Statistical Methods for the Social Sciences Fourth Edition Agresti Solutions

INSTRUCTOR’S MANUAL

to accompany

STATISTICAL METHODS FOR THE SOCIAL SCIENCES Fourth Edition

Alan Agresti and Barbara Finlay

published by Pearson Education

Manual prepared by: Jackie Miller 404 Cockins Hall Department of Statistics The Ohio State University Columbus, OH 43210

Instructors: Please notify Alan Agresti of any errors in this manual or the text so they can be corrected for future printings. Please send e-mail to [email protected].

Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

Table of Contents 1. Introduction

1

2. Sampling and Measurement

2

3. Descriptive Statistics

5

4. Probability Distributions

20

5. Statistical Inference: Estimation

29

6. Statistical Inference: Significance Tests

39

7. Comparison of Two Groups

50

8. Analyzing Association Between Categorical Variables

64

9. Linear Regression and Correlation

71

10. Introduction to Multivariate Relationships

90

11. Multiple Regression and Correlation

95

12. Comparing Groups: Analysis of Variance (ANOVA) Methods

109

13. Combining Regression and ANOVA: Quantitative and Categorical Predictors

116

14. Model Building with Multiple Regression

123

15. Logistic Regression: Modeling Categorical Responses

139

16. An Introduction to Advanced Methodology

145

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Chapter 1 1.1. (a) An individual Prius (automobile). (b) All Prius automobiles used in the EPA tests. (c) All Prius automobiles that are or may be manufactured. 1.2. (a) All 7 million voters. (b) A statistic is the 56.5% who voted for Schwarzenegger from the exit poll sample of size 2705; a parameter is the 55.9% who actually voted for Schwarzenegger. 1.3. (a) All students at the University of Wisconsin. (b) A statistic, since it’s calculated only for the 100 sampled students. 1.4. A statistic, since it is based on the approximately 1200 Floridians in the sample. 1.5. (a) All adult Americans. (b) Proportion of all adult Americans who would answer definitely or probably true. (c) The sample proportion 0.523 estimates the population proportion. (d) No, it is a prediction of the population value but will not equal it exactly, because the sample is only a very small subset of the population. 1.6. (a) The most common response was 2 hours per day. (b) This is a descriptive statistic because it describes the results of a sample. 1.7. (a) A total of 85.7% said ―yes, definitely‖ or ―yes, probably.‖ (b) In 1998, a total of 85.8% said ―yes, definitely‖ or ―yes, probably.‖ (c) A total of 74.4% said ―yes, definitely‖ or ―yes, probably.‖ The percentages of yes responses were higher for HEAVEN than for HELL. 1.8. (a) Statistics, since they’re based on a sample of 60,000 households, rather than all households. (b) Inferential, predicting for a population using sample information. 1.9. (a) 1.10. Race white black white Hispanic white

Age Sentence Felony? 19 23 38 20 41

2 1 10 2 5

no no yes no yes

Prior Prior Arrests Convictions 2 1 0 0 8 3 1 1 5 4

1.14. (a) A statistic is a numerical summary of the sample data, while a parameter is a numerical summary of the population. For example, consider an exit poll of voters on election day. The proportion voting for a particular candidate is a statistic. Once all of the votes have been counted, the proportion of voters who voted for that candidate would be known (and is the parameter). (b) Description deals with describing the available data (sample or population), whereas inference deals with making predictions about a population using information in the sample. For example, 1 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

consider a sample of voters on election day. One could use descriptive statistics to describe the voters in terms of gender, race, party, etc., and inferential statistics to predict the winner of the election. 1.15. If you have a census, you do not need to use the information from a sample to describe the population since you have information from the population as a whole. 1.16. (a) The descriptive part of this example is that the average age in the sample is 24.1 years. (b) The inferential part of this example is that the sociologist estimates the average age of brides at marriage for the population to between 23.5 and 24.7 years. (b) The population of interest is women in New England in the early eighteenth century. 1.17. (a) A statistic is the 45% of the sample of subjects interviewed in the UK who said yes. (b) A parameter is the true percent of the 48 million adults in the UK who would say yes. (c) A descriptive analysis is that the percentage of yes responses in the survey varied from 10% (in Bulgaria) to 60% in Luxembourg). (d) An inferential analysis is that the percentage of adults in the UK who would say yes falls between 41% and 49%.

Chapter 2 2.1. (a) Discrete variables take a finite set of values (or possible all nonnegative integers), and we can enumerate them all. Continuous variables take an infinite continuum of values. (b) Categorical variables have a scale that is a set of categories; for quantitative variables, the measurement scale has numerical values that represent different magnitudes of the variable. (c) Nominal variables have a scale of unordered categories, whereas ordinal variables have a scale of ordered categories. The distinctions among types of variables are important in determining the appropriate descriptive and inferential procedures for a statistical analysis. 2.2. (a) Quantitative (b) Categorical (c) Categorical (d) Quantitative (e) Categorical (f) Quantitative (g) Categorical (h) Quantitative (i) Categorical 2.3. (a) Nominal (b) Nominal (c) Interval (d) Nominal (e) Nominal (f) Ordinal (g) Interval (h) Ordinal (i) Nominal (j) Interval (k) Nominal 2.4. (a) Nominal (b) Nominal (c) Ordinal (d) Interval (e) Interval (f) Interval (g) Ordinal (h) Interval (i) Nominal (j) Interval 2.5. (a) Interval (b) Ordinal (c) Nominal 2.6. (a) State of residence. (b) Number of siblings. (c) Social class (high, medium, low). (d) Student status (full time, part time). (e) Number of cars owned. (f) Time (in minutes) needed to complete an exam. (g) Number of siblings.

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2.7. (a) Ordinal, since there is a sense of order to the categories. (b) Discrete. (c) These values are statistics since them come from a sample. 2.8. Ordinal. 2.9. (b), (c), (d) 2.10. (a), (c), (e), (f) 2.11. Students numbered 10, 22, 24. 2.12. Number names 00001 to 52000. First five that are selected are 15011, 46573, 48360, 39975, 06907. 2.13. Observational study (b) Experiment (c) Observational study (d) Experiment 2.14. (a) Experimental study, since the researchers are assigning subjects to treatments. (b) An observational study could look those who grew up in nonsmoking or smoking environments and examine incidence of lung cancer. 2.15. (a) Sample-to-sample variability causes the results to vary. (b) The sampling error for the Gallup poll is –2.4% for Gore, 0.1% for Bush, and 1.3% for Nader. 2.16. (a) This is a volunteer sample because viewers chose whether to call in. (b) Randomly sample the population. 2.17. The first question is confusing in its wording. The second question has clearer wording. 2.18. (a) Skip number is k = 52,000/5 = 10,400. Randomly select one of the first 10,400 names and then skip 10,400 names to get each of the next names. For example, if the first name picked is 01536, the other four names are 01536 + 10400 = 11936, 11936 + 10400 = 22336, 22336 + 10400 = 32736, 32736 + 10400 = 43136. (b) We could treat the pages as clusters. We would select a random sample of pages, and then sample every name on the pages selected. Its advantage is that it is much easier to select the sample than it is with random sampling. A disadvantage is as follows: Suppose there are 100 ―Martinez‖ listings in the directory, all falling on the same page. Then with cluster sampling, either all or none of the Martinez families would end up in the sample. If they are all sampled, certain traits which they might have in common (perhaps, e.g., religious affiliation) might be over-represented in the sample. 2.19. Draw a systematic sample form the student directory, using skip number k = 5000/100 = 50. 2.20. (a) This is not a simple random sample since the sample with necessarily have 40 women and 40 men. A simple random sample may or may not have exactly 40 men and 40 women. (b) This is stratified random sampling. You ensure that neither men nor women are over-sampled. 3 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

2.21. (a) The clusters. (b) The subjects within every stratum. (c) The main difference is that a stratified random sample uses every stratum, and we want to compare the strata. By contrast, we have a sample of clusters, and not all clusters are represented—the goal is not to compare the clusters but to use them to obtain a sample. 2.22. (a) Categorical are GE, VE, AB, PI, PA, RE, LD, AA; quantitative are AG, HI, CO, DH, DR, NE, TV, SP, AH. (b) Nominal are GE, VE, AB, PA, LD, AA; ordinal are PI and RE; interval are AG, HI, CO, DH, DR, NE, TV, SP, AH. 2.24. (a) Draw a systematic sample from the student directory, using skip number k = N/100, where N = number of students on the campus. (b) High school GPA on a 4-point scale, treated as quantitative, interval, continuous; math and verbal SAT on a 200 to 800 scale, treated as quantitative, interval, continuous; whether work to support study (yes, no), treated as categorical, nominal, discrete; time spent studying in average day, on scale (none, less than 2 hours, 2-4 hours, more than 4 hours), treated as quantitative, ordinal, discrete. 2.25. This is nonprobability sampling; certain segments may be over- or under-represented, depending on where the interviewer stands, time of day, etc. Quota sampling fails to incorporate randomization into the selection method. 2.26. Responses can be highly dependent on nonsampling errors such as question wording. 2.27. (a) This is a volunteer sample, so results are unreliable; e.g., there is no way of judging how close 93% is to the actual population who believe that benefits should be reduced. (b) This is a volunteer sample; perhaps an organization opposing gun control laws has encouraged members to send letters, resulting in a distorted picture for the congresswoman. The results are completely unreliable as a guide to views of the overall population. She should take a probability sample of her constituents to get a less biased reaction to the issue. (c) The physical science majors who take the course might tend to be different from the entire population of physical science majors (perhaps more liberal minded on sexual attitudes, for example). Thus, it would be better to take random samples of students of the two majors from the population of all social science majors and all physical science majors at the college. (d) There would probably be a tendency for students within a given class to be more similar than students in the school as a whole. For example, if the chosen first period class consists of college-bound seniors, the members of the class will probably tend to be less opposed to the test than would be a class of lower achievement students planning to terminate their studies with high school. The design could be improved by taking a simple random sample of students, or a larger random sample of classes with a random sample of students then being selected from each of those classes (a two-stage random sample). 2.28. A systematic sample with a skip number of 7 (or a multiple of 7) would be problematic since the sampled editions would all be from the same day of the week (e.g., Friday). The day of the week may be related to the percentage of newspaper space devoted to news about entertainment. 2.29. Because of skipping names, two subjects listed next to each other on the list cannot both be in the sample, so not all samples are equally likely.

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2.30. If we do not take a disproportional stratified random sample, we might not have enough Native Americans in our sample to compare their views to those of other Americans. 2.31. If a subject is in one of the clusters that is not chosen, then this subject can never be in the sample. Not all samples are equally likely. 2.33. The nursing homes can be regarded as clusters. A systematic random sample is taken of the clusters, and then a simple random sample is taken of residents from within the selected clusters. 2.34. (b) 2.35. (c) 2.36. (c) 2.37. (a) 2.38. False. This is a convenience sample. 2.39. False. This is a voluntary response sample. 2.40. An annual income of $40,000 is twice the annual income of $20,000. However, 70 degrees Fahrenheit is not twice as hot as 35 degrees Fahrenheit. (Note that income has a meaningful zero and temperature does not.) IQ is not a ratio-scale variable.

Chapter 3 3.1. (a) Place of Birth Relative Frequency Europe 13.7% Asia 25.4% Caribbean 9.6% Central America 37.6% South America 6.1% Other 7.6%

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(b) 40.00

Percent

30.00

20.00

10.00

0.00 C America

Asia

Europe

Caribbean

Other

Place of Birth

(c) ―Place of birth‖ is categorical. (d) The mode is Central America. 3.2. (a) Religion Relative Frequency Christianity 41.2% Islam 25.5% Hinduism 17.6% Confucianism 7.8% Buddhism 7.8%

6 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

S America

(b)

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0.00 Christianity

Islam

Hinduism

Buddhism

Confucianism

Religion

(c) The mode of these five religions is Christianity. Christianity is also the mode of all religions. 3.3. (a) There are 33 students. The minimum score is 65, and the maximum score is 98. Histogram (b)

12.5

Frequency

10.0

7.5

5.0

2.5 Mean =82.88 Std. Dev. =8.947 N =33 0.0 60

70

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90

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Midterm_Score

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3.4. (a) Number Persons Relative Frequency 1 27.1% 2 33.3% 3 16.0% 4 13.8% 5 or more 9.8% (b) 40.00

Relative Frequency

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0.00 1

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5 or more

Number of Persons

(c) The median household size is 2 persons, and the mode is also 2 persons. 3.5. (a)

Valid

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Frequency 3

6.0

Cumulative Percent 6.0

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96.0

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98.0

13

1

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(b)

Histogram

10

Frequency

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2 Mean =4.8 Std. Dev. =2.571 N =50 0 0

2.5

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MU_noDC

The distribution appears to be bimodal and skewed to the right. (c) Stem 1 2 3 4 5 6 7 8 9 10 11 12 13

Leaves 000 00000000 000000000 00 00000000 000000000 00000 00 00 0 0

The stem-and-leaf plot shows the same bimodality and right skew that the histogram does. 3.6. (a) GDP is rounded to the nearest thousand Stem (10 thousands) 2 2 3 3 4 4 5 5 6 6 7

Leaves (thousands 023 58899 00011122233 8

0

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(b)

Histogram

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2.5 Mean =32.00 Std. Dev. =9.615 N =23 0.0 20.00

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RoundGDP

(c) The outlier in each plot is Luxembourg. 3.7. (a) The mean is (26 + 17 + 236 + 2 + 6)/5 = 287/5 = 57.4 abortions per 1000 women 15 to 41 years of age. (b) The median is 17 abortions per 1000 women 15 to 41 years of age. The mean and median are so different because California is an extreme outlier in this small data set. 3.8. (a) The mean is (0.3 + 1.8 + 2.3 + 1.2 + 1.4 + 0.7 + 9.9 + 20.1)/8 = 37.7/8 = 4.7 metric tons per person. The median is 1.6 metric tons per person. (b) The United States appears to be an outlier, since it is far greater than any other data value. (Without the United States, the mean is 2.5 and the median is 1.4.) 3.9. (a) The response ―not far enough‖ is the mode. (b) We cannot compute and mean or median with these data since they are categorical. 3.10. (a) Stem 0 1 2 3 4

Leaves 4679 133 0 9 4

(b) The mean is 16.6 days, and the median is 12 days. (c) Leaves 25 years ago 875 440 21 0 5

Stem 0 1 2 3 4 5

Leaves 4679 133 0 9 4

For the data from 25 years ago, the mean was 27.6 days, and the median was 24 days. The mean has decreased by 11 days, and the median has decreased by 12 days since 25 years ago. (d) Of the 10 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

11 observations, the median is 13 days. We cannot calculate the mean, but substituting 40 for the censored observation gives a mean of 18.7 days. 3.11. (a) TV Hours Frequency Relative Frequency 0 79 4.0 1 422 21.2 2 577 29.0 3 337 17.0 4 226 11.4 5 136 6.8 6 99 5.0 7 23 1.2 8 34 1.7 9 4 0.2 10 23 1.2 12 14 0.7 13 1 0.1 14 7 0.4 15 2 0.1 18 2 0.1 24 1 0.1 Total 1987 100.0 (b) The distribution is unimodal and right skewed. (c) The median is the 994th data value, which is 2. (d) The mean is larger than 2 because the data is skew right by a few high values. 3.12. Central America 8540 85210 82

Stem 4 5 6 7 8 9

Western Europe 488 003678 1268 567 0

Female economic activity seems greater, on average, in Western Europe than in Central America. Most of the values in Western Europe exceed the highest value in Central America. There appear to be more women in the labor force (per 100 men) in Western Europe than in Central America. 3.13. Since the mean is much greater than the median, the distribution of 2000 household income in Canada is most likely skewed to the right. 3.14. (a) The median is ―2 or 3 times a month.‖ The mode is ―not at all.‖ The data are centered around the respondents having sex about 2 or 3 months in the past 12 months. The most frequent answer to the question is ―not at all.‖ (b) The sample mean is 4.1, which means that, on average, the respondents had sex about 4 times a month in the past 12 months. 3.15. (a) The mode is ―every day.‖ The median is ―a few times a week.‖ (b) The mean is 3.7 times per week, which is lower than the 4.4 times a week in 1994.

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3.16. (a) For each gender, the distribution of earnings is skewed to the right, since each mean is greater than its respective median. (b) The overall mean income is ($39,890×73.8 + $56,724×83.4)/(73.8 + 83.4) = $7674663.6/157.2 = $48,821. 3.17. (a) The response variable is median family income, and the explanatory variable is race. (b) We cannot find the median income for the combined groups since we do not know how many families are in each group. (c) We would need to know how many families were in each group. 3.18. (a) The distribution is skewed to the right. (b) The Empirical Rule only applies to bellshaped distributions, so it does not apply here. (c) The median is 0. If the 500 observations were to shift from 0 to 6, the median would remain zero, since half of the data values fall below 0 and half fall above 0. This illustrates the resistance of the median to skewness and extreme values. 3.19. (a) Median: $10.13; mean: $10.18; range: $0.46; standard deviation: $0.22. (b) Median: $10.01; mean: $9.17; range: $5.31; standard deviation: $2.26. The median is resistant to outliers, but the mean, range, and standard deviation are highly impacted by outliers. 3.20. (a) Mean: 30; standard deviation: 9.0. (b) Minimum: 13; lower quartile: 25.5; median: 31; upper quartile: 36; maximum: 42. 3.21. The mean is 28.7, and the standard deviation is 12.5. The 2006 HDI ratings for the top 10 nations vary greatly. 3.22. (a) The life expectancies in Africa vary more than the life expectancies in Western Europe, because the life expectancies for the African countries are more spread out than those for the Western European countries. (b) The standard deviation is 1.1 for the Western European nations and 7.1 for the African nations. 3.23. (a) (i) $40,000 to $60,000; (ii) $30,000 to $70,000; (iii) $20,000 to $80,000. (b) A salary of $100,000 would be unusual because it is 5 standard deviations above the mean. 3.24. (a) Approximately 68% of the values are contained in the interval 32 to 38 days; approximately 95% of the values are contained in the interval 29 to 41 days; all or nearly all of the values are contained in the interval 26 to 44 days. (b) (i) The mean would decrease if the observation for the U.S. was included. (ii) The standard deviation would increase if the observation for the U.S. was included. (c) The U.S. observation is 5.3 standard deviations below the mean. 3.25. (a) 88.8% of the observations fall within one standard deviation of the mean. (b) The Empirical Rule is not appropriate for this variable, since the data are highly skewed to the right. 3.26. 10 is realistic; –20 is impossible since the standard deviation cannot be negative; 0 implies that every student scored 76 on the exam, which is highly improbable; 50 is too large (it is half of the possible range of scores). 3.27. (a) The most realistic value is 0.4, because the range is 5 times the length of this value. (b) The value of –10.0 is impossible since the standard deviation cannot be negative. 3.28. (d)

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3.29. (a) Since the range is 43.5 standard deviations above the mean, the distribution is most likely skewed to the right. (b) The distribution probably has outliers (take the maximum usage, for example). 3.30. The distribution is most likely skewed to the right since the minimum water consumption (0 thousands of gallons) is less than one standard deviation below the mean. 3.31. (a) The range is $28,700, which is the difference between the mean salary for secondary school teachers in Illinois (highest mean) and in South Dakota (lowest mean). (b) The interquartile range is $9600 and represents the spread of the mean salaries for the middle 50% of the states. 3.32. (a)

Salary

600 00

500 00

400 00

(b) The box plot suggests that the data are skewed to the right. (c) 7000 is the most plausible standard deviation, since the range of the data is about 4 standard deviations. The values 100 and 1000 are too small for the spread that we see, and 25,000 is just slightly over the value for the range. 3.33. The mean, standard deviation, maximum, and range all decrease, because the observation for D.C. was a high outlier. Note that these statistics are not resistant to outliers. On the other hand, the median, Q3, Q1, the interquartile range, and the mode remain the same, as these are all resistant to outliers. The minimum remains the same since D.C. was a high outlier and not a low outlier. 3.34. (a) The Empirical Rule does not apply to this distribution because the standard deviation is much larger than the mean, suggesting a right-skewed distribution. (b) The five-number summary confirms that the distribution is skewed to the right, since the distance between Q3 and the medians is larger than the distance between the median and Q1 and the maximum is so large. (c) IQR = Q3 – Q1 = 1105 – 256 = 849. Low outliers would be observations less than Q1 – 1.5(IQR) = 256 – 1.5(849) = –1017.5. There are no values that are low outliers. High outliers would be observations greater than Q3 + 1.5(IQR) = 1105 + 1.5(849) = 2378.5. At least the maximum is a high outlier.

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3.35. (a) The sketch should show a right-skewed distribution. (b) The sketch should show a rightskewed distribution. (c) The sketch should show a left-skewed distribution. (d) The sketch should show a right-skewed distribution. (e) The sketch should show a left-skewed distribution. 3.36. (a) Skewed to the left (b) Bell shaped (c) Skewed to the right (d) Skewed to the left (e) Skewed to the left (f) Skewed to the right (g) U shaped 3.38. A box plot using only the five-number summary follows: 

EU unemployme nt

12.0

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8.0

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4.0

This box plot shows us that the maximum is an outlier. Since the mean and the median are the same, the distribution may be slightly more symmetric than the box plot implies. It is important to note that only the five-number summary was used to produce this box plot. 3.39. (a) Minimum = 0, Q1 = 20, median = 30, Q3 = 50, maximum = 14. (b) Same as part (a). (c) The observations with values 12 and 14 are outliers. (d) The standard deviation is 3. The value 0.3 is too small for the distribution, the value 13 is almost equal to the range, and the value 23 exceeds the range. 3.40. Side-by-side box plots are not very informative as shown below:

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The infant mortality rates in Africa are much higher than the infant mortality rates in Western Europe. In addition, since Q1, the median, and Q3 are all equal for Western Europe, we do not actually see a ―box‖ in the box plot. Infant mortality rates in Africa are skewed to the right, while infant mortality rates in Western Europe are symmetric. 3.41. (a)

No health insura nce

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(b) The distribution appears to be skewed to the right. 3.42. (a) The range is 92.3 – 78.3 = 14. The interquartile range is 88.8 – 83.6 = 5.2. (b) Low outliers would be observations less than Q1 – 1.5(IQR) = 83.6 – 1.5(5.2) = 75.8. There are no values that are low outliers. High outliers would be observations greater than Q3 + 1.5(IQR) = 88.8 + 1.5(5.2) = 96.6. There are no values that are high outliers. 3.43. (a) Minimum = 1, Q1 = 3, median = 5, Q3 = 6, maximum = 13. (b) 

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Louisiana appears to be a mild outlier. (c) Minimum = 1, Q1 = 3, median = 5, Q3 = 6, maximum = 44.

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

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Louisiana is still a mild outlier, and D.C. is an extreme outlier. Only the maximum changes in the five-number summary when the observation for D.C. is added to the data set. 3.44. The IQR is most likely 350. The value of 1500 is the range. The value of –10 is impossible for an IQR, and the 0 would only occur if all data values were the same. Given the minimum, median, and maximum, the value 10 is too small to be the IQR. 3.45. (a) Luxembourg’s observation is 3.9 standard deviations above the mean. (b) Sweden’s observation is 0.8 standard deviations below the mean. (c) (i) Canada’s observation is 2.5 standard deviations above the mean of the EU countries. (ii) The U.S. observation is 3.6 standard deviations above the mean of the EU countries (but not as high as Luxembourg). 3.46. (a) Italy’s observation is 0.4 standard deviations below the mean. (b) The U.S. observation is 3.4 standard deviations above the mean of the EU countries. (c) We expect almost all of the values in a bell shaped distribution to be within 3 standard deviations of the mean. Thus, the U.S. observation would be considered a high outlier. 3.47. (a) Response variable: opinion about national health insurance (favor, oppose); explanatory variable: political party (Democrat, Republican). (b) The data could be summarized in a contingency table with political party as the rows and opinion about national health insurance as the columns. 3.48. (a) Response variable: happiness; explanatory variable: religious attendance. (b) For those who attend religious services nearly every week or more, 44.5% reported being very happy. For those who attend religious services never or less than once a year, 23.2% reported being very happy. (c) There appears to be an association between happiness and religious attendance since the percentages that reported being very happy differed greatly by attendance at religious services. 3.49. (a) United States: predicted fertility = 3.2 – 0.04(50) = 1.2; Yemen: predicted fertility = 3.2 – 0.04(0) = 3.2. (b) The negative value implies that the fertility rate decreases as Internet use increases.

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3.50. (a) Points in a scatterplot for these data should have a negative association and be fairly tightly clustered in a linear pattern. (b) Contraceptive use is more strongly associated with fertility than is Internet use because –0.89 is a stronger linear association than is –0.55. 3.51. (a) Based on the plot (see next page), the correlation should be positive, since higher values of GDP tend to go with higher values of CO2 (and vice versa). (b) Luxembourg has a GDP of 69,961 and CO2 of 22.0, both of which are extreme values. 25.0

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70,000

GDP

3.52. The number of physicians is more strongly correlated with carbon dioxide emissions than is female economic activity, since the absolute value of the correlation for number of physicians and CO2 emissions is closer to 1 than is the correlation for female economic activity and CO2 emissions. 3.53. (a) y is a sample statistic (sample mean) used to estimate the population mean µ. (b) s is a sample statistic (sample standard deviation) used to estimate the population standard deviation . 3.54. (a) The mean is 1232.2 miles, with standard deviation 1681.7 miles. The median is 640 miles. The minimum distance from home is 0 miles, while the maximum distance is 8000 miles. The histogram shows that the distribution of the distance from home is skewed to the right.

17 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

Histogram

40

Frequency

30

20

10

Mean =1232.2 Std. Dev. =1681.748 N =60 0 0

2000

4000

6000

8000

DHome

(b) The mean is 7.3 hours, with standard deviation 6.7 hours. The median is 6 hours. The minimum from home is 0 hours,Histogram while the maximum is 37 hours. The histogram shows that the distribution of the hours of watching television is skewed to the right. 20

Frequency

15

10

5

Mean =7.27 Std. Dev. =6.717 N =60 0 0.0

10.0

20.0

30.0

40.0

TVhours

3.56. Report should include graphical displays and summary statistics. The summary statistics are: mean = 2.7, standard deviation = 2.1, minimum = 0.1, Q1 = 1.5, median = 2.1, Q3 = 3.6, maximum = 9.4. The U.S. is an outlier with 9.4 gun deaths per 100,000 people. 3.57. Report should state that the explanatory variable is the percentage with income below the poverty level and the response variable is the violent crime rate. The correlation coefficient is 0.496. D.C. appears to be an outlier.

18 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

2,000

Violent Crime Rate

1,500

1,000

500

0 6.0

8.0

10.0

12.0

14.0

16.0

18.0

20.0

Poverty Level

3.59. The distribution of cost for New York and Boston are similar, and both cities have high and low outliers. The distributions for all three cities are roughly symmetric. The distribution for cost in London is higher than the distributions in both New York and Boston, with 75% of the costs in London being higher than all costs in Boston and almost all costs in New York. 3.60. All associations are positive. Quality of food rating and service rating have the highest correlation of 0.81, which is fairly strong. Quality of food rating is moderately positively associated with décor rating (0.61) and with cost rating (0.53). 3.62. The mean salary is $7,095,078. Salaries are typically right-skewed. There will be a few very high paid players and more ―modestly‖ paid players. 3.63. The median overall new worth is $86,100. The distribution of overall net worth will be skewed to the right, so the median will be less than the mean. 3.64. The median is not impacted by gains made by the wealthiest Americans because the wealthiest Americans are at the high end of net worth, and the median is calculated from values at the center of the data. 3.65. When comparing countries to each other, the mean number of children makes sense. For example, the fertility rate in Ireland is similar to the fertility rate in the U.S. The fertility rate of Mexico is almost twice that of Italy or Spain. We are looking at rates across the countries and not the number of children each individual woman in the country has. 3.66. (a) The median male height is between 69 and 70 inches. (b) A rough approximation for the standard deviation can be found by dividing the range by 6, since almost all of the data in a bellshaped distribution falls within 3 standard deviations of the mean. Thus, the standard deviation is approximately 20/6 = 3.3 inches. 3.67. One example is the type of pet that people prefer. The mode might be ―dog,‖ and it makes no sense to talk about the mean or the median. 19 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

3.68. (a) Heights (in inches) of adult women. (b) Incomes in a large city. (c) Scores on an easy exam. (d) Heights (in inches) of adults in the U.S. (e) Number of cigarettes smoked in a week. 3.69. (a) The median is preferred over the mean when the data are skewed and/or there are outliers that will affect the mean. One example is incomes in the large city. (b) The mean is preferred over the median when the distribution is very highly discrete, such as the number of times you have been married. 3.70. (a) The standard deviation s is generally preferred over the range because it is calculated from all of the data and will not be impacted as much as the range when there are outliers. (b) The IQR is preferred to the standard deviation s when the distribution is very highly skewed, because the IQR is more robust to skewness than s is. 3.71. (a) False (b) False (c) True (d) True 3.72. (c) 3.73. (c) 3.74. (a) 3.75. The standard deviation is incorrectly recorded. It exceeds the range of the data. 3.76. Florida has the larger overall mean income ($35,100) compared to Alabama ($29,600). 3.77. Population sizes vary by state; the overall rate gives more weight to states with larger population sizes, whereas the mean of the 50 measurements gives the same weight to each state. 3.78. (a) The mean is now 77, while the standard deviation stays at 20. (b) The mean is 200,000£, and the standard deviation is 60,000£. 3.79.

 y  y    y   y   y  n   y n    y   y  0 i

i

i

i

i

i

3.80. (a) 1/4, 1/9, 1/100. (b) 25% maximum versus 5% in bell-shaped case. Most distributions have much less than 25% of the distribution falling more than two standard deviations from the mean.

 y  c  0 implies that  y  nc or c    y 

3.81. f   c   2

i

i

i

n y.

Chapter 4 4.1. 907/1127 = 0.805 4.2. (a) 1 – 0.85 = 0.15 (b) (0.85)(0.84) = 0.714 4.3. (a) 96 is the total number of members of an environmental group, and 1117 is the total number of subjects who answered both questions. (b) (i) 30/96 = 0.312 (ii) 88/1021 = 0.086. (c) (i) 30/1117 = 0.027 (ii) (0.086)(0.312) = 0.027. (d) (30 + 933)/1117 = 963/1117 = 0.862. 20 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

4.4. (a) The number of languages in which a person is fluent (y) is discrete, since it must take on integer values. (b) y 0 1 2 Probability 0.02 0.81 0.17 yP  y   0  0.02 1 0.81  2  0.17   1.15 (c) 0.02 + 0.81 = 0.83. (d)



4.5. (a) The probabilities of each y value need to be taken into account (the mean is a weighted yP  y   0  0.91 1 0.06  2  0.02  3 0.01  0.13 average). (b)



4.6. y 0 1,000,000 Probability 0.9999999 0.0000001

 yP  y   0 0.9999999 1,000,000 0.0000001  0.10 or $0.10 4.7. (a)

(b)



y 0 1 2 3 4 5 6 7 8 9 Probability 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 0.10 yP  y   0  0.10 1 0.10  9  0.10  0.45 (c) The standard deviation  is 2.9,

since 0.4 is too small to be the standard deviation, 7.0 is too large to be the standard deviation (almost equal to the range), and 12.0 is an impossible value for the standard deviation (exceeds the range). 4.8. (a) P  Z  1  1  P  Z  1  1  0.8413  0.1587

(b) P  Z  1  0.1587

(c) P  Z  0.67  1  P  Z  0.67   1  0.7486  0.2514

4.9. (a) P      X       P  1  Z  1  0.8413  0.1587  0.6826

(b) P   1.96  X    1.96   P  1.96  Z  1.96  0.9750  0.0250  0.95

(c) P    3  X    3   P  3  Z  3  0.9987  0.0013  0.9974

(d) P    0.67  X    0.67   P  0.67  Z  0.67  0.7486  0.2514  0.4972 4.10. (a) 2.33 (b) 1.96 (c) 1.64 (d) 1.28 (e) 0.67 (f) 0 4.11. (a) 0.67 (b) 1.64 (c) 1.96 (d) 2.33 (e) 2.58 4.12. (a) 1.28 (b) 1.64 (c) 2.06 (d) 2.33 4.13. If the interval   z to   z contains 90% of a normal distribution, there is 5% below   z and 5% above   z . Thus,   z equals the 95th percentile. 4.14. (a) (i) 75th percentile (ii) 25% percentile. (b) z = 0.67. (c) Plug z = 0.67 into the equations in part (a) to get that   0.67 is the upper quartile of a normal distribution and   0.67 is the lower quartile of a normal distribution. 21 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

4.15. (a) 0.0179 (b) 0.0179 (c) 0.9821 (d) 0.9642 4.16. Right-tail probability of 0.01 has z = 2.33. 4.17. (a) The 98th percentile is 2.05 standard deviations above the mean. (b) The IQ score for the 98th percentile is 100 + 2.05(16) = 132.8, or about 133. 4.18. 40 hours per week has z = (40 – 45)/15 = –0.33, and is 0.33 standard deviations below the mean. The proportion of self-employed individuals who averaged more than 40 hours per week is 0.6293. 4.19. (a) An MDI of 120 has z = (120 – 100)/16 = 1.25, and is 1.25 standard deviations above the mean. The proportion of children with an MDI of 120 or more is 0.1056. (b) The MDI score that is the 90th percentile is 1.28 standard deviations above the mean, so this score is 100 + 1.28(16) = 120.48, or 120. (c) The lower quartile is 0.67 standard deviations below the mean, which gives a lower quartile of 100 – 0.67(16) = 89.28, or 89. Similarly, the upper quartile is 0.67 standard deviations above the mean, which gives an upper quartile of 100 + 0.67(16) = 110.72, or 111. Since the MDI scores are approximately normal, the median will be equal to the mean of 100. 4.20. (a) 258 days has z = (258 – 281.9)/11.4 = –2.10, and is 2.10 standard deviations below the mean. The proportion of babies that would be born prematurely is 0.0179. (b) Since 0.036 (the actual proportion of premature babies) is twice the proportion we would expect if gestation times were normally distributed, the actual distribution is probably skewed to the left, so the left-tail probability more than 2.1 standard deviations below the mean is larger than the right-tail probability more than 2.1 standard deviations above the mean. 4.21. (a) 20 gallons per week has z = (20 – 16)/5 = 0.8, and is 0.8 standard deviations above the mean. The proportion of adults who use more than 20 gallons per week is 0.2119. (b) The 95 th percentile is 1.645 standard deviations above normal. We need to solve for x where 1.645 = (x – 16)/5. The value of x is 24.225. So, the mean would need to be about 24.2 gallons so that only 5% of adults use more than 20 gallons per week. (c) If the distribution of gasoline use is not actually normal, we should expect it to be right-skewed, since there will be fewer adults with high gasoline usages that will cause the distribution to be right-skewed. 4.22. 80 has z = (80 – 83)/5 = –0.6, and is 0.6 standard deviations below the mean. 90 has z = (90 – 83)/5 = 1.4, and is 1.4 standard deviations above the mean. The proportion of students who earn a B is approximately 0.9192 – 0.2743 = 0.6449. 4.23. An SAT score of 600 is (600 – 500)/100 = 1.0 standard deviations above the mean. An ACT score of 29 is (29 – 21)/4.7 = 1.70 standard deviations above the mean. Relatively speaking, an ACT score of 29 is higher than an SAT score of 600. 4.24. (a) z = (5000 – 2500)/1500 = 1.67. (b) About 0.0475 (4.75%) of the property taxes exceed $5000. (c) If the true distribution is not normal, it is probably skewed to the right. There will be a few very expensive homes that have high property taxes that will cause the distribution to be right-skewed. 4.25. (a) 1000 kWh is z = (1000 – 673)/556 = 0.59 standard deviations above the mean. If the distribution were normal, about 27.76% of the households had use above 1000 kWh. (b) The distribution is probably right skewed, due to a few very large homes that have high electricity usage. 22 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

4.26. (a) The probability distribution is y 0 1 2 Probability 0.30 0.60 0.10 (b) The sampling distribution of the sample proportion p of the students selected who are female is p 0 0.5 1 Probability 0.30 0.60 0.10 4.27. (a) The sampling distribution of the sample proportion of heads for flipping a balanced coin once is p 0 1 Probability 0.50 0.50 (b) The sampling distribution of the sample proportion of heads for flipping a balanced coin twice is p 0 0.5 1 Probability 0.25 0.50 0.25 (c) The sampling distribution of the sample proportion of heads for flipping a balanced coin three times is p 0 1/3 2/3 1 Probability 0.125 0.375 0.375 0.125 (d) The sampling distribution of the sample proportion of heads for flipping a balanced coin four times is p 0 0.25 0.50 0.75 1 Probability 0.0625 0.25 0.375 0.25 0.0625 (e) As the number of flips increases, the sampling distribution of the sample proportion of heads seems to be getting more normal, with the probabilities concentrating more around 0.50. 4.28. (a) The 36 possible pairs are (1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6). (b) The sampling distribution for the sample mean y is p 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 Probability 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36 (c) (i) The histogram of the probability distribution for each roll is uniform. (ii) The shape is triangular, but starting to approach a bell shape, compared to the uniform distribution for Y. (d) The mean of the probability distribution for each roll is

1 1 1 1 1 1   Y  P Y   1   2    3   4    5    6    3.5 . 6

6

6

6

6

6

The mean of the probability distribution for y is

   y  P  y   1

1 2 3  2 1  1.5    2      5.5    6    3.5 .   36   36   36   36   36 

The means of the two distributions are the same because they are both symmetric about the mean of 3.5. (e) There are more  y1 , y2  pairs that have a sample mean closer to the true mean for the average of two rolls than for the value of one roll. The spread of the probability distribution for y is less than that for the probability distribution of a single roll.

23 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

4.29. (a)

y 

 n



0.5  0.0104 . (b) If actually 50% of the population voted for DeWine, 2293

it would be surprising to obtain 44% in this exit poll, since 44% is 6% lower than 50%, and the standard error for the sampling distribution is 1.04%; that is, the sample proportion of 0.44 is nearly 6 standard errors below 0.50. (c) Based on the information from the exit poll, I would be willing to predict that Sherrod Brown would win the Senatorial election. 4.30. (a) The mean is 13.6, and the standard error is and the standard error is

y 

 n



y 

 n



y 

 n



3.0  1.0 . (b) The mean is 13.6, 9

3.0  0.5 . (c) The mean is 13.6, and the standard error is 36

3.0  0.3 . As n increases, the standard error decreases by a factor of n . 100

4.31. (a) The mean is 0.10, and the standard error is

y 

 n



316.23  0.32 . (b) It is 1,000,000

very unlikely to come out ahead. The z-score for $1 is (1 – 0.10)/0.32 = 2.15. The probability of winning more than $1 is therefore 0.0158, which is 1.58%. 4.32. (a) y does not have a normal distribution since the standard deviation is the same as the mean. This implies that y has a distribution that is skewed to the right. (b) The sampling distribution of y is approximately normal with mean 1.1 and standard error

y 

 n



1.1  0.0225 . (c) The sample mean would almost surely fall within 3 standard 2400

errors of the sample mean, which is the interval 1.03 to 1.17. 4.33. (a) The probability that PDI is below 90 is

90 100   P Y  90  P  Z   P  Z  0.67   0.2514 . 15  

(b) The probability that the sample mean PDI is below 90 is

 90 100  P Y  90  P  Z    P  Z  3.33  0.00135 . 15 25   (c) An individual PDI of 90 is not surprising, since the probability is 0.2514. However, a sample mean PDI of 90 would be surprising since this value would happen almost never. (d) The sketch of the sampling distribution should be less spread out and have a taller peak and thinner tails than the sketch of the population distribution. 4.34. (a) P





  10  Y   10  P  20010100  Z  20010100   P  0.5  Z  0.5 .

  The probability is then 0.6815  0.3085  0.3730 . (b) If the actual population standard deviation is larger than 200, the probability would be smaller than the probability found in part (a).

24 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

4.35. (a) The variable y is the number of people in a household in the U.S. (b) The center of the population distribution is 2.6 people, with standard deviation 1.5 people. (c) The center of the sample data distribution is 2.4 people, with standard deviation 1.4 people. (d) The center of the sampling distribution of the sample mean for 225 homes is 2.6 people, with standard error

225  0.1 . This distribution describes the theoretical distribution for the sample mean.

1.5

4.36. (a) The population distribution is skewed to the right with mean 5.2 and standard deviation 3.0. (b) The sample data distribution based on the sample of 36 families and is skewed to the right with mean 4.6 and standard deviation 3.2. (c) The sampling distribution of y is approximately normal with mean 5.2 and standard error 3.0

36  0.5 . This distribution describes the

theoretical distribution for the sample mean.

4.37. (a)

 0.5 0.5  P    0.5  Y    0.5  P  Z    P  1  Z  1 . This 3.0 36   3.0 36

probability is 0.8413 – 0.1587 = 0.6826. (b) P









   0.5  Y    0.5  P  3.00.5100  Z  3.00.5100   P  1.67  Z  1.67 . This

probability is 0.9525 – 0.0475 = 0.9050. The probability is larger than in part (a), because the standard error is smaller (since the sample size is larger). (c) If the sample were truly random, then the probability that y would be 4.0 or less is

 4.0  5.2  P Y  4.0  P  Z    P  Z  4  0.0000317 . This would be a surprising 3.0 100   result. 4.38. (a) Let y = 1 if the student is female and y = 0 if the student is male. (b) The population

distribution of gender at this university has P Y  1  0.60 and P Y  0  0.40 . (c) The sample data distribution of gender has P Y  1  0.52 and P Y  0  0.48 . (d) In a random sample of size 50, we expect the sampling distribution of the sample proportion of females in the sample to be approximately normal with mean 0.60 and standard error 0.07. 4.39. (a) The population distribution is skewed to the left with mean 60 years and standard deviation 16 years. (b) The sample data distribution is skewed to the left with mean 58.3 years and standard deviation 15.0 years. (c) The sampling distribution of y is approximately normal with mean 60 years and standard error 1.6 years. This distribution specifies probabilities for the possible values of y for all the possible samples. (d) The probability of observing a person of

age 40 in Sunshine City is

40  60   P Y  40  P  Z   P  Z  1.25  0.1056 . 16   The probability of observing a sample mean of 40 for a random sample of size 100 is 25 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

 40  60  P  y  40  P  Z    P  Z  12.5  0 . 16 100   Thus, it is not unusual to observe an individual of age 40 in Sunshine City, but it is very unusual to observe a random sample of size 100 in Sunshine City with an average age of 40. 4.40. (a) The sampling distribution of y for a random sample of size n = 1 is exactly the same as the population distribution. (b) If you sample all 50,000 residents in Sunshine City, there will not be a sampling distribution, and you will know that the population mean is 60 years and the population standard deviation is 16 years. 4.41 (b) Even though the population distribution is not normal (there are only two possible values), the sample proportions for the 1000 samples of size 100 each should have a histogram with an approximately bell shape. 4.42 (a) The population distribution is skewed, but the empirical distribution of sample means probably has a bell shape, reflecting the Central Limit Theorem. (b) The Central Limit Theorem applies to relatively large random samples, but here n = 2 for each sample. 4.44. (a) A stem-and-leaf plot of the population is 2|3 2|578899 3|0123 3|6677899 4|011233 4|556789 5|0014 5|677 6|02234 6|6778 7|01 7|6 8|1 (c) The mean of the y -values in a long run of repeated samples of size 9 should be approximately 47.18. (c) The standard deviation of the y -values in a long run of repeated samples of size 9 should be approximately 4.9. 4.45. (a) The probability distribution is y 0 1 Probability 0.50 0.50 The mean is 0.5. (d) (i) The mean should be 0.5. (ii) The standard deviation should be 0.16. 4.46. (a) The sample data distribution tends to resemble the population distribution more closely than the sampling distribution. A random sample of data from a population should be 26 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

representative of the population, and its distribution should be similar to the population distribution. (b) The sample data distribution is the distribution of data that we actually observe. The sampling distribution of y is the probability distribution for the possible values of the sample statistic y . 4.47. (a) A lower bound for the mean is

   yP  y   1 0.01  2  0.10  3 0.09  4  0.31  5  0.19  6 0.29  4.41 .

(b) Since we know the category of ideal number of children that falls at the 50% point, we can find the median. The median is 4 children. 4.48. (a)

P  y    0.67   P  z  0.67  0.7514 Thus, the upper quartile equals

  0.67 . (b) The IQR for a normal distribution is   0.67     0.67   1.34 . This gives us 1.5(IQR) = 1.51.34  2.01 . The 1.5(IQR) criterion would tell us that an outlier lies above

  0.67  2.01    2.68 , which is about 2.7 standard deviations above the

mean. The probability that data from a normal distribution would fall above this point is

P  y    2.7   P  z  2.7  0.0035 . Note that outliers will also fall below   2.7 ,

which has area 0.0035 (by symmetry). Therefore, only 0.7% of data are outliers using the 1.5(IQR) criterion. 4.49. The standard error for the poll, assuming that the true proportion is 0.5 is

y 

0.5 0.5   0.014 , or 1.4%. Since the 67% statistic from the exit poll is more than 12 n 1336

standard deviations above the expected mean, I would be willing to predict that Hillary Clinton would be the winner of the 2006 Senatorial election in New York based on this exit poll.

0.5 0.5   0.05 . The interval 0.35 to 0.65 is the interval within n 100 0.5  0.016 . which the sample proportion is almost certain to fall. When n = 1000,  y  1000 4.50. When n = 100,

y 

The interval 0.453 to 0.547 is the interval within which the sample proportion is almost certain to fall. When n = 10,000,

y 

0.5  0.005 . The interval 0.485 to 0.515 is the interval within 10,000

which the sample proportion is almost certain to fall. 4.51. (a) 4.52. (c)

27 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

4.53. False. As the sample size increases, the standard error of the sampling distribution of y decreases, since

y 



decreases as n increases.

n

 

4.54. (a) Group A: P  y  400  P  z 

400  500   P  z  1  0.1587 . Almost 16% of 100 

students from Group A are not admitted to Lake Wobegon Junior College. Group B:

400  450   P  y  400  P  z   P  z  0.5  0.3085 . Almost 31% of students from 100   Group B are not admitted to Lake Wobegon Junior College. (b) Of the students who are not admitted, 0.3085/(0.3085 + 0.1587) = 0.3085/0.4672 = 0.6603, or about 66%, are from Group B. (c) If the new policy is implemented, the proportion of students from Group A that are not admitted would be 0.0228, while the proportion of students from Group B that are not admitted would be 0.0668. In this case, about 75% of the students who are not admitted would be from Group B. Relatively speaking, this policy would hurt students from Group B more than the current policy. 4.55. (a)



 y    P  y   0  0.5 0.5  1 0.5 0.5  2

2

2

0.25  0.5 . (b)

   yP  y   0 1   1    ;

 0   2 1    1  2     2   3    2 2   3     2   1   .



(c)

The standard error for a sample proportion for a random sample of size n is

 n



 1   

4.56.

f    c 

n



 1    n

.

f    c 

Since

2   c     2 2   c 2 2 2 1 1 e  e   2 2

and

2   c    2 2   c 2 2 2 1 1 e  e   are equal. Thus, f    c   f    c  2 2

and the curve is symmetric.

30,000  300 30,000 1  0.99  0.995 . (b) If n = N, the finite population correction is  N  N   N 1  0 , so  y  0 . (c) When n 4.57. (a) The finite population correction is

= 1, the finite population correction is

 N 1  N 1  1 , so  y 

 n



 1

  . Thus, the

sampling distribution of y and its standard error are identical to the population distribution and its standard deviation.

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Chapter 5 5.1. ˆ  412,878 577,006  0.716

2  3  2  1  0  1  4  3 16   2.0 hours per day spent watching TV. (b) The 8 8 s 1.309 margin of error is t  2.365  1.10 . This represents the amount that is added to and n 8 5.2. (a) y 

subtracted from the point estimate to form a confidence interval.

5.3. The estimated standard error is

5.4. The estimated standard error is

5.5. The margin of error is z

ˆ 1  ˆ  n

ˆ 1  ˆ 

ˆ 1  ˆ  n

n



0.74  0.26  0.017 . 644



0.54  0.46  0.011 . 2003

0.51 0.49  0.031 , or 3.1%. 1008

 1.96

5.6. (a) Democrat: ˆ  90 142  0.634 ; Republican: ˆ  26 102  0.255 . (b) We are 95% confident that the population proportion of yes responses falls in the interval 0.55 to 0.71 for Democrats. We are also 95% confidence that the population proportion of yes responses falls in the interval 0.17 to 0.34 for Republicans. It appears that more Democrats feel that the government is responsible for reducing the income differences between the rich and the poor.

5.7. (a) The estimated standard error in 2004 is margin of error is z

ˆ 1  ˆ  n

ˆ 1  ˆ  n



0.36  0.64  0.016 . (b) The 833

 1.96  0.016  0.03 , or 3%. (c) The 95% confidence interval

is 0.36 – 0.03 = 0.33 to 0.36 + 0.03 = 0.39. We are 95% confident that the population proportion of people agreeing that it is much better for everyone involved if the man is the achiever outside the home and the woman takes care of the home and family falls in the interval 0.33 to 0.39. 5.8. ˆ  366 598  0.612 The 99% confidence interval is

ˆ  z

ˆ 1  ˆ  n

 0.612  2.576

0.612  0.388  0.56 to 0.66 . 598

5.9. (a) ˆ  229 1200  0.191 The 95% confidence interval is 29 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

ˆ  z

ˆ 1  ˆ  n

 0.191  1.96

0.191 0.809  0.17 to 0.21 . 1200

We are 95% confidence that the interval 0.17 to 0.21 contains the population proportion of those who believe that environmental regulations are too strict. (b) The 99% confidence interval is

0.191  2.576

0.191 0.809  0.16 to 0.22 . We are 99% confidence that the interval 16% to 1200

22% contains the population proportion of those who believe that environmental regulations are too strict. Note that the 99% confidence interval is wider than the 95% confidence interval. 5.10. A 99% confidence interval would be wider because we need to have more possible values in the interval in order to be more confident. 5.11. (a) 2.326 (b) 1.645 (c) 0.67 (d) 3.0 5.12. (a) ―Sample prop‖ = 1885/2815 = 0.6696. (b) Since we are 95% confident that the interval 0.652 to 0.687 contains the population proportion of American adults who are in favor of the death penalty and the entire interval exceeds 50%, it is reasonable to conclude that more than half of all American adults are in favor of the death penalty. (c) A 95% confidence interval for the proportion of American adults who opposed the death penalty is 0.313 to 0.348. 5.13. (a) The proportion that said legal is 0.364; the proportion that said not legal is 0.636. (b) The 95% confidence interval is 0.364  1.96

0.364  0.636  0.331 to 0.397 . We are 95% 802

confident that the interval 0.331 to 0.397 contains the population proportion that thinks marijuana should be made legal. Since this interval is entirely below 50%, we can conclude that a minority of Americans felt this way. (c) The proportion that said marijuana should be legal dropped until 1990 and has increased each year since.

5.14. The 99% confidence interval is 0.538  2.576

0.538  0.462  0.499 to 0.577 . We are 1095

99% confident that the interval 49.9% to 57.7% contains the population proportion of those who think that it is probably or definitely not true that human beings developed from earlier species of animals. Since this interval dips slightly below 50%, we cannot conclude that a majority of Americans felt this way.

5.15. The 99% confidence interval is 0.255  2.576

0.255  0.745  0.250 to 0.260 . We are 42,000

99% confidence that the interval 0.25 to 0.26 contains the population proportion of smokers.

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5.16. The 95% confidence interval is 0.565  1.96

0.565  0.435  0.546 to 0.584 . We are 2705

95% confidence that the interval 0.546 to 0.584 contains the population proportion of those who voted for Schwarzenegger. Assuming that the sample used for the exit poll is representative of the population of voters, there is enough evidence to conclude that Schwarzenegger would win the election, since the entire confidence interval exceeds 0.50. 5.17. (a) The 99% confidence interval is 0.40  2.576

0.40  0.60  0.337 to 0.463 . Since 400

the entire confidence interval is below 50%, it would appear that Jones would not win the election. (b) The 99% confidence interval is 0.40  2.576

0.40  0.60  0.20 to 0.60 . Since 40

the confidence interval contains 0.50, there appears to be no clear winner. The smaller sample size makes the confidence interval so wide that it is almost useless. (Note that a sample size that is too big will make a confidence interval so narrow that it is almost useless as well—see, for example, Exercise 5.15.) 5.18. If the sample size had been one-fourth as large, the confidence interval would be twice as wide and would be 0.23 to 0.31. 5.19. (a) 2.776 (b) 2.145 (c) 2.064 (d) 2.060 (e) 2.787 5.20. (a) The 95% confidence interval is y  t

s 10  70  2.776  57.6 to 82.4 . We are n 5

95% confident that the interval 57.6 to 82.4 contains the population mean. (b) The 95% confidence interval is y  t

s 10  70  2.093  65.3 to 74.7 . We are 95% confident that n 20

the interval 65.3 to 74.7 contains the population mean.

5.21. (a) The standard error is

s 52.554   1.656 . (b) We are 95% confident that the interval n 1007

21.5 to 28.0 contains the population mean number of female partners males have had sex with since their eighteenth birthday. (c) The mean is quite high compared to the median and the mode, which means that there were a few male respondents with a very large number of female sex partners. In addition, the standard deviation is more than twice the mean, confirming the right skew of the distribution of the number of female sex partners. A confidence interval based on the mean does not seem to be the best idea.

5.22. (a) The point estimate is 3.02 children. (b) The standard error is

s 1.81   0.081 . (c) n 497

We are 95% confident that the interval 2.9 to 3.2 contains the population mean ideal number of

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children for a family to have. (d) Since the confidence interval is entirely above 2.0, it does not seem plausible that the population mean equals 2.0 children.

5.23. (a) The standard error is

y t

s 1.77   0.089 . (b) The 95% confidence interval is n 397

s 1.77  2.89  1.96  2.7 to 3.1. We are 95% confident that the interval 2.7 to 3.1 n 397

contains the population mean ideal number of children that males think a family should have. The ―95% confidence‖ means that we have constructed this confidence interval in such a way that 95% of the 95% confidence intervals would contain the true population mean. 5.24. (a)

y

11  11  6  9  14  3  0  7  22  5  4  13  13  9  4  6  11 124   7.29 ; 17 17

s

 y  y  n 1

2

 51.596  7.18 . (b) The standard error is

s 7.18   1.74 . (c) The tn 17

score that is in the df = 16 row and t0.025 column is 2.120. (d) The 95% confidence interval is

y t

s 7.18  7.29  2.120  3.6 to 11.0 . We are 95% confident that the interval 3.6 to 11.0 n 17

pounds contains the population mean change in weight for this therapy. 5.25. A confidence is not about any one subject or about 95% of the subjects, it is an interval estimate for our population parameter. The correct interpretation is that we are 95% confident that the interval 2.60 to 2.93 hours is the population mean number of hours of TV watched on the average day. 5.26. (a) y  1.93 , s = 1.53, se =

1.53  0.396 . (b) The 95% confidence interval is 15

1.93  2.145  0.396  1.1 to 2.8 . We are 95% confident that the interval 1.1 to 2.8 hours contains the population mean daily hours watching TV spent by Buddhists. (c) Since the standard deviation is almost as large as the mean, it appears that the population distribution is not normal. The t procedures are robust against violations of normality, but we should still be cautious about the validity of the results since the sample size is so small. 5.27. (a) The population distribution is probably not normal, since the standard deviation is almost equal to the mean (which implies that the population distribution is skewed to the right). (b) Because our sample size is so large, the t procedures will be robust against this violation of normality. The 99% confidence interval is 20.3  2.576

18.2  19.1 to 21.5 . We are 99% 1415

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confident that the interval 19.1 to 21.5 contains the population mean length that residents have lived in the city, town, or community where they live now. 5.28. (a) The 95% confidence interval is 1.81  1.96

1.98  1.67 to 1.95 . We are 95% 816

confident that the interval 1.67 to 1.95 contains the population mean number of days in the past 7 days that women have felt sad. (b) Since the standard deviation is larger than the mean, the variable is most likely skewed to the right. Since t procedures are robust against violations of normality and our sample size is large, our findings in part (a) are probably okay, unless there are extreme outliers. 5.29. (a) We are 95% confident that the interval 1.09 to 1.18 contains the population mean number of sex partners that people had in the previous 12 months. (b) The standard deviation is almost as large as the mean, indicating that the distribution is probably skewed to the right. Since t procedures are robust against violations of normality and our sample size is large, our findings in part (a) are probably okay, unless there are extreme outliers. 5.30. (a) We are 95% confident that the interval 3.32 to 4.88 contains the mean number of times a week University of Florida students read a newspaper. (b) Since the standard deviation is almost as large as the mean, the distribution is probably skewed to the right. (c) ―Robust‖ means that the t procedures are not affected by violations against normality where other procedures may be. This means that our calculations and interpretations are probably okay, unless there are extreme outliers. 5.31. (a) The 99% confidence interval is 4.23  2.576

1.39  4.13 to 4.33 . (b) (i) A 95% 1294

confidence interval would be narrower, since we do not need to include as many possible values in the confidence interval when we are less confident. (ii) A 99% confidence interval for only the strong Democrats would be wider, since the standard deviation is larger and the sample size is smaller (thus causing the standard error to be larger). (c) When we use the sample mean and standard deviation, we are assuming that the measurement scale for political ideology is interval scale. 5.32. (a) y = 1.5 days. (b) The 95% confidence interval is 1.5  1.96

2.21 = 1.4 to 1.6. We are 1450

95% confident that the interval 1.4 to 1.6 contains the population mean number of days in the past 7 days that people have felt lonely. 5.33. (a) Based on the stem-and-leaf plot, the population distribution might be skewed to the right.

33 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

5|69 6|04 7|003789 8|33456 9|022346 10|008 11|12 12|024 13|9 (b) y  90 , s = 20.67 (c) The 95% confidence interval is 90  2.045

20.67  82.3 to 97.7 . We 30

are 95% confident that the interval $8230 to $9770 contains the population mean annual family income for families living in public housing in Chicago. 5.34. (a) The confidence interval is 4.3 to 6.3. We are 95% confident that the interval 4.3 to 6.3 days contains the population mean length of stay for all inpatients in that hospital. (b) If the administrator wants the confidence interval to be half as wide, she needs to take a random sample of 400 records. 2

2

 z   1.645  5.35. n   1      0.30  0.70     156.89 The necessary sample size is 157. M   0.06  2

2

 z   1.96  5.36. (a) n   1      0.50  0.50     600.25 The necessary sample size is M   0.04  600. (b) To have a margin of error of 0.02, they should sample 4(600.25) = 2401 people. 2

2

 z   1.96  5.37. (a) n   1       0.10  0.90    864.36 , or 864. (b) Using   0.50 , n = M   0.02  2401. This sample size is a little less than three times the sample size in part (a). If we can make an educated guess about the value of π, we can collect a smaller sample size; the closer π is to 0 or 1, the smaller the sample size needed. 2

 1.96    1534.2 The sample size was about 1534.  0.025 

5.38. n  0.48  0.52 

2

 1.96  5.39. n  0.83  0.17     602.3 The sample size was about 602.  0.03 

34 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

2

2

 z   1.96   5002     96.04 The sample size should be about 96 farms. (b) M   100  s 300  1.96  60.0 . The margin of error is actually z n 96

5.40. (a) n   2 

5.41. We estimate the standard deviation to be (18 – 0)/6 = 3. The sample size calculation is 2

2

 z   1.96  n      32    34.57 , so a sample of size 35 is needed. M   1  2

5.42. We estimate the standard deviation to be (50,000 – 6,000)/4 = 11,000. The sample size 2

2

 z   2.576  calculation is n      11,0002    802.9 , so a sample of size 803 is needed. M   1000  2

5.43. We cannot use the ordinary large-sample formula because 3 < 15. An appropriate confidence

0.147  1.96

interval

uses

ˆ  5 34  0.147 and the 95% confidence interval is

0.147  0.853  0.028 to 0.266 . We are 95% confident that the interval 0.028 34

to 0.266 contains the population proportion of children who died before reaching adulthood. 5.44. (a) ˆ  0 5  0 , se 

ˆ 1  ˆ  n



0 1  0 . (b) Since the number in each category (0 5

like tofu and 5 do not like tofu) is less than 15, we cannot use the large-sample formula for a confidence interval. An appropriate confidence interval uses ˆ  2 9  0.222 and the 95% confidence interval is 0.222  1.96

0.222  0.778  0.0 to 0.49 . We are 95% confident that 9

the interval 0 to 0.49 contains the population proportion of students who like tofu. 5.45. For n = 30, the endpoints of a 95% confidence interval have indices

n 1 30  1  n  30  15.5  5.5  10 to 21. 2 2 The confidence interval consists of the 10th smallest and 21st smallest values. The 95% confidence interval is 79 to 96. We are 95% confident that the interval $7900 to $9600 contains the population median income of the public housing residents. 5.46. For n = 54, the endpoints of a 95% confidence interval have indices

n 1 54  1  n  54  27.5  7.3  20.2 to 34.8 . 2 2

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The confidence interval consists of the 20th smallest and 35th smallest values. The 95% confidence interval is 2 to 6. We are 95% confident that the interval 2 to 6 years contains the population median time since a book was last checked out. 5.49. (a) SPSS gives us the following output: Statistic Mean

7.267

95% Confidence Interval for Mean

Lower Bound Upper Bound

Std. Error .8672

5.531 9.002

We are 95% confident that the mean weekly number of hours spent watching TV is between 5.5 and 9.0 hours. (b) SPSS gives us the following output: Statistic AfterLife

Proportion

0.52

95% Confidence Interval for Proportion

Lower Bound

Std. Error 0.065

0.39

Upper Bound

0.65

We are 95% confident that the proportion of students who believe in life after death is between 0.39 and 0.65. 5.52. The report should include the following elements. The explanatory variable is political ideology, and the response variable is whether a person lived with her/his husband/wife before they got married. About 33% of politically liberal respondents lived with their spouse prior to marriage, compared to about 16% of politically conservative respondents. A 95% confidence interval for politically liberal respondents is 26.1% to 40.2%. A 95% confidence interval for politically conservative respondents is 11.6% to 20.1%. Note that the confidence interval for politically conservative respondents is entirely below the confidence interval for politically liberal respondents. 5.53. The report should include the following elements. About 15% of subjects agree that women should take care of running their homes and leave running the country up to men (the 95% confidence interval is 13.4% to 16.7%). About 34% of subjects agree that it is better for everyone involved if the man is the achiever outside the home and the woman takes care of the home and the family (the 95% confidence interval is 32% to 36.3%). About 42.4% of subjects agree that a preschool child is likely to suffer if her mother works (the 95% confidence interval is 40.1% to 44.7%). 5.54. A 95% confidence interval for the population mean is 0 to 30.4 (actually, the lower bound is –10.43, but we report this as 0). Outliers have a tremendous impact on confidence intervals for means, since they affect both the mean and the standard deviation.

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5.55. (a) An estimator is unbiased if its sampling distribution centers around the parameter it is estimating (i.e., the parameter is the mean of the distribution). (b) The sample range will always be biased, since it is improbable that our sample would include the minimum and maximum of the population. For this reason, the sample range will typically be less than the population range. The largest that the sample range could be is the population range, which only occurs when both the minimum and maximum value in the population are included in the sample. 5.56. We form a confidence interval for a parameter because it gives us a range of values that we believe to be plausible for the parameter. A point estimate is only one number and can vary greatly from sample to sample. If there is a particular value that we think the parameter might be, we can check to see if it is inside or outside of the confidence interval (for example, will a majority of people vote for a particular candidate?). 5.57. (a) The sample mean is always the midpoint of the confidence interval. In this case, the sample mean is 4.8 dates. If we only knew that the mean is 4.8 dates and did not know the sample standard deviation or the sample size, we could not construct a confidence interval for the population mean. (b) The sample mean is 4.8 dates. The sample standard deviation can be found by solving M  t

s . Here M = 0.8, n = 50, and t = 2.010. Thus, the sample standard deviation n

is 2.8 dates. 5.58. (a) Larger confidence levels involve larger t- and z-scores, which make the margin of error wider and the confidence interval itself wider. To be more confident about our estimate, the confidence interval must include more values that are plausible for the population parameter. (b) Smaller sample sizes make the standard error larger, the margin of error larger, and the confidence interval wider. 5.59. (a) To have a 99.9999% confidence interval, we would have to go approximately 3.9 standard errors on either side of the sample statistic to get our confidence interval. The confidence interval is about 1.5 times as wide as a 99% confidence interval but does not add much confidence. It is not worth the reduced precision for the added confidence. (b) A 25% confidence interval tells us that we used a process that would capture the population parameter about 25% of the time. The accuracy of this process is not high enough for us to use a 25% confidence interval in practice. 5.61. We need a larger sample size to offset the variability in a population that is heterogeneous. Examples will vary. 5.63. It does not make sense to construct a confidence interval since we have the population parameter. The sample and the population are the same, and π = 0.05.

37 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

5.64. (a) 1.96 error is z

0.40  0.60 0.60  0.40  1.96  0.030 , or 3.0 points. (b) Since the margin of 1000 1000

ˆ 1  ˆ  n

, it will change as the actual percentages, ˆ ’s change. (c) Multiplication

does not depend on order. Thus, as seen in part (a), the margin of error is the same for 40% and 60%. (d) It is more difficult to estimate a population proportion when it is near 0.5 since the confidence limits are much wider (because the standard error is larger) than they are when the population proportion is near 0 or 1. 5.65. (a) 30(0.50) = 15 (b) 50(0.30) = 15 (c) 150(0.10) = 15 5.66. (a) 5.67. (a) 5.68. (b) 5.69. (b) and (e) are correct 5.70. (a) A confidence interval for the mean is about the population mean, not the sample mean. (b) A confidence interval for the mean is about the population mean, not individuals. (c) We can actually be 100% confident that the sample mean is in the interval we construct (it is the midpoint of the interval). (d) This statement implies that the population mean changes. 5.71. We are 95% confident that the interval 21.5 to 23.0 years contains the mean age at first marriage of women in a certain country.

5.72. When π = 0.50, 2se  2

0.50  0.50 1  . Since π = 0.50 will give the largest margin of n n

1 is the largest value of the standard error. Since the margin of error M in estimating a n 1 1 1 1 population proportion is no larger than , we set M  and find M 2   n  2 . n M n n error,

5.73. Since y 

 y ,  y  ny . If we know  n 1 of the observations, we can add these up

n and subtract from ny to find the value of the remaining observation.

38 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

  0 , the true se 

5.74. For

0 1 1 0  0 . For   1, the true se   0 . If the n n

population proportion is either 0 or 1, then necessarily the sample proportion is exactly the same. 5.75. ˆ  0 This estimate does not seem sensible, especially since our sample size is so small. 5.76. (a) If one flipped a coin twice, the possible responses would be (H,H), (H,T), (T,H), (T,T), each with probability 1/4. Thus, the possible outcomes (H,H) and (H,T) each have probability 1/4. Of the half who flip a tail the first time, the probability equals π of reporting heads for the second flip (since this is the probability of yes) and (1 – π) of reporting tails. Since this happens for half the population, the overall probabilities are π/2 and (1 – π)/2. (b) The expected proportion of heads for the second response equals 0.25 + π/2. If we set p  0.25   2 and solve for π, we get the estimate ˆ  2 p  0.5 . (c) (i) p = 50/200 = 0.25, so ˆ = 2(0.25) – 0.50 = 0; (ii) p = 70/200 = 0.35, so ˆ = 2(0.35) – 0.50 = 0.20; (iii) p = 100/200 = 0.50, so ˆ = 2(0.50) – 0.50 = 0.50; (iv) p = 150/200 = 0.75, so ˆ = 2(0.75) – 0.50 = 1.

5.77. Given

ˆ  0 and n = 20, 0    1.96

 1    20

. Squaring both sides gives us

1.962 1.962 2    0.19208  0.19208 2 . This equation simplifies to 20 20 1.19208 2  0.19208   1.19208  0.19208  0 . The roots that solve this equation are

2 

  0 and   0.161. Chapter 6 6.1. (a) null hypothesis (b) alternative hypothesis (c) alternative hypothesis (d) H0 :   0.50 ;

Ha :   0.24 ; Ha :   100 .

6.2. (a) Let µ = the ideal number of children. H0 :   2 and Ha :   2 . (b) The test statistic is t

t

=

20.80.

This

test

statistic

was

obtained

with

the

following:

y  0 2.49  2   20.80 . (c) The P-value is the probability, assuming the null se 0.850 1302

hypothesis is true, that the test statistic equals the observed value or a value even more extreme in the direction predicted by the alternative hypothesis. In this case, the P-value = 0, which means that if the true mean were 2, we would see results as extreme as or more extreme than we did almost never.

39 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

6.3. (a) P  2P t  1.04  2P  z  1.04  2  0.1492  0.2984 . If the true mean were 0, we would see results are extreme as we did about 30% of the time. (b) Now,

P  2P t  2.50  2P  z  2.50  2  0.0062  0.0124 . Since the P-value is smaller than

the P-value calculated in part (a), we have stronger evidence against the null hypothesis. (c) (i) 0.1492 (ii) 0.8508. 6.4. (a) (i) t = ±2.064 (ii) t = 1.711 (iii) t = -1.711. (b) This P-value provides weaker evidence against the null hypothesis than P = 0.01, since it is larger than P = 0.01. 6.5. (a) t 

y  0 103 100   1.5 ; P  2P  t  1.5  2P  z  1.5  2 0.0668  0.1336. se 40 400

If the mean were 100, we would see results as extreme as we did about 13.4% of the time. (b)

t

y  0 103 100   3.0 ; se 40 1600

P  2P  t  3.0  2P  z  3.0  2 0.00135  0.0027 .

When n is larger, the test statistic is more extreme, and the P-value is smaller. Keeping everything else constant, a larger sample size will provide more evidence against the null hypothesis than will a smaller sample size. 6.6. ―Upper‖ = 11.0; ―t-value‖ = 4.19; ―df‖ = 16. 6.7. (a) Assumptions: The sample is randomly selected. Income is a quantitative variable. We do not know if the population of incomes is normally distributed, but the t test is robust against violations of normality, especially for two-sided alternatives. Hypotheses: Let µ denote the population mean weekly income for female employees. The null hypothesis is H0 :   $500 , and the alternative hypothesis is Ha :   $500 . Test statistic: We are given y  410 and s = 90. The estimated standard error of the sampling distribution of y is se 

s 90   30 . The value of the test statistic is n 9 y  0 410  500 t   3.0 . se 30

The sample mean falls 3.0 standard errors below the null hypothesis value of the mean. The df value is 9 – 1 = 8. P-value: The P-value is the two-tail probability, presuming H0 is true, that t would exceed 3.0 in absolute value. From the t distribution with df = 8, this two-tail probability is between 0.01 and 0.02 (the calculator value is 0.017). If the population mean weekly income for female employees were $500, then the probability equals 0.017 that a sample mean of n = 9 subjects would fall at least as far from $500 as the observed y value of $410. Conclusion: The P-value of P = 0.017 is small, so it contradicts H0 . If H0 were true, the data we observed would be unusual. It appears that the population mean weekly income for female employees differs from $500. 40 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

(b) The P-value for Ha :   $500 is between 0.005 and 0.01 (the calculator value is 0.009). If the population mean weekly income for female employees were $500, then the probability equals 0.009 that a sample mean of n = 9 subjects would fall at least as far below $500 as the observed y value of $410. (c) The P-value for Ha :   $500 is between 0.99 and 0.995 (the calculator value is 0.991). If the population mean weekly income for female employees were $500, then the probability equals 0.009 that a sample mean of n = 9 subjects would fall at $410 or above. 6.8. (a) Assumptions: The sample is randomly selected. Discharge is a quantitative variable. We do not know if the population of discharge is normally distributed, but the t test is robust against violations of normality. Hypotheses: Let µ denote the population discharge of waste water (in gallons per hour). The null hypothesis is H0 :   500 , and the alternative hypothesis is Ha :   500 . Test statistic: We are given y  1000 and s = 400. We are also given the estimated standard error of the sampling distribution of y : se = 200. The value of the test statistic is

t

y  0 1000  500   2.5 . se 200

The sample mean falls 2.5 standard errors above the null hypothesis value of the mean. The df value is 4 – 1 = 3. P-value: The P-value is the one-tail probability, presuming H0 is true, that t would exceed 2.5. From the t distribution with df = 3, this one-tail probability is between 0.025 and 0.05 (the calculator value is 0.044). If the population mean discharge of waste water were 500 gallons per hour, then the probability equals 0.044 that a sample mean of n = 3 hours would fall at least as far above 500 as the observed y value of 1000. Conclusion: The P-value of P = 0.044 is small, so it contradicts H0 . If H0 were true, the data we observed would be somewhat unusual. It appears that the population mean discharge of waste water is greater than 500 gallons per hour. (b) The t procedures are robust against violations of normality especially when the sample size is large and when we are testing a two-sided alternative hypothesis. In this case, the sample size is very small and we have a one-sided alternative hypothesis, so we should be skeptical about making sweeping decisions based on the findings. (c) Since the null hypothesis includes all of the values not in the alternative hypothesis, the alternative hypothesis of Ha :   500 implicitly tests the broader null hypothesis that

  500 .

6.9. (a) The null hypothesis is H0 :   0 , and the alternative hypothesis is Ha :   0 . (b)

t

y  0 0.052  0   1.310 . The P-value is P = 0.190. Since P > 0.05, we fail to reject se 0.0397

the null hypothesis. There is not enough evidence to conclude that the mean score differs from the neutral value of zero. (c) We cannot accept H0 :   0 , since failing to reject the null hypothesis does not prove the null hypothesis to be true. We just did not have enough evidence to reject the null hypothesis. (d) A 95% confidence interval is 41 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

y t

s  0.052  1.96  0.0397   0.13 to 0.03 . Note that zero falls within our confidence n

interval. When the null hypothesis value falls within the confidence interval, we fail to reject the null hypothesis; when the null hypothesis value falls outside of the confidence interval, we reject the null hypothesis. 6.10. (a) The sample mean would shift down by 4, from 4.075 to 0.075. The sample standard deviation would remain the same at 1.512. (b) The test statistic would remain the same at 0.68. Finally, the P-value would remain the same at P = 0.50. We would have the same interpretation that we did in Example 6.2. 6.11. Results of 99% confidence intervals for means are consistent with results of two-sided tests at α = 1 – 0.99 = 0.01 level. 6.12. (a) P = 0.1492 (b) P = 0.2983 (c) P = 0.8508 (d) None of the P-values in (a), (b), or (c) give strong evidence against H0 , since all exceed any reasonable alpha level. 6.13. (a) z 

ˆ   0 se0



0.35  0.50  3.0 (b) P  P  z  3.0  0.00135 If the population 0.50  0.50 100

proportion were 0.50, then the probability equals 0.00135 that a sample proportion of n = 100 subjects would fall at 0.35 or below. (c) Since P = 0.00135 < 0.05, we reject the null hypothesis. There is enough evidence to conclude that the population proportion is less than 0.50. (d) If the decision in (c) was in error, this would be a Type I error. To reduce a type I error, we could reduce the significance level.

6.14. (a) se0 

0.50  0.50 ˆ   0 0.55  0.50  0.0158 (b) z    3.16 The sample 1000 se0 0.0158

proportion falls 3.16 standard errors in absolute value away from the null hypothesis value of the proportion. (c) P  2P  z  3.16  2  0.0008  0.0016 If the population proportion were

0.50, then the probability equals 0.0016 that a sample proportion of n = 1000 subjects would fall as extreme as or more extreme than 0.55. We reject the null hypothesis. There is enough evidence to conclude that the population proportion of Canadian adults who believe the bill should stand differs from 0.50. 6.15. (a) H0 :   0.50 and Ha :   0.50 . (b) z = –3.91 The sample proportion falls 3.91 standard errors below the null hypothesis value of the proportion. (c) P = 0.000 If the null hypothesis were true, then there is almost no chance that we would see results as extreme as or more extreme than our sample results. There is enough evidence to conclude that the proportion of Americans who would be willing to pay higher taxes in order to protect the environment differs from 0.50. (d) The confidence interval is entirely below 0.50, which leads us to believe 42 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

that a minority of Americans would be willing to pay higher taxes in order to protect the environment. 6.16. The test statistic is z 

0.60  0.50  6.93 . The P-value is P = 0.000. There is enough 0.50  0.50 1201

evidence to conclude that the proportion of adults who think affirmative action programs designed to increase the number of black and minority students on college campuses are good differs from 0.50.

6.17.

ˆ 

0.345 1 3 40  0.26 . The P-value is  0.345 The test statistic is z  116 1 3 2 3 116

P  P  z  0.26  0.3974 . We fail to reject the null hypothesis at α = 0.05 level. There is not enough evidence to conclude that the astrologers were correct with their predictions more than 1/3 of the time. 6.18. (a) A Type I error would be concluding that the astrologers were correct with their predictions more than 1/3 of the time when they really were not. (b) A Type II error would be failing to conclude that the astrologers were correct with their predictions more than 1/3 of the time when they really were. 6.19. (a) H0 :   0.50 and Ha :   0.50 ; z 

0.575  0.50  3.0 ; P = 0.00135; reject the 0.50  0.50 400

null hypothesis. There is enough evidence to conclude that a majority of the population would vote for the candidate. (b) Here z 

0.575  0.50  0.95 and P = 0.1711; fail to reject the null 0.50  0.50 40

hypothesis. There is not enough evidence to conclude that a majority of the population would vote for the candidate. 6.20. ˆ = 0/300 = 0.0, se0 

0.06  0.94  0.0137 , z = (0 – 0.06)/0.0137 = –4.38, P < 300

0.0001. If the sentences in the document could be regards as a random sample of sentences written by the author, it is very unlikely that no sentences would begin with whereas. There is very strong evidence against H0 , that is, against the hypothesis that the author was Levine.

43 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

H0 :   0.25

and

Ha :   0.25 .

(b)

ˆ

6.21.

(a)

=

125/400

=

0.3125,

se0 

0.25  0.75  0.0217 , z = (0.3125 – 0.25)/0.0217 = 2.89, P = 0.0019. We reject the null 400

hypothesis. There is enough evidence to conclude that more people answer the question correctly than would be expected just due to chance. 6.22. (a) (i) A Type I error would be concluding that the mean weight change was positive when it was not. (ii) A Type II error would be failing to conclude that the mean weight change was positive when it really was. (b) This would be a Type I error. (c) If α = 0.01, we would fail to reject the null hypothesis. If this decision were in error, it would be a Type II error.

y  0 519.5  500   1.95 , P  2P t  1.95  2P  z  1.95  0.0512 ; se 10.0 519.7  500 Smith: t   1.97 , P  2P t  1.97  2P  z  1.97  0.0488 . (b) Jones’s result 10.0

6.23. (a) Jones: t 

is not statistically significant, but Smith’s result is. (c) These two studies give such similar results that they should not yield different conclusions. Reporting the actual P-value shows that each study has moderate evidence against H0 and shows that the results are very similar in practical terms.

0.50  0.50  0.025 ; 400 ˆ   0 0.55  0.50 Jones: z    2.0 , P  2P  z  2.0  0.0456 ; se0 0.025 ˆ   0 0.5475  0.50 Smith: z    1.90 , P  2P  z  1.90  0.0574 . (b) Jones’s result is se0 0.025 6.24. (a) se0 

not statistically significant, but Smith’s result is. (c) These two studies give such similar results that they should not yield different conclusions. Reporting the actual P-value shows that each study has moderate evidence against H0 and shows that the results are very similar in practical terms. (d) The two confidence intervals almost entirely overlap, showing that the results of Jones and Smith are very similar. 6.25. t 

y  0 497  500   3.0 , P  2P t  3.0  2P  z  3.0  0.0027 Since the se 100 10,000

P-value is so small, we reject the null hypothesis and find the results to be highly statistically significant. However, the difference between the sample mean of 497 and the hypothesized mean of 500 is not important in a practical sense.

44 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

6.26. With large sample sizes, it is not unusual to find that small results (e.g., the difference between 3.51 for females and 3.55 for males) are statistically significant. However, these results are certainly not different enough from each other in a practical sense. 6.27. Given that H0 is rejected, P(Type I error) = 40/(40 + 140) = 40/180 = 0.22. 6.28. (a) If the true mean were 5, the probability that we would fail to reject H0 :   0 is 0.17. (b) If the test used α = 0.01, the Type II error would increase. When we have a smaller significance level, the evidence must be more extreme to reject H0 , and the rejection region is smaller, having smaller probability. (c) If µ = 10, the probability of a Type II error would be less than 0.17. The farther the true value is from H0 , in the direction of H a , the greater the chance that the test statistic falls in the rejection region. 6.29. (a) H0 is rejected when P ≤ 0.05, which happens when z ≥ 1.64, that is when the sample proportion falls at least 1.64 standard errors above the null hypothesis value of 0.50. This region is ˆ  0.50  1.64

0.50  0.50  0.664 . (b) z = (0.664 – 0.60)/0.10 = 0.64; We fail to reject if 25

ˆ  0.664 , which happens with probability 0.74. 6.30. (a) H0 is rejected when P ≤ 0.05, which happens when t ≥ 1.699, that is when the sample mean falls at least 1.699 standard errors above the null hypothesis value of 0. This region is

5.583 10 18  1.344 ; We fail to reject if y  5.583 , which  5.583 . t  18 30 30 5.6 10  1.339 ; We fail happens with probability 0.095, so P(Type II error) = 0.095. (b) t  18 30 to reject if y  5.6 , which happens with probability 0.095, so P(Type II error) = 0.095. (c) y  0  1.699

P(Type II error) = P(t < –1.33) = 0.097. (d) P(Type II error) = 0.10. 6.31. (a) t 

5.583  5  0.177 ; We fail to reject if y  5.583 , which happens with probability 18 30

0.57, so P(Type II error) = 0.57. Type II error increases as the alternative parameter value gets closer to H0 . (b) H0 is rejected when P ≤ 0.01, which happens when t ≥ 2.462, that is when the sample mean falls at least 2.462 standard errors above the null hypothesis value of 0. This region is y  0  2.462

8.09 10 18  0.581 ; We fail to reject if y  8.09 , which  8.09 . t  18 30 30

happens with probability 0.28, so P(Type II error) = 0.28. Type II error increases as α decreases. (c) Type II error decreases as the sample size increases.

45 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

6.32. (a) Binomial probability of outcome 0 when n = 12, π = 0.53, and (1 – π) = 0.47 is

 0.4712  0.0001 .

(b) Under random sampling, the probability π that a particular person

selected is female equals 0.53. We have H0 :   0.53 and Ha :   0.53 . The P-value is the probability of the observed outcome or even less likely outcomes, which equals the probability that there are 0 women, which is 0.0001. 6.33. (a) Let π = probability she guesses correctly on a particular flip. We test H0 :   0.50 and

Ha :   0.50 . The null says she actually does no better than random guessing, whereas the alternative says she does better than that. (b) Find the right-tail probability for the binomial distribution with n = 5 and π = 0.50. That is, the P-value is P(4) + P(5) = 0.15625 + 0.03125 = 0.1875. This outcome is not unusual if she does not actually possess ESP. It can be explained by chance. We cannot reject H0 , and thus her claim is not convincing. 6.34. (a) We have a fixed number of observations (1336), each of which falls into one of two categories (voted for Clinton, did not vote for Clinton). The probability of falling in each category is the same for every observation. The outcomes of successive observations are independent (no

  n  1336  0.50  668 ;

person’s vote impacted any other person’s vote). (b)

  n 1    1336 0.50 0.50  18.28 . (c) We would almost certainly expect x to fall in the interval 613 to 723. (d) Since x = 895 exceeds the upper bound of the interval in part (c), it appears that the population proportion of voters who voted for Hillary Clinton in the 2006 Senatorial election in New York is greater than 0.50. 6.35. (a) We have a fixed number of observations (1 million), each of which falls into one of two categories (died in motor vehicle accident, did not die in motor vehicle accident). The probability of falling in each category is the same for every observation. The outcomes of successive observations are independent (the result of one motor vehicle accident does not impact the results of

another

motor

vehicle

  n  1,000,000  0.0001  100 ;

accident).

  n 1    1,000,000 0.00010.9999  10.0 (b) It would be very surprising if x = 0, since 0 is 10 standard deviations below the expected value. (c) We expect x to fall in the interval

80

to

120

with

probability

0.95.

(d)

  1,000,000  0.0002  200 ;

  1,000,000 0.0002 0.9998  14.1 The expected value for males is twice that for females, but the standard deviation is not quite 1.5 times the standard deviation for females. 6.37. (a) H0 :   4 , Ha :   4 SPSS output appears below: N PolIdeology

60

Mean 3.03

Std. Deviation 1.636

Std. Error Mean 0.211

46 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

Test Value = 4 95% Confidence Interval of the Difference t PolIdeology

df

-4.577

Sig. (2-tailed) 59

0.000

Mean Difference -0.967

Lower -1.39

Upper -0.54

There is enough evidence to conclude that the mean political ideology differs from 4.0. (b) H0 :   0.5 , Ha :   0.5 SPSS output appears below: N Abortion

60

Proportion 0.78

Std. Error 0.054

z = (0.78 – 0.50)/0.054 = 5.19; P < 0.0001. There is enough evidence to conclude that the proportion of students who favor legalizing abortion differs from 0.50. 6.39. The report should include y = 4, s = 2, H0 :   0 , Ha :   0 , t = (4 – 0)/1 = 4.0, P = 0.014. 6.40. With a P-value of 0.1985, we fail to reject the null hypothesis. There is not enough evidence to conclude that a majority of students prefer one drink over the other. 6.41. The report should include H0 :   0.422 , Ha :   0.422 , ˆ = 0.79, z = 12.06, P < 0.0001. There is strong evidence of possible bias, since results like these would rarely occur by chance alone. 6.43. y = 2.39, s = 6.45, H0 :   0 , Ha :   0 , t = 1.993, P = 0.028. If the observation 20.9 was misrecorded and should have been 2.9, the test statistic decreases from 2.22 to 1.99, and the P-value increases from 0.02 to 0.03. 6.44. This is a Type I error, and is controlled by the choice of α. To make the chance low, only reject H0 when P is a very small value, such as P ≤ 0.01 (i.e., set α = 0.01). 6.45. (a) A Type I error occurs when one convicts the defendant, when he or she is actually innocent; a Type II error occurs when one acquits the defendant even though he or she is actually guilty. Most people would consider convicting an innocent defendant to be more serious. (b) To decrease the chance of a Type I error, one gives the defendant additional rights and makes it more difficult to introduce evidence that may be inadmissible in some way; this makes it more likely that the defendant will not be convicted, hence the relatively more guilty parties will be incorrectly acquitted. (b) If this strategy is used, α = 0.000000001, which means that the jury would almost never find the defendant guilty. It would be almost impossible to convict a defendant, and the many guilty defendants would be acquitted. This strategy goes way too far in terms of ―beyond a reasonable doubt.‖ 47 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

6.46. Type I error: The diagnostic test is positive, indicating that the disease is present, even though the disease is actually absent. Type II error: The diagnostic test is negative, indicating that the disease is absent, even though the disease is actually present. 6.47. (a) ―Significant‖ means that the observed difference was unlikely to have occurred by chance alone. In this case, if there were no difference in religious attitudes over time, the probability of seeing the differences that we did would happen less than 5% of the time. (b) Had the authors provided the P-value, we would know the true probability of finding results as extreme as they did, assuming that there were no difference in religious attitudes over time. The P-value gives us the strength of the evidence against the null hypothesis instead of just a cut-anddry decision. 6.48. No, we cannot accept a null hypothesis. Though it is plausible that the voting rates are identical for men and women, a confidence interval would show that other values are also plausible. 6.49. Stating that P = 0.4173545 is problematic in that it seems more precise than it truly is (because the sampling distribution used to get the P-value is merely an approximation for the exact sampling distribution) and that it might confuse the reader. Stating that P = 0.42 is enough to get the sense of the P-value and does not pretend to be more accurate than it is. 6.50. If H0 is true every time, the expected number of times one would get P ≤ 0.05 just by chance is 60(0.05) = 3, which is the mean of the binomial distribution with n = 60 and π = 0.05; thus, the 3 cases could all simply be Type I errors. 6.51. (a) For each test, the probability equals 0.05 of falsely rejecting H0 and committing a Type I error. This policy encourages the publishing of Type I errors. (b) Of all the studies conducted, the one with the most extreme or unusual results is the one that gets substantial attention. That result may be an unusual sample, with the sample far from the actual population mean. Further studies in later research would reveal that the true mean is not so extreme. 6.52. (a) 6.53. (b) and (e) are both correct 6.54. (b) 6.55. (a) and (c) are both correct 6.56. (b) and (d) are both correct

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6.57. (a) False; P(Type I error) is a single value, the fixed α value (such as 0.05), whereas P(Type II error) decreases as the true parameter value falls farther from the H0 value in the direction of values in H a . (b) True; If we reject using α = 0.01, then P ≤ 0.01. Thus, P ≤ 0.05 also, so we also reject using α = 0.05. (c) False; The P-value is the probability of obtaining a sample statistic as extreme as or more extreme than we did, given H0 is true. (d) True; Since P = 0.063 is greater than 0.05, we fail to reject H0 at that level and do not conclude that µ differs from 0. Similarly, the 95% confidence interval would contain 0, indicating that 0 is plausible for µ. The confidence interval shows precisely which values are plausible. 6.58. With a one-sided alternative hypothesis, values far from the hypothesized parameter in only one direction are considered unusual. With a two-sided alternative hypothesis, values far from the hypothesized parameter in both directions are considered unusual. The P-value for a two-sided alternative hypothesis is twice that for a one-sided alternative hypothesis (when the one-sided Pvalue is less than 0.50). 6.59. The value in H0 is only one of many plausible values for the parameter. A confidence interval displays a range of possible values for the parameter. The terminology ―accept H0 ‖ makes it seem as if the null value is the only plausible one. 6.60. (a) The null hypothesis is a statement about the parameter that we believe to be true. In this case, the null hypothesis would be that 50% of college students feel that the legal age for drinking alcohol should be reduced. The alternative hypothesis is the research statement that we are trying to garner evidence for. In this case, the alternative hypothesis would be that more than 50% of college students feel that the legal age for drinking alcohol should be reduced. (b) The P-value is the probability of seeing results as extreme as we did, assuming that 50% of college students feel that the legal age for drinking alcohol should be reduced. (c) The α-level is a set level below which we would reject the null hypothesis. The most typical α-level is 0.05. This means that the probability of seeing results as extreme as ours would have to be less than 0.05 to reject the null hypothesis. (d) A Type II error is failing to reject a false null hypothesis. In this case, a Type II error would be concluding that 50% of college students feel that the legal age for drinking alcohol should be reduced when, in reality, more than 50% of college students feel this way. 6.61. (a) H0 either is, or is not, correct. It is not a variable, so one cannot phrase probability statements about it. (b) If H0 is true, the probability that y ≥ 120 or that y ≤ 80 (i.e., that y is

0 = 100, so that |z| is at least as large as observed) is 0.057. (c) This is true if the statement ―   100 ‖ is substituted for ―   100 .‖ (d) The probability of a Type I error at least 20 from

equals α (which is not specified here), not the P-value. The P-value is compared to α in determining whether one can reject H0 . (e) It would be better to say ―We do not reject H0 at the α = 0.05 level.‖ (f) No, we need P ≤ 0.05 to be able to reject H0 .

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6.62. (a) If α equals P or any number larger than it, then P ≤ α, so one can reject H0 . If α equals 0.057 or any larger number (such as α = 0.10), then one can reject H0 . The P-value is the smallest level at which one can reject H0 , since if α is smaller than the P-value, then one cannot reject H0 . (b) For any α-level below 0.057, we cannot reject H0 ; thus, confidence intervals with error probability below 0.057, and hence confidence 0.943 or higher, would contain the null hypothesis value. The 94.3% interval is the narrowest of these. 6.63. (a) Binomial, n = 100, π = 0.05. (b) Yes. Since the binomial has µ = 100(0.05) = 5, rejecting H0 in 5 of the tests would not be unusual, since that is the expected value. 6.64. (a) The number that contain the parameter has the binomial distribution with n = 20 and π = 0.95. The probability that all intervals contain the mean is the probability that such a binomial variable equals 20, which is (0.95)20 = 0.36. (b) 1 – 0.36 = 0.64. 6.65. If 0 were in the category of interest, se 

ˆ 1  ˆ  n  0 , which means that the value of

the test statistic z would be ∞. For tests, we are assuming that the null hypothesis is true. Because of the relationship between the mean and standard deviation in the binomial setting, if we assume H0 :    0 , we should use the value  0 for both the mean and the standard deviation of the null distribution. Thus, we should use se0 

ˆ0 1  ˆ0  n for the hypothesis test.

6.66. If we fail to reject H0 when H a is actually true, we are committing a Type II error. As π gets closer to 0.50, it becomes more difficult to reject H0 , thereby increasing the probability of a Type II error. At the null hypothesis value, the probability of not rejecting the null is 0.95 (since the probability of rejecting the null is the probability of Type I error, which is 0.05), and as the parameter value increases from 0.50. 6.67. (a) Only X = 5 gives P ≤ 0.05 and leads to rejecting H0 . (b) No values give P ≤ 0.01. The smallest possible P-value, obtained with X = 5, is (0.50)5 = 0.031. (c) The probability of a Type I error equals the probability the P-value is less than or equal to 0.05. The value X = 5 is the only one that give this small a P-value. For discrete distributions, only some P-values can occur, so it is artificial to use values for α (such as 0.05) that P may not be able to equal.

Chapter 7 7.1. These are independent samples since the subjects in the two samples are different, with no matching between one sample with the other sample. 7.2. (a) (i) These results would be based on independent samples if the subjects in the two samples were different and no matching was done between the two samples. (ii) These results 50 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

would be based on dependent samples if the samples included either the same people or matching was done between the two samples. (b) The explanatory variable is year, and the response variable is the percent of European adults who expressed a positive attitude about President George W. Bush’s handling of international affairs. 7.3. (a) ˆ2  ˆ1  0.215  0.419  0.204 , which is an estimated 20.4% decrease. (b) The standard error of the difference is se 

 0.0202   0.0202  0.028 .

7.4. (a) ˆ2  ˆ1  0.52  0.14  0.38 . (b) se  7.5. (a) y2  y1  164 140  24 ; se 

 0.01572   0.1102  0.111.

 22   22  2.83 . (b) 164/140 = 1.17, which is a

17% increase. (c) y2  y1  191 166  25 ; There was an estimated 25 pound increase in the mean weight of adult American men between 1962 and 2002. 191/166 = 1.15; There was an estimated 15% increase in the mean weight of adult American men between 1962 and 2002. 7.6. (a) The response variable is median net worth in 2002. The explanatory variable is race (white, black). (b) (i) $89,000 - $6,000 = $83,000; There is an estimated difference of $83,000 in median net worth between white and black households in 2002. (ii) $89,000/$6,000 = 14.8; The median net worth in white households in 2002 was estimated to be 14.8 times the median net worth in black households in 2002. 7.7. (a) 832/58 = 14.3; The proportion of men incarcerated in the nation’s prisons was 14.3 times the proportion of women. (b) 0.00832 – 0.00058 = 0.00774; The difference in proportions is extremely small. (c) The relative risk is more informative, especially since the proportions are close to 0. 7.8. (a) 0.00164 – 0.00015 = 0.00149; This estimated difference of 0.001 seems very small. (b) 0.00164/0.00015 = 10.9; The probability that a black male is a homicide victim is 10.9 times the probability that a white male is a homicide victim, which is a very large effect. (c) The relative risk better summarizes results, especially when both proportions are very small. 7.9. (a) We are 95% confident that the interval 0.18 to 0.26 contains the difference between the proportion of teens who listen to lots of degrading sexual music and have intercourse and the proportion of teens who listen to little or no degrading music and have intercourse. (b) If the two proportions were equal, it would be very unlikely to observe a difference as large as we did. It appears that the proportion of teens who listen to lots of degrading sexual music and have intercourse is greater than the proportion of teens who listen to little or no degrading music and have intercourse. 7.10. (a) Since 0 is contained in the confidence interval, there may have been no change in support. (b) If there was a decrease in the proportion of approval, the decrease may have been as 51 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

much as 0.07 (7%). (c) If there was an increase in the proportion of approval, the increase was most likely small, only as much as 0.01 (1%). 7.11. (a)

se  ˆ1 1  ˆ1  n1  ˆ2 1  ˆ2  n2  0.399  0.601 12,708  0.482 0.518 8783  0.0069 (b)

ˆ2  ˆ1   z  se  0.482  0.399 1.96 0.0069  0.083  0.014  0.069 to 0.097

We are 95% confident that interval 0.069 to 0.097 contains the difference in the proportion of college students who drink to get drunk in 2001 and the proportion of college students who drink to get drunk in 1993. 7.12. (a) We have two independent random samples with at least 10 observations in each category for each group. Let 1 = the proportion of students in 1993 who said they engaged in unplanned sexual activities because of drinking alcohol and

 2 = the proportion of students in 2001 who

said they engaged in unplanned sexual activities because of drinking alcohol. The null hypothesis is H0 :  2  1  0 , and the alternative hypothesis is Ha :  2  1  0 . (b) If the true difference in proportions were really 0, the chance of seeing results as extreme as or more extreme than we did would happen about 0.02% of the time. In other words, these results are unlikely to have occurred by chance alone. (c) We are 95% confident that the true difference in proportions between 2001 and 1993 falls in the interval 0.009 to 0.033. The lower bound of this interval is very close to 0, and the upper bound is not strikingly different from 0 in a practical sense. Thus, while our hypothesis test results are statistically significant, they do not appear to be practically important. 7.13. se  0.55  0.45 1219  0.74 0.26  733  0.022 ;

0.74  0.55 1.96 0.022  0.147 to 0.233 We are 95% confident that the interval 0.147 to 0.233 contains the true difference in proportions that working women and men spend time cooking and washing up during a typical day. 7.14. (a) ˆ1 = 989/1503 = 0.66;

ˆ2 = 704/1968 = 0.36; ˆ1  ˆ2  0.66  0.36  0.30 ;

se  0.66 0.34  1503  0.36 0.64  1968  0.0163 . (b) 0.30 ±1.96(0.0163) = 0.27 to 0.33; We are 95% confident that the interval 0.27 to 0.33 contains the true difference in proportions who agree that ―it is much better if the man is the achiever outside the home and the woman takes care of the home and family‖ between 1977 and 2006. (c) The se would not change in the confidence interval. The sample proportions for each sample would be the respective proportions in part (a) subtracted from 1. Thus, the 95% confidence interval for comparing the proportions who did not agree in the two years would be 0.67 to 0.73.

1 = the proportion of males who agree with the statement and  2 = the proportion of females who agree with the statement. The null hypothesis is H0 :  2  1  0 , and the 7.15. (a) Let

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alternative hypothesis is

Ha :  2  1  0 . (b) ˆ2  ˆ1 = 0.352 – 0.372 = –0.020;

ˆ  153 166  411  472  319 883  0.36 ; se0  ˆ 1  ˆ  n1  ˆ 1  ˆ  n2  0.36  0.64 411  0.36 0.64  472  0.032 ;

z

ˆ2  ˆ1 se0



0.020  0.62 . (c) P = 0.54; If there were no difference between the proportion 0.032

of men who agree with the statement and the proportion of women who agree with the statement, we would see results as extreme as our sample results with probability 0.54. (d) Let 1 = the proportion of those having less education than a college degree who agree with the statement and  2 = the proportion of those having at least a college degree who agree with the statement. The null hypothesis is H0 :  2  1  0 , and the alternative hypothesis is Ha :  2  1  0 . ˆ2  ˆ1 = 0.256 – 0.40 = –0.144; ˆ  320 / 883  0.36 ; se0  0.032 ; z 

ˆ2  ˆ1 se0



0.144  4.50 . 0.032

P = 0.000007; It appears that opinion differed more between the most and least educated than between men and women.

1 = the proportion of female senior high school students who have ever used marijuana and  2 = the proportion of male senior high school students who have ever used marijuana. H0 :  2  1  0 , Ha :  2  1  0 . (b) We are 95% confident that the proportion of

7.16. (a) Let

male senior high school students who have used marijuana is between 0.1% and 8.9% higher than the proportion of female senior high school students who have used marijuana. (c) If the difference between the true proportions of female and male senior high school students who have ever used marijuana were actually 0, we would see results like these with probability 0.02. It appears that the proportion of male senior high school students who have ever used marijuana is greater than the proportion of female senior high school students who have ever used marijuana.

1 = the proportion of men judged to be compulsive buyers and  2 = the proportion of H0 :  2  1  0 , Ha :  2  1  0 . women judged to be compulsive buyers. ˆ1  44 / 800  0.055 ; ˆ2  90 /1501  0.060 ; ˆ  134 2301  0.058 ; se0  0.010 ;

7.17. Let

z

0.060  0.055  0.48 ; P = 0.63. It appears that there is no difference between the proportion 0.010

of men judged to be compulsive buyers and the proportion of women judged to be compulsive buyers. A 95% confidence interval for this difference is –0.015 to 0.025.

1 = the proportion of females who believe in an afterlife and  2 = the proportion of males who believe in an afterlife. H0 :  2  1  0 , Ha :  2  1  0 . ˆ1  435 / 582  0.747 ;

7.18. Let

ˆ2  375 / 509  0.737 ; ˆ  810 1091  0.742 ; se0  0.027 ; z 

0.742  0.737  0.40 ; P = 0.027

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0.69. It appears that there is no difference between the proportion of females who believe in an afterlife and the proportion of males who believe in an afterlife. If an error has been made with this test, it would be a Type II error (failing to reject the null hypothesis when it is actually false). 7.19. (a) We are 95% confident that the mean number of close friends for males is between 1.5 close friends below and 2/7 close friends above the mean number of close friends for males. (b) It does not appear that the distribution of the number of close friends is normal for either sex, since the standard deviations exceed their respective means (suggesting right-skewed distributions). Because the sample sizes are large, the t procedures will be robust against the violation of normality. 7.20. (a) Let

1 = the mean number of hours spent on housework per week for men and 2 = the

mean number of hours spent on housework per week for women. The estimated difference is

s12 s22   y2  y1 =12.8 – 8.4 = 4.4 hours. (b) se  n1 n2

9.52  11.62 292

391

 0.81 . (c) The 99%

confidence interval is  y2  y1   t  se = 4.4 ± 2.576(0.81) = 2.3 to 6.5. We are 99% confident that the mean number of hours spent on housework per week is between 2.3 hours and 6.5 hours greater for women than for men. 7.21. (a) The estimated difference between the HONC means for smokers and ex-smokers is 4.9. We are 95% confident that the HONC means for smokers is between 4.1 and 5.7 more than the HONC means for ex-smokers. (c) THE HONC sample data distribution for ex-smokers appears to be right-skewed, since the standard deviation is greater than the mean. If there are no outliers, the inference might not be affected too much, since t procedures are robust against violations of normality and the sample size is large. 7.22. (a) Let

1 = the mean total credit card balance for non-compulsive buyers and 2 = the

mean total credit card balance for compulsive buyers. y2  y1 = $3399 – $2837 = $562;

55952   63352

 $580.4 . H0 : 2  1  0 , Ha : 2  1  0 ; (b) 100 1682 y y 562 z 2 1   0.97 ; P = 0.33. It is plausible that there is no difference in the mean total se 580.4

se 

credit card balance for compulsive buyers and the mean total credit card balance for noncompulsive buyers. 7.23. We are 95% confident that the mean number of days in the past 7 days that women have felt sad is between 0.2 and 0.6 greater than the mean number of days in the past 7 days that men have felt sad. Both the confidence interval and the hypothesis test lead us to believe that there is a difference in the mean number of days in the past 7 days that women have felt sad and the mean

54 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

number of days in the past 7 days that men have felt sad. These results are statistically significant, but one needs to decide if the confidence interval indicates practical significance.

1 = the mean number of hours a day that females watch TV and 2 = the mean number of hours a day that males watch TV. y2  y1 = 2.86 – 2.99 = –0.13; se  0.10 . H0 : 2  1  0 , Ha : 2  1  0 ; z  1.30 ; P = 0.19. It appears that there is no difference

7.24. (a) Let

in the mean number of hours a day that females watch TV and the mean number of hours a day that males watch TV. (b) Since we failed to reject H0 : 2  1  0 at the α = 0.05 level, a 95% confidence interval comparing the means would contain 0. (d) The distribution of TV watching does not appear to be normal, since the standard deviations are almost as large as their respective means. Since the sample sizes are large and the t procedures are robust against violations of normality, the inferences are probably fine. 7.25 (a) We are 95% confident that the mean number of hours that blacks watch TV per day is between 0.77 and 2.23 hours more than the mean number of hours that whites watch TV per day. Since the entire confidence interval is above 0, it appears that the mean number of hours that blacks watch TV per day is higher than the mean number of hours that whites watch TV per day. (b) The P-value of 0.000 says that, if the means were equal, it would be almost impossible to observe results as extreme as the results that we have. (c) The result of the significance test and the result of the confidence interval both lead us to believe that the mean number of hours that blacks watch TV per day is higher than the mean number of hours that whites watch TV per day. 7.26. df = 29 – 1 = 28. The one-sided P-value is between 0.005 and 0.01 (P = 0.006), and the twosided P-value is between 0.01 and 0.02 (P = 0.012). It appears that those who are not adult children of alcoholics have a higher mean well-being than those who are adult children of alcoholics. 7.27. The estimated standard error is 49.4

12  14.3 , and the test statistic is t = 70.1/14.3 = 4.9

with df = 11, which has P < 0.01. 7.28. (a) We should treat these as dependent samples, since the same subjects are used in each sample. (b) We are 95% confidence that the mean number of times students went to movies was between 7.6 below and 15.6 above the mean number of times students went to sporting events. (c) t = 4.0/5.112 = 0.78. P = 0.454. It appears that there is no significant difference between the mean number of times students went to movies and the mean number of times students went to sporting events. 7.29. (a) If there was no difference between the mean number of times students went to parties was between and the mean number of times students went to sporting events, we would see results as extreme as ours with probability 0.106. There does not appear to be a difference between mean attendance at parties and mean attendance at sporting events. (b) Since 0 is in the

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confidence interval, we would fail to reject H0 and come to the same conclusion that we did in part (a).

 2102   2 8.72

1 1   7.64 . (b) The 3 3 4 90% confidence interval is  y2  y1   t  se  20  2.132  7.64  3.7 to 36.3 . We are 90%

7.30. (a) sA = 10, sB = 8.7, s 

 9.35 , se  9.4

confident that the mean improvement scores for patients receiving therapy B is between 3.7 and 36.3 points higher than the mean improvement scores for patients receiving therapy A. (c) The effect size is (40 – 20)/9.35 = 2.1. The difference between the sample means is more than 2 standard deviations, a relatively large difference. 7.31. (a) (i) 40 – 20 = 20 (ii) 20; The difference between the means is equal to the mean difference. (b)

s = 5.0;

se  5 3  2.887 . (c) The 95% confidence interval is

20  4.303 2.887  7.6 to 32.4 . We are 95% confident that the mean difference between patients receiving therapy B and patients receiving therapy A is between 7.6 and 32.4. (d) t = 20/2.887 = 6.93; df = 3 – 1 = 2; 0.02 < P < 0.05 (P = 0.0202). There appears to be a difference between the mean scores for patients receiving therapy B and those receiving therapy A.

7.32. (a) s 

12 2.12  16 3.22

 2.78 , se  2.78

1 1   1.025 . The 95% confidence 13 17

28 interval is 2.8  2.048 1.025  0.7 to 4.9 . We are 95% confident that the mean family cohesion for nonabused students is between 0.7 and 4.9 higher than the mean family cohesion for sexually abused students. (b) With P = 0.007, there appears to be a significant difference between the mean family cohesion for nonabused students and the mean family cohesion for sexually abused students. 7.33. The sample standard deviations are quite different, so we might not trust the results based on assuming equal variances. The approximate test without that assumption shows very strong evidence (P = 0.0066) that the means differ; the data suggest that vegetarians are more liberal, on average. 7.34. Let

1 = the mean number of hours a week that females spend on the WWW and 2 = the

mean number of hours a week that males spend on the WWW. y2  y1 = 6.2 – 4.9 = 1.3;

s

1568 8.62  1195 9.92

 9.18 , se  9.18

1 1   0.35 . H0 : 2  1  0 , 1569 1196

2763 Ha : 2  1  0 ; t = 3.69; P = 0.0002. It appears that there is a difference in the mean number of

hours a week that females spend on the WWW and the mean number of hours a week that males spend on the WWW.

56 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

7.35. (a) SPSS output gives the following: Course A

DropScore

N

B

Mean

Std. Deviation

5

2.00

1.225

0.548

5

6.00

1.871

0.837

t DropScore

Std. Error Mean

Df

Sig. (2-tailed)

Equal variances assumed

-4.000

8

0.004

Equal variances not assumed

-4.000

6.897

0.005

There is very strong evidence that the mean drop was higher for Course B. A 95% confidence interval for the difference in means is: (2.0 – 6.0) ± 2.306(1) = –6.3 to –1.7. We are 95% confident that the mean drop for Course B was between 1.7 and 6.3 points higher than the mean drop for Course A. (b) SPSS output for the two-sided Wilcoxon test is: Course A

DropScore

N

B Total

Mean Rank

Sum of Ranks

5

3.20

16.00

5

7.80

39.00

10

Test Statistics(b) DropScore Mann-Whitney U

1.000

Wilcoxon W

16.000

Z

-2.447

Asymp. Sig. (2-tailed)

.014

Exact Sig. [2*(1-tailed Sig.)]

.016(a)

a Not corrected for ties. b Grouping Variable: Course

P = 0.014; There is very strong evidence that the mean drop was higher for Course B. (c) The effect size is (6.0 – 2.0)/1.581 = 2.5. The difference between the sample means is about 2.5

standard deviations, a relatively large difference. (d) The effect size P  yB  yA   0.92 . The

estimated probability that a student in Course B has a larger drop in math phobia score than a student in Course A is 0.92. This is a relatively strong effect. 7.36. (a) Believe in heaven Yes No 833 2 Believe Yes in hell No 125 160

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1 = the proportion who believe in heaven and  2 = the proportion who believe in hell. ˆ1 = 958/1120 = 0.855; ˆ2 = 835/1120 = 0.746. (c) H0 :  2  1  0 , Ha :  2  1  0 .

(b) Let

McNemar z 

n12  n21 2 135   11.4 . P = 0.000. There is enough evidence to n12  n21 2  135

conclude that the proportion of people who believe in heaven differs from the proportion of ˆ1  ˆ2 people who believe in hell. (d) = 0.109;

se 

 n  n   n  n  n  n   2 135  2 135 1120  1120  0.0098 2

12

21

12

2

21

.

A

95% confidence interval is 0.109 ± 1.96(0.0098) = 0.09 to 0.13; We are 95% confident that the interval 0.09 to 0.13 contains the true difference between the proportion of people who believe in heaven and the proportion of people who believe in hell. 7.37. (a) Let 1 = the proportion who favor law enforcement spending and  2 = the proportion who favor health spending. ˆ1 = 306/340 = 0.90; ˆ2 = 319/340 = 0.94. (b) H0 :  2  1  0 ,

Ha :  2  1  0 . McNemar z 

25 14  1.76 . P = 0.078. There is not enough evidence to 25  14

conclude that the proportion of people who favor law enforcement spending differs from the = –0.04; proportion of people who favor health spending. (c) ˆ1  ˆ2

se 

 25 14  25 14 340  340  0.018 2

. A 95% confidence interval is –0.04 ±

1.96(0.018) = –0.08 to –0.01; We are 95% confident that the interval –0.08 to 0 contains the true difference in the proportion of people who favor law enforcement spending and the proportion of people who favor health spending.

138  217  4.2 , so the P-value is < 0.0001. The proportion living with 138  217 a child is significantly higher at the later time. ˆ1 = 0.162; ˆ2 = 0.185. The confidence interval is

7.38. McNemar z 

0.012 to 0.034. We infer that the proportion living with a child is between 0.012 and 0.034 higher after four years. 7.39. (a) Fisher’s exact test is used because of the small sample sizes and the fact that two cells have counts less than 5. (b) There is not enough evidence to conclude that bisexual/gay/lesbian orientation is more likely for children with lesbian mothers. 7.40. Using software, Fisher’s exact test has one-sided P-value – 0.022. There is strong evidence that the probability is higher for those raised with lesbian mothers.

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7.41. (a) The following is a bar chart:

Summary statistics for political ideology: pa

n Mean Std. Dev. Std. Err. Median Range Min Max Q1 Q3

Dem 21 Rep

15

2

0.84

0.18

2

3

1

4

2

2

5.13

1.36

0.35

5

5

2

7

5

6

(b) Software gives us the following for a hypothesis test: Hypothesis test results: μ1 : mean of Dem μ2 : mean of Rep H0 : μ1 - μ2 = 0 HA : μ1 - μ2 ≠ 0 (without pooled variances) Difference Sample Mean Std. Err. T-Stat P-value μ1 - μ2

-3.13

0.39

-7.94 SSE2. (h) True, a 10-unit increase in x2 is estimated to correspond to a 0.40(10) = 4.0 (i.e., $4000) increase in mean income. (i) False, yˆ = 10 + 1.0(10) = 20. If s = 8, then an income of $70,000 is (70 – 20)/8 = 6.25 standard deviations above the predicted value, which would be very unusual. (j) True, since then ryx1  b sx1 sy = 1.0(3.6/12) = 0.30. (k) False.





At x1 = 13, yˆ = 10 + 1.0(13) = 23. Since the line passes through the point imply that y = 23 rather than 20. 9.59. (b) 9.60. (b), (d), and (g) 9.61. (c), (f), and (g) 88 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

 x , y  , this would

9.62. yˆ = 28,910 – 270x





9.63. b  r sy sx = b(1/1) = b, so slope = correlation. The formula for the y-intercept is

a  y  bx = 0 – 1(0) = 0, so the prediction equation is yˆ = 0 +rx = rx. 9.64. (a) For r = 0.8338, T(r) = (1/2)log(1.8338/0.1662) = 1.20. The standard error is 1 97 = 0.1015. (b) 1.20 ± 1.96(0.1015) = 1.00 to 1.40. (c) The lower endpoint is

e 

2 1.00 2

 e 

2 1.00

1





 e 

1 = 0.76, and the upper endpoint is e21.40 1

2 1.40



1 = 0.89. (d)

2

(0.76 , 0.89 ) = 0.58 to 0.79; we conclude that the true association is quite strong. 9.65. T1 = (1/2)log(1.96/0.04) = 1.946, T2 = (1/2)log(1.76/0.24) = 0.996. The standard error of the difference is 1 8  1 86  0.370 , so the test statistic equals z = (1.946 – 0.996)/0.370 = 2.57, for which P = 2(0.0051) = 0.0102. There is enough evidence to conclude that the true correlations differ. 9.66. (a) Since a  y  bx , it follows that a  bx  y . Thus, yˆ  a  bx  y ; the predicted value at x  x is yˆ  y , and the line passes through

 x, y  .

(b) Since yˆ  a  bx and

a  y  bx , yˆ  y  bx  bx . This equation can be expressed as yˆ  y  b  x  x  . (c)





When sx  sy , b  r sy sx  r 1  r . Thus, we can substitute 0.70 into the equation from part (b) to get yˆ  y  0.70  x  x  . 9.67. (a) Interchange x and y in the formula, and one gets the same value. (b) If the units of measurement change, the z-score does not. For instance, if the values are doubled, then the deviation of an observation from the mean doubles, but so does the standard deviation, and the ratio of the deviation to the standard deviation does not change.

 

2 9.68. The new mean is cy , and the variance is s  

2 cy  cy    n 1 , which is c2 times 

the original variance. Taking the square root, the standard deviation is the original one multiplied by |c|. The numerator of b changes to  x  x cy  cy  , which is c times as large, so the



slope multiples by a factor of c. Thus, since the slope and the standard deviation of y multiply by c, the correlation is now r   cb  sx

cs   b  s y

x

sy  , the same as the original correlation.

9.69. SPSS output for the regression of selling price on size is Model Summary Model 1

R 0.834(a)

R Square 0.695

Adjusted R Square 0.692

Std. Error of the Estimate 56190.324

a Predictors: (Constant), Size

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Coefficients(a)

Unstandardized Coefficients Model 1

B -50926.255

Std. Error 14896.373

126.594 a Dependent Variable: Price

8.468

(Constant) Size

(b)

Standardized Coefficients

t

Sig.

Beta

B -3.419

Std. Error 0.001

14.951

0.000

0.834

yˆ  50,926.255 126.594  2000  $202,261.75 ; A 95% prediction interval is

 x  x   202,261.75 1.96 56,190.324 1 1   2000 1629.28 1 yˆ  ts 1     n   x  x 2 100 44,036,312.2 2

2

= 202,261.75 ± 110,853.21 = 91,409 to 313,115, or $91,409 to $313,115. (b) A 95% prediction interval is

 x  x   202,261.75 1.96 56,190.324 1   2000 1629.28 = 1 yˆ  ts    n   x  x 2 100 44,036,312.2 2

2

202,261.75 ± 12,615.36 = 189,646 to 214,877, or $189,646 to $214,877. (c) As the sample size increases, the width of the confidence interval for the mean goes to 0, and we can estimate the mean nearly perfectly. However large the sample size, even if we know the true mean, we cannot predict individual observations, which fluctuate around the mean with a certain variability that does not depend on the sample size. (d) (i) For instance, the width of the prediction interval is the same at an x value that is c units above x as it is at an x value that is c units below x . On the other hand, if the variability increases, it should be wider above the mean than below the mean. (ii) If the response variable is highly discrete, it is not approximately normal.

Chapter 10 10.1. The three criteria for causation are: 1) association between the variables; 2) an appropriate time order; and 3) the elimination of alternative explanations. Examples will vary. 10.2. (a) HAS YOUNG CHILDREN? HAIR COLOR Yes No Gray 0 4 Not gray 5 0 (b) While the young child notices an association, clearly hair color does not cause a woman to have babies. It is possible that any or all of the gray-haired women have children, just not young children who are living at home. 10.3. (a) No. Without as many firefighters, the damage could potentially be even worse. (b) Larger fires tend to have more firefighters at the fire and have more costly damage. 10.4. No. If one wanted to infer a causal relationship, Y might cause X. However, the size of the city might be a common cause of X and Y.

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10.5. (b) A third variable dealing with a factor such as the subject’s natural curiosity or inquisitiveness could be positively associated with both variables. Students who tend to be higher in this characteristic might tend to have higher GPAs and to be more likely to experiment with marijuana. 10.6. Keep its value constant while studying the association between the other variables. For a sample of students of various ages from a school district, the association between X = height and Y = score on vocabulary exam, controlling for age. 10.7. (a) The positive correlation between shoe size and number of books read is explained by age, which is strongly positively associated with each of these. 10.8. Per capita GNP could influence both birth rate and television ownership, higher values of GNP tending to occur with lower birth rates and with higher levels of television ownership. 10.9. (a) Older, perhaps elderly, people need more sleep than younger people. They are also at risk for more factors that cause death. (b) Age is more likely to be a common cause. 10.10. Perhaps both are affected by parents’ socioeconomic status (SES). Compare the mean math scores for the two groups at various fixed levels of parents’ SES. 10.11. To support this explanation, the association between race and arrest rates would have to disappear. 10.12. Perhaps more women are applying to departments that have stricter admissions standards and more men are applying to departments that have less strict admissions standards. 10.13. (a) Yes, the percentage of white collar occupations is 26.5% for Democrats and 73.5% for Republicans. White Collar Blue Collar Democrat 265 735 Republican 735 265 (b) No, the conditional distributions are identical in each partial table; the differences of proportions = 0 and odds ratio = 1 in each table. Controlling for income, the variables are independent. (c) Income tends to be higher for Republicans than Democrats, and it tends to be higher for white-collar than blue-collar occupations. (d) Occupation affects income, which affects party choice. (e) Income jointly affects occupation and party choice. (d) is more appropriate, since it is more plausible that occupation affects income than the reverse. 10.14. (a) Victim = White Victim = Black DEATH PENALTY DEATH PENALTY DEFENDANT Yes No DEFENDANT Yes No White 19 132 White 0 9 Black 11 52 Black 6 97 For white victims, 12.6% of white defendants receive the death penalty, whereas 17.5% of black defendants receive the death penalty. For black victims, 0% of white defendants receive the death penalty, whereas 5.8% of black defendants receive the death penalty.

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(b) DEFENDANT Yes No White 19 141 Black 17 149 Ignoring the race of the victim, 11.9% of white defendants receive the death penalty, whereas 10.2% of black defendants receive the death penalty. (c) The proportions are higher for black defendants in each partial table, reflecting a greater chance of the death penalty for black defendants (controlling for victim’s race), but the proportion is higher for white defendants for the bivariate table, indicating the reverse association when victim’s race is ignored rather than controlled. (d) In terms of the difference of proportions between white defendants and black defendants receiving the death penalty, it is negative when we control for victim’s race and positive when we ignore it. (e) Any or the four models could be reasonable, depending on one’s reasoning. 10.15. Size of home is positively correlated with selling price and with number of bedrooms; it overlaps with number of bedrooms in explaining variability in selling price. Also, controlling (holding fixed) size of home, as the number of bedrooms increases, then the number of other rooms or the size of those rooms would tend to decrease, so increasing number of bedrooms would not necessarily have such a strong positive effect. Number of bedrooms has a direct effect on selling price, and in indirect effect through its effect on size of home. 10.16. This is possibly a spurious relationship, whereby the percent living in metropolitan areas is positively related both to crime rate and to percent with a high school education. Even if there is no association between high school graduation rate and crime rate at each fixed level of percent living in metropolitan areas, overall there will be a positive association because of the joint positive association between percent living in metropolitan areas and each of the other variables. 10.17. Let A = attitude about legal availability of abortion, R = religiosity, S = sexual attitudes. More than one diagram is plausible. One possible diagram follows. RSA 10.18. (a) The response variable is salary, and the control variable is academic rank. (b) Mean salary is 12.1 thousand dollars higher for men than women. (c) At each academic rank, the mean salary is higher for men than women, though by a relatively small amount at levels below professor. (d) A chain relationship in which gender affects academic rank, which itself affects income, would satisfy this. This does not seem plausible, since men have a higher mean income at each academic rank (in fact, considerably higher at the professor level). (e) If there are more men at the professor level and more women at the lower academic ranks, including instructor, the overall difference could be larger than the difference for each academic rank. 10.19. Ignoring the subject of the exam, there is no association, but there is a substantial association in each partial table. 10.20. If the percentage of the tract used for business is positively correlated with median tax bill, then higher percentages of the tract used for business are associated with higher median tax bills. If the percentage of the tract used for business is negatively correlated with median lot size, then higher percentages of the tract used for business are associated with smaller median lot sizes. Thus, if we control for the percentage of the tract used for business, these correlations will be suppressed by this control.

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10.21. (a) White men have a higher median income than white women. (b) Black men have a higher median income than black women. There is an interaction because white men have a higher median income than white women and black men have a higher median income than black women. There is an interaction between race and gender. 10.22. The intercepts are the same, so we can focus in on the slopes in the regression equations. The slope for men is greater than the slope for women, which means that the annual income for men increases at a higher rate than the annual income for women, as number of years on the job increases. 10.23. (a) The response variable is whether the subject has cancer, and the explanatory variable is whether the subject is a smoker. The control variable is age. (b) Yes, the association is stronger for older subjects. Smoking seems to have a stronger effect for older subjects, who have presumably been smoking for a longer time period. 10.24. (a) Increased frequency of church attendance tends to be associated with less favorable opinion about legalizing marijuana, particularly for females. (b) Interaction. (c) A 95% confidence interval is

 0.287   0.581 1.96 0.081  0.091 2

2

= 0.294 ± 0.122 =

0.172 to 0.416. We are 95% confident that the difference in population gamma values is between 0.172 and 0.416, a moderate difference. 10.25. LifeAfterDeath

Count Abortion

n Row N %

Column N %

Count

u Row N %

Column N %

Count

y Row N %

Column N %

n

1

7.7%

7.7%

0

0.0%

0.0%

12

92.3%

38.7%

y

12

25.5%

92.3%

16

34.0%

100.0%

19

40.4%

61.3%

Religiosity most/every week LifeAfterDeath

Abortion

Count

n Row N %

Column N %

Count

n

0

0.0%

0.0%

0

y

0

0.0%

0.0%

2

u Row N %

y Row N %

Column N %

Count

Column N %

0.0%

0.0%

8

100.0%

57.1%

25.0%

100.0%

6

75.0%

42.9%

Religiosity never/occasionally LifeAfterDeath

Count Abortion

n Row N %

Column N %

Count

u Row N %

Column N %

Count

y Row N %

Column N %

n

1

20.0%

7.7%

0

0.0%

0.0%

4

80.0%

23.5%

y

12

30.8%

92.3%

14

35.9%

100.0%

13

33.3%

76.5%

The first table shows that those who do not favor legalized abortion are more likely to believe in life after death than those who favor legalized abortion. This might be a spurious association, with religiosity affecting both variables. However, the partial tables still show the same type of

93 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

association, though the sample sizes are so small that we cannot precisely estimate the true association. 10.27. Answers will vary, depending on how students define the most conservative and most liberal subjects. 10.29. (a) The mean number of children is higher for families for which English is the primary language. (b) For each level of Province, the mean number of children is higher for families for which French is the primary language. (c) Most French-speaking families are in Quebec, where the means are lower regardless of language, and most English-speaking families are in other provinces. 10.30. (a) The response variable is the mean eight-grade math scores on the NAEP, and the control variable is race. (b) Nebraska has more whites than it does blacks and other nonwhite students, compared to New Jersey. 10.31. (b) Possible confounding means that if those praying did not pray in a heartfelt manner this might interfere with (confound) the results of complications for coronary surgery patients. 10.32. Age may be associated with both exercising and illnesses. For example, perhaps older people are not as able to exercise and are more susceptible to illnesses. 10.33. SES may be a common cause of birth defects and of buying bottled water. 10.34. Suppose that at a fixed age, there is no difference in rates between now and the beginning of the century. Because women tend to live longer now and because the cancer rate is higher at higher ages, the overall rate of breast cancer would be higher now than at the beginning of the century. 10.35. This is possible, because death rates are higher at higher ages, and the U.S. nay have relatively more people with higher ages than does Mexico. 10.36. This is possible, because death rates are higher at higher ages, and Colorado has relatively more people with higher ages, so the overall death rate is higher in Colorado. 10.37. (a) For females, mean GPA is higher for those with an employed mother. For males, mean GPA is about the same for those with an employed mother as for those with an unemployed mother. Since the results differ according to gender, there is evidence of interaction. (b) The two means are about equal for males, yet for females the mean is higher for those with an employed mother. 10.38. (a) Y = test score, X1 = height, X2 = age, for students aged between 5 and 17. (b) Y = subject’s income, X1 = parent’s education, X2 = subject’s education. (c) Y = income, X1 = gender, X2 = number of years of experience. Mean income is higher for men than women, but part of that difference may be due to greater average number of years on the job. (d) Y = political ideology, X1 = gender, X2 = race. (e) Y = income, X1 = gender, X2 = race. There is a large difference in mean income between men and women when race = white, but a small difference in mean income between men and women when race = black. (f) Y = cost of damages of a fire, X1 = number of firefighters at the fire, X2 = size of the fire.

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10.39. Income masked any potential association between compulsive buying and mean total credit card balances. If we controlled for income, we might find that mean total credit card balances are higher for compulsive buyers than for noncompulsive buyers. 10.40. False. First, association is not causation. Second, there may be other variables that confound the relationship between degree of depression and saturated fat intake. 10.41. False. 10.42. (b) 10.43. (b) 10.44. (a) 10.45. (a)

Chapter 11 11.1. (a) (i) E(y) = 0.20 + 0.50(4.0) + 0.002(800) = 3.8; (ii) E(y) = 0.20 + 0.50(3.0) + 0.002(300) = 2.3. (b) E(y) = 0.20 + 0.50x1 + 0.002(500) = 1.20 + 0.50x1. (c) E(y) = 0.20 + 0.50x1 + 0.002(600) = 1.40 + 0.50x1. (d) For instance, consider x1 = 3 for which E(y) = 1.70 + 0.002x2. By contrast, when x1 = 2, E(y) = 1.20 + 0.002x2, which has a different y-intercept but the same slope of 0.002. 11.2. (a) yˆ = –10,536 + 53.8x1 + 2.84x2 = –10,536 + 53.8(1240) + 2.84(18,000) = $107,296; The residual is y  yˆ = $145,000 – $107,296 = $37,704. The home sold for $37,704 more than we would have predicted based on the multiple regression model. (b) For a fixed lot size, the predicted house selling price is predicted to increase by $53.80 for each square foot increase in home size. 11.3. (a) For a fixed home size, the lot size would need to increase by 53.8/2.84 = 18.94, or about 19 square feet, to have the same impact as a one square foot increase in home size. (b) If house selling prices are changed from dollars to thousands of dollars, we need to divide all of coefficients by 1000 to obtain the proper prediction equation. 11.4. (a) D.C. appears to be an outlier in each of the partial regression plots.

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Partial Regression Plot

Dependent Variable: MU 30

MU

20

10

0

-10 -6.0

-3.0

0.0

3.0

6.0

HS

Partial Regression Plot

Dependent Variable: MU 40

30

MU

20

10

0

-10 -4.0

-2.0

0.0

2.0

4.0

6.0

8.0

10.0

PO

(b) yˆ = –60.498 + 0.588x1 + 1.605x2. For fixed poverty rate, the murder rate is predicted to increase by 0.588 for each additional percent increase of high school graduates. For fixed percentage of graduates, the murder rate is predicted to increase by 1.65 for each addition percent increase in the poverty rate. Model Summary(b) Model 1

R

R Square

0.636(a) 0.405 a Predictors: (Constant), PO, HS b Dependent Variable: MU

Adjusted R Square

Std. Error of the Estimate

0.380

4.764

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Model 1

(Constant)

Coefficients(a) Unstandardized Standardized Coefficients Coefficients

t

Sig.

B

B

Std. Error

Std. Error

Beta

-60.498

24.615

-2.458

0.018

HS

0.588

0.260

0.351

2.256

0.029

PO

1.605

0.301

0.830

5.332

0.000

a Dependent Variable: MU

(c) With D.C. removed, the predicted effect of poverty rate is reduced by a factor of about 5. In addition, the predicted effect of percent of high school graduates now has a negative partial coefficient. Model Summary(b) Model 1

R 0.582(a)

R Square 0.338

Adjusted R Square 0.310

Std. Error of the Estimate 2.136

a Predictors: (Constant), PO_noDC, HS_noDC b Dependent Variable: MU_noDC Coefficients(a) Unstandardized Standardized Coefficients Coefficients

t

(Constant)

B 18.912

Std. Error 12.437

B 1.521

Std. Error 0.135

HS_noDC

-0.196

0.130

-0.278

-1.510

0.138

0.304 a Dependent Variable: MU_noDC

0.164

0.340

1.846

0.071

Model 1

PO_noDC

Beta

Sig.

11.5. (a) yˆ = –3.601 + 1.2799x1 + 0.1021x2. (b) yˆ = –3.601 + 1.2799(10) + 0.1021(50) = 14.3. (c) (i) yˆ = –3.601 + 1.2799x1 + 0.1021(0) = –3.601 + 1.2799x1; (ii) yˆ = –3.601 + 1.2799x1 + 0.1021(100) = 6.609 + 1.2799x1; On average, for each increase of $1000 in GDP, the percentage of people who use the Internet increases by 1.28. (d) Since the slope for GDP is the same for each of the two models in part (c), the lines are parallel, and there is no interaction between cell-phone use and GDP. 11.6. (a) R-squared = (TSS – TSE)/TSS = (12,959.3 – 2642.5)/12,959.3 = 10,316.8/12,959.3 = 0.796. (b) Since cell-phone use and GDP are strongly positively correlated, adding cell-phone use to the model will not improve the model much beyond only using GDP as a predictor of Internet use. 11.7. (a) Positive, since crime appears to increase as income increases. (b) Negative, since controlling for percent urban, the trend appears to be negative. (c) yˆ = –11.5 + 2.6x1; the predicted crime rate increases by 2.6 (per 1000 residents) for every $1000 increase in median income. (d) yˆ = 40.3 – 0.81x1 + 0.65x2; the predicted crime rate decreases by 0.8 (per 1000 residents) for each $1000 increase in median income, controlling for level of urbanization. Compared to (c), the effect is weaker and has a different direction. (e) Urbanization is highly positively correlated both with income and with crime rate. This makes the overall bivariate association between income and crime rate more positive than the partial association. (f) (i) yˆ =

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40.3 – 0.81x1; (ii) yˆ = 73 – 0.81x1; (iii) yˆ = 105 – 0.81x1. The slope stays constant, but at a fixed level of x1, the crime rates are higher at higher levels of x2. 11.8. (a) 13 125

CrimeRate

100

75

50

25

0

40.0

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35.0

30.0

25.0

20.0

15.0

100.0

80.0

Urban

60.0

40.0

20.0

0.0

98 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

Scatter diagrams are:

125

CrimeRate

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0 15.0

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125

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UrbanRegression Plot Partial

Partial regression plots are: Dependent Variable: CrimeRate 60

CrimeRate

40

20

0

-20

-40 -5.0

0.0

5.0

10.0

Income

99 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

Partial Regression Plot

Dependent Variable: CrimeRate 80

60

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40

20

0

-20

-40

-80.0

-60.0

-40.0

-20.0

0.0

20.0

40.0

Urban

The scatter diagrams indicate a positive association between crime rate and income, as well as between crime rate and urbanization. The partial regression plots indicate a negative partial effect for income and a positive partial effect for urbanization. (b) (i) yˆ = –11.61 + 2.61x1 , yˆ = 24.54 + 0.56x2 Coefficients(a) Unstandardized Standardized Coefficients Coefficients Model 1

(Constant) Income

B -11.606

Std. Error 16.786

2.611

0.673

Beta 0.434

t

Sig.

B -0.691

Std. Error 0.492

3.881

0.000

a Dependent Variable: CrimeRate

Model 1

(Constant) Urban

Coefficients(a) Unstandardized Standardized Coefficients Coefficients

t

Sig.

B

B

Std. Error

Std. Error

24.541

4.539

0.562

0.076

Beta 0.677

5.406

0.000

7.424

0.000

a Dependent Variable: CrimeRate

2.61 = change in predicted crime rate for 1 unit change in income, and 0.56 = change in predicted crime rate for 1 unit change in urbanization, ignoring other variables. (ii) yˆ = 40.0 – 0.79x1 + 0.64x2

Model 1

Coefficients(a) Unstandardized Standardized Coefficients Coefficients

T

Sig.

B

B

Std. Error

Std. Error

Beta

(Constant)

39.972

16.354

2.444

0.017

Income

-0.791

0.805

-0.131

-0.982

0.330

Urban

0.642

0.111

0.773

5.784

0.000

a Dependent Variable: CrimeRate

–0.79 = change in predicated crime rate for 1 unit change in income, controlling for urbanization, and 0.64 = change in predicted crime rate for 1 unit change in urbanization, controlling for income. 100 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

(c) R2 = 0.683, a 68.3% reduction in error using income and urbanization to predict crime rate, 2 instead of using y . By contrast, ryx2 = 0.677 is nearly as large, so once urbanization is in the model, it does not help much to add income as a predictor. 11.9. (a) –0.13 = change in predicted birth rate for 1 unit increase in women’s economic activity. (b) Parallel lines, so the effect of x1 is the same at each level of x2, but at fixed x1, predicted birth rates are lower at higher levels of x2. (c) x1 and x2 are moderately positively correlated and explain some of the same variation in y. Controlling for x2, the effect of x1 weakens. 11.10. (a) R = 0.787 describes the association between carbon dioxide use and the two explanatory variables. It is the correlation between the observed and predicted y-values. (b) It did not help much to add life expectancy to the model, since the multiple correlation is about the same as the correlation of carbon dioxide use and GDP alone. This does not mean that life expectancy is very weakly correlated with carbon dioxide use, it means that life expectancy does not add much information to carbon dioxide use when GDP is already in the model. 11.11. (a) yˆ = –498.7 + 32.6x1 + 9.1x2; 32.6 = change in predicted violent crime rate for 1 unit change in poverty rate, controlling for percent living in urban areas, and 9.1 = change in predicted violent crime rate for 1 unit change in percent living in urban areas, controlling for poverty rate. (b) yˆ = –498.7 + 32.6(10.7) + 9.1(96.2) = 725.5; residual = 805 – 725.5 = 79.5, so the observed violent crime rate is somewhat higher than the model predicts. (c) (i) yˆ = –498.7 + 32.6x1, (ii) yˆ = –43.1 + 32.6x1, (iii) yˆ = 412.5 + 32.6x1. The lines have the same slope (32.6) for poverty rate. The violent crime rate is higher for states having higher percents living in urban areas. (d) The violent crime rate tends to increase as poverty rate increases or as the percent in urban areas increases; there is a weak negative association between poverty rate and the percent living in urban areas. (e) R2 = 0.57 = PRE in using x1 and x2 together to predict y, R  0.57  0.76 = sample correlation between observed and predicted y values. 11.12. (a) Using R2, F = [0.571/2]/[(1 – 0.571)/(50 – 3)] = 31.2, df1 = 2, df2 = 47, P = 0.0001; there is very strong evidence that at least one predictor has an effect on y. (b) t = 32.62/6.68 = 4.89, df = 50 – 3 = 47, P = 0.0001; there is very strong evidence of a partial effect of x1 on y, controlling for x2. (c) 32.62 ± 2.01(6.677) = 19 to 46; controlling for x2, we can be 95% confident that the change in the mean of y for a 1 unit change in x1 falls between 19 and 46. (d) These inferences are irrelevant in this situation since we have data from the entire population. 11.13. (a) yˆ = –1197.5 + 18.3x1 + 7.7x2 + 89.4x3; 18.3 = change in predicted violent crime rate for 1 unit change in poverty rate, controlling for percent living in urban areas and percentage of single-parent families, 7.7 = change in predicted violent crime rate for 1 unit change in percent living in urban areas, controlling for poverty rate and percentage of single-parent families, and 89.4 = change in predicted violent crime rate for 1 unit change in poverty rate, controlling for percent living in urban areas and percentage living in urban areas. (b) x1 and x3 are highly positively correlated and explain some of the same variability in y; controlling for x3, the effect of x1 weakens. 11.14. (a) yˆ = 5.25 – 0.24x1 + 0.02x2; –0.24 = change in predicted number of children for 1 unit change in MEDUC, controlling for FSES, and 0.02 = change in predicted number of children for 1 unit change in FSES, controlling for MEDUC. (b) SSE = 199.3; the sum of squared residuals (prediction errors) would be greater than 199.3 if we used any other linear prediction equation with these two variables. (c) No, the partial effect of x1, controlling for x2, is negative, not 101 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

positive, since the coefficient of x1 is negative. (d) No, the partial slopes have the same sign as partial correlations, and they may have different signs from Pearson correlations. 11.15. (a) yˆ = 135.3 – 14.07x1 – 2.95x2; –14.07 = change in predicted feelings toward liberals for 1 unit change in political ideology, controlling for religious attendance. (b) yˆ = 135.3 – 14.07(7) – 2.95(9) = 10.3; residual = 10 – 10.3 = –0.3. (c) R2 = 0.799; there is an 80% reduction in error by predicting y using x1 and x2 instead of y . (d) t = –14.07/3.16 = –4.45, df = 10 – 3 = 7, P < 0.01 for a two-sided test. It is better to show the actual P-value; for instance, 0.049 is not practically different from 0.051, but the * system would make it seem like the effect with P = 0.049 is important but the one with P = 0.051 is not. (e) Using R2, F = [0.799/2]/[(1 – 0.599)/(10 – 3)] = 13.9, df1 = 2, df2 = 7, P < 0.01; there is strong evidence that at least one predictor has an effect on y. (f) Ideology appears to have a stronger partial effect than religion; a standard deviation increase in ideology has a 0.79 standard deviation predicted decrease in feelings, controlling for religion. 11.16. Test statistic = 3.35. (a) P = 0.002 for the two-sided test, giving strong evidence of dependence between mental impairment and SES, controlling for life events. (b) P = 0.001 for the one-sided test, giving strong evidence of a negative relationship between mental impairment and SES, controlling for life events. 11.17. (a) Model sum of squares = 813.3, df values are 5, 60, 65, model mean square = 813.3/5 = 162.7, mean square error = 2940.0/60 = 49, F = 162.7/49 = 3.3 with df1 = 5, df2 = 60the P-value (Sig) is 0.01, R2 = 0.217, Root MSE = 7.0, t values are 2.22, –2.00, 0.50, –0.80, 2.40, with Pvalues (Sig) 0.03, 0.05, 0.62, 0.43, 0.02. (b) No, one could probably drop x3 or x4, or both, since the P-values are large for their partial tests. (c) The test of H0 : 1  2  3  4  5  0 . (d) Test H0 : 1  0 ; P = 0.03, so there is considerable evidence that x1 has an effect on y, controlling for the other xs.

11.18. (a) The confidence interval for 1 is 0.10 ± 0.09; controlling for all other explanatory variables in the model, the change in the mean of y for a 1 unit increase in x1 falls between 0.01 and 0.19, with 95% confidence. (b) 50(0.01, 0.19) = (0.5, 9.5); controlling for all other explanatory variables in the model, the change in the mean of y for a 50 unit increase in x1 falls between 0.5 and 9.5, with 95% confidence. 11.19. (a) The scatterplots are:

102 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

600,000

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400,000

300,000

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100,000

0 1,000

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0 1

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3

4

4

Baths

There is a moderately strong positive linear relationship between the selling price of a home and its size (in square feet). In the scatterplots of selling price versus either number of bedrooms or number of bathrooms, we see stacking at the integer values. 103 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

(b) yˆ = –27,290 + 130x1 – 14,466x2 + 6890x3; 130 = change in predicted selling price of a home for 1 unit change in size, controlling for number of bedrooms and number of bathrooms. Coefficients(a) Standardized Unstandardized Coefficients Coefficients Model 1

Sig.

B -0.966

Std. Error 0.336

B -27290.075

Std. Error 28240.514

130.434

11.951

0.859

10.914

0.000

-14465.770

10583.489

-0.093

-1.367

0.175

6890.267 a Dependent Variable: Price

13539.977

0.039

0.509

0.612

(Constant) Size Beds Baths

Beta

t

(c) (i) The size of the house has the strongest correlation with the selling price of the house. (ii) The number of bedrooms has the weakest correlation with the selling price of the house. Correlations Price Price

Pearson Correlation

1

Size 0.834(**)

Beds 0.394(**)

Baths 0.558(**) 0.000

Sig. (2-tailed) N Size

Pearson Correlation Sig. (2-tailed) Pearson Correlation Sig. (2-tailed) N

Baths

0.000

100

100

100

0.834(**)

1

0.545(**)

0.658(**)

0.000

N Beds

0.000 100

Pearson Correlation Sig. (2-tailed) N

0.000

0.000

100

100

100

100

0.394(**)

0.545(**)

1

0.492(**)

0.000

0.000

100

100

100

100

0.558(**)

0.658(**)

0.492(**)

1

0.000

0.000

0.000

100

100

100

0.000

100

** Correlation is significant at the 0.01 level (2-tailed).

(d) R2 = 0.701 for the full model; r2 = 0.695 for the simpler model using x1 alone as the predictor. Once x1 is in the model, x2 and x3 do not add much more information for predicting selling price. 11.20. (a) t = 0.509; P = 0.612; there is no evidence that the number of bathrooms has an effect on selling price, controlling for home size and number of bedrooms. (b)

ryx3 x2 



ryx3  ryx2 rx2 x3



1  ryx2 2 1  rx22 x3





0.558  0.394  0.492

1 0.3942 1 0.4922 

 0.455 ; the number of bathrooms

have a relatively weak positive association with selling price, controlling for the number of bedrooms. (b) b1* = 0.859, b2* = –0.093, b3* = 0.039; each coefficient described the predicted change in selling price, in standard deviation units, for a one standard deviation increase in xi, controlling for the other explanatory variables. (d) zˆy  0.859zx1  0.093zx2  0.039zx3 . 11.21. (a) As the percentage living in urban areas increases, the effect of poverty rate increases, since the coefficient for the interaction term is positive. (b) When x2 = 0, yˆ  158.9 14.72x1 . When x2 = 50, yˆ  94.4  23.28x1 . When x2 = 100, yˆ  29.9  61.28x1 . As the poverty rate

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increases, the predicted violent crime rate increases both in terms of the y-intercept and the slope of the regression equation. 11.22. Yes, the positive coefficient of x1x2 means that the slope of the effect for x1 increases as x2 increases. 11.23. (a) yˆ = 106,640 + 127x1 – 59,071x2 – 55,238x3 + 21,107x2x3. (i) yˆ = –11,502 + 127x1 – 13,024x3; (ii) yˆ = –59,074 + 127x1 + 4250x2. (c) t = 2.035, P = 0.045; there is moderately strong evidence that the interaction term is significant in the model. Coefficients(a) Standardized Unstandardized Coefficients Coefficients Model 1

B (Constant)

Std. Error

Beta

t

Sig.

B

Std. Error

106639.565

71447.153

1.493

0.139

126.599

11.910

0.834

10.629

0.000

Beds

-59071.421

24270.019

-0.380

-2.434

0.017

Baths

-55238.487

33314.273

-0.309

-1.658

0.101

21106.826 a Dependent Variable: Price

10373.182

0.578

2.035

0.045

Size

bedbath

11.24. F = [(20 – 19)/2]/[19/(195 – 5)] = 0.5/0.1 = 5.0, df1 = 2, df2 = 190, P < 0.01. The complete model is significantly better. 11.25. (a) (i) –0.612, (ii) –0.819, (iii) 0.757, (iv) 2411.4, (v) 585.4, (vi) 29.27, (vii) 5.41, (viii) 10.47, (ix) 0.064, (x) –2.676, (xi) 0.0145, (xii) 0.007, (xiii) 31.19, (xiv) 0.0001. (b) yˆ = 61.71 – 0.171x1 – 0.404x2; 61.71 = predicted birth rate at ECON = 0 and LITERACY = 0 (may not be useful), –0.171 = change in predicted birth rate for 1 unit change in ECON, controlling for LITERACY, –0.404 = change in predicted birth rate for 1 unit change in LITERACY, controlling for ECON. (c) –0.612; there is a moderate negative association between birth rate and ECON; –0.819; there is a strong negative association between birth rate and LITERACY. (d) R2 = (2411.4 – 585.4)/2411.4 = 0.76; there is a 76% reduction in error in using these two variables (instead of y ) to predict birth rate. (e) R  0.76  0.87 = the correlation between observed y values and the predicted y values. (f) F = [0.757/2]/[(1 – 0.757)/20] = 31.2, df1 = 2, df2 = 20, P = 0.0001; at least one of ECON and LITERACY has a significant effect. (g) t = –0.171/0.064 = –2.676, df = 20, P = 0.0145; there is strong evidence of a relationship between birth rate and ECON, controlling for LITERACY. 11.26. (a) ryx1x2   –0.612 –  –0.819 0.421

1   0.8192  1  0.4212  = –0.51;   

controlling for x2, there is a moderate tendency for y to decrease as x1 increases. Its square equals 0.26, which is the proportion of variation in y explained by x1, out of that part unexplained by x2. (b) s = 585.4 20  29.27  5.4 = estimated standard deviation of y values at fixed values of x1 and x2. (c) –0.171(19.87/10.47) = –0.325; birth rate is predicted to decrease 0.325 standard deviations for each standard deviation increase in economic activity, controlling for literacy. (d) zˆy  0.325zx1  0.682zx2 ; birth rate is predicted to decrease 0.325 standard deviations for each standard deviation increase in economic activity, controlling for literacy, and birth rate is predicted to decrease 0.682 standard deviations for each standard deviation increase in literacy, 105 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

controlling for economic activity. (e) –0.325(1) –0.682(1) = –1.0; the country is predicted to be one standard deviation below the mean in birth rate. 11.27. Urbanization is highly positively correlated with both variables. Even though there is a weak association between crime rate and high school graduation rate at a fixed level of urbanization (since partial correlation = –0.15), as urbanization increases, so do both of these variables tend to increase, thus producing an overall moderate positive association (Pearson correlation = 0.47) between crime rate and high school graduation rate. 11.28. (a) rVH  A  0.65   0.85 0.75

1   0.852  1   0.752  = 0.036. There is a very    weak association between vocabulary and height, controlling for age. (b) For H0 : VH  A  0 , the

test statistic is t = 0.036

1  0.0362 = 0.355, df = 97; the null hypothesis is plausible. (c) Yes, 100  3

from parts (a) and (b), the effect of height on vocabulary is insignificant after age is taken into account. Both height and vocabulary increase with age. 11.29. (a) zˆ y  0.75zx1  0.125zx2  0.30zx3  0.20zx4 . (b) Number of police officers (per 100,000 residents) has the greatest partial effect on y since its estimated standardized regression coefficient is the largest (in absolute value). (c) zˆy  0.75 1  0.125 1  0.30 1  0.20 1 = –0.7, so the city is predicted to be 0.7 standard deviations below the mean in murder rate. 11.30. (a) 0.473 = violent crime rate is predicted to increase 0.473 standard deviations for each standard deviation increase in poverty rate, controlling for urbanization; 0.668 = violent crime rate is predicted to increase 0.668 standard deviations for each standard deviation increase in urbanization, controlling for poverty rate. (b) zˆ y  0.473zx1  0.668zx2 . Each coefficient describes the predicted change in violent crime rate, in standard deviation units, for a one standard deviation increase in xi, controlling for the other explanatory variable. 11.43. (b) r2 = 0.50 for the simple linear regression of life satisfaction on GDP. (c) Test statistics for life expectancy and GDP have the smallest P-values out of all P-values in the analysis. (d) The coefficient is small because it is per capita, in dollars. (e) No, this means that after GDP, life expectancy, and political freedom are in the model, education does not add much more information to predicting quality of life. 11.44. Since there is essentially no effect for political conservatives and a considerably positive effect for political liberals, there is an interaction between number of years of education and political ideology. One would add an interaction term to the multiple regression model. 11.45. Estimating standardized coefficients allows one to make comparisons of the impact of each variable on the response variable, controlling for all other variables, without having to consider units. 0.21 = attitudes toward homosexuality is predicted to increase 0.21 standard deviations for each standard deviation increase in educational level, controlling for all other variables in the model. 11.46. False. We could make this conclusion if these were standardized coefficients, but not as unstandardized coefficients. 106 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

11.47. (b) The partial change is 0.34, not 0.45, which is the overall change ignoring rather than controlling the other variables. (c) Cannot tell, since variables have different units of measurement. (d) False, since this cannot exceed 0.38, the R2 value for the model with x3 and other variables in the model. 11.48. (h) R2 cannot exceed 1. (i) This is the multiple correlation, which cannot be negative. (j) No, this is only true when df1 = 1, in which case F = t2. (k) We need to compare the absolute values of standardized coefficients to determine which explanatory variable has a stronger effect. (n) The slopes and intercepts do change when we examine different values of the explanatory variables. 11.49. (b), since (20 – 10)3 = 30. 11.50. (d) is correct. X3 has the greatest partial slope, but the effects are not comparable since the variables may have different units. 11.51. (c), since the coefficient of x3 is –8 and since this partial slope has the same sign as the partial correlation. We need the sign of the slope for the bivariate model to tell the sign of the correlation in (a) and the sign of the partial slope for x3 for the model with it and x1 alone as predictors to tell the sign of the partial correlation in (b). 11.52. (a), (b), and (c) . (d) is incorrect because one cannot use this F test to compare two models that are not nested; in this case, neither model is a special case of the other, since each has a predictor that the other does not have. 11.53. For quantitative variables, the correlation describes association between two variables, the multiple correlation describes association between a response variable and predictions based on a set of predictor variables, and the partial correlation describes association between two variables while controlling for one or more other variables. 11.54. The correlation is close to 1.0 for x1 and x2, and it is a large positive number for y and x2. The multiple correlation R is close to the maximum of the correlations of each of x1 and x2 with y; that is, once one of these predictors is in the model, the other would explain hardly any additional variation. For this reason also, the partial correlation would be close to 0, since its square has in 2 the numerator the difference between R2 and ryx1 , which would be tiny. 11.55. For a sample of children of various ages, y = score on vocabulary achievement test, x1 = height, x2 = age. 11.56. (a) H0 :   0 . (b) H0 :   0 (population multiple correlation = 0) (c) H0 :  yx2 x1  0 . 11.57. x1 and x2 do not overlap in their explained variation in y. 11.58. Case (a), since the explanatory variables explain separate portions of the variability in y. In fact, because of the result in Exercise 11.57, for case (a), R2 = (0.4)2 + (0.4)2 = 0.32. By contrast, in case (c), R2 = 0.16, because x1 and x2 together are no better than either one alone, since they are perfectly correlated.

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2 2 2 11.59. ryx2 x1 = 0, since its numerator is R  ryx1 , which is 0 when R  ryx1 . Hence its square

root also equals 0. 11.60. (a) Type I SS = variability explained by the variable, controlling for variables already in the model. For x1, there is no other predictor yet, so the variability explained is the model sum of squares. For x2, only one variable (x1) is already in the model, so the amount of variation explained, controlling for x1, is SSE1 – SSE12. (b) For the last variable, variables previously entered into the model and all other variables in the model are the same set. 2 11.61. (a) 0.150, 0.303, 0.325, 0.338. (b) Set up the inequality Radj  0 and solve for R2.

11.62. The numerator is the additional variation in y explained by xk beyond that explained by x1, x2,…, xk–1. The denominator is the maximum value this additional explained variation can be, because of the upper limit of 1 for R2. The ratio is the square of the partial correlation. 11.63. (a) ry2 x2 x1  = 0.3392 – (0.3722)2 = 0.20, which is the proportion of the variation in y that is explained by x2, beyond that explained by x1 (i.e., after x1 is already in the model). Similarly, ry2 x2 x1  = 0.3392 – (–0.3986)2 = 0.18, which is the proportion of the variation in y that is explained by x1, beyond that explained by x2. (b) Note that in the squared partial correlation, the squared semipartial correlation is divided by a quantity less than or equal to 1. Therefore the value for the squared partial correlation must be at least as large as the squared semipartial correlation. 11.64. Using the same equation obtained with k predictors, by giving coefficient 0 to the new predictor, yields the same predictions and thus the same value of R. So, we can always do at least this well, and even better if a non-zero coefficient for the new predictor results in even better predictions. 11.65. (a) Let b denote the minimum of the two standardized estimates, and let B denote the maximum. Then bi*bi*  bB  BB  B2 and bi*bi*  bB  bb  b2 . Thus, since the partial correlation has the same sign as the standardized coefficient, it falls between b and B. (b) Because the partial correlation must fall between –1 and +1, and its square must fall between 0 and 1.

yˆ  26.1  72.6x1 19.6x2 ; Older homes: yˆ  26.1 72.6x1 ; new homes: yˆ  6.5  72.6x1 . (b) When x2 = 0, the y-intercept equals the y-intercept of the full model.

11.66. (a)

When x2 = 1, the y-intercept is 19.6 higher than the y-intercept of the full model, which is the difference of mean selling prices between new and older homes, controlling for house size. 11.67. (a) Older homes: yˆ  16.6  66.6x1 ; New homes: yˆ  15.2  96x1 . New homes cost more, on average, than older homes. In addition, the selling price for a new home increases at a higher rate as the size increases than does the selling price for an older home. (b) Older homes: yˆ  16.6  66.6x1 ; New homes: yˆ  7.6  71.6x1 . With the outlier removed, there is not as much difference in selling price between new and older homes, and, as the size increases, the selling price for a new home increases at a slower rate than the model with the outlier, when compared to an older home.

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Chapter 12 12.1. The response variable is the number of firefights reported. The explanatory variable was branch of service and deployment venue combination. The null hypothesis is that the means are the same for each branch of service and deployment venue combination. The alternative hypothesis is that at least one mean differs. 12.2. (a) H0 : 1  2    12 ; H a : at least two of the population means are unequal. (b) If the two estimates of the variance were equal, we would expect F to be 1. Since F = 0.61 does not differ from 1 by much, it does not appear that we have strong evidence against H0 . (c) If all 12 population means were equal (e.g., the mean number of good friends was not associated with astrological sign), we would see results as extreme as those observed with probability 0.82. There is not enough evidence to show an association between the mean number of good friends and astrological sign. 12.3. (a) H0 : 1  2  3  4  5 ; H a : at least two of the population means are unequal. (b) If the two estimates of the variance were equal, we would expect F to be 1. Since F = 0.80 does not differ from 1 by much, it does not appear that we have strong evidence against H0 . (c) If all 5 population means were equal (e.g., the mean number of good friends was not associated with marital status), we would see results as extreme as those observed with probability 0.53. There is not enough evidence to show an association between the mean number of good friends and marital status. 12.4. (a) H0 : 1  2  3  4 ; H a : at least two of the population means are unequal. (b) F = 5.48, P-value = 0.001. There is enough evidence to conclude at least two of the population means are not equal, that is, that the mean number of children for a family differs by religion. (c) No. We can only conclude that at least two of the means differ. 12.5. (a) (i) H0 : 1  2  3 ; H a : at least two of the population means are unequal. (ii) F = 3.03; (iii) P = 0.049; (iv) We reject the null hypothesis at the 0.05 significance level. (b) Yes. The largest standard deviation is more than double the smallest standard deviation. 12.6. (a) H0 : 1  2  3 ; H a : at least two of the population means are unequal. F = 2.50, P = 0.18. There is not enough evidence to conclude that the mean quiz scores differ for the three groups. (b) The F statistic will be smaller, because the between groups variation would not be as large, so the numerator of the F statistic would be smaller. (c) The F statistic will be larger, since the within groups variation is smaller. (d) The F statistic would be larger, since larger sample sizes make differences between the means more pronounced. (e) The P-values would be larger for part (b), smaller for part (c), and smaller for part (d). 12.7. SPSS output for the ANOVA is: ANOVA QuizScore

Between Groups

Sum of Squares 30.000

df 2

Mean Square 15.000 6.000

Within Groups

30.000

5

Total

60.000

7

F 2.500

109 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

Sig. 0.177

12.8. (a) Within-groups SS = 12.0; within estimate = 12/3 = 4.0. (b) Between-groups SS = 148.0; between estimate = 148/2 = 74.0. (c) H0 : 1  2  3 ; H a : at least two of the population means are unequal. F = 18.5, P = 0.02. There is enough evidence to conclude that the mean damages differ for the three bumper types. SPSS output for the ANOVA is: ANOVA Damage Sum of Squares Between Groups Within Groups Total

df

Mean Square

148.000

2

74.000

12.000

3

4.000

160.000

5

F 18.500

Sig. 0.021

12.9. Each confidence interval is an ordinary 94% confidence interval, and has plus or minus part

1 1 2.951 4.0     5.9 ; only A and B are not significantly different  2 2 2 3 13 A B C 12.10. (a) F = 13.00/0.47 = 27.6; df1 = 2; df2 = 297; P = 0.000. There is sufficient evidence to conclude that the population mean customer satisfaction ratings differ between the cities. (b) Since all of the sample sizes are the same, the margin of error for separate 95% confidence intervals will be the same for comparing the population means for each pair of cities.

1   1 1.972 0.47     0.19 . We are 95% confident that the interval –0.39 to –0.01  100 100  contains the true difference in mean customer satisfaction ratings between San Jose and Toronto. We are 95% confident that the interval 0.31 to 0.69 contains the true difference in mean customer satisfaction ratings between San Jose and Bangalore. We are 95% confident that the interval 0.51 to 0.89 contains the true difference in mean customer satisfaction ratings between Toronto and Bangalore. (c) The Bonferroni method uses error probability 0.05/3 = 0.0167 for each interval, leading to a larger margin of error than for the individual confidence intervals. The Tukey method is similar in theory. The advantage of this approach is that the overall error probability is 0.05 instead of 0.05 for each interval. (d) For San Jose, z1 = 1 and z2 = 0, so yˆ = 7.6. For Toronto, z1 = 0 and z2 = 1, so yˆ = 7.8. For Bangalore, z1 = 0 and z2 = 0, so yˆ = 7.1. 12.11. Because 0 is not included in the confidence interval for comparing the population means of blacks and whites, we can conclude that there is a significant difference between the two population means. We are 95% confident that blacks watch TV between 1.1 and 1.7 hours more each day on the average than do whites. Because 0 is not included in the confidence interval for comparing the population means of blacks and the other category, we can conclude that there is a significant difference between the two population means. We are 95% confident that blacks watch TV between 0.9 and 2.1 hours more on the average each day than do those in the other category. 12.12. (a) Does the population mean number of good friends differ for those who consider themselves very happy, pretty happy, and not too happy? (b) F = 3.47, P = 0.032; At the 0.05 significance level there appears to be a difference in the population mean number of good friends 110 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

for the happiness categories. (c) Since 0 is contained in the Tukey intervals for the comparison of very happy and not too happy, these two groups are not significantly different. Similarly, the pretty happy and not too happy groups are not significantly different. Since the Tukey interval comparing very happy and pretty happy is entirely above zero, these two groups are significantly different, and that the mean number of good friends is higher for those who are very happy than those who are pretty happy. 12.13. (a) For 10 groups, there are 10(9)/2 = 45 comparisons. Using the Bonferroni method with error probability 0.20/45 = 0.0044 for each guarantees at least 80% confidence for the entire set. Since the sample sizes are large, the t score is close to the t score with single-tail probability 0.0022, which is 2.88. (b) For 5 groups, there are 10 comparisons. The Bonferroni method with error probability 0.02 for each uses the t score with single-tail probability 0.01, which is 2.35. The t score increases, and the confidence intervals tend to be wider, as the number of groups increases. 12.14. H0 : 1  2  3 ; H a : at least two of the population means are unequal. Betweengroups SS = 36.0; within-groups SS = 45.333, F = 0.79, P = 0.48. There is not enough evidence to conclude that the mean amount of time of REM sleep differs for the three groups. The

1 1    13.95 .  4 4

minimum significant difference is 2.93 45.333 

12.15. (a) E  y     1z1  2 z2 , with z1 = 1 for group 1 and 0 otherwise, z2 = 1 for group 2 and 0 otherwise. H0 : 1  2  3 is equivalent to H0 : 1  2  0 . (b) For group 1, yˆ = 18 – 6(1) – 3(0) = 12.0. For group 2, yˆ = 18 – 6(0) – 3(1) = 15.0. For group 3, yˆ = 18 – 6(0) – 3(0) = 18.0. These are the means listed at the bottom of the SAS printout. 12.16. (a) E  y     1z1  2 z2 , with z1 = 1 for Bumper A and 0 otherwise, z2 = 1 for Bumper B and 0 otherwise. (b) H0 : 1  2  3 is equivalent to H0 : 1  2  0 . (c) The prediction equation is yˆ  13 11z1 10z2 . 12.17. Sex does not appear to be a significant factor in the comparison of mean number of hours a day watching TV (P = 0.55, and the means for each sex are comparable within race). Race does appear to be a significant factor in the comparison of mean number of hours a day watching TV (P = 0.000, and the means for each race are quite different, with blacks having a higher mean than whites). 12.18. (a) Gender does not appear to be a significant factor in the comparison of mean ideal number of kids for a family (P = 0.55). Race does appear to be a significant factor in the comparison of mean ideal number of kids for a family (P = 0.000). (b) The estimated mean for black females is yˆ = 2.42 + 0.04(1) + 0.37(1) = 2.83. The estimated mean for black males is yˆ = 2.42 + 0.04(1) + 0.37(0) = 2.46. The estimated mean for white females is yˆ = 2.42 + 0.04(0) + 0.37(1) = 2.79. The estimated mean for white males is yˆ = 2.42 + 0.04(0) + 0.37(0) = 2.42. These means are close enough to each other that there does not appear to be any interaction between gender and race.

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12.19. Females (s = 1): 3.79, 3.96, 4.50; Males (s = 0): 3.87, 4.04, 4.58. For instance, for female Democrats, yˆ = 4.58 – 0.71(1) – 0.54(0) – 0.08(1) = 3.79. The means for females and males at each level of party ID are very similar, indicating that there is a lack of interaction. 12.20. yˆ = 30.2 + 0.76g + 0.62r; white females = 30.2 + 0.76(0) + 0.62(0) = 30.2; white males = 30.2 + 0.76(1) + 0.62(0) = 30.96; nonwhite females = 30.2 + 0.76(0) + 0.62(1) = 30.82; nonwhite males = 30.2 + 0.76(1) + 0.62(1) = 31.58. 12.21. (a) The mean number of hours per day watching TV is 0.5 hours more than for those whose religious affiliation is none or other. (b) E  y     1z1  2 z2  3 z3  4 z4 For y to be independent of religious affiliation for each sex,

2  3  4  0 .

12.22. (a) The response variable is population mean hourly wage. The two factors are sex and whether the job is classified as white-collar, blue-collar, or service jobs. (b) The mean hourly wages are White-collar Blue-collar Service Males $22 $14 $11 Females $15 $10 $8 (c) (i) For white-collar jobs, the mean hourly wage is $7 more for males than for females. (ii) For blue-collar jobs, the mean hourly wage is $4 more for males than for females. Since this difference is not the same for both job categories, there is evidence of interaction. The disparity between mean hourly wages for males and females increases from $3 for service jobs to $4 for blue-collar jobs to $7 for white-collar jobs. 12.23. (a) The response variable is population median income. The two factors are sex and race. Black White Males $30,886 $40,350 Females $25,736 $29,661 (b) The difference in median income between black males and black females is $5150, while the difference in median income between white males and white females is $10,689. Since this difference in median income is not equal for both races, there is interaction present. (c) One possible set of median incomes would be $35,000 for black males and white males and $30,000 for black females and white females. 12.24. ―No interaction‖ means that the differences in political ideology are consistent across religion and also between the sexes. The large difference in sample mean political ideology between Jewish women and Jewish men (compared to the other groups, for which the means are relatively similar) suggests interaction is present. 12.25. n = 206, SSE = 3600, df values are 1, 2, 2, 200, mean square values are 100, 100, 50, 18, F values are 5.6, 5.6, 2.8, P-values (Sig) are P < 0.05, P < 0.01, P > 0.05. 12.26. (b) SSE for interaction model = 1565.886; SSE for no interaction model = 1569.525 Partial SS for interaction = 3.64 = 1569.525 – 1565.886 (c) Predicted means = sample means 12.27. Predicted means are 16 for black women, 19 for white women, 18 for black men, 29 for white men; if the mean increases by 8 for black men or white women or decreases by 8 for black women or white men, the interaction disappears. 112 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

12.28. (a) The overall sample means for females and males are more similar than they are for each race. (b) A two-way ANOVA shows the interaction between gender and race, which means that gender is important in the model. However, a one-way ANOVA for gender does not show a significant effect. 12.29. (a) Tests of Within-Subjects Effects Source Influence Error(Influence)

Type III Sum of Squares

df

Mean Square

2.889

2

1.444

12.444

22

0.566

F 2.554

Sig. 0.101

(b) This assumes that the relative distances of the responses differ. The conclusions are sensitive to the choice of scores. While the decision for the hypothesis test remains the same, Tests of Within-Subjects Effects Source Influence Error(Influence)

Type III Sum of Squares 7.722 42.278

df 2 22

Mean Square 3.861 1.922

F 2.009

Sig. 0.158

12.30. (a) (i) response, (ii) fixed, (iii) random. (b) F = 9.25, df1 = 2, df2 = 18, P = 0.002. There is very strong evidence that the means differ for at least two of the issues. 12.31. Between-subjects: gender; within-subjects: issue. Interaction: F = 1.23, df1 = 2, df2 = 16, P = 0.32. There is no evidence of interaction. Main effect of gender: F = 0.26, df1 = 1, df2 = 8, P = 0.62. There is no evidence of a difference in means between males and females, for each issue. Main effect of issue: F = 9.48, df1 = 2, df2 = 16, P = 0.002. There is very strong evidence that the means differ for at least two of the issues. 12.32. During a given period, the same subjects (rather than independent samples) evaluated liberals and conservatives. Time is between-subjects, and group rated is within subjects. 12.33. (a) F = 5.42, df1 = 2, df2 = 69, P = 0.006. There is strong evidence of an interaction between treatment and time. (b) Since there is evidence of an interaction between treatment and time, testing for main effects is meaningless since the main effects cannot be separated out from the interaction. 12.34. Results should match the results on page 396. 12.35. (a) ANOVA Sum of Squares Between Groups

df

Mean Square

72.607

1

72.607

Within Groups

812.376

58

14.006

Total

884.983

59

F 5.184

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Sig. 0.027

(b) Model with interaction: Tests of Between-Subjects Effects Source Model

Type III Sum of Squares 116.522

3

Mean Square 38.841

Error

768.462

56

13.723

Total

884.983

59

Source Gender

df

Type III Sum of Squares 2.241

df 1

Mean Square 2.241

F 2.830

Sig. 0.047

F .163

Sig. 0.688

Vegetarian

26.815

1

26.815

1.954

0.168

Gender * Vegetarian

29.608

1

29.608

2.158

0.147

Model with no interaction: Tests of Between-Subjects Effects Source Model

Type III Sum of Squares 86.914

df 2

Mean Square 43.457 14.001

Error

798.069

57

Total

884.983

59

Source Gender Vegetarian

Type III Sum of Squares 63.617 14.307

df

F 3.104

Sig. 0.053

1

Mean Square 63.617

F 4.544

Sig. 0.037

1

14.307

1.022

0.316

12.36. The linear regression model assumes a linear relationship between the two variables, whereas the ANOVA model looks for any difference among the means. For example, if the population means on y were 3 when x = 0, 1 when x =1, and 3 when x = 2, there is not a linear trend and the regression model would probably not detect the fact that the population means of y are different. 12.38. There is strong evidence of a relationship between the mean number of good friends and how often subjects attend religious services. There is not much difference between the mean number of good friends for those who attend religious services with low or medium frequency, but those who attend religious services with high frequency have a higher mean number of good friends than those in either of the other two groups. 12.39. (a) If F = 0, each of the four sample means equaled 60. (b) If F = ∞, the observations in each group were identical. 12.40. The P-value had to be less than 0.05/35 = 0.0014 for a given test to be significant. 12.41. Neither men nor women as a group have a stronger verbal memory for abstract words or for concrete words than those of the opposite gender. 12.42. (a) With the single-comparison approach, in the long run 95% of the intervals will contain the true differences in means. With the multiple comparisons method, in the long run 95% of the 114 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

time the entire set of comparisons is correct, and any one comparison has confidence greater than 95%. (b) Suppose the sample means equal 8 for A, 15 for B, and 22 for C. 12.43. There is no evidence of an interaction between race and gender, so we can examine the main effects of each factor. It appears that both race and gender are statistically significant. There is strong evidence that the means differ between the sexes and (separately) between the races. 12.44. The response variable is mean number of dates in the past three months. One factor is gender and the other factor is attractiveness. These data do appear to show interaction between gender and attractiveness, because the mean number of dates was very similar for the two attractiveness groups for males but very different for females. 12.45. a. 10 10 b. 10 20 c. 10 20 d. 10 10 20 20 30 40 30 60 10 10 12.46. No. For instance, at each level of a factor B, the mean at the first level of factor A could be higher than the mean at the second level of A, yet overall (ignoring B) the means could be equal at both levels of A. 12.47. No, not unless at each level of B the sample sizes are equal at each level of A. A main effect may not be significant in a two-way ANOVA yet attain strong significance in a one-way ANOVA. 12.48. (a) Science Humanities Women 80,000 (5) 66,000 (25) Men 79,000 (30) 65,000 (20) The overall means are $68,333 for women and $73,400 for men. The mean is higher overall for men even though it is lower in each division. (b) Controlling for division, women have the higher mean, whereas for a one-way analysis ignoring division, men have the higher mean. 12.49. Model with interaction: Tests of Between-Subjects Effects Source Model

Type III Sum of Squares

df

Mean Square

68.893

3

22.964

Error

48.222

22

2.192

Total

117.115

25

Source Church Religion Church * Religion

Type III Sum of Squares

df

Mean Square

F 10.477

F

Sig. 0.000

Sig.

33.212

1

33.212

15.152

0.001

11.778

1

11.778

5.373

0.030

0.709

1

0.709

0.323

0.575

115 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

Model with no interaction: Tests of Between-Subjects Effects Source Model

Type III Sum of Squares 68.185

2

Mean Square 34.092

Error

48.931

23

2.127

Total

117.115

25

Source Church

Type III Sum of Squares 36.843

Religion

11.069

df

df

F 16.025

Sig. 0.000

1

Mean Square 36.843

F 17.318

Sig. 0.000

1

11.069

5.203

0.032

12.50. False, since the differences between mean annual medical expenses is small between smokers and nonsmokers under 50 and is large between smokers and nonsmokers over 50, there is evidence of an interaction. 12.51. (a), (b), (c), and (d) 12.52. (c) 12.53. (c), (e), and (f) 12.54. (c) and (d) are both correct 12.55. You have enough information. The sample standard deviations can be combined to obtain the pooled (within) estimate. The between estimate can be calculated from the separate sample means, since the overall mean is a weighted average of their values. 12.56. (a) (i) (0.95)5 = 0.77, (ii) 1 – 0.77 = 0.23. (b) (0.9898)5 = 0.95. The confidence coefficient for each interval in the Bonferroni method is (1 – 0.05/5) = 0.99. 12.57. (a) This is the sample variance of the sample mean values. (b) Multiply both sides by n in part (a).

Chapter 13 13.1. (a) Use the first equation (ignoring father’s education) to get 11 + 2(1) = 13 for whites and 11 for nonwhites, for which the difference = 2. (b) The lines are parallel to one another, with slope 0.8 and intercept for whites = 2.4 and nonwhites = 3.0. (c) The coefficient of z in the second equation, which is –0.6. (d) Substitute x = 12 into the second equation, yielding 12.0 for whites (z = 1) and 12.6 for nonwhites (z = 0). 13.2. (a) 10.5 = value of predicted percentage who use the Internet when z = 0; 30.4 = the difference in predicted percentage who use the Internet between a Western nation and a nonWestern nation. (b) 8.3 = value of predicted percentage who use the Internet when x = 0 and z = 0; 0.73 = the increase in predicted percentage who use the Internet for each 1 unit increase in GDP, controlling for nation; 10.9 = the difference in predicted percentage who use the Internet between a Western nation and a non-Western nation, controlling for GDP. 116 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

13.3. (a) The predicted proportion of pro-choice votes was 0.167 lower for Democrats, controlling for other predictors. (b) Ideology seems to be, by far, the most important predictor of proportion of pro-choice votes. A standard deviation increase in ideology corresponds to a 0.83 standard deviation predicted increase in the response, controlling for the other variables in the model. 13.4. (a) The predicted number of assembly defects decreases by 0.78 for each additional hour in assembly, controlling for facility type. The predicted number of assembly defects decreases by 36.2 if the facility is Japanese, controlling for assembly time. (b) The bivariate effect of time to assemble each vehicle has a positive coefficient, meaning that the predicted number of assembly defects increases with time in assembly. However, once facility type is controlled, the partial effect of time is negative, meaning that the predicted number of assembly defects decreases with time in assembly. 13.5. (a) yˆ  8.3  9.8 f  5.3s  7.0m1  2.0m2 1.2m3  0.501x (b) yˆ = 8.3 + 9.8(1) + 7.0(1) + 0.501(10) = 30.1. 13.6. (a) t = –5.3/1.6 = –3.3, P = 0.001; the mean alcohol consumption is lower for females than for males, controlling for all other variables. (b) –5.3 ± 1.96(1.6) = –8.4 to –2.2; the mean alcohol consumption is between 2.2 and 8.4 drinks per month lower for females than males, controlling for other variables. (c) It compares the mean for the divorced group to the baseline group (married), controlling for other predictors. (d) To compare all four levels of marital status, we need to conduct the F test comparing this model to the simpler model without these three dummy variables. 13.7. (a) yˆ  40,230.9 116.1s  57,736.3n ; for new homes, yˆ  17,505.4  116.1s and for old homes, yˆ  40,230.9 116.1s . (b) (i) yˆ = 17,505.4 + 116.1(3000) = $365,901, (ii) yˆ = –40,230.9 + 116.1(3000) = $308,165. 13.8. (a) (i) yˆ  100,755.3 166.4s , (ii) yˆ  22,227.8  104.4s . (b) (i) yˆ = –100,755.3 + 166.4(3000) = $398,445, (ii) yˆ = –22,227.8 + 104.4(3000) = $291,086. (c) yˆ = –100,755.3 + 166.4(1500) = $148,845, (ii) yˆ = –22,227.8 + 104.4(1500) = $134,429. The difference in predicted prices increases as the size of the home increases for the model allowing interaction. 13.9. Answers should match those in the chapter. 13.10. (a) Do percentage of adults registered to vote, racial-ethnic representation, or both impact the percentage of adults voting? (b) Anglos: yˆ = –4.09 + 0.74x; Blacks: yˆ = –5.63 + 0.74x; Mexican-Americans: yˆ = –2.78 + 0.74x. (c) Anglos: yˆ = –1.27 + 0.70x; Blacks: yˆ = 1.56 + 0.60x; Mexican-Americans: yˆ = –8.24 + 0.88x. (d) F = [(673.17 – 619.38)/2]/18.22 = 1.48, df1 = 2, df2 = 34, P > 0.05, so there is insubstantial evidence of interaction. (e) F = [(713.25 – 673.17)/2]/18.7 = 1.07, df1 = 2, df2 = 36, P > 0.05, so there is insubstantial evidence of a racialethnic effect, controlling for percent registered to vote. (f) F = 123.9, df1 = 1, df2 = 36, P < 0.001, there is very strong evidence of a positive association between these two variables, controlling for race-ethnicity. (g) It appears that only the percentage of adults registered to vote has an effect on the percentage of adults voting. There is no interaction between percentage of adults registered to vote and racial-ethnic representation, and there is no effect of a racial-ethnic effect, controlling for percent registered to vote. 117 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

13.11. (a) Anglos: –4.09 + 0.74(60.4) = 40.6. The unadjusted mean is very different from the adjusted mean. Controlling for percent registered decreases the mean. 13.12.

The

analysis

of

covariance

model

equation

is

yˆ  18.178  4.625x 10.726z1  4.668z2 1.232z3 , where the zs are dummy variables for

black, Hispanic, and Asian. For a given racial-ethnic group, the estimated mean income increases by 4.625 thousand dollars for each additional year of education. The coefficients of the dummy variables are the estimated differences between mean income of each group and whites, controlling for education. The difference is largest (10.7 thousand) between blacks and whites and smallest (1.2 thousand) between Asian Americans and whites. The separate prediction lines for each of the four groups are Asian: yˆ  19.41  4.625x Black: yˆ  28.904  4.625x Hispanic: yˆ  22.846  4.625x White: yˆ  18.178  4.625x 13.13. Interaction model: F = 1.43, df1 = 3, df2 = 82, P = 0.24 Model with no interaction: F = 2.06, df1 = 3, df2 = 85, P = 0.11 Test for effect of education, controlling for racial-ethnic group: t = 7.9, df = 85, P = 0.0001 13.14. (a) 12.0 and 13.6, compared to 13 and 11. (b) The unadjusted mean is higher for whites, but the adjusted mean is higher for nonwhites. (This is an example of Simpson’s paradox.) 13.15. Tests of Between-Subjects Effects Type III Sum of Squares

Source Model

df

Mean Square

71.434

3

23.811

Error

86.499

56

1.545

Total

157.933

59

F 15.416

Sig. 0.000

Parameter Estimates Parameter Intercept

B 2.370

0.793

2.988

0.004

[Vegetarian=n]

-0.484

0.836

-0.579

0.565

[Vegetarian=y]

0(a)

.

.

.

-0.111

0.507

-0.219

0.827

[Vegetarian=n] * RelServices

1.246

0.537

2.320

0.024

[Vegetarian=y] * RelServices

0(a)

.

.

.

RelServices

Std. Error

t

Sig.

a This parameter is set to zero because it is redundant.

There is no significant evidence of interaction. So we next use the simpler model without the interaction term.

118 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

Tests of Between-Subjects Effects Source Model

Type III Sum of Squares

df

Mean Square

63.123

2

31.561

Error

94.810

57

1.663

Total

157.933

59

F

Sig.

18.975

0.000

Parameter Estimates Parameter Intercept

B

Std. Error

t

Sig.

0.887

0.487

1.821

0.074

[Vegetarian=n]

1.151

0.468

2.461

0.017

[Vegetarian=y]

0(a)

.

.

.

1.001

0.172

5.810

0.000

RelServices

a This parameter is set to zero because it is redundant.

Political ideology tends to be more conservative for the more religious and for nonvegetarians.

yˆ  2.04  0.01x  0.01x2  0.38x4  0.15x5 ; Political nonconservatives: yˆ  0.94  0.13x  0.01x2  0.38x4  0.15x5 ; since the coefficient for

13.18. (a) Political conservatives:

education is very small for the political conservatives, it appears that education does not have as much of an effect for political conservatives as for political nonconservatives, and that political nonconservatives tend to be more accepting of homosexuality as education increases, controlling for the other variables. (b) Attitudes toward homosexuality increase (become less negative) by 0.09 for each one year increase in education, controlling for all other variables. Attitudes toward homosexuality decrease (become more negative) by 0.01 for each one year increase in age, controlling for all other variables. Attitudes toward homosexuality decrease (become more negative) by 0.49 for political conservatives, controlling for all other variables. Attitudes toward homosexuality decrease (become more negative) by 0.39 for religious fundamentalists, controlling for all other variables. Attitudes toward homosexuality decrease (become more negative) by 0.15 for those who live in the same city as when age 16, controlling for all other variables. 13.20. (a) (i) Least permissive in sexual attitudes are old white females with low levels of education from the South who are fundamentalist Protestants and who are intolerant of freedom of speech. (ii) Most permissive are young black males with high levels of education from the nonSouth who are Jewish and tolerant of freedom of speech. 13.21. The interaction model is Tests of Between-Subjects Effects Dependent Variable: Fertility Source Model

Type III Sum of Squares 88.235

df 5

Mean Square 17.647 1.789

Error

82.284

46

Total

170.519

51

F 9.865

119 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

Sig. 0.000

Parameter Estimates Dependent Variable: Fertility Parameter Intercept

B 7.372

Std. Error 0.958

t 7.693

Sig. 0.000

Education

-0.242

0.142

-1.701

0.096

[rural_m=0]

-1.719

0.796

-2.159

0.036

[urban_m=0]

-0.105

0.805

-0.130

0.897

[urban_n=0]

0(a)

.

.

.

[rural_m=0] * Education

0.017

0.124

0.140

0.890

[urban_m=0] * Education

-0.023

0.111

-0.203

0.840

[urban_n=0] * Education

0(a)

.

.

.

a This parameter is set to zero because it is redundant.

The no interaction model is Tests of Between-Subjects Effects Dependent Variable: Fertility Source Model

Type III Sum of Squares

df

Mean Square

88.004

3

29.335

Error

82.516

48

1.719

Total

170.519

51

F

Sig.

17.064

0.000

F 194.266

Sig. 0.000

F 393.710

Sig. 0.000

Parameter Estimates Dependent Variable: Fertility Parameter Intercept

B 7.399

Std. Error 0.596

t 12.405

Sig. 0.000

Education

-0.242

0.045

-5.325

0.000

[rural_m=0]

-1.641

0.450

-3.648

0.001

[urban_m=0]

-0.239

0.465

-0.515

0.609

[urban_n=0]

0(a)

.

.

.

a This parameter is set to zero because it is redundant.

13.22. (a) The model with interaction is Tests of Between-Subjects Effects Dependent Variable: Price Source Model

Type III Sum of Squares 155811.606

df 3

Mean Square 51937.202 267.351

Error

23794.261

89

Total

179605.867

92

Source Size

Type III Sum of Squares 105258.928

df 1

Mean Square 105258.928

New

1233.323

1

1233.323

4.613

0.034

New * Size

3439.395

1

3439.395

12.865

0.001

120 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

Parameter Estimates Dependent Variable: Price Parameter Intercept

B

Std. Error

-48.426

t

Sig.

13.454

-3.599

0.001

Size

95.996

7.315

13.123

0.000

[New=0]

31.826

14.818

2.148

0.034

-29.392

8.195

-3.587

0.001

[New=0] * Size

The interaction term is statistically significant (P = 0.001) and is needed in the model. (b) The model with the outlier removed is Tests of Between-Subjects Effects Dependent Variable: Price_no_outlier Source Model

Type III Sum of Squares 114604.677

df 3

Mean Square 38201.559 232.709

Error

20478.433

88

Total

135083.110

91

F 164.160

Sig. 0.000

Tests of Between-Subjects Effects Dependent Variable: Price_no_outlier Source Size_no_outlier

Type III Sum of Squares

df

Mean Square

F

Sig.

44299.727

1

44299.727

190.365

0.000

New_no_outlier

61.168

1

61.168

0.263

0.609

New_no_outlier * Size_no_outlier

57.372

1

57.372

0.247

0.621

Parameter Estimates Dependent Variable: Price_no_outlier Parameter Intercept

B -7.601

Std. Error 16.569

t -0.459

Sig. 0.648

Size_no_outlier

71.577

[New_no_outlier=0]

-8.999

9.404

7.612

0.000

17.553

-0.513

[New_no_outlier=0] * Size_no_outlier

0.609

-4.973

10.015

-0.497

0.621

The interaction term is not statistically significant (P = 0.621) and is therefore not needed in the model. With the one outlier removed, the interaction becomes insignificant. 13.23. (a) Without including D.C. in the model, we have Tests of Between-Subjects Effects Dependent Variable: VI_noDC Source Model

Type III Sum of Squares 441576.893

Error Total

df 3

Mean Square 147192.298

1075889.987

46

23388.913

1517466.880

49

F 6.293

121 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

Sig. 0.001

Source PO_noDC

Type III Sum of Squares

df

Mean Square

F

Sig.

30132.368

1

30132.368

1.288

0.262

zSouth

267399.493

1

267399.493

11.433

0.001

zSouth * PO_noDC

206193.944

1

206193.944

8.816

0.005

Parameter Estimates Dependent Variable: VI_noDC Parameter Intercept

B 713.399

Std. Error 161.677

t 4.413

Sig. 0.000

PO_noDC

-17.592

13.080

-1.345

0.185

[zSouth=0]

-676.259

200.003

-3.381

0.001

56.957

19.183

2.969

0.005

[zSouth=0] * PO_noDC

The interaction term is statistically significant (P = 0.005) and is needed in the model. (b) Including D.C. in the model, we have Tests of Between-Subjects Effects Dependent Variable: VI Source Model

Type III Sum of Squares 785727.939

Error Total

Source PO

Type III Sum of Squares 290673.508

df 3

Mean Square 261909.313

2156899.982

47

45891.489

2942627.922

50

df

F 5.707

Sig. 0.002

1

Mean Square 290673.508

F 6.334

Sig. 0.015

zSouth_withDC

34481.290

1

34481.290

0.751

0.390

zSouth_withDC * PO

15436.205

1

15436.205

0.336

0.565

Parameter Estimates Dependent Variable: VI Parameter PO

B

[zSouth_withDC=0] [zSouth_withDC=0] * PO

Std. Error

t

Sig.

24.620

16.125

1.527

0.134

-229.350

264.590

-0.867

0.390

14.745

25.424

0.580

0.565

The interaction term is not statistically significant (P = 0.565) and is therefore not needed in the model. With D.C. added, the interaction becomes insignificant. 13.24. Fit an analysis of covariance model containing x and a dummy variable for the two groups as predictors. Fit also the more complex model containing an interaction between these two terms. Then use the F test based on comparing SSE values for the two models, or simply use a t test based on dividing the coefficient of the cross-product term by its standard error; the F statistic equals the square of the t statistic. 13.25. E  y     1x  2 z  3  xz  , where x = frequency of going to bars, z = 1 for married subjects and z = 0 for unmarried subjects. 122 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

13.27. See Figure 13.3b. 13.28. Since the adjusted means are identical, the same line with positive slope relating income to age applies to each marital status. The value of that line at a group’s mean age equals the unadjusted mean income for that group. 13.29. (a) and (d) are both correct 13.30. (a) 13.31. A one-way ANOVA describes the effect of a qualitative predictor on y, by analyzing differences among means of y for different categories of that predictor. An analysis of covariance describes effects of qualitative and quantitative predictors on y. 13.32. 2 is difference between the means of y for the two groups when x falls at its mean. Without the centering, it is the difference between the means when x = 0.

Chapter 14 14.1. (a) Backward elimination gives the no interaction model with LIFE and SES predictors. Coefficients(a) Unstandardized Standardized Coefficients Coefficients Model 2

B

Std. Error

Beta

t

Sig.

(Constant)

28.230

2.174

12.984

0.000

LifeEvents

0.103

0.032

0.428

3.177

0.003

-0.097

0.029

-0.451

-3.351

0.002

SES

a Dependent Variable: MImpairment

(b) Forward selection gives the no interaction model with LIFE and SES predictors. Coefficients(a) Unstandardized Standardized Coefficients Coefficients Model 2

(Constant) SES

B 28.230 -0.097

Std. Error 2.174 0.029

LifeEvents

0.103

0.032

Beta -0.451

t 12.984 -3.351

Sig. 0.000 0.002

0.428

3.177

0.003

a Dependent Variable: MImpairment

14.2. (a) In backward elimination, (i) gender equality would be eliminated first since it has the smallest (in absolute value) t statistic, (ii) political freedom would be eliminated second since it has the second smallest (in absolute value) t statistic. (b) In forward selection, life expectancy would be added first since it has the largest (in absolute value) t statistic. 14.3. (a) BEDS, because it is least significant in the model with all the predictors. (b) SIZE, because it is most highly correlated with selling price and would have the smallest P-value. (c) BEDS can be predicted well knowing SIZE, BATHS, and NEW, so it does not uniquele explain much variability in selling price. 123 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

14.4. (a) Adjusted R-squared = 0.8637 is highest for model with predictors size, bath, new. (b) PRESS = 27860 is lowest for model with predictors. (c) C(p) = 3.4878 for model with predictors size, bath, new selects the same model as does adjusted R-squared and PRESS criteria. 14.5. (a) The backward elimination model includes M = percent in metropolitan areas, W = percent white, H = percent high school graduates, P = percent with income below the poverty level, S = percent of families headed by a single parent. Coefficients(a) Unstandardized Standardized Coefficients Coefficients Model 2

B (Constant)

Std. Error

-5.363

4.039

M_noDC

0.050

0.013

W_noDC

-0.093

P_noDC S_noDC

0.416 1.028

Beta

t

Sig.

-1.328

0.191

0.271

3.785

0.000

0.027

-0.257

-3.376

0.002

0.071 0.217

0.448 0.381

5.864 4.734

0.000 0.000

t 0.429

Sig. 0.670

a Dependent Variable: MR_noDC

(b) The forward selection model includes all five variables. Coefficients(a) Unstandardized Standardized Coefficients Coefficients Model 5

B 3.049

Std. Error 7.100

1.137 -0.117 0.047 -0.078

0.228 0.081 0.013 0.029

0.421 -0.165 0.256 -0.216

4.993 -1.433 3.585 -2.690

0.000 0.159 0.001 0.010

0.288 a Dependent Variable: MR_noDC

0.114

0.310

2.535

0.015

(Constant) S_noDC H_noDC M_noDC W_noDC P_noDC

Beta

(c) The stepwise regression model excludes only H = percent high school graduates. Coefficients(a) Unstandardized Standardized Coefficients Coefficients Model 6

B (Constant)

Std. Error

-5.363

4.039

1.028

0.217

M_noDC

0.050

0.013

W_noDC P_noDC

-0.093 0.416

0.027 0.071

S_noDC

Beta

t

Sig.

-1.328

0.191

4.734

0.000

0.271

3.785

0.000

-0.257 0.448

-3.376 5.864

0.002 0.000

0.381

a Dependent Variable: MR_noDC

(d) The methods look at the additional information provided by the variable with the other variables included in the model. In backward elimination, high school education is not statistically significant once the other four variables are in the model. In forward selection, high school education is statistically significant when only the percent of families headed by a single parent is in the model. 124 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

(e) The backward elimination model includes M = percent in metropolitan areas, H = percent high school graduates, P = percent with income below the poverty level. The forward selection model includes H, M, P as well (and agrees with the backward elimination model). An outlier has an impact of which variables are included or excluded from the model in automatic selection procedures. 14.6. This plot suggests that the variability increases as the predicted y value increases. This is called heteroscedasticity. 14.7. (a) Observation 25 is an outlier with studentized residual = 3.06. (b) Observations 38 and 39 have the most noticeable large leverage values. (c) No observations have both large studentized residuals and large leverage. (d) Observation 39 has a very large value of DFFIT (1.09). (e) Observation 39 has a large (–0.83) DFBETA value for literacy, but a small (0.01) DFBETA value for female economic activity. Scatterplot 14.8. (a) D.C. is a clear regression outlier.

Regression Studentized Residual

Dependent Variable: VI

4

2

0

-2 0

500

1000

1500

2000

VI

(b) No observations have noticeable leverage. (c) Since D.C. is not a leverage point, it may not be particularly influential on the regression. (d) D.C. has a very strong influence on the fitted values. (e) D.C. has a very strong influence on the parameter estimate. (f) The predicted equation with D.C. in the model is yˆ  98.893  7.265x . When D.C. is removed from the model, yˆ  64.337  4.714x . Both the intercept and the slope are very different for the regression model without D.C. 14.9. (a) For the complete data set, yˆ  47.992  62.263size  20.072baths  18.371new . For the data set without the three outliers (observations 5, 89, and 93), yˆ  31.625  55.105size 17.248baths 17.454new . The R2 value is 0.868 for the model with the outliers and 0.847 for the model without the outliers. The standard errors are slightly higher for the model with the outliers. Thus, including the outliers does not change the R2 value by much and it drops the standard errors slightly. (b) The VIF values for the model with the outliers excluded are

125 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

Coefficients(a) Model

VIF Size_no_outlier

1.676

Baths_no_outlier

1.642

New_no_outlier

1.055

a Dependent Variable: Price_no_outlier

The standard error for whether new is not affected much by correlation with the other predictors, but the other two standard errors multiply by a factor of about 1.3. 14.10. (a) The interaction term is highly significant with P = 0.001. Coefficients(a) Unstandardized Standardized Coefficients Coefficients Model 1

B -16.600

Std. Error 6.210

Size

66.604

3.694

New

-31.826 29.392

(Constant)

SizeNew

Beta

t -2.673

Sig. 0.009

0.792

18.033

0.000

14.818

-0.332

-2.148

0.034

8.195

0.571

3.587

0.001

a Dependent Variable: Price

(b) Observation 5 has a high studentized residual, high leverage, and high DFFIT and DFBETA. (c) The interaction term is not significant (P = 0.621) with observation 5 removed. Coefficients(a) Unstandardized Standardized Coefficients Coefficients Model 1

B -16.600

(Constant)

Std. Error 5.794

Beta

t -2.865

Sig. 0.005

Size_no_outlier

66.604

3.446

0.862

19.328

0.000

New_no_outlier

8.999

17.553

0.107

0.513

0.609

SizeNew_no_outlier

4.973

10.015

0.105

0.497

0.621

a Dependent Variable: Price_no_outlier

(d) Using a gamma distribution for y, P = 0.263 for the interaction, and we find that the interaction is not statistically significant. 14.11. Partial correlations, like partial regression coefficients, have large standard errors when there is multicollinearity. 14.12. (a) No. Both x1 and x2 would be highly positively correlated with y. However, rx1x2 would be very close to 1.0, and once one of them is in the model, the other would be redundant for practical purposes. Thus, the partial effects might well be nonsignificant. (b) No, this test would probably have a very small P-value, since R2, like either ryx1 or ryx2 , would be large. (c) Select the model using x2 alone as the predictor. It would be selected first, since it is most highly correlated with y, but once it is in the model, x1 would not make a significant contribution toward explaining additional variation, so it would not be added to the model.

126 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

14.13. (a) The variability increases as yˆ increases, which the gamma distribution allows but the ordinary normal model (i.e., normal distribution with constant standard deviation) does not. (b) For the gamma model, the estimated education effect is a bit weaker and there is stronger evidence of a difference between Hispanics and whites. (c) (i) 20 0.117  6.8 thousand dollars, (ii) 17.1 thousand dollars. 14.14. The quadratic GLM with a gamma distribution for fertility rate gives the equation yˆ  3.235  0.0000983x  0.00000000147 x2 . The minimum occurs at x = 33,435. (Note this solution uses GDP in dollars rather than 10,000s of dollars as in the text. In the alternative units, the coefficient is -0.983 for GPD and 0.147 for its square) Parameter Estimates Parameter

(Intercept) GDP sqGDP

B

Std. Error

3.235

0.2409

-9.83E-005

2.524E-005

15.148

1

0.000

1.47E-009

5.864E-010

6.304

1

0.012

(Scale)

0.088(a) Dependent Variable: Fert Model: (Intercept), GDP, sqGDP a Maximum likelihood estimate.

The

least

Hypothesis Test Wald ChiSquare df Sig. 180.282 1 0.000

squares

0.0196

model

with a quadratic term for GDP yˆ  3.278 1.054x  0.163x (see Example 14.7 on page 466).

gives

the

equation

2

14.15. (a) (iv) is the correct response, since the function increases up to its maximum value at x1 = 0.2/[2(0.001)] = 100. (b) (iii) is correct, since the slope for x1 decreases from 0.07 when x2 = 0 to 0.01 when x2 = 100. 14.16. (a) 30

a

25

20

15

10 0.0

1.0

2.0

3.0

4.0

x

127 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

(b) 50.00

b

40.00

30.00

20.00

10.00 0.0

1.0

2.0

3.0

4.0

3.0

4.0

3.0

4.0

x

(c) 14.00

c

13.00

12.00

11.00

10.00 0.0

1.0

2.0

x

(d) 10.00

d

5.00

0.00

-5.00

-10.00 0.0

1.0

2.0

x

128 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

(e) 10.00

e

9.00

8.00

7.00

6.00 0.0

1.0

2.0

3.0

4.0

3.0

4.0

x

(f) 10.00

f

0.00

-10.00

-20.00

-30.00 0.0

1.0

2.0

x

14.17. (a) yˆ  5507.551  65.156x  0.014x2 ; this model is continually increasing and ―bowl shaped.‖ (b) (i) yˆ = 5507.551 + 65.156(1000) + 0.014(1000)2 = $84,664, (ii) $191,820, (iii) $326.976; with the increasing bowl shape, the curve keeps climbing more quickly as size increases. 14.18. (a) The degree of nonlinearity is minor, since R2 is only slightly larger than r2. The linear association is moderately strong, since r2 is relatively large. (b) t = 1.740, P = 0.085, the quadratic model does not give a significantly better fit than the straight line model. 14.19. (a) yˆ  33.37  4.50x  0.177 x2 ; The test statistic is t = 3.84, df = 48, P = 0.0004; there is very strong evidence of a quadratic effect. (b) yˆ  4.84  0.20x  0.024x2 ; The test statistic is t = 1.19, df = 47, P = 0.24; it is plausible that there is no quadratic effect, that is, that a straight line model is adequate. (c) For D.C., the studentized residual = 16.1, hat-value = 0.34, DFFITS = 11.5, and DFBETAS = 8.4 for the coefficient of x2. No other studentized residual exceeds 3 in absolute value, no other DFFITS value is above 2.1 in absolute value, and no other DFBETAS 129 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

value for the quadratic term exceeds 1.5 in absolute value. Clearly, D.C. is a very unusual observation. 14.20. (a) (i) yˆ = 20.34(1.77)0 = 20.34 million, (ii) yˆ = 20.34(1.77)6 = 625.45 million. (b) The estimated increase in the number of people (in millions) worldwide using the Internet is 77% per year after 1995. (c) 106,647 million, which is over 107 billion people, which exceeds the current world population by a significant factor. Extrapolation is dangerous, because you cannot guarantee that the relationship between x and y continues outside of the range of x values. (d) The linear regression model only allows for an estimated increase in the number of people (in millions) worldwide using the Internet of 81 million people per year after 1995. During this period, there was an explosion in the number of people worldwide using the Internet, and the linear equation cannot model this exponential growth. 14.21. (a) 100,000 = predicted number of articles January 1, 2003, and predicted number is multiplied by 2.1 for each successive year. (b) (i) 100,000(2.1)5 = 4,084,101, (ii) 166,798,810; extrapolation is particularly dangerous with an exponential model, since at some point there will be a leveling off of the growth, whereas the exponential model continues to grow by leaps and bounds. 14.22. (a) 1.014 = 1 + 0.014; the 0.014 corresponds to a 1.4% increase each year. (b) (i) 1.4193(1.014)50 = 2.8443, which is approximately double the initial size, (ii) 1.4193(1.014) 100 = 5.6999, which is approximately quadruple the initial size. (c) The model of log(y) and x seems more appropriate, because the linear association between the two variables is stronger than that between y and x. 14.23. (a) 12.00

11.00

a

10.00

9.00

8.00

7.00

6.00 0

10

20

30

40

x

(b) Predicted world population size (in billions) x years from now if there is a 5% decrease in population size each year.

130 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

6.00

5.00

b

4.00

3.00

2.00

1.00

0.00 0

10

20

30

40

x

(c)  > 1 is ever increasing,  < 1 is ever decreasing 14.24. (a)

125.00

a

100.00

75.00

50.00

25.00

0.00 0

1

2

3

4

5

x

(b) The intercept is 1.386, and the slope is 0.6931.

131 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

5.00

b

4.00

3.00

2.00

1.00 0

1

2

3

4

5

60

70

80

x

14.25. (a) 120

100

DeathRate

80

60

40

20

0 30

40

50

x

132 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

5.00

logDR

4.00

3.00

2.00

1.00 30

40

50

60

70

80

An exponential regression model seems reasonable, since there appears to be roughly a linear relation between age and the log of death rate. (c) log  E  y   1.146  0.747 x . (d) yˆ = 0.318(1.0776)x. The death rate increases by 7.8% for each additional year of age. 14.26. (a) (i) Parameter Estimates Parameter

(Intercept) GDP

B

Std. Error

Hypothesis Test

1.148

0.0764

Wald Chi-Square 226.211

-2.06E-005

5.205E-006

15.653

(Scale)

0.828(a) Dependent Variable: Fert Model: (Intercept), GDP a Maximum likelihood estimate.

df 1

Sig. 0.000

1

0.000

0.1876

(ii) Parameter Estimates

Parameter

(Intercept) GDP (Scale)

B

Std. Error

Hypothesis Test

1.112

0.0736

Wald Chi-Square 228.528

df 1

Sig. 0.000

-1.77E-005

3.165E-006

31.340

1

0.000

0.098(a) Dependent Variable: Fert Model: (Intercept), GDP a Maximum likelihood estimate.

0.0219

133 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

(b) The exponential model provides curvature but with a continual decrease of fertility, which seems more appropriate than a parabolic relationship in which fertility eventually increases. 14.27. (a) For example consider using backward elimination with predictors gender, age, news, TV, abortion, political affiliation (with dummy variables for Republican and Democratic affiliation), religiosity, and affirmative action on the response political ideology. The result is: ANOVA(g) Model 6

Regression

Sum of Squares 100.597

Residual

df

57.335

3

Mean Square 33.532

55

1.042

F 32.167

Sig. 0.000(f)

Total

157.932 58 f Predictors: (Constant), RelServices, zDem, zRep g Dependent Variable: PolIdeology Coefficients(a) Unstandardized Standardized Coefficients Coefficients Model 6

B

Std. Error

(Constant)

2.191

0.262

zRep

2.027

0.384

zDem -0.540 RelServices 0.457 a Dependent Variable: PolIdeology

0.309 0.166

Beta

t

Sig.

8.377

0.000

0.540

5.278

0.000

-0.158 0.265

-1.745 2.749

0.087 0.008

(b) For example, consider using stepwise selection with predictors gender, age, high school GPA, TV, and sports. The result is: Coefficients(a) Unstandardized Standardized Coefficients Coefficients Model 1

B

Std. Error

(Constant)

2.749

0.322

HSGPA

0.213

0.096

Beta 0.278

t

Sig.

8.536

0.000

2.207

0.031

a Dependent Variable: CGPA

14.29. For example, consider using backward elimination with predictors murder rate, percent in metropolitan areas, percent white, percent high school graduates, percent below the poverty level. The result is: ANOVA(e) Model 4

Sum of Squares

df

Mean Square

Regression

865006.800

2

432503.400

Residual

652460.080

47

13882.129

1517466.880

49

Total

F 31.155

d Predictors: (Constant), ME_noDC, MU_noDC e Dependent Variable: VI_noDC

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Sig. 0.000(d)

Coefficients(a) Unstandardized Coefficients Model 4

Standardized Coefficients

B -55.200 44.511

Std. Error 84.575 6.643

ME_noDC 3.402 a Dependent Variable: VI_noDC

1.146

(Constant) MU_noDC

Beta 0.650

t -0.653 6.700

Sig. 0.517 0.000

0.288

2.967

0.005

14.30. Backward elimination using the main effects, quadratic terms, and interaction, results in the model with only the main effects. Residuals and diagnostics do not show any marked abnormality. 14.31. Let x = year – 1830.

log_pop

16.00

14.00

12.00

10.00 0

50

100

150

200

Since log of population is approximately linear in x, the exponential model seems appropriate. Model Information Dependent Variable

pop

Probability Distribution

Normal

Link Function

Log Parameter Estimates

Parameter

B

Std. Error

Hypothesis Test

11.243

0.1686

Wald Chi-Square 4445.564

0.032

0.0011

901.308

177973833307.578(a) Dependent Variable: pop Model: (Intercept), x a Maximum likelihood estimate.

59324611102.5192

(Intercept) x (Scale)

df

135 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

1

Sig. 0.000

1

0.000

Thus, log  E  y   11.243  0.032x , which corresponds to the exponential regression fit yˆ = 76,343.6(1.0325)x. This predicts a 3.3% rate of growth per year. 14.32. (a) 90.0

80.0

Life

70.0

60.0

50.0

40.0 0

10,000

20,000

30,000

40,000

50,000

9.00

10.00

11.00

GDP 90.0

80.0

Life

70.0

60.0

50.0

40.0 6.00

7.00

8.00

log_GDP

One can use log GDP to predict life expectancy. Model Summary Model 1

R

R Square

0.770(a) 0.592 a Predictors: (Constant), log_GDP

Adjusted R Square

Std. Error of the Estimate

0.581

5.8110

136 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

ANOVA(b) Model 1

Sum of Squares

df

Mean Square

Regression

1815.211

1

1815.211

Residual

1249.387

37

33.767

Total

3064.597

38

F

Sig.

53.757

0.000(a)

t 6.712

Sig. 0.000

7.332

0.000

a Predictors: (Constant), log_GDP b Dependent Variable: Life Coefficients(a) Unstandardized Standardized Coefficients Coefficients Model 1

(Constant)

B 36.546

Std. Error 5.445

4.404

0.601

log_GDP

Beta 0.770

a Dependent Variable: Life

There appear to be two regression outliers (Nigeria and South Africa). After removing these two outliers, we get the following regression results: Model Summary Model 1

R

R Square

0.897(a) 0.804 a Predictors: (Constant), log_GDP

Adjusted R Square

Std. Error of the Estimate

0.799

2.6668

ANOVA(b) Model 1

Sum of Squares Regression Residual Total

df

Mean Square

1023.181

1

1023.181

248.912

35

7.112

1272.092

36

F

Sig.

143.872

0.000(a)

t 17.306

Sig. 0.000

11.995

0.000

a Predictors: (Constant), log_GDP b Dependent Variable: Life Coefficients(a) Unstandardized Standardized Coefficients Coefficients Model 1

(Constant)

B 46.002

Std. Error 2.658

3.482

0.290

log_GDP

Beta 0.897

a Dependent Variable: Life

(b) Consider a stepwise selection, for example. Model Summary Model 1

R 0.964(a)

R Square 0.930

Adjusted R Square 0.921

Std. Error of the Estimate 0.3338

a Predictors: (Constant), Cont

137 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

ANOVA(b)

Model 1

Sum of Squares 11.865

Regression

1

Mean Square 11.865

0.891

8

0.111

12.756

9

Residual Total

df

F 106.496

Sig. 0.000(a)

a Predictors: (Constant), Cont b Dependent Variable: Fert

Model 1

(Constant) Cont

Coefficients(a) Unstandardized Standardized Coefficients Coefficients

t

Sig.

B

B

Std. Error

Std. Error

6.049

0.312

-0.053

0.005

Beta -0.964

19.393

0.000

-10.320

0.000

a Dependent Variable: Fert

We get extremely good predictions R2 = 0.964 using only contraceptive use. 14.33. y = height, x1 = length of left leg, x2 = length of right leg. 14.35. In the U.S., for each of the past 60 years, y = cumulative federal deficit, x = year. This might well be modeled by an exponential regression model. 14.36. (a) There would probably be a lot of variability for younger ages and less variability for older ages. (b) Coital frequency would probably increase for a while and then decrease as age increases. A quadratic model would be more suitable. 14.37. Precision improves (i.e., the standard error of bj decreases) when (a) multicollinearity, as 2 described by R j , decreases, (b) conditional variability of the response variable increases, (c) the variability of xj increases, (d) the sample size increases. 14.38. (a) 100,000(1.042)10 = 150,896. (b) 50.9% growth over the decade. 14.39. (a) A 1.27% growth rate per year corresponds to a multiplicative effect after 10 years of (1.0127)10 = 1.1345, or 13.45%. Or, to find the yearly multiplicative factor corresponding to a 10year multiplicative effect of 1.1345, set 1.1345 = 10, and solve for ; then log(1.1345) = 10log(), so log() = log(1.1345)/10 = 0.01262, and  = e0.01262 = 1.0127. (b) If the growth rate is 1.27% per year, then after x years, the multiplicative effect is (1.0127)x. 14.40. (a) 1000(1.10)x. (b) Approximately 8 years, since (1.1)8 = 2.1; more precisely, setting (1.1)x = 2.0 yields the solution x = (log2)/(log1.1) = 7.3. 14.41. (b), (c), and (d) 14.42. (b), (c), and (d) 14.43. (b)

138 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

14.44. (a), (c), and (d) 14.45. (a) True (b) True (c) False (d) False 14.46. c, e, f, i, a, h, j, g 2 14.47. E  y     1x1  2 x2  3 x1 x2  4 x1 x2

14.48. The binomial probability of outcome zero (i.e., entering none of the predictors into the model) when the chance is 0.05 of entering any particular one equals (0.05) 0(0.95)10 = 0.599. Thus, the probability that at least one is selected is 1 – 0.599 = 0.401.

Chapter 15 15.1. (a) The estimated probability of voting Republican increases as income increases. (b) (i) At 1.00.0210

x = 10, e

1  e1.00.0210  = 0.31; (ii) 0.73. (c) (i) The estimate of –α/ is 1.00/0.02 =  

50 thousand. (ii) Since  > 0, it is greater than 0.50 for x values above 50 thousand. (d)

ˆ 1    = 0.02(0.50)(0.50) = 0.005. (e) Odds multiply by e0.02 = 1.02 for each thousand dollar 

increase in family income. 15.2. (a) (i) 0.42, (ii) 0.37. (b) (i) odds = 0.42/0.58 = 0.724; (ii) odds = 0.37/0.63 = 0.587, so odds ratio = 0.724/0.587 = 1.233 = e0.2. (c) The odds multiply by e0.2 = 1.22 with the change from female to male, controlling for x1 and x2. (d) (i) The odds multiply by e0.08(3) = 1.27 for a standard deviation increase in x2; (ii) The odds multiply by e0.02(25) = 1.65 for a standard deviation increase in x1; the effect is larger for x1. 15.3. (a) (i) Pˆ  y  1 = 0.5 at x  ˆ ˆ = 2.043/0.282 = 7.2. (ii) Since the effect is negative, 2.0430.282 20 1  e2.0430.28220  = the probability is less than 0.5 for x above 7.2. (b) Pˆ  y  1  e



0.027. (c) Pˆ  y  1 = 0.847 – 0.051(20) = –0.17; no, probability cannot be negative. (d) z =



–0.282/0.101 = –2.80, Wald = (–2.80)2 = 7.8, P = 0.005. There is strong evidence of an association between WAIS and senility, and the sign of the coefficient of x suggests that it is negative. 15.4. (a) logit  Pˆ  y  1  5.427  0.772 x or Pˆ  y  1  e5.4270.772 x 1  e5.4270.772 x    Variables in the Equation B Step 1(a)

score Constant

-0.772

5.427 a Variable(s) entered on step 1: score.

S.E.

Wald

df

Sig.

Exp(B)

0.358

4.656

1

0.031

0.462

2.628

4.264

1

0.039

227.368

5.4270.772 0 1  e5.4270.7720  = 0.996; (ii) 0.00004. (c) Pˆ  y  1 = (b) (i) At x = 0, Pˆ  y  1  e





0.5 at x = 5.427/0.772 = 7.0. Since the estimate of  < 0, the probability is less than 0.5 for x above 7.0. (d) Odds multiply by e–0.772 = 0.46 for each one-unit increase in x. 139 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

e 15.5. When wife’s earnings = $20,000, Pˆ  y  1 

2.8700.56950.306 20.03930.37310.02716

1  e2.8700.56950.30620.03930.37310.02716

=

0.60. This increases to 0.95 when wife’s earnings = $100,000. 15.6. (a) Using scores x = (1, 2, 3, 4, 5, 6, 7) for ideology and treating it as a quantitative predictor, the predicted log odds of being a Democrat instead of Republican are 4.325 – 1.0302x. The predicted probabilities are (i) 0.964, (ii) .053; i.e., an extremely liberal person is very likely to be a Democrat whereas an extremely conservative person is very unlikely to be a Democrat. (b) Wald chi-squared statistic = 33.12, likelihood-ratio chi-squared statistic = 52.25, each with df = 1 give extremely strong evidence of an association (P-value < 0.0001). (c) (i) e.–1.0302 = 0.357. The estimated odds a liberal person is a Democrat equal 0.357 times the estimated odds an extremely liberal person is a Democrat. (ii) e.(7 – 1)(–1.0302) = 0.0021. The estimated odds an extremely conservative person is a Democrat equal 0.002 times the estimated odds an extremely liberal person is a Democrat. (d) Exponentiate the endpoints of –1.0302 ± 1.96(0.179), which gives (0.25, 0.51). We can be 95% confident that the odds a liberal person is a Democrat equal between 0.25 and 0.51 times the odds an extremely liberal person is a Democrat. 15.7. (a) The probability the child obtains a high school degree increases with mother’s education. Estimated odds of degree multiplied by e0.09 = 1.09 for each one-unit increase in mother’s education, controlling for other variables. (b) Probability of degree is lower when mother is employed. Estimated odds of degree when mother is employed equal e–0.92 = 0.40 times odds when mother is not employed. (c) e0.21 = 1.23, which corresponds to a 23% increase in the odds. 15.8. (a) ˆ = e0.16 = 1.17. (b) Estimated odds ratio = e0.47 = 1.60. The odds that a Democrat supports legalized abortion are estimated to be 1.6 times the odds for Independents. (c) Estimated odds ratio = e0.47 – (–1.67) = e2.14 = 8.5. The odds that a Democrat supports legalized abortion are estimated to be 8.5 times the odds for Republicans. (d) (i) e0.110.160.47 1  e0.110.160.47  = 0.68; (ii) e2.22 1  e2.22  = 0.098. 15.9. (a) Let r = 1 for black, 0 for white, let a = 1 for AZT = yes and 0 for AZT = no. Prediction

equation is logit  Pˆ  y  1  1.074  0.7195a  0.0555r . (b) The estimated probability of





AIDS symptoms decreases with AZT use (given race) and is slightly higher for blacks, given AZT use. (c) Pˆ  y  1  e1.074 1  e1.074  = 0.25. (d) e–0.720 = 0.49, so the estimated odds of AIDS symptoms for those using AZT are 0.49 times the estimated odds for those not using AZT, controlling for race. (e) Wald statistic is (–0.720/0.279)2 = 6.65, df = 1, P = 0.01; there is strong evidence that AIDS symptoms are less likelt for those using AZT, controlling for race. 15.10. (a) Let a = 1 for alcohol = yes, 0 for alcohol = no, let c = 1 for cigarette use = yes and 0

for cigarette use = no. Prediction equation is logit  Pˆ  y  1  5.309  2.986a  2.848c . (b)





The second category of each predictor is redundant, since there are only two categories. (c) (i) e5.309 1  e5.309  = 0.005; (ii) e5.3092.9862.848 1  e5.3092.9862.848  = 0.63. (d) (i) e2.986 = 19.8; controlling for cigarette use, the estimated odds of using marijuana for those who have used alcohol are 19.8 times the odds for those who have not used alcohol. (ii) e2.848 = 17.3; controlling for alcohol use, the estimated odds of using marijuana for those who have used cigarettes are 17.3 140 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

times the odds for those who have not used cigarettes. Both effects are very strong, since the estimated odds ratios are far above 1.0, the value corresponding to no effect. 15.11. (a) Men Women Pos Neg Pos Neg Yes 306 61 110 13 Report injecting drugs No 581 2983 87 275 (b) The model is the logistic regression model using dummy variables for S and H to predict the log odds of the probability that D = 1. (c) Letting dummy variables be 1 for males and 0 for females for s = sex and 1 for HCV positive and 0 for HCV negative for HCV status, the estimated log odds of injecting drugs equal –3.027 – 0.868s + 3.255 HCV. The estimated odds ratio is the exponential of 3.255, or 25.9, an extremely strong effect. 15.12. (a) The marginal table has counts Admitted Yes No M 686 1180 F 468 1259 and an odds ratio of 1.56. The estimated odds of a male being admitted are 1.56 times the estimated odds of a female being admitted. (b) Letting s = 1 for males and s = 0 for females and setting up dummy variables d1, d2, d3, d4 for the first four departments, the predicted log odds of being admitted equal –2.6918 + 0.0307 s + 3.205 d1 + 2.065 d2 + 2.011 d3 + 1.592 d4. The conditional odds ratio is e0.0307 = 1.03. There is essentially no association between sex and admissions, whereas it seemed there was a moderate association when we analyzed the marginal table in (a). (c) They differ so much because males applied more to department 1 than any other, and the admissions rate was high there (regardless of gender), whereas females applied in greater numbers to the other departments, and the admissions rate was lower in them (regardless of gender). A note to instructors: For these data shown here, Simpson's paradox does not quite hold, as the estimated odds ratio in (b) is not less than 1. However, the original data set contained one other department, with males having 512 admitted and 313 not admitted and females having 89 admitted and 19 not admitted. When these data are added to the data set (as they should have been when this exercise was formulated), the sample odds ratio in the marginal table is 1.84, and the estimated odds ratio in the logistic model is 0.90. The extra department is one also (like D1 in Table 15.21) in which males applied in relatively large numbers and females in small numbers yet the admissions rate was high regardless of gender. 15.13. (a) There are 3 categories of happiness and 2 cumulative probabilities in the model. The logit for each cumulative probability has its own intercept term. (b) ˆ = 0.418, so happiness tends to increase with income. (c) Wald statistic = (0.418/0.223)2 = 3.53, df = 1, P = 0.06 for Ha :   0 and 0.03 for Ha :   0 . (d)  2 =3.816, df = 4, P = 0.43. The cumulative logit model treats the response as ordinal and predictor as quantitative and results in much stronger evidence of an effect than the ordinary Pearson chi-squared test of independence, which treats both variables as nominal. 15.14. Answers should match those in Section 15.4. 141 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall

ˆ = –0.54 for gender, 0.77 for location, and –0.82 for seat belt use. logit  Pˆ  y  1  1.97  0.54g  0.76l  0.82s . Injury is more likely for females, accidents

15.15. (a)

in rural locations, and for those not wearing a seat belt. (b) e–0.824 = 0.44; for those wearing seat belts, the estimated odds ratio of injury more serious than any fixed category of location and gender are 0.44 times the estimated odds for those not wearing seat belts. (c) A 95% confidence interval is –0.82 ± 1.96(0.0277), or –0.87 to –0.76 for the true coefficient of seat belt use. This has exponentiated endpoints 0.42 to 0.46, which form the interval for this odds ratio. (d) The Wald statistic = 891.5, df = 1. This is extremely strong evidence of an effect of seat belt use, controlling for gender and location. 15.16. The likelihood-ratio statistic comparing the models equals 40,082.2 – 40,076.0 = 6.2, with df = 3, which has P = 0.10. The evidence on interaction is minor, especially given the enormous sample size. 15.17. (a) e0.3 = 1.35, so the estimated odds of voting Republican (given that one votes either Republican or Independent) multiply by 1.35 for each $10,000 increase in annual income. (b)

log  Pˆ  y  2 Pˆ  y  3 = (1.0 + 0.3x) – (3.3 – 0.2x) = –2.3 + 0.5x. Given that one votes

Republican or Democrat, the odds of voting Republican increase with income, by a multiplicative factor of e0.5 = 1.65 for each $10,000 increase in annual income. 15.18. Examine a baseline-category logit model using y as the choice of shopping venue with the other variables as explanatory variables. 15.19. (a) Using PROC LOGISTIC in SAS, treating ideology as quantitative with scores 1, 2, 3, 4, 5, 6, 7, the estimates are: Parameter Intercept Intercept ideology ideology

party 1 2 1 2

DF 1 1 1 1

Estimate 3.7325 2.5424 -0.8854 -0.4961

Standard Error 0.6103 0.5782 0.1435 0.1269

Wald Chi-Square 37.4006 19.3317 38.0589 15.2965

Pr > ChiSq