient is not particularly large, there is a fairly weak negative relationship between sweetness index and pectin.

550

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b.

To determine if there is a negative association between the sweetness index and the amount of pectin, we test: H0: ρs = 0 Ha: ρs < 0 The test statistic is rs = −.4854 Reject H0 if rs < −rs,α where α = .01 and n = 24. Reject H0 if rs < −.485 (from Table XVII, Appendix B) Since the observed value of the test statistic falls in the rejection region (rs = −.4854 < −.485), H0 is rejected. There is sufficient evidence to indicate there is a negative association between the sweetness index and the amount of pectin at α = .01.

14.66

a.

Some preliminary calculations are: Parent 643 381 342 251 216 208 192 141 131 128 124

Rank, u 11 10 9 8 7 6 5 4 3 2 1

rs = 1 −

Subsid 2,617 1,724 1,867 1,238 890 681 1,534 899 492 579 672

6∑ di2 n( n − 1) 2

=1−

Rank, v 11 9 10 7 5 4 8 6 1 2 3

Difference di 0 1 -1 1 2 2 -3 -2 2 0 -2

di2 0 1 1 1 4 4 9 4 4 0 4 2 ∑ di = 32

6(32) = 1 − .145 = .855 11(112 − 1)

Since this correlation coefficient is fairly close to 1, it indicates that there is a relatively strong positive relationship between the number of parent companies and the number of subsidiaries. To determine if the number of parent companies is positively related to the number of subsidiaries, we test: H0: ρs = 0 Ha: ρs > 0 The test statistic is rs = .855.

Nonparametric Statistics

551

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From Table XVI, Appendix B, rs,.05 = .523, with n = 11. The rejection region is rs > .523. Since the observed value of the test statistic falls in the rejection region (rs = .855 > .523), H0 is rejected. There is sufficient evidence to indicate that the number of parent companies is positively related to the number of subsidiaries at α = .05. b.

We must assume: 1. The sample is randomly selected. 2. The probability distributions of both of the variables are continuous. The actual number of companies and subsidiaries are not continuous. However, since the numbers of companies/subsidiaries are very large, this assumption is basically met. From the information given, we cannot tell whether the sample was random or not.

14.68

b.

Some preliminary calculations:

Involvement

1 2 3 4 5 6 7 8 9 10 11

rs = 1 −

6∑ d i2 n(n − 1) 2

ui

vi

Differences di = ui − vi

8 6 10 2 5 9 1 4 7 11 3

9 7 10 1 5 8 2 4 6 11 3

−1 −1 0 1 0 1 −1 0 1 0 0

=1−

d i2

∑ di2

1 1 0 1 0 1 1 0 1 0 1 =6

6(6) = .972 11(112 − 1)

To determine if a positive relationship exists between participation rates and cost savings rates, we test: H0: ρs = 0 Ha: ρs > 0 The test statistic is rs = .972. From Table XVII, Appendix B, rs,.01 = .736, with n = 11. The rejection region is rs > .736.

552

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Since the observed value of the test statistic does falls in the rejection region (rs = .972 > .736), H0 is rejected. There is sufficient evidence to indicate that a positive relationship exists between participation rates and cost savings rates at α = .01. c.

In order for the above test to be valid, we must assume: 1. 2.

The sample is randomly selected. The probability distributions of both of the variables are continuous.

In order to use the Pearson correlation coefficient, we must assume that both populations are normally distributed. It is very unlikely that the data are normally distributed. 14.70

The appropriate test for this completely randomized design is the Kruskal-Wallis H-test. Some preliminary calculations are: Sample 1 18 32 43 15 63

Rank 4.5 6 9 3 12

Sample 2 12 33 10 34 18

Rank Sample 3 12 87 7 53 1 65 8 50 4.5 64 77 R2 = 22.5

R1 = 34.5

Rank

16 11 14 10 13 15 R3 = 79

To determine whether at least two of the populations differ in location, we test: H0: The three probability distributions are identical Ha: At least two of the three probability distributions differ in location 2

Rj 12 The test statistic is H = − 3( n + 1) ∑ n( n + 1) nj =

⎡ (34.5) 2 (22.5) 2 (79) 2 ⎤ 12 + + ⎢ ⎥ − 3(16 + 1) 16(16 + 1) ⎣ 5 5 6 ⎦

= 60.859 − 51 = 9.859 The rejection region requires α = .05 in the upper tail of the χ2 distribution with df = p − 1 = 3 2 − 1 = 2. From Table VII, Appendix B, χ.05 = 5.99147. The rejection region is H > 5.99147. Since the observed value of the test statistic falls in the rejection region (H = 9.859 > 5.99147), reject H0. There is sufficient evidence to indicate a difference in location for at least two of the three probability distributions at α = .05.

Nonparametric Statistics

553

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14.72

The appropriate test for two independent samples is the Wilcoxon rank sum test. Some preliminary calculations are: Sample 1 1.2 1.9 .7 2.5 1.0 1.8 1.1

Rank 4 8.5 1 10 2 7 3 T1 = 35.5

Sample 2 1.5 1.3 2.9 1.9 2.7 3.5

Rank 6 5 12 8.5 11 13

T2 = 55.5

To determine if there is a difference between the locations of the probability distributions, we test: H0: The two sampled populations have identical probability distributions Ha: The probability distribution for population 1 is shifted to the left or right of that for 2 The test statistic is T2 = 55.5. Reject H0 if T2 ≤ TL or T2 ≥ TU where α = .05 (two-tailed), n1 = 7 and n2 = 6: Reject H0 if T2 ≤ 28 or T2 ≥ 56 (from Table XV, Appendix B). Since T2 = 55.5 ≤/ 28 and T2 = 55.5 ≥/ 56, do not reject H0. There is insufficient evidence to indicate a difference between the locations of the probability distributions for the sampled populations at α = .05. 14.74

a.

To determine whether the median biting rate is higher in bright, sunny weather, we test: H0: η = 5 Ha: η > 5

b.

( S − .5) − .5n (95 − .5) − .5(122) = = 6.07 .5 n .5 122 (where S = number of observations greater than 5)

The test statistic is z =

The p-value is p = P(z ≥ 6.07). From Table IV, Appendix B, p = P(z ≥ 6.07) ≈ 0.0000. c.

554

Since the observed p-value is less than α (p = 0.0000 < .01), H0 is rejected. There is sufficient evidence to indicate that the median biting rate in bright, sunny weather is greater than 5 at α = .01.

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14.76

Some preliminary calculations are: Difference Highway 1 − Highway 2 −25 4 −23 −16 −16

Rank of Absolute Differences 5 1 4 2.5 2.5 T+ = 1

To determine if the heavily patrolled highway tends to have fewer speeders per 100 cars than the occasionally patrolled highway, we test: H0: The two sampled populations have identical probability distributions Ha: The probability distribution for highway 1 is shifted to the left of that for highway 2 The test statistic is T+ = 1. The rejection region is T+ ≤ 1 from Table XVI, Appendix B, with n = 5 and α = .05. Since the observed value of the test statistic falls in the rejection region (T+ = 1 ≤ 1), H0 is rejected. There is sufficient evidence to indicate the probability distribution for highway 1 is shifted to the left of that for highway 2 at α = .05. b.

Some preliminary calculations are: Day

1 2 3 4 5

Difference Highway 1 − Highway 2 25 4 −23 −16 −16

d=

∑ di = −76 5

n

∑

di2

= −15.2

( ∑ di ) −

2

n = n −1 sd = 131.7 = 11.4761

sd2 =

(−76) 2 5 5 −1

1682 −

To determine if the mean number of speeders per 100 cars differ for the two highways, we test: H0: μ1 = μ2 Ha: μ1 ≠ μ2 The test statistic is t =

Nonparametric Statistics

d −0 −15.2 = = − 2.96 s d / n 11.4761 5

555

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The rejection region requires α/2 = .05/2 = .025 in each tail of the t-distribution with df = n − 1 = 5 − 1 = 4. From Table VI, Appendix B, t.025 = 2.776. The rejection region is t > 2.776 and t < −2.776. Since the observed value of the test statistic falls in the rejection region (t = −2.96 < −2.776), H0 is rejected. There is sufficient evidence to indicate the mean number of speeders per 100 cars differ for the two highways at α = .05. We must assume that the population of differences is normally distributed and that a random sample of differences was selected. 14.78

a.

Since only 70 of the 80 customers responded to the question, only the 70 will be included. To determine if the median amount spent on hamburgers at lunch at McDonald's is less than $2.25, we test: H0: η = 2.25 Ha: η < 2.25 S = number of measurements less than 2.25 = 20. The test statistic is z =

( S − .5) − .5n .5 n

=

(20 − .5) − .5(70) .5 70

= −3.71

No α was given in the exercise. We will use α = .05. The rejection region requires α = .05 in the lower tail of the z-distribution. From Table IV, Appendix B, z.05 = 1.645. The rejection region is z > 1.645. Since the observed value of the test statistic does not fall in the rejection region (z = −3.71 >/ 1.645), H0 is not rejected. There is insufficient evidence to indicate that the median amount spent on hamburgers at lunch at McDonald's is less than $2.25 at α = .05.

556

b.

No. The survey was done in Boston only. The eating habits of those living in Boston are probably not representative of all Americans.

c.

We must assume that the sample is randomly selected from a continuous probability distribution.

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14.80

Some preliminary calculations: 1

Urban 4.3 5.2 6.2 5.6 3.8 5.8 4.7

2 3 Rank Suburban Rank Rural Rank 4.5 5.9 14 5.1 9 10.5 6.7 17 4.8 7 15.5 7.6 19 3.9 2 12 4.9 8 6.2 15.5 1 5.2 10.5 4.2 3 13 6.8 18 4.3 4.5 6 R1 = 62.5 R2 = 86.5 R3 = 41

To determine if there is a difference in the level of property taxes among the three types of school districts, we test: H0: The three probability distributions are identical Ha: At least two of the three probability distributions differ in location 2

The test statistic is H =

Rj 12 − 3( n + 1) ∑ n( n + 1) nj

⎛ 62.52 86.52 412 ⎞ 12 + + ⎜ ⎟ − 3(20) = 65.8498 − 60 19(19 + 1) ⎝ 7 6 6 ⎠ = 5.8498 =

The rejection region requires α = .05 in the upper tail of the χ2 distribution with df = p − 1 = 2 = 5.99147. The rejection region is H > 5.99147. 3 − 1 = 2. From Table VII, Appendix B, χ.05 Since the observed value of the test statistic does not fall in the rejection region (H = 5.8498 >/ 5.99147), H0 is not rejected. There is insufficient evidence to indicate that there is a difference in the level of property taxes among the three types of school districts at α = .05. 14.82

a. Some preliminary calculations are: Truck Static Weight of Truck (ui) 1 3 2 4 3 10 4 1 5 6 6 8 7 2 8 5 9 7 10 9 55

Nonparametric Statistics

Weigh-in-Motion Prior (vi) 3 4 9 1.5 6 8 1.5 5 7 10 55

Weigh-in-Motion After (wi) 3 4 10 2 6 8 1 5 7 9 55

uivi

9 16 90 1.5 36 64 3 25 49 90 383.5

uiwi

9 16 100 2 36 64 2 25 49 81 384

557

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∑ ui ∑ vi = 383.5 − 55(55)

SSuv =

∑ ui vI −

SSuw =

∑ u i wi −

SSuu =

∑ ui2 −

SSvv =

∑

SSww =

∑

rs1 = rs2 =

vi2

n ( ∑ ui ∑ wi )

( ∑ ui ) n

= 385 −

SSuu SSvv

n

=

= 384.5 −

SSuw SSuu SSww

=

2

= 385 −

81 82.5(82)

= 81

55(55) = 81.5 10

552 = 81.5 10

2

( ∑ wi ) −

SSuv

= 384 −

2

n

( ∑ vi ) −

wi2

n

10

552 = 82 10 552 = 82.5 10

= .9848

81.5 = .9879 82.5(82.5)

The correlation coefficient for x and y1 is rs1 = .9848. Since rs1 > 0, the relationship between static weight and weigh-in-motion prior to adjustment is positive. Because the value is close to 1, the relationship is very strong. It is larger than r1 = .965 found in Exercise 10.89. The correlation coefficient for x and y2 is rs2 = .9879. Since rs2 > 0, the relationship between static weight and weigh-in-motion after the adjustment is positive. Because the value is close to 1, the relationship is very strong. It is smaller than r2 = .996 found in Exercise 10.89. b.

In order for rs to be exactly 1, the rankings for the static weight and the weigh-in-motion must be the same for each truck. In order for rs to be exactly 0, the rankings for one of the variables (static weight) must be equal to 11 minus ranking of the other variable (weigh-in-motion) for each truck.

14.84

a.

To determine if the median level differs from the target, we test: H0: η = .75 Ha: η ≠ .75

b.

S1 = number of observations less than .75 and S2 = number of observations greater than .75. The test statistic is S = larger of S1 and S2. The p-value = 2P(x ≥ S) where x is a binomial random variable with n = 25 and p = .5. If the p-value is less than α = .10, reject H0.

558

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c.

A Type I error would be concluding the median level is not .75 when it is. If a Type I error were committed, the supervisor would correct the fluoridation process when it was not necessary. A Type II error would be concluding the median level is .75 when it is not. If a Type II error were committed, the supervisor would not correct the fluoridation process when it was necessary.

d.

S1 = number of observations less than .75 = 7 and S2 = number of observations greater than .75 = 18. The test statistic is S = larger of S1 and S2 = 18. The p-value = 2P(x ≥ 18) where x is a binomial random variable with n = 25 and p = .5. From Table II, p-value = 2P(x ≥ 18) = 2(1 − P(x ≤ 17)) = 2(1 − .978) = 2(.022) = .044 Since the p-value = .044 < α = .10, H0 is rejected. There is sufficient evidence to indicate the median level of fluoridation differs from the target of .75 at α = .10.

e.

A distribution heavily skewed to the right might look something like the following:

One assumption necessary for the t-test is that the distribution from which the sample is drawn is normal. A distribution which is heavily skewed in one direction is not normal. Thus, the sign test would be preferred. 14.86

Some preliminary calculations are: Hours

Rank

1 2 3 4 5 6 7 8

1 2 3 4 5 6 7 8

Nonparametric Statistics

Fraction Defective .02 .05 .03 .08 .06 .09 .11 .10

Rank

1 3 2 5 4 6 8 7

di

0 −1 1 −1 1 0 −1 1

d i2

∑ di2

0 1 1 1 1 0 1 1 =6

559

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To determine if the fraction defective increases as the day progresses, we test: H0: ρs = 0 Ha: ρs > 0 The test statistic is rs = 1 −

6∑ di2 n(n − 1) 2

=1−

6(6) = 1 − .071 = .929 8(82 − 1)

Reject H0 if rs > rs,α where α = .05 and n = 8: Reject H0 if rs > .643 (from Table XVII, Appendix B). Since rs = .929 > .643, reject H0. There is sufficient evidence to indicate that the fraction defective increases as the day progresses at α = .05. 14.88

a.

The design utilized was a completely randomized design.

b.

Some preliminary calculations are: Site 1 34.3 35.5 32.1 28.3 40.5 36.2 43.5 34.7 38.0 35.1

Rank 6 11 3 1 19 12 23 8 15 9 R1 = 107

Site 2 39.3 45.5 50.2 72.1 48.6 42.2 103.5 47.9 41.2 44.0

Rank 17 25 28 29 27 21 30 26 20 24 R2 = 247

Site 3 34.5 29.3 37.2 33.2 32.6 38.3 43.3 36.7 40.0 35.2

Rank 7 2 14 5 4 16 22 13 18 10 R3 = 111

To determine if the probability distributions for the three sites differ, we test: H0: The three sampled population probability distributions are identical Ha: At least two of the three sampled population probability distributions differ in location 2

Rj 12 The test statistic is H = − 3( n + 1) − 3(n + 1) ∑ n( n + 1) nj =

560

12 ⎡107 2 247 2 1112 ⎤ + + ⎢ ⎥ − 3(31) = 109.3923 − 93 30(31) ⎣ 10 10 10 ⎦ = 16.3923

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The rejection region requires α = .05 in the upper tail of the χ2 distribution with df = 2 = 5.99147. The rejection region is p − 1 = 3 − 1 = 2. From Table VII, Appendix B, χ.05 H > 5.99147. Since the observed value of the test statistic falls in the rejection region (H = 16.3923 > 5.99147), H0 is rejected. There is sufficient evidence to indicate the probability distributions for at least two of the three sites differ at α = .05. c.

Since H0 was rejected, we need to compare all pairs of sites. Some preliminary calculations are: Site 1 34.3 35.5 32.1 28.3 40.5 36.2 43.5 34.7 38.0 35.1

Site 2 39.3 45.5 50.2 72.1 48.6 42.2 103.5 47.9 41.2 44.0

Rank 3 6 2 1 10 7 13 4 8 5 T1 = 59 Rank 9 15 18 19 17 12 20 16 11 14 T2 = 151

Site 2 39.3 45.5 50.2 72.1 48.6 42.2 103.5 47.9 41.2 44.0

Rank 9 15 18 19 17 12 20 16 11 14 T2 = 151 Site 3 34.5 29.3 37.2 33.2 32.6 38.3 43.3 36.7 40.0 35.2

Site 1 34.3 35.5 32.1 28.3 40.5 36.2 43.5 34.7 38.0 35.1

Rank 6 11 3 1 18 12 20 8 15 9 T1 = 103

Site 3 34.3 29.3 37.2 33.2 32.6 38.3 43.3 36.7 40.0 35.2

Rank 7 2 14 5 4 16 19 13 17 10 T3 = 107

Rank 4 1 7 3 2 8 13 6 10 5 T3 = 59

For each pair, we test: H0: The two sampled population probability distributions are identical Ha: The probability distribution for one site is shifted to the right or left of the other. The rejection region for each pair is T ≤ 79 or T ≥ 131 from Table XV, Appendix B, with n1 = n2 = 10 and α = .05.

Nonparametric Statistics

561

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For sites 1 and 2: The test statistic is T1 = 59. Since the observed value of the test statistic falls in the rejection region, (TA = 59 ≤ 79), H0 is rejected. There is sufficient evidence to indicate the probability distribution for site 1 is shifted to the left of that for site 2 at α = .05. For sites 1 and 3: The test statistic is T1 = 103. Since the observed value of the test statistic does not fall in the rejection region (T1 = 103 / 131), H0 is not rejected. There is insufficient evidence to indicate the probability distribution for site 1 is shifted to the right or left of that for site 3 at α = .05. For sites 2 and 3: The test statistic is T2 = 151. Since the observed value of the test statistic falls in the rejection region (T2 = 151 ≥ 131), H0 is rejected. There is sufficient evidence to indicate the probability distribution for site 2 is shifted to the right of that for site 3 at α = .05. Thus, the income for those at site 2 is significantly higher than at the other two sites. d.

The necessary assumptions are: 1. 2. 3.

The three samples are random and independent. There are five or more measurements in each sample. The three probability distributions from which the samples are drawn are continuous.

For parametric tests, the assumptions are: 1. 2. 3.

562

The three populations are normal. The samples are random and independent The three population variances are equal.

Chapter 14

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14.90

Using MINITAB, the results of the Wilcoxon Rank Sum Test (Mann-Whitney Test) for each of the Variables are: Mann-Whitney Test and CI: CREATIVE-S, CREATIVE-NS CREATIVE-S CREATIVE-NS

N 47 67

Median 5.0000 4.0000

Point estimate for ETA1-ETA2 is 1.0000 95.0 Percent CI for ETA1-ETA2 is (0.9999,1.0000) W = 3734.5 Test of ETA1 = ETA2 vs ETA1 not = ETA2 is significant at 0.0000 The test is significant at 0.0000 (adjusted for ties)

Mann-Whitney Test and CI: INFO-S, INFO-NS INFO-S INFO-NS

N 47 67

Median 5.000 5.000

Point estimate for ETA1-ETA2 is 0.000 95.0 Percent CI for ETA1-ETA2 is (-0.000,1.000) W = 2888.5 Test of ETA1 = ETA2 vs ETA1 not = ETA2 is significant at 0.2856 The test is significant at 0.2743 (adjusted for ties)

Mann-Whitney Test and CI: DECPERS-S, DECPERS-NS DECPERS-S DECPERS-NS

N 47 67

Median 3.000 2.000

Point estimate for ETA1-ETA2 is -0.000 95.0 Percent CI for ETA1-ETA2 is (-0.000,1.000) W = 2963.5 Test of ETA1 = ETA2 vs ETA1 not = ETA2 is significant at 0.1337 The test is significant at 0.1228 (adjusted for ties)

Mann-Whitney Test and CI: SKILLS-S, SKILLS-NS SKILLS-S SKILLS-NS

N 47 67

Median 6.0000 5.0000

Point estimate for ETA1-ETA2 is 1.0000 95.0 Percent CI for ETA1-ETA2 is (0.9999,1.9999) W = 3498.5 Test of ETA1 = ETA2 vs ETA1 not = ETA2 is significant at 0.0000 The test is significant at 0.0000 (adjusted for ties)

Nonparametric Statistics

563

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Mann-Whitney Test and CI: TASKID-S, TASKID-NS N 47 67

TASKID-S TASKID-NS

Median 5.000 4.000

Point estimate for ETA1-ETA2 is 1.000 95.0 Percent CI for ETA1-ETA2 is (-0.000,1.000) W = 3028.0 Test of ETA1 = ETA2 vs ETA1 not = ETA2 is significant at 0.0614 The test is significant at 0.0566 (adjusted for ties)

Mann-Whitney Test and CI: AGE-S, AGE-NS AGE-S AGE-NS

N 47 67

Median 47.000 45.000

Point estimate for ETA1-ETA2 is 1.000 95.0 Percent CI for ETA1-ETA2 is (-1.000,4.001) W = 2891.5 Test of ETA1 = ETA2 vs ETA1 not = ETA2 is significant at 0.2779 The test is significant at 0.2771 (adjusted for ties)

Mann-Whitney Test and CI: EDYRS-S, EDYRS-NS EDYRS-S EDYRS-NS

N 47 67

Median 13.000 13.000

Point estimate for ETA1-ETA2 is -0.000 95.0 Percent CI for ETA1-ETA2 is (0.000,-0.000) W = 2664.0 Test of ETA1 = ETA2 vs ETA1 not = ETA2 is significant at 0.8268 The test is significant at 0.8191 (adjusted for ties)

A summary of the tests above and the t-tests from Chapter 7 are listed in the table: Variable CREATIVE INFO DECPERS SKILLS TASKID AGE EDYRS

Wilcoxon Test Statistic, T2 3734.5 2888.5 2963.5 3498.5 3028.0 2891.5 2664.0

p-value 0.000 0.274 0.123 0.000 0.057 0.277 0.819

t 8.847 1.503 1.506 4.766 1.738 0.742 -0.623

p-value 0.000 0.136 0.135 0.000 0.087 0.460 0.534

The p-values for the Wilcoxon Rank Sum Tests and the t-tests are similar and the decisions are the same. Since the sample sizes are large (n = 47 and n = 67), the Central Limit Theorem applies. Thus, the t-tests (or z-tests) are valid. One assumption for the Wilcoxon Rank Sum test is that the distributions are continuous. Obviously, this is not true. There are many ties in the data, so the Wilcoxon Rank Sum tests may not be valid.

564

Chapter 14