SOLUTIONS MANUAL for Introduction to Statistics and Data Analysis 6th Edition by Roxy Peck, Chris Ol by digitaldownload87

Complete Solutions Manual to Accompany

Introduction to Statistics and Data Analysis 6th Edition

Roxy Peck

California Polytechnic State University, San Luis Obispo

Tom Short

West Chester University of Pennsylvania

Chris Olsen Grinnell College

Prepared by

Stephen Miller Winchester Thurston School, Pittsburgh, PA

Australia • Brazil • Mexico • Singapore • United Kingdom • United States

ISBN-13: 978-1-337-79417-6 ISBN-10: 1-337-79417-1

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Contents

Chapter 1: The Role of Statistics and the Data Analysis Process ..................................................1 Chapter 2: Collecting Data Sensibly ............................................................................................14 Chapter 3: Graphical Methods for Describing Data ....................................................................28 Chapter 4: Numerical Methods for Describing Data ...................................................................70 Chapter 5: Summarizing Bivariate Data ......................................................................................88 Chapter 6: Probability ................................................................................................................136 Chapter 7: Random Variables and Probability Distributions.....................................................167 Chapter 8: Sampling Variability and Sampling Distributions ...................................................215 Chapter 9: Estimation Using a Single Sample ...........................................................................226 Chapter 10: Hypothesis Testing Using a Single Sample ...........................................................248 Chapter 11: Comparing Two Populations or Treatments ..........................................................287 Chapter 12: The Analysis of Categorical Data and Goodness-of-Fit Tests ...............................343 Chapter 13: Simple Linear Regression and Correlation: Inferential Methods ...........................361 Chapter 14: Multiple Regression Analysis ................................................................................410 Chapter 15: Analysis of Variance ..............................................................................................454 Chapter 16: Nonparametric (Distribution-Free) Statistical Methods...........................................488

Chapter 1 The Role of Statistics and the Data Analysis Process 1.1

Descriptive statistics is the branch of statistics that involves the organization and summary of the values in a data set. Inferential statistics is the branch of statistics concerned with reaching conclusions about a population based on the information provided by a sample.

1.2

The population is the entire collection of individuals or objects about which information is required. A sample is a subset of the population selected for study in some prescribed manner.

1.3

The proportions are stated as population values (although they were very likely calculated from sample results).

1.4

The sample is the set of 2121 children used in the study. The population is the set of all children between the ages of one and four.

1.5

The population of interest is the set of all 15,000 students at the university.

The sample is the 200 students who are interviewed.

1.6

The estimates given were computed using data from a sample.

1.7

The population is the set of all 7000 property owners. The sample is the 500 owners included in the survey.

1.8

The population is the set of all 2019 Toyota Camrys. The sample is the set of six cars that are tested.

1.9

The population is the set of 5000 used bricks. The sample is the set of 100 bricks she checks.

1.10

The researchers wanted to know whether the new surgical approach would improve memory functioning in Alzheimer’s patients. They hoped that the negative effects of the disease could be reduced by toxins being drained from the fluid filled space that cushions the brain.

First, it is not stated that the patients were randomly assigned to the treatments (new approach and standard care); this would be necessary in a well designed study. Second, it would help if the experiment could have been designed so that the patients did not know whether they were receiving the new approach or the standard care; otherwise, it is possible that the patients’ knowledge that they were receiving a new treatment might in itself have brought about an improvement in memory. Third, as stated in the investigators’ conclusion, it would have been useful if the experiment had been conducted on a sufficient number of patients so that any difference observed between the two treatments could not have been attributed to chance.

The researchers wanted to find out whether taking a garlic supplement reduces the likelihood that you will get a cold. They wanted to know whether a significantly lower proportion of people who took a garlic supplement would get a cold than those who did not take a garlic supplement.

1.11

1.12

1.13

1.14

1.15

1.16

Chapter 1: The Role of Statistics and the Data Analysis Process b

It is necessary that the participants were randomly assigned to the treatment groups. If this was the case, it seems that the study was conducted in a reasonable way.

Numerical (discrete)

Categorical

Numerical (continuous)

Categorical

Numerical (discrete)

Numerical (continuous)

Categorical

Numerical (continuous)

Discrete

Continuous

Discrete

Continuous

Discrete

For example: a Ford, Toyota, Ford, General Motors, Chevrolet, Chevrolet, Honda, BMW, Subaru, Nissan. b c

3.23, 2.92, 4.0, 2.8, 2.1, 3.88, 3.33, 3.9, 2.3, 3.56, 3.32, 2.4, 2.8, 3.9, 3.12. 4, 2, 0, 6, 3, 3, 2, 4, 5, 0, 8, 2, 5, 3, 4, 7, 3, 2, 0, 1

50.27, 50.67, 48.98, 50.58, 50.95, 50.95, 50.21, 49.70, 50.33, 49.14, 50.83, 49.89

In minutes: 10, 10, 18, 0, 17, 17, 0, 17, 12, 19, 12, 13, 15, 15, 15

Chapter 1: The Role of Statistics and the Data Analysis Process 1.17

1.18

1.19

Gender of purchaser, brand of motorcycle, telephone area code

Number of previous motorcycles

Bar chart

Dotplot

Over half of the responses (57%) were from people who indicated that the best long-term investments were real estate (35%) and stocks & mutual funds (22%). The remaining 43% of respondents indicated that gold (17%), savings (15%), bonds (7%), and other (4%) were the best long-term investments.

The dotplot below shows the concussion rate (concussions per 10,000 athletes).

1.4

2.8

4.2

5.6

7.0

8.4

9.8

11.2

Concussion Rate (per 10,000 athletes) b The dotplot below shows the concussion rate (concussions per 10,000 athletes), with different symbols for boys and girls.

1.4

2.8

4.2

5.6

7.0

8.4

9.8

11.2

Girls Boys

Concussion Rate (per 10,000 athletes) The sport with an unusually high (compared to all the other sports) concussion rate is football. Without considering football, the concussion rates for girls’ sports is essentially the same as the concussion rate for boys’ sports. However, if we consider football, the concussion rate for girls’ sports tends to be lower than that for boys’ sports.

4 1.20

Chapter 1: The Role of Statistics and the Data Analysis Process Dotplots (for parts a and b) drawn on the same scale for the 2014 and 2015 sales data are shown below. 2014 2015

1.21

150 200 250 300 350 400 450 500 550 600 650 700 750 800 850 900 950 Sales (millions of dollars)

A typical sales figure for 2014 was around 215 million dollars, with sales figures ranging from around 150 to around 350 million dollars. The greatest density of points was at the lower end of the distribution. There were no extreme observations in 2014.

A typical sales figure for 2015 was around 200 million dollars, with sales figures ranging from around 155 to around 937 million dollars. The greatest density of points was at the lower end of the distribution. There are two extreme observations in 2015, namely, 936.7 million dollars and 652.3 million dollars.

Sales figures were generally speaking higher in 2015 than in 2014. There were two extreme observations in 2015, and no extreme observations in 2014. If the extreme sales figures are taken into account, the variation in the sales figures (among the top 20 movies) was far greater in 2015 than in 2014. If the extreme sales figures are disregarded, the variation was still greater in 2015 than in 2014, but by not nearly as much. The distributions are similar in shape, with the greatest density of points being at the lower end of the distribution in both cases.

The most common response was “A lot of thought”, accounting for 130 (or 65%) of the students who started college but did not complete a degree. The next two most common responses were “Some thought” and “No thought”, accounting for 48 (or 24%) and 18 (or 9%), respectively, of the students who started college but did not complete a degree. Finally,

Chapter 1: The Role of Statistics and the Data Analysis Process

4 of the 200 respondents (2%) indicated that don’t know how much thought they have given to going back to school. 1.22

1.23

Categorical

Since the variable being graphed is categorical, a dotplot would not be suitable.

If you add up the relative frequencies you get 107%. This total should be 100%, so a mistake has clearly been made.

The dotplot shows that there were three sites that received far greater numbers of visits than the remaining 6 sites. Also, it shows that the distribution of the number of visits has the greatest density of points for the smaller numbers of visits, with the density decreasing as the number of visits increases.

It is clear from the dotplot that there were two sites that were used by far greater numbers of individuals (unique visitors) than the remaining 7 sites. However, these two sites are less far above the others in terms of the number of unique visitors than they are in terms of the total number of visits. As with the distribution of the total number of visits, the distribution of the number of unique visitors has the greatest density of points for the smaller numbers of visitors, with the density decreasing as the number of unique visitors increases. This is the case even when only the 7 less popular sites are considered.

The statistic “visits per unique visitor” tells us how heavily the individuals are using the sites. The table tells us that the most popular sites (Facebook and YouTube) in terms of total visits and unique visitors do not have the highest value of this statistic. The dotplot of visits per unique visitor shows that there are two individual sites are far ahead of the rest in this respect (Pinterest and Twitter).

1.24

Rating A+ A B C D F

Relative Frequency 11/20 = 0.55 2/20 = 0.10 2/20 = 0.10 3/20 = 0.15 0/20 = 0.00 2/20 = 0.10

1.25

Chapter 1: The Role of Statistics and the Data Analysis Process

b Seventy-five percent (75%) of the dry weather ratings are B or higher, and 70% of wet weather ratings are B or higher, indicating that dry weather ratings are higher than wet weather ratings. Note that the wet weather ratings are only A+ or F, so the wet weather ratings are more extreme than dry weather ratings. If we only consider A+ ratings, then the wet weather ratings tend to be better than dry weather ratings because only 55% of dry weather ratings are A+, compared with 70% of wet weather ratings being A+.

a Region

1.26

E M W

1.27

40 Wireless %

Looking at the dotplot we can see that Eastern states have, on average, lower wireless percents than states in the other two regions. The West and Middle states regions have, on average, roughly equal wireless percents.

6 8 10 12 14 16 Number of Violent Crimes Reported

Three schools seem to stand out from the rest, these being, in increasing order of number of crimes, University of Central Florida (14 crimes reported), Florida International University (15 crimes reported), and Florida State University (20 crimes reported).

Chapter 1: The Role of Statistics and the Data Analysis Process

b Violent Crime Rate Per 1000 Students Florida A&M University 0.60435 Florida Atlantic University 0.19750 Florida Gulf Coast University 0.20225 Florida International University 0.30131 Florida South Western State College 0.00000 Florida State University, Tallahassee 0.48984 New College of Florida 1.16144 Pensacola State College 0.00000 Santa Fe College 0.00000 Tallahassee Community College 0.16071 University of Central Florida 0.22239 University of Florida 0.19745 University of North Florida 0.19139 University of South Florida, St. Petersburg 0.00000 University of South Florida, Tampa 0.19017 University of West Florida 0.78351 University/College

0.00

0.16

0.32 0.48 0.64 0.80 Violent Crimes per 1000 Students

0.96

1.12

The colleges that stand out in violent crimes per 1000 students are, in increasing order of crime rate, Florida State University, Florida A&M University, University of West Florida, and New College of Florida. Only Florida State University stands out in both dotplots.

1.28

For the number of violent crimes, there are three schools that stand out by having high numbers of crimes, with the majority of the schools having similar, and low (10 or fewer), numbers of crimes. There seems to be greater consistency for crime rate (per 1000 students) among the 16 schools than there is for number of crimes, with four schools standing out as having high crime rates, and four schools with crime rates that stand out as being noticeably low.

When ranking the airlines according to delayed flights, one airline would be ranked above another if the probability of a randomly chosen flight being delayed is smaller for the first airline than it is for the second airline. These probabilities are estimated using the rate per 10,000 flights values, and so these are the data that should be used for this ranking. (Note that the total number of flights values are not suitable for this ranking. Suppose that one airline had a larger number of delayed flights than another airline. It is possible that this could be accounted for merely through the first airline having more flights than the second.)

There are two airlines, ExpressJet and Continental, which, with 4.9 and 4.1 of every 10,000 flights delayed, stand out as the worst airlines in this regard. There are two further airlines that stand out above the rest: Delta and Comair, with rates of 2.8 and 2.7 delayed flights per

Chapter 1: The Role of Statistics and the Data Analysis Process 10,000 flights. All the other airlines have rates below 1.6, with the best rating being for Southwest, with a rate of only 0.1 delayed flights per 10,000.

1.29

1.30

Most public community college graduates have no debt at all, and a debt of $10,000 or less accounts for 79% of the graduates. Among the 21% of the graduates who have a debt of more than $10,000, nearly 43% (9% of all graduates) have a debt of more than $20,000.

a Frequency 600 500 400 300 200 100 0

am C

s pu

am C

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s ok

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eb w

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s ok

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n Re

ted

tex

o tbo

os M

tly

oo eB

dn Di

y bu

tex

o tbo

ff O Where Books Purchased

By far the most popular place to buy books is the campus bookstore, with half of the students in the sample buying their books from that source. The next most popular sources are online bookstores other than the online version of the campus bookstore and off-campus bookstores,

Chapter 1: The Role of Statistics and the Data Analysis Process with these two sources accounting for around 35% of students. Purchasing mostly eBooks was the least common response. 1.31

Relative Frequency

How Often More than once a week Once a week Every other week Every three weeks Less often than every three weeks

0.10 0.53 0.26 0.05 0.06

1.32

The relative frequency distribution is: Rating G PG PG-13 R

Relative Frequency 1/25 = 0.04 6/25 = 0.24 13/25 = 0.52 5/25 = 0.20

Chapter 1: The Role of Statistics and the Data Analysis Process

PG-13 is the rating with the highest relative proportion (0.52), followed by PG (0.24), R (0.20), and G (0.04). Seventy-two percent (72%) of the top 25 movies of 2015 are PG-13 or R, and the remaining 28% are rated G or PG.

1.33

The dotplot for Los Angeles County is shown below.

12 18 24 30 Percent Failing (Los Angeles County)

A typical percent of tests failing for Los Angeles County is around 16. There is one value that is unusually high (43), with the other values ranging from 2 to 33. There is a greater density of points toward the lower end of the distribution than toward the upper end. b

The dotplot for the other counties is shown below.

18 24 30 Percent Failing (Other Counties)

A typical percent of tests failing for the other counties is around 3. There is one extreme result at the upper end of the distribution (40); the other values range from 0 to 17. The density of points is highest at the left hand end of the distribution and decreases as the percent failing values increase. c

The typical value for Los Angeles County (around 16) is greater than for the other counties (around 3) and, disregarding the one extreme value in each case, there is a greater variability in the values for Los Angeles County than for the other counties. In the distribution for Los Angeles County the points are closer to being uniformly distributed than in the distribution for the other counties, where there is a clear tail-off of density of points as you move to the right of the distribution.

Chapter 1: The Role of Statistics and the Data Analysis Process 1.34

Categorical

b Relative Frequency (%) 50

Below Basic

Basic Intermediate Literacy Level

Proficient

No, since dotplots are used for numerical data.

1.35 Frequency 140 120 100 80 60 40 20 0

Strongly disagree

Disagree

Not sure Response

Agree

Strongly agree

12 1.36

Chapter 1: The Role of Statistics and the Data Analysis Process a Relative Frequency 0.4

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ce an n te a in M

ht ig Fl

io at er

ns do ar az

ls ia er t ma

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H Type of Violation

1.37

By far the most frequently occurring violation categories were security (43%) and maintenance (39%). The least frequently occurring violation categories were flight operations (6%) and hazardous materials (3%).

25 30 35 Acceptance Rate (%)

A typical acceptance rate for these top 25 schools is around 30, with the great majority of acceptance rates being between 19 and 39. There are no particularly extreme values. The pattern of the points is roughly symmetrical.

Chapter 1: The Role of Statistics and the Data Analysis Process 1.38 Frequency

90 80 70 60 50 40 30 20 10 0

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Chapter 2 Collecting Data Sensibly

2.1

This is an observational study. The treatments (length of stay) were determined by the condition of the patients. (In an experiment the patients would be assigned to the various lengths of stay by the investigators, usually using random assignment.)

2.2

This was an experiment, since the investigators (not the students) determined which discussion sections received the chocolate and which did not.

2.3

This was an experiment, since the professor (not the students) determined who was supplying a buying price and who was supplying a selling price.

2.4

This is an observational study.

No, it is not reasonable to conclude that getting less than 8 hours of sleep on school nights causes teenagers to fall asleep during school and to consume more caffeine. This is an observations study, so cause-and-effect conclusions cannot be drawn.

This is an experiment since it was decided by the researchers (in this case by random assignment) which participants would receive which treatments.

Yes. Since the participants were randomly assigned to the treatments it is reasonable to conclude that receiving either real or fake acupuncture was the cause of the observed reductions in pain.

This is an observational study.

Yes. Since the researchers looked at a random sample of publically accessible MySpace web profiles posted by 18-year-olds, it is reasonable to generalize the stated conclusion to all 18year-olds with publically accessible MySpace profiles.

No, it is not reasonable to generalize the stated conclusion to all 18-year-old MySpace users since no users without publically accessible profiles were included in the study.

No, it is not reasonable to generalize the stated conclusion to all MySpace users with publically accessible profiles since only 18-year-olds were included in the study.

This is an experiment.

Yes. Since the participants were randomly assigned to the treatments the researcher can reasonably claim that hearing the cork pop was the cause of the higher rating.

2.5

2.6

2.7

2.8

It is quite possible, for example, that 3- and 4-year olds who drink something sweet once or twice a day generally speaking consume larger amounts of fat than those who do not, and that it is the fat that is causing the weight problems a year later, rather than the consumption of the sweet drinks.

Chapter 2: Collecting Data Sensibly

2.9

We are told that moderate drinkers, as a group, tended to be better educated, wealthier, and more active than nondrinkers. It is therefore quite possible that the observed reduction in the risk of heart disease amongst moderate drinkers is caused by one of these attributes and not by the moderate drinking.

2.10

No. It is quite possible, for example, that women who choose to watch Oprah generally speaking have a more health oriented outlook than those who watch other daytime talk shows, and it is this health oriented outlook that causes the decrease in craving for fattening foods, not the watching of Oprah.

Neither generalization would be reasonable since the survey was conducted on the DietSmart.com website. It is unlikely that users of this website would be representative of the population of women in the United States or of the population of women who watch daytime talk shows.

The data would need to be collected from a simple random sample of all adult American Internet users.

No. Since the survey included only adult Americans who are internet users, the result cannot be generalized to all adult Americans.

2.11

2.12

I do agree with the statement. Observational studies cannot be used to draw cause-and-effect conclusions.

2.13

The following is one possible method. Use a computer list of all the students at the college. Assign to each student a 10-digit decimal number. Sort the students according to the numbers assigned, smallest to largest. The first 100 students on the sorted list will form the sample.

2.14

Method 1: Using a computer list of the graduates, number the graduates 1-140. Use a random number generator on a calculator or computer to randomly select a whole number between 1 and 140. The number selected represents the first graduate to be included in the sample. Repeat the number selection, ignoring repeated numbers, until 20 graduates have been selected. Method 2: Using a computer list of the graduates, number the graduates 001-140. Take the first three digits from the left hand end of a row from a table of random digits. If the three-digit number formed is between 001 and 140 inclusive, the graduate with that number should be the first graduate in the sample. If the number formed is not between 001 and 140 inclusive, the number should be ignored. Repeat the process described for the next three digits in the random number table, and continue in the same way until 20 graduates have been selected. (Three-digit numbers that are repeats of numbers previously selected should be ignored.)

2.15

The following is one possible method. Number the signatures 1-500. Use a random number generator on a calculator or computer to randomly select a whole number between 1 and 500. The number selected represents the first signature to be included in the sample. Repeat the number selection, ignoring repeated numbers, until 30 signatures have been selected.

2.16

The 716 bicycle fatalities constitute all bicycle fatalities in 2008, and so the group represents a census.

The average age of 41 years is a population characteristic, since it is the average for all bicycle fatalities in 2008.

16 2.17

2.18

2.19

Chapter 2: Collecting Data Sensibly a

The population is all American women.

No. The sample included women only from Maryland, Minnesota, Oregon, and Pennsylvania. It is quite possible that the relationship between exercise and cognitive impairment is different for women in other states.

No. As mentioned in Part (b), it is quite possible that the relationship between exercise and cognitive impairment is different for women in other states. For example, other states might contain a different racial mix from the four included in the study, and it is possible that women of different racial origins respond differently to exercise in terms of cognitive impairment.

The inadequate sampling method used in this study is an example of selection bias. Women in the other 46 states were excluded from the study.

Cluster sampling

Stratified random sampling

Convenience sampling

Simple random sampling

Systematic sampling

Using the list, first number the part time students 1-3000. Use a random number generator on a calculator or computer to randomly select a whole number between 1 and 3000. The number selected represents the first part time student to be included in the sample. Repeat the number selection, ignoring repeated numbers, until 10 part time students have been selected. Then number the full time students 1-3500 and select 10 full time students using the same procedure.

No. With 10 part time students being selected out of a total of 3000 part time students, the probability of any particular part time student being selected is 10/3000 = 1/300. Applying a similar argument to the full time students, the probability of any particular full time student being selected is 10/3500 = 1/350. Since these probabilities are different, it is not the case that every student has the same chance of being included in the sample.

2.20

Convenience samples are, by nature, very unlikely to be representative of the population.

2.21

The pages of the book have already been numbered between 1 and the highest page number in the book. Use a random number generator on a calculator or computer to randomly select a whole number between 1 and the highest page number in the book. The number selected will be the first page to be included in the sample. Repeat the number selection, ignoring repeated numbers, until the required number of pages has been selected.

Pages that include exercises tend to contain more words than pages that do not include exercises. Therefore, it would be sensible to stratify according to this criterion. Assuming that 20 non-exercise pages and 20 exercise pages will be included in the sample, the sample should be selected as follows. Use a random number generator to randomly select a whole number between 1 and the highest page number in the book. The number selected will be the

Chapter 2: Collecting Data Sensibly

first page to be included in the sample. Repeat the number selection, ignoring repeated numbers and keeping track of the number of pages of each type selected, until 20 pages of one type have been selected. Then continue in the same way, but ignore numbers corresponding to pages of that type. When 20 pages of the other type have been selected, stop the process. c

Randomly select one page from the first 20 pages in the book. Include in your sample that page and every 20th page from that page onwards.

Roughly speaking, in terms of the numbers of words per page, each chapter is representative of the book as a whole. It is therefore sensible for the chapters to be used as clusters. Using a random number generator randomly choose three chapters. Then count the number of words on each page in those three chapters.

2.22

Answers will vary.

2.23

Answers will vary.

2.24

It is not the proportion of voters that is important, but the number of voters in the sample – and 1000 is an adequate number.

2.25

The researchers should be concerned about nonresponse bias. Only a small proportion (20.7%) of the selected households completed the interview, and it is quite possible that those households who did complete the interview are different in some relevant way concerning Internet use from those who did not.

2.26

This was a cluster sample, with each team being a cluster. Note that all student-athletes on a particular team are surveyed.

The student-athletes on the randomly selected teams are unlikely to be representative of the set of all college students in the U.S. Also, since the survey was about illegal drug use (a sensitive subject), we cannot be sure that the students were giving truthful answers to the questions.

2.27

First, the participants in the study were all students in an upper-division communications course at one particular university. It is not reasonable to consider these students to be representative of all students with regard to their truthfulness in the various forms of communication. Second, the students knew during the week’s activity that they were surveying themselves as to the truthfulness of their interactions. This could easily have changed their behavior in particular social contexts and therefore could have distorted the results of the study.

2.28

a b

2.29

No. The parents in the study were not randomly selected and therefore are not likely to be representative of the population of all parents. No. The sample is not likely to have been representative of the population of all parents and therefore it would not be reasonable to generalize the researchers’ conclusion to all parents.

First, the people who responded to the print and online advertisements might be different in some way relevant to the study from the population of people who have online dating profiles. Second, only the Village Voice and Craigslist New York City were used for the recruitment. It is quite possible that people who read that newspaper or access those websites differ from the population

Chapter 2: Collecting Data Sensibly in some relevant way, particularly considering that they are both New York City based publications.

2.30

It is important to know if the sample of 800 college students was randomly selected, and from what population of college students was the sample selected. For example, were the students randomly selected from colleges in one particular state or region, or were all colleges considered in the sampling scheme.

2.31

Yes. It is possible that students of a given class standing tend to be similar in the amounts of money they spend on textbooks.

Yes. It is possible that students who pursue a certain field of study tend to be similar in the amounts of money they spend on textbooks.

No. It is unlikely that stratifying in this way will produce groups that are homogeneous in terms of the students’ spending on textbooks.

2.32

The individuals within each stratum should on the whole be similar in terms of the topic of the study. This is true of the proposed strata in Scheme 2, since it is likely that college students will on the whole be similar in their opinions regarding the possible tax increase; likewise nonstudents who work full time will on the whole be similar in their opinions regarding the possible tax increase, and nonstudents who do not work full time will on the whole be similar in their opinions regarding the possible tax increase. Scheme 1, however, is not suitable since we have no reason to believe that people within the proposed first-letter-of-last-name strata will be similar in terms of their attitudes to the possible tax increase. Similarly the suggested stratification in Scheme 3 is very unlikely to produce homogeneous groups.

2.33

It is not reasonable to generalize these results to the population of U.S. adults since the people who sent their hair for testing did so voluntarily. It is quite possible that people who would choose to participate in a study of this sort differ in their mercury levels from the population as a whole.

2.34

Different subsets of the population might have responded by different methods. For example, it is quite possible that younger people (who might generally be in favor of continuing the parade) chose to respond via the Internet while older people (who might on the whole be against the parade) chose to use the telephone to make their responses.

2.35

Binding strength

Type of glue

The extraneous variables mentioned are the number of pages in the book and whether the book is bound as a hardback or a paperback. Further extraneous variables that might be considered include the weight of the material used for the cover and the type of paper used.

Use two IQ tests (Test 1 and Test 2) of equal levels of difficulty. Randomly select 50 students from the school. Randomly assign the 50 students to two groups, Group A and Group B. The experiment will be conducted over two days. On the first day, students in Group A will do an IQ test without listening to any music and students in Group B will do an IQ test after listening to a Mozart piano sonata. On the second day the activities of the two groups will be switched. For each student decide randomly whether he/she will take Test 1 on the first day

2.36

Chapter 2: Collecting Data Sensibly

and Test 2 on the second day, or vice versa. All conditions (temperature, time of day, amount of light, etc. – everything apart from the presence of the music or not) should be kept equal between the two days and between the two groups. In particular, the students taking the test without listening to the music should nonetheless sit quietly in a room for the length of time of the piano sonata. At the end of the experiment the after-music IQ scores should be compared to the no-music IQ scores. b

Yes, the fact that all conditions are kept the same is direct control.

By having each student take IQ tests under both experimental conditions we are using a matched pairs design, and matched pairs is a form of blocking.

The students were randomly assigned to the two orders of the treatments (no music then music, or music then no music). Also it was decided randomly for each student whether Test 1 or Test 2 would be taken first.

2.37

The following is one possible method. Write the names of the subjects on identical slips of paper, and place the slips in a hat. Mix the slips, and randomly select ten slips from the hat. The names on those ten slips are the people who will dry their hands by the first method. Randomly select a further ten slips from the hat. The names on those ten slips are the people who will dry their hands by the second method. The remaining ten people will dry their hands by the third method.

2.38

Random assignment should have been used to determine, for each cyclist, which drink would be consumed during which break.

2.39

Blocking

Direct control

2.40

We rely on random assignment to produce comparable experimental groups. If the researchers had hand-picked the treatment groups, they might unconsciously have favored one group over the other in terms of some variable that affects the subjects’ ability to deal with multiple inputs.

2.41

The figure shows that comparable groups in terms of age have been formed. (Differences between the age distributions in the groups can be seen: there are one or two children in the LR group who are younger than all those in the OR group, and also there is a greater number of children over 14 years old in the LR group. It is inevitable that there will be some differences between the groups under random assignment.)

2.42

If the participants had been able to choose their own avatars, then it is quite possible, for example, that people with a lot of self confidence would tend to choose the attractive avatar while those with less self confidence would tend to choose the unattractive avatar. Then, if the same result was obtained as the one described in the paper, it would be impossible to tell whether the greater closeness achieved by those with the attractive avatar came about as a result of the avatar or as a result of those people’s greater self confidence.

Chapter 2: Collecting Data Sensibly

Attractive avatar

2.43

Random Assignment

Participants

Measure closeness Compare closeness for attractive avatar vs. unattractive avatar

Unattractive avatar

Measure closeness

2.44

Random Assignment

Subjects

Take notes using laptop

Take notes by hand in notebook

Test on material from TED Talk

Compare performance on fact recall questions and conceptualapplication questions

Chapter 2: Collecting Data Sensibly

2.45

Random Assignment

30 trials

Additive 1

Measure distance traveled

Additive 2

Measure distance traveled

Additive 3

Measure distance traveled

Compare distances traveled for the three additives

2.46

How many bottles of water were used in the experiment? Were the bottles identical? Was there a control group in which the bottles had identical labels to those used in the treatment group but without the positive words? Were the bottles randomly assigned to the groups? Was the state of the water measured both before and after the experiment? Did the people who measured the water’s structure know which bottles were in which group?

2.47

This is an experiment. The researchers assigned the students to the treatments.

No. The subjects were randomly assigned to the treatments, but there was no random selection from a population of interest.

Yes.

If the performance was significantly better for the group that read the material in the landscape orientation then, yes, for the set of subjects used in the experiment, reasoning was improved by turning the screen to landscape orientation. The subjects were randomly assigned to the treatment groups, and a significant difference in the results of the two groups means that the observed difference is unlikely to have occurred by chance.

No. The students used in the study were all undergraduates taking psychology at a large university, and therefore cannot be considered to be representative of any larger population.

The treatments are the names – Ann Clark and Andrew Clark – given to the participants.

The response variables are the participants’ answers to the questions given.

2.48

2.49

2.50 Selecting a random sample of 1161 voters and giving them the female name, and then selecting a second random sample 1139 voters and giving them the male name, is exactly equivalent to selecting a random sample of 2300 voters, and then randomly assigning 1161 of them to the female name and the remainder to the male name. (It is assumed here that in the study given in the question it was ensured that there was no overlap between the two samples.)

22 2.51

Chapter 2: Collecting Data Sensibly a

Red wine, yellow onions, black tea

Absorption of flavonols

Alcohol tolerance, amount of flavonols in the diet (other than from red wine, yellow onions, and black tea), gender

2.52

Suppose that an experiment is conducted in which people are given either a drug or a placebo, and that those who are given the drug do significantly better, on average, than those who are given the placebo. Since both groups experience the placebo effect (the psychological effect of taking a tablet) we are able to attribute the greater improvement of those who took the drug to the chemicals in the drug. However we are unable to tell just how much of a placebo effect is being experienced by all the subjects. By adding a control group (a group that is given nothing) and comparing the results for this group with the results for the placebo group we can measure the extent of the placebo effect.

2.53

“Blinding” is ensuring that the experimental subjects do not know which treatment they were given and/or ensuring that the people who measure the response variable do not know who was given which treatment. When this is possible to implement, it is useful that the subjects do not know which treatments they were given since, if a person knows what treatment he/she was given, this knowledge could influence the person’s perception of the response variable, or even, through psychological processes, have a direct effect on the response variable. If the response variable is to be measured by a person other than the experimental subjects it is useful if this person doesn’t know who received which treatment since, if this person does know who received which treatment, then this could influence the person’s perception of the response variable.

2.54

Answers will vary.

2.55

Allowing study participants to choose which group they want to be in could introduce systematic differences between the two experimental conditions (knee replacement surgery with exercise and exercise therapy alone), resulting in potential confounding. Those who chose knee replacement surgery plus exercise might, in some way, be different from those who chose exercise therapy alone. We would not know if differences in pain relief between the two groups were due to the knee replacement surgery with exercise, or due to some inherent differences in the subjects who chose their experimental groups .

The researchers likely did not include a control group because the study participants needed some relief from their pain. Because the purpose of this experiment is to determine

whether knee replacement surgery with exercise provided more pain relief than exercise therapy alone, a control group would not allow the study participants to have the opportunity to experience pain relief . 2.56

It is very possible that the nurses might have preconceptions about the two forms of surgery in terms of the amount of pain and nausea caused. Therefore, if the nurses know which children have been given which form of surgery, this might affect the amounts of medication they give. By making sure that the nurses do not have this knowledge, this possible effect is avoided.

Chapter 2: Collecting Data Sensibly b

2.57

Since the incisions made under the two procedures are different the patients and/or their parents would know which method had been used.

We will assume that only four colors will be compared, and that only headache sufferers will be included in the study. Prepare a supply of “Regular Strength” Tylenol in four different colors: white (the current color of the medication, and therefore the “control”), red, green, and blue. Recruit 20 volunteers who suffer from headaches. Instruct each volunteer not to take any pain relief medication for a week. After that week is over, issue each volunteer a supply of all four colors. Give each volunteer an order in which to use the colors (this order would be determined randomly for each volunteer). Instruct the volunteers to use one fixed dose of the medication for each headache over a period of four weeks, and to note on a form the color used and the pain relief achieved (on a scale of 0-10, where 0 is no pain relief and 10 is complete pain relief). At the end of the four weeks gather the results and compare the pain relief achieved by the four colors.

2.58

Randomly assigning 852 children to the book group and the rest to the control group consists of randomly selecting 852 to be in the book group and putting the remaining children in the control group.

If no control group had been included in the study, then the only results available to the researchers would be the reading scores of the children who had been given the reading books. There would be no way of telling whether these scores were any better than the scores for children who were not given reading books.

2.59

Of the girls, randomly assign 350 to the book group and 350 to the no-book group. (You could do this by starting with a computer list of the names of the 700 girls. Assign to each name a random 10-digit number. Sort the names according to the numbers, from smallest to largest. The first 350 names on the sorted list are assigned to the book group, and the remainder to the no-book group.) Using a similar method, randomly assign 315 of the boys to the book group and 315 to the nobook group.

2.60

Suppose that the dog handlers and/or the experimental observers had known which patients did and did not have cancer. It would then be possible for some sort of (conscious or unconscious) communication to take place between these people and the dogs so that the dogs would pick up the conditions of the patients from these people rather than through their perception of the patients’ breath. By making sure that the dog handlers and the experimental observers do not know who has the disease and who does not it is ensured that the dogs are getting the information from the patients.

2.61

If the judges had known which chowder came from which restaurant then it is unlikely that Denny’s chowder would have won the contest, since the judges would probably be conditioned by this knowledge to choose chowders from more expensive restaurants.

In experiments, if the people measuring the response are not blinded they will often be conditioned to see different responses to some treatments over other treatments, in the same way as the judges would have been conditioned to favor the expensive restaurant chowders. It is therefore necessary that the people measuring the response should not know which subject received which treatment, so that the treatments can be compared on their own merits.

Chapter 2: Collecting Data Sensibly

2.62

This describes the placebo effect. The implication is that the experiments have included patients who have been given a placebo in place of the antidepressants.

2.63

A placebo group would be necessary if the mere thought of having amalgam fillings could produce kidney disorders. However, since the experimental subjects were sheep the researchers do not need to be concerned that this would happen.

A resin filling treatment group would be necessary in order to provide evidence that it is the material in the amalgam fillings, rather than the process of filling the teeth, or just the presence of foreign bodies in the teeth, that is the cause of the kidney disorders. If the amalgam filling group developed the kidney disorders and the resin filling group did not, then this would provide evidence that it is some ingredient in the amalgam fillings that is causing the kidney problems.

Since there is concern about the effect of amalgam fillings it would be considered unethical to use humans in the experiment.

2.64

Answers will vary.

2.65

This is an observational study.

In order to evaluate the study, we need to know whether the sample was a random sample.

No. Since the sample used in the Healthy Steps study was known to be nationally representative, and since the paper states that, compared with the HS trial, parents in the study sample were disproportionately older, white, more educated, and married, it is clear that it is not reasonable to regard the sample as representative of parents of all children at age 5.5 years.

The potential confounding variable mentioned is what the children watched.

The quotation from Kamila Mistry makes a statement about cause and effect and therefore is inconsistent with the statement that the study can’t show that TV was the cause of later problems.

2.66

2.67

Answers will vary.

2.68

Answers will vary.

2.69

There are many possible designs. We will describe here a design that blocks for the day of the week and the section of the newspaper in which the advertisement appears. For the sake of argument we will assume that the mortgage lender is interested in advertising on only two days of the week (Monday and Tuesday) and that there are three sections in the newspaper (A, B, and C). We will refer to the three types of advertisement as Ad 1, Ad 1, and Ad 3. The experimental units are 18 issues of the newspaper (that is, 18 dates) consisting of Mondays and Tuesdays over 9 weeks. Use a random process to decide which three Mondays will receive advertisements in Section A, which three Mondays will receive advertisements in Section B, and which three Mondays will receive advertisements in Section C. Do the same for the nine Tuesdays. We have now effectively split the 18 issues into the six blocks shown below. (There are 3 issues in each block.)

Chapter 2: Collecting Data Sensibly

Mon, Sect A Tue, Sect A

Mon, Sect B Tue, Sect B

Mon, Sect C Tue, Sect C

Now randomly assign the three issues in each block to the three advertisements. (Ad 1 is then appearing on three Mondays, once in each section, and on three Tuesdays, once in each section. The same applies to Ad 2 and Ad 3.) The response levels for the three advertisements can now be compared (as can the three different sections and the two different days). b

2.70

Within each block, the three issues of the newspaper were randomly assigned to the three advertisements.

Study 1 1. Observational study 2. No 3. No 4. No. The fact that calcium takers were more common among the heart attack patients implies mathematically that calcium takers were more likely to be heart attack patients than noncalcium takers. However, it is quite possible that people who take a calcium supplement very often also take another supplement, and it is this other supplement that is causing heart attacks, not the calcium. 5. No. The hospital at which this study was conducted cannot be considered to be representative of any larger population. Study 2 1. Observational study 2. Yes 3. No 4. No. It is quite possible that people who take a calcium supplement very often also take another supplement, and it is this other supplement that is causing heart attacks, not the calcium. 5. The conclusions can only be generalized to the population of people living in Minneapolis who receive Social Security. Study 3 1. Experiment 2. Yes 3. No 4. No 5. No reasonable conclusion can be drawn from the study. Study 4 1. Experiment 2. No 3. Yes 4. Yes 5. No. The participants were volunteers.

2.71

By randomly selecting the phone numbers, calling back those where there are no answers, and asking for the adult in the household with the most recent birthday the researchers are avoiding selection bias. However, selection bias could come about as a result of the fact that not all

Chapter 2: Collecting Data Sensibly Californians have phones. Also, by selecting the adult with the most recent birthday certain birth dates are being favored, and it could be suggested that different attributes of people born at different times of the year could introduce further selection bias. Further to that, there is always the concern that people might not answer truthfully. This is response bias.

2.72

Answers will vary.

2.73

2.74

Observational study

There are many reasons the results of the study may not generalize to the population of all U.S. men. Here are two possible reasons. First, the study participants might not have been randomly selected. Second, if they were randomly chosen, we don’t know from what population they were selected.

2.75

This is an observational study based on results of a survey (no consumers were assigned to different experimental conditions).

2.76

Nonresponse bias: it is possible that those who responded differed in some important way from those who did not. Also there is a possibility of response bias in that those who did respond might not have been answering truthfully.

2.77

The design is good in that it includes several desired experimental design components. Specifically, the design includes a control group, random assignment of the subjects to the treatments, and includes blinding.

Perhaps rather than taking photos of the top of the head of all the women, the expert who determined the change in hair density should have the opportunity to evaluate the women in person. Additionally, there should have been more than one expert doing the change in hair density evaluation.

Answers will vary. The most basic experimental design would be as follows. First number the 100 locations in the kiln. Randomly assign 50 of the locations to receive the first type of clay. The remaining 50 locations will receive the second type of clay. Fire the tiles in the kiln. After firing is complete, compare the proportions of cracked tiles for the two types.

2.78

Alternatively, select 50 pairs of locations where the temperature is expected to be the same for the two locations in each pair. For each pair, randomly assign one location to the first type of clay and the other location to the other type of clay. Fire the tiles and compare the proportions of cracked tiles for the two types. b

In the first design, the extraneous variable temperature is dealt with by randomly assigning the locations to the clay types. In the second design it is dealt with by blocking by temperature.

Chapter 2: Collecting Data Sensibly

Online Exercises 2.79

“forests are being destroyed…80 acres per minute”

“vanishing tropical forests”

“man-made extinction”

“destruction of tropical forests”

2.80

Answers will vary.

2.81

Answers will vary.

2.82

Answers will vary.

2.83

Answers will vary.

2.84

Answers will vary.

Chapter 3 Graphical Methods for Describing Data

3.1 T oned and fit 18.0% Out of shape 27.0%

Flabby 55.0%

3.2

Segmented bar graph

3.3

The second and third categories (“Permitted for business purposes only” and “Permitted for limited personal use” were combined into one category (“No, but some limits apply”).

b Response Category Don’t know Permitted for personal use Limited personal use Business purposes only Prohibited

1.0

0.8

0.6

0.4

0.2

0.0

Responses

Pie chart, regular bar graph

Chapter 3: Graphical Methods for Describing Data

3.4 No Answer 4.0%

Yes 41.0%

3.5

No 55.0%

a Relative Frequency (%)

30 25 20 15 10 5 0

Strongly disagree Disagree

Agree

Strongly agree Don't know

Response

3.6

“Majority of Students Don’t Want Textbooks in E-format”

Overall score is a numerical variable and grade is a categorical variable.

The shaded regions representing the various grades have areas that are proportional to the relative frequencies for those grades, just as in a segmented bar graph.

3.7 35

60 65 Overall Score

One possibility would be to require 72 or higher for “top of the class,” 63 or higher for “passing,” 59 or higher for “barely passing,” and below 59 for “failing.” This makes a clear separation between the jurisdictions in the four grade categories. (Since this makes it easier to be rated as “top of the class,” it could be considered more appropriate to leave the grade boundaries as they are.)

Chapter 3: Graphical Methods for Describing Data

3.8 Percentage 80 70 60 50 40 30 20 10 0

3.9

3.10

Japan

France

UK Country

Canada

Were the surveys carried out on random samples of married women from those countries? How were the questions worded?

In one country, Japan, the percentage of women who say they never get help from their husbands is far higher than the percentages in any of the other four countries included. The percentages in the other four countries are similar, with Canada showing the lowest percentage of women who say they do not get help from their husbands.

a Relative Frequency 0.6

Men Women

0.5 0.4 0.3 0.2 0.1 0.0 Rating

Highest 10%

Above average

Average

Below average

Lowest 10%

If frequencies were used in place of relative frequencies then, for example, the bar for “highest 10%” for women would be taller than the bar for men, even though men are much more likely than women to respond in this way.

The great majority of the college seniors who responded consider themselves to be at “above average” or in the highest 10% of fitness levels, with “above average” being the most common response for both men and women. Men were more likely than women to respond that they were in the highest 10% and women were more likely than men to respond in the

Chapter 3: Graphical Methods for Describing Data

“above average” and “average” categories. Very few of the students considered themselves to be “below average” or in the lowest 10%. 3.11

a Relative Frequency (%) 30 25 20 15 10 5 0 Response

3.12

2011 2012

Strongly disagree

Disagree

Agree

Strongly agree

Don't know

The distribution of responses was very similar for the two years. However, there was a small shift toward wanting textbooks in digital form from 2011 to 2012.

Family or social expectations 4.0% Learning and knowledge 5.0%

Other/No response 2.0% Location 28.0%

Good job or career 19.0%

School reputation and fit 20.0%

Access/Affordability 22.0%

Chapter 3: Graphical Methods for Describing Data

b 30

Percent

25 20 15 10 5 0

n io at c Lo A

/A ss e cc

y lit bi a rd ffo

r ol ho

n io at t u ep

d an

fit

od Go

b jo

e ar c or

ng ni ar

d an

ow kn

e dg le

ily

l cia o s

ns tio a ct pe x e

he Ot

No r/

se on p s

Response

3.13

The bar graph is preferable because it is easier to distinguish differences in percents of the response categories, particularly between response percents that are close in number.

a Relative Frequency 1.0 0.8 0.6 0.4 0.2 Loan Status

0.0

o Go

g lt in f au d e an S t In D d

b Pu

o Go

g lt in f au d e an St In D d

lic

iv Pr

ate

n No

f ro -p

o Go

g lt in f au d e an St I n D d

it Fo

r r -p

it of

Chapter 3: Graphical Methods for Describing Data b

The “in default” bar is taller in the “for-profit” category than in either of the other two categories.

3.14

The largest deviations in relative frequency between “Major Strength” and “Not a Major Strength” occur within Engineering and Business. Humanities, Social Science, and Biological Science are much closer in relative frequencies. 3.15 9 10 11 12 13 14 15 16

345 0457 01456889 00022333334445666677899 124689 12479 3 9

Stem: Ones Leaf: Tenths

A typical number of births per thousand of the population is around 12.4, with most birth rates concentrated in the 11.0 to 13.9 range. The distribution has just one peak (at the 12-13 class). There are no extreme values. The distribution is approximately symmetrical.

3.16

Chapter 3: Graphical Methods for Describing Data

a 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

3.17

2233344556677999 5557789 114599 0012245 001368 8 2 159 2 3 5

The distribution has just one peak (at the lowest class) and tails off for values above that class. There are four unusual observations (9296, 10310, 12596, and 14689) at the high end of the data set.

No. Since these four states have large populations, the large values given do not necessarily reflect a problem relative to other states.

No. It would be more sensible to use data giving the number of cancer deaths due to cigarette smoking per 1000 people in each state.

a 6H 7L 7H 8L 8H 9L 9H

3.18

Stem: Thousands Leaf: Hundreds

8 0044 67889 01233444 5555667777788 0000012223334444 5777

Stem: Tens Leaf: Ones

There is state-to-state variability in percent of drivers wearing seat belts, and seat belt use percentages tend to be relatively high. A typical seat belt use percentage is around 86 or 87%.

a No Enforcement 8 9876400 75433210 4

Enforcement 6 7 8 9

48 4455566777788 0000012223334445777

Stem: Tens Leaf: Ones

Although there is state-to-state variability in both the states with seat belt enforcement and the states without seat belt enforcement, the seat belt use percentages tend to be higher in states with enforcement.

Chapter 3: Graphical Methods for Describing Data

3.19

a 2H 3L 3H 4L 4H 5L 5H

569 01114 67789 0011122333 55566677777788999 00011124 566

Stem: Tens Leaf: Ones

A typical percentage of households with only a wireless phone is around 45. b West

98 21 98877 210 6

East 2H 3L 3H 4L 4H 5L 5H

569 01114 677 012 55799 Stem: Tens Leaf: Ones

A typical percentage of households with only a wireless phone for the West is around 48, which is greater than for the East (around 37). There is a slightly greater spread of values in the East than in the West, with values in the West ranging from 36 to 56 (a range of 18) and values in the East ranging from 25 to 49 (a range of 24). The distributions for both the West and the East are roughly symmetrical. Neither distribution has any outliers 3.20

3.21

The largest is 43.5 and the smallest is 18.5.

If the stems were ones then there would be 26 stems, which would be too many. One sensible strategy would be for the stems to be tens, splitting each into two, one taking the lower leaves (0-4) and the other taking the higher leaves (5-9). This would produce 6 stems: 1H, 2L, 2H, 3L, 3H, 4L.

a 0 1 2 3 4 5

9 4667788899 00113333334556667789 000000011223333457 4 1

Stem: Tens Leaf: Ones

The center is approximately 26 cents per gallon, and most states have a tax that is near the center value, with tax values ranging from 9 cents per gallon to 51 cents per gallon. The distribution is also approximately symmetric .

No The only value that might be considered unusual is the 51 cents per gallon tax in

Pennsylvania. Although not an obvious outlier, it is approximately 7 cents per gallon higher than the next lower gasoline tax. There are no other states that make such a big

Chapter 3: Graphical Methods for Describing Data

jump to the next higher tax. The lowest gasoline tax is 9 cents per gallon (Alaska) and the highest is 51 cents per gallon (Pennsylvania). 3.22

a Very Large Urban Areas

8 99 8711 9730 2 3 b

Large Urban Areas 1 2 3 4 5 6 7 8 9

023478 369 0033589 0366 012355

Stem: tens Leaf: ones

The statement is not correct, since the extra travel time is not ordered strictly according to the size of the urban area. However, it is certainly true to say that a typical travel time for an urban area classified as “very large” is greater than a typical travel time for an urban area classified as “large.”

3.23 0f 0s 0* 1 1t 1f

4455555 66677777 888888888999999999 00001111 222223 445

Stem: Tens Leaf: Ones

The stem-and-leaf display shows that the distribution of high school dropout rates is approximately symmetric. A typical dropout rate is 9%. The dropout rates range from 4% to 15%.

Chapter 3: Graphical Methods for Describing Data

3.24

a 14 12

Frequency

10 8 6 4 2 0

3.25

40 50 Percentage of People

A typical percentage of people age25 to 34 with a 4-year degree is around 42. The smallest value is 21.8% and the largest is 60.6%, giving a range of 60.6 – 21.8 = 38.8%. The distribution is roughly symmetric.

a Frequency 14 12 10 8 6 4 2 0

40 45 50 Max Wind Speed (m/s)

The distribution of maximum wind speeds is positively skewed

The distribution is bimodal, with peaks at the 35-40 and 60-65 intervals.

a 35 30

Percent

25 20 15 10 5 0

200

300

400

500 SAT Score

600

700

800

200

300

400

500 SAT Score

600

700

800

b 35 30 25 Percent

3.26

Chapter 3: Graphical Methods for Describing Data

20 15 10 5 0

The centers of the distributions for the males and females are approximately equal, between 400 and 500. Both distributions are roughly symmetrical. The spreads of the two distributions are roughly equal.

Chapter 3: Graphical Methods for Describing Data

3.27

a Frequency 18 16 14 12 10 8 6 4 2 0

5 10 15 20 Percentage of Workers who Belong to a Union

10 15 20 Percentage of Workers who Belong to a Union

A typical percentage of workers belonging to a union is around 11, with values ranging from 3.5 to 24.9. There are three states whose percentages stand out as being higher than those of the rest of the states. The distribution is positively skewed. c

The dotplot is more informative as it shows where the data points actually lie. For example, in the histogram we can tell that there are 3 observations in the 20 to 25 interval, but we don’t see the actual values and miss the fact that these values are actually considerably higher than the other values in the data set.

Chapter 3: Graphical Methods for Describing Data

3.28 Frequency 12 10 8 6 4 2 0

2.5

5.0

7.5 10.0 12.5 15.0 17.5 20.0 22.5 Percentage of Workers who Belong to a Union

25.0

The histogram in part (a) could be taken to imply that there are states with a percent of workers belonging to a union near zero. It is clear from this second histogram that this is not the case. Also, the second histogram shows that there is a gap at the high end and that the three largest values are noticeably higher than those of the other states. This fact is not clear from the histogram in part (a). 3.29

a Density 0.25

0.20

0.15

0.10

0.05

0.00

21 Hours on Social Networks

Chapter 3: Graphical Methods for Describing Data

b Density 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0

Hours Playing Video and Computer Games

3.30

Generally speaking, the students spent longer on social media than on games, with the center for social media being between 1 and 6 hours, and the center for games being between 0 and 1 hour. There was a greater spread of time spent on social networks than there was playing games, as a result of the fact that a majority of the students spent between 0 and 1 hour playing games and there was no time interval containing a majority of students for social media use. Both distributions are positively skewed.

In order to show necessary detail at the lower end of the distribution it was necessary to use class intervals as narrow as 5 minutes. However, to use this class width throughout would result in too large a number of intervals. Therefore wider intervals were used for the higher commute times where the results were more scarce.

b Commute Time 0 to <5 5 to <10 10 to <15 15 to <20 20 to <25 25 to <30 30 to <35 35 to <40 40 to <45 45 to <60 60 to <90 90 to <120

Relative Density Frequency

0.056 0.156 0.177 0.155 0.147 0.061 0.121 0.015 0.024 0.040 0.030 0.018

0.0112 0.0312 0.0354 0.0310 0.0294 0.0122 0.0242 0.0030 0.0048 0.0027 0.0010 0.0006

Chapter 3: Graphical Methods for Describing Data

c 0.04

Density

0.03

0.02

0.01

0.00

5 10 15 20 25 30 35 40 45

Commute Time (minutes)

120

A typical commute time is between 10 and 15 minutes, and the great majority of commute times lie between 5 and 35 minutes. The distribution is positively skewed, and has an unexpected peak, in the 30-35 class occurring more frequently than the two adjacent classes. 3.31

a Commute Time 0 to <5 5 to <10 10 to <15 15 to <20 20 to <25 25 to <30 30 to <35 35 to <40 40 to <45 45 to <60 60 to <90 90 to <120

Relative Cumulative Relative Frequency Frequency

0.056 0.156 0.177 0.155 0.147 0.061 0.121 0.015 0.024 0.040 0.030 0.018

0.056 0.212 0.389 0.544 0.691 0.752 0.873 0.888 0.912 0.952 0.982 1

Chapter 3: Graphical Methods for Describing Data

Cumulative Relative Frequency

1.0

0.8

0.6

0.4

0.2

0.0 0

3.32

40 60 80 Commute Time (minutes)

100

120

i 0.93 ii 1 − 0.62 = 0.38 iii approximately 19 minutes

a Relative Frequency (%) 50

Tuition and Fees (Thousands)

A typical value for tuition and fees for a student at a public four-year college is between $6,000 and $9,000. Tuition and fees for these colleges range from close to zero to $42,000. The distribution is positively skewed.

Chapter 3: Graphical Methods for Describing Data

b Relative Frequency (%) 50

Tuition and Fees (Thousands) A typical value for tuition and fees for a student at a private not-for-profit four-year college is between $30,000 and $33,000. Tuition and fees at these colleges range from around $3,000 to $48,000. The distribution is slightly negatively skewed, and seems to have peaks between $3,000 and $6,000, between $24,000 and $36,000, and between $42,000 and $45,000.

3.33

Tuition and fees tend to be higher at private not-for-profit four-year colleges than at public four-year colleges. Tuition and fees are also more variable among private not-for-profit fouryear colleges than among public four-year colleges. The two distributions have very different shapes, with a simple positively skewed distribution for public four-year colleges and a distribution that could be considered to have three peaks for private not-for-profit four-year colleges.

The distribution is likely to be negatively skewed. There will be a high density of scores close to 100, with no score greater than 100, and the density of scores tailing off to the left.

So long as there is no cutoff at the lower end the distribution is likely to be roughly symmetrical. For example, the greatest density of points might be around 65, with the density tailing off equally to the left and to the right of that value. (It could also be suggested that the distribution would be positively skewed, since certain students will always get high scores however tough the exam.)

In this case a bimodal distribution would be likely. There would a clustering of points around, say, 65 for the students with less math knowledge and another clustering around, say, 95, for those who have had calculus.

Chapter 3: Graphical Methods for Describing Data

3.34 Density 0.4

0.3

0.2

0.1

0.0

3.35

4 5 6 7 8 Number of attempts

a Rainfall (inches) Frequency 10 to <12 12 to <14 14 to <16 16 to <18 18 to <20 20 to <22 22 to <24 24 to <26 26 to <28 28 to <30 30 to <32

2 4 8 6 16 10 6 1 0 4 2

Relative Frequency

Cumulative Relative Frequency

0.034 0.068 0.136 0.102 0.271 0.169 0.102 0.017 0 0.068 0.034

0.034 0.102 0.237 0.339 0.610 0.780 0.881 0.898 0.898 0.966 1.000

Chapter 3: Graphical Methods for Describing Data

b Relative Frequency (percent) 30 25 20 15 10 5 0

20 24 Rainfall (inches)

The histogram shows a distribution that is slightly positively skewed with a trough between 24 and 28. 3.36

(The cumulative relative frequencies are shown in the table in Part (a).) Cumulative Relative Frequency 1.0

0.8

0.6

0.4

0.2

0.0 10

i 0.20 ii 0.89 iii 0.89 − 0.31 = 0.58

20 25 Rainfall (inches)

Chapter 3: Graphical Methods for Describing Data

3.37

a Cumulative Relative Frequency 1.0

0.8

0.6

0.4

0.2

0.0 0

3.38

8 10 Years Survived

i 0.53 ii 0.62 iii 1 − 0.68 = 0.32

a Years Survived 0-<2 2-<4 4-<6 6-<8 8 - < 10 10 - < 12 12 - < 14 14 - < 16

Relative Frequency .10 .42 .02 .10 .04 .02 .02 .28

Chapter 3: Graphical Methods for Describing Data

b Relative frequency (percent)

3.39

6 8 10 Years survived

The histogram shows a bimodal distribution, with peaks at the 2-4 year and 14-16 year intervals. All the other survival times were considerably less common than these two.

We would need to know that the set of patients used in the study formed a random sample of all patients younger than 50 years old who had been diagnosed with the disease and had received the high dose chemotherapy.

Answers will vary. One possibility for each part is shown below. a Class Interval Frequency

100 to <120 5

120 to <140 10

140 to <160 40

160 to <180 10

180 to <200 5

Class Interval Frequency

100 to <120 20

120 to <140 10

140 to <160 4

160 to <180 25

180 to <200 11

Class Interval Frequency

100 to <120 33

120 to <140 15

140 to <160 10

160 to <180 7

180 to <200 5

Class Interval Frequency

100 to <120 5

120 to <140 7

140 to <160 10

160 to <180 15

180 to <200 33

Chapter 3: Graphical Methods for Describing Data

3.40

a 900 875

APEAL Rating

850 825 800 775 750 40

3.41

60 Quality Rating

No. Customer satisfaction does not seem to be related to car quality. The scatterplot shows what seems to be a random scatter of points. (It could be argued that there is a slight tendency for APEAL rating to be higher for higher quality ratings.)

a Calories 700 600 500 400 300 200 100 0

Fat

As expected there is a positive relationship between the amount of fat and the number of calories. The relationship is weak.

Chapter 3: Graphical Methods for Describing Data

b Calories 700 600 500 400 300 200 100 200

400

600

800

1000 Sodium

1200

1400

1600

1800

As with the relationship between calories and fat, there is a weak positive relationship between calories and sodium. The second relationship is, if anything, slightly stronger than the first, particularly if three points ((770, 660), (750, 560), and (1090, 120)) are disregarded. c Sodium 1800 1600 1400 1200 1000 800 600 400 200 0

Fat

There does not appear to be a relationship between fat and sodium. d

The region containing points with x coordinates less than 3 and y coordinates less than 900 corresponds to the healthier food choices since these are the low-fat, low-sodium options.

Chapter 3: Graphical Methods for Describing Data

3.42

Percent with Wireless Only

50 40 30 20 10 0

m ce

13 20 r be

ne Ju

14 20 De

m ce

14 20 r be

ne Ju

15 20 c De

15 20 r be em

ne Ju

16 20 c De

16 20 r be em

The graph shows a gradual upward trend in the percentage of homes with only a wireless phone service from December 2013 to December 2016. b a 90

Overall Score

3.43

The increase has been at a roughly steady rate..

40 0

100

150

200

250

Price

There appears to be a weak positive relationship between price and overall score. The scatterplot only weakly supports the statement.

Chapter 3: Graphical Methods for Describing Data

3.44

There is a relatively weak positive relationship between price and overall score. More expensive laptops tend to earn higher overall scores, but this doesn’t hold true for all laptops in this data set.

3.45

The time series plot is consistent with the statement of having seen steady growth in recycling and composting because those trends are both increasing. However, the statement about amounts landfilled have generally declined might be somewhat misleading, at least when the conclusion is based on the graph shown, with the given scale. Rather, it might be more appropriate to state that the amounts landfilled have remained roughly constant, with a slight decrease in the given time period.

3.46

a 22.5

Percentage who Smoke

20.0 17.5 15.0 12.5 10.0 7.5 5.0 2000 2001 2002 2003 2004 2005 2006 2007 2008 2009 2010 2011 2012 2013 2014

Year

There has been a downward trend in the percent of grade 12 students who smoke daily, from a high of 20.6% in 2000 down to 6.7% in 2014. b

Percentage who Smoke

0 00 01 02 03 04 05 06 07 08 09 10 11 12 13 14 20 20 20 20 20 20 20 20 20 20 20 20 20 20 20 Year

Grade 8 Grade 10 Grade 12

Chapter 3: Graphical Methods for Describing Data

There has been a downward trend in the percent of students who smoke regardless of grade level. For all the years in the data set, grade 8 students had a lower percentage of smokers than the other grades. The percentage of grade 10 students who smoke was between the percentages of grade 8 and grade 12, and the percentage of grade 12 students who smoke was the highest for all years in the data set. In addition to the maximum and minimum indicated in part (a), the percentages of grade 10 students who smoke fell from 14% in 2000 to 3.2% in 2014. Similarly, the percentage of grade 8 students who smoke fell from 7.4% in 2000 to 1.4% in 2014.

3.47

According to the 2001 and 2002 data, there are seasonal peaks at weeks 4, 9, and 14, and seasonal lows at weeks 2, 6, 10-12, and 18.

3.48

For 12-year-olds to 17-year-olds, the percentage owning cell phones increased with age at each of the years 2004, 2006, and 2008. Also, in each age group the percentage owning cell phones increased from 2004 to 2006 and from 2006 to 2008. Amongst 12-year-olds the percentage increased from 18% to 51% between 2004 and 2008, and amongst 17-year-olds the percentage increased from 64% to 84% over the same period.

3.49

a 1.0

Unknown/Other Native American African American Hispanic/Latino Asian American White

0.8

0.6

0.4

0.2

0.0

Enrollment

The graphical display created in Part (a) is more informative, since it gives an accurate representation of the proportions of the ethnic groups.

The people who designed the original display possibly felt that the four ethnic groups shown in the segmented bar section might seem to be underrepresented at the college if they used a single pie chart.

3.50

Chapter 3: Graphical Methods for Describing Data

a Relative Frequency Men Women

Nation

World

Local Community

The USA Today graph is constructed so that the sleeves and the hands together have heights that are in proportion to the relative frequencies. However many readers could think that the sleeves (which could be thought to look like the bars of a bar chart) are supposed to represent the relative frequencies, and the lengths of the sleeves are not in the correct proportion for these quantities. Also, given that the six hands all have equal areas and that the heights of the sleeve-hand combinations are in proportion to the relative frequencies, the areas of the sleeve-hand combinations will not be proportional to the relative frequencies, whereas in the traditional comparative bar graph the areas of the bars are in proportion to the relative frequencies.

3.51

The first graphical display is not drawn appropriately. The Z’s have been drawn so that their heights are in proportion to the percentages shown. However, the widths and the perceived depths are also in proportion to the percentages, and so neither the areas nor the perceived volumes of the Z’s are proportional to the percentages. The graph is therefore misleading to the reader. In the second graphical display, however, only the heights of the cars are in proportion to the percentages shown. The widths of the cars are all equal. Therefore the areas of the cars are in proportion to the percentages, and this is an appropriately drawn graphical display.

3.52

The display is misleading. The cones are drawn so that their heights are in proportion to the relative frequencies. However this means that their areas are not in proportion to the relative frequencies. A more suitable display is shown below.

Chapter 3: Graphical Methods for Describing Data

Relative Frequency (percent) Men Women

Preference

Cup

Cone

Sundae

Sandwich

Other

3.53

The piles of cocaine have been drawn so that their heights are in proportion to the percentages shown. However, the widths are also in proportion to the percentages, and therefore neither the areas (nor the perceived volumes) are in proportion to the percentages. The graph is therefore misleading to the reader.

3.54

One would want to use relative frequencies when constructing a comparative bar chart to compare ideal distance for students and parents because the number of students (8,347) and number of parents (2,087) are not equal to each other.

Students tend to want to be farther away from home than their parents would like them to be. In fact, 37% of students want to be over 500 miles from home, whereas only 18% of parents want their children to be over 500 miles from home.

Chapter 3: Graphical Methods for Describing Data

3.55

Average Critical Reading SAT Score

500

400

300

200

100

English and another language

A language other than English

Reading Math English and another language

Reading Math A language other than English

First Language Learned

a 600

500

Average Score

3.56

English

400

300

200

100

Reading Math English

For the group whose first language is English the average scores in the two sections of the SAT were the same. In the other two groups the math scores were higher than the verbal scores, with the difference being most marked in the “language other than English” group. The average math scores were roughly the same for the three groups, while the verbal scores were lower for the “English and another” group than for the “English” group, and still lower for the “language other than English” group.

Chapter 3: Graphical Methods for Describing Data

3.57 20

Frequency

10 15 20 Percent Uninsured Drivers

The distribution of the percent of uninsured drivers is positively skewed, centered in the 1015% interval. The percent uninsured drivers varies between a low of 3.9% and a high of 25.9%. Approximately 75% of states have 15% or less uninsured drivers. 3.58

a -2 -1 -0 0 1 2 3 4 5 6 7 8 9 10 11 12 b

1 43 256679 0188999 224677 005 344578 3467 1589 47 57 08 02578 8

Stem: Units Leaf: Tenths

The distribution is positively skewed. The center of the distribution is roughly 4.5%, and the range of the distribution is 14.9 percentage points. Two values (-2.1% and 12.8%) are possible outliers.

Chapter 3: Graphical Methods for Describing Data

c Eastern 1 34 976652 99880 77422 00 843 43 985 7

2 8

Western -2 -1 -0 0 1 2 3 4 5 6 7 8 9 10 11 12

19 6 5 457 67 1 4 57 08 0578 Stem: Units Leaf: Tenths

The distribution of percent change for the Eastern states is positively skewed, whereas the distribution for the percent change for Western states is negatively skewed. The center of the Eastern states percent change (approximately 3%) is less than that of the Western states (approximately 6%). The spread of percent change values is greater for Eastern states (range is 14.9%) than Western states (range is 9.7%). 3.59

High graft weight ratios are clearly associated with low body weights (and vice versa), and the relationship is not linear. (In fact there seems to be, roughly speaking, an inverse proportionality between the two variables, apart from a small increase in the graft weight ratios for increasing body weights amongst those recipients with the greater body weights. This is interesting in that an inverse proportionality between the variables would imply that the actual weights of transplanted livers are chosen independently of the recipients’ body weights.)

A likely reason for the negative relationship is that the livers to be transplanted are probably chosen according to whatever happens to be available at the time. Therefore, lighter patients are likely to receive livers that are too large and heavier patients are likely to receive livers that are too small.

Chapter 3: Graphical Methods for Describing Data

3.60

a Percent of Households with a Computer

90 80 70 60 50 40 30 20 10 0 84 85 8 6 87 88 89 90 9 1 92 93 94 9 5 9 6 97 98 99 0 0 01 02 03 04 0 5 06 07 08 09 1 0 11 12 13 1 9 19 19 19 1 9 1 9 19 19 19 1 9 19 19 19 19 1 9 19 20 20 2 0 2 0 20 20 20 2 0 20 20 20 20 2 0 20

Year

3.61

The percentage of households with computers has increased over time, from a low of approximately 8% in 1984 to nearly 84% in 2013. The rate of increase of the percentage of households with a computer increased over time from 1984 to 2001, and the rate of increase remained roughly constant from 2001 to 2013.

Disney

Other

975332100 0 765 1 920 2 3 4 4 5

0001259 156 0 Stem: Hundreds Leaf: Tens

On average, the total tobacco exposure times for the Disney movies are higher than the others, with a typical value for Disney being around 90 seconds and a typical value for the other companies being around 50 seconds. Both distributions have one peak and are positively skewed. There is one extreme value (548) in the Disney data, and no extreme value in the data for the other companies. There is a greater spread in the Disney data, with values ranging from 6 seconds to 540 seconds, than for the other companies, where the values range from 1 second to 205 seconds.

3.62

Chapter 3: Graphical Methods for Describing Data

a Average T ransportation Expenditure 400

300

200

100

0 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000 Year

Percentage of Household Expenditures for T ransportation 18 16 14 12 10 8 6 4 2 0 1990

1991

1992

1993

1994

1995 Year

1996

1997

1998

1999

2000

Yes. The first time-series plot shows that average expenditure on transportation has increased over time at a roughly constant rate, whereas the second graph shows that the percentage of household expenditures for transportation, whilst varying from year to year, has remained at roughly the same level.

Chapter 3: Graphical Methods for Describing Data

3.63 25000

Average Cost

22500

20000

17500

15000 2002 2003 2004 2005 2006 2007 2008 2009 2010 2011 2012 2013 Year

There is a strong, positive trend in the average cost of per year for tuition, fees, and room and board for four-year public institutions in the U.S. The average cost has steadily increased from a low of $14,439 per year in 2002 to a high of $ 24,706 per year in 2013. 3.64

3.65

Yes. The first histogram gives the impression of a distribution that is negatively skewed, while the second histogram gives an impression of a distribution that is positively skewed. Also, the second histogram shows a decrease in density of points in the second and third class intervals as compared to the first and fourth intervals: this feature of the shape of the distribution is not at all evident in the first histogram.

According to the first histogram the proportion of observations less than 800 is approximately (6  7  (0.5)10) 27  0.667 . According to the second histogram this proportion is approximately (6  5  5  (1/ 3)6) 27  0.667 , also. The true proportion of observations less than 800 is 18 27  0.667 . The approximations according to the two histograms are both correct.

a Skeletal Retention 0.15 to <0.20 0.20 to <0.25 0.25 to <0.30 0.30 to <0.35 0.35 to <0.40 0.40 to <0.45 0.45 to <0.50 0.50 to <0.55 0.55 to <0.60

Frequency 4 2 5 21 9 9 4 0 1

Chapter 3: Graphical Methods for Describing Data

Frequency 20

0.15

0.20

0.25

0.30

0.35 0.40 0.45 Skeletal retention

0.50

0.55

0.60

The histogram is centered at approximately 0.34, with values ranging from 0.15 to 0.5, plus one extreme value in the 0.55-0.6 range. The distribution has a single peak and is slightly positively skewed.

Chapter 3: Graphical Methods for Describing Data

Cumulative Review Exercises CR3.1 No. It is quite possible, for example, that men who ate a high proportion of cruciferous vegetables generally speaking also had healthier lifestyles than those who did not, and that it was the healthier lifestyles that were causing the lower incidence of prostate cancer, not the eating of cruciferous vegetables. CR3.2 Neither the conclusion nor the headline is appropriate. Since this is an observational study, we can only conclude that there is an association between religious activity and blood pressure, not that religious activity causes reduced blood pressure. It is quite possible, for example, that people who attend a religious service once a week and pray or study the Bible at least once a day also have more exercise than people who do not engage in these activities, and that it is the exercise not the religious activity that is causing the reduced blood pressure. CR3.3 Very often those who choose to respond generally have a different opinion on the subject of the study from those who do not respond. (In particular, those who respond often have strong feelings against the status quo.) This can lead to results that are not representative of the population that is being studied. CR3.4 The survey was conducted using people who chose to attend a budget workshop and so we have no reason to think that this set of people was representative of the population as a whole. (Also, there is no reason to think that people who are not exposed to the content of the workshop would have the same opinion regarding the proposed sales tax increase.) CR3.5 Only a small proportion (around 11%) of the doctors responded, and it is quite possible that those who did respond had different opinions regarding managed care from the majority who did not. Therefore the results could have been very inaccurate for the population of doctors in California. CR3.6 a

Answers will vary. A completely randomized design (no blocking) should include randomly choosing which exiting cars should have a car waiting for the space and which should not.

The response variable is the time taken to exit the parking space.

Possible extraneous variables include the time of day, the gender of the exiting driver, and the location of the parking space.

CR3.7 Suppose, for example, the women had been allowed to choose whether or not they participated in the program. Then it is quite possible that generally speaking those women with more social awareness would have chosen to participate, and those with less social awareness would have chosen not to. Then it would be impossible to tell whether the stated results came about as a result of the program or of the greater social awareness amongst the women who participated. By randomly assigning the women to participate or not, comparable groups of women would have been obtained. CR3.8 a

Randomly assign the 100 female volunteers to two groups, A and B. (This could be done by writing the names of the volunteers on slips of paper, placing the slips in a hat, mixing them, and randomly picking one slip at a time. The first 50 names picked will go into Group A and the remainder into Group B.) At the beginning of the experiment, by means of a questionnaire, the PMS symptoms of the volunteers will be assessed. Then the women in Group A will be given the nasal spray containing the pheromones and the women in Group B

Chapter 3: Graphical Methods for Describing Data will be given a nasal spray that in appearance and sensation is identical to the first one but contains no active ingredient. (No volunteer will be told which treatment she is receiving.) Then all the women will be instructed to use the spray over a period of a few months. At the end of the experiment the women’s PMS symptoms will be reassessed and the reduction in PMS symptoms for the two groups will be compared. b

Yes. If Group B were to receive no spray at all, and if there were a greater improvement in symptoms for Group A than for Group B, then it would be impossible to tell whether this improvement took place as a result of the pheromones or as a result of the psychological effect of using the spray.

The design is single blind, since the women are not told what treatment they are receiving. A double blind design would be one in which, additionally, the people who measure the PMS symptoms do not know who received which treatment. In the design given in part (a) the women are required to assess their own symptoms, however it would certainly be useful if the people who collate the information given in the questionnaires were not to know who received which treatment.

CR3.9 a Pass Rate (%) 90 80

District San Luis Obispo High School San Luis Obispo County State of California

70 60 50 40 30 20 10 0 2002

2003

2004

Between 2002 and 2003 and between 2003 and 2004 the pass rates rose for both the high school and the state, with a particularly sharp rise between 2003 and 2004 for the state. However, for the county the pass rate fell between 2002 and 2003 and then rose between 2003 and 2004.

Chapter 3: Graphical Methods for Describing Data

CR3.10 a Not sure Hardly anyone Almost everyone 2.0% 1.0% 10.0% A small number 20.0%

A lot of people 28.0%

A moderate number 39.0%

Not sure 6.0% Neither 15.0% Democrats 36.0%

Both equally 10.0%

Republicans 33.0%

Chapter 3: Graphical Methods for Describing Data

b Relative Frequency Not sure Almost everyone A lot of people A moderate number A small number Hardly anyone

1.0

0.8

0.6

0.4

0.2

0.0

Response

Relative Frequency Not sure Neither Both equally Republicans Democrats

1.0

0.8

0.6

0.4

0.2

0.0

Response

In both cases, but particularly for the first data set, since there is a relatively large number of categories the segmented bar graph does a better job of presenting the data. Also, in the first data set the order in which the six possible responses are listed is important. This order is clearly reflected in the segmented bar graph but not in the pie chart.

CR3.11 a 0 1 2 3 4 5

123334555599 00122234688 1112344477 0113338 37 23778

Stem: Thousands Leaf: Hundreds

Chapter 3: Graphical Methods for Describing Data

The stem-and-leaf display shows a positively skewed distribution with a single peak. There are no extreme values. A typical total length is around 2100 and the great majority of total lengths lie in the 100 to 3800 range. b Frequency 12 10 8 6 4 2 0

1000

2000

3000 T otal Length

4000

5000

6000

The number of subdivisions that have total lengths less than 2000 is 12 + 11 = 23, and so the proportion of subdivisions that have total lengths less than 2000 is 23/47 = 0.489. The number of subdivisions that have total lengths between 2000 and 4000 is 10 + 7 = 17, and so the proportion of subdivisions that have total lengths between 2000 and 4000 is 17/47 = 0.361.

CR3.12 a Class Interval 0 to <6 6 to <12 12 to <18 18 to <24 24 to <30 30 to <36

Frequency 2 10 21 28 22 6

Relative Frequency Cumulative Relative Frequency 0.022 0.022 0.112 0.135 0.236 0.371 0.315 0.685 0.247 0.933 0.067 1.000

The relative frequencies and the cumulative relative frequencies are shown in the table in Part (a).

(Relative frequency for 12 to <18) = (cum rel freq for 12 to <18) − (cum rel freq for 6 to <12) = 0.371 − 0.135 = 0.236.

i ii

1 − 0.135 = 0.865. 0.236 + 0.315 = 0.561

Chapter 3: Graphical Methods for Describing Data

e Cumulative Relative Frequency 1.0

0.8

0.6

0.4

0.2

0.0 0

i ii

10 20 30 T ime to First Malfunction (months)

20 months 29 months

CR3.13 a

The histogram shows a smooth positively skewed distribution with a single peak. There are around three values that could be considered extreme, with those values lying in the 650 to 750 range. The majority of time differences lie between 50 and 350 seconds.

A typical time difference between the two phases of the race is 150 seconds.

Estimating the frequencies from the histogram we see that approximately 920 runners were included in the study and that approximately 8 of those runners ran the late distance more quickly than the early distance (indicated by a negative time difference). Therefore the proportion of runners who ran the late distance more quickly than the early distance is approximately 8/920 = 0.009.

Generally speaking, states with higher poverty rates have higher dropout rates, but the relationship is not strong. There are two states with relatively low poverty rates (between 10% and 15%) but high dropout rates (over 15%).

There is a weak positive relationship between poverty rate and dropout rate.

CR3.14

CR3.15 There is a strong negative linear relationship between racket resonance frequency and sum of peak-to-peak accelerations. There are two rackets whose data points are separated from the remaining data points. Those two rackets have very high resonance frequencies and their peak-topeak accelerations are lower than those of all the other rackets.

Chapter 3: Graphical Methods for Describing Data

CR3.16 a Percentage 80

Good T ime Bad T ime

70 60 50 40 30 20 1980

1985

1990 Date

1995

2000

2005

The new line has been added to the plot in Part (a).

The time series plot is more effective as its horizontal axis shows an actual time scale, which is not present in the bar graph.

Chapter 4 Numerical Methods for Describing Data

4.1

x  159.00  199.00  157.00  127.65  123.99  126.00  6 = $148.77. To calculate the median, we first list the data values in order of size: 123.99 126.00 127.65 157.00 159.00 199.00 Because there are an even number of data values, the median is the average of the two middle values in this list, 127.65 and 157.00. Therefore, the median is $142.33.

4.2

4.3

The mean is larger than the median since the distribution of these six values is positively skewed. The largest value is greatly separated from the remaining five values.

The median is better as a description of a typical value since it is not influenced by the extreme value.

x   9.5  10.0  10.0  10.0  11.4  8.9  9.5  9.1 8 = 9.8 mg/oz.

The caffeine concentration of top-selling energy drinks is much greater than that of colas.

The mean caffeine concentration for the brands of coffee listed is

140  195  155  115  195  180  110  110  130  55  60  60 = 125.417 mg/cup. 12 Therefore the mean caffeine concentration of the coffee brands in mg/oz is (125.417)/8 = 15.677. This is significantly greater than the mean caffeine concentration of the energy drinks given in the previous exercise. 4.4

120 150 180 Sodium Content (mg)

210

240

The distribution is roughly symmetrical. One brand of peanut butter has an unusually high sodium content (250). If this value is disregarded, then the distribution is negatively skewed. b

x  (120   110) 11  134.091. The values listed in order are: 50 65 110 120 Median = 6th value = 140.

120

140

150

170

250

The values of the mean and the median are similar because the distribution is roughly symmetrical.

Chapter 4: Numerical Methods for Describing Data 4.5

4.6

The large difference between the mean cost and the median cost along with the fact that the mean is greater than the median tells us that there are some large outliers in the distribution of wedding costs in 2012.

I agree that the average cost is misleading because 50% of the weddings in 2012 cost less than $18,086, which is much lower than the mean wedding cost.

I agree with this statement. The mean is strongly influenced by outliers, and large outliers will pull the mean toward larger values. Given that the mean is so much larger than the median (which divides the distribution of wedding costs in half), less than 50% of the wedding costs must be greater than or equal to the mean.

Fifty percent (50%) of incomes for single person households are less than $60,691, and 50% of single person household incomes are greater than $60,691.

Statement 1 is incorrect because the extremely low income category refers to 30% of the county median income (which is $18,207.30), not 30% of the households. Statement 2 is incorrect because, once again, the 50% is referring to 50% of the median income (which is $30,345.50), not 50% of households. Statement 3 is incorrect because there could be some outliers in the distribution of incomes for single person households. The 120% (or $72,829.20) refers to the income and not the number of households.

4.7

The dotplot of circulation numbers (reproduced below) shows that the distribution is positively skewed with a large outlier. The mean should be used to describe a typical value of symmetric distributions, and therefore should not be used to describe the center of this distribution.

4.8

4.9

4.10

500000

1000000

1500000 2000000 Circulation

2500000

3000000

3500000

The number in the sample who are registered to vote is 12, in a sample of 20 students. So the proportion of successes is 12/20 = 0.6.

This generalization could be made if the sample were a simple random sample of seniors at the university.

The sum of the values given is 6631, and so the mean is 6631/15 = 442.07.

The median is the middle value in the ordered list, so the median is 356.

This sample represents the 15 states with the highest number of speed-related fatalities, and so it is not reasonable to generalize from this sample to the other 35 states.

Both means are greater than the respective medians, suggesting that both distributions are positively skewed. The fact that the median is greater for angioplasty and the mean is greater for bypass surgery tells us that the median-to-mean increase is greater in the case of bypass surgery, suggesting that the positive skew is more pronounced for bypass surgery than for angioplasty.

Chapter 4: Numerical Methods for Describing Data

4.11

Neither statement is correct. Regarding the first statement it should be noted that, unless the “fairly expensive houses” constitute a majority of the houses selling, these more costly houses will not have an effect on the median. Turning to the second statement, we point out that the small number of very high or very low prices will have no effect on the median, whatever the number of sales. Both statements can be corrected by replacing the median with the mean.

4.12

Here the word “average” refers to the mean and, yes, it is possible that 65% of the residents would earn less than that value. If the distribution of wages is positively skewed then the mean will be greater than the median, which tells us that more than 50% (and therefore possibly 65%) of residents will earn less than the mean.

4.13

The two possible solutions are x5  32 and x5  39.5.

4.14

pˆ  7 10  0.7.

x  (1  1  0  1  1  1  0  0  1  1) 10  0.7. The values of x and p̂ are the same.

The sample size is now 10 + 15 = 25. We need the number of successes to be (0.8)(25) = 20. Therefore we need 13 of these 15 observations to be successes.

4.15

The only measure of center discussed in this section that can be calculated is the median. To find the median we first list the data values in order: 170 290 350 480 570 790 860 920 1000+ 1000+ The median is the mean of the two middle values: (570 + 790)/2 = 680 hours.

4.16

The total score after 19 students is (19)(70) = 1330. For the average for all 20 students to be 71, we need the total for all 20 students to be (20)(71) = 1420. So the last student needs to score 1420 − 1330 = 90. For the average for all 20 students to be 72, we need the total for all 20 students to be (20)(72) = 1440. So the last student needs to score 1440 − 1330 = 110, which is not possible.

4.17

x  (29  62  37  41  70  82  47  52  49) 9  52.111.

Variance 

(29  52.111)2 

 (49  52.111)2 8

 279.111.

s  279.111  16.707. b

4.18

The addition of the very expensive cheese would increase the values of both the mean and the standard deviation.

Cereals rated very good: Median = 47.5, upper quartile = 62, lower quartile = 41, iqr = 21. Cereals rated good: Median = 53, upper quartile = 60.5, lower quartile = 36.5, iqr = 24.

Chapter 4: Numerical Methods for Describing Data

The median cost per serving is higher for cereals rated good than for cereals rated very good. The interquartile range is higher for cereals rated good than for cereals rated very good, reflecting a greater spread in the distribution for cereals rated good. 4.19

4.20

The complete data set, listed in order, is: 19 28 30 41 43 46 48 49 53 53 62 67 71 77 Lower quartile = 4th value = 41. Upper quartile = 12th value = 62. Iqr = 21.

The iqr for cereals rated good (calculated in exercise 4.18) is 24. This is greater than the value calculated in Part (a).

This question can be answered by comparing standard deviations (or variances) or interquartile ranges. The standard deviation for the high caffeine energy drinks (8.311) is much larger than the standard deviation for the top selling energy drinks (0.667). The high caffeine drinks are the more variable with respect to caffeine per ounce. Likewise, the interquartile range for the high caffeine energy drinks (31.3 − 15.0 = 16.3) is much larger than the interquartile range for the top selling energy drinks (10.0 − 9.05 = 0.95). This, too, tells us that the high caffeine drinks are the more variable with respect to caffeine per ounce.

4.21

x  (44.0   21.4) 18  51.333 A typical amount poured into a tall, slender glass is 51.333 ml.

s

(44.0  51.333) 2 

 (21.4  51.333) 2

 15.216 17 A typical deviation from the mean amount poured is 15.216 ml. 4.22

x  (89.2   57.8) 18  59.233 A typical amount poured into a short, wide glass is 59.233 ml.

s

(89.2  59.233)2 

 (57.8  59.233)2

 16.715 17 A typical deviation from the mean amount poured is 16.715 ml.

4.23

The mean amount for a short, wide glass (59.233 ml) is greater than the mean amount for a tall, slender glass (51.333 ml). This suggests that bartenders tend to pour more into a short, wide glass than into a tall, slender glass.

x  (730 

 520) / 8  6380 / 8  797.5

Variance  ((730  797.5)2 

 (520  797.5)2 ) 7  22735.7 .

Standard deviation  22735.7  150.78 b

The large value of the standard deviation tells us that there is considerable variation between prices of these highly rated smart phones.

Chapter 4: Numerical Methods for Describing Data

4.24

For the lowest rated smartphones, mean = $142.90, variance = 1298.8, and standard deviation = $36.0. The mean cost for the lowest rated smartphones is much less than for the highly rated smartphones, showing a lower typical purchase price for the lowest rated smartphones. The standard deviation for the lowest rated smartphones is considerably less than for the highly rated ones, showing a lower purchase price variability for the lowest rated phones.

4.25

x  (1014  1009 

Variance 

 860  1108) 10  1005.6 milliseconds

(1014  1005.6)2 

 (1108  1005.6)2

9 s  4922.0  70.2 milliseconds

4.26

 4922.0

a Listening Not Listening

875

910

945 980 1015 1050 Reaction time (milliseconds)

1085

The range of reaction times for those listening to audiobooks (1108 – 806 = 248 milliseconds) is greater than the range of reaction times for those not listening to audiobooks (1041 – 904 = 137). Graphically, there is less variability in reaction times for those not listening to audiobooks than for those listening to audiobooks. As such, one can conclude that the standard deviation for those not listening to audiobooks is less than the standard deviation for those listening to audiobooks.

x  (961 

s

 991) 10  988.1 milliseconds

(961  988.1) 2 

 (991  988.1) 2

 37.2 milliseconds (note that this value is smaller 9 than the standard deviation calculated in the previous question).

4.27

The mean reaction time for drivers listening to audiobooks is higher than that for drivers not listening to audiobooks (1005.6 vs. 988.1 milliseconds). There is more variability in reaction times for drivers listening to audiobooks than for those not listening to audiobooks (70.2 vs. 37.2 milliseconds).

The data values, listed in order, are: 0 130 236

0 142 236

0 142 306

0 165

0 177

0 189

59 189

71 201

83 212

106 224

Lower quartile = average of 6th and 7th values = (0 + 0)/2 = 0. Upper quartile = average of 19th and 20th values = (189 + 201)/2 = 195. Interquartile range = 195 − 0 = 195. b

The lower quartile is equal to the minimum value for this data set because there are a large number of equal values (zero in this case) at the lower end of the distribution. In most data

Chapter 4: Numerical Methods for Describing Data

sets this is not the case and therefore, generally speaking, the lower quartile is not equal to the minimum value. 4.28

Answers will vary.

4.29

The standard deviation of percent return is a reasonable measure of unpredictability because it measures how much, on average, individual returns deviate from the mean return of the fund. A smaller standard deviation indicates smaller deviations (on average) from the mean return, and therefore less risk.

A fund with a small standard deviation can still lose money if the average percent return is small relative to the standard deviation. Recall that the standard deviation is a typical deviation from the mean (either above or below the mean), and if the mean is smaller than the standard deviation, the return could be negative, resulting in the fund losing money.

x  (141 

4.30

 70) 10  147.5.

Variance  ((141  147.5)2 

 (70  147.5)2 ) 9  2505.83333.

Standard deviation  2505.83333  50.058. b

The Memorial Day data are a great deal more consistent than the New Year’s Day data, and therefore the standard deviation for Memorial Day would be smaller than the standard deviation for New Year’s Day.

The standard deviations are given in the table below. Holiday New Year's Day Memorial Day July 4th Labor Day Thanksgiving Christmas

Standard Deviation 50.058 18.224 47.139 17.725 15.312 52.370

The standard deviations for Memorial Day, Labor Day, and Thanksgiving are 18.224, 17.725, and 15.312, respectively. The standard deviations for the other three holidays are 50.058, 47.139, and 52.370. The standard deviations for the same day of the week holidays are all smaller than all of the standard deviations for the holidays that can occur on different days. There is less variability for the holidays that always occur on the same day of the week. 4.31

i. The lower quartile must be less than the median.

iii. Only 10% of the bypass surgeries took longer than 42 days to be completed, and so the upper quartile must be between the median (13) and this value. iv. The number of days for which only 5% of the bypass surgery wait times were longer must be greater than the number of days for which 10% of the bypass surgery wait times were longer, which was 42.

Chapter 4: Numerical Methods for Describing Data

4.32

The mean of the values given is 747.370 and the standard deviation is 606.894. Thus the maximum amount that could be awarded under the “2-standard deviations” rule is 747.370 + 2(606.894) = 1961.158 thousands of dollars.

4.33

a Mean Sample 1 Sample 2

4.34

7.81 49.68

Standard Coef. of Deviation Variation 0.398 5.102 1.739 3.500

The values of the coefficient of variation are given in the table in Part (a). The fact that the coefficient of variation is smaller for Sample 2 than for Sample 1 is not surprising since, relative to the actual amount placed in the containers, it is easier to be accurate when larger amounts are being placed in the containers.

Since the mean is greater than the median, the distribution is likely to be positively skewed.

100 T ravel T ime (minutes)

150

200

There are clearly no outliers at the lower end. There are extreme outliers at the upper end if there are travel times more than 3(iqr) above the upper quartile. Here (upper quartile) + 3(iqr) = 31 + 3(31 − 7) = 31 + 3(24) = 103. Therefore, since we are assuming that the maximum travel time is 205 minutes and 205 is greater than 103, there are extreme outliers in this data set, and knowing these data came from a large sample we can be confident that there are mild outliers, also.

4.35

Median = 1.62, lower quartile = 1.28, upper quartile = 1.92

Outliers at the high end of the distribution are values greater than upper quartile + 1.5(iqr). The iqr = 1.92 – 1.28 = 0.64, so values greater than 1.92 + 1.5(0.64) = 2.88 are considered outliers. Since the value for Connecticut (5.10%) is greater than 2.88, it is an outlier.

Chapter 4: Numerical Methods for Describing Data

2 3 4 Percentage of Drivers over Age 85

The distribution is positively skewed with one outlier on the high end of the scale. The median is 1.62%, and the lower and upper quartiles are 1.28% and 1.92%, respectively. The middle 50% of the data values range between these quartiles, and is approximately symmetric. Excluding the outlier, the distribution of the remaining data values is approximately symmetric. 4.36

Lower quartile = 13th value = 20.9 Upper quartile = 39th value = 31.4 Interquartile range = 31.4 – 20.9 = 10.5 (Lower quartile) – 1.5(iqr) = 20.9 – 1.5(10.5) = 5.15 (Upper quartile) + 1.5(iqr) = 31.4 + 1.5(10.5) = 47.15 Since 9 is greater than 5.15, Alaska’s value is not an outlier. Since 51.4 is greater than 47.15, Pennsylvania’s value is an outlier.

30 40 Gasoline Tax (cents per gallon)

The boxplot shows that a typical gasoline tax is around 27 cents per gallon, that the middle 50% of gasoline taxes lie between 20.9 and 31.4 cents per gallon, and that the distribution is positively skewed. There is one outlier in the distribution.

4.37

Chapter 4: Numerical Methods for Describing Data

Region

East

Middle

West

4.38

40 45 Wireless %

The Eastern region has the smallest median (37.2%), and the Middle and Western regions have medians that are much closer to each other (47.1% and 48.3%, respectively). The distribution of wireless percent for the West is negatively skewed, and the other two regions are positively skewed. The Eastern region has the most variability in wireless percent, and the Middle region has the least variability.

The fiber data listed in order are:

7 10

7 12

8 13

8 14

Median = average of 9th and 10th values = (8 + 8)/2 = 8 Lower quartile = 5th value = 7 Upper quartile = 14th value = 12 Interquartile range = 12 − 7 = 5

4.39

The minimum and lower quartile for the fiber data set are both equal to 7 since the five smallest values (that is, more than a quarter of the data values) are all equal to 7.

The sugar data listed in order are:

0 11

5 13

6 14

6 17

9 18

9 19

Median = average of 9th and 10th values = (10 + 10)/2 = 10 Lower quartile = 5th value = 6 Upper quartile = 14th value = 13 Interquartile range = 13 − 6 = 7

Chapter 4: Numerical Methods for Describing Data

For the sugar data, (Upper quartile) + 1.5(iqr) = 13 + 1.5(7) = 23.5 (Lower quartile) − 1.5(iqr) = 6 − 1.5(7) = −4.5 There are no values less than −4.5 or greater than 23.5 and therefore there are no outliers.

4.40 Fiber Content

Sugar Content

Generally speaking, there are more grams per serving of sugar content than of fiber content in these cereals, with the median sugar content being 10 and the median fiber content being 8. The sugar content is also more variable, with both the interquartile range and the range (7 and 19, respectively) for the sugar data being larger than for the fiber data (5 and 7, respectively). The distribution of the sugar content is roughly symmetrical, whereas the fiber content distribution seems to be positively skewed. Neither distribution contains outliers. 4.41

Lower quartile: 51.3; Upper quartile: 67.45; Interquartile range = 67.45 – 51.3 = 16.15. Outliers are observations that are smaller than 51.3  1.5(16.15)  27.075 and larger than 67.45  1.5(16.15)  91.675 . South Korea’s observation (25.0) is an outlier.

50 60 Glass Ceiling Index

The Nordic countries are located in the whisker at the right side of the boxplot.

80 4.42

4.43

4.44

Chapter 4: Numerical Methods for Describing Data a

Value 1 standard deviation above mean = 40 min. Value 1 standard deviation below mean = 30 min. Value 2 standard deviations above mean = 45 min. Value 2 standard deviations below mean = 25 min.

The values 25 and 45 are 2 standard deviations from the mean. Therefore, by Chebyshev’s Rule, at least 100(1 − 1/22)% = 75% of the times are between these two values.

The values 20 and 50 are both 3 standard deviations from the mean, so by Chebyshev’s Rule at most 100(1/32)% = 11.1% of the times are either less than 20 min or more than 50 min.

Assuming that the distribution of times is normal, approximately 95% of the times are between 25 and 45 min., approximately 0.3% of the times are either less than 20 min. or greater than 50 min., and approximately 0.15% of the times are less than 20 min.

Since the values given are 1 standard deviation above and below the mean, roughly 68% of speeds would have been between those two values.

(1 − 0.68)/2 = 0.16. Roughly 16% of speeds would exceed 57 mph.

a Density 0.035 0.030 0.025 0.020 0.015 0.010 0.005 0.000

100 T ravel T ime (minutes)

200

The center of the distribution of travel times is around 20 minutes, with the great majority of travel times being between 5 and 45 minutes. The distribution is positively skewed.

No. The distribution of travel times is not approximately normal and therefore the Empirical Rule is not applicable.

4.45 Since the mean of the distribution is 27 minutes and the standard deviation is 24 minutes, 0 is just over 1 standard deviation below the mean. Therefore, if the distribution were approximately normal, then just under (100 − 68)/2 = 16% of travel times would be less than 0, which is clearly not the case. Thus the distribution cannot be well approximated by a normal curve.

Chapter 4: Numerical Methods for Describing Data 4.46

4.47

Since the mean of the distribution is 27 and the standard deviation is 24, 75 is 2 standard deviations above the mean and the point 2 standard deviations below the mean is −21. Therefore by Chebyshev’s Rule at least 75% of travel times are between these two values. However, since travel times cannot be less than zero, we can conclude that at least 75% of travel times are between 0 and 75 minutes.

A travel time of 54 minutes is (54 − 27)/24 = 1.125 standard deviations above the mean, and a travel time of 0 minutes is 1.125 standard deviations below the mean ((0 − 27)/24 = −1.125). Therefore, we can conclude that at least 100(1 − 1/1.1252)% = 21% of travel times are between 0 and 54 minutes.

The actual percentage of travel times between 0 and 75 minutes is approximately (100 − 2 − 2.5)% = 95.5%, which is a lot more than the 75% minimum given by Chebyshev’s Rule. The actual percentage of travel times between 0 and 54 minutes is approximately (100 − 2 − 5 − 4)% = 89%, which is again a lot more than the 21% minimum given by Chebyshev’s Rule.

The values given are two standard deviations below and above the mean. Therefore by Chebyshev’s Rule at least 75% of observations must lie between those two values.

By Chebyshev’s Rule at least 89% of observations must lie within 3 standard deviations of the mean. So the required interval is 36.92  3(11.34)  ( 2.90, 70.94) .

If the distribution were approximately normal then roughly 2.5% of observations would be more than 2 standard deviations below the mean. However, here x  2s is negative, and so this cannot be the case. Therefore the distribution cannot be approximately normal.

4.48

First note that 0 is 0.34/0.86 = 0.395 standard deviations below the mean. If the distribution were approximately normal then the Empirical Rule would tell us that considerably more than (100 − 68)/2 = 16% of adults drank regular soda less than zero times per day. This is obviously not possible, so the distribution cannot be approximately normal. Therefore use of the Empirical Rule to calculate the proportion of adults who drink regular soda more than 1.20 times per day (on average) is not appropriate.

4.49

For the first test z = (625 − 475)/100 = 1.5 and for the second test z = (45 − 30)/8 = 1.875. Since the student’s z score in the second test is higher than in the first, the student did better relative to the other test takers in the second test.

4.50

Of students taking the SAT, 83% scored at or below her score on the verbal section and 94% scored at or below her score on the math section.

4.51

At least 10% of the students owe nothing. In fact, at least 25% of the students owe nothing. 50% of the students owe $11,000 or less. 75% of the students owe $24,600 or less. 90% of the students owe $39,300 or less.

4.52

a Because , we are looking for the data value that is in approximately the 71st position in the ordered list of observations. Therefore, the 54th percentile is approximately 100 meters.

Chapter 4: Numerical Methods for Describing Data

b Because , we are looking for the data value that is in approximately the 106th position in the ordered list of observations. Therefore, the 80th percentile is approximately 2000 meters. c Because , we are looking for the data value that is in approximately the 121st position in the ordered list of observations. Therefore, the 92nd percentile is approximately 5,000 meters. 4.53

We require the proportion of observations between 49.75 and 50.25. At 49.75, z = (49.75 − 49.5)/0.1 = 2.5. Chebyshev’s Rule tells us that at most 1 2.52  0.16 of observations lie more than 2.5 standard deviations from the mean. Therefore, since we know nothing about the distribution of weight readings, the best conclusion we can reach is that at most 16% of weight readings will be between 49.75 and 50.25.

4.54

My score was 2.2 standard deviations above the mean. Less than 2.5% of students did better than I did.

My score was 0.4 standard deviations above the mean. I was easily in the top half of the scores.

My score was 1.8 standard deviations above the mean. I was a little under the 97th percentile on this exam.

My score was 1 standard deviation above the mean. I was around the 84th percentile on this exam. (The Empirical Rule tells us that approximately 68% of scores are within 1 standard deviation of the mean. Therefore approximately 32% of scores are more than 1 standard deviation away from the mean, and approximately 16% of scores are more than 1 standard deviation above the mean. This puts the score roughly at the 84th percentile.)

My score was roughly equal to the mean and the median on this exam.

4.55

The value of the standard deviation tells us that a typical deviation of the number of answers changed from right to wrong from the mean of this variable is 1.0. However, 0 is only 0.9 below the mean and negative values are not possible, and so for a typical deviation to be 1.0 there must be some values more than 1.0 above the mean, that is, values above 1.9. This suggests that the distribution is positively skewed. The value 3 is the lowest whole number value more than 3 standard deviations above the mean. Therefore, using Chebyshev’s Rule, we can conclude that at most 1/32 = 1/9 of students, that is, at most 72/9 = 8 students, changed at least three answers from correct to incorrect.

4.56

z = (320 − 450)/70 = −1.857

z = (475 − 450)/70 = 0.357

z = (420 − 450)/70 = −0.429

z = (610 − 450)/70 = 2.286

Chapter 4: Numerical Methods for Describing Data

4.57

a Per Capita Expenditure 10 - < 20 20 - < 30 30 - < 40 40 - < 50 50 - < 60 60 - < 70

Frequency 6 12 15 7 10 1

16 14

Frequency

12 10 8 6 4 2 0

i ii iii iv v

30 40 50 Per Capita Operating Expenditures

The 50th percentile is between per capita operating expenditures of 30 and 40. The 70th percentile is between per capita operating expenditures of 40 and 50. The 10th percentile is between per capita operating expenditures of 10 and 20. The 90th percentile is between per capita operating expenditures of 50 and 60. The 40th percentile is between per capita operating expenditures of 30 and 40.

4.58

For the first stimulus, z = (4.2 − 6)/1.2 = −1.5. For the second stimulus, z = (1.8 − 3.6)/0.8 = −2.25. Since a lower z score is obtained for the second stimulus, the person is reacting relatively more quickly to the second stimulus.

4.59

There is no lower whisker because the minimum value and the lower quartile were both 1.

The minimum, the lower quartile, and the median are all equal because more than half of the data values were equal to the minimum value.

The boxplot shows that 2 is between the median and the upper quartile. Therefore between 25% and 50% of patients had unacceptable times to defibrillation.

(Upper quartile) + 3(iqr) = 3 + 3(2) = 9. Since 7 is less than 9, 7 must be a mild outlier.

The sample doesn’t offer a good representation of young adults since it consisted only of students taking one particular course at one particular university.

4.60

4.61

Chapter 4: Numerical Methods for Describing Data b

By reporting the mean, the author is giving information about the center of the distribution of cell phone use times. If the standard deviation had been quoted then we would also have an idea of the spread of the distribution.

x  (345  292  334  276  248) 5  299 . The five deviations are: 46, -7, 35, -23, -51

The sum of the rounded deviations is 0.

Variance  (345  299) 2 



 (248  299) 2

 4  1630

s  1630  40.37 4.62

4.63

4.64

100000 200000 300000 Sulphur Dioxide Emissions (tons)

400000

The median Sulphur dioxide emissions is 38,533 tons, with the middle 50% of the data extending from 12,423 to 80,541 (giving an interquartile range of 80,541 – 12,423 = 68,118). The distribution is positively skewed with six outliers at the upper end.

This is a correct interpretation of the median.

Here the word “range” is being used to describe the interval from the minimum value to the maximum value. The statement claims that the median is defined to be the midpoint of this interval, which is not true.

If there is no home below $300,000 then certainly the median will be greater than $300,000 (unless more than half of the homes cost exactly $300,000).

x  (62 

Variance  (62  48.364)2 

 83) 11  48.364 cm.

 (83  48.364)2  10  327.055 cm2.

s  327.055  18.085 cm. The variance can be interpreted as the mean squared deviation of the distance at which a bat first detected a nearby insect from the mean distance at which a bat first detected a nearby

Chapter 4: Numerical Methods for Describing Data

insect. The standard deviation can be interpreted as a typical deviation of the distance at which a bat first detected a nearby insect from the mean distance at which a bat first detected a nearby insect. 4.65

The new mean is x  (52   73) 11  38.364. The new values and their deviations from the mean are shown in the table below. Value 52 13 17 46 42 24 32 30 58 35

4.66

Deviation 13.636 −25.364 −21.364 7.636 3.636 −14.364 −6.364 −8.364 19.636 −3.364

The deviations are the same as the deviations in the original sample. Therefore the value of s 2 for the new values is the same as for the old values. In general, subtracting (or adding) the same number from/to each observation has no effect on s 2 or on s, since the mean is decreased (or increased) by the same amount as the values, and so the deviations from the mean remain the same. x  (620   830) 11  483.636 cm.

Variance  (620  483.636)2 

 (830  483.636)2  10  32705.455.

s  32705.455  180.846. The value of s for the new values is 10 times the value of s for the old values. More generally, when each observation is multiplied by the positive constant c, the value of s is multiplied by c, also. 4.67

30 40 Number of Movies Made

The distribution of number of movies made is slightly positively skewed. This tells us that there is less variability in the lower 50% of number of movies made than in the upper 50%.

4.68

Chapter 4: Numerical Methods for Describing Data

Lower quartile: 233.5 millions of dollars; Upper quartile: 458.5 millions of dollars; Interquartile range = 458.5 – 233.5 = 225.0 millions of dollars. Outliers are observations that are smaller than 233.5  1.5(225.0)  104 millions of dollars and larger than 458.5  1.5(225.0)  796.0 . There are no values below zero or above 796.0 million dollars, so there are no outliers.

100

4.69

200 300 400 500 600 Inflation-Adjusted Gross Income (millions of dollars)

700

The boxplot shows that the distribution of inflation-adjusted gross incomes is positively skewed. As such, we would expect the mean to be greater than the median.

x  (18 

 18) 20  22.15.

Variance  (18  22.15)2 

 (18  22.15)2  19  129.187.

s  129.187  11.366. b

The data values, in order, are: 17 19

18 20

18 21

18 23

18 25

18 28

18 69

The lower quartile is the average of the 5th and 6th values = (18 + 18)/2 = 18. The upper quartile is the average of the 15th and 16th values = (20 + 21)/2 = 20.5. Interquartile range = 20.5 − 18 = 2.5. c

An outlier at the lower end of the distribution will be a value less than 18 − 1.5(2.5) = 14.25. Therefore there are no outliers at the lower end of the distribution. At the upper end a mild outlier will be a value greater than 20.5 + 1.5(2.5) = 24.25, and an extreme outlier will be a value greater than 20.5 + 3(2.5) = 28. Thus the values 25 and 28 are mild outliers and 69 is an extreme outlier.

Chapter 4: Numerical Methods for Describing Data

4.70

40 Age (years)

The distribution is roughly symmetrical and 0.84 = 1 − 0.16, and so the 84th percentile is the same distance above the mean as the 16th percentile is below the mean. The 16th percentile is 20 units below the mean and so the 84th percentile is 20 units above the mean. Therefore the 84th percentile is 120.

The proportion of scores below 80 is 16% and the proportion above 120 is 16%. Therefore the proportion between 80 and 120 is 100 − 2(16) = 68%. So by the Empirical Rule 80 and 120 are both 1 standard deviation from the mean, which is 100. This tells us that the standard deviation is approximately 20.

z = (90 − 100)/20 = −0.5.

A score of 140 is 2 standard deviations above the mean. By the Empirical Rule approximately 5% of scores are more than 2 standard deviations from the mean. So approximately 5/2 = 2.5% of scores are greater than 140. Thus 140 is at approximately the 97.5th percentile.

A score of 40 is 3 standard deviations below the mean, and so the proportion of scores below 40 would be approximately (100 − 99.7)/2 = 0.15%. Therefore there would be very few scores below 40.

Chapter 5 Summarizing Bivariate Data

5.1

5.2

5.3

5.4

5.5

Scatterplot 1

(i) Yes

(ii) Yes

(iii) Positive

Scatterplot 2

(i) Yes

(ii) Yes

(iii) Negative

Scatterplot 3

(i) Yes

(ii) No

(iii) -

Scatterplot 4

(i) Yes

(ii) Yes

(iii) Negative

Positive. As temperatures increase, cooling costs are likely to increase.

Negative. As interest rates rise, fewer people are likely to apply for loans.

Close to zero. The points in the scatterplot will form an inverted “U” shape, making a correlation close to zero.

Positive. Husbands and wives tend to come from similar backgrounds, and therefore have similar expectations in terms of income.

Close to zero. There is no reason to believe that there is an association between height and IQ.

Positive. People with large feet tend to be taller than people with small feet.

Positive. People who are smart and/or well educated tend to do well on both sections, with those lacking these attributes doing less well on both sections.

Negative. Those who spend a lot of time on their homework are likely to spend little time watching television, and vice versa.

No. A correlation coefficient of 0 implies that there is no linear relationship between the variables. There could still be a nonlinear relationship.

Chapter 5: Summarizing Bivariate Data 5.6

Scatterplot for which r = 1: y

5.7

Scatterplot for which r = -1: y

a 900 875

Satisfaction Rating

5.8

850 825 800 775 750 40

60 Quality Rating

There does not appear to be a relationship between quality rating and satisfaction rating.

5.10

5.11

5.12

r = 0.426. The linear relationship between quality rating and satisfaction rating is weak and positive.

Using a calculator or statistical software package we get r = 0.204. There is a weak positive linear relationship between cost per serving and fiber per serving.

Using a calculator or statistical software package we get r = 0.241. This correlation coefficient is slightly greater than the correlation coefficient for the per serving data.

There is a weak negative linear relationship between average hopping height and arch height. The negative correlation tells us that large arch heights are (weakly) associated with small hopping heights, and vice versa.

Yes. Since all five correlation coefficients are close to zero, the results imply that there is very little relationship between arch height and the motor skills measured.

The value of the correlation coefficient is negative, which suggests that students who use a cell phone for more hours per day tend to have lower GPA’s.

The relationship between texting and GPA has the same direction as the direction for cell phone use and GPA, so the correlations must have the same sign. The relationship between texting and GPA is not as strong as the relationship between cell phone use and GPA, so the correlation coefficient must be closer to zero. Therefore, r = -0.10 is the only option that satisfies both of the criteria.

Since it is reasonable to believe that texts sent would be approximately equal to texts received, there would be a positive association between these items. In addition, given that the two texting items are nearly perfectly correlated, the correlation coefficient must be close to +1 or –1. Therefore, the correlation coefficient must be close to +1.

Using a calculator or statistical software package we get r = 0.726.

b Consumer Debt (percent of personal income)

5.9

Chapter 5: Summarizing Bivariate Data

7.0

6.5

6.0

5.5

5.0

4.5 5.0

5.2 5.4 5.6 5.8 Household Debt (percent of personal income)

6.0

Chapter 5: Summarizing Bivariate Data

Yes, it is reasonable to conclude that there is no strong relationship, linear or otherwise, between household debt and consumer debt. The correlation coefficient of r = 0.726 indicates that there is no strong linear relationship (just a moderate relationship), and the scatterplot below confirms this observation 5.13

Percent Change in Stock Price

100

75 50

0 -25

-50 115

120 125 130 Median Worker Pay (thousands of dollars)

135

r = 0.578; There is a moderately strong, positive association between the percent change in stock price and median worker pay.

The conclusion is justified based on these data because there is a positive association between the variables. In general, as median worker pay increases, so does the percent change in stock price.

It is not reasonable to generalize conclusions based on these data to all U.S. companies because these data were not randomly selected. These are the top 13 highest paying companies in the U.S.

5.14

(88.8)(86.1) 85.057 39 r   0.935. 2 2 (9.263)(9.824) 88.8 86.1 288  286.6  39 39 281.1 

There is a strong positive linear relationship between the concentrations of neurolipofuscin in the right and left eye storks. 5.15

No, because “artist” is a categorical variable.

92 5.16

Chapter 5: Summarizing Bivariate Data The time needed is related to the speed by the equation

time 

distance , speed

where the distance is constant. Using this relationship, and plotting the times (over a fixed distance) for various feasible speeds, a scatterplot is obtained like the one below. T ime

Speed

These points show a strong negative correlation, and therefore the correlation coefficient is most likely to be close to −0.9. 5.17

Scatterplot 1 seems to show a linear relationship between x and y, while Scatterplot 2 shows a curved relationship between the two variables. So it makes sense to use the least squares regression line to summarize the relationship between x and y for the data set in Scatterplot 1, but not for the data set in Scatterplot 2.

5.18

The equation of the least squares regression line is yˆ  11.482  0.970 x , where x = miniWright meter reading and y = Wright meter reading.

When x = 500, ŷ  496.454 . The least squares regression line predicts a Wright meter reading of 496.454 for a person whose mini-Wright meter reading was 500.

When x = 300, ŷ  302.465 . The least squares regression line predicts a Wright meter reading of 302.465 for a person whose mini-Wright meter reading was 500. However, 300 is well outside the range of mini-Wright meter readings in the data set, so this prediction involves extrapolation and is therefore not reliable.

The equation of the least squares regression line is yˆ  51.305  0.1633x , where ŷ is the predicted percentage of alumni who would strongly agree and x is the ranking.

The predicted percentage of alumni who would strongly agree for a ranking of 50 is yˆ  51.305  0.1633(50)  59.47 .

The rankings in the data set used to create the least squares regression line range between 28 and 98. Since the ranking of 10 is outside this range, the least squares regression line should not be used to predict the percentage of alumni who would strongly agree because there is no

5.19

Chapter 5: Summarizing Bivariate Data

guarantee that the linear pattern will continue outside this range. This is the danger of extrapolation.

5.20

a Net Directionality 0.3

0.2

0.1

0.0

-0.1

-0.2 5.0

7.5

10.0

12.5 15.0 Mean T emperature

17.5

20.0

There is one point, (8.06, 0.25), which is separated from the general pattern of the data. If this point is disregarded then there is a somewhat strong positive linear relationship between mean temperature and net directionality. Even if this point is included, there is still a moderate linear relationship between the two variables.

5.21

5.22

5.23

Using a calculator or statistical software package we find that the equation of the leastsquares regression line is yˆ  0.14282  0.016141x , where x = mean water temperature and y = net directionality.

When x = 15, yˆ  0.14282  0.016141(15)  0.0993 .

The scatterplot and the least-squares line support the fact that, generally speaking, the higher the temperature the greater the proportion of larvae that were captured moving upstream.

Approximately the same number of larvae moving upstream as downstream is represented by a net directionality of zero. According to the least-squares line this will happen when the mean temperature is approximately 8.8°C.

The dependent variable is the number of fruit and vegetable servings per day, and the predictor variable is the number of hours of television viewed per day.

Negative. As the number of hours of TV watched per day increases, the number of fruit and vegetable servings per day (on average) decreases.

Negative. As the patient-to-nurse ratio increases we would expect nurses’ stress levels to increase and therefore their job satisfaction to decrease.

Chapter 5: Summarizing Bivariate Data b

Negative. As the patient-to-nurse ratio increases we would expect patient satisfaction to decrease.

Negative. As the patient-to-nurse ratio increases we would expect quality of care to decrease.

5.24 The scatterplot (shown below) indicates that there is a linear relationship between the on-time arrival percentage and the number of complaints per 100,000 passengers. Therefore, it is appropriate to find the least squares regression equation and use it to predict the number of complaints that were not reported. The least squares regression equation is yˆ  12.417  0.13729 x , where ŷ is the predicted number of complaints per 100,000 passengers, and x is the on-time arrival percentage. The least squares regression line is also shown on the scatterplot. The predicted number of complaints for Spirit airlines, which has an on-time arrival percentage of 69, is yˆ  12.417  0.13729(69)  2.94 complaints per 100,000 passengers. The predicted number of complaints for Virgin America, which has an on-time arrival percentage of 80, is yˆ  12.417  0.13729(80)  1.43 complaints per 100,000 passengers.

Complaints per 100,000 Passengers

0 75

80 85 On-Time Arrival Percentage

Chapter 5: Summarizing Bivariate Data 5.25

The response variable is the acrylamide concentration, and the predictor variable is the frying time.

Acrylamide Concentration (micrograms per kilogram)

b 280 260 240 220 200 180 160 140 120 100 150

175

200 225 250 Frying Time (seconds)

275

300

There is a weak, positive association (r = 0.379) between frying time and acrylamide concentration. 5.26

The equation of the least squares regression line is yˆ  86.9  0.359 x , where y is the acrylamide concentration (in micrograms per kilogram), and x is the frying time in seconds.

Yes, the least squares regression line equation does support the researcher’s conclusion that higher frying times tend to be paired with higher acrylamide concentrations. The association between acrylamide concentration and frying time is positive, which indicates that longer frying times tend to result in higher acrylamide concentrations.

The predicted acrylamide concentration is yˆ  86.9  0.359(225)  167.675 micrograms per kilogram.

No, I would not use the least squares regression equation to predict the acrylamide concentration for a frying time of 500 seconds. The data set that was used to create the least squares regression line was based on frying times that vary between 150 and 300 seconds; the value 500 seconds is far outside that range of values. There is no guarantee that the observed trend will continue as far as 500 seconds. This is a danger of extrapolation.

5.27

Chapter 5: Summarizing Bivariate Data

a Survival Rate (percent) 90 80 70 60 50 40 30 20 10 0 2

6 8 10 Mean Call-to-Shock T ime (minutes)

There is a strong negative relationship between mean call-to-shock time and survival rate. The relationship is close to being linear, particularly if the point with the highest x value is disregarded. If that point is included then there is the suggestion of a curve. b

yˆ  101.32847  9.29562 x , where x is the mean call-to-shock time and y is the survival rate.

When x = 10, yˆ  101.32847  9.29562(10)  8.372.

5.28

Since the slope of the least-squares line is −9.30, we can say that every extra minute waiting for paramedics to arrive with a defibrillator lowers the chance of survival by 9.3 percentage points.

5.29

The slope would be −4000, since the slope of the least-squares line is the increase in the average y value for each increase of one unit in x. Here the average home price decreases by $4000 for each increase of 1 mile in the distance east of the bay.

5.30

Using a calculator or a statistical software package we find that the correlation coefficient between sale price and size is 0.700. There is a moderate linear relationship between sale price and size.

Using a calculator or a statistical software package we find that the correlation coefficient between sale price and land-to-building ratio is −0.332. There is a weak negative linear relationship between sale price and land-to-building ratio.

Size is the better predictor of sale price since the absolute value of the correlation between sale price and size is closer to 1 than the absolute value of the correlation between sale price and land-to-building ratio. Using a calculator or statistical software package we find that the least-squares regression line for predicting y = sale price from x = size is yˆ  1.3281  0.0053x .

5.31

The least-squares line is based on the x values contained in the sample. We do not know that the same linear relationship will apply for x values outside this range. Therefore the least-squares line should not be used for x values outside the range of values in the sample.

Chapter 5: Summarizing Bivariate Data

5.32

5.33

Since the x value is 2 standard deviations above the mean, the y value will be r  2  (0.75)(2)  1.5 standard deviations above the mean.

r  1 1.5  0.667.

We know (as stated in the text) that b  r ( s y / sx ) , where s y and s x are the standard deviations of the y values and the x values, respectively. Since standard deviations are always positive we know that b and r must always have the same sign.

5.34

There is a strong positive linear relationship between user fees collected and operating costs.

Operating Costs (thousands of dollars)

4000

3000

2000

1000

0 0

100

200 300 400 500 600 700 User Fees Collected (thousands of dollars)

800

900

The equation of the least squares regression line is yˆ  19.1  3.0675 x , where ŷ is the

predicted operating costs, and x is the user fees collected.

5.35

The slope is positive, which is consistent with the description in part a.

The correlation coefficient would be greater than 0.5, because the relationship is strong.

The equation of the least squares regression line is yˆ  9.071  1.571x , where x = years of schooling and y = median hourly wage gain. When x = 15, ŷ  14.5 . The least squares regression line predicts a median hourly wage gain of 14.5 percent for the 15th year of schooling.

The actual wage gain percent is very close to the value predicted in Part (a).

5.36

Chapter 5: Summarizing Bivariate Data

a Median Distance Walked 750

700

650

600

550 5.0

7.5

10.0 12.5 Representative Age

15.0

17.5

The scatterplot shows a linear pattern between the representative ages of 10 and 17, but there is a greater increase in the median distance walked between the representative ages of 7 and 10 than there is between any other two consecutive age groups. b

Using a calculator or statistical software package we find that the equation of the leastsquares regression line is yˆ  492.79773  14.76333x , where x is the representative age and y is the median distance walked.

c Representative Median Distance Age (x) Walked (y) 4 544.3 7 584 10 667.3 13.5 701.1 17 727.6

Predicted y

Residual

551.851 596.141 640.431 692.103 743.774

-7.551 -12.141 26.869 8.997 -16.174

Residual 30

-10

-20 5.0

7.5

10.0 12.5 Representative Age

15.0

17.5

Chapter 5: Summarizing Bivariate Data d

5.37

The residual plot reflects the sharp increase in the median distance walked between the representative ages of 7 and 10, with a clear negative residual at x  7 and large positive residual at x  10 .

a Median Six-Minute Walk Distance 680 660 640 620 600 580 560 540 520 500 5.0

7.5

10.0 12.5 Representative Age

15.0

17.5

In the last three age categories, the median distance walked for girls increases only slightly, whereas for boys it increases at a rate only slightly smaller than the rate that applies to the first two age categories. Also, the scatterplot for the girls shows clearer evidence of a curved relationship between the variables than does the scatterplot for the boys.

yˆ  479.997  12.525 x , where x is the representative age and y is the median distance walked.

d Representative Median Distance Age (x) Walked (y) 4 492.4 7 578.3 10 655.8 13.5 657.6 17 660.9

Predicted y

Residual

530.095 567.669 605.243 649.079 692.915

-37.695 10.631 50.557 8.521 -32.015

100

Chapter 5: Summarizing Bivariate Data

Residual 50

-25

-50 5.0

10.0 12.5 Representative Age

15.0

17.5

The authors’ decision to use a curve is supported by the clear curved pattern in the residual plot.

5.38

No, the pattern in the scatterplot does not look linear. There are two unusual values (airline quality ratings of 1 and 2) that make the pattern look curved. Without those two points, the scatterplot would look linear. 90

On-Time Arrival Percentage

5.39

7.5

70 0

6 8 Airline Quality Rating

The regression equation is yˆ  85.615  0.9341x , where ŷ is the predicted on-time arrival percentage and x is the airline quality rating.

Chapter 5: Summarizing Bivariate Data

101

c Airline Quality Rating (x) 5 10 3 12 9 11 4 2 7 6 13 8 1

On-Time Arrival Percentage (y) 86 80 86 74 78 73 88 76 80 80 69 78 80

Predicted On-Time Arrival Percentage ( ŷ ) 80.9451 76.2747 82.8132 74.4066 77.2088 75.3407 81.8791 83.7473 79.0769 80.0110 73.4725 78.1429 84.6813

Residual ( y  yˆ ) 5.05495 3.72527 3.18681 -0.40659 0.79121 -2.34066 6.12088 -7.74725 0.92308 -0.01099 -4.47253 -0.14286 -4.68132

5.0

Residual

2.5

0.0

-2.5

-5.0 -7.5 0

6 8 Airline Quality Rating

The residual plot shows a curved pattern that would call into question the appropriateness of using a linear model to describe the relationship between airline quality rating and on-time arrival percentage.

102

a Acrylamide Concentration (micrograms per kilogram)

5.40

Chapter 5: Summarizing Bivariate Data

280 260 240 220 200 180 160 140 120 100 150

175

200 225 250 Frying Time (seconds)

275

300

yˆ  86.9  0.359 x , where y is the acrylamide concentration (in micrograms per kilogram), and x is the frying time in seconds. The predicted acrylamide concentration for a frying time of 270 seconds is yˆ  86.9  0.359(270)  183.83 micrograms per kilogram. The residual associated with the observation (270, 185) is y  yˆ  185  183.83  1.17 .

5.41

5.42

The observation (150,155) is potentially influential because that point has an x value that is far away from the rest of the data set.

yˆ  44  0.83(270)  180.1; This prediction is smaller than that made in the previous exercise.

The equation of the least-squares regression line is yˆ  232.258  2.926 x , where x = rock surface area and y = algae colony density.

r 2  0.300 . Thirty percent of the variation in algae colony density can be attributed to the approximate linear relationship between rock surface area and algae colony density.

se  63.315 . This is a typical deviation of an algae colony density value from the value predicted by the least-squares regression line.

r = −0.547. The linear relationship between rock surface area and algae colony density is moderate and negative.

Chapter 5: Summarizing Bivariate Data

5.43

103

a Percent T ransported 80 70 60 50 40 30 20 0

5.44

5.45

5000

10000 T otal Number

15000

20000

Yes, there appears to be a strong linear relationship between the total number of salmon in the stream and the percent of salmon killed by bears that are transported away from the stream.

The equation of the least-squares regression line is yˆ  18.483  0.00287x , where x is the total number of salmon in a creek and y is the percent of salmon killed by bears that were transported away from the stream prior to the bear eating. The regression line has been drawn on the scatterplot in Part (a).

The point (3928, 46.8) is unlikely to be influential as its x value does not differ greatly from the others in the data set.

The two points are not influential since the least-squares line provides a good fit for the remaining 8 points. Removing the two points will make only a small change in the regression line.

se  9.16217. This is a typical deviation of a percent transported value from the value predicted by the regression line.

r 2  0.832. This is a large value of r 2 , and means that 83.2% of the variation in the percent transported values can be attributed to the approximate linear relationship between total number and percent transported.

a Height (x) 188 4 188 178 183 180

Weight (y) 95 4 91 72 93 78

Predicted Weight ( ŷ ) 88.66 2.916 88.66 84 86.33 84.932

Residuals ( y  yˆ ) 6.34 1.084 2.34 -12 6.67 -6.932

104

Chapter 5: Summarizing Bivariate Data

Residual

-5

-10

100 Height (cm)

150

200

The residual plot has one observation (the Lego Batman, at 4 cm and 4 kg) that is noticeably different from the others. There is a strong pattern in the residual plot.

5.46

5.47

5.48

This point is influential in determining both the slope and the intercept. That point is far removed from the other points in both the x and y directions. In fact, by removing the Lego Batman point, the equation of the least squares regression line changes to yˆ  220.43  1.6643x .

Removing Jupiter Networks would have the greatest impact on the equation of the least squares regression line.

An outlier in a bivariate data set is a point that has a large residual. Outliers fall far away from the least squares regression line. An influential observation has an x value that is far away from the rest of the data.

The value of r 2 would be 0.154.

No, since the r 2 value for y = first year college GPA and x = SAT II score was 0.16, which is not large. Only 16% of the variation in first year college GPA could be attributed to the approximate linear relationship between SAT II score and first year college GPA.

r 2  0.303 ; Approximately 30.3% of the variability in the percentage of alumni who strongly agree can be explained by the linear relationship between the percentage of alumni who strongly agree and ranking.

se  5.34668 (obtained using software); se tells us that the percentage of alumni who strongly agree will deviate from the least squares regression line by 5.34668, on average.

Chapter 5: Summarizing Bivariate Data c

The relationship between the percentage of alumni who strongly agree and ranking is moderate and positive. It is positive because the slope of the least squares regression line is positive, and it is moderate because the correlation coefficient (which has the same sign as the slope) is r 

5.49

105

r 2  0.303  0.550 .

a Number of Employees 150

125

100

5.50

100000 200000 300000 400000 500000 600000 T otal Park Size (Acres)

700000

Using a graphing calculator or computer software package we see that the equation of the least-squares line is yˆ  85.334  0.0000259 x , where x is the total park size in acres and y is the number of employees.

r 2 = 0.016. With only 1.6% of the variation of the number of employees being attributable to the least-squares line, the line will not give accurate predictions.

Deleting the point (620231, 67), the equation of the least-squares line is now yˆ  83.402  0.0000387 x . Yes, removal of the point does greatly affect the equation of the line, since it changes the slope from negative to positive.

The equation of the least-squares regression line is yˆ  27.709  0.201x, where y = percent of operating costs covered by park revenues and x = number of employees.

Since r 2  0.053, we know that only 5.3% of the variability in operating costs is attributable to the least-squares regression line. Therefore the least-squares line does not do a good job of describing the relationship. Also, with se  24.148 being of the same order of size as the y values themselves, we see again that the line does not do a good job of describing the relationship.

The points (69, 80), (87, 97), and (112, 108) are outliers. No, the three observations with the largest residuals are not for the three districts with the largest numbers of employees. However, generally speaking districts with larger numbers of employees tend to have larger residuals.

106

5.51

5.52

Chapter 5: Summarizing Bivariate Data The coefficient of determination is r 2  1  SSResid SSTo   1  1235.470 25321.368   0.951. This tells us that 95.1% of the variation in hardness is attributable to the approximate linear relationship between time elapsed and hardness. a

For a given value of r 2 a scatterplot can be constructed with any value of se you care to specify. For example, suppose you have a scatterplot with r 2  0.984 and se  0.42. If you multiply the y values by 10 you obtain a scatterplot with r 2  0.984 and se  4.2. If you multiply the original y values by 100 you obtain a scatterplot with r 2  0.984 and se  42. Likewise, by multiplying the y values by a positive number less than 1 you obtain a scatterplot with a smaller value of se . Now, suppose that you were using a least-squares line to predict the prices, y, of objects that cost in the region of $3. A value of se of around 0.8 would be considered large, since this means that the deviations of the true prices from predicted prices tend to be around 80 cents, which is quite a large amount when the prices of the objects are around $3. If, however, the least-squares line were being used to predict the prices of objects that cost in the region of $300, then a value of se of around 0.8 would be considered small. Thus we can conclude that, for a given value of r 2 , however large or small, it is possible to have both large and small values of se .

5.53

As explained in Part (a), for any value of r 2 it is possible to have a values of se that are large or small.

We want r 2 to be large (close to 1) because, if it is, we know that most of the variation in y is accounted for by the straight line relationship. In other words, there is very little else, apart from x, that need be considered when predicting values of y. We want se to be small because, if it is, when we use the least-squares line to predict values of y, the errors involved are small when expressed as percentages of the true y values.

The value of r that makes se  s y is 0. The least-squares line is then ŷ  y .

For values of r close to 1 or −1, se will be much smaller than s y .

se  1  r 2 s y  1  0.82  2.5   1.5 .

We now let x = 18-year-old height and y = 6-year-old height. The slope is b  r s y sx  0.8 1.7 2.5  0.544. So the equation is yˆ  a  bx  a  0.544 x . The line





passes through ( x , y )  (70, 46) , so 46  a  0.544(70), from which we find that a  46  0.544(70)  7.92. Hence the equation of the least-squares line is yˆ  7.92  0.544 x . Also, se  1  r 2 s y  1  0.82 1.7   1.02 .

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a Moisture Content (%) 18 16 14 12 10 8 6 4 2 0

Frying Time (seconds)

b 5.55

No. The pattern in the scatterplot is curved.

a Log(y) 1.2 1.0 0.8 0.6 0.4 0.2 0.0 0.6

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The relationship between log(x) and log (y) seems to be linear. b

The value of r2 is large (0.973) and the value of se is relatively small (0.0657), indicating that a line is a good fit for the data.

The regression equation is

Predicted(log( y ))  2.01780  1.05171  log( x ) . When x = 35, log(x) = 1.544. So then,

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Predicted(log( y ))  2.01780  1.05171  1.544  0.394 So yˆ  100.394  2.477 . The predicted moisture content for a frying time of 35 seconds is 2.477%. 5.56

a y 70 60 50 40 30 20 10 0

0 1 2 3 4 5 6 There seems to be a curved relationship between the two variables.

7 x

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The curved pattern is not as pronounced as in the scatterplot from Part (a), but there is still curvature in the plot. c

Answers will vary. One possible answer is to use y '  log( y ) and x '  log( x) . (See scatterplot below.)

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log(y) 2.00 1.75 1.50 1.25 1.00 0.75 0.50

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This transformation would be preferred over the transformation in Part (b) because this scatterplot is closer to being linear. 5.57

Answers will vary. Scatterplot (b) shows a curve that tends to follow the pattern of points in the scatterplot.

5.58

a Success (%) 100 90 80 70 60 50 40 30

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The relationship appears to be nonlinear. b

yˆ  22.48  34.36 x . The residuals are: −6.36, 1.46, 7.78, 5.5, −8.38.

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Yes. There is a curved pattern in the residual plot. 5.59

The residual plot for the square root transformation is shown below.

The residual plot for the natural log transformation is shown below.

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The residual plot for the square root transformation shows a curved pattern of points, while the residual plot for the natural log transformation shows what would seem to be a random scatter of points. So the log transformation is more successful in producing a linear relationship. The regression equation is yˆ  62.417  43.891  ln( x) .

When x = 1.75, yˆ  62.417  43.891  ln(1.75)  86.979 . The predicted success level is 86.979%. When x = 0.8, yˆ  62.417  43.891  ln(0.8)  52.623 . The predicted success level is 52.623%.

a 2800 2700 Heart Transplants

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2600 2500 2400 2300 2200 2100 1

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Chapter 5: Summarizing Bivariate Data From 2006 to 2015, the number of heart transplants increased, with the number increasing by a greater amount each year (i.e., increasing at a nonlinear rate). b

The equation of the least-squares regression line is yˆ  2014.73  66.376 x , where ŷ is the predicted number of heart transplants and x is the year using 1 to represent 2006, 2 to represent 2007, and so on.

c Number of Transplants (y) 2193 2209 2163 2211 2332 2322 2378 2531 2655 2804

Year (x) 1 2 3 4 5 6 7 8 9 10

Predicted Number of Transplants ( ŷ ) 2081.11 2147.48 2213.86 2280.24 2346.61 2412.99 2479.36 2545.74 2612.12 2678.49

Residuals ( y  yˆ ) 111.891 61.515 -50.861 -69.236 -14.612 -90.988 -101.364 -14.739 42.885 125.509

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The residual plot shows a curved pattern, suggesting that the relationship between year and number of heart transplants is nonlinear and would be better described by a curve rather than a line.

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One transformation to straighten the plot is x = year2 and y = number of heart transplants.

Number of Heart Transplants

2800 2700 2600 2500 2400 2300 2200 2100 0

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100

The equation of the least-squares regression line is yˆ  2139.87  6.2321x* , where ŷ is the *

predicted number of heart transplants, and x is the year (since 2006) squared. The predicted * 2 number of heart transplants in 2016 (the 11th year since 2006, or x  11  121 ) is yˆ  2139.87  6.2321(121)  2893.9541 . Therefore, it is predicted that there will be approximately 2894 heart transplants in 2016. c

5.62

We have to be confident that the pattern observed between 2006 and 2015 will continue up to 2016. This is reasonable so long as circumstances remain basically the same. To expect the same pattern to continue to 2026, however, might be unreasonable.

a y 160 140 120 100 80 60 40 20 0 0

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Yes. The scatterplot shows a positive relationship between the two variables. b y 160 140 120 100 80 60 40 20 0

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Yes. The relationship between y and x2 seems to be linear. c log(y) 2.25 2.00 1.75 1.50 1.25 1.00 0.75 0.50

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This transformation straightens the plot. In addition, the variability in log(y) does not seem to increase as x increases, as was the case in Part (b).

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d log(y) 2.25 2.00 1.75 1.50 1.25 1.00 0.75 0.50

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The relationship between age and canal length is not linear. A transformation that makes the plot roughly linear is x  1 x (with y   y ). The resulting scatterplot and residual plot are shown below.

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Any image plotted between the dashed lines would be associated with Cal Poly by roughly the same percentages of enrolling and non-enrolling students.

The images that were more commonly associated with non-enrolling students than with enrolling students were “Average,” “Isolated,” and “Back-up school,” with “Back-up school” being the most common of these amongst non-enrolling students. The images that were more commonly associated with enrolling students than with non-enrolling students were (in increasing order of commonality amongst enrolling students) “Excitingly different,” “Personal,” “Selective,” “Prestigious,” “Exciting,” “Intellectual,” “Challenging,” “Comfortable,” “Fun,” “Career-oriented,” “Highly respected,” and “Friendly,” with this last image being marked by over 60% of students who enrolled and over 45% of students who didn’t enroll. The most commonly marked image amongst students who didn’t enroll was “Career-oriented.”

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5.65 Fuel Efficiency (mpg) 30

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As illustrated in the graph above, it is quite possible that the relationship between speed and fuel efficiency is modeled by a curve (in particular, by a quadratic curve), and that for greater speeds the fuel efficiency is negatively related to the speed. 5.66

r  0.89  0.943. (Note that r is 0.943 rather than −0.943 since the slope of the leastsquares line is positive.) There is a very strong positive linear relationship between assault rate and lead exposure 23 years prior. No, we cannot conclude that lead exposure causes increased assault rates, since the value of r close to 1 tells us that there is a strong linear association between lead exposure and assault rate, but tells us nothing about causation.

The equation of the least-squares regression line is yˆ  24.08  327.41x , where y is the assault rate and x is the lead exposure 23 years prior. When x = 0.5, yˆ  24.08  327.41(0.5)  139.625 assaults per 100,000 people.

89% of the year-to-year variability in assault rates can be explained by the relationship between assault rate and gasoline lead exposure 23 years earlier.

The two time series plots, generally speaking, move together. That is, generally when one goes up the other goes up and when one goes down the other goes down. Thus high assault rates are associated with high lead exposures 23 years earlier and low assault rates are associated with low lead exposures 23 years earlier.

5.67

Consider the scatterplot for the combined group. Between x = 0 and x = 1 the points form an approximately linear pattern with a slope of 6.93, while between x = 1 and x = 10 the points form an approximately linear pattern with a slope of just over 3. Clearly this pattern could not be modeled with a single straight line; a curve would be more appropriate.

5.68

r = −0.981. This suggests a very strong linear relationship between the amount of catalyst and the resulting reaction time.

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b Reaction T ime 50

25 1

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The word linear does not provide the most effective description of the relationship. There are curves that would provide a much better fit. r  0.944. There is a strong, positive linear relationship between depression rate and sugar consumption.

No. Since this was an observational study, no conclusion relating to causation may be drawn.

Yes. Since the set of countries used was not a random sample from the set of all countries (nor do we have any reason to think that these countries are representative of the set of all countries), we cannot generalize the conclusions to countries in general.

a Exam Score 80

40 0

10 15 T est Anxiety

There is one point, (0, 77), that is far separated from the other points in the plot. There is a clear negative relationship between scores on the measure of test anxiety and exam scores.

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There appears to be a very strong negative linear relationship between test anxiety and exam score. (However, without the point (0, 77) the relationship would be significantly less strong.)

r = −0.912. This is consistent with the observations given in Part (b).

No, we cannot conclude that test anxiety caused poor exam performance. Correlation measures the strength of the linear relationship between the two variables, but tells us nothing about causation.

When x = 25, yˆ  62.9476  0.54975(25)  49.204 . So the residual is y  yˆ  70  49.204  20.796 .

r   0.57  -0.755 (The correlation coefficient is negative since the slope of the leastsquares regression line is negative.)

We know that r 2  1  SSResid SSTo . Solving for SSResid we get

SSResid  SSTo(1  r 2 )  2520(1  0.57)  1083.6 . Therefore se  SSResid (n  2)  1083.6 8  11.638 . 5.72

The equation of the least-squares regression line is yˆ  6.873  5.078 x , where ŷ is the predicted depression score change and x is the BMI change.

Using software, we find that r 2  0.235 and se  5.36559 . Therefore, only 23.5% of the variation in depression score change can be explained by the linear relationship between these variables. In addition, the value of se is large relative to the y values. The values of r2 and se indicate that the least squares regression line does not do a good job describing the relationship between depression score change and BMI change

The point (0.5, -1) is the only obvious outliers. No, the observation with the largest residual does not correspond to the patient with the largest change in BMI.

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a Percentage of Eagles in the Air 70 60 50 40 30 20 10 0 0

3 4 Salmon Availability

There seems to be a curved relationship between percentage of eagles in the air and salmon availability. b

The scatterplot in Part (c) shows that the transformation is partially successful in finding a linear relationship.

c sqrt(x) 2.5

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sqrt(y)

Yes, the plot is straighter than the one in Part (a). d

Taking the cube roots of both variables might be successful in straightening the plot. (In fact, this transformation does seem to result in a more linear relationship, as shown in the scatterplot below.)

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cube root (x) 2.0

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r  0.835 . The absolute value of this correlation is greater than the absolute value of the correlation calculated in Part (a). This suggests that the transformation was successful in straightening the plot.

a Moisture Content 18 16 14 12 10 8 6 4 2 0 0

30 Frying Time

No, the relationship does not seem to be linear.

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b Frying Time (x) 5 10 15 20 25 30 45 60

Moisture Content (y) 16.3 9.7 8.1 4.2 3.4 2.9 1.9 1.3

y' = log(y) 1.212 0.987 0.908 0.623 0.531 0.462 0.279 0.114

Log(y) 1.2 1.0 0.8 0.6 0.4 0.2 0.0 0

30 Frying Time

Yes, the transformed relationship is closer to being linear than the relationship shown in Part (a). c

The least-squares line predicts y   1.143  0.0192 x , where x = frying time and y  is log(moisture content).

For x = 35, the least-squares line predicts y   log y  1.143  0.0192(35)  0.472. So the predicted moisture content is 100.472  2.964% .

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a y 70 60 50 40 30 20 10 0

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log(y) 1.8 1.7 1.6 1.5 1.4 1.3 1.2 1.1 1.0 -0.5

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Plotting log( y ) against log( x) does the best job of producing an approximately linear relationship. The least-squares line of log( y ) on log( x) is log( yˆ )  1.61867  0.31646log( x) . So when x = 25, log( yˆ )  1.61867  0.31646log(25)  1.17629 . Therefore

yˆ  101.17629  15.007 . The predicted lead content is 15.007 parts per million. 5.77

The correlation coefficient of 0.12 indicates a weak relationship between price and weight of the wine bottles.

r2 = 0.0144. Approximately 1.44% of the variation in price of the wine can be explained by the linear relationship between price and weight of the wine bottles.

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5.79

Initially the scatterplot might suggest a linear relationship between the variables, but it becomes clear that this relationship is dependent on one highly influential point. If that point is removed the scatterplot suggests very little relationship between the variables.

The equation of the least-squares line is yˆ  11.37  1.0906 x , where x = treadmill consumption and y = fire-suppression simulation consumption. When x = 40, yˆ  11.37  1.0906(40)  32.254 ml per kg per min.

Since r 2  0.595, we know that 59.5% of the variation in fire-suppression simulation consumption is accounted for by the approximate straight line relationship between the two variables. This is a reasonably high percentage, but it should be remembered that this linear relationship is entirely dependent on the one influential point.

The new regression equation is yˆ  36.418  0.198 x and the new value of r 2 is 0.027. The large change in the slope of the regression line shows how very influential the point for Firefighter 1 was, and the very small new value of r 2 tells us that there is effectively no linear relationship between the two variables when the data point for Firefighter 1 is omitted.

r 0

For example, adding the point (6, 1) gives r  0.510 . (Any y-coordinate greater than 0.973 will work.)

For example, adding the point (6, −1) gives r  0.510 . (Any y-coordinate less than −0.973 will work.)

Online Exercises 5.80

The least-squares line for the transformed data is y   4.587  0.0658 x , where x is the peak intake. Thus the logistic regression equation is e4.5870.0658 x p . 1  e4.5870.0658 x For x  40, the equation predicts

p

5.81

e4.587 0.0658(40)  0.876. 1  e4.587 0.0658(40)

Calculating the least-squares line for y  ln  p (1  p)  against x = high school GPA we get y   2.89399  1.70586 x . Thus the logistic regression equation is

p For x  2.2 the equation predicts

p 5.82

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e2.893991.70586 x . 1  e2.893991.70586 x

e2.893991.70586(2.2)  0.702. 1  e2.893991.70586(2.2)

The logistic regression equation is

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p b

5.83

e0.91710.10716 x . 1  e0.91710.10716 x

For x  15 the equation predicts e0.91710.10716(15) p  0.074. 1  e0.91710.10716(15)

a Lowland Proportion 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 0

4 5 Exposure (days)

Mid-Elevation Proportion 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 0

Yes, the plots have roughly the shape you would expect from “logistic” plots.

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b Exposure (days) (x) 1 2 3 4 5 6 7 8

Cloud Forest Proportion (p) 0.75 0.67 0.36 0.31 0.14 0.09 0.06 0.07

y' = ln(p/(1−p)) 1.09861 0.70819 -0.57536 -0.80012 -1.81529 -2.31363 -2.75154 -2.58669

The least-squares line relating y and x (where x is the exposure time in days) is y   1.51297  0.58721x . The negative slope reflects the fact that as exposure time increases the hatch rate decreases. c

The logistic regression equation is

p

e1.512970.58721x . 1  e1.512970.58721x

For x  3 the equation predicts

p

e1.51297 0.58721(3)  0.438. 1  e1.51297 0.58721(3)

For x  5 the equation predicts

p

e1.51297 0.58721(5)  0.194. 1  e1.51297 0.58721(5)

When p = 0.5, y  ln  p (1  p)   ln  0.5 (1  0.5)   0 . So, solving 1.51297  0.58721x  0 we get x  1.51297 0.58721  2.577 days.

5.84

As elevation increases, the species becomes less common. This is made clear in the table by the fact that the proportion of plots with the lichen decreases as the elevation values increase.

b Proportion of Plots Elevation with Lichen (p) 400 0.99 600 0.96 800 0.75 1000 0.29 1200 0.077 1400 0.035 1600 0.01

y' = ln(p/(1−p)) 4.595 3.178 1.099 -0.895 -2.484 -3.317 -4.595

The least-squares line is y   7.537  0.00788 x , where x = elevation.

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The logistic regression equation is

p

e7.5370.00788 x . 1  e7.5370.00788 x

For x  900 the equation predicts

p 5.85

e7.537 0.00788(900)  0.609. 1  e7.537 0.00788(900)

a Concentration (g/cc) 0.10 0.15 0.20 0.30 0.50 0.70 0.95

Number of Mosquitoes 48 52 56 51 47 53 51

Number Killed 10 13 25 31 39 51 49

Proportion Killed 0.208333 0.250000 0.446429 0.607843 0.829787 0.962264 0.960784

y' = ln(p/(1−p)) -1.33500 -1.09861 -0.21511 0.43825 1.58412 3.23868 3.19867

Proportion Killed 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1

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The least-squares line relating y and x (where x is the concentration in g/cc) is yˆ   1.55892  5.76671x . The positive slope reflects the fact that as the concentration increases the proportion of mosquitoes that die increases. When p = 0.5, y  ln  p (1  p)   ln  0.5 (1  0.5)   0 . So, solving 1.55892  5.76671x  0 we get x  1.55892 5.76671  0.270 . LD50 is estimated to be around 0.270 g/cc.

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a Proportion Failing

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y' = ln(p/(1−p)) -3.892 -3.178 -1.386 -1.208 -0.754 -0.663 -0.282

The least-squares line is y   3.579  0.03968 x , where x = load applied. c

The logistic regression equation is

p

e3.5790.03968 x . 1  e3.5790.03968 x

For x  60 the equation predicts e3.579 0.03968(60) p  0.232. 1  e3.579 0.03968(60) d

For p  0.05, y   ln(0.05 0.95)  2.944. So we need 3.579  0.03968 x  2.944. Solving for x we get x  (2.944  3.579) 0.03968  15.989 lb/sq in.

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Cumulative Review Exercises CR5.1 Here is one possible design. Gather a number of volunteers (around 50, for example) who are willing to take part in an experiment involving exercise. Establish some measure of fitness, involving such criteria as strength, endurance, and muscle mass. Measure the fitness of each person. Randomly assign the 50 people to two groups, Group A and Group B. (This can be done by writing the names of the 50 people on identical slips of paper, placing the slips of paper in a hat, mixing them, and picking 25 names at random. Those 25 people will be put into Group A and the remainder will be put into Group B.) People in Group A should be instructed on a program of exercise that does not involve the sort of activity one would engage in at the gym, and this exercise should be undergone wearing the new sneakers. People in Group B should be instructed on an equivalent program of exercise that primarily involves gym-based activities, and this exercise should be undergone without the wearing of the new sneakers. At the end of the program the fitness of all the participants should be measured and a comparison should be made regarding the increase in fitness of the people in the two groups. This is an experiment since the participants are assigned to the groups by the experimenters. CR5.2 The first histogram below shows the distribution of the amount planned to spend for respondents age 50–64, and the second shows the distribution of the amount planned to spend for respondents age 65 and older. Density 0.0020

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The center of the distribution for the younger age group (around 300) is larger than the center of the distribution for the older age group (around 220). As a result of the way the information has been summarized it is difficult to compare the spreads of the two distributions, however it seems likely, looking at the histograms, that the spread is a little greater for the younger age group. Both distributions are positively skewed. CR5.3 The peaks in rainfall do seem to be followed by peaks in the number of E. coli cases, with rainfall peaks around May 12, May 17, and May 23 being followed by peaks in the number of cases on May 17, May 23, and May 28th. (The incubation period seems to be more like 5 days than the 3 to 4 days mentioned in the caption.) Thus the graph does show a close connection between unusually heavy rainfall and the incidence of the infection. The storms may not be responsible for the increased illness levels, however, since the graph can only show us association, not causation. CR5.4 For broadband users the mean is only very slightly larger than the median telling us that the distribution of amounts paid is roughly symmetrical (maybe slightly positively skewed). However, in the case of dial-up users the mean is significantly larger than the median, suggesting a positively skewed distribution.

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CR5.5 a

r = 0.394

r = -0.664

There is a stronger relationship between Happiness Index and response to Statement 2.

There is a weak, positive association between the Happiness Index and the response to Statement 1. In contrast, there is a moderate, negative association between the Happiness Index and the response to Statement 2.

CR5.6 The correlation coefficient will be positive and strong. This is apparent from the time series plots because the plots follow the same general trend. Both amount spent on science and amount spent on pets increase over time. CR5.7 r = 0.975; this value is consistent with the previous answer, because the correlation coefficient is large and positive (close to 1), which indicates a strong positive association between the amount spent on science and the amount spent on pets. CR5.8 The fact that the mean is so much larger than the median tells us that the distribution of amounts spent was strongly positively skewed. CR5.9 The data, in ascending order, are: 8457 8758 9116 10047 10108 10426 10478 10591 10680 10788 10815 10912 11040 11092 11145 11495 11644 11663 11673 11778 11781 12148 12353 12605 12962 13226 Lower quartile = 7th value = 10478 mg/kg. Upper quartile = 20th value = 11778 mg/kg. Interquartile range = 11778 − 10478 = 1300 mg/kg. CR5.10 a

x  (3099 

 3700) 10  2965.2 .

Variance   (3099  2965.2)2 

 (3700  2965.2)2  9  294416.622 .

s  294416.622  542.602 . The data values listed in order are: 2297

2401

2510

2682

2824

3068

3099

3112

3700

3959

Lower quartile = 3rd value = 2510. Upper quartile = 8th value = 3112. Interquartile range = 3112 − 2510 = 602. b

The interquartile range for the chocolate pudding data (602) is less than the interquartile range for the tomato catsup data (1300). So there is less variability in sodium content for the chocolate pudding data than for the tomato catsup data.

Chapter 5: Summarizing Bivariate Data

CR5.11 a

Mean = (98 

133

 40) 13  59.846.

The data, listed in order, are: 40 75

48 98

Median = 7th value = 55. Since the mean is greater than the median, the distribution of extra travel hours is likely to be positively skewed. b

60 70 80 Extra Hours per Traveler

100

There is one outlier (Los Angeles, with 98 extra travel hours per traveler). The median number of extra hours of travel per traveler is 55, and, apart from the one outlier, the values range from 40 to 75. The distribution is positively skewed. CR5.12 a

x  (4.8 

 3.7) 20  4.93.

The data, listed in order are: 0.4 4.8

0.9 5

1.4 5

1.4 5.4

2.1 6.1

2.4 7.5

2.9 10.8

3.3 13.8

3.4 14.8

3.5

3.7

Median = average of 10th and 11th = (3.5 + 3.7)/2 = 3.6. b

The mean is greater than the median. This is explained by the fact that the distribution of blood lead levels is positively skewed.

Whites

African Americans

2 4 6 8 10 12 14 Blood Lead Level (micrograms per decilliter)

134

Chapter 5: Summarizing Bivariate Data

The median blood lead level for the African Americans (3.6) is slightly higher than for the Whites (3.1). Both distributions seem to be positively skewed. There are two outliers in the data set for the African Americans. The distribution for the African Americans shows a greater range than the distribution for the Whites, even if you discount the two outliers.

r  0.730.

CR5.13

Outpatient Cost-to-Charge Ratio 75 70 65 60 55 50 45 50

70 80 Inpatient Cost-to-Charge Ratio

100

Based on the correlation coefficient value of 0.730 and the scatterplot, there is a moderate linear relationship between the cost-to-charge ratio for inpatient and outpatient services. Looking at the scatterplot, it is clear that one outlier is affecting the correlation. If that point were removed, there would be a strong linear relationship between the variables. b

As mentioned above, there is an outlier (the point for Harney District).

If the outlying point were removed then the linear relationship would be much stronger, and the value of r would therefore be greater.

The y intercept is −147.

The slope is 6.175. For every 1-cm increase in the snout-vent length the predicted clutch size increases by 6.175.

It would be inadvisable to use the least-squares line to predict the clutch size when the snoutvent length is 22, since 22 is well outside the range of the x values in the original data set, and we have no reason to think that the observed linear relationship applies outside this range.

CR5.14

Chapter 5: Summarizing Bivariate Data

135

CR5.15 a

This value of r 2 tells us that 76.64% of the variability in clutch size can be attributed to the approximate linear relationship between snout-vent length and clutch size.

Using r 2  1  SSResid SSTo we see that SSResid  SSTo(1  r 2 ) . So here SSResid  43951(1  0.7664)  10266.9536. Therefore

SSResid 10266.9536   29.250. This is a typical deviation of an observed clutch n2 12 size from the clutch size predicted by the least-squares line. se 

Chapter 6 Probability

6.1

A chance experiment is any activity or situation in which there is uncertainty about which of two or more possible outcomes will result. For example, a random number generator is used to select a whole number between 1 and 4, inclusive.

6.2

The sample space is the collection of all possible outcomes of a chance experiment. The sample space for Exercise 6.1 is {1, 2, 3, 4}.

6.3

{AA, AM, MA, MM}

b A

A M A M M

6.4

6.5

6.6

136

i B = {AA, AM, MA} ii C = {AM, MA} iii D = {MM}. D is a simple event.

B and C = {AM, MA} B or C = {AA, AM, MA}

A = {Head oversize, Prince oversize, Slazenger oversize, Wimbledon oversize, Wilson oversize}

B = {Wimbledon midsize, Wilson midsize, Wimbledon oversize, Wilson oversize}

not B = {Head midsize, Prince midsize, Slazenger midsize, Head oversize, Prince oversize, Slazenger oversize}

B or C = {Head midsize, Prince midsize, Wimbledon midsize, Wilson midsize, Head oversize, Prince oversize, Wimbledon oversize, Wilson oversize}

B and C = {Wilson midsize, Wilson oversize}

Chapter 6: Probability

137

c Midsize Head

Oversize

Midsize Prince Oversize Midsize Slazenger Oversize Midsize Wimbledon Oversize

Midsize Wilson

6.7

Oversize

a 50 100 10

150 200 50 100

12 150 200 50 100 15

150 200

AC = {(15, 50), (15, 100), (15, 150), (15, 200)}

138

Chapter 6: Probability

ii iii c

6.8

10, 50  , 10, 100  , 10, 150  , 10, 200  , 12, 50     A B    12, 100  , 12, 150  , 12, 200  , 15, 50  , 15, 100    

A  B  10, 50  , 10, 100  , 12, 50  , 12, 100 

A and C are not disjoint events. B and C are disjoint events.

a 2-hr reserve

Overnight check out 2

3 4 1 3

4 1 2

4 1 4

2 3

A = {(1, 2), (1, 4), (2, 1), (2, 3), (2, 4), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)}

B = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 1), (3, 2), (4, 1), (4, 2)}

Chapter 6: Probability

6.9

139

a 3

4 5

3 4 5

6.10

6.11

6.12

6.13

A = {(3), (4), (5)}

C = {1 2 5, 1 5, 2 1 5, 2 5, 5}

For example: N, DN, DDN, DDDN, DDDDN

There is an infinite number of outcomes.

E = {DN, DDDN, DDDDDN, DDDDDDDN, …}

A = {NN, DNN}

B = {DDNN}

There are an infinite number of outcomes.

The 27 possible outcomes are (1,1,1), (1,1,2), (1,1,3), (1,2,1), (1,2,2), (1,2,3), (1,3,1), (1,3,2), (1,3,3), (2,1,1), (2,1,2), (2,1,3), (2,2,1), (2,2,2), (2,2,3), (2,3,1), (2,3,2), (2,3,3), (3,1,1), (3,1,2), (3,1,3), (3,2,1), (3,2,2), (3,2,3), (3,3,1), (3,3,2), (3,3,3).

A = {(1,1,1), (2,2,2), (3,3,3)}

B = {(1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1)}

C = {(1,1,1), (1,1,3), (1,3,1), (1,3,3), (3,1,1), (3,1,3), (3,3,1), (3,3,3)}

B C  {(1,1,1), (1,1,2), (1,1,3), (1,2,1), (1,2,2), (1,3,1), (1,3,3), (2,1,1), (2,1,2), (2,2,1), (2,2,2), (2,2,3), (2,3,2), (2,3,3), (3,1,1), (3,1,3), (3,2,2), (3,2,3), (3,3,1), (3,3,2), (3,3,3)}

140

Chapter 6: Probability

b c d e 6.14

C C  {(1,1,2), (1,2,1), (1,2,2), (1,2,3), (1,3,2), (2,1,1), (2,1,2), (2,1,3), (2,2,1), (2,2,2), (2,2,3), (2,3,1), (2,3,2), (2,3,3), (3,1,2), (3,2,1), (3,2,2), (3,2,3), (3,3,2)} A  B  {(1,1,1), (2,2,2), (3,3,3), (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1)} A B   A  C  {(1,1,1), (3,3,3)}

Chapter 6: Probability

6.15

6.16

The event ( A  B )C is shaded in the Venn diagram below.

141

142

Chapter 6: Probability The event AC  B C is shaded in the Venn diagram below.

The two events are the same event. b

The event ( A  B )C is shaded in the Venn diagram below.

The event AC  B C is shaded in the Venn diagram below.

The two events are the same event. 6.17

The sample space is {expedited overnight delivery, expedited second business day delivery, standard delivery, delivery to the nearest store for customer pick-up}.

6.18

a b c

1 − 0.1 − 0.3 − 0.4 = 0.2 0.1 + 0.3 = 0.4 0.4 + 0.2 = 0.6

6.19

0.45 + 0.25 + 0.1 = 0.8

P(at most 3)  0.45  0.25  0.1  0.8. This probability is the same as the one in Part (a).

Chapter 6: Probability

6.20

6.21

6.22

6.23

6.24

6.25

6.26

143

0.07 + 0.03 = 0.1

0.45 + 0.25 = 0.7

P (more than 2)  0.1  0.1  0.07  0.03  0.3. P (more than 2)  1  P (2 or fewer)  1  0.7  0.3.

The sample space is {fiction hardcover, fiction paperback, fiction digital, fiction audio, nonfiction hardcover, nonfiction paperback, nonfiction digital, nonfiction audio}

No. For example, a customer is more likely to buy a paperback than a hardcover book.

0.15 + 0.08 + 0.45 + 0.04 = 0.72

0.1 + 0.02 + 0.1 + 0.06 = 0.28; 1 − 0.72 = 0.28

0.15 + 0.45 + 0.1 + 0.1 = 0.8

i A = {(C, N), (N, C), (N, N)}. ii P ( A)  0.09  0.09  0.01  0.19.

i B = {(C, C), (N, N)}. ii P ( B )  0.81  0.01  0.82.

P(Red) = 18/38 = 0.474.

No. If the segments were rearranged, there would still be 18 red segments out of a total of 38 segments, and so the probability of landing in a red segment would still be 18/38 = 0.474.

1000(0.474) = 474. So, if the wheel is balanced you would expect the resulting color to be red on approximately 474 of the 1000 spins. If the number of spins resulting in red is very different from 474, then you would suspect that the wheel is not balanced.

P(Phoenix is final destination) = 1800/8000 = 0.225.

P(Phoenix is not final destination) = 1 – 0.225 = 0.775.

P(Connecting and missed flight) = 480/8000 = 0.06.

P(Connecting and did not miss flight) = (6200 – 480)/8000 = 0.715.

P(Either had Phoenix as final destination or delayed overnight) = (1800 + 75)/8000 = 0.234.

We know that 50 of the 8000 passengers will be selected, and 50/8000 = 1/160. In other words, 1 in every 160 passengers will be selected. So, of the 75 passengers who were delayed overnight, you would expect 75(1/160) = 0.47 to be included in the survey. So it is very likely that none, or just one, of the passengers who were delayed overnight will be included in the survey, and so the airline has little need to worry.

144 6.27

Chapter 6: Probability a

The number of ways of selecting two problems from the five is 10. (The easiest way to see this is to list the possibilities. Calling the problems A, B, C, D, E, the possibilities are: AB, AC, AD, AE, BC, BD, BE, CD, CE, DE.) Suppose now that you completed problems A, B, and C. The number of ways of selecting two of these problems to be graded is 3. (The possibilities are: AB, AC, BC.) So the probability that the two problems selected are from the three you completed is 3/10 = 0.3.

6.28

No. The problems to be graded will be selected at random, so the probability of being able to turn in both of the selected problems is the same whatever your choice of three problems to complete.

If you complete four problems, the number of ways of selecting two from the ones you completed is 6. (Suppose you completed problems A, B, C, and D. The possibilities for the two selected are: AB, AC, AD, BC, BD, CD.) As in Part (a), there are 10 ways of selecting the two to be graded from the 5 problems assigned. So the probability that you are able to turn in both problems is now 6/10 = 0.6.

500000/42005100 = 0.0119

100/42005100 = 0.00000238

505100/42005100 = 0.0120

6.29

If the events F and I were mutually exclusive, we could add the probabilities to obtain P ( F  I ) . Since P ( F )  P ( I )  0.71  0.52  1.23 is greater than 1 (and probabilities cannot be greater than 1), we know that F and I cannot be mutually exclusive.

6.30

P(the selected student is female) 

P(the selected student is male) 

P(received a degree in the life sciences) 

P(received a degree not life or physical science) 

P(did not receive a degree in physical sciences) 

6.31

613, 034 1, 003,329

390, 295 1, 003,329

 0.611

 0.389

12,504 54, 070

 0.231256 31, 704 54, 070

 0.586407

(54, 070  9,859) 54, 070

 0.817662

Chapter 6: Probability

6.32

6.33

6.34

6.35

6.36

145

P( just spades)  1287 2598960  0.000495. Since there are 1287 hands consisting entirely of spades, there are also 1287 hands consisting entirely of clubs, 1287 hands consisting entirely of diamonds, and 1287 hands consisting entirely of hearts. So the number of possible hands consisting of just one suit is 4(1287) = 5148. Thus, P(single suit)  5148 2598960  0.00198.

P(entirely spades and clubs with both suits represented)  63206 2598960  0.024.

The two suits could be spades and clubs, spades and diamonds, spades and hearts, clubs and diamonds, clubs and hearts, or diamonds and hearts. So there are six different combinations of two suits, with 63,206 possible hands for each combination. Therefore, P(exactly two suits)  6(63206) 2598960  0.146.

The 24 outcomes (including the given outcome) are: (1, 2, 3, 4), (1, 2, 4, 3), (1, 3, 2, 4), (1, 3, 4, 2), (1, 4, 2, 3), (1, 4, 3, 2), (2, 1, 3, 4), (2, 1, 4, 3), (2, 3, 1, 4), (2, 3, 4, 1), (2, 4, 1, 3), (2, 4, 3, 1), (3, 1, 2, 4), (3, 1, 4, 2), (3, 2, 1, 4), (3, 2, 4, 1), (3, 4, 1, 2), (3, 4, 2, 1), (4, 1, 2, 3), (4, 1, 3, 2), (4, 2, 1, 3), (4, 2, 3, 1), (4, 3, 1, 2), (4, 3, 2, 1).

The event is {(1, 2, 4, 3), (1, 4, 3, 2), (1, 3, 2, 4), (4, 2, 3, 1), (3, 2, 1, 4), (2, 1, 3, 4)}. The required probability is 6/24 = 1/4.

The outcomes associated with this event are: (1, 3, 4, 2), (1, 4, 2, 3), (3, 2, 4, 1), ( 4, 2, 1, 3), (2, 4, 3, 1), (4, 1, 3, 2), (2, 3, 1, 4), (3, 1, 2, 4). The required probability is 8/24 = 1/3.

It is not possible that exactly three receive their own books, because if three students have their own books, then the fourth student must receive his/her own, also. Thus the required probability is 0.

P (at least 2 get the correct books)  P(2 get the correct books)  P(4 get the correct books)  1 4  1 24  7 24.

BC, BM, BP, BS, CM, CP, CS, MP, MS, PS

1/10 = 0.1

4/10 = 0.4

There are three outcomes in which both representatives come from laboratory science subjects: BC, BP, and CP. Thus P(both from laboratory science)  3 10  0.3.

Denoting the three math majors as M1, M2, and M3, and the two statistics majors by S1 and S2, the ten possible outcomes for the pair of students chosen are: M1M2, M1M3, M1S1, M1S2, M2M3, M2S1, M2S2, M3S1, M3S2, S1S2. There is only one outcome for which both students chosen are statistics majors, and so the required probability is 1/10 = 0.1.

Three of the ten outcomes satisfy this criterion, so the required probability is 0.3.

146

6.37

Chapter 6: Probability c

Seven of the ten outcomes satisfy this criterion, so the required probability is 0.7.

Six of the ten outcomes satisfy this criterion, so the required probability is 0.6.

From p  2 p  p  2 p  p  2 p  1 we see that 9 p  1 , and so p  1 9 . Therefore, P(O1 )  P(O3 )  P(O5 )  1 9 and P(O2 )  P(O4 )  P(O6 )  2 9 .

P(odd)  P(O1 )  P(O3 )  P(O5 )  1 9  1 9  1 9  1 3 . P(at most 3)  P(O1 )  P(O2 )  P(O3 )  1 9  2 9  1 9  4 9.

From c  2c  3c  4c  5c  6c  1 we see that 21c  1 , and so c  1 21 . Thus P(O1 )  1 21, P(O2 )  2 21, P(O3 )  3 21  1 7, P(O4 )  4 21, P(O5 )  5 21, P(O6 )  6 21  2 7. Therefore, P(odd)  P(O1 )  P(O3 )  P(O5 )  1 21  3 21  5 21  9 21 and P(at most 3)  P(O1 )  P(O2 )  P(O3 )  1 21  2 21  3 21  6 21  2 7.

6.38

6.39

6.40

P( E | F ) 

P( E  F ) 0.54   0.9. P( F ) 0.6

P( F | E ) 

P( F  E ) 0.54   0.771. P( E ) 0.7

P(tattoo)  24 100  0.24

P(tattoo | age 18-29)  18 50  0.36

P(tattoo | age 30-50)  6 50  0.12

P(age 18-29 | tattoo)  18 24  0.75

The row and column totals are given in the table below.

Male Female Total a

Hybrid 77 34 111

Not Hybrid 117 83 200

Total 194 117 311

194 311  0.624

ii 111 311  0.357 iii 77 194  0.397 iv 34 117  0.291 v 34 111  0.306 b

If a person who purchased a Honda Civic is chosen at random, the probability that this person is a male is 0.624.

Chapter 6: Probability

147

If a person who purchased a Honda Civic is chosen at random, the probability that this person purchased a hybrid is 0.357. iii If a male who purchased a Honda Civic is chosen at random, the probability that he purchased a hybrid is 0.397. iv If a female who purchased a Honda Civic is chosen at random, the probability that she purchased a hybrid is 0.291. v If a person who purchased a hybrid Honda Civic is chosen at random, the probability that this person is female is 0.306. 6.41

P (hybrid | male)  0.397. P (male | hybrid)  0.694. These probabilities are not equal. The first is the proportion of males who bought hybrids, and the second is the proportion of hybrid buyers who were males; they are different quantities.

6.42

0.72

The value 0.45 is the conditional probability that the selected individual drinks 2 or more cups a day given that he or she drinks coffee. We know this because the percentages given in the display add to 100, and yet we know that only 72% of Americans drink coffee. So the percentages given in the table must be the proportions of coffee drinkers who drink the given amounts.

P( A) 

P ( D) 

6.43

c d 6.44

922 2, 096 1,174

 0.4399

 0.5601 2, 096 693 P( A | F )   0.5699 1, 216 229 P( A | F C )   0.2602 880 i ii

If a person has been out of work for 1 month then the probability that the person will find work within the next month is 0.3. If a person has been out of work for 6 months then the probability that the person will find work within the next month is 0.19.

148

Chapter 6: Probability

b P(E|T i) 0.30 0.25 0.20 0.15 0.10 0.05 0.00

6.45

6.46

The longer you have been unemployed the less likely you are to find a job during the next month, with the amount by which the likelihood decreases becoming smaller as the number of months out of work increases. It seems from the information given that, after a time, the probability of finding a job during the next month stabilizes at about 0.18.

Yes. The probabilities given are consistent with the comments in the article that a baby is more likely to have Down Syndrome if it is born to an older mother and that younger women are more fertile.

No. According to the quote from the article P( D | Y C ) is greater than P ( D | Y ) . In this statement these two probabilities are equal.

No. The statement “ P (Y )  0.4 ” is not consistent with the comment in the article that younger women are more fertile.

P( A | B ) is larger. P( A | B ) is the probability that a randomly chosen professional basketball player is over six feet tall – a reasonably large probability. P ( B | A) is the probability that a randomly chosen person over six feet tall is a professional basketball player – a very small probability.

Chapter 6: Probability

6.47

6.48

149

P(female | predicted female)  432 562  0.769.

P(male | predicted male)  390 438  0.890.

Since the conditional probabilities in (a) and (b) are not equal, we see that a prediction that a baby is male and a prediction that a baby is female are not equally reliable.

There are 18,639 participants in the study, and of those, 12,392 reported their weight within 3 12,392  0.665 . pounds of their actual weight, so P(weight reported within 3 pounds)  18, 639 Therefore, approximately 66.5% of women reported their weight within 3 pounds of their actual weight. Define an “older woman” as ages 65+ years, and a “younger woman” as ages <45 years. The proportion of older women who under report their weight by at least four 263  444  0.215 . The proportion of pounds is P(older women under report weight)  3, 292 younger women who under report their weight by at least four pounds is 297  373 P(younger women under report weight)   0.271 . The proportion of older 2, 471 women who over report their weight by at least four pounds is 300 P(older women over report weight)   0.091 . The proportion of younger women who 3, 292 over report their weight by at least four pounds is 207 P(younger women over report weight)   0.084 . These probabilities are consistent 2, 471 with the conclusion given in the paper.

6.49 The following probabilities can be used to justify the report’s conclusion that females are more likely to wear seatbelts than males in both urban and rural areas. 871 P(urban wear seat belt | male)   0.871 871  129 928 P(urban wear seat belt | female)   0.928 928  72 770 P(rural wear seat belt | male)   0.77 770  230 837 P(rural wear seat belt | female)   0.837 837  163 The differences in percentages of females and males who wear seatbelts are below, which support the report’s conclusion that the difference in the proportion of females and the proportion of males who wear seatbelts is greater for rural areas. Urban: 0.928 – 0.871 = 0.057 Rural: 0.837 – 0.77 = 0.067

150 6.50

6.51

6.52

Chapter 6: Probability a

P (uses seat belt)  0.423  0.452  0.875

P(uses seat belt | male) 

P(does not use a seat belt | female) 

P(female | does not use a seat belt) 

No, the probabilities in parts (a) and (b) are not equal. Part (a) is the proportion of female drivers who do not use a seat belt, and part (b) is the proportion of drivers who do not use a seat belt who are female. There is no reason to believe that these proportions should be equal.

i ii

0.423 0.423  0.077

 0.846 0.048

0.452  0.048

0.048 0.077  0.048

 0.096

 0.384

P( S )  456 600  0.76. P(S | A)  215 300  0.717.

iii P( S | B)  241 300  0.803. iv Treatment B appears to be better, since the probability of survival given that the patient receives treatment B is greater than the probability of survival given that the patient receives treatment A.

6.53

i P(S )  140 240  0.583. ii P( S | A)  120 200  0.6. iii P( S | B)  20 40  0.5. iv Treatment A appears to be better, since the probability of survival given that the patient receives treatment A is greater than the probability of survival given that the patient receives treatment B.

P(S )  316 360  0.878. i ii P( S | A)  95 100  0.95. iii P(S | B)  221 260  0.85. iv Treatment A appears to be better, since the probability of survival given that the patient receives treatment A is greater than the probability of survival given that the patient receives treatment B.

Women respond better to these treatments than men do. Moreover, treatment A was given to far more men than women and treatment B was given to far more women than men. As a result, even though treatment A is the better one for both men and women, when the results are combined the greater number of women receiving treatment B artificially increases the apparent efficacy of treatment B and the greater number of men receiving treatment A artificially decreases the apparent efficacy of treatment A. The combined results are in this way distorted in favor of treatment B.

Eighty-five percent of all calls to the station are for medial assistance.

0.15

Chapter 6: Probability

6.54

(0.85)(0.85) = 0.7225

(0.85)(0.15) = 0.1275

P(exactly one for medical assistance)  P(1st for medical  2nd not for medical)  P(1st not for medical  2nd for medical)  (0.85)(0.15)  (0.15)(0.85)  0.255. It would seem reasonable to assume that the outcomes of successive calls (for medical assistance or not) do not affect each other. However, it is likely that at certain times of the day calls are more likely to be for medical assistance than at other times of the day. Therefore, if a call is chosen at random and found to be for medical assistance, then it becomes more likely that this call was received at one of these times. The next call, being at roughly the same time of the day, then has a more than 0.85 probability of being for medical assistance. Therefore it is not reasonable to assume that the outcomes of successive calls are independent.

6.55

151

Convention Medications Usually Help Convention Medications Usually Do Not Help Total b

Does Not Use Complementary Therapies

Does Use Complementary Therapies

Total

0.758

0.122

0.879

0.096

0.025

0.121

0.853

0.147

The cell entry 0.122 represents the probability of (conventional medicines usually help and does use complementary therapies). The cell entry 0.096 represents the probability of (conventional medicines usually do not help and does not use complementary therapies). The cell entry 0.025 represents the probability of (conventional medicines usually do not help and does use complementary therapies).

P(CH  CT ) 0.12163   0.829, and P(CH )  0.879. Since these two P(CT ) 0.14670 probabilities are not equal the two events are not independent. P(CH | CT ) 

6.56

P ( L)  P ( F )  (0.58)(0.5)  0.29  P ( L  F ). Therefore the events L and F are not independent.

6.57

Given that the selected graduate finished college with no student debt, the probability that the selected graduate strongly agrees that education was worth the cost is 0.49.

Given that the selected graduate finished college with high student debt, the probability that the selected graduate strongly agrees that education was worth the cost is 0.18.

We are given that P ( A)  0.38 , and P ( A | H )  0.18 . The events A and H are not independent because P ( A | H )  P ( A) .

152 6.58

6.59

6.60

6.61

6.62

Chapter 6: Probability a

P(selected in each of the next two years)  (0.15)(0.15)  0.0225.

P(selected in each of the next three years)  (0.15)(0.15)(0.15)  0.003375.

(0.1) (0.1) (0.1) = 0.001. We have to assume that she deals with the three errands independently.

P(remembers at least one)  1  P(forgets them all)  1  0.001  0.999.

P(remembers 1st, forgets 2nd, forgets 3rd)  (0.9)(0.1)(0.1)  0.009.

P(1-2 subsystem works)  (0.9)(0.9)  0.81.

P (1-2 subsystem doesn't work)  1  0.81  0.19. P(3-4 subsystem doesn't work)  0.19.

P(system won't work)  (0.19)(0.19)  0.0361. P(system will work)  1  0.0361  0.9639.

P(system won't work)  (0.19)(0.19)(0.19)  0.006859. So P (system will work)  1  0.006859  0.993141.

The probability that one particular subsystem will work is now (0.9) (0.9) (0.9) = 0.729. So the probability that the subsystem won’t work is 1  0.729  0.271 . Therefore the probability that neither of the two subsystems works (and so the system doesn’t work) is (0.271) (0.271) = 0.073441. So the probability that the system works is 1  0.073441  0.926559.

i ii iii

P ( F )  0.33

P ( F | L )  0.48 P ( F | H )  0.21

No, F and L are not mutually exclusive because one can name pro football as a favorite sport and have a household income of $75,000 - <$100,000.

Yes, H and L are mutually exclusive because a household income cannot be both $75,000 <$100,000 and $100,000+.

No, F and H are not independent because P ( F )  0.33  P ( F | H )  0.21 .

6.63

The reason that P(T) is not the average of the two given conditional probabilities is because there are different numbers (or different proportions) of people in the two given age groups (19 to 36 and 37 or older).

6.64

The expert was assuming that there was a 1 in 12 chance of a valve being in any one of the 12 clock positions and that the positions of the two air valves were independent.

Chapter 6: Probability

6.65

6.66

Since the car’s wheels are probably the same size, if one of the wheels happens to have its air valve in the same position as before then the other wheel is likely also to have its air valve in the same position as before. Thus the positions of the two air valves are not independent.

Assuming independence makes the probability smaller.

1/144 is smaller than the correct probability.

(0.7)(0.8)(0.6) = 0.336

(0.7)(0.8) = 0.56

(0.3)(0.2) = 0.06

P(each person wins one match)  P (A beats B, C beats A, and B beats C)  P (B beats A, A beats C, and C beats B)  (0.7)(0.2)(0.6)  (0.3)(0.8)(0.4)  0.18.

P( B1  S )  P( B1 ) P( S )  (0.4)(0.3)  0.12. The required probabilities are given in the table below.

B1 B2 6.67

S M 0.12 0.2 0.18 0.3

L 0.08 0.12

P(both correct)  (50 800)(50 800)  0.00391.

P(2nd is correct |1st is correct)  P(2nd is correct  1st is correct) P(1st is correct). So P(1st is correct  2nd is correct)  P(1st is correct)  P(2nd is correct |1st is correct)

  50 800  49 799   0.00383. This probability is slightly smaller than the one in Part (a). 6.68

6.69

6.70

153

P(both correct)  (50 100)(50 100)  0.25.

P(both correct)  (50 100)(49 99)  0.247.

6/10 = 0.6

P( F | E )  5 9 .

P( F | E )  P( F  E ) P( E ). So P( E and F )  P( F  E )  P( E )  P( F | E )  (6 10)(5 9)  1 3.

(0.4)(0.3) = 0.12

(0.6)(0.7) = 0.42

154

6.71

Chapter 6: Probability c

P (at least one successful)  1  P(neither is successful)  1  0.42  0.58.

P( E  F )  P ( E )  P ( F )  P ( E  F )  0.4  0.3  0.15  0.55.

The Venn diagram is shown below.

P(needn't stop at either light)  1  P( E  F )  1  0.55  0.45. c

P(stops at exactly one)  0.25  0.15  0.4.

P (stops at just the first light)  0.25.

6.72

The event E  F is the event that the randomly chosen registered voter has signed the petition and votes in the recall election. P( E  F )  P ( E | F )  P ( F )  (0.8)(0.1)  0.08.

6.73

P(at least one food allergy and severe reaction) = (0.08)(0.39) = 0.0312.

P(allergic to multiple foods) = (0.08)(0.3) = 0.024.

6.74

Chapter 6: Probability

P(blue | identified as blue)  6.75

155

P(blue and identified as blue) (0.15)(0.8)   0.414 . P(identified as blue) (0.15)(0.8)  (0.85)(0.2)

For a randomly selected customer, let T be the event that the customer subscribes to cable TV service, I be the event that the customer subscribes to Internet service, and P be the event that the customer subscribes to telephone service. The three events are shown in the Venn diagram below, and letters have been used to denote the probabilities of four events. (Please note that x represents the probability that the customer subscribes to the phone and Internet services but not to the cable TV service. The quantities y and z are similarly defined.)

The information we have been given can be summarized as follows:

P(T )  0.8 P( I )  0.42 P( P )  0.32 w  z  0.25 w  y  0.21 w  x  0.23 w  0.15 Using the last four results we can conclude that x  0.08, y  0.06, and z  0.1. a

P (cable TV only)  0.8  ( y  z  w)  0.8  (0.06  0.1  0.15)  0.49. The probability that a randomly chosen customer subscribes only to the cable TV service is 0.49.

P( I  T ) 0.25   0.3125. P(T ) 0.8 If a customer is chosen at random from those who subscribe to the cable TV service, the probability that this customer will subscribe to the Internet service is 0.3125. P(exactly two services)  x  y  z  0.08  0.06  0.1  0.24. The probability that a randomly chosen customer subscribes to exactly two services is 0.24. P(Internet | cable TV)  P( I | T ) 

P (Internet and cable TV only)  z  0.1. The probability that a randomly chosen customer subscribes only to the Internet and cable TV services is 0.1.

156 6.76

6.77

6.78

6.79

6.80

Chapter 6: Probability a

Yes. Since it is a large cable company, meaning that it has many customers, the outcome for the first customer (cable TV or not) will have little effect on the outcome for the second customer.

P(C1  C2 )  (0.8)(0.8)  0.64.

P ( H )  (0.261)(0.375)  (0.739)(0.073)  0.152. P (C  H )  (0.261)(0.375)  0.098. P(C  H ) 0.098 P(C | H )    0.645. P( H ) 0.152

P( H )  (0.495)(0.537)  (0.505)(0.252)  0.393. P (C  H )  (0.495)(0.537)  0.266. P(C  H ) 0.266 P(C | H )    0.676. P( H ) 0.393 The result is a little higher for faculty than for students. A faculty member who has high confidence in a correct diagnosis is slightly more likely actually to be correct that a student who has the same high confidence.

(37  14  16  11) 79  0.987

(37  14) 79  0.646

(14  16) 79  0.380

(14  16) (14  16  37  11)  30 78  0.385

30 30  1

P (T )  0.307.

ii iii iv v

P(T C )  0.693. P (C | T )  0.399. P ( L | T )  0.275. P(C  T )  (0.307)(0.399)  0.122.

30.7% of faculty members use Twitter. 69.3% of faculty members do not use Twitter. 39.9% of faculty members who use Twitter also use it to communicate with students. 27.5% of faculty members who use Twitter also use it as a learning tool in the classroom. 12.2% of faculty members use Twitter and use it to communicate with students.

By the law of total probability, P(C )  P(C  T )  P(C  T C ). However, faculty members who do not use Twitter cannot possibly use it to communicate with students. Therefore P(C  T C )  0. Thus P(C )  P(C  T )  0.122.

Chapter 6: Probability b

6.81

157

Since faculty members who use Twitter as a learning tool must use Twitter, P( L)  P( L  T )  P(T )  P( L | T )  (0.307)(0.275)  0.084.

High School or Less College – 1 to 4 years College – 5 or more years Total

Uses Alternative Therapies

Does Not Use Alternative Therapies

Total

315

7005

7320

393

4400

4793

120

975

1095

828

12380

13208

b Uses Alternative Therapies High School or Less College – 1 to 4 years College – 5 or more years Total c 6.82

6.83

Does Not Use Alternative Therapies Total

0.024

0.530

0.554

0.030

0.333

0.363

0.009

0.074

0.083

0.063

0.937

1.000

i ii

0.083 0.063 0.009 120   0.110 i 0.083 1095 0.024 315   0.043 ii 0.554 7320 0.024 315   0.380 iii 0.063 828 0.030  0.009 393  120   0.087 iv 0.363  0.083 4793  1095

P( A | H )  0.043 and P( A)  0.063. Since these two probabilities are different, the events A and H are not independent.

The reason that P (C ) is not the average of the three conditional probabilities is that there are different numbers of people driving the three different types of vehicle (and also that there are some drivers who are driving vehicles not included in those three types).

158

6.84

Chapter 6: Probability

P (O )  0.31 ; P (O | Y )  0.43 ; P (O | F )  0.23

P(Y | O) 

6.85

P(Y  O) P(O)



P(O | Y )  P(Y ) P(O)



(0.43)(0.40) 0.31

 0.555

Radiologist 1 Predicted Predicted Male Female Baby Is Male 74 12 Baby Is Female 14 59 Total 88 71

Total 86 73 159

P(prediction is male | baby is male) = 74/86 = 0.860.

P(prediction is female | baby is female) = 59/73 = 0.808.

Yes. Since the answer to (a) is greater than the answer to (b), the prediction is more likely to be correct when the baby is male.

The radiologist was correct for 74 + 59 = 133 of the 159 babies in the study. So P(correct) = 133/159 = 0.836.

6.86

Radiologist 2 Predicted Predicted Male Female Baby Is Male 81 8 Baby Is Female 7 58 Total 88 66

Total 89 65 154

For Radiologist 1, P(prediction is male | baby is male) = 0.860. For Radiologist 2, P(prediction is male | baby is male) = 81/89 = 0.910. So Radiologist 2 is better than Radiologist 1 at predicting the gender when the baby is male. For Radiologist 1, P(prediction is female | baby is female) = 0.808. For Radiologist 2, P(prediction is female | baby is female) = 58/65 = 0.892. So Radiologist 2 is also better than Radiologist 1 at predicting the gender when the baby is female. Indeed, Radiologist 2’s overall success rate is (81 + 58)/154 = 0.903, which is higher than Radiologist 1’s success rate of 0.836. 6.87

i ii iii iv

0.99 0.01 0.99 0.01

P(TD)  P(TD  C )  P(TD  D)

 P(C )  P(TD | C )  P( D)  P(TD | D)  (0.99)(0.01)  (0.01)(0.99)  0.0198.

Chapter 6: Probability

6.88

6.89

P(C | TD)  P(C  TD) P(TD)  (0.99)(0.01) 0.0198  0.5. Yes. The quote states that half of the dirty tests are false. The above probability states that half of the dirty tests are on people who are in fact clean. This confirms the statement in the quote. 1, 628 P(complaint about baggage handling | 2014)   0.143 11, 365

6.91

9, 754

P(complaint not about flight problems | 2015) 

P(two complaints were about flight problems | 2015) 

P(complaint about flight problems or customer service | 2014) 

15, 260

 0.639 5,506



5,505

15, 260 15, 259

 0.130

4,304  1, 201 11,365

 0.484

a Probation Yes No Total

6.90

159

High School GPA 2.5 to <3.0 3.0 to <3.5 3.5 or Above 0.10 0.11 0.06 0.09 0.27 0.37 0.19 0.38 0.43

Total 0.27 0.73 1.00

b c

0.27 0.43

P (3.5 or above  probation)  0.06. P(3.5 or above)  P (probation)  (0.43)(0.27)  0.1161. Since P(3.5 or above  probation)  P(3.5 or above)  P(probation), the two outcomes are not independent.

P(probation | 2.5 to 3.0)  0.1 0.19  0.526.

P(probation | 3.5 or above)  0.06 0.43  0.140.

The total number of students listed is 18000. The total number of males listed is 11200. So P(male)  11200 18000  0.622.

3000/18000 = 0.167

2100/18000 = 0.117

The number of males not from Agriculture is 11200  2100  9100. So the required probability is 9100/18000 = 0.506.

160

Chapter 6: Probability

6.92

Monterey San Luis Obispo Santa Barbara Ventura Total

6.93

6.94

Caucasian 163,000 180,000 230,000 430,000 1,003,000

Hispanic 139,000 37,000 121,000 231,000 528,000

Black Asian 24,000 39,000 7,000 9,000 12,000 24,000 18,000 50,000 61,000 122,000

American Indian Total 4,000 369,000 3,000 236,000 5,000 392,000 7,000 736,000 19,000 1,733,000

736/1733 = 0.425

231/736 = 0.314

231/528 = 0.4375

9/1733 = 0.005

(122 + 180 + 37 + 7 + 3)/1733 = 349/1733 = 0.201

(39 + 24 + 50 + 180 + 37 + 7 + 3)/1733 = 340/1733 = 0.196

(1003/1733) (1003/1733) = 0.335. [Note: It could be argued that this calculation should be (1003000/1733000)( 1002999/1732999). However, since the given calculation yields a result that is very close to the result of this second calculation, and since the population figures given are clearly approximations, the given calculation will suffice.]

Proportion who are not Caucasian  1  1003 1733  730 1733. Therefore, when two people

are selected at random, P(neither is Caucasian)   730 1733 730 1733  0.177. (1003/1733)(730/1733) + (730/1733)( 1003/1733) = 0.488

(369/1733)( 369/1733) + (236/1733)( 236/1733) + (392/1733)( 392/1733) + (736/1733)( 736/1733) = 0.295

P(same ethnic group)  1003 17331003 1733   528 1733 528 1733   61 1733 61 1733  122 1733122 1733  19 173319 1733  0.434. So P (different ethnic groups)  1  0.434  0.566.

6.95

Answers will vary.

6.96

The simulation could be designed as follows. Number the 20 companies/individuals applying for licenses 01–20, with the applications for 3 licenses coming from 01–06, the applications for 2 licenses coming from 07–15, and the applications for single licenses coming from 16–20. Assume that the individual with whom this question is concerned is numbered 16. One run of the simulation is conducted as follows. Use two consecutive digits from a random number table to select a number between 01 and 20. (Ignore pairs of digits from the random number table that give numbers outside this range.) The number selected indicates the company/individual whose request is granted. Repeat the number selection, ignoring repeated

Chapter 6: Probability

161

numbers, until there are no licenses left. Make a note of whether or not individual 16 was awarded a license. Perform the above simulation a large number of times. The probability that this particular individual is awarded a license is approximated by (Number of runs in which 16 is awarded a license) (Total number of runs). b

This does not seem to be a fair way to distribute licenses, since any given company applying for multiple licenses has roughly the same chance of obtaining all of its licenses as an individual applying for a single license has of obtaining his/her license. It might be fairer for companies who require two licenses to submit two separate applications and companies who require three licenses to submit three separate applications (with individuals/companies applying for single licenses submitting applications as before). Then 10 of the applications would be randomly selected and licenses would be awarded accordingly.

6.97

Results of the simulation will vary. The correct probability that the project is completed on time is 0.8504.

6.98

Results of the simulation will vary. The correct probability that the project is completed on time is 0.8468.

6.99

Results of the simulation will vary. The correct probability that the project is completed on time is 0.6504.

Jacob’s change makes the bigger change in the probability that the project will be completed on time.

6.100 The events are dependent, because knowing that a woman is over age 40 changes the probability that the woman actually has breast cancer. 6.101 a 0.05

A1 0.95

0.5

0.08 0.3

N R

0.2

0.92

0.1

0.9

P( A1  R)  (0.5)(0.05)  0.025.

P(R)  (0.5)(0.05)  (0.3)(0.08)  (0.2)(0.1)  0.069.

162

Chapter 6: Probability

6.102 a

P ( E )  0.4.

P (C )  0.3.

P( L)  0.25.

P ( E and C )  0.23.

6.103 a

P ( E or C or L)  1  0.51  0.49.

P ( E | L)  0.88.

P ( L | C )  0.7.

6.104 a

First, P ( E and L)  P ( L) P ( E | L)  (0.25)(0.88)  0.22, and P (C and L)  P (C ) P ( L | C )  (0.3)(0.7)  0.21. Now, P ( E  C  L )  P ( E )  P ( C )  P ( L )  P ( E  C )  P ( E  L )  P (C  L )  P ( E  C  L ) So, P( E and C and L)  P( E  C  L)  P( E )  P (C )  P ( L)  P( E  C )  P ( E  L)  P (C  L)

 0.49  0.4  0.3  0.25  0.23  0.22  0.21  0.2. b

P ( E and L)  0.22.

P (C and L)  0.21.

P(C | ( E and L)) 

P(C  E  L) 0.2   0.909. P ( E  L) 0.22

6.105 The events selected adult exercises at least 30 minutes 3 times per week and is a millennial are dependent events. A millennial is more likely to exercise at least 30 minutes 3 times per week (57.1%) when compared with adults over age 35 (51.1%). The age of the adult (millennial or over age 35) has an impact on the length of time the selected adult exercises. 6.106 To better answer the questions, add “totals” column and row to the table.

Age Group 18 to 29 30 to 49 50 to 64 65 and older Total

P(smoker) 

1,102 5, 987

 0.184

Smoking Status Smoker Nonsmoker 174 618 333 1,115 384 1,445 211 1,707 1,102 4,885

Total 792 1,448 1,829 1,918 5,987

Chapter 6: Probability

792  1, 448

P(under age 50) 

P(smoker age 65 or older) 

P(smoker or age 65 or older) 

6.107 a b

5,987

163

 0.374

211 5,987

 0.035

1,102  1, 918  211 5, 987

 0.469

P ( D )  0.148 , P (T )  0.374 , and P ( D  T )  0.106 P ( D  T )  P ( D )  P (T )  P ( D  T )  0.148  0.374  0.106  0.416 Therefore, P (neither a change in diagnosis nor change in treatment)  1  P ( D  T )

 1  0.416  0.584 c

P ( D  T )  P ( D )  P (T )  P ( D  T )  0.148  0.374  0.106  0.416

6.108 (0.4)(0.02) = 0.008 6.109 a 0.02

A1 0.98

0.4

0.01 0.5

LC L

0.1

0.99

0.05

0.95

P( L)  (0.4)(0.02)  (0.5)(0.01)  (0.1)(0.05)  0.018. b c d 6.110 a b

P( A1 | L)  P( A1  L) P( L)  (0.4)(0.02) 0.018  0.444. P( A2 | L)  P( A2  L) P( L)  (0.5)(0.01) 0.018  0.278. P( A3 | L)  P( A3  L) P( L)  (0.1)(0.05) 0.018  0.278.

P( A wins in 5 games)  (0.3)5  0.00243. P(5 games to obtain champion)  (0.3)5  (0.2)5  0.00275.

164

Chapter 6: Probability c

A random number table could be used. The digits 0–2 will represent A winning the game, 3–4 will represent B winning, and 5–9 will represent the game resulting in a draw. Use digits one at a time from the random number table, noting the results of the simulated games, and keeping a tally of scores according to the plan given in the question. When one of the competitors reaches a total score of 5, that person is the winner. If both competitors reach 5 at the same time, then the championship ends in a draw. Repeat the simulation above a large number of times. The probability that A wins the championship is estimated by the fraction

number of runs in which A wins . total number of runs 6.111 In Parts (a)–(c) below, examples of possible simulation plans are given. a

Use a single-digit random number to represent the outcome of the game. The digits 0–7 will represent a win for seed 1, and digits 8–9 will represent a win for seed 4.

Use a single-digit random number to represent the outcome of the game. The digits 0–5 will represent a win for seed 2, and digits 6–9 will represent a win for seed 3.

Use a single-digit random number to represent the outcome of the game. If seed 1 won game 1 and seed 2 won game 2, the digits 0–5 will represent a win for seed 1, and digits 6–9 will represent a win for seed 2. If seed 1 won game 1 and seed 3 won game 2, the digits 0–6 will represent a win for seed 1, and digits 7–9 will represent a win for seed 3. If seed 4 won game 1 and seed 2 won game 2, the digits 0–6 will represent a win for seed 2, and digits 7–9 will represent a win for seed 4. If seed 4 won game 1 and seed 3 won game 2, the digits 0–5 will represent a win for seed 3, and digits 6–9 will represent a win for seed 4.

Answers will vary.

6.112 a

Answers will vary.

The estimated probabilities from Parts (e) and (f) will differ because they are based on different sets of simulations. The estimate from Part (f) is likely to be the better one, since it is based on more runs of the simulation than the estimate from Part (e).

6.113 a b c

P( E C )  1  P( E )  1  0.6  0.4. Since P ( E  F )  0, P ( E  F )  P( E )  P ( F )  0.6  0.15  0.75. P( E C  F C )  1  P( E  F )  1  0.75  0.25.

Chapter 6: Probability

165

6.114 Individuals Chosen Total Number of (by Years of Experience) Years’ Experience 3, 6 9 3, 7 10 3, 10 13 3, 14 17 6, 7 13 6, 10 16 6, 14 20 7, 10 17 7, 14 21 10, 14 24

At least 15 years? No No No Yes No Yes Yes Yes Yes Yes

P(at least 15 years' experience)  6 10  0.6. 6.115

P (at least one)  P( A or B or C )  0.14  0.23  0.37  0.08  0.09  0.13  0.05  0.49.

6.116 a

(179 + 87)/2517 = 0.106

(420 + 323 + 179 + 114 + 87)/2517 = 0.446

1  (600  196  205  139) 2517  0.547

6.117 Since the total number of viewers was 2517 and the number of viewers of R-rated movies was 1140, P( R2 | R1 )  1139 2516 and P( R2 | R1C )  1140 2516 . These probabilities are not equal, so the events R1 and R2 are not independent. However, as a result of the fact that the total number of viewers is large, the two probabilities are very close, and therefore from a practical point of view the events may be regarded as independent. 6.118 a

P( E )  20 25  4 5.

P( F | E )  19 24.

P(G | E  F )  18 23.

P(all good)   20 25 19 24 18 23  0.496.

6.119 a b

 20  19  18  17  P(all 4 are good)        0.383.  25  24  23  22  P(at least one is bad)  1  P (all 4 are good)  1  0.383  0.617.

166

Chapter 6: Probability

6.120 a b

(0.8)(0.8)(0.8)  0.512 The possible sequences and their probabilities are shown in the table below. Sent by Transmitter 1 1 1 1

Sent by Relay 1 0 0 1 1

Sent by Sent by Relay 2 Relay 3 0 1 1 1 0 1 1 1

Probability (0.2)(0.8)(0.2) (0.2)(0.2)(0.8) (0.8)(0.2)(0.2) (0.8)(0.8)(0.8)

P(1 is received)  (0.2)(0.8)(0.2)  (0.2)(0.2)(0.8)  (0.8)(0.2)(0.2)  (0.8)(0.8)(0.8)  0.608.

Chapter 7 Random Variables and Probability Distributions Note: In this chapter, numerical answers to questions involving the normal distribution were found using statistical tables. Students using calculators or computers will find that their answers differ slightly from those given. 7.1

7.2

7.3

7.4

7.5

7.6

7.7

Discrete

Continuous

Discrete

Continuous

Discrete

Continuous

Discrete

The possible y values are the positive integers.

(Answers will vary.)

The random variable x takes values between 0 and

Continuous

The possible values of y are the real numbers between 0 and 100, inclusive.

Continuous

The variable y can take any positive even integer value

Discrete

a b

3, 4, 5, 6, 7 3,  2,  1, 0, 1, 2, 3

2  1.414 .

167

168

7.8

Chapter 7: Random Variables and Probability Distributions c

0, 1, 2

0, 1

0.25

0.02 + 0.03 + 0.09 + 0.25 = 0.39

P ( x  5)  0.02  0.03  0.09  0.25  0.40  0.79.

P(student is taking at least 5 courses)  P( x  5)  0.4  0.16  0.05  0.61.

P(student is taking more than 5 courses)  P( x  5)  0.16  0.05  0.21.

7.9

P (3  x  6)  0.09  0.25  0.4  0.16  0.9. P(3  x  6)  0.25  0.4  0.65. The second probability is smaller (and therefore the two probabilities are not equal) because the second probability does not include the possibility of a student taking 3 or 6 courses, while the first does include this possibility.

7.10

p (4)  1  0.65  0.2  0.1  0.04  0.01.

Over a large number of cartons, 20% will contain 1 broken egg.

P ( y  2)  0.65  0.2  0.1  0.95. Over a large number of cartons, 95% will contain at most 2 broken eggs.

P( y  2)  0.65  0.2  0.85. This probability is smaller than the one in Part (c) since it does not include the possibility of 2 broken eggs.

P(exactly 10 unbroken)  P (exactly 2 broken)  0.1.

P( 10 unbroken)  P( 2 broken)  0.95.

You would expect roughly 450 of the graduates to donate nothing, roughly 300 to donate $10, roughly 200 to donate $25, and roughly 50 to donate $50. The frequencies would be close to, but not exactly, these values. The four frequencies would add to 1000.

The most common value of x is 0.

P(x ≥ 25) = 0.20 + 0.05 = 0.25.

P(x > 0) = 0.30 + 0.20 + 0.05 = 0.55.

The probability that everyone who shows up can be accommodated is P ( x  100)  0.05  0.1  0.12  0.14  0.24  0.17  0.82.

7.11

7.12

7.13

Chapter 7: Random Variables and Probability Distributions

7.14

1  0.82  0.18.

For the person who is number 1 on the standby list to get a place on the flight, 99 or fewer people must turn up for the flight. The probability that this happens is P( x  99)  0.05  0.1  0.12  0.14  0.24  0.65. For the person who is number 3 on the standby list to get a place on the flight, 97 or fewer people must turn up for the flight. The probability that this happens is P ( x  99)  0.05  0.1  0.12  0.27.

(1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5)

b x 2 1 1 1 1 1 1 0 0 0

outcome (1, 2) (1, 3) (1, 4) (1, 5) (2, 3) (2, 4) (2, 5) (3, 4) (3, 5) (4, 5)

The probability distribution of x is shown in the table below. x p(x) 7.15

0 0.3

1 0.6

2 0.1

Results of the simulation will vary.

7.16 outcome DDD DDI DID DII IDD IDI IID III

x 3 2 2 1 2 1 1 0

Probability (0.7)(0.7)(0.7) = 0.343 (0.7)(0.7)(0.3) = 0.147 (0.7)(0.3)(0.7) = 0.147 (0.7)(0.3)(0.3) = 0.063 (0.3)(0.7)(0.7) = 0.147 (0.3)(0.7)(0.3) = 0.063 (0.3)(0.3)(0.7) = 0.063 (0.3)(0.3)(0.3) = 0.027

The probability distribution of x is shown in the table below. x p(x)

0 0.027

1 0.189

2 0.441

3 0.343

169

170 7.17

Chapter 7: Random Variables and Probability Distributions a

The sixteen possible outcomes, their probabilities, and the associated values of x, are shown in the table below. Outcome SSSS SSSF SSFS SSFF SFSS SFSF SFFS SFFF FSSS FSSF FSFS FSFF FFSS FFSF FFFS FFFF

Probability (0.2)(0.2)(0.2)(0.2) = 0.0016 (0.2)(0.2)(0.2)(0.8) = 0.0064 (0.2)(0.2)(0.8)(0.2) = 0.0064 (0.2)(0.2)(0.8)(0.8) = 0.0256 (0.2)(0.8)(0.2)(0.2) = 0.0064 (0.2)(0.8)(0.2)(0.8) = 0.0256 (0.2)(0.8)(0.8)(0.2) = 0.0256 (0.2)(0.8)(0.8)(0.8) = 0.1024 (0.8)(0.2)(0.2)(0.2) = 0.0064 (0.8)(0.2)(0.2)(0.8) = 0.0256 (0.8)(0.2)(0.8)(0.2) = 0.0256 (0.8)(0.2)(0.8)(0.8) = 0.1024 (0.8)(0.8)(0.2)(0.2) = 0.0256 (0.8)(0.8)(0.2)(0.8) = 0.1024 (0.8)(0.8)(0.8)(0.2) = 0.1024 (0.8)(0.8)(0.8)(0.8) = 0.4096

x 4 3 3 2 3 2 2 1 3 2 2 1 2 1 1 0

The probability distribution of x is given in the table below. x p(x)

7.18

0 0.4096

1 0.4096

2 0.1536

There are two most likely values of x: 0 and 1.

P ( x  2)  0.1536  0.0256  0.0016  0.1808.

3 0.0256

4 0.0016

The value of w for each pair of slips picked is shown in the table below. $1 $1 $1 1 $1 1 1 $1 $10 10 10 $25 25 25

$1 $10 $25

10 25

Each of the ten outcomes shown above is equally likely. Thus the probability distribution of w is as given in the table below. w p(w) 7.19

1 0.3

10 0.3

25 0.4

The smallest possible y value is 1, and the corresponding outcome is S. The second smallest y value is 2, and the corresponding outcome is FS.

Chapter 7: Random Variables and Probability Distributions b

The set of positive integers

P ( y  1)  P (S)  0.7 P ( y  2)  P (FS)  (0.3)(0.7)  0.21. P ( y  3)  P (FFS)  (0.3)(0.3)(0.7)  0.063. P( y  4)  P(FFFS)  (0.3)(0.3)(0.3)(0.7)  0.0189. P ( y  5)  P(FFFFS)  (0.3)(0.3)(0.3)(0.3)(0.7)  0.00567.

171

The formula is p( y)  (0.3) y 1 (0.7), for y  1, 2, 3, 7.20

The probability distribution of y (with probabilities in terms of k) is as shown below. y p(y)

1 k

2 2k

3 3k

4 4k

5 5k

Since all the probabilities must add to 1 we have k  2k  3k  4k  5k  1, and so 15k  1. Thus k  1 15 . b

Since k  1 15 , the probability distribution of y is as shown below. y p(y)

1 1/15

2 2/15

3 3/15

4 4/15

Thus, P(at most three forms are required)  P ( y  3) 

5 5/15

1 2 3    0.4. 15 15 15

2 3 4    0.6. 15 15 15

P(between 2 and 4 inclusive)  P(2  y  4) 

If p( y )  y 2 50 then the probability distribution of y would be as shown below. y p(y)

1 1/50

2 4/50

3 9/50

4 16/50

5 25/50

However, the probabilities would then not add to 1, and so the given probability distribution is not possible. 7.21

W 2nd Magazine

T F S

W y=0 prob = 0.16 y=1 prob = 0.12 y=2 prob = 0.08 y=3 prob = 0.04

1st Magazine T F y=1 y=2 prob = 0.12 prob = 0.08 y=1 y=2 prob = 0.09 prob = 0.06 y=2 y=2 prob = 0.06 prob = 0.04 y=3 y=3 prob = 0.03 prob = 0.02

S y=3 prob = 0.04 y=3 prob = 0.03 y=3 prob = 0.02 y=3 prob = 0.01

172

Chapter 7: Random Variables and Probability Distributions The probability distribution of y is shown below. y p(y)

7.22

0 0.16

7.23

7.24

P( x  5)  1 10  5  0   0.5.

P(3  x  5)  1 10  5  3  0.2.

1 0.33

2 0.32

3 0.19

Chapter 7: Random Variables and Probability Distributions

7.25

173

P(2  x  3)  P(2  x  3)  1 10  (3  2)  0.1. P( x  2)  1 10  (2  0)  0.2. P( x  7)  1 10  (10  7)  0.3. Thus, P (2  x  3)  P (2  x  3)  P ( x  2)  P( x  7).

7.26

The height is

1 400

 0.0025 ; Note that the area of the rectangular region under the

density curve and above the x-axis should equal 1. The area of a rectangular region is (height)(base), so (0.0025)(400)  1 , as expected. 7.27

7.28

P ( x  100)  (100)(0.0025)  0.25

P (200  x  300)  (0.0025)(300  200)  (0.0025)(100)  0.25

P (50  x  150)  (0.0025)(150  50)  (0.0025)(100)  0.25

These probabilities are the same because the density curve is uniform (always the same height) and the x-values of interest span the same range (the change in x is 100 in both cases). The rectangular regions under the density curve are the same size.

1 3   1  1  P  x    1     (1)  . 2 2 2 4     

1 1 3   P x    P x    . 2 2 4  

1   1  3  3  7 P  x    1       . 4   2  4  2  16

174

7.29

7.30

7.31

Chapter 7: Random Variables and Probability Distributions

1 1 1 3 7 5 1   P  x    P x    P x      . 2 2 4  4 16 16 4  

1   1  1  1  We need P  x       (1)  . 2   2  2  4 

1   1  3  3  9  We need P  x         . 4   2  4  2  16 

P ( x  10)  0.05(10  0)  0.5. P( x  15)  0.05(20  15)  0.25. P(7  x  12)  0.05(12  7)  0.25.

0.05(c  0)  0.9, so c  0.9 0.05  18.

Area 

P( w  20) 

1 P( w  10)  (10)(0.025)  0.125. 2

1 P( w  30)  (10)(0.025)  0.125. 2

1  40  0.05  1 , as required. 2 1  20  0.05  0.5. 2

7.32

P (10  w  30)  1  P ( w  10)  P( w  30)  1  0.125  0.125  0.75.

7.33

μ x = (1)(.140) + (2)(.175) + (3)(.220) + (4)(.260) + (5)(.155) + (6)(.025) + (7)(.015) + (8)(.005) + (9)(.004) + (10)(.001) = 3.306.

7.34

 x  0(0.54)  1(0.16)  2(0.06)  3(0.04)  4(0.2)  1.2.

P( x   x )  P( x  1.2)  P( x  2)  0.06  0.04  0.2  0.3.

 y  0(0.65)  1(0.2)  2(0.1)  3(0.04)  4(0.01)  0.56. Over a large number of cartons, the

7.35

mean number of broken eggs per carton will be 0.56. b

P( y   y )  P( y  0.56)  P( y  0)  0.65. In 65% of cartons, the number of broken eggs will be less than  y . This result is greater than 0.5 because the distribution of y is positively skewed.

Chapter 7: Random Variables and Probability Distributions

175

The mean is not (0  1  2  3  4) 5 since, for example, far more cartons have 0 broken eggs than 4 broken eggs, and so 0 needs a much greater weighting than 4 in the calculation of the mean.

7.36

z  12  y , so  z  12   y  12  0.56  11.44.

7.37

0.02 + 0.03 + 0.09 = 0.14.

  2  4.66  2(1.2)  2.26 ;   2  4.66  2(1.2)  7.06. The values of x more than 2 standard deviations away from the mean are 1 and 2.

7.38

c 0.02 + 0.03 = 0.05. Let x be the number of flaws in a randomly selected panel from the first supplier, and let y be the number of flaws in a randomly selected panel from the second supplier. As shown in Examples 7.11 and 7.12,  x  1,  y  1,  x  1, and  y  0.632. Therefore, in terms of the mean number of flaws per panel, the two suppliers are equal. However, since the standard deviation for the second supplier is smaller than that for the first, the second supplier is more consistent in terms of the number of flaws per panel than the first supplier. It could be argued that greater consistency is desirable in this context, and so for this reason the second supplier should be recommended.

7.39

 x  0(0.05)  1(0.1) 

 5(0.1)  2.8.



 (5  2.8)2 (0.1)  1.66.

2 2 x  (0  2.8) (0.5) 

So  x  1.66  1.288. b

 y  0(0.5)  1(0.3)  2(0.2)  0.7.

 y2  (0  0.7)2 (0.5)  (1  0.7)2 (0.3)  (2  0.7) 2 (0.2)  0.61. So  y  0.61  0.781. 7.40

The amount of money collected in tolls from cars is (number of cars)($3) = 3x. 3 x  3x  3(2.8)  8.4.

 32x  32  x2  9(1.66)  14.94. b

The amount of money collected in tolls from buses is (number of buses)($10) = 10y. 10 y  10 y  10(0.7)  7.

 102 y  102  y2  100(0.61)  61. 7.41

z   x  y   x   y  2.8  0.7  3.5.

 z2   x2 y   x2   y2  1.66  0.61  2.27. b

w  3 x 10 y  3 x  10 y  8.4  7  15.4.

 w2   32x 10 y   32x   102 y  14.94  61  75.94. 7.42

x  1(0.05)  2(0.1)  3(0.12)  4(0.3)  5(0.3)  6(0.11)  7(0.01)  8(0.01)  4.12.

176

Chapter 7: Random Variables and Probability Distributions

 x2  (1  4.12)2 (0.05)  (2  4.12)2 (0.1)  (3  4.12)2 (0.12)  (4  4.12)2 (0.3) (5  4.12)2 (0.3)  (6  4.12)2 (0.11)  (7  4.12)2 (0.01)  (8  4.12)2 (0.01)  1.9456.

 x  1.9456  1.395. The mean squared deviation of the number of systems sold in a month from the mean number of systems sold in a month is 1.9456. A typical deviation of the number of systems sold in a month from the mean number of systems sold in a month is 1.395. c

7.43

 x   x  4.12  1.395  2.725.  x   x  4.12  1.395  5.515.

So we need P (2.725  x  5.515)  P(3  x  5)  0.12  0.3  0.3  0.72.  x  2 x  4.12  2(1.395)  1.33.  x  2 x  4.12  2(1.395)  6.91. So we need P ( x  1.33)  P ( x  6.91)  P ( x  1)  P ( x  7)  0.05  0.01  0.01  0.07.

E ( x)  15(0.1)  30(0.3)  60(0.6)  46.5 seconds.

The probability distribution of y is shown below. y p(y)

500 0.1

800 0.3

1000 0.6

E ( y )  500(0.1)  800(0.3)  1000(0.6)  890. The average amount paid is $890. 7.44

Since  x  1000(0.05)  5000(0.3)  10000(0.4)  20000(0.25)  10550, the author expects to make slightly more under the royalty plan than by accepting the flat payment of $10,000. However, since the royalty plan could result in payments as low as $1000 and as high as $20,000, the author might opt for the less risky (less variable) option of the flat payment.

7.45

For the first distribution shown below,  x  3 and  x  1.414, while for the second distribution  x  3 and  x  1.

7.46

x p(x)

1 0.2

2 0.2

3 0.2

4 0.2

5 0.2

x p(x)

1 0.1

2 0.1

3 0.6

4 0.1

5 0.1

x  13.5(0.2)  15.9(0.5)  19.1(0.3)  16.38.

 x2  (13.5  16.38)2 (0.2)  (15.9  16.38) 2 (0.5)  (19.1  16.38)2 (0.3)  3.9936.  x  3.9936  1.998. b

Price  25(16.38)  8.5  401. The mean of the price paid by the next customer is $401.

Chapter 7: Random Variables and Probability Distributions

177

 Price  25(1.998)  49.960. The standard deviation of the price paid by the next customer is $49.960.

7.47

7.48

Whether y is positive or negative tells us whether or not the peg will fit into the hole.

 y   x2   x1  0.253  0.25  0.003.

 y   x21   x22  (0.006)2  (0.002)2  0.00632.

Since 0 is less than half a standard deviation from the mean in the distribution of y, it is relatively likely that a value of y will be negative, and therefore that the peg will be too big to fit the hole.

7.49

1 1 4 4 If a student guesses the answer to every question, we would expect 1/5 of the questions to be answered correctly and 4/5 to be answered incorrectly. Under this scheme, the student is awarded one point for every correct answer and −1/4 of a point for every incorrect answer. So if the test contains 50 questions then if the student guesses completely at random we expect 10 correct answers and 40 incorrect answers, and thus a total score of 0.

 y   x1   x2  10  (40)  0.

Since x1  50  x2 , the value of x1 is completely determined by the value of x2 , and thus x1 and x2 are not independent. The formulas in this section for computing variances and standard deviations of combinations of random variables require the random variables to be independent.

xR  11 6 

 6 1 6  3.5.

 x2R  (1  3.5)2 1 6  

 (6  3.5)2 1 6   2.917.

 xR  2.917  1.708.

7.50

 xB  3.5,  x2B  2.917, and  xB  1.708.

w1   xR   xB  7  3.5  3.5  7  0.

 w1   x2R   x2B  2.917  2.917  2.415. b

w2  3xR  3xB w2  3 xR  3 xB  3(3.5)  3(3.5)  0.

 w2  9 x2R  9 x2B  9(2.917)  9(2.917)  7.246. c

For both of the games the mean of the money gained is 0. The choice of games is therefore governed by the amount of risk the player is willing to undergo. If you are willing to win or lose larger amounts, Game 2 should be chosen. If less risk is desired, Game 1 should be chosen.

178

7.51

Chapter 7: Random Variables and Probability Distributions

The distribution of x is binomial with n = 10 and p = 4% = 0.05. a

P ( x  3)  P ( x  0)  P ( x  1)  P ( x  2) 10! 10!  (0.96)10  (0.04)1 (0.96)9  (0.04) 2 (0.96)8 1!9! 2!8! = 0.993786

P( x  3)  P( x  0)  P( x  1)  P( x  2)  P( x  3)  (0.96)10 

10! 10! 10! (0.04)1 (0.96)9  (0.04) 2 (0.96)8  (0.04)3 (0.96)7 1!9! 2!8! 3!7!

= 0.999557

7.52

7.53

P( x  4)  1  P ( x  3)  1  0.999557  0.000443

P (1  x  3)  P ( x  1)  P ( x  2)  P ( x  3) 10! 10! 10!  (0.04)1 (0.96)9  (0.04)2 (0.96)8  (0.04)3 (0.96)7 1!9! 2!8! 3!7!  0.334725

Let x be the number of vehicles out of 15 that involve a single vehicle. Then x is binomially distributed with n = 15 and p = 0.7.

15! (0.7) 4 (0.3)11  0.001 . 4!11!

P( x  4) 

Using Appendix Table 9, P( x  4)  0.001  0.000 

P(exactly 6 involve multiple vehicles)  P ( x  9) 

p(4)  P( x  4) 

4!(6  4)! samples of 6 passengers will contain exactly 4 passengers who travel with a smart phone.

p(6)  P( x  6) 

P ( x  4)  p (4)  p (5)  p (6) . From parts (a) and (b), p (4)  0.2458 and p (6)  0.2621 .

p(5) 

6! 5!(6  5)!

6!(6  6)!

p (8)  0.302.

15! (0.7)9 (0.3)6  0.147 . 9!6!

(0.8) 4 (1  0.8) 2  0.2458 . In the long run, 24.58% of random

(0.8)6 (1  0.8)0  0.2621

 0.85 1  0.81  0.3932 . Therefore,

P ( x  4)  0.2458  0.2621  0.3932  0.9011 .

7.54

 0.000  0.001 .

Chapter 7: Random Variables and Probability Distributions

7.55

7.56

179

P ( x  7)  1  P( x  8)  1  (0.302  0.268  0.107)  0.323.

P (more than half rested or slept)  P ( x  6)  0.088  0.201  0.302  0.268  0.107  0.966.

p(2)  5! (2!3!)  (0.25)2 (0.75)3  0.264.

P( x  1)  p(0)  p(1)  0.23730  0.39551  0.633.

P(2  x )  P( x  2)  1  P( x  1)  1  0.633  0.367.

P ( x  2)  1  P( x  2)  1  0.264  0.736.

0, 1, 2, 3, 4, 5.

The probabilities in the table below were calculated using the formula 5! p( x )  (0.25) x (0.75)5 x . x!(5  x)! x 0 1 2 3 4 5

p(x) 0.237 0.396 0.264 0.088 0.015 0.001

c p(x) 0.4

0.3

0.2

0.1

0.0 0

3 x

180

7.57

Chapter 7: Random Variables and Probability Distributions

P ( x  10)  p (11)  p (12)  

20! 11!(20  11)!

 p (19)  p (20)

 0.6 11 1  0.6 9 



20! 20!(20  20)!

 0.6 20 1  0.6 0

 0.755

P ( x  15)  p (16)  p(17)  p(18)  p (19)  p(20) 

20! 16!(20  16)!

 0.6 16 1  0.6 4 



20! 20!(20  20)!

 0.6 20 1  0.6 0

 0.051 and

P ( x  5)  p (0)  p (1)  p (2)  p (3)  p (4) 

20! 0!(20  0)!

 0.6 0 1  0.6 20 



20! 4!(20  4)!

 0.6 4 1  0.6 16

 0.000317 Therefore, more than 15 have security solutions is more likely than fewer than 5 having security solutions. 7.58

The probabilities in the table below are calculated using the formula 5! P( x  r )  (0.5)r (0.5)5r . r !(5  r )! x p(x)

7.59

0 0.03125

1 0.15625

2 0.3125

3 0.3125

4 0.15625

5 0.03125

The probability distribution of x is geometric, with success probability p = 0.44. The distribution is geometric because we are waiting until we find someone that washes sheets at least once a week.

P( x  3)  (1  0.44) 2 (0.44)  (0.56) 2 (0.44)  0.138

P( x  4)  p(1)  p(2)  p(3)  0.44  (0.56)(0.44)  (0.56) 2 (0.44)  0.824

P ( x  3)  1  P ( x  3)  1   p (1)  p (2)  p (3)   1  0.44  (0.56)(0.44)  (0.56) 2 (0.44)   0.176

Chapter 7: Random Variables and Probability Distributions 7.60

7.61

7.62

n  20, p  0.05. Using Appendix Table 9, P(acceptance)  p(0)  p(1)  0.358  0.377  0.735.

n  20, p  0.1. Using Appendix Table 9, P(acceptance)  p(0)  p(1)  0.122  0.270  0.392.

n  20, p  0.2. Using Appendix Table 9, P(acceptance)  p(0)  p(1)  0.012  0.058  0.070.

Expected number showing damage  2000(0.1)  200.

Standard deviation  2000(0.1)(0.9)  13.416.

181

Let x = number of cars failing the inspection. a n  15, p  0.3. Using Appendix Table 9, P( x  5)  0.206  0.219  0.170  0.092  0.031  0.005  0.723. b

n  15, p  0.3. Using Appendix Table 9, P (5  x  10)  0.206  0.147  0.081  0.035  0.012  0.003  0.484.

Let y = number that pass the inspection. Then y is binomially distributed with n  25 and p  0.7.  x  np  25(0.7)  17.5.  x  np(1  p)  25(0.7)(0.3)  2.291.

 x   x  17.5  2.291  15.209.  x   x  17.5  2.291  19.791. So, using Appendix Table 9, the required probability is P(15.209  x  19.791)  P(16  x  19)  0.134  0.165  0.171  0.147  0.617.

7.63

Binomial distribution with n  100 and p  0.2

Expected score  100(0.2)  20.

 x2  100(0.2)(0.8)  16,  x  16  4.

A score of 50 is (50  20) 4  7.5 standard deviations from the mean in the distribution of x. So a score of over 50 is very unlikely.

7.64

Since 2000/10000 = 0.2, 20% of the population is being sampled, which is greater than 5%. Thus the binomial distribution would not be a good model for the number of invalid signatures.

7.65

Using Appendix Table 9, P (program is implemented)  P ( x  15)  0.012  0.004  0.001  0.017.

Using Appendix Table 9, if p  0.7, P (program is implemented)  P( x  15)  0.091  0.054  0.027  0.011  0.004  0.002  0.189. So P(program is not implemented)  1  0.189  0.811.

182

7.66

7.67

Chapter 7: Random Variables and Probability Distributions

Using Appendix Table 9, if p  0.6, P (program is not implemented)  P( x  15)  0.151  0.120  0.080  0.045  0.020  0.007  0.002  0.425.

The error probability when p  0.7 is now 0.811  p (15)  0.811  0.092  0.903. The error probability when p  0.6 is now 0.424  p(15)  0.424  0.161  0.585.

Let x = number of voters who favor the ban. Then x is binomially distributed with n  25 and p  0.9. a

Using Appendix Table 9, P ( x  20)  0.138  0.226  0.266  0.199  0.072  0.901.

Using Appendix Table 9, P ( x  20)  0.065  0.138  0.226  0.266  0.199  0.072  0.966.

 x  np  25(0.9)  22.5.  x  np(1  p)  25(0.9)(0.1)  1.5.

Assuming that 90% of the populace favors the ban, P( x  20)  1  P( x  20)  1  0.966  0.034, which is small. This tells us that if 90% of the populace favored the ban it would be unlikely that fewer than 20 people in the sample would favor the ban. Thus if fewer than 20 people out of a sample of 25 favored the ban we would reject the assertion that 90% of the populace favors the ban.

For a random variable to be binomially distributed, it must represent the number of “successes” in a fixed number of trials. This is not the case for the random variable x described.

The distribution of x is geometric with p  0.08 . i

p(4)  (0.92)3 (0.08)  0.062.

P( x  4)  p(1)  p(2)  p(3)  p(4)  0.08  (0.92)(0.08)  (0.92) 2 (0.08)  (0.92)3 (0.08)  0.284. iii P ( x  4)  1  P( x  4)  1  0.284  0.716. iv P ( x  4)  p(4)  P( x  4)  0.06230  0.71639  0.779. The process described is for songs to be randomly selected until a song by the particular artist is played. i If the process were to be repeated many times, on 6.2% of occasions exactly four songs would be played up to and including the first song by this artist. ii If the process were to be repeated many times, on 28.4% of occasions at most four songs would be played up to and including the first song by this artist. iii If the process were to be repeated many times, on 71.6% of occasions more than four songs would be played up to and including the first song by this artist. iv If the process were to be repeated many times, on 77.9% of occasions at least four songs would be played up to and including the first song by this artist. The differences between the four probabilities lie in the underlined phrases. ii

7.68

Geometric

(0.9)(0.1) = 0.09

Chapter 7: Random Variables and Probability Distributions

7.69

183

The probability that it takes Sophie more than three tosses to catch the ball is the probability that Sophie fails to catch each of the first three tosses, which is (0.9)(0.9)(0.9) = 0.729.

P( x  2)  p(1)  p(2)  0.05  (0.95)(0.05)  0.0975.

p(4)  (0.95)3 (0.05)  0.043.

P( x  4)  1  P( x  4)  1  (0.05)  (0.95)(0.05)  (0.95) 2 (0.05)  (0.95)3 (0.05)  0.815.





(Alternatively, note that for more than four boxes to be purchased, the first four boxes bought must not contain a prize. So P( x  4)  (0.95)4  0.815. ) 7.70

7.71

7.72

7.73

7.74

0.9599

0.2483

1  0.8849  0.1151.

1  0.0024  0.9976.

0.7019  0.0132  0.6887

0.8413  0.1587  0.6826.

1.0000

0.1003

0.9772  0.1587  0.8185

0.5

1.0000

0.9938  0.0548  0.9390

0.5910

0.9909

0.1093

0.9996  0.8729  0.1267

184 7.75

7.76

7.77

7.78

7.79

7.80

Chapter 7: Random Variables and Probability Distributions a

0.2912  0.2206  0.0706

1  0.9772  0.0228

1  0.0004  0.9996

1.0000

0.5398

0.4602

0.8023  0.6554  0.1469

0.1469

0.8023  0.3446  0.4577

1  0.1056  0.8944

0.0668  (1  0.9938)  0.0730

1.96

2.33

1.645

P ( z  z*)  1  0.02  0.98. So z*  2.05.

P ( z  z*)  1  0.01  0.99. So z*  2.33.

P( z  z*)  0.2 2  0.1. So P ( z  z*)  1  0.1  0.9. Therefore z*  1.28.

If P ( z  z*)  0.03 then P ( z  z*)  0.97. So z*  1.88.

If P( z  z*)  0.01 then P ( z  z*)  0.99. So z*  2.33.

If P ( z  z*)  0.04 then z*  1.75.

If P ( z  z*)  0.1 then z*  1.28.

P( z  z*)  0.05 2  0.025. So P( z  z*)  1  0.025  0.975. So z*  1.96.

P( z  z*)  0.1 2  0.05. So P( z  z*)  1  0.05  0.95. So z*  1.645.

P( z  z*)  0.02 2  0.01. So P( z  z*)  1  0.01  0.99. So z*  2.33.

Chapter 7: Random Variables and Probability Distributions

7.81

7.82

P( z  z*)  0.08 2  0.04. So P( z  z*)  1  0.04  0.96. So z*  1.75.

P ( z  z*)  0.91, so z*  1.34.

P ( z  z*)  0.77, so z*  0.74.

P ( z  z*)  0.5, so z*  0.

P ( z  z*)  0.09, so z*  1.34.

The 30th percentile is negative the 70th percentile.

P( x  5)  P  z  (5  5) 0.2   P( z  0)  0.5.

P( x  5.4)  P  z  (5.4  5) 0.2   P( z  2)  0.9772.

P( x  5.4)  P ( x  5.4)  0.9772.

P(4.6  x  5.2)  P  (4.6  5) 0.2  z  (5.2  5) 0.2   P( 2  z  1)  0.8413  0.0228  0.8185. P( x  4.5)  P  z  (4.5  5) 0.2   P( z  2.5)  1  P( z  2.5)  1  0.0062  0.9938.

P( x  4.0)  P  z  (4.0  5) 0.2   P( z  5)  1  P( z  5)  1  0.0000  1.0000.

P( x  4000)   z  (4000  3500) 600   P( z  0.83)  0.2033.

P(3000  x  4000)  P  (3000  3500) 600  z  (4000  3500) 600   P( 0.83  z  0.83)  0.7967  0.2033  0.5934.

P( x  2000)  P  z  (2000  3500) 600   P( z  2.5)  0.0062.

7.83

185

P( x  5000)  P  z  (5000  3500) 600   P( z  2.5)  0.0062. So P ( x  2000 or x  5000)  0.0062  0.0062  0.0124. 7.84

7 lb = 7(453.59) g = 3175.13 g. P( x  3175.13)  P  z  (3175.13  3500) 600   P( z  0.54)  0.7054.

If P ( z  z*)  0.0005 then z*  3.29. So the most extreme 0.1% of birth weights are at least 3.29 standard deviations above or below the mean. Now 3500 + 3.29(600) = 5474 and 3500  3.29(600)  1526. So the most extreme 0.1% of birth weights consist of those greater than 5474 and those less than 1526 grams.

If y is the weight of a baby in pounds, then y  x 453.59. So the mean of y is 3500 453.59  7.71622 , the standard deviation of y is 600 453.59  1.32278, and y is normally distributed.

186

Chapter 7: Random Variables and Probability Distributions

P( y  7)  P  z  (7  7.71622) 1.32278   P( z  0.54)  0.7054. This is the same as the probability calculated in Part (c). 7.85

If P( z  z*)  0.1 then P( z  z*)  0.9; so z*  1.28. Thus x    ( z*)  1.6  (1.28)(0.4)  2.113. The worst 10% of vehicles are those with emission levels greater than 2.113 parts per billion.

7.86

For a distribution with mean 9.9 and standard deviation 6.2, the value 0 is around 1.6 standard deviations below the mean. Therefore, if the distribution were normal, a substantial proportion of observations would be negative. Clearly you can’t have a negative processing time, and so the normal distribution cannot be an appropriate model for this variable.

7.87

Let the left atrial diameter be x. P( x  24)  P  z  (24  26.4) 4.2   P( z  0.57)  0.2843.

P( x  32)  P  z  (32  26.4) 4.2   P( z  1.33)  0.0918.

P(25  x  30)  P  (25  26.4) 4.2  z  (30  26.4) 4.2   P( 0.33  z  0.86)  0.8051  0.3707  0.4344.

If P( z  z*)  0.2, then P( z  z*)  0.8; so z*  0.84. Thus x    ( z*)  26.4  (0.84)(4.2)  29.928 mm. Let the left atrial diameter be x. a P( x  25)  P  z  (25  28) 4.7   P( z  0.64)  0.2611. d

7.88

7.89

7.90

P( x  32)  P  z  (32  28) 4.7   P( z  0.85)  1  P( z  0.85)  1  0.8023  0.1977.

P(25  x  30)  P  (25  28) 4.7  z  (30  28) 4.7   P( 0.64  z  0.43)  0.6664  0.2611  0.4053.

P( x  26.4)  P  z  (26.4  28) 4.7   P( z  0.34)  1  P( z  0.34)  1  0.3669  0.6331.

Let the number of days to handle graffiti complaints be x.

 

60  114 

P( x  60)  P  z 

P( x  120)  P  z 

 

  P( z  2.7)  0.00347 

120  114  20

  P( z  0.3)  0.382 

Let the diameter of the cork produced be x. P(2.9  x  3.1)  P  (2.9  3) 0.1  z  (3.1  3) 0.1  P( 1  z  1)  0.8413  0.1587  0.6826. So the proportion of corks that are defective is 1  0.6826  0.3174.

Chapter 7: Random Variables and Probability Distributions 7.91

Let the diameter of the cork produced be x. P(2.9  x  3.1)  P  (2.9  3.05) 0.01  z  (3.1  3.05) 0.01  P( 15  z  5)  1.0000. A cork made by the machine in this exercise is almost certain to meet the specifications. This machine is therefore preferable to the one in the Exercise 7.78.

7.92

P( x  30)  P  z 

P(15  x  25)  P 

Let x1 and x2 represent the two randomly selected purchases. Therefore,

 

187

30  21.1  7

  P ( z  1.27143)  0.102 

25  21.1   15  21.1 z   P(0.871  z  0.557)  0.5195 7 7  

P( x1  25  x2  25)  P( x1  25)  P( x2  25)  (0.288715)(0.288715)  0.0834 . 7.93

The fastest 10% of applicants are those with the lowest 10% of times. If P ( z  z*)  0.1, then z*  1.28. The corresponding time is   ( z*)  120  ( 1.28)(20)  94.4. Those with times less than 94.4 seconds qualify for advanced training.

7.94

0.5

90  60   45  60 P(45  x  90)  P  z  P( 1  z  2)  0.9772  0.1587  0.8185. 15   15

7.95

105  60   P( x  105)  P  z   P( z  3)  1  P( z  3)  1  0.9987  0.0013. 15  

7.96

75  60   P( x  75)  P  z   P( z  1)  0.1587. So the probability that both typists have 15   rates exceeding 75 wpm is (0.1587) 2  0.025. P ( z  z*)  0.2, then z  0.84. The corresponding typing speed is   ( z*)  60  ( 0.84)(15)  47.4. People with typing speeds of 47 wpm and below would qualify for the training.

188

7.97

Chapter 7: Random Variables and Probability Distributions

a Fussing T ime 14 12 10 8 6 4 2 0 -2

-1

0 Normal Score

The clear curve in the normal probability plot tells us that the distribution of fussing times is not normal.

The square roots of the data values are shown in the table below.

Chapter 7: Random Variables and Probability Distributions

Fussing Time 0.05 0.10 0.15 0.40 0.70 1.05 1.95 2.15 3.70 3.90 4.50 6.00 8.00 11.00 14.00

Normal Score -1.739 -1.245 -0.946 -0.714 -0.515 -0.333 -0.165 0.000 0.165 0.335 0.515 0.714 0.946 1.245 1.739

189

sqrt(Fussing Time) 0.22361 0.31623 0.38730 0.63246 0.83666 1.02470 1.39642 1.46629 1.92354 1.97484 2.12132 2.44949 2.82843 3.31662 3.74166

sqrt(Fussing T ime) 4

0 -2

-1

0 Normal Score

The transformation results in a pattern that is much closer to being linear than the pattern in Part (a).

190

7.98

Chapter 7: Random Variables and Probability Distributions

a Risk Behavior Score 160 150 140 130 120 110 100 -2

-1

0 Normal Score

-1

0 Normal Score

b PANAS Score 70 65 60 55 50 45 40 35 -2

Yes. Since the patterns in both plots are roughly linear it seems reasonable to assume that both distributions are approximately normal.

Chapter 7: Random Variables and Probability Distributions

7.99

191

a Frequency 7 6 5 4 3 2 1 0

No. The distribution of x is positively skewed.

c Frequency 9 8 7 6 5 4 3 2 1 0

1.68

1.72

1.76 log(x)

1.80

1.84

Yes. The histogram shows a distribution that is slightly closer to being symmetric than the distribution of the untransformed data.

192

Chapter 7: Random Variables and Probability Distributions

7.100 a Frequency 7 6 5 4 3 2 1 0

6.9

7.2

7.5 sqrt(x)

7.8

8.1

Both transformations produce histograms that are closer to being symmetric than the histogram of the untransformed data, but neither transformation produces a distribution that is truly close to being normal.

7.101 a sFasL Level 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 -2

-1

0 Normal Score

The normal probability plot appears curved.

Chapter 7: Random Variables and Probability Distributions

193

c Cube root of x 0.9

0.8

0.7

0.6

0.5

0.4 -2

-1

0 Normal Score

Yes. This normal probability plot appears more linear than the plot for the untransformed data. 7.102 No, it is not reasonable to think that the distribution of 2015 Honda Accord prices in this area is approximately normal because of the curvature in the pattern of the points apparent in the normal probability plot. 7.103 Lifetime (hrs) 450 400 350 300 250 200 150 -2

-1

0 Normal Score

Since the normal probability plot shows a clear curve, the normal distribution is not a plausible model for power supply lifetime. (However, it is worth noting that most of the apparent curved pattern is brought about by the single point with coordinates (1.539, 422.6).)

194

Chapter 7: Random Variables and Probability Distributions

7.104 16.3

Disk Diameter

16.2 16.1 16.0 15.9 15.8 15.7 15.6 -2.5

-2.0

-1.5

-1.0

-0.5 0.0 0.5 Normal Score

1.0

1.5

2.0

2.5

Since the pattern in the normal probability plot is very close to being linear, it is plausible that disk diameter is normally distributed. 7.105 Frequency 9 8 7 6 5 4 3 2 1 0

0.6

0.9

1.2 Cuberoot(precipitation)

1.5

1.8

The cube-root transformation appears to result in the more symmetrical histogram.

Chapter 7: Random Variables and Probability Distributions

7.106 a Frequency 900 800 700 600 500 400 300 200 100 0

75 95 115 Number of Purchases

135

155

The histogram is positively skewed. b Interval

Frequency

√10 to <√20 √20 to <√30 √30 to <√40 √40 to <√50 √50 to <√60 √60 to <√70 √70 to <√80 √80 to <√90 √90 to <√100 √100 to <√110 √110 to <√120 √120 to <√130 √130 to <√140 √140 to <√150 √150 to <√160 √160 to <√170

904 500 258 167 94 56 26 20 13 9 7 6 6 3 0 2

Relative Frequency  Freq. 2071 0.437 0.241 0.125 0.081 0.045 0.027 0.013 0.010 0.006 0.004 0.003 0.003 0.003 0.001 0.000 0.001

Interval Width 1.310 1.005 0.847 0.747 0.675 0.621 0.578 0.543 0.513 0.488 0.466 0.447 0.430 0.415 0.402 0.389

Density Rel. Freq.  Int. Wdth. 0.333 0.240 0.147 0.108 0.067 0.044 0.022 0.018 0.012 0.009 0.007 0.006 0.007 0.003 0.000 0.002

195

196

Chapter 7: Random Variables and Probability Distributions

Density 0.35 0.30 0.25 0.20 0.15 0.10 0.05 0.00

7 9 sqrt(Number of Purchases)

No. The transformation has resulted in a histogram which is still clearly positively skewed. 7.107 a Density 0.05

0.04

0.03

0.02

0.01

0.00 0 0 0 0 1 2 3

0 50

Body Mass (grams)

Yes. If a symmetrical distribution is required, a transformation is desirable. b

The logarithms of the body mass values and the resulting histogram are shown below. Body Mass 7.7 10.1 21.6 8.6 12.0 11.4 16.6 9.4

Log(Body Mass) 0.88649 1.00432 1.33445 0.93450 1.07918 1.05690 1.22011 0.97313

1/sqrt(Body Mass) 0.360375 0.314658 0.215166 0.340997 0.288675 0.296174 0.245440 0.326164

Chapter 7: Random Variables and Probability Distributions 11.5 9.0 8.2 20.2 48.5 21.6 26.1 6.2 19.1 21.0 28.1 10.6 31.6 6.7 5.0 68.8 23.9 19.8 20.1 6.0 99.6 19.8 16.5 9.0 448.0 21.3 17.4 36.9 34.0 41.0 15.9 12.5 10.2 31.0 21.5 11.9 32.5 9.8 93.9 10.9 19.6 14.5

1.06070 0.95424 0.91381 1.30535 1.68574 1.33445 1.41664 0.79239 1.28103 1.32222 1.44871 1.02531 1.49969 0.82607 0.69897 1.83759 1.37840 1.29667 1.30320 0.77815 1.99826 1.29667 1.21748 0.95424 2.65128 1.32838 1.24055 1.56703 1.53148 1.61278 1.20140 1.09691 1.00860 1.49136 1.33244 1.07555 1.51188 0.99123 1.97267 1.03743 1.29226 1.16137

0.294884 0.333333 0.349215 0.222497 0.143592 0.215166 0.195740 0.401610 0.228814 0.218218 0.188646 0.307148 0.177892 0.386334 0.447214 0.120561 0.204551 0.224733 0.223050 0.408248 0.100201 0.224733 0.246183 0.333333 0.047246 0.216676 0.239732 0.164622 0.171499 0.156174 0.250785 0.282843 0.313112 0.179605 0.215666 0.289886 0.175412 0.319438 0.103197 0.302891 0.225877 0.262613

197

198

Chapter 7: Random Variables and Probability Distributions

Frequency 18 16 14 12 10 8 6 4 2 0

0.8

1.2

1.6 2.0 Log(Body Mass)

2.4

2.8

The distribution is closer to being symmetrical than the original distribution, but is nonetheless positively skewed. c

The transformed values are shown in the table in Part (c). The resulting histogram is shown below. Frequency 16 14 12 10 8 6 4 2 0

0.0

0.1

0.2 0.3 1/sqrt(Body Mass)

0.4

The shape in the histogram roughly resembles that of a normal curve. Certainly, this transformation was more successful than the other one in producing an approximately normal distribution. 7.108 Yes. In each case the transformation has resulted in a histogram that is much closer to being symmetric than the original histogram.

Chapter 7: Random Variables and Probability Distributions

7.109 a

100.5  100   99.5  100 P( x  100)  P  z   P( 0.03  z  0.03)  0.5120  0.4880 15 15    0.0240.

110.5  100   P( x  110)  P  z    P( z  0.7)  0.7580. 15  

109.5  100   P( x  110)  P  z    P( z  0.63)  0.7357. 15  

7.110 a

199

125.5  100   74.5  100 P(75  x  125)  P  z   P( 1.7  z  1.7)  0.9554  0.0446 15 15    0.9108. Let x = number of items. P( x  120)  P  z  (120.5  150) 10   P( z  2.95)  0.0016.

P( x  125)  P  z  (124.5  150) 10   P( z  2.55)  1  P( z  2.55)  1  0.0054  0.9946.

P(135  x  160)  P  (134.5  150) 10  z  (160.5  150) 10   P(1.55  z  1.05)  0.8531  0.0606  0.7925.

7.111 Let x = number of cars. 649.5  500   a P( x  650)  P  z    P( z  1.99)  0.0233. 75   b

549.5  500   400.5  500 P(400  x  550)  P  z   P( 1.33  z  0.66) 75 75    0.7454  0.0918  0.6536. 550.5  500   399.5  500 P(400  x  550)  P  z   P( 1.34  z  0.67) 75 75    0.7486  0.0901  0.6585.

7.112 a

P( x  30)  P  (29.5  30) 3.4641  z  (30.5  30) 3.4641  P( 0.14  z  0.14)  0.5557  0.4443  0.1114.

P( x  25)  P  (24.5  30) 3.4641  z  (25.5  30) 3.4641  P( 1.59  z  1.30)  0.0968  0.0559  0.0409.

P( x  25)  P  z  (25.5  30) 3.4641  P( z  1.30)  0.0968.

7.113 a

P(25  x  40)  P  (24.5  30) 3.4641  z  (40.5  30) 3.4641  P(1.59  z  3.03)  0.9988  0.0559  0.9429.

200

Chapter 7: Random Variables and Probability Distributions

7.114 a

P(25  x  40)  P  (25.5  30) 3.4641  z  (39.5  30) 3.4641  P( 1.30  z  2.74)  0.9969  0.0968  0.9001. Since np  60(0.7)  42  10 and n(1  p )  60(0.3)  18  10 the normal approximation to the binomial distribution is appropriate.

 x  60(0.7)  42 and  x  np(1  p)  60(0.7)(0.3)  3.550. 42.5  42   41.5  42 P( x  42)  P  z  P( 0.14  z  0.14)  0.5557  0.4443 i 3.550   3.550  0.1114.

ii iii

41.5  42   P( x  42)  P  z   P( z  0.14)  0.4443. 3.550   42.5  42   P( x  42)  P  z   P( z  0.14)  0.5557. 3.550  

The probability in (i) is an approximation to the probability that x is exactly 42. The probability in (ii) is an approximation to the sum of the probabilities of values of x less than 42. The probability in (iii) is an approximation to the sum of the probabilities of values of x less than or equal to 42. These are different quantities.

7.115 a

When p  0.96, n(1  p )  60(0.04)  2.4, which is less than 10, and so the normal approximation to the binomial distribution cannot be used. Therefore the binomial formula must be used when p  0.96.

For a person who is not faking the test, a score of 42 or less is extremely unlikely (a probability of 0.000000000013 ). Therefore, if someone does get a score of 42 or less, we have convincing evidence that the person is faking the test.

7.116 a

Let x = number of bikes out of 100 that are mountain bikes. x  np  100(0.7)  70, and  x  np(1  p)  100(0.7)(0.3)  4.5826. The normal approximation is valid since np  70  10 and n(1  p )  100(0.3)  30  10.

75.5  70   P( x  75)  P  z   P( z  1.20)  0.8849. 4.5826   b

75.5  70   59.5  70 P(60  x  75)  P  z  P( 2.29  z  1.20)  0.8849  0.0110 4.5826   4.5826  0.8739.

80.5  70   P( x  80)  P  z   P( z  2.29)  0.0110. 4.5826  

P (at most 30 are not mountain bikes)  P(at least 70 are mountain bikes)  P( x  70)

69.5  70    P z   P( z  0.11)  0.5438. 4.5826  

Chapter 7: Random Variables and Probability Distributions

201

7.117 The normal approximation is reasonable since np  100(0.25)  25  10 and n(1  p )  100(0.75)

 75  10 . Also,   np  25 and   np(1  p)  100(0.25)(0.75)  4.3301. a

P(20  x  30)  P  (19.5  25) 4.3301  z  (30.5  25) 4.3301  P(1.27  z  1.27)  0.8980  0.1010  0.7970.

P(20  x  30)  P  (20.5  25) 4.3301  z  (29.5  25) 4.3301  P( 1.04  z  1.04)  0.8508  0.1492  0.7016.

P( x  35)  P  z  (34.5  25) 4.3301  P( z  2.19)  1  P( z  2.19)  1  0.9857  0.0143.

  2  25  2(4.3301)  16.3398, and   2  25  2(4.3301)  33.6602. So P ( x is less than two st. devs. from mean)  P (16.3398  x  33.6602)  P (17  x  33)

 P  (16.5  25) 4.3301  z  (33.5  25) 4.3301  P(1.96  z  1.96)  0.9750  0.0250  0.9500. Therefore the probability that x is more than 2 standard deviations from its mean is 1  0.9500  0.05. 7.118 a

Let x = number of voters out of 225 who favor the waiting period. x  np  225(0.65)  146.25, and  x  np(1  p)  225(0.65)(0.35)  7.1545. The normal approximation is valid since np  146.25  10 and n(1  p )  225(0.35)  78.75  10.

149.5  146.25   P( x  150)  P  z    P( z  0.45)  0.3264. 7.1545   b

150.5  146.25   P( x  150)  P  z    P( z  0.59)  0.2776. 7.1545  

124.5  146.25   P( x  125)  P  z    P( z  3.04)  0.0012. 7.1545  

7.119 a b

No, since np  50(0.05)  2.5  10. Now n  500 and p  0.05, so np  500(0.05)  25  10. So the techniques of this section can be used. Using   np  25 and   np(1  p)  500(0.05)(0.95)  4.8734,

P(at least 20 are defective)  P  z  (19.5  25) 4.8734   P( z  1.13)  1  P( z  1.13)  1  0.1292  0.8708. 7.120 a

Let x = number of mufflers out of 400 that are replaced. x  np  400(0.2)  80.

 x  np(1  p)  400(0.2)(0.8)  8. The normal approximation is valid since np  80  10 and n(1  p )  400(0.8)  320  10.

202

Chapter 7: Random Variables and Probability Distributions

100.5  80   74.5  80 P(75  x  100)  P  z   P( 0.69  z  2.56)  0.9948  0.2451  0.7497. 8 8   b

7.121 a

70.5  80   P( x  70)  P  z    P( z  1.19)  0.1170. 8   49.5  80   P( x  50)  P  z    P( z  3.81)  0.0001. This tells us that fewer than 50 8   mufflers being replaced in a sample of 400 is unlikely, assuming that 20% of all mufflers are being replaced. Therefore, yes, the 20% figure would be questioned. P(250  x  300)  P  (250  266) 16  z  (300  266) 16   P( 1  z  2.125)  0.9832  0.1587  0.8245.

P( x  240)  P  z  (240  266) 16   P( z  1.625)  0.0521.

Sixteen days is 1 standard deviation, so we need P ( 1  z  1)  0.8413  0.1587  0.6826.

P( x  310)  P  z  (310  266) 16   P( z  2.75)  1  P( z  2.75)  1  0.9970  0.0030. This should make us skeptical of the claim, since it is very unlikely that a pregnancy will last at least 310 days. The insurance company will refuse to pay if the birth occurs within 275 days of the beginning of the coverage. If the conception took place after coverage began, then the insurance company will refuse to pay if the pregnancy is less than or equal to 275  14  261 days. P( x  261)  P  z  (261  266) 16   P( z  0.31)  0.3783.

7.122 The distribution of x is binomial with n  15 and p  0.6. Using Appendix Table 9, the required probability is P (5  x  10)  0.024  0.061  0.118  0.177  0.207  0.186  0.773. 7.123 a

P( x  3)  0.1  0.15  0.2  0.25  0.7.

P( x  3)  0.1  0.15  0.2  0.45.

P ( x  3)  1  P ( x  3)  1  0.45  0.55.

7.124 a

P (2  x  5)  0.2  0.25  0.2  0.06  0.71.

If 2 lines are not in use, then 4 lines are in use. If 3 lines are not in use, then 3 lines are in use. If 4 lines are not in use, then 2 lines are in use. So the required probability is P (2  x  4)  0.2  0.25  0.2  0.65.

If 4 lines are not in use, then 2 lines are in use. If 5 lines are not in use, then 1 line is in use. If 6 lines are not in use, then 0 lines are in use.

Chapter 7: Random Variables and Probability Distributions

203

So the required probability is P (0  x  2)  0.1  0.15  0.2  0.45. 7.125 a

x  0(0.1)  1(0.15) 

 6(0.04)  2.64.

 x2  (0  2.64)2 (0.1) 

 (6  2.64)2 (0.04)  2.3704.

 x  2.3704  1.540. b

x  3 x  2.64  3(1.540)  1.98.  x  3 x  2.64  3(1.540)  7.26. Since all the possible values of x lie between these two values, the required probability is 0.

7.126 a

p (2)  (0.8)(0.8)  0.64.

p(3)  P(UAA or AUA)  (0.2)(0.8)(0.8)  (0.8)(0.2)(0.8)  0.256.

For y to be 5, the fifth battery’s voltage must be acceptable. The four outcomes are UUUAA, UUAUA, UAUUA, AUUUA. So p(5)  4(0.2)(0.2)(0.2)(0.8)(0.8)  0.02048.

In order for it to take y selections to find two acceptable batteries, the first y  1 batteries must include exactly 1 acceptable battery (and therefore y  2 unacceptable batteries), and the yth battery must be acceptable. There are y  1 ways in which the first y  1 batteries can include exactly 1 acceptable battery. So p( y )  ( y  1)(0.8)1 (0.2) y 2  (0.8)

 ( y  1)(0.2) y 2 (0.8)2 . 7.127 a

Let x = amount of cheese on a medium pizza. 0.550  0.5   0.525  0.5 P(0.525  x  0.550)  P  z  P(1  z  2)  0.9772  0.8413  0.1359. 0.025   0.025

The required probability is P ( z  2)  1  P( z  2)  1  0.9772  0.0228.

P ( x  0.475)  P ( z  1)  0.8413. So the required probability is (0.8413)3  0.595.

7.128 Let the fuel efficiency of a randomly selected car of this type be x. a

P(29  x  31)  P  (29  30) 1.2  z  (31  30) 1.2   P  0.83  z  0.83  0.7967  0.2033  0.5934.

P( x  25)  P  z  (25  30) 1.2   P  z  4.17   0.0000. Yes, you are very unlikely to select a car of this model with a fuel efficiency of less than 25 mpg.

P( x  32)  P  z  (32  30) 1.2   P  z  1.67   0.0475. So, if 3 cars are selected, the probability that they all have fuel efficiencies more than 32 mpg is (0.0475)3  0.0001.

If P ( z  z*)  0.95, then z*  1.645. So c    ( z*)  30  (1.645)(1.2)  28.026.

204

Chapter 7: Random Variables and Probability Distributions

7.129 a

If the coin is fair then the distribution of x is binomial with n  25 and p  0.5. The probability that it is judged to be biased is P ( x  7)  P ( x  18). Using Appendix Table 9, P ( x  7)  0.014  0.005  0.002  0.021. P( x  18)  0.014  0.005  0.002  0.021. So the probability that the coin is judged to be biased is 0.021  0.021  0.042. Now the distribution of x is binomial with n  25 and p  0.9. The probability that the coin is judged to be biased is P ( x  7)  P ( x  18). Using Appendix Table 9, P ( x  7)  0.000. P( x  18)  0.007  0.024  0.065  0.138  0.226  0.266  0.199  0.072  0.997. So the probability that the coin is judged to be fair is 1  0.997  0.003. When P ( H )  0.1, the distribution of x is binomial with n  25 and p  0.1. The probability that the coin is judged to be biased is P ( x  7)  P ( x  18). Using Appendix Table 9, P ( x  7)  0.007  0.024  0.065  0.138  0.227  0.266  0.199  0.072  0.998. P ( x  18)  0.000. So the probability that the coin is judged to be fair is 1  0.998  0.002.

7.130

When P( H )  0.6, the distribution of x is binomial with n  25 and p  0.6. The probability that the coin is judged to be biased is P ( x  7)  P ( x  18). Using Appendix Table 9, P ( x  7)  0.001. P( x  18)  0.080  0.044  0.020  0.007  0.002  0.153. So the probability that the coin is judged to be fair is 1  0.153  0.847. Likewise, by symmetry, when P( H )  0.4, the probability that the coin is judged to be fair is0.847. These probabilities are larger than those in Part (b), since, when P ( H )  0.6 or P( H )  0.4, the probability of getting a head is closer to 0.5 than when P ( H )  0.9 or P( H )  0.1. Thus it is more likely that the coin will be judged to be fair when P ( H )  0.6 or P( H )  0.4.

It is now more likely that the coin will be judged to be fair, and so the error probabilities are increased. This would seem to make the rule less good, however this rule makes it more likely that the coin will be judged to be fair when in fact it is fair.

 x2  (0  1.2)2 (0.54) 

 (4  1.2)2 (0.2)  2.52.

 x  2.52  1.587. 7.131 a

Let x = amount of time spent. 45  60   P( x  45)  P  z   P( z  1.5)  0.9332. 10   If P ( z  z*)  0.1, z  1.28. So x    ( z*)  60  1.28(10)  72.8 minutes.

Chapter 7: Random Variables and Probability Distributions

Letting y = revenue, we have y  10  50  x 60   10 

205

5 x. So the mean revenue is 6

5 6

 y  10  (60)  60. The mean revenue is $60. 7.132 Let the life of a randomly chosen battery be x. a

P( x  4)  P  z  (4  6) 0.8   P( z  2.5)  0.9938. For the player to function for at least 4 hours, both batteries have to last for at least 4 hours. The probability that this happens is (0.9938)2  0.9876. P( x  7)  P  z  (7  6) 0.8  P( z  1.25)  0.1056. For the player to function for at least 7 hours, both batteries have to last for at least 7 hours. The probability that this happens is (0.1056)2  0.0112. So the probability that the player works for at most 7 hours is 1  0.0112  0.9888. We need the probability that both batteries last longer than c hours to be 0.05. So for either one of the two batteries, we need the probability that it lasts longer than c hours to be 0.05  0.2236 (so that the probability that both batteries last longer than c hours is

(0.2236)2  0.05 ). Now, if P ( z  z*)  0.2236, then z  0.76. So c    ( z*)  6  (0.76)(0.8)  6.608. 7.133 Let x = amount of vitamin E. 4.9  5   P( x  4.9)  P  z   P( z  2)  0.0228. 0.05   5.2  5   P( x  5.2)  P  z   P( z  4)  0.0000. 0.05   7.134 a

No, since 5 ft 7 in. is 67 inches, and if x = height of a randomly chosen women, then P( x  67)  P  z  (67  66) 2   P( z  0.5)  0.6915, which is not more than 94%. About 69% of women would be excluded by the height restriction.

7.135 a Outcome SSSS SSSF SSFS SSFF SFSS SFSF SFFS SFFF FSSS FSSF

y 4 3 2 2 2 1 1 1 3 2

Probability 0.0625 0.0625 0.0625 0.0625 0.0625 0.0625 0.0625 0.0625 0.0625 0.0625

206

Chapter 7: Random Variables and Probability Distributions FSFS 1 0.0625 FSFF 1 0.0625 FFSS 2 0.0625 FFSF 1 0.0625 FFFS 1 0.0625 FFFF 0 0.0625 The probability distribution of y is shown below. y p(y)

 y  0(0.0625) 

0 0.0625

1 0.4375

2 0.3125

3 0.1250

4 0.0625

 4(0.0625)  1.6875.

b Outcome SSSS SSSF SSFS SSFF SFSS SFSF SFFS SFFF FSSS FSSF FSFS FSFF FFSS FFSF FFFS FFFF

y 4 3 2 2 2 1 1 1 3 2 1 1 2 1 1 0

Probability 0.1296 0.0864 0.0864 0.0576 0.0864 0.0576 0.0576 0.0384 0.0864 0.0576 0.0576 0.0384 0.0576 0.0384 0.0384 0.0256

The probability distribution of y is shown below. y p(y)

 y  0(0.0256) 

0 0.0256

1 0.3264

2 0.3456

 4(0.1296)  2.0544.

3 0.1728

4 0.1296

Chapter 7: Random Variables and Probability Distributions

207

c z 4 3 2 2 2 1 2 3 3 2 1 2 2 2 3 4

Outcome SSSS SSSF SSFS SSFF SFSS SFSF SFFS SFFF FSSS FSSF FSFS FSFF FFSS FFSF FFFS FFFF

Probability 0.0625 0.0625 0.0625 0.0625 0.0625 0.0625 0.0625 0.0625 0.0625 0.0625 0.0625 0.0625 0.0625 0.0625 0.0625 0.0625

The probability distribution of z is shown below. z p(z)

z  0(0.125)  7.136

1 0.1250

2 0.5000

3 0.2500

4 0.1250

 4(0.125)  2.375.

1 P( x  1)  P(a or b picked first)  . 2

 1  2   1  2  1 P( x  2)  P(c then not d)  P(d then not c)          .  4  3   4  3  3 1 1 1 P( x  3)  1    . 2 3 6 These results are summarized in the table below. x p(x) 7.137 a b

2 1/3

3 1/6

p(4)  P(KKKK)  P(LLLL)  (0.4) 4  (0.6) 4  0.1552.

p(5)  P( LLLK L)  P( KKKL K)  4(0.6)3 (0.4)(0.6)  4(0.4)3 (0.6)(0.4)  0.2688. in any order

1 1/2

in any order

p(6)  P(LLLKK L)  P(KKKLL K)  5C3 (0.6)3 (0.4) 2 (0.6)  5C3 (0.4)3 (0.6)2 (0.4)  0.29952. in any order

in any order

p(7)  P(LLLKKK L)  P(LLLKKK K)  6C3 (0.6)3 (0.4)3 (0.6)  6C3 (0.6)3 (0.4)3 (0.4) in any order

 0.27648.

in any order

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Chapter 7: Random Variables and Probability Distributions

So the probability distribution of x is as shown below. x p(x) d

4 0.1552

5 0.2688

6 0.29952

7 0.27648

E ( x )  4(0.1552)  5(0.2688)  6(0.29952)  7(0.27648)  5.697.

7.138 If P ( z  z*)  0.15, then z*  1.04. So the lowest score to be designated an A is given by   ( z*)  78  (1.04)(7)  85.28. Since 89  85.28, yes, I received an A. 7.139 Let the pH of the randomly selected soil sample be x. a

P(5.9  x  6.15)  P  (5.9  6) 0.1  z  (6.15  6) 0.1  P( 1  z  1.5)  0.9332  0.1587  0.7745.

P( x  6.10)  P  z  (6.10  6) 0.1  P( z  1)  1  P( z  1)  1  0.8413  0.1587.

P( x  5.95)  P  z  (5.95  6) 0.1  P( z  0.5)  0.3085.

If P ( z  z*)  0.05, then z*  1.645. So the required value is   ( z*)  6  (1.645)(0.1)  6.1645.

7.140 If P ( z  z*)  0.2, then z  0.84. So the lowest 20% of lifetimes consist of those less than   ( z*)  700  ( 0.84)(50)  658. So, if the bulbs are replaced every 658 hours, then only 20% will have already burned out. 7.141 a

No. The proportion of the population that is being sampled is 5000/40000 = 0.125, which is more than 5%. We have n  100 and p  11000 40000  0.275, so   np  100(0.275)  27.5 and

  np(1  p)  100(0.275)(0.725)  4.465. c

No. Since n is being doubled, the standard deviation, which is

np(1  p), is multiplied by

Chapter 7: Random Variables and Probability Distributions

209

Cumulative Review Exercises CR7.1 Obtain a group of volunteers (we’ll assume that 60 people are available for this). Randomly assign the 60 people to two groups, A and B. (This can be done by writing the people’s names on slips of paper, placing the slips in a hat, and drawing 30 slips at random. The people whose names are on those slips should be placed in Group A. The remaining people should be placed in Group B.) Meet with each person individually. For people in Group A offer an option of being given $5 or for a coin to be flipped. Tell the person that if the coin lands heads, he/she will be given $10, but if the coin lands tails, he/she will not be given any money. Note the person’s choice, and then proceed according to the option the person has chosen. For people in Group B, give the person two $5 bills, and then offer a choice of returning one of the $5 bills, or flipping a coin. Tell the person that if the coin lands heads, he/she will keep both $5 bills, but if the coin lands tails, he/she must return both of the $5 bills. Note the person’s choice, and then proceed according to the option the person has chosen. Once you have met with all the participants, compare the two groups in terms of the proportions choosing the gambling options. CR7.2 a

Median = 87 Lower quartile = 69 Upper quartile = 96 Interquartile range = 96 − 69 = 27

100 150 200 250 300 Water Consumption (gallons per person per day)

350

The center of the water consumption distribution is around 87, with the majority of values lying between 70 and 114. There are four outliers, three of which are extreme outliers. If the outliers are included the distribution is positively skewed; without the outliers the distribution is roughly symmetrical. b

x  (49 

sx 

 334) 27  104.222.

(49  104.222)2 

 (334  104.222)2  71.580. 26

If the two largest values were removed the standard deviation would be smaller, since the spread of the data would be reduced.

The mean (104.222) is greater than the median (87). This is consistent with the fact that the distribution is positively skewed.

CR7.3 No. The percentages given in the graph are said to be, for each year, the “percent increase in the number of communities installing” red-light cameras. This presumably means the percent increase in the number of communities with red-light cameras installed, in which case the positive results for all of the years 2003 to 2009 show that a great many more communities had red-light cameras installed in 2009 than in 2002.

210

Chapter 7: Random Variables and Probability Distributions

CR7.4 a

P ( I | W )  0.57, while we can infer from the larger proportions of blacks and Hispanics who rated the Census as “very important” that P( I | W C ) is larger than 0.57. Hence I and W are dependent events.

Answers will vary.

CR7.5 We need P(15.8  b1  b2  15.9). Since b1  b2  8, and  b1   b2  0.2, b1  b2  8  8  16 and  b1  b2  (0.2) 2  (0.2) 2  0.2828. So P(15.8  b1  b2  15.9)  P  (15.8  16) 0.2828  z  (15.9  16) 0.2828  P( 0.71  z  0.35)  0.3632  0.2389  0.1243. CR7.6 Number of Manatee Deaths 200

150

100

0 1975

1980

1985 Year

1990

1995

The trend was for the number of manatee deaths per year to increase. Allowing for year-to-year variations, the number of deaths per year seems to have increased at a roughly constant rate. CR7.7 P(Service 1| Late)  P(Service 1  Late) P(Late)  (0.3)(0.1)  (0.3)(0.1)  (0.7)(0.08)   0.349. So P (Service 2 | Late)  1  0.349  0.651. Thus Service 2 is more likely to have been used. CR7.8 It is not reasonable to think that time playing video or computer games is approximately normal because time cannot be negative, and 0 minutes is 1.054 standard deviations below the mean 0  123.4    1.054  , which indicates that approximately 14.6% of playing times would be z  117.1   negative if the playing times actually followed a normal distribution. CR7.9 1. P ( M )  0.021. 2. P( M | B)  0.043. 3. P ( M | W )  0.07.

Chapter 7: Random Variables and Probability Distributions

211

CR7.10 a

i ii

135/182 = 0.742 (173 + 206)/(210 + 231) = 379/441 = 0.859

b Properly Restrained Iowa City (city)

Not Properly Restrained Properly Restrained

Iowa City (interstate)

Not Properly Restrained

CR7.11 To say that a quantity x is 30% more than a quantity y means that the amount by which x exceeds y is 30% of y; that is, in mathematical notation, that y  x  0.3 y. Dividing both sides by 0.3, we see that this can be written as ( y  x) y  0.3 . Here we are told that drivers who live within one mile of a restaurant are 30% more likely to have an accident than those who do not live within one mile of a restaurant; in other words, that P( A | R) is 30% more than P( A | R C ) . Thus



statement iv is correct: P( A | R)  P( A | R C )

 P( A | R )  0.3. None of the other statements is C

correct. CR7.12 They are dependent events, since the probability that an adult favors stricter gun control given that the adult is female (0.66) is greater than the (unconditional) probability that the adult favors stricter gun control (0.56).

212

Chapter 7: Random Variables and Probability Distributions

CR7.13 0.9

Positive

0.1

Negative

0.05

Positive

0.95

Negative

Drug

0.1

0.9

No Drug

P(Drug  Positive)  (0.1)(0.9)  0.09.

P(No Drug  Positive)  (0.9)(0.05)  0.045.

P(Positive)  0.09  0.045  0.135.

P(Drug | Positive)  P(Drug  Positive) P(Positive)  0.09 0.135  0.667.

(0.1)(0.9)(0.9) = 0.081.

(0.1)(0.9)(0.9) + (0.9)(0.05)(0.05) = 0.08325.

P(drug | positive twice) 

P(not positive twice | drug)  1  P(positive twice | drug)  1  (0.9)(0.9)  0.19.

The retest scheme is preferable in that the probability that someone uses drugs given that the person has tested positive twice is 0.973, which is much greater than the probability that someone uses drugs given a single positive test (0.667). However, under the retest scheme it is more likely that someone who uses drugs will not get caught (a probability of 0.19 under the retest scheme as opposed to 0.1 under the single test scheme).

 x  1(0.2) 

 x2  (1  2.3)2 (0.2) 

CR7.14

CR7.15

P(drug  positive twice) 0.081   0.973. P(positive twice) 0.08325

 4(0.1)  2.3.

 (4  2.3) 2 (0.1)  0.81.

 x  0.81  0.9 CR7.16

y  100  5 x . So  y  100  5 x  100  5(2.3)  88.5 ,

Chapter 7: Random Variables and Probability Distributions

213

and  y  5 x  5(0.9)  4.5 . So  y2  (4.5)2  20.25 . CR7.17 Let x be the number of correct identifications. Assume that the graphologist was merely guessing, in other words that the probability of success on each trial was 0.5. Then, using Appendix Table 9 (with n  15, p  0.5 ), P ( x  6)  0.205  0.117  0.044  0.010  0.001  0.377. Since this probability is not particularly small, this tells us that, if the graphologist was guessing, then it would not have been unlikely for him/her to get 6 or more correct identifications. Thus no ability to distinguish the handwriting of psychotics is indicated. CR7.18 A ball bearing is acceptable if its diameter is between 0.496 and 0.504 inches. Under the new setting, P(0.496  diameter  0.504)  P  (0.496  0.499) 0.002  z  (0.504  0.499) 0.002   P ( 1.5  z  2.5)  0.9938  0.0668  0.9270. So the probability that a ball bearing is unacceptable is 1  0.9270  0.0730. Therefore, 7.3% of the ball bearings will be unacceptable. CR7.19 a

P( x  50)  P  z  (50  45) 5  P( z  1)  0.1587.

If P ( z  z*)  0.9 then z*  1.28. So the required time is   ( z*)  45  (1.28)(5)  51.4 minutes.

If P ( z  z*)  0.25 then z*  0.67. So the required time is   ( z*)  45  ( 0.67)(5)  41.65 minutes.

CR7.20 15

Student Teacher Ratio

14 13 12 11 10 9 -2

-1

0 Normal Score

Yes. Since the pattern in the normal probability plot is relatively straight, normality is plausible.

214

Chapter 7: Random Variables and Probability Distributions

CR7.21 a Survival Time (days) 0 to <300 300 to <600 600 to <900 900 to <1200 1200 to <1500 1500 to <1800 1800 to <2100 2100 to <2400 2400 to <2700

Frequency 8 12 5 5 3 3 4 1 2

Relative Frequency 0.186 0.279 0.116 0.116 0.070 0.070 0.093 0.023 0.047

Relative Frequency (%) 30 25 20 15 10 5 0

600

1200 1800 Survival Time (days)

2400

Positive

Yes. If a symmetrical distribution were required then a transformation would be advisable.

Chapter 8 Sampling Variability and Sampling Distributions Note: In this chapter, numerical answers to questions involving the normal distribution were found using statistical tables. Students using calculators or computers will find that their answers differ slightly from those given. 8.1

A population characteristic is a quantity that summarizes the whole population. A statistic is a quantity calculated from the values in a sample.

8.2

The quantity x is the mean of a sample, while the quantity  is the mean of a population. The quantity s is the standard deviation of a sample, while the quantity  is the standard deviation of a population.

8.3

Population characteristic

Statistic

Population characteristic

Statistic

(8 + 14 + 16 + 10 + 11)/5 = 11.8.

Answers will vary.

8.4

8.5

Answers will vary.

8.6

The x values would be more densely congregated than in Example 8.1.

The histogram would have a similar shape to the one in Example 8.1, and the distribution of the x values would have approximately the same center.

215

216

8.7

Chapter 8: Sampling Variability and Sampling Distributions

a Sample 1, 2 1, 3 1, 4 2, 1 2, 3 2, 4 3, 1 3, 2 3, 4 4, 1 4, 2 4, 3 b

Sample mean 1.5 2 2.5 1.5 2.5 3 2 2.5 3.5 2.5 3 3.5

The sampling distribution of the sample mean, x , is shown below.

x p( x )

1.5 1/6

2 1/6

2.5 1/3

3 1/6

3.5 1/6

Density 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 1.5

2.0

2.5 Sample Mean

3.0

3.5

Chapter 8: Sampling Variability and Sampling Distributions

217

c Sample 1, 1 1, 2 1, 3 1, 4 2, 1 2, 2 2, 3 2, 4 3, 1 3, 2 3, 3 3, 4 4, 1 4, 2 4, 3 4, 4

Sample Mean 1 1.5 2 2.5 1.5 2 2.5 3 2 2.5 3 3.5 2.5 3 3.5 4

The sampling distribution of the sample mean, x , is shown below.

x p( x )

1 1/16

1.5 1/8

1.0

1.5

2.0

2 3/16

2.5 1/4

3 3/16

3.5 1/8

4 1/16

Density 0.5

0.4

0.3

0.2

0.1

0.0

8.8

2.5 3.0 Sample Mean

3.5

4.0

Both distributions are symmetrical, and their means are equal (2.5). However, the “with replacement” version has a greater spread than the first distribution, with values ranging from 1 to 4 in the “with replacement” distribution, and from 1.5 to 3.5 in the “without replacement” distribution. The stepped pattern of the “with replacement” distribution more closely resembles a normal distribution than does the shape of the “without replacement” distribution.

Answers will vary.

218

8.9

Chapter 8: Sampling Variability and Sampling Distributions

a Sample 2, 3, 3* 2, 3, 4 2, 3, 4* 2, 3*, 4 2, 3*, 4* 2, 4, 4* 3, 3*, 4 3, 3*, 4* 3, 4, 4* 3*, 4, 4* b

Sample Mean Sample Median (Max + Min)/2 2⅔ 3 2.5 3 3 3 3 3 3 3 3 3 3 3 3 3⅓ 4 3 3⅓ 3 3.5 3⅓ 3 3.5 3⅔ 4 3.5 3⅔ 4 3.5

The probability distributions for the three statistics as shown below. 3 0.4

3⅓ 0.3

3⅔ 0.2

Sample Median p(Sample Median)

3 0.7

4 0.3

x p( x )

2⅔ 0.1

(Max + Min)/2 p((Max  Min) 2)

8.10

2.5 0.1

3 0.5

3.5 0.4

Using the sampling distributions above, the means of the three statistics are calculated to be E ( x )  3.2, E (Sample median)  3.3, and E  (Max  Min) 2   3.15. Since   3.2 and E ( x )  3.2, we know that, on average, the sample mean will give the correct value for  , which is not the case for either of the two other statistics. Thus, the sample mean would be the best of the three statistics for estimating  . (Also, since the distribution of the sample mean has less variability than either of the other two distributions, the sample mean will generally produce values that are closer to  than the values produced by either of the other statistics.)

9  3.333

15  2.582

36  1.667

50  1.414

100  1

400  0.5

Chapter 8: Sampling Variability and Sampling Distributions

219

8.11

The sampling distribution of x will be approximately normal for the sample sizes in Parts (c)‒(f), since those sample sizes are all greater than or equal to 30.

8.12

The quantity  is the population mean, while the quantity  x is the mean of all possible sample means. The quantity  is the population standard deviation, while the quantity  x is the standard deviation of all the possible sample means.

8.13

 x  40, and  x  

n 5

64  0.625. Since n  64  30, the distribution of x will be

approximately normal. b

Since   0.5  40  0.5  39.5 and   0.5  40  0.5  40.5, the required probability is

P(39.5  x  40.5)  P  (39.5  40) 0.625  z  (40.5  40) 0.625  P(0.8  z  0.8)  0.7881  0.2119  0.5762. c

Since   0.7  40  0.7  39.3 and   0.7  40  0.7  40.7, the probability that x will be within 0.7 of  is P(39.3  x  40.7)  P  (39.3  40) 0.625  z  (40.7  40) 0.625  P( 1.12  z  1.12)  0.8686  0.1314  0.7372. Therefore, the probability that x will be more than 0.7 from  is 1  0.7372  0.2628.

8.14

 x    0.5.  0.289 x    0.07225. n 16

When n  50,  x    0.5 and  x 

Since n  30 the distribution of x is approximately normal.

 n



0.289  0.041. 50

Density 10

0.40

0.45

0.50 Sample Mean

0.55

0.60

0.65

220

8.15

Chapter 8: Sampling Variability and Sampling Distributions

 x  2 and  x  

In each case  x  2 .

n  0.8

When n  20,  x  

9  0.267.

n  0.8

20  0.179.

When n  100,  x   n  0.8 100  0.08. The centers of the distributions of the sample mean are all at the population mean, while the standard deviations (and therefore spreads) of these distributions are smaller for larger sample sizes.

8.16

8.17

The sample size of n  100 is most likely to result in a sample mean close to  , since this is the sample size that results in the smallest standard deviation of the distribution of x .

150 lb

x 

x

156.25  150   P( x  156.25)  P  z    P( z  0.93)  0.1762. 6.75  

 n



27  6.75 lb. 16

2500  156.25 lb. 16

Since the distribution of interpupillary distances is normal, the distribution of x is normal, also.   64  65 67  65   P (64  x  67)  P z  P ( 1  z  2)  0.9772  0.1587  0.8185.  5 25 5 25   









  68  65  P( x  68)  P  z   P ( z  3)  0.0013.  5 25   





Since n  100  30, the distribution of x is approximately normal.

  64  65 67  65  P(64  x  67)  P  z  P ( 2  z  4)  1.0000  0.0228  0.9772.  5 100 5 100      68  65  P( x  68)  P  z   P ( z  6)  0.0000.  5 100   



8.18

x 

 n









10  1. Thus P( x is within 1 of  )  P ( 1  z  1)  0.8413  0.1587  0.6826. 100

Chapter 8: Sampling Variability and Sampling Distributions

221

8.19

Approximately 95% of the time, x will be within 2 x of . Therefore, approximately 95% of the time, x will be within 2 of . ii Approximately 99.7% of the time, x will be within 3 x of . Therefore, approximately 0.3% of the time, x will be farther than 3 from . Given that the true process mean is 0.5, the probability that x is not in the shutdown range is  0.49  0.5 0.51  0.5  P(0.49  x  0.51)  P  z   P( 3  z  3)  0.9987  0.0013  0.9974. 0.2 36   0.2 36 So the probability that the manufacturing line will be shut down unnecessarily is 1  0.9974  0.0026.

8.20

In the sketch given below, the horizontal axis represents x (the number of checks written) and the vertical axis represents density.



16.5  1.65. n 100 Since n  100  30, the distribution of x is approximately normal.

 x    22, and  x 



Density

22 24 Sample Mean

20  22   P( x  20)  P  z   P( z  1.21)  0.1131. 1.65   25  22   P( x  25)  P  z   P( z  1.82)  0.0344. 1.65  

222

Chapter 8: Sampling Variability and Sampling Distributions

8.21

 60  50  P(Total  6000)  P( x  60)  P  z    P ( z  5)  0.0000. 20 100  

8.22

The distribution of x is normal with mean 2 and standard deviation 0.05

2  3 x  2  3(0.0125)  1.9625, 2.0375.

The required probability is P ( z  3)  P ( z  3)  0.0013  0.0013  0.0026.

The distribution of x is now normal with mean 2.05 and standard deviation 0.0125. The probability that the problem will not be detected is P (1.9625  x  2.0375)

16  0.0125.

2.0375  2.05   1.9625  2.05  P z  P( 7  z  1)  0.1587  0.0000  0.1587. 0.0125 0.0125   So the probability that the problem will be detected is 1  0.1587  0.8413. 8.23

8.24

8.25

 pˆ  0.65,  pˆ 

(0.65)(0.35)  0.151. 10

 pˆ  0.65,  pˆ 

(0.65)(0.35)  0.107. 20

 pˆ  0.65,  pˆ 

(0.65)(0.35)  0.087. 30

 pˆ  0.65,  pˆ 

(0.65)(0.35)  0.067. 50

 pˆ  0.65,  pˆ 

(0.65)(0.35)  0.048. 100

 pˆ  0.65,  pˆ 

(0.65)(0.35)  0.034. 200

We need both np and n(1  p ) to be greater than or equal to 10. For p  0.65, this is the case for n  30, 50, 100, 200.

For p  0.2, both np and n(1  p ) to be greater than or equal to 10 for n  50, 100, 200.

mean:  pˆ  0.03 ; standard deviation:  pˆ 

0.03(1  0.03)

 0.017059 100 The sampling distribution is not approximately normal because np  (100)(0.03)  3 is less than the required 10.

Chapter 8: Sampling Variability and Sampling Distributions c

223

The change in sample size does not change the mean of the sampling distribution. However, the standard deviation will decrease to 0.008529. The mean does not change when the sample size is increased because the sampling distribution is always centered at the population value (in this case,  pˆ  0.03 ) regardless of the sample size. The standard deviation of the sampling distribution will decrease as the sample size increases because the sample size (n) is in the denominator of the formula for standard deviation. As sample size increases, standard deviation of the sampling distribution of p̂ decreases.

8.26

The sampling distribution of p̂ is approximately normal because np  (400)(0.03)  12 and n(1  p )  (400)(1  0.03)  388 . Both of these values are at least 10, so we have satisfied the rule of thumb.

mean:  pˆ  0.37 ; standard deviation:  pˆ 

0.37(1  0.37)

 0.04828 100 The sampling distribution is approximately normal because np  (100)(0.37)  37 and n(1  p )  (100)(1  0.37)  63 are both at least 10. The change in sample size does not change the mean of the sampling distribution. However, the standard deviation will decrease to 0.02414. The mean does not change when the sample size is increased because the sampling distribution is always centered at the population value (in this case,  pˆ  0.37 ) regardless of the sample size. The standard deviation of the sampling distribution will decrease as the sample size increases because the sample size (n) is in the denominator of the formula for standard deviation.

8.27

Yes, the sampling distribution of p̂ is approximately normal because np  (400)(0.37)  148 and n(1  p )  (400)(1  0.37)  252 are both at least 10.

 pˆ 

No, since np  100(0.005)  0.5, which is not greater than or equal to 10.

We need both np and n(1  p ) to be greater than or equal to 10, and since p  q it will be

1 (0.005)(0.995)  0.005,  pˆ   0.007. 200 100

sufficient to ensure that np  10. So we need n(0.005)  10, that is n  10 0.005  2000. 8.28

No, the sampling distribution of p̂ based on a random sample of size 10 residents would not be approximately normally distributed because np  10(0.325)  3.25 is less than 10.

mean:  pˆ  0.325 ; standard deviation:  pˆ 

The change in sample size does not change the mean of the sampling distribution. However, the standard deviation will increase to 0.033119. The mean does not change when the sample size is decreased because the sampling distribution is always centered at the population value

0.325(1  0.325) 400

 0.023419 .

224

Chapter 8: Sampling Variability and Sampling Distributions (in this case,  pˆ  0.325 ) regardless of the sample size. The standard deviation of the sampling distribution will increase as the sample size decreases because the sample size (n) is in the denominator of the formula for standard deviation.

8.29

0.8(1  0.8)

If p = 0.8, then  pˆ  0.8 and  pˆ 

If p = 0.7, then  pˆ  0.7 and  pˆ 

When p = 0.8, np  225(0.8)  180 and n(1  p )  225(1  0.8)  45 , which are both greater than 10. When p = 0.7, np  225(0.7)  157.5 and n(1  p )  225(1  0.7)  67.5 , which are also both greater than 10. In each case, np and n(1 – p) are both greater than 10, so the sampling distribution of p̂ is approximately normal for p = 0.8 as well as p = 0.7.

225 0.7(1  0.7) 225

 0.0267 .

 0.0306 .

8.30

Since np  500(0.48)  240  10 and n(1  p )  500(0.52)  260  10, the distribution of p̂ is approximately normal.   0.5  0.48 P( pˆ  0.5)  P  z    P ( z  0.90)  0.1841.  (0.48)(0.52) 500  

8.31

8.32

  0.02  0.05 P(Returned)  P( pˆ  0.02)  P  z    P( z  1.95)  0.9744.  (0.05)(0.95) 200    0.02  0.1  P (Returned)  P( pˆ  0.02)  P  z    P ( z  3.77)  0.9999. So the  (0.1)(0.9) 200   probability that the shipment is not returned is 1  0.9999  0.0001.

Since n  100  30, the distribution of x is approximately normal.  0.1 x    0.8 and  x    0.01. n 100 0.79  0.8   P( x  0.79)  P  z   P( z  1)  0.1587. 0.01  

0.77  0.8   P( x  0.77)  P  z   P( z  3)  0.0013. 0.01   8.33

Since n  100  30, x is approximately normally distributed. Its mean is 50 and its standard deviation is

100  0.1.

50.25  50   49.75  50 P(49.75  x  50.25)  P  z   P( 2.5  z  2.5)  0.9938  0.0062 0.1 0.1    0.9876.

Chapter 8: Sampling Variability and Sampling Distributions

Since  x  50, P( x  50)  0.5.

8.34

Since np  100(0.2)  20  10 and n(1  p )  100(0.8)  80  10, the distribution of p̂ is approximately normal. p(1  p ) (0.2)(0.8)  pˆ  p  0.2 and  pˆ    0.04. n 100 0.25  0.2   P( pˆ  0.25)  P  z   P( z  1.25)  0.1056. 0.04   The probability that the cable company will keep the shopping channel is 0.1056.

8.35

1300  1000   850  1000 Let the index of the specimen be x. P(850  x  1300)  P  z  150 150    P( 1  z  2)  0.9772  0.1587  0.8185.

8.36

225

 950  1000 1100  1000  P(950  x  1100)  P  z   P( 1.05  z  2.11) 150 10   150 10  0.9826  0.1469  0.8357.  850  1000 1300  1000  P(850  x  1300)  P  z   P ( 3.16  z  6.32) 150 10   150 10  1.0000  0.0008  0.9992.

Since np  100(0.4)  20  10 and n(1  p )  100(0.6)  60  10, the distribution of p̂ is approximately normal. p(1  p ) (0.4)(0.6)  pˆ  p  0.4 and  pˆ    0.04899. n 100

0.5  0.4   P( pˆ  0.5)  P  z   P( z  2.04)  0.0207. 0.04899   b

8.37

0.6  0.4    P( z  4.08)  0.0000. If p  0.4, P( pˆ  0.6)  P  z  0.04899   This tells us that, if 40% of people in a state participate in a “self-insurance” plan then it is very unlikely that more than 60 people in a sample of 100 would participate in such a plan. Therefore, if more than 60 people in a sample of 100 were found to participate in this type of plan, we would strongly doubt that p  0.4 for that state.

 106  100  P(Total  5300)  P( x  5300 50)  P( x  106)  P  z    P( z  1.41)  0.0793. 30 50  

Chapter 9 Estimation Using a Single Sample Note: In this chapter, numerical answers to questions involving the normal and t distributions were found using statistical tables. Students using calculators or computers will find that their answers differ slightly from those given. 9.1

Statistics II and III are preferable to Statistic I since they are unbiased (their means are equal to the value of the population characteristic). However, Statistic II is preferable to Statistic III since its standard deviation is smaller. So Statistic II should be recommended.

9.2

An unbiased statistic is generally preferred over a biased one because the unbiased statistic will, on average, give the correct value for the population characteristic being estimated, while the biased one will not.

Unbiasedness alone does not guarantee that the value of the statistic will be close to the true value of the population characteristic, since, if the statistic has a large standard deviation, values of the statistic will tend to be far from the population characteristic.

A biased statistic might be chosen in preference to an unbiased one if the biased statistic has a smaller standard deviation than that of the unbiased statistic, and as a result values of the biased statistic are more likely to be close to the population characteristic than values of the unbiased statistic.

584

9.3

pˆ 

9.4

The value of  is estimated using x  (1.9  2.6 

 3.1) 15  3.120

The value of  is estimated using x  (5.1  4.7 

 3.9) 15  4.307

For those patients who receive biofeedback, the estimated value of  is given by s  0.723

1, 668

 0.350

9.5

The value of p is estimated using pˆ , and the value of p̂ is 14 20  0.7.

9.6

The value of p is estimated using pˆ  245 935  0.262. It is estimated that the proportion of smokers would, when given this treatment, refrain from smoking for at least 6 months is 0.262.

9.7

The value of  is estimated using x  (410 

The value of  2 is estimated using s 2  10414.286.

The value of  is estimated using s  102.050. No, s is not an unbiased statistic for estimating  .

The value of  is estimated using s  1.886. The sample standard deviation was used to obtain the estimate.

9.8

226

 530) 7  421.429.

Chapter 9: Estimation Using a Single Sample

9.9

227

The value of  is estimated using x  (11.3   12) 12  11.967. So the required percentile is estimated to be 11.967  1.28(1.886)  14.381.

The value of J is estimated using x  (103 

The value of  is estimated to be 10000(120.6) = 1,206,000 therms.

The value of p is estimated using pˆ  8 10  0.8.

The population median is estimated using the sample median, which is 120 therms.

 99) 10  120.6 therms.

9.10

For any given sample, the 95% confidence interval is wider than the 90% confidence interval. The interval needs to be wider in order to have greater confidence that the interval contains the true value of p.

9.11

The sample of size n  100 is likely to result in a wider confidence interval. The formula for the pˆ (1  pˆ ) confidence interval is pˆ  1.96 , and so a smaller value of n is likely to result in a wider n confidence interval.

9.12

1.96

1.645

2.58

1.28

1.44

9.13

9.14

In order for the interval to be appropriate, we need np  10 and n (1  p )  10. a Yes, since np  50(0.3)  15  10 and n (1  p )  50(0.7)  35  10. b

No, since np  50(0.05)  2.5, which is not greater than or equal to 10.

No, since np  15(0.45)  6.75, which is not greater than or equal to 10.

No, since np  100(0.01)  1, which is not greater than or equal to 10.

Yes, since np  100(0.7)  70  10 and n(1  p )  100(0.3)  30  10.

Yes, since np  40(0.25)  10  10 and n (1  p )  40(0.75)  30  10.

Yes, since np  60(0.25)  15  10 and n (1  p )  60(0.75)  45  10.

No, since np  80(0.1)  8, which is not greater than or equal to 10.

228

9.15

9.16

Chapter 9: Estimation Using a Single Sample

The larger the confidence level the wider the interval.

The larger the sample size the narrower the interval.

Values of p̂ further from 0.5 give smaller values of pˆ (1  pˆ ). Therefore, the further the value of p̂ from 0.5, the narrower the interval.

Check of Conditions 1. Since npˆ  1,855(0.53)  983  10 and n(1  pˆ )  1,855(1  0.53)  872  10 the sample is large enough. 2. The sample size of n = 1855 is much smaller than 10% of the population size (the number of recruiters and human resource professionals). 3. We are told to assume that the sample is representative of the population of recruiters and human resource professionals in the United States. Having made this assumption it is reasonable to regard the sample as a random sample from the population. Calculation The 95% confidence interval for p is

pˆ (1  pˆ )

0.53(1  0.53)

  0.507, 0.553 n 1855 Interpretation We are 95% confident that the proportion of all recruiters and human resource professionals who have reconsidered a job candidate based on his or her social media profile is between 0.507 and 0.553. pˆ  z *

 0.53  1.96

A 90% confidence interval would be narrower than the 95% confidence interval from part (a) because the z critical value is smaller.

9.17

If a large number of random samples of size 1200 were to be taken, 90% of the resulting confidence intervals would contain the true proportion of all Facebook users who would say it is not OK to “friend” someone who reports to you at work.

9.18

We need to assume that the 1,060 people formed a representative or random sample of parents of teens age 13 to 17.

Check of Conditions 1. Since npˆ  1060(0.61)  646.6  10 and n(1  pˆ )  1060(1  0.61)  413.4  10, the sample is large enough. 2. The sample size of n = 1060 is much smaller than 10% of the population size (the number of parents of teens). 3. We are told that the 1060 people formed a representative sample of parents of teens age 13 to 17. Calculation The 90% confidence interval for p is pˆ (1  pˆ ) 0.61(1  0.61) pˆ  1.645  0.61  1.645  (0.586,0.635). n 1060 Interpretation We are 90% confident that the proportion of all parents of teens age 13 to 17 who have checked which web sites their teens visit is between 0.586 and 0.635.

Chapter 9: Estimation Using a Single Sample

9.19

229

A 99% confidence interval the proportion of all parents of teens age 13 to 17 who have checked which web sites their teens visit would be wider than the interval constructed in Part (b). In order to be 99% (rather than 90%) confident that the interval is capturing the true proportion, the interval needs to be wider.

Let p be the proportion of all coastal residents who would evacuate. Check of Conditions 1. Since npˆ  5046(0.69)  3482  10 and n(1  pˆ )  5046(0.31)  1564  10, the sample size is large enough. 2. The sample size of n = 5046 is much smaller than 10% of the population size (the number of people who live within 20 miles of the coast in high hurricane risk counties of these eight southern states). 3. The sample was selected in a way designed to produce a representative sample. So, it is reasonable to regard the sample as a random sample from the population. Calculation The 98% confidence interval for p is pˆ (1  pˆ ) (0.69)(0.31) pˆ  2.33  0.69  2.33  (0.675, 0.705). n 5046

Interpretation of the Confidence Interval We are 98% confident that the proportion of all coastal residents who would evacuate is between 0.675 and 0.705. Interpretation of the Confidence Level If we were to take a large number of random samples of size 5046, 98% of the resulting confidence intervals would contain the true proportion of all coastal residents who would evacuate.

9.20

Check of Conditions 1. Since npˆ  1003(0.53)  571.71  10 and n(1  pˆ )  1003(1  0.53)  431.29  10, the sample is large enough. 2. The sample size of n = 1003 is much smaller than 10% of the population size (the number of American parents of children age 6 to 11). 3. We must assume that the sample of American parents of children age 6 to 11 was randomly selected. Calculation The 90% confidence interval for p is pˆ (1  pˆ ) (0.53)(0.47) pˆ  1.645  0.53  1.645  (0.504,0.556) n 1003 Interpretation We are 90% confident that the proportion of all American parents of children age 6 to 11 who view science-oriented TV shows as a good way to expose their children to science is between 0.504 and 0.556.

9.21

Check of Conditions 1. Since npˆ  1000(650 1000)  650  10 and n(1  pˆ )  1000(350 1000)  350  10, the sample size is large enough. 2. The sample size of n = 1000 is much smaller than 10% of the population size (the number of dog owners).

230

Chapter 9: Estimation Using a Single Sample 3. The sample was selected in a way designed to produce a representative sample. So, it is reasonable to regard the sample as a random sample from the population. Calculation The 90% confidence interval for p is pˆ (1  pˆ ) 650 (650 1000)(350 1000) pˆ  1.645   1.645  (0.625,0.675) n 1000 1000 Interpretation We are 90% confident that the proportion of dog owners who take more pictures of their dog than of their significant others or friends is between 0.625 and 0.675.

9.22

Check of Conditions 1. Since npˆ  1000(460 1000)  460  10 and n(1  pˆ )  1000(540 1000)  540  10, the sample size is large enough. 2. The sample size of n = 1000 is much smaller than 10% of the population size (the number of dog owners). 3. The sample was selected in a way designed to produce a representative sample. So it is reasonable to regard the sample as a random sample from the population. Calculation The 95% confidence interval for p is pˆ (1  pˆ ) 460 (460 1000)(540 1000) pˆ  1.96   1.96  (0.429,0.491) n 1000 1000 Interpretation We are 95% confident that the actual proportion of dog owners who are more likely to complain to their dog than to a friend is between 0.429 and 0.491.

First, the confidence interval in part (b) has a higher confidence level, which uses a larger z critical value in the computation of the confidence interval. Second, the standard error in part (b) is larger than that in part (a). Both of these reasons contribute to the wider confidence interval in part (b).

Check of Conditions 1. Since npˆ  8347(0.31)  2858.57  10 and n(1  pˆ )  8347(0.69)  5759.43  10, the sample size is large enough. 2. The sample size of n = 8347 is much smaller than 10% of the population size (the number of students applying to college). 3. We are told that the sample was a random sample from the population. Calculation The 90% confidence interval for p is pˆ (1  pˆ ) (0.31)(0.69) pˆ  1.645  0.31  1.645  (0.302,0.318) n 8347 Interpretation We are 90% confident that the true proportion of students applying to college who want to attend a college within 250 miles of home is somewhere between 0.302 and 0.318.

Check of Conditions 1. Since npˆ  2087(0.51)  1064.37  10 and n(1  pˆ )  2087(0.49)  1022.63  10, the sample size is large enough.

Chapter 9: Estimation Using a Single Sample

231

2. The sample size of n = 2087 is much smaller than 10% of the population size (the number of parents of students applying to college). 3. We are told that the sample was a random sample from the population. Calculation The 90% confidence interval for p is pˆ (1  pˆ ) (0.51)(0.49) pˆ  1.645  0.51  1.645  (0.492,0.528) n 2087 Interpretation We are 90% confident that the true proportion of parents of students applying to college who want their child to attend a college within 250 miles from home is somewhere between 0.492 and 0.528. c

The two confidence intervals from (a) and (b) do not have the same width because the two sample sizes are different, and also that the margins of error are based on two different values for p̂ .

9.23

Check of Conditions 1. Since npˆ  1001(0.83)  830.83  10 and n(1  pˆ )  1001(0.17)  170.17  10, the sample size is large enough. 2. The sample size of n = 1001 is much smaller than 10% of the population size (the number of mothers of children under the age of 2). 3. We are told that the sample is representative of the population of mothers of children under the age of 2. Therefore, it is reasonable to regard the sample as a random sample from the population. Calculation The 99% confidence interval for p is pˆ (1  pˆ ) (0.83)(0.17) pˆ  2.58  0.83  2.58  (0.799,0.861) n 1001 Interpretation We are 99% confident that the actual proportion of mothers of children under the age of 2 years who posted pictures of their new baby on social media from the delivery room is between 0.799 and 0.861.

9.24

Check of Conditions 1. Since npˆ  2305(484 2305)  484  10 and n(1  pˆ )  2305(1821 2305)  1821  10, the sample size is large enough. 2. The sample size of n = 2305 is much smaller than 10% of the population size (the number of working adults). 3. We are told that the sample is representative of the population of working adults. Calculation The 95% confidence interval for p is pˆ (1  pˆ ) 484 (484 2305)(1821 2305) pˆ  1.96   1.96  (0.193,0.227) n 2305 2305 Interpretation We are 95% confident that the actual proportion of working adults who are very concerned that their job will be automated, outsourced, or otherwise made obsolete in the next five years is between 0.193 and 0.227.

232

Chapter 9: Estimation Using a Single Sample

9.25

Check of Conditions 1. Since npˆ  1000(0.84)  840  10 and n(1  pˆ )  1000(0.16)  160  10, the sample size is large enough. 2. The sample size of n = 1000 is much smaller than 10% of the population size (the number of adult Americans). 3. We are told that the sample is representative of adult Americans. Calculation The 90% confidence interval for p is pˆ (1  pˆ ) (0.84)(0.16) pˆ  1.645  0.84  1.645  (0.821,0.859) n 1000 Interpretation We are 95% confident that the proportion of all adult Americans who prefer cheese on their burgers is between 0.821 and 0.859.

9.26

Check of Conditions 1. Since npˆ  1000(0.38)  380  10 and n(1  pˆ )  1000(0.62)  620  10, the sample size is large enough. 2. The sample size of n = 1000 is much smaller than 10% of the population size (the number of adults who have traveled by air at least once in the previous year). 3. We are told that the sample was representative of the population of adults who have traveled by air at least once in the previous year. Calculation The 95% confidence interval for p is pˆ (1  pˆ ) (0.38)(0.62) pˆ  1.96  0.38  1.96  (0.350,0.410) n 1000 Interpretation We are 95% confident that the proportion of all adults who have traveled by air at least once in the previous year who have yelled at a complete stranger while traveling is between 0.350 and 0.410.

9.27

Check of Conditions 1. Since npˆ  800(312 800)  312  10 and n(1  pˆ )  800(488 800)  488  10, the sample size is large enough. 2. The sample size of n = 800 is much smaller than 10% of the population size (the number of millennials who drive a car that is more than 5 years old). 3. We are told that the sample is representative of the population. Calculation The 90% confidence interval for p is pˆ (1  pˆ ) 312 (312 800)(488 800) pˆ  1.645   1.645  (0.362,0.418) n 800 800 Interpretation We are 90% confident that the proportion of all millennials who drive a car that is more than 5 years old who have named their cars is between 0.362 and 0.418.

Chapter 9: Estimation Using a Single Sample 9.28

233

Check of Conditions 1. Since npˆ  2096(1216 2096)  1216  10 and n(1  pˆ )  2096(880 2096)  880  10, the sample size is large enough. 2. The sample size of n = 2096 is much smaller than 10% of the population size (the number of U.S. adults). 3. We are told that the sample is representative of the population of U.S. adults. Calculation The 95% confidence interval for p is pˆ (1  pˆ ) 1216 (1216 2096)(880 2096) pˆ  1.96   1.96  (0.559,0.601) n 2096 2096 Interpretation We are 95% confident that the actual proportion of U.S. adults who consider themselves to be football fans is between 0.559 and 0.601.

Check of Conditions 1. Since npˆ  1216(692 1216)  692  10 and n(1  pˆ )  1216(524 1216)  524  10, the sample size is large enough. 2. The sample size of n = 1216 is much smaller than 10% of the population size (the number of football fans). 3. We are told that the sample is representative of the population of football fans. Calculation The 95% confidence interval for p is pˆ (1  pˆ ) 692 (692 1216)(524 1216) pˆ  1.96   1.96  (0.541,0.597) n 1216 1216 Interpretation We are 95% confident that the actual proportion of football fans who think that the new rules have been effective in limiting head injuries is between 0.541 and 0.597.

The confidence intervals do not have the same width because the standard errors, and hence the margins of error, are different.

9.29

It is unlikely that the estimated proportion of adult Americans who have pretended to know what the cloud is or how it works ( pˆ  0.22 ) will differ from the true proportion by more than 0.026.

9.30

We have n  1002 and pˆ  0.25. So the margin of error is 1.96

(0.25)(0.75)  0.027 ; thus the 1002

margin of error is approximately 3 percentage points, as stated. 9.31

Check of Conditions 1. Since npˆ  89(18 89)  18  10 and n(1  pˆ )  89(71 89)  71  10, the sample size is large enough. 2. The sample size of n = 89 is much smaller than 10% of the population size (the number of people under 50 years old who use this type of defibrillator). 3. We are told to assume that the sample is representative of patients under 50 years old who receive this type of defibrillator. Having made this assumption it is reasonable to regard the sample as a random sample from the population.

234

Chapter 9: Estimation Using a Single Sample Calculation The 95% confidence interval for p is pˆ (1  pˆ ) 18 (18 89)(71 89) pˆ  1.96   1.96  (0.119, 0.286). n 89 89 Interpretation We are 95% confident that the proportion of all patients under 50 years old who experience a failure within the first two years after receiving this type of defibrillator is between 0.119 and 0.286. b

Check of Conditions 1. Since npˆ  362(13 362)  13  10 and n(1  pˆ )  362(349 362)  349  10, the sample size is large enough. 2. The sample size of n = 362 is much smaller than 10% of the population size (the number of people age 50 or older who use this type of defibrillator). 3. We are told to assume that the sample is representative of patients age 50 or older who receive this type of defibrillator. Having made this assumption it is reasonable to regard the sample as a random sample from the population. Calculation The 99% confidence interval for p is pˆ (1  pˆ ) 13 (13 362)(349 362) pˆ  2.58   2.58  (0.011, 0.061). n 362 362 Interpretation We are 99% confident that the proportion of all patients age 50 or older who experience a failure within the first two years after receiving this type of defibrillator is between 0.011 and 0.061.

Using the estimate of p from the study, 18 89 , the required sample size is given by 2

 1.96   18  71  1.96  n  p(1  p )         688.685.  B   89  89  0.03  So a sample of size at least 689 is required. 9.32

Assuming a 95% confidence level, the margin of error is pˆ (1  pˆ ) (0.430)(0.570) 1.96  1.96  0.03 n 1000 2

9.33

 1.96   1.96  n  p(1  p )    0.25    2401. A sample size of 2401 is required. B    0.02 

9.34

 1.96   1.96  n  p(1  p )   (0.3)(0.7)     2016.84. A sample size of 2017 is required.  B   0.02 

 2.58  n  p(1  p )    B 

Chapter 9: Estimation Using a Single Sample 2

9.35

235

 2.58   2.58  n  p(1  p )   (0.3)(0.7)     3494.61. A sample size of 3495 is required.  B   0.02 

Assuming a 95% confidence level, and using the preliminary estimate of p  0.63 , 2

 1.96   1.96  n  p(1  p)    0.63(1  0.63)    358.191 . The sample should contain 359  B   0.05 

individuals. Using the conservative estimate of p  0.5 , 2

 1.96   1.96    0.5(1  0.5)    384.16 . The sample should contain 385  M   0.05 

n  p (1  p )  individuals. b

As expected, the sample size computed using the conservative estimate is larger than the sample size computed using the preliminary estimate. The sample size 385 should be used for this study because it will guarantee a margin of error of no greater than 0.05 because the margin of error is largest with p  0.5 . The smaller sample size (359) will only guarantee a margin of error no greater than 0.05 if p  0.63 or if p  1  0.63  0.37 . If p̂ lies between 0.37 and 0.63, which could happen with a different sample, then the margin of error would be greater than 0.05.

9.36

9.37

90%

95%

99%

0.5%

2.12

1.80

2.81

1.71

1.78

2.26

236

9.38

9.39

Chapter 9: Estimation Using a Single Sample

x  (114.4  115.6) 2  115.0 Hertz.

The 99% confidence interval is wider than the 90% confidence interval, since the interval needs to be wider in order that there be greater confidence that the interval contains the population proportion. Thus the 90% interval is (114.4, 115.6) and the 99% interval is (114.1, 115.9).

The width of the first interval is 52.7  51.3  1.4. The width of the second interval is



50.6  49.4  1.2. Since the confidence interval is given by x  (t critical value) s



width of the confidence interval is given by 2  (t critical value) s



n , the

n . Therefore, for samples of

equal standard deviations, the larger the sample size the narrower the interval. Thus it is the second interval that is based on the larger sample size. 9.40

9.41

The two necessary conditions are (1) that the sample of adult Americans is a random sample from the population of interest or the sample is selected in a way that results in a sample being representative of the population, and (2) the sample size is large ( n  30 ) or the population distribution is approximately normal. In this case, the sample is representative of adult Americans. However, we don’t know the shape of the population distribution and the sample is not at least size 30. Therefore, I would not recommend using the one-sample t confidence interval to estimate the population mean.

I would recommend using the one-sample t confidence interval in this case. The sample is representative of the population of adult Americans, and the sample size is larger than 30. Therefore, the two necessary conditions have been satisfied.

If the distribution of volunteer times is approximately normal, for the sample standard deviation of s = 16.54 hours and the sample mean of x  14.76 hours, approximately 18.6% of volunteer times would be negative, which is impossible. Therefore, it is not reasonable to think that the distribution of volunteer times is approximately normal.

The two conditions required to use a one-sample t distribution are that the sample was either randomly selected or is representative of the population, and the population distribution is either normally distributed or that the sample size is at least 30. We are told that the sample is representative of the population, and that the sample size is 500. Both conditions are satisfied in the exercise.

Conditions 1. Since n  500  30, the sample size is large enough. 2. We are told that the sample is representative of South Korean middle school students. Therefore, it is reasonable to regard the sample as a random sample from the population. Calculation The 95% confidence interval for  is s 16.54 x  (t critical value)   14.76  1.965   (13.307,16.213) n 500 Interpretation We are 95% confident that the mean number of hours spent in volunteer activities per year for South Korean middle school children is between 13.307 and 16.213 hours.

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237

Conditions 1. A boxplot of the sample values is shown below.

Wrist Extension

Since the boxplot is roughly symmetrical and there are no outliers in the sample we are justified in assuming that the population distribution of wrist extensions is normal. 2. We need to assume that the 24 students used in the study form a random sample from the population in question. Calculation x  25.917, s  1.954, df  23 The 90% confidence interval for  is

x  (t critical value) 

s 1.954  25.917  1.71  (25.235, 26.599). n 24

Interpretation We are 90% confident that the population mean wrist extension is between 25.235 and 26.599.

9.43

In order to generalize the estimate to the population of Cornell students the assumption would have to be made that the 24 students in the study formed a random sample of Cornell students. To generalize the estimate to all university students you would have to assume that the 24 students in the study formed a random sample of university students.

Yes. The entire confidence interval is above 20, so the results of the study would be very unlikely if the population mean wrist extension were as low as 20.

Conditions 1. Since n  65  30, the sample is large enough. 2. The sample was selected to be representative of the population of first-year students from a large university in England. Calculation The 95% confidence interval for  is

x  (t critical value) 

s 6.44  37.02  1.998   (35.424,38.616) n 65

Interpretation We are 95% confident that the mean procrastination scale for first-year students at this college is between 35.424 and 38.616. b

Forty (40) is not a plausible value for the mean population procrastination scale score because that value is not contained within the confidence interval. Therefore, it seems plausible that, on average, students at this university do not have high levels of procrastination.

238 9.44

Chapter 9: Estimation Using a Single Sample a

Conditions 1. Since n  68  30, the sample is large enough. 2. The sample was selected to be representative of the population of second-year students from a large university in England. Calculation The 95% confidence interval for  is

x  (t critical value) 

s 6.82  41.00  1.996   (39.349,42.651) n 68

Interpretation We are 95% confident that the mean procrastination scale for second-year students at this college is between 39.349 and 42.651.

9.45

The confidence interval for second-year students does not overlap, and has values that are entirely above, the confidence interval for first-year students. Therefore, it seems plausible that second-year students tend to procrastinate more than first-year students at this university.

The 90% confidence interval would be narrower. In order to be only 90% confident that the interval captures the true population mean, the interval does not have to be as wide as it would in order to be 95% confident of capturing the true population mean.

The statement is not correct. The population mean,  , is a constant, and therefore we cannot talk about the probability that it falls within a certain interval.

The statement is not correct. We can say that on average 95 out of every 100 samples will result in confidence intervals that will contain  , but we cannot say that in 100 such samples, exactly 95 will result in confidence intervals that contain  .

9.46

Conditions 1. Since n  50  30, the sample is large enough. 2. There is no indication that the study participants were randomly selected or representative of the population of taxi drivers in Greece. Therefore, we must assume the this condition has been satisfied Calculation The 95% confidence interval for  is s 1.11 x  (t critical value)   3.2  2.01   (2.884,3.516) n 50 Interpretation We are 95% confident that the mean following distance while talking on a mobile phone for the population of taxi drivers in Greece is between 2.884 and 3.516 meters.

9.47

For samples of equal sizes, those with greater variability will result in wider confidence intervals. The 12 to 23 month and 24 to 35 month samples resulted in confidence intervals of width 0.4, while the less than 12 month sample resulted in a confidence interval of width 0.2. So the 12 to 23 month and 24 to 35 month samples are the ones with the greater variability.

For samples of equal variability, those with greater sample sizes will result in narrower confidence intervals. Thus the less than 12 month sample is the one with the greater sample size.

Chapter 9: Estimation Using a Single Sample

9.48

Since the new interval is wider than the interval given in the question, the new interval must be for a higher confidence level. (By obtaining a wider interval, we have a greater confidence that the interval captures the true population mean.) Thus the new interval must have a 99% confidence level.

If the distribution of weight gains is approximately normal, then, for the given sample standard deviation of s = 6.8 pounds and the sample mean of x  5.7 pounds, we should see approximately 16% of the weight gain observations be less than -1.1 pounds (one standard deviation below the mean). It is reported that 6.8% lost more than 1.1 pounds, which is too small a percentage if the normal distribution model was correct. Therefore, it is not reasonable to think that the distribution of weight gains is approximately normal.

The two conditions required to use a one-sample t distribution are that the sample was either randomly selected or is representative of the population, and the population distribution is either normally distributed or that the sample size is at least 30. We are told that the sample is representative, and that the sample size is 103. Therefore, both conditions are satisfied in the exercise.

Conditions 1. Since n  103  30, the sample is large enough. 2. The sample is representative of the population of freshmen at this Midwestern university. Calculation The 95% confidence interval for  is

x  (t critical value) 

s 6.8  5.7  1.983   (4.371,7.029) n 103

Interpretation We are 95% confident that the mean weight gain of freshmen students at this university is between 4.371 and 7.029 pounds. 9.49

239

Conditions 1. Since n  100  30, the sample size is large enough. 2. We are told to assume that the sample was a random sample of passengers. Calculation The t critical value for 99 degrees of freedom (for a 95% confidence level) is between 1.98 and 2.00. We will use an estimate of 1.99. Thus, the 95% confidence interval for  is s 20 x  (t critical value)   183  1.99   (179.02, 186.98). n 100 Interpretation We are 95% confident that the mean summer weight is between 179.02 and 186.98 lb.

Conditions 1. Since n  100  30, the sample size is large enough. 2. We are told to assume that the sample was a random sample. Calculation The 95% confidence interval for  is s 23 x  (t critical value)   190  1.99   (185.423, 194.577). n 100

240

Chapter 9: Estimation Using a Single Sample Interpretation We are 95% confident that the mean winter weight is between 185.423 and 194.577 lb.

9.50

Based on the Frontier Airlines data, neither recommendation is likely to be an accurate estimate of the mean passenger weight, since 190 is not contained in the confidence interval for the mean summer weight and 95 is not contained in the confidence interval for the mean winter weight.

Conditions 1. A boxplot of the sample values is shown below.

255

260

265

270 Airborne Time

275

280

285

Since the boxplot is roughly symmetrical and there are no outliers in the sample we are justified in assuming that the population distribution of airborne times is normal. 2. We are told that the sample was a random sample from the population of flights. Calculation x  270.3, s  9.141, df  9 The 90% confidence interval for  is

x  (t critical value) 

s 9.141  270.3  1.83   ( 265.010, 275.590). n 10

Interpretation We are 90% confident that the mean airborne time is between 265.010 and 275.590 minutes.

9.51

If we were to take a large number of random samples of size 10, 90% of the resulting confidence intervals would contain the true mean airborne time.

The sample mean airborne time is 270.3 and the sample standard deviation is 9.141. Assuming these values for the population mean and standard deviation, and assuming that the population of airborne times is normal, approximately 5% of airborne times will be greater than 270.3  1.96(9.141)  288. The time 290 minutes after 10 a.m. is 2.50 p.m. Thus, if the arrival time is published as 2.50 p.m., we can expect less than 5% of flights to be late. (This is assuming that all flights actually do depart at 10 a.m. Note, also, that many approximations have been made in this calculation. The airline might wish to use a more conservative method of calculation, that is, one that results in a later published arrival time.)

It is not reasonable to construct a confidence interval for the mean mileage rating of 2016 midsize hybrid cars. There is no indication that the cars were randomly selected, which is the first required condition. The second required condition of approximate normality of the population of mileage ratings of the cars been satisfied. Referring to the normal probability plot below, although there is some curvature seen in the plot, there is not enough curvature to question the approximate normality assumption.

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42.5 40.0

Mileage Rating

37.5 35.0 32.5 30.0 27.5 25.0 -1.5

9.52

-1.0

-0.5

0.0 Normal Score

0.5

1.0

1.5

Conditions 1. A boxplot of the sample values is shown below.

15 20 Number of Months

Since the boxplot is roughly symmetrical and there are no outliers in the sample we are justified in assuming that the population distribution of the number of months since the last visit is normal. 2. We are told to assume that the sample was a random sample from the population of students participating in the free checkup program. Calculation x  17, s  9.0277, df  4 The 95% confidence interval for  is s 9.0277 x  (t critical value)   17  2.78   (5.776, 28.224). n 5 Interpretation We are 95% confident that the mean number of months elapsed since the last visit for the population of students participating in the program is between 5.776 and 28.224. 9.53

A reasonable estimate of  is given by (sample range) 4  (700  50) 4  162.5. Thus

 1.96   1.96  162.5  n     1014.4225. 10  B    2

So we need a sample size of 1015.

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Chapter 9: Estimation Using a Single Sample

 1.645  For a 90% confidence level, the formula is n    .  B  2

9.54

 2.33  For a 98% confidence level, the formula is n    .  B  2

9.55

First, we need to know that the information is based on a random sample of middle-income consumers aged 65 and older. Second, it would be useful if some sort of margin of error were given for the estimated mean of $10,235.

9.56

The margins of error are different because the sample sizes are different and the sample proportions are different.

9.57

The paper states that Queens flew for an average of 24.2  9.21 minutes on their mating flights, and so this interval is a confidence interval for a population mean.

Conditions 1. Since n  30  30, the sample size is large enough. 2. We are told to assume that the 30 queen honeybees are representative of the population of queen honeybees. It is then reasonable to treat the sample as a random sample from the population. Calculation The 95% confidence interval for  is

x  (t critical value) 

s 3.47  4.6  2.05   (3.301, 5.899). n 30

Interpretation We are 95% confident that the mean number of partners is between 3.301 and 5.899. 9.58

9.59

A bootstrap 95% confidence interval for the population proportion of those who would reply “No” to the question, p, is (0.774, 0.829). Based on this sample, you can be 95% confident that the actual proportion who would reply “No” to the question “Should you be friends with your boss on Facebook?” is somewhere between 0.774 and 0.829.

In this case, one could argue that this situation would tend to overestimate the true proportion because individuals who respond to anonymous and voluntary surveys tend to have strong feelings about the subject, and therefore might artificially inflate the proportion who would reply “No.”

No, it is not appropriate to use the large-sample confidence interval for a population proportion to estimate the proportion of the transportation services customers who have tried Uber or Lyft at least once. We don’t have at least 10 successes in the sample, which is one of the necessary conditions.

Yes, it is appropriate to use a bootstrap confidence interval for a population proportion to estimate the proportion of the transportation services customers who have tried Uber or Lyft because a random sample of regular customers who are 55 or older was taken.

A bootstrap 95% confidence interval for the population proportion of customers 55 or older who have used Uber or Lyft at least once, p, is (0.000, 0.333). Based on this sample, you can

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243

be 95% confident that the actual proportion of customers 55 or older who have used Uber or Lyft at least once is somewhere between 0.000 and 0.333. d

Yes, the value obtained in the study (21%) is contained within the bootstrap confidence interval. Confidence intervals provide a range of plausible values for the true proportion. The interval contains 0.21, so we know that 0.21 is a plausible value for the true proportion of customers 55 or older who have used Uber or Lyft at least once.

9.60

Different simulations will produce different results, so answers will vary. For one simulation, a bootstrap 95% confidence interval for the population proportion of U.S. businesses who have fired workers for email misuse, p, was (0.230, 0.332). Based on this sample, you can be 95% confident that the actual proportion of U.S. businesses who have fired workers for email misuse is somewhere between 0.230 and 0.332.

9.61

It would not be appropriate to use a large-sample confidence interval for one proportion to estimate Kevin Love’s success rate for three-point shots during the 2016 season because there are only 5 successes in the sample, which is fewer than the required 10 successes to use the large-sample interval.

Different simulations will produce different results, so answers will vary. For one simulation, a bootstrap 90% confidence interval for the population proportion of Kevin Love’s threepoint shot success rate during the 2016 NBA season, p, was (0.105, 0.421). Based on this sample, you can be 90% confident that the actual proportion of Kevin Love’s three-point shot success rate during the 2016 NBA season is somewhere between 0.105 and 0.421.

A bootstrap 95% confidence interval for the population proportion of all U.S. employers who would rank stress as their top health and productivity concern, p, is (0.711, 0.791). Based on this sample, you can be 95% confident that the actual proportion of all U.S. employers who would rank stress as their top health and productivity concern is somewhere between 0.711 and 0.791.

The width of the resulting interval would decrease if the international employers were included in the sample because larger sample sizes result in narrower confidence intervals.

Yes, it would be appropriate to use the large-sample confidence interval for a population proportion to estimate the proportion of residential properties successfully sold at auction in the post-Brexit UK. The necessary conditions (a representative sample of residential properties and at least 10 successes and failures in the sample) have both been satisfied.

It is appropriate to use a bootstrap confidence interval because we have a representative sample of the population.

A bootstrap 95% confidence interval for the population proportion of residential properties successfully sold at auction in the post-Brexit UK, p, is (0.346, 0.731). Based on this sample, you can be 95% confident that the actual proportion of residential properties successfully sold at auction in the post-Brexit UK is somewhere between 0.346 and 0.731.

Yes, the success rate for properties sold at auction throughout the UK during one stretch a year earlier (0.72) is contained within the bootstrap confidence interval. Confidence intervals provide a range of plausible values for the true proportion. The interval contains 0.72, so we

9.62

9.63

244

Chapter 9: Estimation Using a Single Sample know that 0.72 is a plausible value for the true success rate for properties sold at auction throughout the UK during one stretch a year earlier.

9.64

Different simulations will produce different results, so answers will vary. For one simulation, a bootstrap 95% confidence interval for the population proportion of all U.S. households that feel obliged to do business with one or more financial services companies they distrust, p, is (0.282, 0.337). Based on this sample, you can be 95% confident that the actual proportion of all U.S. households that feel obliged to do business with one or more financial services companies they distrust is somewhere between 0.282 and 0.337.

9.65

Different simulations will produce different results, so answers will vary. For one simulation, a bootstrap 95% confidence interval for the population proportion of all brand-name products that are active on Snapchat, p, is (0.617, 0.783). Based on this sample, you can be 95% confident that the actual proportion of all brand-name products that are active on Snapchat is somewhere between 0.617 and 0.783.

9.66

We will find a 95% confidence interval to estimate the population mean gas mileage for electric or plug-in hybrid cars. The given simulation produced a confidence interval of (81.372, 115.253). We are 95% confident that the population mean gas mileage for electric or plug-in hybrid cars lies somewhere between 81.372 and 115.253 mpg.

9.67

The sample size of n = 21 is smaller than 30, so the methods based on the t distribution may not be appropriate.

We will find a 95% confidence interval to estimate the mean time discrimination score for the population of male smokers who abstain from smoking for 24 hours. The given simulation produced a confidence interval of (0.994, 1.082). We are 95% confident that the population mean time discrimination score for the population of male smokers who abstain from smoking for 24 hours lies somewhere between 0.994 and 1.082.

A boxplot (shown below) was constructed.

9.68

-30

-20

-10

Points

Note that the boxplot indicates an outlier in the distribution, so the methods based on the t distribution might not be appropriate. b

We will find a 95% confidence interval to estimate the population mean points difference for NFL teams coming off a bye week. Different simulations will produce different results, so answers will vary. One simulation produced a confidence interval of (-1.906, 4.688). We are 95% confident that the population mean points difference for NFL teams coming off a bye week lies somewhere between -1.906 and 4.688.

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Based on the results from part (b), I do not believe that teams coming off a bye week have a significant advantage in points scored over their opponents. The confidence interval includes zero, so zero is a plausible value for the true mean points difference.

9.69

We will find a 95% confidence interval to estimate the mean annual energy cost for the population of all small televisions. The given simulation produced a confidence interval of (4.579, 5.754). We are 95% confident that the population mean annual energy cost for the population of all small televisions lies somewhere between $4.579 and $5.754.

9.70

The sample size of n = 12 is smaller than 30, so the methods based on the t distribution may not be appropriate.

We will find a 95% confidence interval to estimate the mean price of a Big Mac in Europe. The given simulation produced a confidence interval of (3.499, 4.192). We are 95% confident that the population mean price of a Big Mac lies somewhere between $3.499 and $4.192.

The sample size of n = 15 is smaller than 30, so the methods based on the t distribution may not be appropriate.

We will find a 95% confidence interval to estimate the mean interleague winning percentage for NL teams. Different simulations will produce different results, so answers will vary. One simulation produced a confidence interval of (38, 52). We are 95% confident that the mean interleague winning percentage for NL teams lies somewhere between 38% and 52%.

In Part (b), the mean winning percentage for NL teams was estimated (with 95% confidence) to be between 38% and 52%, which indicates that 50% is a plausible value. Therefore, it is not reasonable to say that the National League or American League performs significantly better than the other in interleague play.

9.71

9.72

Conditions 1. Since npˆ  1001(851 1001)  851  10 and n(1  pˆ )  1001(150 1001)  150  10, the sample size is large enough. 2. The sample size of n = 1001 is much smaller than 10% of the population size (the number of students age 13 to 19 years). 3. We are not told any details about how the sample was collected. We must therefore assume the sample was randomly selected or representative of the population of students age 13 to 19 years. Calculation The 95% confidence interval for p is pˆ (1  pˆ ) 851 (851 1001)(150 1001) pˆ  1.96   1.96  (0.828,0.872) n 1001 1001 Interpretation We are 95% confident that the actual proportion of students age 13 to 19 who would say that they learn best using a mix of digital and non-digital tools is between 0.828 and 0.872. In order for the method used to construct the interval to be valid, we must assume that the sample was randomly selected or is representative of the population of all students age 13 to 19.

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9.73

Intervals constructed when the sample proportions ( p̂ ) are closer to 0.5 are wider than those farther from 0.5 because the product pˆ (1  pˆ ) is largest when pˆ  0.5 . In the previous exercise,

pˆ 

851 1, 001

 0.850 and in this exercise, pˆ 

811 1, 001

 0.810 . The sample proportion in this

exercise is closer to 0.5 than that in the previous exercise. Therefore, the interval in this exercise will be wider than the interval in the previous exercise. 9.74

Check of Conditions 1. Since npˆ  1668(934 1668)  934  10 and n(1  pˆ )  1668(734 1668)  734  10, the sample size is large enough. 2. The sample size of n = 1668 is much smaller than 10% of the population size (the number of adult Americans). 3. The sample is representative of adult Americans. Calculation The 95% confidence interval for p is pˆ (1  pˆ ) 934 (934 1668)(734 1668) pˆ  1.96   1.96  (0.536,0.584) n 1668 1668 Interpretation We are 95% confident that the actual proportion of all adult Americans who would give themselves a grade of A or B on their financial knowledge of personal finance is between 0.536 and 0.584.

Because this confidence interval is entirely above 0.5, the interval is consistent with the statement that a majority of adult Americans would give themselves a grade of A or B.

9.75

Conditions 1. Since npˆ  369(0.52)  191.88  10 and n(1  pˆ )  369(0.48)  177.12  10, the sample size is large enough. 2. The sample size of n = 369 is much smaller than 10% of the population size (the number of homeowners in the western United States). 3. We are told that the sample is representative of the population of homeowners in the western U.S. Calculation The 90% confidence interval for p is pˆ (1  pˆ ) (0.52)(0.48) pˆ  1.645  0.52  1.645  (0.477,0.563) n 369 Interpretation We are 90% confident that the true proportion of all homeowners in western United States who have considered installing solar panels is between 0.477 and 0.563.

9.76

Check of Conditions 1. Since npˆ  533(0.33)  175.89  10 and n(1  pˆ )  533(0.67)  357.11  10, the sample size is large enough. 2. The sample size of n = 533 is much smaller than 10% of the population size (the number of homeowners in the southern United States). 3. We are told that the sample is representative of the population of homeowners in the southern United States Calculation

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The 90% confidence interval for p is pˆ (1  pˆ ) (0.33)(0.67) pˆ  1.645  0.33  1.645  (0.296,0.364) n 533 Interpretation We are 90% confident that the actual proportion of homeowners in the southern states who have considered installing solar panels is between 0.296 and 0.364. b

9.77

The confidence interval in part (a) is narrower than the confidence interval in the previous exercise due to the larger sample size and that the sample proportion ( p̂ ) in this exercise is farther from 0.5 than in the previous exercise.

Assuming a 95% confidence level, and using p = 0.32, 2

 1.96   1.96  n  p(1  p)    (0.32)(0.68)  0.05   334.4. A sample size of 335 is required. B     2

 1.96   1.96  Using the conservative value, p = 0.5, n  p(1  p)   (0.5)(0.5)     384.2.  B   0.05  A sample size of 385 is required. As always, the conservative estimate of p gives the larger sample size. Since the relevant proportion could have changed significantly since 2011, it would be sensible to use a sample size of 385. 2

9.78

 1.96   1.96  n  p(1  p )    (0.5)(0.5)  0.1   96.04. A sample size of 97 is required. B    

9.79

 1.96   1.96  0.8  n     245.862. A sample size of 246 is required.  B   0.1 

9.80

The 99% upper confidence bound for the mean wait time for bypass surgery is



19  2.33 10



539  20.004 days.

9.81

The 95% confidence interval for the population standard deviation of wait time (in days) for angiography is   9 9  1.96   (8.571,9.429).  2(847)   

9.82

The new 95% interval has width (1.75  2.33) 



discussed in the text has width 2(1.96) 

 n   (4.08)  n . The 95% interval n   (3.92)  n . Therefore, the new interval

would not be recommended, since it is wider than the interval discussed in the text.

Chapter 10 Hypothesis Testing Using a Single Sample Note: In this chapter, numerical answers to questions involving the normal and t distributions were found using values from a calculator. Students using statistical tables will find that their answers differ slightly from those given. 10.1

Legitimate hypotheses concern population characteristics; x is a sample statistic.

10.2

Does not comply. The alternative hypothesis must involve an inequality.

Does not comply. The inequality in the alternative hypothesis must refer to the hypothesized value.

Does comply.

Does not comply. The alternative hypothesis must involve an inequality referring to the hypothesized value.

Does not comply, since p̂ is not a population characteristic.

10.3

Because so much is at stake at a nuclear power plant, the inspection team needs to obtain convincing evidence that everything is in order. To put this another way, the team needs not only to obtain a sample mean greater than 100 but, beyond that, to be sure that sample mean is sufficiently far above 100 to provide convincing evidence that the true mean weld strength is greater than 100. Hence an alternative hypothesis of H a :   100 will be used.

10.4

The conclusion is consistent with testing Ho: HGH in addition to IVF does not increase the chance of getting pregnant versus Ha: HGH in addition to IVF does increase the chance of getting pregnant. The null hypothesis is considered a statement of no difference, or equality. In this case, HGH in addition to IVF doesn’t increase the chance of getting pregnant

10.5

Recall that a statistical hypothesis test is capable of demonstrating strong support for the alternative hypothesis (by rejecting the null hypothesis). The statement of “no strong evidence” is referring to no strong evidence in support of the alternative hypothesis. Therefore, the null hypothesis was not rejected.

In this case, consider the population of youths with pre-existing psychological conditions, and the following hypotheses: H0: Video games do not increase antisocial behavior versus Ha: Video games do increase antisocial behavior.

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249

The article states that the researchers found no evidence “that violent video games increase antisocial behavior in youths with pre-existing psychological conditions.” Finding no evidence simply means that the data did not provide strong support for the alternative, not that the null hypothesis is true. Therefore, the title is misleading in that it is implying that the null hypothesis is true. 10.6

H0 : p 

2 3

versus H a : p 

2 3

, where p is the proportion of employers who perform background

checks. 10.7

H0: p = 0.5 versus Ha: p > 0.5, where p = the population proportion of adult Americans who believe they are better off.

10.8

H0:   170 versus Ha:   170.

10.9

Let p be the proportion of all constituents who favor spending money for the new sewer system. She should test H0: p  0.5 versus Ha: p  0.5.

10.10 H0: μ = 10.3 versus Ha: μ > 10.3 10.11

H 0 : p  0.7 versus H a : p  0.7 , where p is the proportion of college students who are Facebook users and log into their Facebook profile at least six times a day.

10.12 a

Type I

A Type I error is coming to the conclusion that cancer is present when, in fact, it is not. Treatment may be started when, in fact, no treatment is necessary.

A Type II error is coming to the conclusion that no cancer is present when, in fact, the illness is present. No treatment will be prescribed when, in fact, treatment is necessary.

Type I error. Using a small significance level (  ) reduces the probability of making a Type I error by making it more difficult to reject the null hypothesis.

10.13 a

This is a Type II error because the statement describes the result of failing to reject the null hypothesis when the null hypothesis is actually false (concluding that the woman does not have breast cancer when, in actuality, she does). This probability is approximately

P(Type II error)  b

 0.077 .

The other error that is possible is a Type I error, in which the null hypothesis is rejected when it should not be. In this scenario, a Type I error is concluding that a woman has cancer when she really does not have cancer. This probability is approximately

P(Type I error)  10.14 a

90 637

 0.141 .

A Type I error is coming to the conclusion that the symptoms are due to disease when in fact the symptoms are due to child abuse.

250

Chapter 10: Hypothesis Testing Using a Single Sample A Type II error is coming to the conclusion that the symptoms are due to child abuse when in fact the symptoms are due to disease. b

10.15 a

b 10.16 a

The doctor considers the presence of child abuse more serious than the presence of disease. Thus, according to the doctor, undetected child abuse is more serious than undetected disease, and a Type I error is the more serious. A Type I error is when it is concluded that a particular man is the father when, in fact, he is not the father. A Type II error is when it is concluded that a particular man is not the father when, in fact, he is the father.

  0 and   0.0001 A Type I error would be obtaining convincing evidence that less than 90% of the TV sets need no repair when in fact (at least) 90% need no repair. The consumer agency might take action against the manufacturer when in fact the manufacturer is not at fault. A Type II error would be not obtaining convincing evidence that less than 90% of the TV sets need no repair when in fact less than 90% need no repair. The consumer agency would not take action against the manufacturer when in fact the manufacturer is making untrue claims about the reliability of the TV sets.

Taking action against the manufacturer when in fact the manufacturer is not at fault could involve large and unnecessary legal costs to the consumer agency. Thus a Type I error could be considered serious, whereas a Type II error would only involve not catching a manufacturer who is making false claims. Therefore, in order to reduce the probability of a Type I error, a procedure using   0.01 should be recommended.

10.17 a

A Type I error is obtaining convincing evidence that more than 1% of a shipment is defective when in fact (at least) 1% of the shipment is defective. A Type II error is not obtaining convincing evidence that more than 1% of a shipment is defective when in fact more than 1% of the shipment is defective.

The consequence of a Type I error would be that the calculator manufacturer returns a shipment when in fact it was acceptable. This will do minimal harm to the calculator manufacturer’s business. However, the consequence of a Type II error would be that the calculator manufacturer would go ahead and use in the calculators circuits that are defective. This will then lead to faulty calculators and would therefore be harmful to the manufacturer’s business. A Type II error would be the more serious for the calculator manufacturer.

At least in the short term, a Type II error would not be harmful to the supplier’s business; payment would be received for a shipment that was in fact faulty. However, if a Type I error were to occur, the supplier would receive back, and not be paid for, a shipment of circuits that was in fact acceptable. A Type I error would be the more serious for the supplier.

10.18 a

A Type I error is obtaining convincing evidence that the mean water temperature is greater than 150°F when in fact it is (at most) 150°F. A Type II error is not obtaining convincing evidence that the mean water temperature is greater than 150°F when in fact it is greater than 150°F.

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251

If a Type II error occurs, then the ecosystem will be harmed and no action will be taken. This could be considered more serious than a Type I error, where a company will be required to change its practices when in fact it is not contravening the regulations. A Type II error is more serious.

10.19 The probability of a Type I error is equal to the significance level. Here the aim is to reduce the probability of a Type I error, so a small significance level (such as 0.01) should be used. 10.20 a

The area will be closed to fishing if the fish are determined to have an unacceptably high mercury content. Thus we should test H0:   5 versus Ha:   5.

If a Type II error occurs, then an unacceptably high mercury level will go undetected, and people will continue to fish in the area. This could be considered more serious than a Type I error, where fishing will be prohibited in an area where the mercury level is in fact acceptable. We thus wish to reduce the probability of a Type II error, and therefore a significance level of 0.1 should be used.

10.21 a

The phrase “No evidence of increased risk of thyroid cancer in areas that were near a nuclear power plant” indicates that the researchers failed to reject the null hypothesis.

If the researchers are incorrect in their conclusion, then they would have failed to reject H0 when H0 was actually false. This is a Type II error.

No. The study did not provide sufficient evidence to reject the null hypothesis. In other words, there was not convincing evidence that the proportion of the population in areas near nuclear power plants who are diagnosed with thyroid cancer during a given year is greater than the proportion of the population in areas where no nuclear power plants are present. This does not mean that there was no effect, only that the data did not provide evidence to reject the null hypothesis.

10.22 a

The conversion will be undertaken only if there is strong evidence that the proportion of defective installations is lower for the robots than for human assemblers. Thus the manufacturer should test H0: p  0.02 versus Ha: p  0.02.

A Type I error would be obtaining convincing evidence that the proportion of defective installations for the robots is less than 0.02 when in fact it is (at least) 0.02. A Type II error would be not obtaining convincing evidence that the proportion of defective installations for the robots is less than 0.02 when in fact it is less than 0.02.

If a Type I error occurred, then people would lose their jobs and the company would install a costly new system that does not perform any better than its former employees. If a Type II error occurred, then people would keep their jobs but the company would maintain a production process less effective than would be achieved with the proposed new system. Since the Type I error, with people losing their jobs, appears to be more serious than a Type II error, the probability of a Type I error should be reduced, and a significance level of 0.01 should be used.

10.23 a

A P-value of 0.0003 means that it is very unlikely (probability = 0.0003), assuming that H0 is true, that you would get a sample result at least as inconsistent with H0 as the one obtained in the study. Thus H0 is rejected.

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A P-value of 0.350 means that it is not particularly unlikely (probability = 0.350), assuming that H0 is true, that you would get a sample result at least as inconsistent with H0 as the one obtained in the study. Thus there is no reason to reject H0.

10.24 The null hypothesis will be rejected if the P-value is less than or equal to 0.05. a The null hypothesis will be rejected. b c

The null hypothesis will be rejected. The null hypothesis will not be rejected.

The null hypothesis will be rejected.

The null hypothesis will not be rejected.

10.25 a

H0 is not rejected.

10.26 a

H0 is rejected.

H0 is not rejected.

10.27 a

P -value  P ( z  1.40)  0.081.

P -value  P ( z  0.93)  0.176.

P -value  P ( z  1.96)  0.025.

P -value  P ( z  2.45)  0.007.

P -value  P ( z  0.17)  0.567.

10.28 a

The large-sample z test is not appropriate since np  25(0.2)  5  10.

The large-sample z test is appropriate since np  210(0.6)  126  10 and n (1  p )  210(0.4)  84  10.

The large-sample z test is appropriate since np  100(0.9)  90  10 and n (1  p )  100(0.1)  10  10.

The large-sample z test is not appropriate since np  75(0.05)  3.75  10.

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10.29 1. p = proportion of all employers in the U.S. who have fired an employee for holiday shopping online while at work 2. H0: p = 0.1 3. Ha: p > 0.1 4.   0.01 pˆ  p pˆ  0.1 5. z   p(1  p)  0.1 0.9  n 2379 6. We are told to assume that it is reasonable to regard the sample as representative of employers in the U.S. The sample size is much smaller than the population size (the number of employers in the U.S.). Furthermore, np  2379  0.1  237.9  10 and

n(1  p)  2379  0.9   2141.1  10 , so the sample is large enough. Therefore the large sample test is appropriate. 262 0.11013  0.10  0.11013 , therefore z   1.64697 7. pˆ  2,379 0.10 1  0.10 

2, 379 8. P-value  P( Z  1.64697)  0.0498 9. Because the P-value of 0.0498 is greater than the selected significance level of   0.01 , we fail to reject the null hypothesis. There is not convincing evidence that more than 10% of employers have fired an employee for shopping online at work. 10.30 a

1. p = proportion of all women who work full time, age 22 to 35, who would be willing to give up some personal time in order to make more money. 2. H0: p = 0.5 3. Ha: p > 0.5 4.   0.01 pˆ  p pˆ  0.5  5. z  p(1  p ) (0.5)(0.5) n 1000 6. The sample was selected in a way that was designed to produce a sample that was representative of women in the targeted group, so it is reasonable to treat the sample as a random sample from the population. The sample size is much smaller than the population size (the number of women age 22 to 35 who work full time). Furthermore, np  1000(0.5)  500  10 and n (1  p )  1000(0.5)  500  10 , so the sample is large enough. Therefore the large sample test is appropriate. 540 1000  0.5  2.52982 7. z  (0.5)(0.5) 1000 8. P -value  P ( Z  2.52982)  0.00571 9. Since P-value  0.00571  0.01 we reject H0. We have convincing evidence that a majority of women age 22 to 35 who work full time would be willing to give up some personal time for more money. No. The survey only covered women age 22 to 35.

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10.31 a

1. p = proportion of all adult Americans who would answer the question correctly 2. H0: p = 0.4 3. Ha: p < 0.4 4.   0.05 pˆ  p pˆ  0.4 5. z   p(1  p ) (0.4)(0.6) n 1000 6. The sample size is much smaller than the population size (the number of adult Americans). Furthermore, np  1000(0.4)  400  10 and n (1  p )  1000(0.6)  600  10 , so the sample is large enough. We are told to assume that the sample is representative of adult Americans. Thus the large sample test is appropriate. 354 1000  0.4 7. z   2.969 (0.4)(0.6) 1000 8. P -value  P ( Z  2.969)  0.001 9. Since P-value  0.001  0.05 we reject H0. We have convincing evidence that the proportion of all adult Americans who would answer the question correctly is less than 0.4. 1. p = proportion of all adult Americans who would select a wrong answer 2. H0: p = 1/3 3. Ha: p > 1/3 4.   0.05 pˆ  p pˆ  1 3 5. z   p(1  p ) 1 3 2 3 n 1000 6. The sample size is much smaller than the population size (the number of adult Americans). Furthermore, np  1000 1 3  33.3  10 and

n(1  p )  1000  2 3  66.7  10 , so the sample is large enough. We are told to assume that the sample is representative of adult Americans. Thus the large sample test is appropriate. 0.378  1 3  2.996 7. z  1 3 2 3 1000 8. P -value  P ( Z  2.996)  0.001 9. Since P-value  0.001  0.05 we reject H0. We have convincing evidence that more than one-third of adult Americans would select a wrong answer. 10.32 a

H0: p = 0.5 versus Ha: p > 0.5, where p = proportion of all adult Americans who would prefer to live in a hot climate rather than a cold climate

Since P-value  0.000001  0.01 we reject H0. We have convincing evidence that a majority of adult Americans would prefer to live in a hot climate over a cold climate.

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10.33 1. p = proportion of all adult Americans who were somewhat interested or very interested in having Web access in their cars 2. H0: p = 0.5 3. Ha: p < 0.5 4.   0.05 pˆ  p pˆ  0.5 5. z   p(1  p ) (0.5)(0.5) n 1005 6. The sample size is much smaller than the population size (the number of adult Americans). Furthermore, np  1005(0.5)  502.5  10 and n(1  p )  1005(0.5)  502.5  10 , so the sample is large enough. We are told to assume that the sample can be considered as representative of adult Americans. Thus the large sample test is appropriate. pˆ  0.5 0.46  0.5 7. z    2.536 (0.5)(0.5) (0.5)(0.5) 1005 1005 8. P -value  P ( Z  2.536)  0.006 9. Since P-value  0.006  0.05 we reject H0. We have convincing evidence that the proportion of all adult Americans who want car Web access is less than 0.5. The marketing manager is not correct in his claim. 10.34 a

H0 : p 

2 3

versus H a : p 

2 3

, where p is the actual proportion of adult Americans who

believe that doctors should be able to end a terminally ill patient’s life if requested to do so by the patient. b

Because the P-value of 0.058 is greater than the significance level of   0.05 , we fail to reject H0. There is not convincing evidence that more than 2/3 of adult Americans believe that doctors should be able to end a terminally ill patient’s life if requested to do so by the patient.

Yes, I would reach a different conclusion if   0.10 . In this case, the P-value of 0.058 is less than the significance level of   0.10 , so we reject H0. There is convincing evidence that more than 2/3 of adult Americans believe that doctors should be able to end a terminally ill patient’s life if requested to do so by the patient.

10.35 H0: p = 0.69 versus Ha: p ≠ 0.69, where p = proportion of teens at the high school who access social media from a mobile phone 10.36 a

Because the null hypothesis was rejected, there is convincing evidence that the proportion of mobile phone users who received at least one nuisance call on their mobile phones within the past month is greater than 0.667.

Yes, it is reasonable to say that the data provide strong support for the alternative hypothesis.

Yes, it is reasonable to say that the data provide strong evidence against the null hypothesis.

10.37

H 0 : p  0.87 versus H a : p  0.87 , where p is the proportion of college students who are Facebook users and log into their Facebook profile at least six times a day.

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10.38 There is strong evidence that the proportion of adult Americans who think that taxing sugary beverages is an acceptable intervention in an effort to reduce consumption of sugary beverages is less than 0.5 10.39 a

Here np = 728(0.25) = 182 ≥ 10 and n(1 – p) = 728(0.75) = 546 ≥ 10. So the sampling distribution of p̂ is approximately normal. The sampling distribution of p̂ has mean p = 0.25 and standard deviation

p(1  p) n  (0.25)(0.75) 728  0.016 .

P( pˆ  0.27)  P(Z  (0.27  0.25) 0.016)  P(Z  1.246)  0.106 . This probability is not particularly small, so it would not be particularly surprising to observe a sample proportion as large as pˆ  0.27 if the null hypothesis were true.

P( pˆ  0.31)  P(Z  (0.31  0.25) 0.016)  P(Z  3.739)  0.0001 . This probability is small, so it would be surprising to observe a sample proportion as large as pˆ  0.31 if the null hypothesis were true.

10.40 If the null hypothesis, H0: p = 0.25, is true, P( pˆ  0.33)  P(Z  (0.33  0.25) 0.016)  P ( Z  4.985)  0 . This means that it is very unlikely that you would observe a sample proportion as large as 0.33 if the null hypothesis is true. So, yes, there is convincing evidence that more than 25% of law enforcement agencies review social media activity as part of background checks. 10.41 a

1. p = proportion of U.S. businesses who monitor employees’ web site visits 2. H0: p = 0.6 3. Ha: p > 0.6 4.   0.01 pˆ  p pˆ  0.6  5. z  p(1  p ) (0.6)(0.4) n 304 6. We are told to regard the sample as representative of U.S businesses, and it is therefore reasonable to treat the sample as a random sample from that population. The sample size is much smaller than the population size (the number of U.S. businesses). Furthermore, np  304(0.6)  182.4  10 and n(1  p )  304(0.4)  121.6  10 , so the sample is large enough. Therefore the large sample test is appropriate. 201 304  0.6  2.178 7. z  (0.6)(0.4) 304 8. P -value  P ( Z  2.178)  0.015 9. Since P-value  0.015  0.01 we do not reject H0. We do not have convincing evidence that more than 60% of U.S. businesses monitor employees’ web site visits. 1. p = proportion of U.S. businesses who monitor employees’ web site visits 2. H0: p = 0.5 3. Ha: p > 0.5 4.   0.01

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8. 9.

257

pˆ  p pˆ  0.5  p(1  p ) (0.5)(0.5) n 304 We are told to regard the sample as representative of U.S businesses, and it is therefore reasonable to treat the sample as a random sample from that population. The sample size is much smaller than the population size (the number of U.S. businesses). Furthermore, np  304(0.5)  152  10 and n(1  p )  304(0.5)  152  10 , so the sample is large enough. Therefore the large sample test is appropriate. 201 304  0.5 z  5.621 (0.5)(0.5) 304 P -value  P ( Z  5.621)  0 Since P-value  0  0.01 we reject H0. We have convincing evidence that a majority of U.S. businesses monitor employees’ web site visits. z

10.42 It is not necessary to carry out a hypothesis test to determine if the proportion of registered voters in California who voted in the 2016 presidential election is less than the national proportion of 0.600 because the 57.8% stated in the article is the proportion of all registered voters who voted in the 2016 election, which is a population characteristic. This tells us that the population proportion is less than 0.600, so there is no need to carry out a hypothesis test. 10.43 a

P-value  P(t8  2.0)  0.040.

P-value  P(t13  3.2)  0.003.

P-value  P(t10  2.4)  0.019.

P-value  P(t21  4.2)  0.000.

10.44 a

P-value  2  P(t15  1.6)  0.130.

P-value  2  P(t15  1.6)  0.130.

P-value  2  P(t15  6.3)  0.000.

10.45 a

P-value  2  P(t9  0.73)  0.484.

P-value  P(t10  0.5)  0.686.

P-value  P(t19  2.1)  0.025.

P-value  P(t19  5.1)  0.000.

P-value  2  P(t39  1.7)  0.097.

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P-value  P(t14  3.2)  0.003. a Since 0.003 < 0.05 we reject H0. We have convincing evidence that the mean reflectometer reading for the new type of paint is greater than 20. b

Since 0.003 < 0.01 we reject H0. We have convincing evidence that the mean reflectometer reading for the new type of paint is greater than 20.

Since 0.003 > 0.001 we do not reject H0. We do not have convincing evidence that the mean reflectometer reading for the new type of paint is greater than 20.

10.47 a

P-value  P(t17  2.3)  0.017  0.05. H0 is rejected. We have convincing evidence that the mean writing time for all pens of this type is less than 10 hours.

P-value  P(t17  1.83)  0.042  0.01. H0 is not rejected. We do not have convincing evidence that the mean writing time for all pens of this type is less than 10 hours.

Since t is positive, the sample mean must have been greater than 10. Therefore, we certainly do not have convincing evidence that the mean writing time for all pens of this type is less than 10 hours. H0 is certainly not rejected.

10.48 a

P-value  2  P(t12  1.6)  0.136. Since 0.136 > 0.05 we do not reject H0. We do not have convincing evidence that the mean diameter of ball bearings of this type is not equal to 0.5.

P-value  2  P(t12  1.6)  0.136. Since 0.136 > 0.05 we do not reject H0. We do not have convincing evidence that the mean diameter of ball bearings of this type is not equal to 0.5.

P-value  2  P(t24  2.6)  0.016. Since 0.016 > 0.01 we do not reject H0. We do not have convincing evidence that the mean diameter of ball bearings of this type is not equal to 0.5.

P-value  2  P(t24  3.6)  0.001. Since the P-value is very small, we reject H0 at any reasonable significance level. We have convincing evidence that the mean diameter of ball bearings of this type is not equal to 0.5.

10.49 a

1.  = mean heart rate after 15 minutes of Wii Bowling for all boys age 10 to 13 2. H0:   98 3. Ha:   98 4.   0.01 x   x  98  5. t  s n s n 6. We are told to assume that it is reasonable to regard the sample of boys as representative of boys age 10 to 13. Under this assumption, it is reasonable to treat the sample as a random sample from the population. We are also told to assume that the distribution of heart rates after 15 minutes of Wii Bowling is approximately normal. So we can proceed with the t test. 101  98  0.74833 7. t  15 14 8. P-value  2  P(t13  0.74833)  0.468

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9. Since P-value  0.468  0.01 we do not reject H0. We do not have convincing evidence that the mean heart rate after 15 minutes of Wii Bowling is not equal to 98 beats per minute. b

1.  = mean heart rate after 15 minutes of Wii Bowling for all boys age 10 to 13 2. H0:   66 3. Ha:   66 4.   0.01 x   x  66  5. t  s n s n 6. We are told to assume that it is reasonable to regard the sample of boys as representative of boys age 10 to 13. Under this assumption, it is reasonable to treat the sample as a random sample from the population. We are also told to assume that the distribution of heart rates after 15 minutes of Wii Bowling is approximately normal. So we can proceed with the t test. 101  66  8.731 7. t  15 14 8. P-value  P(t13  8.731)  0 9. Since P-value  0  0.01 we reject H0. We have convincing evidence that the mean heart rate after 15 minutes of Wii Bowling is greater than 66 beats per minute.

It is known that treadmill walking raises the heart rate over the resting heart rate, and the study provided convincing evidence that Wii Bowling does so, also. Although the sample mean heart rate for Wii Bowling was higher than the known population mean heart rate for treadmill walking, the study did not provide convincing evidence of a difference of the population mean heart rate for Wii Bowling from the known population mean for the treadmill.

10.50 a

The large standard deviation tells us that there is great variability in the number of calories consumed.

Using the sample mean and standard deviation as approximations to the population mean and standard deviation we see that zero is just over one standard deviation below the mean. If the distribution of the number of calories consumed were normal, then a significant proportion of people would be consuming negative numbers of calories. Since this is not possible, we see that the distribution of the number of calories consumed must be positively skewed, and therefore that the assumption of normality is not valid.

1.  = mean number of calories in a New York City hamburger chain lunchtime purchase 2. H0:   750 3. Ha:   750 4.   0.01 x   x  750  5. t  s n s n 6. We are told to regard the sample of 3857 fast-food purchases as representative of all hamburger chain lunchtime purchases in New York City. Also, n  3857  30. So we can proceed with the t test.

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857  750  9.816 677 3857 8. P-value  P(t3856  9.816)  0 9. Since P-value  0  0.01 we reject H0. We have convincing evidence that the mean number of calories consumed is above the recommendation of 750. 7.

10.51 a

t

No. The study was conducted in New York City only, and therefore the results cannot be generalized to the lunchtime fast food purchases of all adult Americans.

If you ask the customers what they purchased, some customers might misremember or might give false answers. By looking at the receipt, you know that you are receiving an accurate response.

Yes. It is possible that knowing that a record of their lunch order was going to be seen might have influenced the amount of food customers ordered. For example, this knowledge might cause customers to order less, out of the fear of embarrassment at people seeing the sizes of their orders.

10.52 1.  = mean salary offering for mathematics and statistics graduates at this university 2. H0:   62985 3. Ha:   62985 4.   0.05 x   x  62985 5. t   s n s n 6. The sample was a random sample from the population. Also, n  50  30. So we can proceed with the t test. 63500  62985 7. t   1.104 3300 50 8. P-value  P(t49  1.104)  0.14 9. Since P-value  0.14  0.05 we fail to reject H0. We do not have convincing evidence that the mean salary offer for mathematics and statistics graduates of this university is higher than the 2016 national average of $62,985. 10.53 1.  = mean July 2016 price of a Big Mac in Europe 2. H0:   5.04 3. Ha:   5.04 4.   0.05 x   x  5.04  5. t  s n s n 6. We are told to assume that it is reasonable to regard the sample as representative of European McDonald’s restaurants, and therefore we can treat the sample as a random sample from that population. The sample values are summarized in the boxplot below.

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2.5

3.0 3.5 4.0 European Big Mac Prices (in U.S. Dollars)

261

4.5

The boxplot is only moderately left-skewed and the sample contains no outliers, so we are justified in assuming that the distribution of Big Mac prices across all McDonald’s restaurants in Europe is approximately normal. Thus we can proceed with the t test. 7.

x  3.86, s  0.64265, n  12, df  11 3.86  5.04 t  6.36 0.64265 12 P-value  P(t11  6.36)  0

8. 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that the mean July 2016 price of a Big Mac in Europe is less than $5.04. 10.54 1.

 = mean credit card balance of college students who pay their credit card balance in full

each month 2. H0:   906 3. Ha:   906 4.   0.01 x   x  906 5. t   s n s n 6. The sample is representative of the population. Also, n  500  30. So we can proceed with the t test. 825  906 7. t   9.056 200 500 8. P-value  P(t499  9.056)  0 9. Since P-value  0  0.01 we reject H0. We have convincing evidence that college students who pay their credit card balance in full each month have a mean balance that is lower than $906. 10.55 a

The fact that the sample standard deviation is greater than the sample mean indicates that there is a quite a bit of variability in the distribution of purchase amounts. In addition, since purchase amounts cannot be negative, the larger standard deviation indicates that the distribution is positively skewed.

1.  = mean amount spent per purchase after the change to increase healthy food options 2. H0:   2.80 3. Ha:   2.80

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Chapter 10: Hypothesis Testing Using a Single Sample 4. 5.

  0.05 x   x  2.80 t  s n s n

6. We are told that the sample is representative of purchases at corner stores in Philadelphia after the stores increase their healthy food options. Also, n  5949  30. So we can proceed with the t test. 2.86  2.80 7. t   0.857 5.40 5949 8. P-value  P(t5948  0.857)  0.196 9. Since P-value=0.196>0.05 we fail to reject H0. We do not have convincing evidence that the population mean amount spent per purchase is greater after the change to increase healthy food options. 10.56 1.  = mean wrist extension for all people using the new mouse design 2. H0:   20 3. Ha:   20 4.   0.05 x   x  20  5. t  s n s n 6. We need to assume that the 24 students used in the study form a random sample from the set of all people using this new mouse design. The sample values are summarized in the boxplot below.

25 26 27 28 Wrist Extension (degrees)

As shown in the boxplot, there is an outlier. The boxplot shows a roughly symmetrical shape. Despite the presence of the outlier, we assume that the distribution of wrist extensions for the population is normal, and proceed with the t test.

x  25.917, s  1.954, n  24, df  23 25.917  20 t  14.836 1.954 24 8. P-value  P(t23  14.836)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that the mean wrist 7.

extension for all people using the new mouse design is greater than 20 degrees. To generalize the result to the population of Cornell students, we need to assume that the 24 students used in the study are representative of all students at the university. To generalize the result to the population of all university students, we need to assume that the 24 students used in the study are representative of all university students. 10.57 Since the sample was large, it was possible for the hypothesis test to provide convincing evidence that the mean score for the population of children who spent long hours in child care was greater

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263

than the mean score for third graders in general, even though the obtained sample mean didn’t differ greatly from the known mean for third graders in general. 10.58 a

It is not reasonable to think the distribution of time spent playing video or computer games for the population of male Canadian high school students is approximately normal. If the distribution were approximately normal with the given mean and standard deviation, roughly 15% of the observed times would be negative, which is not possible.

We would need to know that the sample was randomly selected or is representative of the population of male Canadian high school students. Additionally, we would need to know that the distribution of the population of times is normal or that the sample is large (at least size 30).

10.59 a

 = mean time (in minutes) spent playing video or computer games for male Canadian

high school students 2. H0:   120 3. Ha:   120 4.   0.05 x   x  120 5. t   s n s n 6. The sample is representative of the population. Also, n  500  30. So we can proceed with the t test. 123.4  120 7. t   0.649 117.1 500 8. P-value  P(t499  0.649)  0.258 9. Since P-value  0.258  0.05 we do not reject H0. We do not have convincing evidence that the average time spent playing video or computer games for male Canadian high school students is greater than 2 hours (120 minutes). b

 = mean time (in minutes) spent playing video or computer games for male Canadian

high school students 2. H0:   120 3. Ha:   120 4.   0.05 x   x  120 5. t   s n s n 6. The sample is representative of the population. Also, n  500  30. So we can proceed with the t test. 123.4  120 7. t   2.049 37.1 500 8. P-value  P(t499  2.049)  0.020 9. Since P-value  0.020  0.05 we reject H0. We have convincing evidence that the average time spent playing video or computer games for male Canadian high school students is greater than 2 hours (120 minutes).

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The sample standard deviation of 117.1 minutes in part (a) indicates that there is likely to be more variability in the number of minutes of video or computer game playing than when the sample standard deviation was 37.1 minutes, as in part (b). If the mean of the population of average time spent playing video or computer games for male Canadian high school students was indeed 120 minutes, it is less likely that a sample mean of 123.4 minutes would be obtained when the standard deviation was 37.1 minutes than when the standard deviation was 117.1 minutes. Therefore, the larger standard deviation in part (a) makes it more difficult to detect a difference, resulting in a failure to reject the null hypothesis.

10.60 By saying that listening to music reduces pain levels the authors are telling us that the study resulted in convincing evidence that pain levels are reduced when music is being listened to. (In other words, the results of the study were statistically significant.) By saying, however, that the magnitude of the positive effects was small, the authors are telling us that the effects were not practically significant. 10.61 a

Yes. Since the pattern in the normal probability plot is roughly linear, and since the sample was a random sample from the population, the t test is appropriate. The boxplot shows a median of around 245, and since the distribution is roughly symmetrical distribution, this tells us that the sample mean is around 245, also. This might initially suggest that the population mean differs from 240. But when you consider the fact that the sample is relatively small, and that the sample values range all the way from 225 to 265, you realize that such a sample mean would still be feasible if the population mean were 240. 1.  = mean calorie content for frozen dinners of this type 2. H0:   240 3. Ha:   240 4.   0.05 x   x  240  5. t  s n s n 6. As explained in Part (a), the conditions for performing the t test are met. 7. The mean and standard deviation of the sample values are 244.33333 and 12.38278, 244.33333  240  1.21226. respectively. So t  12.38278 12 8. P-value  2  P(t11  1.21226)  0.251 9. Since P-value  0.251  0.05 we do not reject H0. We do not have convincing evidence that the mean calorie content for frozen dinners of this type differs from 240.

10.62 1.  = mean rate of uptake for cultures with nitrates 2. H0:   8000 3. Ha:   8000 4.   0.1 x   x  8000  5. t  s n s n 6. We need to assume that the cultures used in this study form a random sample from the set of all possible such cultures. The sample values are summarized in the boxplot below.

Chapter 10: Hypothesis Testing Using a Single Sample

6000

7000

8000 Rate of Uptake (dpm)

9000

265

10000

The boxplot is roughly symmetrical and the sample contains no outliers, so we are justified in assuming that the population distribution of rates of uptake is approximately normal. Thus we can proceed with the t test. 7.

x  778.8, s  1002.431, n  15, df  14 778.8  8000 t  0.816 1002.431 15 P-value  P(t14  0.816)  0.214

8. 9. Since P-value  0.214  0.1 we do not reject H0. We do not have convincing evidence that the mean rate of uptake is reduced by the addition of nitrates. 10.63 a b 10.64 a

Increasing the sample size increases the power. Increasing the significance level increases the power. The z statistic should be used because we know the population standard deviation,  .

  P( z  1.8)  0.036.

Since   153, Ha is true, and a Type II error would consist of failing to reject H0. Now, x  150 z .  n So, for z  1.8, x  150 1.8  . 10 50 This gives  10  x  150  1.8    152.546.  50  Thus, H0 will be rejected for values of x greater than 152.546, and H0 will not be rejected for values of x less than 152.546. The required sketch is shown below.

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Density

149

10.65 a

151

152.55

153

155

X-bar

A Type II error occurs when H0 is not rejected, and, as shown in the sketch, H0 will not be rejected for values of x less than 152.546. Thus, 152.546  153   area under standard normal curve to the left of 10 50

 area under standard normal curve to the left of  0.3213  0.374. b

Again, H0 will not be rejected for values of x less than 152.546. Thus, 152.546  160   area under standard normal curve to the left of 10 50

 area under standard normal curve to the left of  5.271  0. c 10.66 a b

Since 152.4 < 152.546, H0 will not be rejected, and a Type II error could have occurred.

  area under standard normal curve to the left of  1.28  0.1. When z  1.28, x  10  ( 1.28)(.1)  9.872. So H0 is rejected for values of x  9.872 . If   9.8, then x is normally distributed with mean 9.8 and standard deviation 0.1. So P( H 0 is rejected)  P( x  9.872)

 area under standard normal curve to left of (9.872  9.8) 0.1  area under standard normal curve to left of 0.72  0.7642. So   P( H 0 not rejected)  1  0.7642  0.2358. c

H0 states that   10 and Ha states that   10 . Since 9.5 is further from 10 (in the direction indicated by Ha),  is less for   9.5 than for   9.8 . For   9.5 ,

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P( H 0 is rejected)  P( x  9.872)  area under standard normal curve to left of (9.872  9.5) 0.1  area under standard normal curve to left of 3.72  0.9999. So   P( H 0 not rejected)  1  0.9999  0.0001. d

10.67 a

Power when   9.8 is 1  0.2358  0.7642. Power when   9.5 is 1  0.0001  0.9999.

102 125  0.75  1.704, and so the P(0.75)(0.25) 125 value is P ( Z  1.704)  0.044, which is less than 0.05. Thus H0 is rejected, and we have convincing evidence that more than 75% of apartments exclude children. Testing H0: p = 0.75 against Ha: p > 0.5 we have z 

This is a one-tailed test with   0.05, so H0 is rejected for values of z  1.6449. Now pˆ  0.75 z , (0.75)(0.25) 125 so when z  1.6449,

(0.75)(0.25)  0.8137. 125 Thus we need P( pˆ  0.8137) when p  0.8. Now, when p  0.8, the distribution of p̂ is approximately normal with mean 0.8 and pˆ  0.75  1.6449

(0.8)(0.2) 125. So we require the area under the standard normal curve 0.8137  0.8 , that is, the area to the right of 0.3831. This area is 0.351. (0.8)(0.2) 125

standard deviation to the right of

10.68 a

0.0372  0.035  0.46565. So P-value  P(t6  0.46565)  0.329  0.05. Therefore, H0 is 0.0125 7 not rejected, and we do not have convincing evidence that   0.035. t

0.04  0.035  0.4. Using Appendix Table 5, for a one-tailed test,   0.05, 6 degrees 0.0125 of freedom, we get   0.75.

Power  1  0.75  0.25.

10.69 a

d

df  10  1  9 52  50 d  0.2. Using Appendix Table 5 for a one-tailed test at   0.05,   0.84. i 10 55  50  0.5. Using Appendix Table 5 for a one-tailed test at   0.05,   0.57. ii d  10

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iii iv b

60  50  1. Using Appendix Table 5 for a one-tailed test at   0.05,   0.10. 10 70  50 d  2. Using Appendix Table 5 for a one-tailed test at   0.05,   0. 10 d

When  increases, each d value decreases, and, looking at the graphs in Appendix Table 5, we see that each value of  will be greater than in Part (a).

10.70 Using Appendix Table 5: 0.52  0.5 a d  1,   0.04. 0.02 b

d

0.48  0.5  1,   0.04. 0.02

d

0.52  0.5  1,   0.24. 0.02

10.71 a

d

0.54  0.5  2,   0. 0.02

d

0.54  0.5  1,   0.04. 0.04

d

0.54  0.5  1,   0.01. 0.04

10.72 The article states that the FDA approved a benchmark in which Liberte could be up to three percentage points worse than Express. Thus, if the reclogging rate for the Express stent is 7%, Boston Scientific has to find convincing evidence that Liberte’s rate of reclogging is less than 10%. Thus the given hypotheses are appropriate. 10.73 a

1. p = proportion of all women who would like to choose a baby’s sex who would choose a girl. 2. H0: p = 0.5 3. Ha: p  0.5 4.   0.05 pˆ  p pˆ  0.5  5. z  p(1  p ) (0.5)(0.5) n 229 6. We need to assume that the sample was a random sample from the population of women who would like to choose the sex of a baby. The sample size is presumably much smaller than the population size (the number of women who would like to choose the sex of a baby). Also, np  229(0.5)  114.5  10 and n(1  p )  229(0.5)  114.5  10 , so the sample is large enough. Therefore the large sample test is appropriate.

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140 229  0.5  3.370 (0.5)(0.5) 229 8. P -value  P ( Z  3.370)  0.0004 9. Since P-value  0.0004  0.05 we reject H0. We have convincing evidence that the proportion of all women who would like to choose a baby’s sex who would choose a girl is not equal to 0.5. (This contradicts the statement in the article.) 7.

z

The survey was conducted only on women who had visited the Center for Reproductive Medicine at Brigham and Women’s Hospital. It is quite possible that women who choose this institution have views on the matter that are different from those of women in general. (This is selection bias.) Also, with only 561 of the 1385 women responding, it is quite possible that the majority who did not respond had different views from those who did. (This is nonresponse bias.) For these two reasons it seems unreasonable to generalize the results to a larger population.

10.74 a

The large-sample hypothesis test for a population proportion cannot be used in this example because the large sample size conditions ( np and n(1  p) both at least 10) are not satisfied.

The randomization test is used to test the hypotheses H 0 : p  0.27 versus H a : p  0.27 , where p is the proportion of lunar astronauts who are at increased risk of CVD. A P-value of 0.301 was obtained. Because the P-value of 0.301 is greater than any reasonable significance level, we fail to reject the null hypothesis. There is not sufficient evidence to conclude that, as a group, lunar astronauts are at increased risk of death caused by CVD.

10.75 a

We should test the hypotheses H 0 : p  0.20 versus H a : p  0.20 , where p is the proportion of hospital patients who had been treated for pneumonia using a respiratory therapist protocol who were readmitted to the hospital within 30 days after discharge.

It is not stated that the sample was randomly selected or that the sample is representative of the population of all hospital patients who had been treated for pneumonia using a respiratory therapist protocol. As such, we must use caution and assume the sample was selected in a reasonable way. The large sample size condition has been satisfied because there are at least 10 successes and failures in the sample.

The output for the exact binomial test indicates that the P-value is 0.000. Since the P-value of 0.000 is less than any reasonable significance level, we reject the null hypothesis. We have sufficient evidence to conclude that the proportion of subjects who will be readmitted to a hospital within 30 days after following a respiratory therapist protocol for treatment of pneumonia is less than 0.20.

In this case, pˆ 

15 162

 0.0926 , and z 

0.0926  0.20 0.20 1  0.20 

 3.42 . This is a lower-tailed test

162 (the inequality in Ha is <), so the P-value is the area under the z curve and to the left of –3.42. Therefore, the P-value is P( z  3.42)  0.0003 . This P-value is larger than the one obtained in part (c), but the same conclusion will be drawn.

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10.76 a

These data should not be analyzed using a large-sample hypothesis test for one proportion because the number of successes and failures are not both at least 10. In this case, the number of successes is 33, but the number of failures is 2, which is less than 10. The exact binomial test gives a P-value of approximately 0. Therefore, we reject the null hypothesis. We have sufficient evidence to conclude that the proportion of all dogs trained using this method who would perform the correct new action is greater than 0.5.

10.77 a

These data should not be analyzed using a large-sample hypothesis test for one population proportion because the number of successes and failures are not both at least 10. In this case, the number of successes is 17, and the number of failures is 8, which is less than 10.

We will use a randomization test to test the hypotheses H 0 : p  0.10 versus H a : p  0.10 , where p is the proportion of people who quit smoking prior to surgery who are smoking again after one year. A P-value of 0.000 was obtained. Because the P-value of 0.000 is less than any reasonable significance level, we reject the null hypothesis. There is sufficient evidence to conclude that the proportion of people who quit smoking prior to surgery who would smoke again after one year is greater than 0.10.

The patients who were smokers were required to quit for their surgery, but maybe they quit against their will. These patients perhaps never wanted to quit, so began smoking again once they were allowed to do so.

10.78 a

We want to determine if there is evidence that a majority of U.S. businesses monitor employees’ web site visits. Therefore, the population characteristic of interest, p, is the proportion of U.S. businesses that monitor employees’ web site visits. Test the hypotheses H 0 : p  0.50 versus H a : p  0.50 .

We are told that it is reasonable to regard the sample as representative of U.S. businesses. In addition, np  304(0.5)  152 and n(1  p )  304(1  0.5)  152 , which are both at least 10. The two required conditions are satisfied.

Because the P-value of 0.000 is less than any reasonable significance level, we reject the null hypothesis. There is convincing evidence that a majority of U.S. businesses monitor employees’ web site visits.

The P-value obtained using the large-sample test was approximately 0, which is what we obtained here. The same conclusion would have been reached.

10.79 a

We should test the hypotheses H 0 : p  0.50 versus H a : p  0.50 , where p is the proportion of coin flip calls that the Patriots win. The exact binomial test gives a P-value of 0.007. Because the P-value of 0.007 is less than any reasonable significance level, we reject the null hypothesis. We have sufficient evidence to conclude that the proportion of coin flip calls that the Patriots win is greater than 0.50.

In order to carry out the exact binomial version of the hypothesis test, there are four necessary conditions. (1) There is a fixed number of trials; (2) Each trial can result in one of only two possible outcomes; (3) Outcomes of different trials are independent; and (4) The success probability is the same for each trial. In this case, all the conditions have been satisfied.

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Rejecting the null hypothesis does not mean that the alternative hypothesis is true. Rather, it means we have evidence that supports the alternative hypothesis based on the data we collected. It could be that we incorrectly rejected the null hypothesis based on this sample (which is a Type I error). 10.80 a

We want to know if there is convincing evidence that the population mean brain size for birds that are relatives of the dodo differs from the established dodo brain size of 4.17. The population characteristic of interest is  the true mean brain size for birds that are relatives of the dodo. Hypotheses: H 0 :   4.17 versus H a :   4.17 . The given randomization test resulted in a P-value of 0.000. Because this P-value of 0.000 is less than any reasonable significance level, we reject the null hypothesis. We have convincing evidence that the population mean brain size for birds that are relatives of the dodo differs from the established dodo brain size of 4.17.

The results of this test provide evidence that the population mean brain size for birds that are relatives of the dodo differs from the established dodo brain size of 4.17.

10.81 a

The sample size of n = 21 is smaller than 30, so the methods based on the t distribution may not be appropriate.

We want to know if there is convincing evidence that the population mean time discrimination score for male smokers who abstain from smoking for 24 hours is significantly greater than 1. The population characteristic of interest is , the true mean time discrimination score for male smokers who abstain from smoking for 24 hours. Hypotheses: H 0 :   1 versus H a :   1 . The given randomization test resulted in a P-value of 0.029. Because this P-value of 0.029 is less than a significance level of 0.05, we reject the null hypothesis. We have convincing evidence that the population mean time discrimination score for male smokers who abstain from smoking for 24 hours is significantly greater than 1.

10.82 a

A boxplot (shown below) was constructed. Note that the boxplot indicates an outlier in the distribution, so the methods based on the t distribution might not be appropriate.

-30

-20

-10

Points

We want to know if there is convincing evidence that the population mean points difference for NFL teams coming off a bye week differs from zero. The population characteristic of interest is , the true mean points difference for NFL teams coming off a bye week. Hypotheses: H 0 :   0 versus H a :   0 . Different simulations will produce different

272

Chapter 10: Hypothesis Testing Using a Single Sample results, so answers will vary. One randomization test resulted in a P-value of 0.355. Because this P-value of 0.355 is greater than a significance level of 0.05, we fail to reject the null hypothesis. We do not have convincing evidence that the population mean points difference for NFL teams coming off a bye week differs from zero. c

Based on the results from part (b), it is not plausible that teams coming off a bye week have a significant advantage in points scored over their opponents. In the hypothesis test, the null hypothesis was not rejected, concluding that we do not have convincing evidence that the population mean points difference for NFL teams coming off a bye week differs from zero.

10.83 a

We want to know if there is convincing evidence that the population mean 2000-meter ergometer time for U.S. junior male sculls rowers differs from the 2007 international standard of 387 seconds. The population characteristic of interest is , the true mean 2000-meter ergometer time for U.S. junior male sculls rowers. Hypotheses: H 0 :   387 versus

H a :   387 . The given randomization test resulted in a P-value of 0.000. Because this Pvalue of 0.000 is less than a significance level of 0.05, we reject the null hypothesis. We have convincing evidence that the population mean 2000-meter ergometer time for U.S. junior male sculls rowers differs from the 2007 international standard of 387 seconds. b

Based on the result of the hypothesis test, we can say that there is evidence to conclude that the mean 2000-meter ergometer time for U.S. junior male sculls rowers differs from the 2007 international standard of 387 seconds. The sample mean time of 394.25 seconds, given in the Shiny App output, is greater than 387 seconds. Therefore, since the null hypothesis was rejected and the sample mean is greater than the hypothesized mean, we can conclude that the U.S. junior male sculls rowers seem to have “caught up,” on average, with the international championship rowers from 2007.

10.84 a

The sample size of n = 12 is smaller than 30, so the methods based on the t distribution may not be appropriate.

We want to know if the sample data provide convincing evidence that the mean July 2016 price of a Big Mac in Europe is less than the reported U.S. price. The population characteristic of interest is , the true mean July 2016 price of a Big Mac in Europe. Hypotheses: H 0 :   5.04 versus H a :   5.04 . The given randomization test resulted in a P-value of 0.000. Because this P-value of 0.000 is less than a significance level of 0.05, we reject the null hypothesis. We have convincing evidence that the population mean July 2016 price of a Big Mac in Europe is less than the reported U.S. price.

10.85 a

The fact that the sample size (n = 15) is smaller than 30 indicates that the methods based on the t distribution might not be appropriate.

We want to know if the sample data provide convincing evidence that the population mean interleague winning percentage for NL teams differs from 50%. The population characteristic of interest is , the true mean interleague winning percentage. Hypotheses: H 0 :   50 versus H a :   50 . Different simulations will produce different results, so answers will vary. One randomization test resulted in a P-value of 0.170. Because this Pvalue of 0.170 is greater than a 5% significance level, we fail to reject the null hypothesis. We do not have convincing evidence that the population mean interleague winning percentage for NL teams differs from 50%.

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In Part (b), the null hypothesis was not rejected, which indicates that there is insufficient evidence to conclude that the mean winning percentage for NL teams differs from 50%. Therefore, it is not reasonable to say that the National League or American League performs significantly better than the other in interleague play.

10.86 1. p = proportion of adult Americans who can identify the correct definition of the Bill of Rights. 2. H0: p = 0.5 3. Ha: p < 0.5 4.   0.01 pˆ  p pˆ  0.5 5. z   p(1  p ) (0.5)(0.5) n 1000 6. We are told that the sample was representative of adult Americans, so it is reasonable to treat the sample as a random sample from the population. The sample size is much smaller than the population size (the number of adult Americans). Furthermore, np  1000(0.5)  500  10 and n (1  p )  1000(0.5)  500  10 , so the sample is large enough. Therefore the large sample test is appropriate. 459 1000  0.5 7. z   2.59 (0.5)(0.5) 1000 8. P -value  P( Z  2.59)  0.005 9. Since P-value=0.005  0.01 we reject H0. We have convincing evidence that less than half of adult Americans can identify the correct definition of the Bill of Rights. 10.87 1. p = proportion of all U.S. adults who believe that rudeness is a worsening problem 2. H0: p = 0.75 3. Ha: p < 0.75 4.   0.05 pˆ  p pˆ  0.75  5. z  p(1  p ) (0.75)(0.25) n 2013 6. We need to assume that the sample was a random sample from the population of U.S. adults. The sample size is much smaller than the population size (the number of U.S. adults). Furthermore, np  2013(0.75)  1509.75  10 and n(1  p )  2013(0.25)  503.25  10 , so the sample is large enough. Therefore the large sample test is appropriate. 1283 2013  0.75  11.671 7. z  (0.75)(0.25) 2013 8. P -value  P ( Z  11.671)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that less than threequarters of all U.S. adults believe that rudeness is a worsening problem.

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10.88 a

1. p = proportion of spins that land heads up 2. H0: p = 0.5 3. Ha: p ≠ 0.5 4.   0.01 pˆ  p pˆ  0.5 5. z   p(1  p ) (0.5)(0.5) n 250 6. We need to assume that the coin-spins in the study formed a random sample of all coinspins of Euro coins of this sort. Also, np  250(0.5)  125  10 and n(1  p )  250(0.5)  125  10 , so the sample is large enough. Therefore the large sample test is appropriate. 140 250  0.5 7. z   1.897 (0.5)(0.5) 250 8. P -value  2  P( Z  1.897)  0.058. 9. Since P-value  0.058  0.01 we do not reject H0. We do not have convincing evidence that the proportion of the time this type of coin lands heads up is not 0.5. With a significance level of 0.05, the conclusion would be the same, since the P-value is greater than 0.05.

10.89 1.  = mean MPV for patients with gastric cancer 2. H0:   7.85 3. Ha:   7.85 4.   0.05 x   x  7.85 5. t   s n s n 6. We are told that the sample was representative of the population. Also, n  31  30. So we can proceed with the t test. 8.31  7.85 7. t   3.28 0.78 31 8. P-value  P(t30  3.28)  0.0013 9. Since P-value=0.0013  0.05 we reject H0. We have convincing evidence that the mean MPV for patients with gastric cancer is greater than 7.85 fL. 10.90 1.  = mean sleep duration for students at this college 2. H0:   8.4 3. Ha:   8.4 4.   0.01 x   x  8.4  5. t  s n s n 6. We are told that the sample was randomly selected from the population. Also, n  236  30. So we can proceed with the t test.

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7.71  8.4  10.29 1.03 236 8. P-value  P(t235  10.29)  0 9. Since P-value  0  0.01 we reject H0. We have convincing evidence that the mean sleep duration for students at this college is less than the recommended number of 8.4 hours. 7.

10.91 a

t

We would need to assume that the sample was randomly selected from the population of white-collar workers in the U.S. or that the sample is representative of the population.

1.  = mean time spent checking work email 2. H0:   4 3. Ha:   4 4.   0.05 x  x 4 5. t   s n s n 6. We are told to assume that the sample was randomly selected from the population. Also, n  1004  30. So we can proceed with the t test. 4.1  4 7. t   2.437 1.3 1004 8. P-value  P(t1003  2.437)  0.0075 9. Since P-value=0.0075  0.05 we reject H0. We have convincing evidence that the mean time spent checking work email for white-collar workers in the U.S. is more than half of the 8-hour work day.

10.92 a

1. p = proportion of adult Americans who disapprove of the National Security Agency collecting records of phone and internet data 2. H0: p = 0.5 3. Ha: p > 0.5 4.   0.05 pˆ  p pˆ  0.5  5. z  p(1  p) (0.5)(0.5) n 1000 6. We are told that the sample was randomly selected. Also, np  1000(0.5)  500  10 and n(1  p )  1000(0.5)  500  10 , so the sample is large enough. Therefore the large sample test is appropriate. 0.54  0.5  2.53 7. z  (0.5)(0.5) 1000 8. P -value  P( Z  2.53)  0.0057 9. Since P-value  0.0057  0.05 we reject H0. We have convincing evidence that a majority of adult Americans disapprove of the National Security Agency collecting records of phone and internet data.

Using a sample size much larger than 1,000 would result in a smaller P-value. The test statistic (z) would be larger, and therefore the sample result would be more standard deviations above the hypothesized value. As such, the P-value would be smaller.

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10.93 1. p = proportion of adult Americans ages 26 to 32 years who own a fitness band 2. H0: p = 0.25 3. Ha: p >0.25 4.   0.05 pˆ  p pˆ  0.25 5. z   p(1  p) (0.25)(0.75) n 500 6. We are told that the sample of size 500 was selected to be representative of adult Americans ages 26 to 32 years. The sample size is much smaller than the population size (the number of adult Americans ages 26 to 32 years). Furthermore, np  500(0.25)  125  10 and n(1  p )  500(0.75)  375  10 , so the sample is large enough. Therefore the large sample test is appropriate. 0.27  0.25 7. z   1.03 (0.25)(0.75) 500 8. P -value  P ( Z  1.03)  0.15 9. Since P-value  0.15  0.05 we fail to reject H0. We do not have convincing evidence that more than one-quarter of all adult Americans ages 26 to 32 years own a fitness band. 10.94 a

1. p = proportion of students at this college who are Facebook users and update their

status at least two times a day 2. H0: p = 0.8 3. Ha: p > 0.8 4.   0.05 pˆ  p pˆ  0.8  5. z  p(1  p) (0.8)(0.2) n 261 6. We are told to assume that the sample is representative of the students at this university. Also, np  261(0.8)  208.8  10 and n(1  p )  261(0.2)  52.2  10 , so the sample is large enough. Therefore the large sample test is appropriate. 0.87  0.8  2.83 7. z  (0.8)(0.2) 261 8. P -value  P( Z  2.83)  0.0023 9. Since P-value  0.0023  0.05 we reject H0. We have convincing evidence that more than 80% of the students at this college who are Facebook users update their status at least two times a day. b

No, it would not be reasonable to generalize the conclusion from the test in Part (a) to all college students in the U.S. This sample was taken to be representative of the population of students at this university, so that is the only population to which we can generalize the results.

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10.95 1. p = proportion of all U.S. adults who approve of casino gambling 2. H0: p = 2/3 3. Ha: p > 2/3 4.   0.05 pˆ  p pˆ  2 3 5. z   p(1  p ) (2 3)(1 3) n 1523 6. We need to assume that the sample selected at random from households with telephones was a random sample from the population of U.S. adults. The sample size is much smaller than the population size (the number of U.S. adults). Furthermore, np  1523(2 3)  1015  10 and n(1  p)  1523(1 3)  508  10 , so the sample is large enough. Therefore the large sample test is appropriate. 1035 1523  2 3 7. z   1.06902 (2 3)(1 3) 1523 8. P -value  P ( Z  1.06902)  0.143 9. Since P-value  0.143  0.05 we do not reject H0. We do not have convincing evidence that more than two-thirds of all U.S. adults approve of casino gambling. 10.96 1. p = proportion of all local residents who oppose hunting on Morro Bay 2. H0: p = 0.5 3. Ha: p > 0.5 4.   0.01 pˆ  p pˆ  0.5 5. z   p(1  p ) (0.5)(0.5) n 750 6. We are told that the sample was a random sample of local residents. The sample size is much smaller than the population size (the number of local residents). Furthermore, np  750(0.5)  375  10 and n(1  p )  750(0.5)  375  10 , so the sample is large enough. Therefore the large sample test is appropriate. 560 750  0.5  13.510 7. z  (0.5)(0.5) 750 8. P -value  P ( Z  13.510)  0 9. Since P-value  0  0.01 we reject H0. We have convincing evidence that a majority of local residents oppose hunting on Morro Bay. 10.97 1. p = proportion of all people who would respond if the distributor is fitted with an eye patch 2. H0: p = 0.4 3. Ha: p > 0.4 4.   0.05 pˆ  p pˆ  0.4  5. z  p(1  p ) (0.4)(0.6) n 200 6. We have to assume that the sample was random sample from the population of people who could be approached with a questionnaire. The sample size is much smaller than the

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Chapter 10: Hypothesis Testing Using a Single Sample population size. Furthermore, np  200(0.4)  80  10 and n (1  p )  200(0.6)  120  10 , so the sample is large enough. Therefore the large sample test is appropriate. 109 200  0.4 7. z   4.186 (0.4)(0.6) 200 8. P -value  P ( Z  4.186)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that more than 40% of all people who could be approached with a questionnaire will respond when the distributor wears an eye patch.

10.98 1.  = mean fuel efficiency under these circumstances 2. H0:   30 3. Ha:   30 4.   0.05 x   x  30  5. t  s n s n 6. We need to assume that the five drives in the study form a random sample from the set of all possible such drives. We are told to assume that fuel efficiency is normally distributed under these circumstances. Thus we can proceed with the t test. 7. x  29.333, s  1.408, n  6, df  5 29.333  30 t  1.160 1.408 6 8. P-value  P(t5  1.160)  0.149 9. Since P-value  0.149  0.05 we do not reject H0. We do not have convincing evidence that the mean fuel efficiency under these circumstances is less than 30 miles per gallon. 10.99 1.  = mean daily revenue since the change 2. H0:   75 3. Ha:   75 4.   0.05 x   x  75  5. t  s n s n 6. The sample was a random sample of days. In order to proceed with the t test we must assume that the distribution of daily revenues since the change is normal. 70  75  5.324 7. t  4.2 20 8. P-value  P(t19  5.324)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that the mean daily revenue has decreased since the price increase.

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10.100 1.  = mean time to 100°F 2. H0:   15 3. Ha:   15 4.   0.05 x   x  15  5. t  s n s n 6. The sample was a random sample from the population. Since the sample was relatively small, we need to assume that the population distribution of times to 100°F is normal. 17.5  15  5.682 7. t  2.2 25 8. P-value  P(t24  5.682)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that the mean time to 100°F is greater than 15.

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Cumulative Review Exercises CR10.1 Gather a set of volunteer older people with knee osteoarthritis (we will assume that 40 such volunteers are available). Have each person rate his/her knee pain on a scale of 1-10, where 10 is the worst pain. Randomly assign the volunteers to two groups, A and B, of equal sizes. (This can be done by writing the names of the volunteers onto slips of paper. Place the slips into a hat, and pick 20 at random. These 20 people will go into Group A, and the remaining 20 people will go into Group B.) The volunteers in Group A will attend twice weekly sessions of one hour of tai chi. The volunteers in Group B will simply continue with their lives as they usually would. After 12 weeks, each volunteer should be asked to rate his/her pain on the same scale as before. The mean reduction in pain for Group A should then be compared to the mean reduction in pain for Group B. CR10.2 No. It seems that the designer of the display is intending the lines without the arrow heads to represent the percentages. However, a person looking at the display could think that the proportions are represented by the lines including the arrow heads, and as a result the display is misleading. For example, when the arrow heads are included the length of the arrow representing 2% is much more than 2/13 of the length of the arrow representing 13%. CR10.3 a

28 42 56 70 Number of flights delayed more than 3 hours

There are three airlines that stand out from the rest by having large numbers of delayed flights. These airlines are ExpressJet, Delta, and Continental, with 93, 81, and 72 delayed flights, respectively. b

0.0

0.7

1.4

2.1 2.8 Rate per 100,000 flights

3.5

4.2

4.9

A typical number of flights delayed per 10,000 flights is around 1.1, with most rates lying between 0 and 1.6. There are four airlines that standout from the rest by having particularly high rates, with two of those four having particularly high rates. c

The rate per 100,000 flights data should be used, since this measures the likelihood of any given flight being late. An airline could standout in the number of flights delayed data purely as a result of having a large number of flights.

CR10.4 Median = 19.2, Lower Quartile = 14.4, Upper Quartile = 23.4, IQR = 9. Lower Quartile − 1.5(IQR) = 0.9 Upper Quartile + 1.5(IQR) = 36.9 Upper Quartile + 3(IQR) = 50.4

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Since 49.6 is greater than 36.9, this value of 49.6 (for Boston) is an outlier. There are no other values greater than 36.9 or less than 0.9, so there are no other outliers. The boxplot is shown below.

30 Average Wait Time

CR10.5 a

The number of people in the sample who change their passwords quarterly is binomially distributed with n  20 and p  0.25. So, using Appendix Table 9, p (3)  0.134.

Using Appendix Table 9, P (more than 8 change passwords quarterly)  0.027  0.010  0.003  0.001  0.041.

 x  np  100(0.25)  25 ,  x  np(1  p)  100(0.25)(0.75)  4.330.

Since np  100(0.25)  25  10 and n(1  p )  100(0.75)  75  10 the normal approximation to the binomial distribution can be used. Thus, 19.5  25   P( x  20)  P  z   P( z  1.27017)  0.102. 4.33013  

CR10.6 2

 1.96   1.96  n  p(1  p )    0.25    2401. A sample size of 2401 is required.  B   0.02  CR10.7 a

P (O )  0.4.

Anyone who accepts a job offer must have received at least one job offer, so P( A)  P(O  A)  P ( A | O ) P (O )  (0.45)(0.4)  0.18.

P(G )  0.26.

P ( A | O )  0.45.

Since anyone who accepts a job offer must have received at least one job offer, P (O | A)  1.

P ( A  O )  P ( A)  0.18.

Check of Conditions 1. Since npˆ  200(0.173)  34.6  10 and n(1  pˆ )  200(0.827)  165.4  10, the sample is large enough.

CR10.8

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Chapter 10: Hypothesis Testing Using a Single Sample 2. The sample size of n = 200 is much smaller than 10% of the population size (the number of adults in the U.S.). 3. The sample was said to be selected in a way that would produce a set of people representative of adults in the U.S., so we are justified in assuming that the sample was a random sample from that population. Calculation The 95% confidence interval for p is pˆ (1  pˆ ) (0.173)(0.827) pˆ  1.96  0.173  1.96  (0.121, 0.225). n 200 Interpretation We are 95% confident that the proportion of all U.S. adults who planned to buy a Valentine’s Day gift for their pet was between 0.121 and 0.225. b

The confidence interval would be narrower.

1. p = proportion of all U.S. adults who planned to buy a Valentine’s Day gift for their pet in 2010. 2. H0: p = 0.15 3. Ha: p > 0.15 4.   0.05 pˆ  p pˆ  0.15 5. z   p(1  p ) (0.15)(0.85) n 200 6. The sample size is much smaller than the population size (the number of adults in the U.S. in 2010). Furthermore, np  200(0.15)  30  10 and n (1  p )  200(0.85)  170  10 , so the sample is large enough. The sample was said to be selected in a way that would produce a set of people representative of adults in the U.S., so we are justified in assuming that the sample was a random sample from that population. Therefore the large sample test is appropriate. 0.173  0.15  0.911 7. z  (0.15)(0.85) 200 8. P -value  P ( Z  0.911)  0.181 9. Since P-value  0.181  0.05 we do not reject H0. We do not have convincing evidence that the proportion of adults in the U.S. in 2010 who planned to buy a Valentine’s Day gift for their pet was greater than 0.15.

Check of Conditions 1. Since npˆ  115(38 115)  38  10 and n(1  pˆ )  2002(77 2002)  77  10, the sample size is large enough. 2. The sample size of n = 115 is much smaller than 10% of the population size (the number of U.S. medical residents). 3. We are told to regard the sample as a random sample from the population. Calculation The 95% confidence interval for p is

CR10.9

Chapter 10: Hypothesis Testing Using a Single Sample

pˆ  1.96

283

pˆ (1  pˆ ) 38 (38 115)(77 115)   1.96  (0.244, 0.416). n 115 115

Interpretation We are 95% confident that the proportion of all U.S. medical residents who work moonlighting jobs is between 0.244 and 0.416. b

Check of Conditions 1. Since npˆ  115(22 115)  22  10 and n(1  pˆ )  115(93 2002)  93  10, the sample size is large enough. 2. The sample size of n = 115 is much smaller than 10% of the population size (the number of U.S. medical residents). 3. We are told to regard the sample as a random sample from the population. Calculation The 90% confidence interval for p is pˆ (1  pˆ ) 22 (22 115)(93 115) pˆ  1.645   1.645  (0.131, 0.252). n 115 115 Interpretation We are 90% confident that the proportion of all U.S. medical residents who have credit card debt of more than $3000 is between 0.131 and 0.252.

The interval in Part (a) is wider than the interval in Part (b) because the confidence level in Part (a) (95%) is greater than the confidence level in Part (b) (90%) and because the sample proportion in Part (a) (38/115) is closer to 0.5 than the sample proportion in Part (b) (22/115).

CR10.10 a

Check of Conditions 1. Since npˆ  3000(0.9)  2700  10 and n(1  pˆ )  3000(0.1)  300  10, the sample size is large enough. 2. The sample size of n = 3000 is much smaller than 10% of the population size (the number of people age 18 to 24 in these nine countries). 3. We need to assume that the 3000 respondents formed a random sample of people age 18 to 24 in those nine countries. Calculation The 90% confidence interval for p is pˆ (1  pˆ ) (0.9)(0.1) pˆ  1.645  0.9  1.645  (0.891, 0.909). n 3000 Interpretation We are 90% confident that the proportion of people age 18 to 24 in those nine countries who can identify their own country on a blank world map is between 0.891 and 0.909.

As stated above, we need to assume that the 3000 respondents formed a random sample of people age 18 to 24 in those nine countries.

Having made the assumption stated above, it would be reasonable to generalize the confidence interval to people age 18 to 24 in those nine countries.

CR10.11 A reasonable estimate of  is given by (sample range) 4  (20.3  19.9) 4  0.1. Thus

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 1.96   1.96  0.1  n     384.16.  B   0.01  2

So we need a sample size of 385. CR10.12 1. p = proportion of all adult Americans who plan to alter their shopping habits if gas prices remain high 2. H0: p = 0.75 3. Ha: p > 0.75 4.   0.05 pˆ  p pˆ  0.75 5. z   p(1  p ) (0.75)(0.25) n 1813 6. We are told to regard the sample as representative of adult Americans, so it is reasonable to treat the sample as a random sample from that population. The sample size is much smaller than the population size (the number of adult Americans). Furthermore, np  1813(0.75)  1359.75  10 and n(1  p )  1813(0.25)  453.25  10 , so the sample is large enough. Therefore the large sample test is appropriate. 1432 1813  0.75 7. z   3.919 (0.75)(0.25) 1813 8. P -value  P ( Z  3.919)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that more than threequarters of adult Americans plan to alter their shopping habits if gas prices remain high. CR10.13 1. p = proportion of all baseball fans who believe that the designated hitter rule should either be expanded to both baseball leagues or eliminated 2. H0: p = 0.5 3. Ha: p > 0.5 4.   0.05 pˆ  p pˆ  0.5  5. z  p(1  p ) (0.5)(0.5) n 394 6. The sample was a random sample from the population. The sample size is much smaller than the population size (the number of baseball fans). Furthermore, np  394(0.5)  197  10 and n (1  p )  394(0.5)  197  10 , so the sample is large enough. Therefore the large sample test is appropriate. 272 394  0.5  7.557 7. z  (0.5)(0.5) 394 8. P -value  P ( Z  7.557)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that a majority of baseball fans believe that the designated hitter rule should either be expanded to both baseball leagues or eliminated.

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285

CR10.14 1. p = proportion of adult Americans who do not use the internet 2. H0: p = 0.1 3. Ha: p > 0.1 4.   0.05 pˆ  p pˆ  0.1 5. z   p(1  p) (0.1)(0.9) n 600 6. The sample was chosen to be representative of the population of adult Americans. The sample size is much smaller than the population size (the number of adult Americans). Furthermore, np  600(0.1)  60  10 and n(1  p )  600(0.9)  540  10 , so the sample is large enough. Therefore, the large sample test is appropriate. 0.13  0.1 7. z   2.45 (0.1)(0.9) 600 8. P -value  P ( Z  2.45)  0.007 9. Since P-value  0.007  0.05 , we reject H0. We have convincing evidence that the proportion adult Americans who do not use the internet is greater than 0.10 (10%). CR10.15 a

CR10.16 a

With a sample mean of 14.6, the sample standard deviation of 11.6 places zero just over one standard deviation below the mean. Since no teenager can spend a negative time online, to get a typical deviation from the mean of just over 1, there must be values that are substantially more than one standard deviation above the mean. This suggests that the distribution of online times in the sample is positively skewed. 1.  = mean weekly time online for teenagers 2. H0:   10 3. Ha:   10 4.   0.05 x   x  10  5. t  s n s n 6. The sample was a random sample of teenagers. Also, n  534  30. Therefore we can proceed with the t test. 14.6  10  9.164. 7. t  11.6 534 8. P-value  P(t533  9.164)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that the mean weekly time online for teenagers is greater than 10 hours. 1.  = mean number of hours Canadian parents of teens think their teens spend online 2. H0:   10 3. Ha:   10 4.   0.05

286

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5. 6. 7. 8. 9.

x   x  10  s n s n We are told that the sample formed a random sample from the population of Canadian parents of teens. Also, n  676  30. So we can proceed with the t test. 6.5  10 t  10.581 8.6 676 P-value  P(t675  10.581)  0 Since P-value  0  0.05 we reject H0. We have convincing evidence that the mean number of hours Canadian parents of teens think their teens spend online is less than 10 per week. t

In the previous exercise, we found that there was convincing evidence that the mean of the time the teens say they spend online is greater than 10. In Part (a) of this exercise we found that there was convincing evidence that the mean time the parents say they think their teens spend online is less than 10. Thus it is clear that either the parents have an unrealistic view of their teens’ Internet use, or the teens themselves are unclear as to the actual time they spend online, or a combination of the two.

Chapter 11 Comparing Two Populations or Treatments Note: In this chapter, numerical answers to questions involving the normal and t distributions were found using values from a calculator. Students using statistical tables will find that their answers differ slightly from those given. 11.1

Since n1 and n2 are large, the distribution of x1  x2 is approximately normal. Its mean is

1  2  30  25  5 and its standard deviation is 11.2

11.3

12 n1



 22 n2



22 32   0.529. 40 50

H0: 1  2  10 versus Ha: 1  2  10

H0: 1  2  10 versus Ha: 1  2  10

1 = mean reading level for health-related pages for Wikipedia  2 = mean reading level for health-related pages for WebMD H0: 1  2  0 Ha: 1  2  0   0.05 ( x  x )  (hypothesized value) ( x1  x2 )  0 t 1 2  s12 s22 s12 s22   n1 n2 n1 n2

2. 3. 4. 5.

6. We are told that the samples were independent and selected to be representative of the two populations. Also n1  40  30 and n2  40  30 , so we can proceed with the two-sample t test.  26.7  43.9   0 7. t   4.5359 2 2 14.1 19.4      40 40 8. df = 71.214

P-value  2  P(t71.214  4.5359)  0 9. Since P-value  0  0.05 , we reject H0. We have convincing evidence that the mean reading level for health-related pages differs for Wikipedia and WebMD. 11.4

287

Check of Conditions We are told that the samples were independent and selected to be representative of the two populations. Also n1  40  30 and n2  40  30 , so we can proceed with construction of a two-sample t interval. Calculation df = 71.214. The 90% confidence interval for 1  2 is

288

Chapter 11: Comparing Two Populations or Treatments

( x1  x2 )   t critical value 

 26.7  43.9   1.66653

s12 n1

14.12 40

 

s22 n2

19.42 40

17.2  1.66653 3.792 17.2  6.31949

 23.5195, 10.8805 Interpretation We are 90% confident that the difference in mean Flesch reading ease score for health-related pages on Wikipedia and health-related pages on WebMD is between -23.5195 and -10.8805. Both endpoints of this interval are negative, so you can estimate that the mean Flesch reading ease score for WebMD is greater than the mean Flesch reading ease score for Wikipedia by somewhere between 10.8805 and 23.5195. b

11.5

This interval indicates that, on average, WebMD pages have a more difficult reading levels. This is consistent with the conclusion reached in the previous exercise because the null hypothesis of no difference in reading levels was rejected, and that the t test statistic was negative when the mean WebMD score subtracted from the mean Wikipedia score.

a 2009

1999

6 7 8 T ime per day using electronic media

We need to assume that the population distributions of time per day using electronic media are normal. Since the boxplots are roughly symmetrical and since there is no outlier in either sample this assumption is justified, and it is therefore reasonable to carry out a two-sample t test. b

1. 2. 3. 4. 5.

1 = mean time using electronic media for all kids age 8 to 18 in 2009  2 = mean time using electronic media for all kids age 8 to 18 in 1999 H0: 1  2  0 Ha: 1  2  0   0.01 ( x  x )  (hypothesized value) ( x1  x2 )  0 t 1 2  s12 s22 s12 s22   n1 n2 n1 n2

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289

6. We are told to assume that it is reasonable to regard the two samples as representative of kids age 8 to 18 in each of the two years when the surveys were conducted. We can then treat the samples as random samples from their respective populations. Also, as discussed in Part (a), the boxplots show that it is reasonable to assume that the population distributions are normal. So we can proceed with a two-sample t test. 7. x1  7.6 s1  1.595 x2  5.933 s2  1.100 7.6  5.933 t  3.332 1.5952 1.1002  15 15 8. df = 24.861 P-value  P(t24.861  3.332)  0.001 9. Since P-value  0.001  0.01 we reject H0. We have convincing evidence that the mean number of hours per day spent using electronic media was greater in 2009 than in 1999. c

As explained in Parts (a) and (b), the conditions for the two-sample t test or interval are satisfied. A 98% confidence interval for 1  2 is

( x1  x2 )  (t critical value)

s12 s22  n1 n2

 (7.6  5.93333)  2.48605

1.5952 1.1002  15 15

 (0.423, 2.910) We are 98% confident that the difference between the mean number of hours per day spent using electronic media in 2009 and 1999 is between 0.423 and 2.910. 11.6

1. 2. 3. 4. 5.

1 = mean number of minutes of sleep adults in Mexico get on a typical work night  2 = mean number of minutes of sleep adults in the U.S. get on a typical work night H0: 1  2  0 Ha: 1  2  0   0.05 ( x  x )  (hypothesized value) ( x1  x2 )  0 t 1 2  s12 s22 s12 s22   n1 n2 n1 n2

6. We are told that the samples were independent and selected to be representative of the two populations. Also n1  250  30 and n2  250  30 , so we can proceed with the two-sample t test.  426  391  0 7. t   11.068 2 2 40 30      250 250 8. df = 461.795 P-value  P(t461.795  11.068)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that, on average, adults in the U.S. get less sleep on work nights than adults in Mexico. The conclusion in the report is reasonable.

290

11.7

Chapter 11: Comparing Two Populations or Treatments

Check of Conditions We are told that the samples were independent and selected to be representative of the two populations. Also n1  250  30 and n2  250  30 , so we can proceed with construction of a two-sample t interval. Calculation df = 482.316. The 95% confidence interval for 1  2 is

( x1  x2 )   t critical value 

 423  409   1.96489 

s12 n1

352 250



s22 n2

422 250

14  1.96489  3.45774 14  6.79409

 7.206, 20.794  Interpretation We are 95% confident that the difference in the mean amount of sleep on a work night for adults in Canada and adults in England between 7.206 and 20.794 minutes. Both endpoints of this interval are positive, so you can estimate that the mean amount of sleep on a work night for adults in Canada is greater than the mean amount of sleep on a work night for adults in England by somewhere between 7.206 and 20.794 minutes. b

The confidence interval allows us to conclude that there is evidence of a difference in the mean amount of sleep on a work night for the two countries because zero is not contained in the confidence interval.

11.8

The treatment groups were small (10 people in each group) and so in order to use the two-sample t test it is necessary to assume that the population distributions of estimated times for those given the easy font and for those given the difficult font are normal. However, in both samples zero is about one-and-a-half standard deviations below the mean. If we assume that the populations have means and standard deviations equal to the sample means and standard deviations, this would imply that a significant proportion of people in each population have negative estimated times. Since no-one will, in fact, estimate the time to be negative, this tells us that the populations are unlikely to be approximately normally distributed. Thus the two-sample t test is not appropriate.

11.9

1. 2. 3. 4. 5.

1 = mean number of minutes per day spent on Facebook by females  2 = mean number of minutes per day spent on Facebook by males H0: 1  2  0 Ha: 1  2  0   0.05 ( x  x )  (hypothesized value) ( x1  x2 )  0 t 1 2  s12 s22 s12 s22   n1 n2 n1 n2

Chapter 11: Comparing Two Populations or Treatments

291

6. We are told that the two samples are representative of their populations of interest, and the samples are independent. Also n1  195  30 and n2  66  30 , so we can proceed with the two-sample t test. 7. x1  159.61 s1  100 x2  102.31 s2  100

t

159.61  102.31  0  4.02368 100 2 100 2

 195 66 8. df = 112.142 P -value  2  P (t  4.02368)  0.0001 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that the mean time spent on Facebook is not the same for males and for females. b

It is not reasonable to generalize the conclusion from the hypothesis test in part (a) to all male and all female college students in the U.S. because the sample was taken from one large university in Southern California, and is therefore not representative of all college students in the US.

11.10 a

Check of Conditions We are told that the samples were independent and selected to be representative of the two populations. Also n1  195  30 and n2  66  30 , so we can proceed with construction of a two-sample t interval. Calculation df = 112.142. The 95% confidence interval for 1  2 is

( x1  x2 )   t critical value 

s12

159.61  102.31  1.98134 

1002



195

s22 n2



1002 66

57.3  1.98134 14.2407 57.3  28.2157

 29.0843, 85.5157  Interpretation We are 95% confident that the difference in the mean time spent on Facebook for male college students and female college students in Southern California is between 29.0843 and 85.5157 minutes. Both endpoints of this interval are positive, so you can estimate that the mean amount of time spent on Facebook by female students is greater than the mean amount of time spent on Facebook by male students by somewhere between 29.0843 and 85.5157 minutes. b

The confidence interval implies that, on average, female students in Southern California spend more time on Facebook than male students. This is consistent with the conclusion in the hypothesis test, in which we rejected the claim of no difference in the amount of time spent on Facebook by male and female students.

292

Chapter 11: Comparing Two Populations or Treatments

11.11 1. 2. 3. 4. 5.

1 = mean time spent using a computer per day for males  2 = mean time spent using a computer per day for females H0: 1  2  0 Ha: 1  2  0   0.05 ( x  x )  (hypothesized value) ( x1  x2 )  0 t 1 2  s12 s22 s12 s22   n1 n2 n1 n2

6. It is necessary to assume that the samples were random samples of male and female students. Also n1  46  30 and n2  38  30 , so we can proceed with the two-sample t test.

t

45.8  39.4

 0.486 63.32 46  57.32 38 8. df = 81.283 P-value  P(t81.283  0.486)  0.314 9. Since P-value  0.314  0.05 we do not reject H0. We do not have convincing evidence that the mean time per day male students at this university spend using a computer is greater than the mean time for female students. 7.

11.12 Check of Conditions Since n1  311  30 and n2  620  30, and it is believed that the two samples were representative of the two populations of interest, we can proceed with construction of a two-sample t interval. Calculation df = 421.543. The 90% confidence interval for 1  2 is

( x1  x2 )  (t critical value)  (8.39  5.39)  1.648

s12 s22  n1 n2

14.832 8.762  311 620

 (1.497,4.503) Interpretation We are 90% confident that the difference in mean number of servings of chocolate per month for people who would and would not screen positive for depression is between 1.497 and 4.503. 11.13 Check of Conditions We are told that the samples were independent and selected to be representative of the two populations. We need to assume that the population distributions of sit-stand-sit times are normal. Since the boxplots (shown below) are roughly symmetrical and since there is no outlier in either sample this assumption is justified, and it is therefore reasonable to carry out a two-sample t test.

Chapter 11: Comparing Two Populations or Treatments

293

Biofeedback

No Biofeedback

2.0

2.5

3.0 3.5 4.0 4.5 Sit-Stand-Sit Times (seconds)

5.0

5.5

Calculation df = 26.2468. The 95% confidence interval for 1  2 is

s12

( x1  x2 )   t critical value 

 3.120  4.307    2.05459 

0.7232 15

 

s22 n2 0.5552 15

1.187   2.05459  0.235337 1.187  0.483521

 1.6705, 0.7035 Interpretation We are 90% confident that the difference in the mean sit-stand-sit time for the population of stroke patients who receive biofeedback weight training for 8 weeks and the population of stroke patients who did not receive biofeedback weight training for 8 weeks is between 1.6705 and 0.7035 seconds. Both endpoints of this interval are negative, so you can estimate that the mean sit-stand-sit time for the population of stroke patients who receive biofeedback weight training for 8 weeks is smaller than the mean sit-stand-sit time for the population of stroke patients who did not receive biofeedback weight training for 8 weeks by somewhere between 0.7035 and 1.6705 seconds. 11.14 1. 2. 3. 4. 5.

1 = mean number of hours spent studying per day for Facebook users  2 = mean number of hours spent studying per day students who are not Facebook users H0: 1  2  0 Ha: 1  2  0   0.01 ( x  x )  (hypothesized value) ( x1  x2 )  0 t 1 2  s12 s22 s12 s22   n1 n2 n1 n2

6. The samples were selected to be representative of the two populations. Also n1  141  30 and n2  68  30 . So we can proceed with the two-sample t test.

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Chapter 11: Comparing Two Populations or Treatments

1.47  2.76

 9.286 0.832 141  0.992 68 8. df = 113.861 P-value  P(t113.861  9.286)  0 9. Since P-value  0  0.01 we reject H0. We have convincing evidence that the mean time spent studying for Facebook users at this university is less than the mean time spent studying by students at the university who do not use Facebook. 7.

t

11.15 a

If the distributions were both approximately normal, then 22.8% of the male Internet Addiction scores would be negative, and 25.6% of the female Internet Addiction scores would be negative. Assuming that the scores must be positive, these percentages are too large to make it reasonable to believe that the distributions are approximately normal.

It would be appropriate to use the two-sample t test to test the null hypothesis that there is no difference in the mean internet addiction scores for males and female Chinese 6 th grade students. The sample sizes are both much larger than 30, so the Central Limit Theorem states that the sampling distribution of sample means is approximately normal, which allows us to carry out the test.

1. 2. 3. 4. 5.

1 = mean Internet Addiction score for female Chinese 6 th grade students  2 = mean Internet Addiction score for male Chinese 6 th grade students H0: 1  2  0 Ha: 1  2  0   0.05 ( x  x )  (hypothesized value) ( x1  x2 )  0 t 1 2  s12 s22 s12 s22   n1 n2 n1 n2

6. We are told that the two samples are representative of their populations of interest, and the samples are independent. Also n1  879  30 and n2  836  30 , so we can proceed with the two-sample t test. 7. x1  1.07 s1  1.63 x2  1.51 s2  2.03

t

1.07  1.51  0  4.93419 1.632  2.032

 879 836 8. df = 1600.7 P-value  P(t  4.93419)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that the mean Internet Addiction score is greater for male Chinese 6th grade students than for female Chinese 6th grade students. 11.16 a

Check of Conditions We are told that the subjects were randomly assigned to one of two experimental treatment groups (clear glass mug or white mug). We are also told to assume that the distribution of quality ratings for each of the two treatments is approximately normal. We can proceed with construction of a two-sample t interval.

Chapter 11: Comparing Two Populations or Treatments

295

Calculation df = 21.2555. The 95% confidence interval for 1  2 is

( x1  x2 )   t critical value 

 61.48  50.35    2.07809 

s12 n1

16.692

12 11.13   2.07809  7.55747

 

s22 n2 20.17 2 12

11.13  15.7051

 4.575, 26.835  Interpretation We are 95% confident that the difference in the mean quality rating for this coffee when served in a white mug and when served in a glass mug is between -4.575 and 26.835. Because the endpoints of the confidence interval have opposite signs, zero is included in the interval, and there may be no difference in the mean quality rating for this coffee when served in a white mug and when served in a glass mug. b

11.17 a

No, there is no reason to be convinced that the color of the mug makes a difference in the mean quality rating. Because the endpoints of the confidence interval have opposite signs, zero is included in the interval, and there may be no difference in the mean quality rating for this coffee when served in a white mug and when served in a glass mug. 1. 2. 3. 4. 5.

1 = mean percentage of time playing with police car for male monkeys  2 = mean percentage of time playing with police car for female monkeys H0: 1  2  0 Ha: 1  2  0   0.05 ( x  x )  (hypothesized value) ( x1  x2 )  0 t 1 2  s12 s22 s12 s22   n1 n2 n1 n2

6. We are told that that it is reasonable to regard these two samples of 44 monkeys as representative of the populations of male and female monkeys. It is therefore reasonable to regard them as random samples. Also n1  44  30 and n2  44  30 , so we can proceed with the two-sample t test. 18  8  10.359 7. t  52 42  44 44 8. df = 82.047 P-value  P(t82.047  10.359)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that the mean percentage of the time spent playing with the police car is greater for male monkeys than for female monkeys.

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Chapter 11: Comparing Two Populations or Treatments

1. 2. 3. 4. 5.

1 = mean percentage of time playing with doll for male monkeys  2 = mean percentage of time playing with doll for female monkeys H0: 1  2  0 Ha: 1  2  0   0.05 ( x  x )  (hypothesized value) ( x1  x2 )  0 t 1 2  s12 s22 s12 s22   n1 n2 n1 n2

6. We are told that that it is reasonable to regard these two samples of 44 monkeys as representative of the populations of male and female monkeys. It is therefore reasonable to regard them as random samples. Also n1  44  30 and n2  44  30 , so we can proceed with the two-sample t test. 9  20 7. t   16.316 22 42  44 44 8. df = 63.235 P-value  P(t63.235  16.316)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that the mean percentage of the time spent playing with the doll is greater for female monkeys than for male monkeys. c

1. 2. 3. 4. 5.

1 = mean percentage of time playing with furry dog for male monkeys  2 = mean percentage of time playing with furry dog for female monkeys H0: 1  2  0 Ha: 1  2  0   0.05 ( x  x )  (hypothesized value) ( x1  x2 )  0 t 1 2  s12 s22 s12 s22   n1 n2 n1 n2

6. We are told that that it is reasonable to regard these two samples of 44 monkeys as representative of the populations of male and female monkeys. It is therefore reasonable to regard them as random samples. Also n1  44  30 and n2  44  30 , so we can proceed with the two-sample t test. 25  20  4.690 7. t  52 52  44 44 8. df = 86 P-value  2  P(t86  4.690)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that the mean percentage of the time spent playing with the furry dog is not the same for male monkeys as it is for female monkeys. d

The results do seem to provide convincing evidence of a gender basis in the monkeys’ choices of how much time to spend playing with each toy, with the male monkeys spending

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significantly more time with the “masculine toy” than the female monkeys, and with the female monkeys spending significantly more time with the “feminine toy” than the male monkeys. However, the data also provide convincing evidence of a difference between male and female monkeys in the time they choose to spend playing with a “neutral toy.” It is possible that it was some attribute other than masculinity/femininity in the toys that was attracting the different genders of monkey in different ways. e

11.18 a

The given mean time playing with the police car and mean time playing with the doll for female monkeys are sample means for the same sample of female monkeys. The two-sample t test can only be performed when there are two independent random samples. Check of Conditions We are told that the subjects were randomly assigned to one of two experimental treatment groups (the video treatment group or the control group). We are also told to assume that the distribution of body esteem scores for each of the two treatments is approximately normal. We can proceed with construction of a two-sample t interval. Calculation df = 46.4451. The 90% confidence interval for 1  2 is

( x1  x2 )   t critical value 

 4.43  5.08  1.67833

s12 n1

1.022 23



s22 n2

0.982 29

0.65  1.67833 0.279914 0.65  0.469789

 1.11979, 0.180211 Interpretation We are 90% confident that the difference in the mean body esteem score for women who watch the video and those who do not is between -1.11979 and -0.180211. Both endpoints of the confidence interval are negative, so you can estimate that the mean body esteem score in the control (no video) treatment group is greater than the mean body esteem score in the video treatment group by somewhere between 0.180211 and 1.11979. b

Based on this confidence interval, and the fact that the study was a designed experiment, it appears as if watching the video does have an effect on body esteem. The women who would watch the video have, on average, lower body esteem scores than those who don’t.

11.19 a

Since the samples are small it is necessary to know, or to assume, that the distributions from which the random samples were taken are normal. However, in this case, since both standard deviations are large compared to the means, it seems unlikely that these distributions would have been normal.

Now, since the samples are large, it is appropriate to carry out the two-sample t test, whatever the distributions from which the samples were taken.

1 = mean fumonisin level for corn meal made from partially degermed corn  2 = mean fumonisin level for corn meal made from corn that has not been degermed

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Chapter 11: Comparing Two Populations or Treatments 2. H0: 1  2  0 3. Ha: 1  2  0 4.   0.01 ( x  x )  (hypothesized value) ( x1  x2 )  0 5. t  1 2  s12 s22 s12 s22   n1 n2 n1 n2 6. We are told that the samples were random samples from the populations. Also n1  50  30 and n2  50  30 , so we can proceed with the two-sample t test.

0.59  1.21

 2.207 1.012 1.712  50 50 8. df = 79.479 P-value  2  P(t79.479  2.207)  0.030 9. Since P-value  0.030  0.01 we do not reject H0. We do not have convincing evidence that there is a difference in mean fumonisin level for the two types of corn meal. 11.20 a

t

1 = mean IQ for those who average fewer than 10 headers per game  2 = mean IQ for those who average 10 or more headers per game H0: 1  2  0 Ha: 1  2  0   0.05 ( x  x )  (hypothesized value) ( x1  x2 )  0 t 1 2  s12 s22 s12 s22   n1 n2 n1 n2

2. 3. 4. 5.

6. We are given that n1  35  30 . The second sample size, 25, is less than 30, but it is somewhat close. Additionally, we have no particular reason to think that the distribution of IQs for those who average 10 or more headers per game is not normal. The 60 players used in the study were selected randomly, and so it is acceptable to treat the sets of 35 and 25 players as random samples from their respective populations. Thus we can proceed with the two-sample t test. 112  103  3.867 7. t  102 82  35 25 8. df = 57.185 P-value  P(t57.185  3.867)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that the mean IQ for those who average fewer than 10 headers per game is greater than the mean IQ for those who average 10 or more headers per game. b

11.21 a

No. Since this was an observational study, we cannot conclude that heading the ball causes lower IQ. 1.

1 = mean oxygen consumption for noncourting pairs  2 = mean oxygen consumption for courting pairs

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2. H0: 1  2  0 3. Ha: 1  2  0 4.   0.05

( x1  x2 )  (hypothesized value)

t

sp 



( x1  x2 )  0

, where s p 

(n1  1) s12  (n2  1) s22 n1  n2  2

s 2p s 2p s 2p s 2p   n1 n2 n1 n2 6. We need to assume that the samples were random samples from the populations, and that the population distributions are normal. Additionally, the similar sample standard deviations justify our assumption that the populations have equal standard deviations. 10(0.0066) 2  14(0.0071) 2  0.00690, t  24

0.072  0.099 0.006902 0.006902  11 15

 9.863

8. df = 24 P-value  P(t24  9.863)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that the mean oxygen consumption for courting pairs is higher than the mean oxygen consumption for noncourting pairs. b

For the two-sample t test, t  9.979, df  22.566, and P-value  0. Thus the conclusion is the same.

11.22 a

Gather a number of volunteers who suffer from glaucoma in both eyes. Measure the pressure in each eye of each participant before the treatments begin. For each experimental subject, randomly assign the treatments to the eyes (so that one eye receives the standard treatment and the other receives the drug that is being evaluated). Once the patients have applied the treatments for a fixed time, measure again the pressures in all the eyes, and compare the mean reduction in pressure for the eyes that have received the new treatment with the mean reduction in pressure for those that have received the standard treatment.

Yes, since, for example, the reduction in pressure for the eye that receives the new treatment for the first patient is paired with the reduction in pressure for the eye that receives the standard treatment for the first patient.

The design given in Part (a) could be used, with the modification that each participant receives the same treatment in both eyes, and the treatments are randomly assigned to the participants. This design is not as informative as the paired design, since, in the paired design, the set of eyes to which the new treatment is applied is almost identical to the set of eyes to which the standard treatment is applied, which is not the case in the other design. What is more, the analysis of the results for the paired design takes into account, for example, that the result for the new treatment for the first patient is paired with the result for the standard treatment for the first patient.

11.23 Runner 1 2

Motivational Music 535 533

No Music

Difference

467 446

68 87

300

Chapter 11: Comparing Two Populations or Treatments 3 4 5 6 7 8 9 10 11

527 524 431 498 555 396 539 542 523

482 573 562 592 473 496 552 500 524

45 -49 -131 -94 82 -100 -13 42 -1

1. d = mean difference in the paired means (motivational music – no music) 2. H0: d  0 3. Ha: d  0 4.   0.05 x  hypothesized value 5. t  d sd n 6. We are told to assume that it is reasonable to regard these 11 triathletes as representative of the population of experienced triathletes. Because the sample size is small ( n  11 ), we must be willing to assume that the distribution of differences for the population of times to exhaustion is at least approximately normal. The following boxplot of the eleven sample differences is approximately symmetric and there are no outliers, so it is not unreasonable to think that the population distribution could be at least approximately normal. These observations suggest that it is appropriate to use the paired-sample t test

-150

-100 -50 0 50 Differences (time with motivational music minus time with no music)

100

xd  5.8 and sd  78.0 seconds

t

5.8  0  0.247 78.0

11 8. df = 10 P-value  P(t10  0.247)  0.595 9. Since P-value  0.595  0.05 we do not reject H0. We do not have convincing evidence that the mean time to exhaustion for experienced triathletes is greater when they run while listening to motivational music. 11.24 1.

d = mean difference in the paired means (motivational music – neutral music)

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2. H0: d  0 3. Ha: d  0 4.   0.05 x  hypothesized value 5. t  d sd n 6. We are told to assume that it is reasonable to regard these 11 triathletes as representative of the population of experienced triathletes and that the population difference distribution is approximately normal. These observations suggest that it is appropriate to use the pairedsample t test. 7. xd  7 and sd  80.0

t

7  0  0.290 80

11 8. df = 10 P-value  2  P(t10  0.290)  0.778 9. Since P-value  0.778  0.05 we do not reject H0. We do not have convincing evidence that the mean time to exhaustion for experienced triathletes running to motivational music differs from the mean time to exhaustion when running to neutral music. 11.25 Check of Conditions We are told to assume that it is reasonable to regard these 11 triathletes as representative of the population of experienced triathletes. Because the sample size is small ( n  11 ), we must be willing to assume that the distribution of differences for the population of times to exhaustion is at least approximately normal. The following boxplot of the eleven sample differences is approximately symmetric and there are no outliers, so it is not unreasonable to think that the population distribution could be at least approximately normal. These observations suggest that it is appropriate to construct a paired-sample t interval.

-150

-100 -50 0 50 Differences (time with motivational music minus time with no music)

100

Calculation df = 10. The 95% confidence interval for d is s 78 xd  (t critical value)  d  5.8  2.228   (58.1978,46.5978) n 11 Interpretation We are 95% confident that the difference in mean time to exhaustion for experienced triathletes when running to motivational music and the mean time when running with no music is somewhere between 58.1978 and 46.5978 seconds. Because the endpoints of the confidence

302

Chapter 11: Comparing Two Populations or Treatments interval have opposite signs, zero is included in the interval, and there may be no difference in the mean time to exhaustion for experienced triathletes when running to motivational music and the mean time when running with no music.

11.26 a b

The samples are paired because two readings were taken on each dog owner (one reading before interaction with their dog, and one reading after interaction with their dog). 1. d = the difference in the paired means (before interaction – after interaction) 2. H0: d  0 3. Ha: d  0 4.   0.05 x  hypothesized value 5. t  d sd n 6. We are told to assume that it is reasonable to regard the 22 dog owners who participated in this study as representative of dog owners in general. We are not given any information regarding the shape of the distribution of population differences, so we must assume that the population difference distribution is approximately normal. These observations suggest that it is appropriate to use the paired-sample t test. 27  0  4.221 7. t  30

22 8. df = 21 P-value  P(t21  4.221)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that there is an increase in mean oxytocin level of dog owners after 30 minutes of interaction with their dogs. 11.27 a

1. d = the difference in the paired means (reported weight – actual weight) 2. H0: d  0 3. Ha: d  0 4.   0.05 x  hypothesized value 5. t  d sd n 6. The exercise has us assume that the sample was representative of male college students. Additionally, the sample size is greater than 10 ( n  634 ). Therefore, it is appropriate to use the paired-sample t test. 1.2  0  5.29 7. t  5.71

634 8. df = 633 P-value  2  P(t633  5.29)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence of a significance difference in the mean reported weight and the mean actual weight for male college students. 1. d = the difference in the paired means (reported height – actual height) 2. H0: d  0

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3. Ha: d  0 4.   0.05 x  hypothesized value 5. t  d sd n 6. The exercise has us assume that the sample was representative of male college students. Additionally, the sample size is greater than 10 ( n  634 ). Therefore, it is appropriate to use the paired-sample t test. 0.6  0  18.885 7. t  0.8

634 8. df = 633 P-value  2  P(t633  18.885)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence of a significance difference in the mean reported height and the mean actual height for male college students. c

11.28 a

Both of the hypothesis tests in Parts (a) and (b) resulted in rejecting the null hypothesis. Additionally, the t test statistics are both positive, which tells us that the sample mean differences are greater than zero for reported value minus actual value. As such, we have convincing evidence that male college students tend to over-report both height and weight. 1. d = the difference in the paired means (reported weight – actual weight) 2. H0: d  0 3. Ha: d  0 4.   0.05 x  hypothesized value 5. t  d sd n 6. The exercise has us assume that the sample was representative of female college students. Additionally, the sample size is greater than 10 ( n  1052 ). Therefore, it is appropriate to use the paired-sample t test. 0.6  0 7. t   4.054 4.8 8. 9.

1. 2. 3. 4. 5.

1, 052 df = 1051 P-value  2  P(t1051  4.054)  0 Since P-value  0  0.05 we reject H0. We have convincing evidence of a significance difference in the mean reported weight and the mean actual weight for female college students. d = the difference in the paired means (reported height – actual height) H0: d  0 Ha: d  0   0.05 x  hypothesized value t d sd n

304

Chapter 11: Comparing Two Populations or Treatments 6. The exercise has us assume that the sample was representative of female college students. Additionally, the sample size is greater than 10 ( n  1052 ). Therefore, it is appropriate to use the paired-sample t test. 0.2  0 7. t   8.109 0.8

1, 052 8. df = 1051 P-value  2  P(t1051  8.109)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence of a significance difference in the mean reported height and the mean actual height for female college students. c

11.29

Both of the hypothesis tests in Parts (a) and (b) resulted in rejecting the null hypothesis. Additionally, the t test statistics are both negative, which tells us that the sample mean differences are less than zero for reported value minus actual value. As such, we have convincing evidence that female college students tend to under-report both height and weight.

1. d = mean difference in the paired means (no distraction – talking on mobile phone) 2. H0: d  0 3. Ha: d  0 4.   0.05 x  hypothesized value 5. t  d sd n 6. We are told to assume that it is reasonable to regard the sample as representative of Greek taxi drivers. Because the sample size is large ( n  50 ), we have satisfied the large sample requirement. These observations suggest that it is appropriate to use the paired-sample t test. 7. xd  0.47 and sd  1.22

t

0.47  0  2.724 1.22 50

8. df = 49 P-value  P(t49  2.724)  0.004 9. Since P-value  0.004  0.05 we reject H0. We have convincing evidence to support the claim that the mean following distance for Greek taxi drivers is greater when there are no distractions that when the driver is talking on a mobile phone. 11.30

1. d = mean difference in the paired means (no distraction – reading text messages) 2. H0: d  0 3. Ha: d  0 4.   0.05 x  hypothesized value 5. t  d sd n

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6. We are told to assume that it is reasonable to regard the sample as representative of Greek taxi drivers. Because the sample size is large ( n  50 ), we have satisfied the large sample requirement. These observations suggest that it is appropriate to use the paired-sample t test. 7. xd  1.3 and sd  1.54

t

1.3  0  5.969 1.54

50 8. df = 49 P-value  P(t49  5.969)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence to support the claim that the mean following distance for Greek taxi drivers is greater when there are no distractions that when the driver is texting. 11.31 Check of Conditions We are told to assume that it is reasonable to regard the sample as representative of Greek taxi drivers. Because the sample size is large ( n  50 ), we have satisfied the large sample requirement. These observations suggest that it is appropriate to construct a paired-sample t interval. Calculation df = 49. The 95% confidence interval for d is s 1.54 xd  (t critical value)  d  1.3  2.00958   (0.862,1.738) n 50 Interpretation We are 95% confident that the mean following distance for Greek taxi drivers while driving with no distractions and while driving and texting is between 0.862 and 1.738 meters. Both endpoints of this interval are positive, so you can estimate that the mean following distance for Greek taxi drivers while driving with no distractions and while driving and texting somewhere between 0.862 and 1.738 meters. 11.32 Cyclist 1 2 3 4 5 6 7 8 9

Chocolate Milk 24.85 50.09 38.3 26.11 36.54 26.14 36.13 47.35 35.08

Carbohydrate Replacement 10.02 29.96 37.4 15.52 9.11 21.58 31.23 22.04 17.02

Difference 14.83 20.13 0.9 10.59 27.43 4.56 4.9 25.31 18.06

d = mean of the time to exhaustion difference (chocolate mile − carbohydrate replacement) 2. H0: d  0 3. Ha: d  0 4.   0.05 1.

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t

xd  hypothesized value sd n

15 Difference

The boxplot shows that the distribution of the differences in the sample is roughly symmetrical and has no outliers, so we are justified in assuming that the population distribution of differences is normal. Additionally, we need to assume that the set of cyclists used in the study was randomly selected. 7. xd  14.079, sd  9.475

14.079  0  4.458 9.475 9 8. df = 8 P-value  P(t8  4.458)  0.001 9. Since P-value  0.001  0.05 we reject H0. We have convincing evidence that the mean time to exhaustion is greater after chocolate milk than after carbohydrate replacement drink. t

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

Water 0.9 0.92 1 1.1 1.2 1.25 1.25 1.3 1.35 1.4 1.4 1.5 1.65 1.7 1.75 1.8 1.8 1.85 1.9 1.95

Guar Syrup 0.92 0.96 0.95 1.13 1.22 1.2 1.26 1.3 1.34 1.41 1.44 1.52 1.58 1.7 1.8 1.76 1.84 1.89 1.88 1.95

Difference −0.02 −0.04 0.05 −0.03 −0.02 0.05 −0.01 0 0.01 −0.01 −0.04 −0.02 0.07 0 −0.05 0.04 −0.04 −0.04 0.02 0

1. d = mean swimming velocity difference (water − guar syrup) 2. H0: d  0

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3. Ha: d  0 4.   0.01 x  hypothesized value 5. t  d sd n 6.

-0.050

-0.025

0.000 0.025 Difference

0.050

0.075

The boxplot shows that the distribution of the differences is roughly symmetrical and has no outliers, so we are justified in assuming that the population distribution of differences is normal. Additionally, we need to assume that this set of differences forms a random sample from the set of differences for all swimmers. 7. xd  0.004, sd  0.035

0.004  0  0.515 0.035 20 8. df = 19 P-value  2  P(t19  0.515)  0.612 9. Since P-value  0.612  0.01 we do not reject H0. We do not have convincing evidence of a difference between the mean swimming speeds in water and guar syrup. The given data are consistent with the authors’ conclusion. t

11.34 1. d = mean rating difference ($90 − $10) 2. H0: d  0 3. Ha: d  0 4.   0.01 x  hypothesized value 5. t  d sd n 6.

-1

Difference

The boxplot shows that the sample distribution of differences is roughly symmetrical and has no outliers, so we are justified in assuming that the population of differences is normal distributed. Additionally we need to assume that the participants in the experiment formed a random sample from the population in question. 7. xd  1.6, sd  1.353

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1.6  0  5.287 1.353 20 8. df = 19 P-value  P(t19  5.287)  0 9. Since P-value  0  0.01 we reject H0. We have convincing evidence that the mean rating assigned when the cost is described as $90 is greater than the mean rating assigned when the cost is described as $10. t

11.35 Subject

Clockwise

1 2 3 4 5 6 7 8 9 10 11 12 13 14

57.9 35.7 54.5 56.8 51.1 70.8 77.3 51.6 54.7 63.6 59.2 59.2 55.8 38.5

Counterclockwise 44.2 52.1 60.2 52.7 47.2 65.6 71.4 48.8 53.1 66.3 59.8 47.5 64.5 34.5

Difference 13.7 −16.4 −5.7 4.1 3.9 5.2 5.9 2.8 1.6 −2.7 −0.6 11.7 −8.7 4

1. d = mean of the neck rotation difference (clockwise − counterclockwise) 2. H0: d  0 3. Ha: d  0 4.   0.01 x  hypothesized value 5. t  d sd n 6.

-20

-15

-10

-5

Difference

The boxplot shows that the distribution of the differences in the sample is roughly symmetrical and has no outliers, so we are justified in assuming that the population distribution of differences is normal. Additionally, we are told to assume that the 14 subjects are representative of the population of adult Americans. 7. xd  1.343, sd  7.866 1.343  0 t  0.639 7.866 14

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8. df = 13 P-value  P(t13  0.639)  0.267 9. Since P-value  0.267  0.01 we do not reject H0. We do not have convincing evidence that mean neck rotation is greater in the clockwise direction than in the counterclockwise direction. 11.36 a

Eosinophils. Here virtually every child showed a reduction in eosinophils percentage, with many of the reductions by large amounts. (The other two graphs seem to show less conclusive results.)

FENO. Here, while the majority seems to have shown a reduction, there are several who showed increases, and some of those by non-negligible amounts. (The graph for PEF seems to show more conclusive results.)

11.37 a

1. 2. 3. 4. 5.

d = mean difference between profile height and actual height (profile − actual) H0: d  0 Ha: d  0   0.05 x  hypothesized value t d sd n

6. We are told to assume that the sample is representative of male online daters, and therefore we are justified in treating it as a random sample. Therefore, since n  40  30, we can proceed with the paired t test. 0.57  0  4.451 7. t  0.81 40 8. df = 39 P-value  P(t39  4.451)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that, on average, male online daters overstate their height in online dating profiles. Check of Conditions We are told to assume that the sample is representative of female online daters, and therefore we are justified in treating it as a random sample. Therefore, since n  40  30, we can proceed with the paired t interval. Calculation df = 39. The 95% confidence interval for d is s 0.75 xd  (t critical value) d  0.03  2.023  ( 0.210, 0.270) n 40 Interpretation We are 95% confident that the difference between the mean online dating profile height and mean actual height for female online daters is between −0.210 and 0.270. 1.

m = mean height difference (profile − actual) for male online daters  f = mean height difference (profile − actual) for female online daters

2. H0: m   f  0 3. Ha: m   f  0 4.

  0.05

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t

( xm  x f )  (hypothesized value) 2 sm2 s f  nm n f



( xm  x f )  0 2 sm2 s f  nm n f

6. We are told to assume that the samples were representative of the populations, and therefore we are justified in assuming that they are random samples. Also nm  40  30 and n f  40  30 , so we can proceed with the two-sample t test.

t

0.57  0.03

 3.094 0.812 0.752  40 40 8. df = 77.543 P-value  P(t77.543  3.094)  0.001 9. Since P-value  0.001  0.05 we reject H0. We have convincing evidence that m   f  0 . 7.

In Part (a), the male profile heights and the male actual heights are paired (according to which individual has the actual height and the height stated in the profile), and with paired samples we use the paired t test. In Part (c) we were dealing with two independent samples (the sample of males and the sample of females), and therefore the two-sample t test was appropriate.

11.38 The differences applicable to Parts (a) and (b) are shown in the table below.

Player 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19

Dominant Arm 30.31 44.86 22.09 31.26 28.07 31.93 34.68 29.1 25.51 22.49 28.74 27.89 28.48 25.6 20.21 33.77 32.59 32.6 29.3

Nondominant Arm 32.54 40.95 23.48 31.11 28.75 29.32 34.79 28.87 27.59 21.01 30.31 27.92 27.85 24.95 21.59 32.48 32.48 31.61 27.46

Difference -2.23 3.91 -1.39 0.15 -0.68 2.61 -0.11 0.23 -2.08 1.48 -1.57 -0.03 0.63 0.65 -1.38 1.29 0.11 0.99 1.84

Pitcher 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

Dominant Arm 27.63 30.57 32.62 39.79 28.5 26.7 30.34 28.69 31.19 36 31.58 32.55 29.56 28.64 28.58 31.99 27.16

Nondominant Arm 24.33 26.36 30.62 33.74 29.84 26.71 26.45 21.49 20.82 21.75 28.32 27.22 28.86 28.58 27.15 29.46 21.26

Difference 3.3 4.21 2 6.05 -1.34 -0.01 3.89 7.2 10.37 14.25 3.26 5.33 0.7 0.06 1.43 2.53 5.9

Chapter 11: Comparing Two Populations or Treatments a

311

Check of Conditions

0.0

2.5

5.0

7.5 Difference

10.0

12.5

15.0

The boxplot shows that the distribution of the differences (dominant − nondominant, for pitchers) is positively skewed and has an outlier. Having assumed that the sample was a random sample of pitchers, we will nonetheless proceed with calculation of the confidence interval. Calculation xd  4.066, sd  3.955, n  17, df  16 The 95% confidence interval for d is

xd  (t critical value)

sd 3.955  4.066  2.120  ( 2.033, 6.100) n 17

Interpretation We are 95% confident that the true average difference in translation between dominant and nondominant arms for pitchers is between 2.033 and 6.100 mm. Check of Conditions

-3

-2

-1

1 Difference

The boxplot shows that the distribution of the differences (dominant − nondominant, for players) is slightly skewed and has no outliers. The skewness is not extreme, so we are justified in assuming that the population distribution of differences is normal. We need to assume that the sample was a random sample of position players. Calculation xd  0.233, sd  1.603, n  19, df  18 The 95% confidence interval for d is s 1.603 xd  (t critical value) d  0.233  2.101  (0.540,1.005) n 19 Interpretation We are 95% confident that the true average difference in translation between dominant and nondominant arms for position players is between −0.540 and 1.005 mm. c

Yes. Since the confidence intervals calculated in Parts (a) and (b) do not intersect, and the sample mean difference was greater for pitchers than for position players, we would appear to have strong evidence that the population mean difference is greater for pitchers than for position players.

312

Chapter 11: Comparing Two Populations or Treatments

11.39 a

1. d = mean difference in wrist extension (type A − type B) 2. H0: d  0 3. Ha: d  0 4.   0.05 x  hypothesized value 5. t  d sd n 6. We are told to assume that the sample is representative of the population of computer users, and therefore we are justified in treating it as a random sample from that population. However, in order to proceed with the paired t test we need to assume that the population of differences is normally distributed. 8.82  0  4.321 7. t  10 24 8. df = 23 P-value  P(t23  4.321)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that the mean wrist extension for mouse type A is greater than for mouse type B.

8.82  0  1.662, and so P-value  P(t23  1.662)  0.055 . Since 26 24 P-value  0.055  0.05 we do not reject H0. We do not have convincing evidence that the mean wrist extension for mouse type A is greater than for mouse type B. A lower standard deviation in the sample of differences means that we have a lower estimate of the standard deviation of the population of differences. Assuming that the mean wrist extensions for the two mouse types are the same (in other words, that the mean of the population of differences is zero), a sample mean difference of as much as 8.82 is much less likely when the standard deviation of the population of differences is around 10 than when the standard deviation of the population of differences is around 26. Now t 

11.40

Athlete X-ray Ultrasound Difference 1 5 4.75 0.25 2 7 3.75 3.25 3 9.25 9 0.25 4 12 11.75 0.25 5 17.25 17 0.25 6 29.5 27.5 2 7 5.5 6.5 −1 8 6 6.75 −0.75 9 8 8.75 −0.75 10 8.5 9.5 −1 11 9.25 9.5 −0.25 12 11 12 −1 13 12 12.25 −0.25 14 14 15.5 −1.5 15 17 18 −1 16 18 18.25 −0.25

Chapter 11: Comparing Two Populations or Treatments

313

Check of Conditions

-2

-1

Difference

The boxplot shows that the distribution of the differences has an outlier, while the distribution is roughly symmetrical if the outlier is ignored. The outlier should lead us to question the normality of the population distribution. However, the sample is relatively small, and outliers can occur in samples from normal distributions when the sample is small. So we will proceed with the assumption that the population distribution of differences is normal. Additionally, we are told to assume that the 16 athletes who participated in this study are representative of the population of athletes. Calculation df = 15. The 95% confidence interval for d is

xd  (t critical value) 

 0.09375  2.131

1.217  (-0.743,0.555) 16

n Interpretation We are 95% confident that the difference in mean body fat percentage measurement for the two methods (X-ray – ultrasound) is between −0.743 and 0.555. 11.41 1.

2. 3. 4. 5.

p1 = proportion of people receiving liquid nitrogen treatment for whom wart is successfully removed p2 = proportion of people receiving duct tape treatment for whom wart is successfully removed H0: p1  p2  0 Ha: p1  p2  0   0.01 pˆ1  pˆ 2 z pˆ c (1  pˆ c ) pˆ c (1  pˆ c )  n1 n2

6. We are told that the patients were randomly assigned to the treatments. Also n1 pˆ1  100(60 100)  60  10, n1 (1  pˆ1 )  100(40 100)  40  10, n2 pˆ 2  104(88 104)  88  10, and n2 (1  pˆ 2 )  104(16 104)  16  10, so the treatment groups are large enough. n pˆ  n2 pˆ 2 60  88   0.725 7. pˆ c  1 1 n1  n2 100  104

60 100  88 104  3.938 (0.725)(0.275) (0.725)(0.275)  100 104 8. P -value  P ( Z  3.938)  0 9. Since P-value  0  0.01 we reject H0. We have convincing evidence that the proportion of people for whom the wart is successfully removed is less for the liquid nitrogen treatment than for duct tape treatment. z

314

Chapter 11: Comparing Two Populations or Treatments

11.42 1. 2. 3. 4. 5.

p1 = proportion of adult Americans ages 33 to 49 who rate a landline phone in the top three p2 = proportion of adult Americans ages 50 to 68 who rate a landline phone in the top three H0: p1  p2  0 Ha: p1  p2  0   0.05 pˆ1  pˆ 2 z pˆ c (1  pˆ c ) pˆ c (1  pˆ c )  n1 n2

6. We are told that the samples are independent and representative of their respective populations . Also n1 pˆ1  (600)(0.31)  186  10 , n1 1  pˆ1   (600)(1  0.31)  414  10 ,

n2 pˆ 2  (650)(0.48)  312  10 , and n2 1  pˆ 2   (650)(1  0.48)  338  10 , so the samples are large enough. n pˆ  n2 pˆ 2 (600)(0.31)  (650)(0.48) 7. pˆ C  1 1   0.3984 n1  n2 600  650

0.31  0.48

z

 6.134 0.3984(1  0.3984) 0.3984(1  0.3984)  600 650 8. P -value  P( z  6.134)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that the proportion of adult Americans ages 33 to 49 who rate a landline phone in the top three is less than this proportion for adult Americans ages 50 to 68. 11.43 a

1. 2. 3. 4. 5.

p1 = proportion of Gen Y respondents who donated by text message p2 = proportion of Gen X respondents who donated by text message H0: p1  p2  0 Ha: p1  p2  0   0.01 pˆ1  pˆ 2 z pˆ c (1  pˆ c ) pˆ c (1  pˆ c )  n1 n2 We are told to regard the samples as representative of the Gen Y and Gen X populations, so it is reasonable to treat them as independent random samples from the populations. Also n1 pˆ1  400(0.17)  68  10, n1 (1  pˆ1 )  400(0.83)  332  10, n2 pˆ 2  400(0.14)  56  10, and n2 (1  pˆ 2 )  400(0.86)  344  10, so the samples are large enough. n pˆ  n pˆ 400(0.17)  400(0.14) pˆ c  1 1 2 2   0.155 n1  n2 400  400

0.17  0.14  1.172 (0.155)(0.845) (0.155)(0.845)  400 400 P-value  P ( Z  1.172)  0.121 z

Chapter 11: Comparing Two Populations or Treatments

315

9. Since P-value  0.121  0.01 we do not reject H0. We do not have convincing evidence that the proportion of those in Gen Y who donated to Haiti relief via text message is greater than the proportion of those in Gen X. b

Check of Conditions See Part (a). Calculation The 99% confidence interval for p1  p2 is

( pˆ1  pˆ 2 )  ( z critical value)  (0.17  0.14)  2.576

pˆ1 (1  pˆ1 ) pˆ 2 (1  pˆ 2 )  n1 n2

(0.17)(0.83) (0.14)(0.86)  400 400

 ( 0.036, 0.096) Interpretation of Interval We are 99% confident that the difference between the proportion of Gen Y and the proportion of Gen X who made a donation via text message is between −0.036 and 0.096. Interpretation of Confidence Level In repeated sampling with random samples of size 400, 99% of the resulting confidence intervals would contain the true difference in proportions who donated via text message. 11.44 1.

p1 = proportion of all male likely voters who think the U.S. should have a military draft p2 = proportion of all female likely voters who think the U.S. should have a military draft

2. H0: p1  p2  0 3. Ha: p1  p2  0 4.   0.01 pˆ1  pˆ 2 5. z  pˆ c (1  pˆ c ) pˆ c (1  pˆ c )  n1 n2 6. We are told that the samples are representative of male and female likely voters. Also, n1 pˆ1  500(0.36)  180  10 , n1 1  pˆ1   500(1  0.36)  320  10 ,

n2 p2  500(0.21)  105  10 , and n2 1  pˆ 2   500(1  0.21)  395  10 , so the samples are large enough. n pˆ  n2 pˆ 2 (500)(0.36)  (500)(0.21)   0.285 7. pˆ C  1 1 n1  n2 500  500

z

0.36  0.21

 5.25 0.285(1  0.285) 0.285(1  0.285)  500 500 8. P -value  2  P ( z  5.25)  0 9. Since P-value  0  0.01 we reject H0. We have convincing evidence that the proportion of male likely voters who think the U.S. should have a military draft is different from this proportion for female likely voters. 11.45 No; the large-sample test for a difference in population proportions is not appropriate because this question is phrased as a one-proportion hypothesis test, not a difference in proportions.

316

Chapter 11: Comparing Two Populations or Treatments

11.46 a

1. 2. 3. 4. 5.

p1 = proportion of American teenage girls who say that newspapers are boring p2 = proportion of American teenage boys who say that newspapers are boring H0: p1  p2  0 Ha: p1  p2  0   0.05 pˆ1  pˆ 2 z pˆ c (1  pˆ c ) pˆ c (1  pˆ c )  n1 n2 The samples were representative of the populations, so it is reasonable to treat them as independent random samples from the populations. Also n1 pˆ1  58(0.41)  24  10, n1 (1  pˆ1 )  58(0.59)  34  10, n2 pˆ 2  41(0.44)  18  10, and n2 (1  pˆ 2 )  41(0.56)  23  10, so the samples are large enough. n pˆ  n pˆ 58(0.41)  41(0.44) pˆ c  1 1 2 2   0.422 n1  n2 58  41

0.41  0.44  0.298 (0.422)(0.578) (0.422)(0.578)  58 41 8. P -value  2  P ( Z  0.298)  0.766 9. Since P-value  0.766  0.05 we do not reject H0. We do not have convincing evidence that the proportion of girls who say that newspapers are boring is different from the proportion of boys who say that newspapers are boring. z

Since the samples are larger than in Part (a), the conditions for performing the test are also satisfied here. The calculations will change to the following: n pˆ  n pˆ 2000(0.41)  2500(0.44) pˆ c  1 1 2 2   0.427 n1  n2 2000  2500

0.41  0.44  2.022 (0.427)(0.573) (0.427)(0.573)  2000 2500 P -value  2  P ( Z  2.022)  0.043 Since P-value  0.043  0.05 we reject H0. We have convincing evidence that the proportion of girls who say that newspapers are boring is different from the proportion of boys who say that newspapers are boring. z

Assuming that the population proportions are equal, you are much less likely to get a difference in sample proportions as large as the one given when the samples are very large than when the samples are relatively small, since large samples are likely to give more accurate estimates of the population proportions. Therefore, when the given difference in sample proportions was based on larger samples, this produced stronger evidence of a difference in population proportions.

11.47 Check of Conditions We are told that the samples were representative of the populations in question. Also, n1 pˆ1  1200(0.122)  146.4  10, n1 (1  pˆ1 )  1200(0.878)  1053.6  10,

Chapter 11: Comparing Two Populations or Treatments

317

n2 pˆ 2  1000(0.148)  148  10, and n2 (1  pˆ 2 )  1000(0.852)  852  10, so the samples are large enough. Calculation The 90% confidence interval for p1  p2 is

( pˆ1  pˆ 2 )  ( z critical value)  (0.122  0.148)  1.645

pˆ1 (1  pˆ1 ) pˆ 2 (1  pˆ 2 )  n1 n2

(0.122)(0.878) (0.148)(0.852)  1200 1000

 (0.050,  0.002) Interpretation of Interval We are 90% confident that the difference in the proportions living in poverty for men and women (men – women) is between −0.050 and −0.002. 11.48 No. It is not appropriate to use the two-sample z test because the groups are not large enough. We are not told the sizes of the groups, but we know that each is at most 81. The sample proportion for the fish oil group is 0.05, and 81(0.05) = 4.05, which is less than 10. So the conditions for the two-sample z test are not satisfied. 11.49 a

2. 3. 4. 5.

p1 = proportion of young adults who think that their parents would provide financial support for marriage p2 = proportion of parents who say they would provide financial support for marriage H0: p1  p2  0 Ha: p1  p2  0   0.05 pˆ1  pˆ 2 z pˆ c (1  pˆ c ) pˆ c (1  pˆ c )  n1 n2 We are told that they were independent random samples. Also n1 pˆ1  600(0.41)  246  10, n1 (1  pˆ1 )  600(0.59)  354  10, n2 pˆ 2  300(0.43)  129  10, and n2 (1  pˆ 2 )  300(0.57)  171  10, so the samples are large enough. n pˆ  n pˆ 600(0.41)  300(0.43) pˆ c  1 1 2 2   0.417 n1  n2 600  300

0.41  0.43  0.574 (0.417)(0.583) (0.417)(0.583)  600 300 8. P -value  2  P ( Z  0.574)  0.566 9. Since P-value  0.566  0.05 we do not reject H0. We do not have convincing evidence of a difference between the proportion of young adults who think that their parents would provide financial support for marriage and the proportion of parents who say they would provide financial support for marriage. z

p1 = proportion of young adults who think that their parents would help with buying a house or apartment

318

Chapter 11: Comparing Two Populations or Treatments

2. 3. 4. 5.

p2 = proportion of parents who say they would help with buying a house or apartment H0: p1  p2  0 Ha: p1  p2  0   0.05 pˆ1  pˆ 2 z pˆ c (1  pˆ c ) pˆ c (1  pˆ c )  n1 n2 We are told that they were independent random samples. Also n1 pˆ1  600(0.41)  246  10, n1 (1  pˆ1 )  600(0.59)  354  10, n2 pˆ 2  300(0.43)  129  10, and n2 (1  pˆ 2 )  300(0.57)  171  10, so the samples are large enough. n pˆ  n pˆ 600(0.37)  300(0.27) pˆ c  1 1 2 2   0.337 n1  n2 600  300

0.37  0.27  2.993 (0.337)(0.663) (0.337)(0.663)  600 300 8. P -value  P ( Z  2.993)  0.001 9. Since P-value  0.001  0.05 we reject H0. We have convincing evidence that the proportion of parents who say they would help with buying a house or apartment is less than the proportion of young adults who think that their parents would help. z

11.50 Check of Conditions We are told that the samples were representative of the populations in question. Also, n1 pˆ1  825(0.52)  429  10, n1 (1  pˆ1 )  825(0.48)  396  10, n2 pˆ 2  586(0.60)  352  10, and n2 (1  pˆ 2 )  586(0.40)  234  10, so the samples are large enough. Calculation The 90% confidence interval for p1  p2 is

( pˆ1  pˆ 2 )  ( z critical value)  (0.52  0.60)  1.96

pˆ1 (1  pˆ1 ) pˆ 2 (1  pˆ 2 )  n1 n2

(0.52)(0.48) (0.60)(0.40)  825 586

 (0.133, 0.028) Interpretation of Interval We are 95% confident that the difference in the proportions of working fathers who find balancing work and family difficult, and this proportion for working mothers, is between −0.133 and −0.028. 11.51 Check of Conditions We are told that the two samples are independently selected representative samples. Also, n1 pˆ1  500(0.038)  19  10, n1 (1  pˆ1 )  500(0.962)  481  10, n2 pˆ 2  500(0.031)  15.5  10, and n2 (1  pˆ 2 )  500(0.969)  484.5  10, so the samples are large enough. Calculation The 95% confidence interval for p1  p2 is

Chapter 11: Comparing Two Populations or Treatments

 pˆ1  pˆ 2    z critical value   0.038  0.031  1.96

pˆ1 (1  pˆ1 ) n1

0.038(1  0.038) 500



319

pˆ 2 (1  pˆ 2 ) n2

0.031(1  0.031) 500

 0.0156, 0.02962  Interpretation of Interval We are 95% confident that the actual difference in the proportion of college graduates who were unemployed in October 2013 and the proportion of college graduates who were unemployed in October 2014 is somewhere between –0.0156 and 0.02962. Because the endpoints of the confidence interval have opposite signs, zero is included in the interval, and there may be no difference in the proportion of college graduates who were unemployed in October 2013 and the proportion who were unemployed in October 2014. 11.52 a

Check of Conditions We are told that the samples are representative of the populations high school graduates in October 2013 and October 2014. Also, n1 pˆ1  400(0.073)  29.2  10, n1 (1  pˆ1 )  400(0.927)  370.8  10, n2 pˆ 2  400(0.057)  22.8  10, and n2 (1  pˆ 2 )  400(0.943)  377.2  10, so the samples are large enough. Calculation The 99% confidence interval for p1  p2 is

 pˆ1  pˆ 2    z critical value   0.073  0.057   2.58

pˆ1 (1  pˆ1 ) n1

0.073(1  0.073) 400



pˆ 2 (1  pˆ 2 ) n2

0.057(1  0.057) 400

 0.02895, 0.06095  Interpretation of Interval We are 99% confident that the actual difference in the proportion of high school graduates who were unemployed in October 2013 and high school graduates who were unemployed in October 2014 is somewhere between –0.02895 and 0.06095. Because the endpoints of the confidence interval have opposite signs, zero is included in the interval, and there may be no difference in the proportion of high school graduates that were unemployed in October 2013 and the proportion that were unemployed in October 2014. b

11.53 a

Wider, because the confidence level in part (a) is greater than the confidence level in the previous exercise, and the sample sizes are smaller.

Yes. Let pˆ 1  0.48 and pˆ 2  0.38 , n1  178 , and n2  427 , where the 1 subscript indicates the age 18-29 group and the 2 subscript indicates the age 50-64 group. Since n1 pˆ1  178(0.48)  85.44 , n1 1  pˆ1   178(1  0.48)  92.56 , n2 pˆ 2  427(0.38)  162.26 , and

n2 1  pˆ 2   427(1  0.38)  264.74 are all at least 10, the sample sizes are large enough to use the large-sample confidence interval.

320

Chapter 11: Comparing Two Populations or Treatments b

Check of Conditions We are told that the two samples are representative of their populations. Also, from part (a), we know that the samples are large enough to use the large-sample confidence interval. Calculation The 90% confidence interval for p1  p2 is

 pˆ1  pˆ 2    z critical value   0.48  0.38  1.645

pˆ1 (1  pˆ1 ) n1

0.48(1  0.48) 178



pˆ 2 (1  pˆ 2 ) n2

0.38(1  0.38) 427

 0.0273, 0.173 Interpretation of Interval We are 99% confident that the actual difference in the proportion of adult Americans

ages 18 to 29 who believe the foods made with genetically modified ingredients are bad for their health and the corresponding proportion for adult Americans ages 50 to 64 is somewhere between 0.0273 and 0.173 . c

Zero is not included in the confidence interval. Therefore, because both endpoints of the confidence interval are positive, we believe that the percent of adult Americans ages 18 to 29 who believe the foods made with genetically modified ingredients are bad for their health is greater than the corresponding proportion for adult Americans ages 50 to64 by somewhere between 2.73 and 17.3 percentage points.

11.54 Check of Conditions We are told that the two samples were representative of high school students in the two income groups. In addition, n1 pˆ1  500(0.14)  70  10 , n1 1  pˆ1   500(1  0.14)  430  10 ,

n2 p2  500(0.24)  120  10 , and n2 1  pˆ 2   500(1  0.24)  380  10 , so the samples are large enough. Calculation The 95% confidence interval for p1  p2 is

 pˆ1  pˆ 2    z critical value   0.14  0.24   1.96

pˆ1 (1  pˆ1 ) n1

0.14(1  0.14) 500



pˆ 2 (1  pˆ 2 ) n2

0.24(1  0.24) 500

 0.148, 0.0518  Interpretation of Interval We are 95% confident that the actual difference in the proportion of low income students who participate in a science club and the proportion of higher income students who participate in a science club is somewhere between –0.148 and –0.0518. Because both endpoints of the confidence interval are negative, we believe that the percent of higher income students who participate in a science club is greater than this percent for low income students by somewhere between 5.18 and 14.8 percentage points. 11.55 a

It is quite possible that a patient’s knowledge of being given one of the treatments will itself contribute positively to the health of the patient, and that the effect of the knowledge of being

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given an injection might be greater than the corresponding effect of the knowledge of being given a nasal spray (or vice versa). Therefore, if the patients were given only one treatment or the other, then the greater effect of the knowledge of being given one treatment over the knowledge of being given the other might be confounded with the differing effects of the two ways of giving the vaccine. b

Check of Conditions We need to assume that the 8000 children were randomly assigned to the two treatments. Also, we have n1 pˆ1  4000(0.086)  344  10, n1 (1  pˆ1 )  4000(0.914)  3656  10, n2 pˆ 2  4000(0.039)  156  10, and n2 (1  pˆ 2 )  4000(0.961)  3844  10, so the treatment groups are large enough. Calculation The 99% confidence interval for p1  p2 is

( pˆ1  pˆ 2 )  ( z critical value)  (0.086  0.039)  2.576

pˆ1 (1  pˆ1 ) pˆ 2 (1  pˆ 2 )  n1 n2

(0.086)(0.914) (0.039)(0.961)  4000 4000

 (0.033, 0.061) Interpretation We are 99% confident that the proportion of children who get sick with the flu after being vaccinated with an injection minus the proportion of children who get sick with the flu after being vaccinated with the nasal spray is between 0.033 and 0.061. c

11.56 a

Yes. Since zero is not included in the interval, we have convincing evidence at the 0.01 significance level that the proportions of children who get the flu are different for the two vaccination methods. If pˆ1  sample proportion of Americans reporting major depression  0.09, then n pˆ  n pˆ 5183(0.09)  3505(0.082) pˆ c  1 1 2 2   0.087. Thus n1  n2 5183  3505

0.09  0.082  1.300, and so P -value  P ( Z  1.300)  0.097. (0.087)(0.913) (0.087)(0.913)  5183 3505 Since this P-value is greater than 0.05, the difference in proportions is not significant, which is as stated in the article. Therefore, yes, the sample proportion of Americans reporting major depression could have been as large as 0.09. z

If pˆ1  sample proportion of Americans reporting major depression  0.1, then n pˆ  n pˆ 5183(0.1)  3505(0.082) pˆ c  1 1 2 2   0.093. Thus n1  n2 5183  3505

0.1  0.082  2.838, and so P-value  P( Z  2.838)  0.0023. (0.093)(0.907) (0.093)(0.907)  5183 3505 Since this P-value is less than 0.05, the difference in proportions is significant, which is not z

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Chapter 11: Comparing Two Populations or Treatments as is stated in the article. Therefore, no, the sample proportion of Americans reporting major depression could not have been as large as 0.10.

11.57 Check of Conditions We are told that the samples are representative of the populations of adults and teens age 13-17. In addition, n1 pˆ1  2, 463(0.26)  640.38  10 , n1 1  pˆ1   2, 463(1  0.26)  1,822.62  10 ,

n2 p2  510(0.19)  96.9  10 , and n2 1  pˆ 2   510(1  0.19)  413.1  10 , so the samples are large enough. Calculation The 95% confidence interval for p1  p2 is

 pˆ1  pˆ 2    z critical value   0.26  0.19   1.96

pˆ1 (1  pˆ1 ) n1

0.26(1  0.26) 2, 463



pˆ 2 (1  pˆ 2 ) n2

0.19(1  0.19) 510

 0.0318, 0.1082  Interpretation of Interval We are 95% confident that the actual difference in the proportion of adults and teens age 13-17 who believe in reincarnation is somewhere between 0.0318 and 0.1082. Because both endpoints of the confidence interval are positive, we believe that the percent of adults who believe in reincarnation is greater than the proportion of teens age 13-17 who believe in reincarnation by somewhere between 3.18% and 10.82%. 11.58 Letting p1  proportion of women receiving a mastectomy who survive for 20 years and p2  proportion of women receiving a lumpectomy who survive for 20 years, the hypotheses tested would have been H0: p1  p2  0 versus Ha: p1  p2  0. The null hypothesis was not rejected.

p1 = proportion of Gen Xers who do not pay off their credit cards each month p2 = proportion of millennials who do not pay off their credit cards each month 2. H0: p1  p2  0

11.59 1.

3. Ha: p1  p2  0 4.   0.05 pˆ1  pˆ 2 5. z  pˆ c (1  pˆ c ) pˆ c (1  pˆ c )  n1 n2 6. We are told that the samples are representative of the population of Gen Xers and the population of millennials. In addition, n1 pˆ1  (300)(0.61)  183 , n1 1  pˆ1   (300)(1  0.61)  117 , n2 pˆ 2  (450)(0.47)  211.5 , and

n2 1  pˆ 2   (450)(1  0.47)  238.5 , which are all at least 10. Therefore, the samples are

large enough. n pˆ  n2 pˆ 2 (300)(0.61)  (450)(0.47)   0.526 7. pˆ C  1 1 n1  n2 300  450

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0.61  0.47

z

 3.76 0.526(1  0.526) 0.526(1  0.526)  300 450 8. P-value  P ( z  3.76)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that the proportion of Gen Xers who do not pay off their credit cards each month is greater than this proportion for millennials. 11.60 a

Random assignment is important in order to create comparable treatment groups in terms of the participants’ attitudes to money. Also, random assignment is required for the hypothesis test used to analyze the results.

t

( x1  x2 )  0 s12 n1

df 



s22

0.77  1.34



 s12 s22      n1 n2 

s n   s n  2 1

n1  1

 2.122

0.742 1.022  22 22

2 2

n2  1



 0.742 1.022     22   22

 0.74 22   1.02 22  2

 38.311

Hence, P-value  P(t38.311  2.122)  0.020, which is less than 0.05, as stated. c

Since neither of the treatment groups was of size greater than 30, we need to assume that the two population distributions of donation amounts are approximately normal. This seems to be an invalid assumption in both cases, since the distance of zero below the sample mean is as little as 1.0 sample standard deviations for the “money primed” group and as little as 1.3 sample standard deviations for the control group, and since negative donations are not possible.

11.61 a

First hypothesis test p1 = proportion of those receiving the intervention whose references to sex decrease to zero

p2 = proportion of those not receiving the intervention whose references to sex decrease to zero H0: p1  p2  0 Ha: p1  p2  0 (Note: We know that the researchers were using two-sided alternative hypotheses, otherwise the P-value greater than 0.5 in the second hypothesis test would not have been possible for the given results.) Since P-value = 0.05, H0 is rejected at the 0.05 level. Second hypothesis test p1 = proportion of those receiving the intervention whose references to substance abuse decrease to zero p2 = proportion of those not receiving the intervention whose references to substance abuse decrease to zero

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Chapter 11: Comparing Two Populations or Treatments H0: p1  p2  0 Ha: p1  p2  0 Since P-value = 0.61, H0 is not rejected at the 0.05 level. Third hypothesis test p1 = proportion of those receiving the intervention whose profiles are set to “private” at follow-up p2 = proportion of those not receiving the intervention whose profiles are set to “private” at follow-up H0: p1  p2  0 Ha: p1  p2  0 Since P-value = 0.45, H0 is not rejected at the 0.05 level. Fourth hypothesis test p1 = proportion of those receiving the intervention whose profiles show any of the three protective changes p2 = proportion of those not receiving the intervention whose profiles show any of the three protective changes H0: p1  p2  0 Ha: p1  p2  0 Since P-value = 0.07, H0 is not rejected at the 0.05 level. b

11.62 a

If we want to know whether the email intervention reduces (as opposed to changes) adolescents’ display of risk behavior in their profiles, then we use one-sided alternative hypotheses and the P-values are halved. If that is the case, using a 0.05 significance level, we are convinced that the intervention is effective with regard to reduction of references to sex and that the proportion showing any of the three protective changes is greater for those receiving the email intervention. Each of the other two apparently reduced proportions could have occurred by chance. 1. 2. 3. 4. 5.

1 = mean cohabitation endorsement for emerging adult men  2 = mean cohabitation endorsement for emerging adult women H0: 1  2  0 Ha: 1  2  0   0.05 ( x  x )  (hypothesized value) ( x1  x2 )  0 t 1 2  s12 s22 s12 s22   n1 n2 n1 n2

6. We need to assume that the samples of men and women were selected in such a way as to be representative of men and women at these five colleges (which then justifies our treating the samples as random samples from these populations). Also n1  307  30 and n2  481  30 , so we can proceed with the two-sample t test. 3.75  3.39  4.125 7. t  1.212 1.172  307 481

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8. df = 635.834 P-value  P(t635.834  4.125)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that the mean cohabitation endorsement for emerging adult women is less than the mean for emerging adult men, for students at these five colleges. b

In order to generalize the conclusion to all college students we would need to know, in addition to the assumption in Step 6 above, that students at the five colleges were representative of students in general.

11.63 Test the hypotheses H 0 : 1  2  0 versus H a : 1   2  0 . Different simulations will produce different results, so answers will vary. For one randomization test, results indicated a P-value of 0.001. Because this P-value of 0.001 is less than the significance level of 0.05, the null hypothesis is rejected. There is convincing evidence of a difference in the population mean heart rate percentage for drivers in a two-firefighter team is greater than the mean heart rate for drivers in a team with more than two firefighters. 11.64 Different simulations will produce different results, so answers will vary. One simulation resulted in a confidence interval of (9.678, 23.450). You can be 95% confident that the actual difference in population mean heart rate percentage for drivers in a two-firefighter team is greater than the mean heart rate for drivers in a team with more than two firefighters by somewhere between 9.678 and 23.450. 11.65 Test the hypotheses H 0 : 1  2  0 versus H a : 1   2  0 . Different simulations will produce different results, so answers will vary. For one randomization test, results indicated a P-value of 0.001. Because this P-value of 0.001 is less than a significance level of 0.05 or 0.01, the null hypothesis is rejected at either of these significance levels. There is convincing evidence that the mean Personal Meaning Score for patients taking a high dose is greater than the mean score for patients taking a low dose. 11.66 Different simulations will produce different results, so answers will vary. One simulation resulted in a confidence interval of (-3.222, -0.738). You can be 95% confident that the actual difference (Low – High) in mean Personal Meaning Score is greater for patients taking a high dose is greater than for patients taking a low dose by somewhere between 0.738 and 3.222. 11.67 a

Because the sample sizes for each treatment group are smaller than 30, we must be wary of using the two-sample t methods. In addition, the boxplots for ADHD Severity Scores show that the data are skewed.

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ONTRAC

Control

6 8 ADHD Severity Score

Test the hypotheses H 0 : 1  2  0 versus H a : 1   2  0 , where the subscripts 1 and 2 refer to the ONTRAC and control treatment groups, respectively. Different simulations will produce different results, so answers will vary. For one randomization test, results indicated a P-value of 0.000. Because this P-value of 0.000 is less than a significance level of 0.05, the null hypothesis is rejected. There is convincing evidence that the mean one-year improvement in ADHD Severity Score for the ONTRAC treatment is different than the mean one-year improvement in ADHD Severity Score for the control treatment.

11.68 Different simulations will produce different results, so answers will vary. One simulation resulted in a confidence interval of (2.065, 8.236). You can be 95% confident that the actual difference in mean one-year improvement in ADHD Severity Scores for the ONTRAC and control treatments is somewhere between 2.065 and 8.236. Because both endpoints of the interval are positive, we can conclude that the mean one-year improvement in ADHD Severity Score for the ONTRAC treatment is greater than the mean ADHD Severity Score for the control treatment by somewhere between 2.065 and 8.236. 11.69 a

The sample size is not greater than 30 and a dotplot of the difference data suggests that it may not be reasonable to assume that the difference distribution is normal.

Test the hypotheses H 0 :  d  0 versus H a :  d  0 . Different simulations will produce different results, so answers will vary. For one randomization test, results indicated a P-value of 0.003. Because this P-value of 0.003 is less than any reasonable significance level, the null hypothesis is rejected. There is convincing evidence that the mean difference in movement, OFF – CL, is greater than 0.

11.70 Different simulations will produce different results, so answers will vary. One simulation resulted in a confidence interval of (5.120, 29.659). You can be 95% confident that the actual mean difference in movement is somewhere between 5.120 and 29.659. 11.71 a

In order to use the large-sample hypothesis test for the difference in two population proportions, the number of successes and failures in each sample must be at least 10. In this

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case, the number of successes in one of the groups is n1 pˆ1  (21)(0.05)  1.05 , which is less than 10. b

We want to determine if there is convincing evidence that the proportion of Tasmanian devils with the genetic marker was greater after DFTD than before DFTD. The proportions are p1 , the proportion of Tasmanian devils with the genetic maker before DFTD, p2 , the proportion of Tasmanian devils with the genetic maker after DFTD, and p1  p2 , the difference in proportions. Test the hypotheses H 0 : p1  p2  0 versus H a : p1  p2  0 . For the given simulation, results indicate a P-value of 0.002. Because this P-value of 0.002 is less than any reasonable significance level, we reject the null hypothesis. We have convincing evidence that the proportion of Tasmanian devils with the genetic marker was greater after DFTD than before DFTD.

We will find a 95% confidence interval to estimate the difference in the rates of occurrence of the specific genetic marker in the genes of Tasmanian devils, before and after DFTD. The given simulation produced a confidence interval of (-0.548, -0.214). We are 95% confident that the true difference in the rates of occurrence of the specific genetic marker in the genes of Tasmanian devils, before and after DFTD, lies somewhere between -0.548 and -0.214. Because both endpoints of the confidence interval are negative, we believe that the rate of occurrence of the genetic marker in the genes of Tasmanian devils after DFTD is greater than the rate before DFTD by somewhere between 21.4% and 54.8%.

11.72 a

We want to determine if there is convincing evidence that the proportion who require students to submit papers through plagiarism-detection software is different for full-time faculty and part-time faculty. The proportions are p1 , the proportion of full-time faculty who require students to submit papers through plagiarism-detection software, p2 , the proportion of part-time faculty who require students to submit papers through plagiarism-detection software, and p1  p2 , the difference in proportions. Test the hypotheses H 0 : p1  p2  0 versus H a : p1  p2  0 . The given randomization test resulted in a P-value of 0.502. Because this P-value of 0.502 is greater than any reasonable significance level, we fail to reject the null hypothesis. We do not have convincing evidence that the proportion who require students to submit papers through plagiarism-detection software is different for full-time faculty and part-time faculty.

We will find a 95% confidence interval to estimate the difference in the population proportions of full-time college faculty and part-time college faculty who require students to submit papers through plagiarism-detection software. For the given simulation, a confidence interval of (-0.037, 0.080) was obtained. We are 95% confident that the actual difference in the population proportions of full-time college faculty and part-time college faculty who require students to submit papers through plagiarism-detection software is somewhere between –0.037 and 0.080. Because the endpoints of the confidence interval have opposite signs, zero is included in the interval, and there may be no difference in in the population proportions of full-time college faculty and part-time college faculty who require students to submit papers through plagiarism-detection software.

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11.73 a

Data from this study should not be analyzed using a large-sample hypothesis test for a difference in two population proportions because the number of successes in each group (5 in the high-intensity group and 0 in the regular-intensity group) are both less than 10.

We want to determine if there is convincing evidence of a difference in the population proportions of patients in the two exercise groups (high-intensity and regular-intensity) who die within 1.5 years. The proportions are p1 , the proportion of patients in the high-intensity group who would die within 1.5 years, p2 , the proportion of patients in the regular-intensity group who would die within 1.5 years, and p1  p2 , the difference in proportions. Test the hypotheses H 0 : p1  p2  0 versus H a : p1  p2  0 . Different simulations will produce different results, so answers will vary. For one randomization test, results indicated a P-value of 0.017. Because this P-value of 0.017 is less than any reasonable significance level, we reject the null hypothesis. We have convincing evidence of a difference in the population proportions of patients in the two exercise groups (high-intensity and regular-intensity) who die within 1.5 years.

We will find a 95% bootstrap confidence interval to estimate the difference in the population proportions of patients who die within 1.5 years for the two exercise groups. Different simulations will produce different results, so answers will vary. One simulation resulted in a confidence interval of (0.123, 0.133). We are 95% confident that the actual difference in the population proportions of patients who die within 1.5 years for the two exercise groups is somewhere between 0.123 and 0.133. Because both endpoints of the confidence interval are positive, we believe that the proportion of patients who die within 1.5 years for the highintensity group is greater than the proportion of patients who die within 1.5 years for the regular-intensity group by somewhere between 12.3% and 13.3%.

11.74 a

The data in this exercise should not be analyzed using a large-sample hypothesis test for the difference in two population proportions because the number of successes and failures for each group are not all at least 10.

We want to determine if there is convincing evidence that supports the statement that Williams is a better shooter behind the arc than from the free throw line. The proportions are p1 , the proportion of free-throws made from behind the arc, p2 , the proportion of freethrows made from the free throw line, and p1  p2 , the difference in proportions. Test the hypotheses H 0 : p1  p2  0 versus H a : p1  p2  0 . The given randomization test indicated a P-value of 0.994. Because this P-value of 0.994 is greater than any reasonable significance level, we fail to reject the null hypothesis. We do not have convincing evidence of a difference in the proportions of free-throws made from behind the arc and free-throws made from the free throw line. Therefore, we cannot support the statement that Williams is a better shooter from behind the arc than from the free throw line.

We will find a 95% bootstrap confidence interval to estimate the difference in the proportions of made free-throws and made three-point shots by Kenrich Williams for the 2016-2017 season. The simulation resulted in a confidence interval of (-0.514, 0.457). We are 95% confident that the actual difference in the proportions of made free-throws and made threepoint shots by Kenrich Williams for the 2016-2017 season is somewhere between -0.514 and 0.457. Because the interval endpoints have opposite signs, zero is included in the confidence

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interval. Therefore, there may be no difference in the proportion of made free-throws and made three-point shots by Kenrich Williams for the 2016-2017. 11.75 a

We want to determine if there is convincing evidence that the proportion who think it is OK to use a cell phone at a restaurant is higher for the 18 to 29 age group than for the 30 to 49 age group. The proportions are p1 , the proportion who think it is OK to use a cell phone at a restaurant for the 18 to 29 age group, p2 , the proportion who think it is OK to use a cell phone at a restaurant for the 30 to 49 age group, and p1  p2 , the difference in proportions. Test the hypotheses H 0 : p1  p2  0 versus H a : p1  p2  0 . The given randomization test resulted in a P-value of 0.000. Because this P-value of 0.000 is less than any reasonable significance level, we reject the null hypothesis. We have convincing evidence that the proportion who think it is OK to use a cell phone at a restaurant is higher for the 18 to 29 age group than for the 30 to 49 age group.

We will find a 90% bootstrap confidence interval to estimate the difference in the proportions of people ages 18 to 29 and those ages 30 to 49 who would say that it is acceptable to use a cell phone in a restaurant. The given simulation resulted in a confidence interval of (0.058, 0.139). We are 90% confident that the actual difference in the proportions of people ages 18 to 29 and those ages 30 to 49 who would say that it is acceptable to use a cell phone in a restaurant is somewhere between 0.058 and 0.139. Because both endpoints of the confidence interval are positive, we believe that the proportion of people ages 18 to 29 who would say that it is acceptable to use a cell phone in a restaurant is greater than the proportion of those ages 30 to 49 who would respond the same way by somewhere between 5.8% and 13.9%.

No, the interpretation would not change. The endpoints of the confidence interval change slightly (from Example 11.11, the confidence interval is (0.061, 0.139)), but the conclusion and interpretation remain the same.

11.76 a

The number of successes in each group is at least 10, but the number of failures in the two groups are less than 10 (4 in the Apple Watch group, and 8 in the Fitbit group). The number of successes and failures in each group must be at least 10.

We want to determine if there is convincing evidence that the proportion of accurate results for people wearing an Apple Watch is greater than this proportion for those wearing a Fitbit Charge. The proportions are p1 , the proportion of accurate results for people wearing an Apple watch, p2 , the proportion of accurate results for people wearing a Fitbit Charge, and

p1  p2 , the difference in proportions. Test the hypotheses H 0 : p1  p2  0 versus H a : p1  p2  0 . Different simulations will produce different results, so answers will vary. One randomization test resulted in a P-value of 0.083. Because this P-value of 0.083 is greater than a 5% significance level, we fail to reject the null hypothesis. We do not have convincing evidence that the proportion of accurate results for people wearing an Apple Watch is greater than this proportion for those wearing a Fitbit Charge. c

We will find a 95% bootstrap confidence interval to estimate the difference in the population proportions of accurate results for people wearing an Apple Watch and those wearing a Fitbit Charge. Different simulations will produce different results, so answers will vary. One simulation resulted in a confidence interval of (-0.040, 0.200). We are 95% confident that the

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Chapter 11: Comparing Two Populations or Treatments actual difference in the population proportions of accurate results for people wearing an Apple Watch and those wearing a Fitbit Charge is somewhere between -0.040 and 0.200. Because the interval endpoints have opposite signs, zero is included in the confidence interval. Therefore, there may be no difference in the population proportions of accurate results for people wearing an Apple Watch and those wearing a Fitbit Charge.

11.77 a

Check of Conditions

There is no indication as to how the samples were selected. As such, we must assume that the samples were either representative of their respective populations or they were randomly selected. In addition, the 20 to 39 age group had 168 successes and 90 failures and the 40 to 49 group had 61 successes and 69 failures, which are all greater than 10, so the samples are large enough. Calculation The 90% confidence interval for p1  p2 is

 pˆ1  pˆ 2    z critical value   0.651  0.473  1.645

pˆ1 (1  pˆ1 ) n1

0.651(1  0.651) 258



pˆ 2 (1  pˆ 2 ) n2

0.473(1  0.473) 129

 0.0908, 0.2652  Interpretation We are 90% confident that the difference in the population proportions of cell phone users ages 20 to 39 and those ages 40 to 49 who say that they sleep with their cell phones is between 0.0908 and 0.2652. Because both endpoints of the confidence interval are positive, we believe that the proportion of cell phone users ages 20 to 39 who sleep with their cell phones is greater than the proportion of cell phone users ages 40 to 49 who sleep with their cell phones by somewhere between 9.08% and 26.52%. b

The given simulation resulted in a bootstrap confidence interval of (0.093, 0.267). Therefore, we are 90% confident that the difference in the population proportions of cell phone users ages 20 to 39 and those ages 40 to 49 who say that they sleep with their cell phones is between 0.093 and 0.267. Because both endpoints of the confidence interval are positive, we believe that the proportion of cell phone users ages 20 to 39 who sleep with their cell phones is greater than the proportion of cell phone users ages 40 to 49 who sleep with their cell phones by somewhere between 9.3% and 26.7%.

The confidence intervals in Parts (a) and (b) are very similar. The lower endpoints are 0.0908 and 0.093, respectively, and the upper endpoints are 0.2652 and 0.267, respectively. The interpretations are essentially the same, with the only differences due to the slight differences in the endpoints.

11.78 a

p1 = proportion of low-income adults who would never consider buying an electric car p2 = proportion of high-income adults who would never consider buying an electric car 2. H0: p1  p2  0 3. Ha: p1  p2  0 4.   0.05 1.

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331

pˆ1  pˆ 2 pˆ c (1  pˆ c ) pˆ c (1  pˆ c )  n1 n2 6. We are told that the samples were independently selected and representative of their respective populations. In In addition, n1 pˆ1  (300)(0.65)  195 , 5.

z

n1 (1  pˆ1 )  300(1  0.65)  105 , n2 pˆ 2  (300)(0.59)  177 , and n2 (1  pˆ 2 )  300(1  0.59)  123 are all greater than 10, so the large sample size condition has been satisfied. n pˆ  n2 pˆ 2 (300)(0.65)  (300)(0.59) 7. pˆ c  1 1   0.62 n1  n2 300  300

z

0.65  0.59 0.62 1  0.62 

0.62 1  0.62 

 1.5139

 300 300 8. P -value  2  P( z  1.5139)  0.13 9. Since P-value  0.13  0.05 we do not reject H0. We do not have convincing evidence to conclude that the proportion low-income and high-income adults who would never consider buying an electric car differ. b

We want to determine if there is convincing evidence to conclude that the proportion lowincome and high-income adults who would never consider buying an electric car differ. The population proportions are p1 , the proportion of low-income adults who would never consider buying an electric car, p2 , the proportion high-income adults who would never consider buying an electric car, and p1  p2 , the difference in treatment proportions. Test the hypotheses H 0 : p1  p2  0 versus H a : p1  p2  0 . The given randomization test gave a Pvalue of 0.148. Because the P-value of 0.148 is greater than the selected significance level of   0.05 , we fail to reject the null hypothesis. There is not convincing evidence to conclude that the proportion low-income and high-income adults who would never consider buying an electric car differ.

In both Parts (a) and (b), we failed to reject the null hypothesis because the P-values were greater than the selected significance level. As such we drew the same conclusion using the large-sample test and the randomization test.

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Chapter 11: Comparing Two Populations or Treatments

11.79 Initial Treatment Control

11.4 9.6 10.1 8.5 10.3 10.6 11.8 9.8 10.9 10.3 10.2 11.4 9.2 10.6 10.8 8.2 a

After 9 Days Treatment Control

9.1 8.7 9.7 10.8 10.9 10.6 10.1 12.3 8.8 10.4 10.9 10.4 11.6 10.9

138.3 104 96.4 89 88 103.8 147.3 97.1 172.6 146.3 99 122.3 103 117.8 121.5 93

9.3 8.8 8.8 10.1 9.6 8.6 10.4 12.4 9.3 9.5 8.4 8.7 12.5 9.1

Difference (Init − 9 Day, for Trtmnt Grp) -126.9 -94.4 -86.3 -80.5 -77.7 -93.2 -135.5 -87.3 -161.7 -136 -88.8 -110.9 -93.8 -107.2 -110.7 -84.8

Difference (Init − 9 Day, for Cntrl Grp) -0.2 -0.1 0.9 0.7 1.3 2 -0.3 -0.1 -0.5 0.9 2.5 1.7 -0.9 1.8

d = mean difference in selenium level (initial level − 9-day level) for cows receiving

supplement 2. H0: d  0 3. Ha: d  0 4.   0.05 x  hypothesized value 5. t  d sd n 6.

-160

-150

-140

-130

-120 -110 Difference

-100

-90

-80

-70

The boxplot shows a distribution of sample differences that is negatively skewed, but in a small sample (along with the fact that there are no outliers) this is nonetheless consistent with an assumption of normality in the population of differences. Additionally, we need to assume that the cows who received the supplement form a random sample from the set of all cows. 7. xd  104.731, sd  24.101 104.731  0 t  17.382 24.101 16 8. df = 15 P-value  P(t15  17.382)  0

Chapter 11: Comparing Two Populations or Treatments

333

9. Since P-value  0  0.05 we reject H0. We have convincing evidence that the mean selenium concentration is greater after 9 days of the selenium supplement. b

d = mean difference in selenium level (initial level − 9-day level) for cows not

receiving supplement 2. H0: d  0 3. Ha: d  0 4.   0.05 x  hypothesized value 5. t  d sd n 6.

-1

1 Difference

Since the boxplot is roughly symmetrical and there are no outliers we are justified in assuming a normal distribution for the population of differences. Additionally, we need to assume that the cows who did not receive the supplement form a random sample from the set of all cows. 7. xd  0.693, sd  1.062 0.693  0 t  2.440 1.062 14 8. df = 13 P-value  2  P(t13  2.440)  0.030 9. Since P-value  0.030  0.05 we reject H0. At the 0.05 level the results are inconsistent with the hypothesis of no significant change in mean selenium concentration over the 9day period for cows that did not receive the supplement. c

No, the paired t test would not be appropriate since the treatment and control groups were not paired samples.

p1 = proportion of resumes with “white-sounding” names that receive responses p2 = proportion of resumes with “black-sounding” names that receive responses 2. H0: p1  p2  0

11.80 1.

3. Ha: p1  p2  0 4.   0.05 pˆ1  pˆ 2 5. z  pˆ c (1  pˆ c ) pˆ c (1  pˆ c )  n1 n2 6. We need to assume that the 5000 jobs applied for were randomly assigned to the names used. Also, n1 pˆ1  2500(250 2500)  250  10, n1 (1  pˆ1 )  2500(2250 2500)  2250  10,

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Chapter 11: Comparing Two Populations or Treatments

n2 pˆ 2  2500(167 2500)  167  10, and n2 (1  pˆ 2 )  2500(2333 2500)  2333  10, so the samples are large enough. 250  167 417  7. pˆ c  2500  2500 5000 250 2500  167 2500 z  4.245 (417 5000)(4583 5000) (417 5000)(4583 5000)  2500 2500 8. P -value  P ( Z  4.245)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that the proportion eliciting responses is higher for “white-sounding” first names. 11.81 a

Check of Conditions We are told that the people were randomly assigned to the two groups. Also, n1 pˆ1  169(25 169)  25  10, n1 (1  pˆ1 )  169(144 169)  144  10, n2 pˆ 2  170(24 170)  24  10, and n2 (1  pˆ 2 )  170(146 170)  146  10, so the samples are large enough. Calculation The 90% confidence interval for p1  p2 is

( pˆ1  pˆ 2 )  ( z critical value)

pˆ1 (1  pˆ1 ) pˆ 2 (1  pˆ 2 )  n1 n2

(25 169)(144 169) (24 170)(146 170)  25 24     1.645   169 170  169 170   ( 0.056, 0.070) b

Interpretation of Interval: We are 90% confident that the proportion of people receiving insulin who develop diabetes minus the proportion of people not receiving insulin who develop diabetes is between −0.056 and 0.070. Interpretation of Confidence Level: Consider the process of randomly selecting 339 people at random from the population of people thought to be at risk of developing type I diabetes, randomly assigning them to groups of the given sizes, performing the experiment as described, and then calculating the confidence interval by the method shown in Part (a). Ninety percent of the time, this process will result in a confidence interval that contains p1  p2 , where p1 is the proportion of all such people who would develop diabetes having been given the insulin treatment, and p2 is the proportion of all such people who would develop diabetes having not been given insulin.

Since zero is included in the interval, we do not have convincing evidence (even at the 0.1 level) of a difference between the proportions of people developing diabetes for the two treatments. In fact, the great closeness of the proportions for the two groups tells us that virtually no significance level (however large) will give us a significant result. We have virtually no evidence whatsoever that the insulin treatment is more effective than no treatment at all.

11.82 a

1 = mean elongation for a square knot for Maxon thread  2 = mean elongation for a Duncan loop for Maxon thread

Chapter 11: Comparing Two Populations or Treatments 2. H0: 1  2  0 3. Ha: 1  2  0 4.   0.05 ( x  x )  (hypothesized value) ( x1  x2 )  0 5. t  1 2  s12 s22 s12 s22   n1 n2 n1 n2 6. We are told that the types of knot and suture material were randomly assigned to the specimens. We are also told to assume that the relevant elongation distributions are approximately normal. 10  11 7. t   11.952 0.12 0.32  10 15 8. df = 18.266 P-value  2  P(t18.266  11.952)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that the mean elongations for the square knot and the Duncan loop for Maxon thread are different. b

1. 2. 3. 4. 5.

1 = mean elongation for a square knot for Ticron thread  2 = mean elongation for a Duncan loop for Ticron thread H0: 1  2  0 Ha: 1  2  0   0.05 ( x  x )  (hypothesized value) ( x1  x2 )  0 t 1 2  s12 s22 s12 s22   n1 n2 n1 n2

6. We are told that the types of knot and suture material were randomly assigned to the specimens. We are also told to assume that the relevant elongation distributions are approximately normal. 2.5  10.9  68.803 7. t  0.062 0.42  10 11 8. df = 10.494 P-value  2  P(t10.494  68.803)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that the mean elongations for the square knot and the Duncan loop for Ticron thread are different. c

1 = mean elongation for a Duncan loop for Maxon thread  2 = mean elongation for a Duncan loop for Ticron thread 2. H0: 1  2  0 3. Ha: 1  2  0 4.   0.05 1.

335

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Chapter 11: Comparing Two Populations or Treatments

8. 9.

t

( x1  x2 )  (hypothesized value)



( x1  x2 )  0

s12 s22 s12 s22   n1 n2 n1 n2 We are told that the types of knot and suture material were randomly assigned to the specimens. We are also told to assume that the relevant elongation distributions are approximately normal. 11  10.9 t  0.698 0.32 0.42  15 11 df = 17.789 P-value  2  P(t17.789  0.698)  0.494 Since P-value  0.697  0.05 we do not reject H0. We do not have convincing evidence that the mean elongations for the Duncan loop for Maxon thread and Ticron thread are different.

11.83 a

Yes. The p value for this test is given to be less than 0.001.

Yes. The p value for this test is given to be less than 0.05.

No. We are only given information concerning the distance apart of the means for male trial and nontrial lawyers and the distance apart of the means for female trial and nontrial lawyers. We have no information concerning a comparison of the two sexes. In order to conduct the test we would need the values of the sample means for male and female trial lawyers, along with the associated sample standard deviations.

11.84 Subject 1 hr later 24 hr later Difference

1 14 10 4

2 12 4 8

3 18 14 4

4 5 7 11 6 9 1 2

6 7 9 16 6 12 3 4

8 15 12 3

1. d = mean difference in number of objects remembered (1 hr − 24 hours) 2. H0: d  3 3. Ha: d  3 4.   0.01 x  hypothesized value 5. t  d sd n 6.

4 Difference

Chapter 11: Comparing Two Populations or Treatments

337

The boxplot is roughly symmetrical but there is one outlier. Nonetheless we will assume that the population distribution of differences is normal. We are told that the eight students were selected at random from the large psychology class. 7. xd  3.625, sd  2.066 3.625  3 t  0.856 2.066 8 8. df = 7 P-value  P(t7  0.856)  0.210 9. Since P-value  0.210  0.01 we do not reject H0. We do not have convincing evidence that the mean number of words recalled after 1 hour exceeds the mean number recalled after 24 hours by more than 3. 11.85 a

In each case the standard deviation of the differences is large compared to the magnitude of the mean. This can be the case, since differences can be negative. (Also, the large standard deviations reflect the fact that shoppers’ consumption can very considerably from week to week.) Check of Conditions We are told that the individuals in the study were randomly selected. Additionally we need to assume that the population of differences is normally distributed. Calculation df = 95 The 95% confidence interval for d is

xd  (t critical value)

sd 5.52  0.38  1.985  (0.738,1.498). n 96

Interpretation We are 95% confident that the mean drink consumption for credit-card shoppers in 1994 minus the mean drink consumption for credit-card shoppers in 1995 is between −0.738 and 1.498. Since zero is included in this interval, we do not have convincing evidence that the mean number of drinks decreased. c

1. 2. 3. 4. 5. 6. 7. 8.

d = mean difference in number of drinks consumed for non-credit card customers (1994 − 1995) H0: d  0 Ha: d  0   0.05 x  hypothesized value t d sd n We are told that the shoppers used in the summary were randomly selected. Therefore, since n  850  30, we can proceed with the paired t test. 0.12  0 t  0.764 4.58 850 df = 849 P-value  2  P(t849  0.764)  0.445

338

Chapter 11: Comparing Two Populations or Treatments 9. Since P-value  0.445  0.05 we do not reject H0. We do not have convincing evidence of a change in the mean number of drinks between 1994 and 1995 for non-credit card shoppers. The P-value tells us that if the mean number of drinks consumed by non-credit card customers were the same for 1994 as for 1995, then the probability of obtaining a sample mean difference as far from zero as the one obtained in this study would be 0.445.

11.86 1. 2. 3. 4. 5.

8. 9.

11.87 a

p1 = proportion of high school seniors exposed to the drug program who use marijuana p2 = proportion of high school seniors not exposed to the drug program who use marijuana H0: p1  p2  0 Ha: p1  p2  0   0.05 pˆ1  pˆ 2 z pˆ c (1  pˆ c ) pˆ c (1  pˆ c )  n1 n2 We are told that the samples were random samples from the populations. Also n1 pˆ1  288(141 288)  141  10, n1 (1  pˆ1 )  288(147 288)  147  10, n2 pˆ 2  335(181 335)  181  10, and n2 (1  pˆ 2 )  335(154 335)  154  10, so the samples are large enough. 141  181 322 pˆ c   288  335 623 141 288  181 335 z  1.263 (322 623)(301 623) (322 623)(301 623)  288 335 P -value  P( Z  1.263)  0.103 Since P-value  0.103  0.05 we do not reject H0. We do not have convincing evidence that the proportion using marijuana is lower for students exposed to the DARE program. 1. 2. 3. 4. 5.

p1 = proportion of games with stationary bases where a player suffers a sliding injury p2 = proportion of games with breakaway bases where a player suffers a sliding injury H0: p1  p2  0 Ha: p1  p2  0   0.01 pˆ1  pˆ 2 z pˆ c (1  pˆ c ) pˆ c (1  pˆ c )  n1 n2 We need to assume that the 2500 games used in the study were randomly assigned to the two treatments (stationary bases and breakaway bases). Also, we have n1 pˆ1  1250(90 1250)  90  10, n1 (1  pˆ1 )  1250(1160 1250)  1160  10, n2 pˆ 2  1250(20 1250)  20  10, and n2 (1  pˆ 2 )  1250(1230 1250)  1230  10, so the samples are large enough. 90  20 110 pˆ c   1250  1250 2500

Chapter 11: Comparing Two Populations or Treatments

339

90 1250  20 1250  6.826 (110 2500)(2390 2500) (110 2500)(2390 2500)  1250 1250 8. P -value  P ( Z  6.826)  0 9. Since P-value  0  0.01 we reject H0. We have convincing evidence that the use of breakaway bases reduces the proportion of games with a player suffering a sliding injury. z

11.88 1. 2. 3. 4. 5.

Since the conclusion states that use of breakaway bases reduces the proportion of games with a player suffering a sliding injury (that is, causes a lower proportion of games with sliding injuries), we need to assume that this was an experiment in which the 2500 games were randomly assigned to the two treatments (stationary bases and breakaway bases). It seems more likely that the treatments were assigned by league (or by region of the country), in which case it would be difficult to argue that each treatment group was representative of all games in terms of sliding injuries.

1 = mean number of goals in games with Gretzky  2 = mean number of goals in games without Gretzky H0: 1  2  0 Ha: 1  2  0   0.01 ( x  x )  (hypothesized value) ( x1  x2 )  0 t 1 2  s12 s22 s12 s22   n1 n2 n1 n2

6. We are told to treat the samples of games as random samples from the populations. Also, n1  41  30 . However, n2  17 , which is less than 30, so we will need to assume that the population distribution of goals scored in games without Gretzky is approximately normal in order to proceed with the two-sample t test. 4.73  3.88  2.429 7. t  1.292 1.182  41 17 8. df = 32.586 P-value  P(t32.586  2.429)  0.0104 9. Since P-value  0.0104  0.01 we do not reject H0. We do not have convincing evidence that the mean number of goals scored was higher for games in which Gretzky played than for games in which he did not play. 11.89 a

Check of Conditions We are told to assume that the peak loudness distributions are approximately normal, and that the participants were randomly assigned to the conditions. Calculation df = 17.276. The 95% confidence interval for 1  2 is

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Chapter 11: Comparing Two Populations or Treatments

( x1  x2 )  (t critical value)  (63  54)  2.107

s12 s22  n1 n2

132 162  10 10

 ( 4.738, 22.738) Interpretation We are 95% confident that the difference in mean loudness for open mouthed and closed mouthed eating of potato chips is between −4.738 and 22.738. b

1. 2. 3. 4. 5.

1 = mean loudness for potato chips (closed-mouth chewing)  2 = mean loudness for tortilla chips (closed-mouth chewing) H0: 1  2  0 Ha: 1  2  0   0.01 ( x  x )  (hypothesized value) ( x1  x2 )  0 t 1 2  s12 s22 s12 s22   n1 n2 n1 n2

6. We are told to assume that the peak loudness distributions are approximately normal, and that the participants were randomly assigned to the conditions. 54  53 7. t   0.140 162 162  10 10 8. df = 18 P-value  2  P(t18  0.140)  0.890 9. Since P-value  0.890  0.01 we do not reject H0. We do not have convincing evidence of a difference between potato chips and tortilla chips with respect to mean peak loudness (closed-mouth chewing). c

1. 2. 3. 4. 5.

1 = mean loudness for stale tortilla chips (closed-mouth chewing)  2 = mean loudness for fresh tortilla chips (closed-mouth chewing) H0: 1  2  0 Ha: 1  2  0   0.05 ( x  x )  (hypothesized value) ( x1  x2 )  0 t 1 2  s12 s22 s12 s22   n1 n2 n1 n2

6. We are told to assume that the peak loudness distributions are approximately normal, and that the participants were randomly assigned to the conditions. 53  56  0.446 7. t  162 142  10 10 8. df = 17.688

Chapter 11: Comparing Two Populations or Treatments

341

P-value  P(t17.688  0.446)  0.330 9. Since P-value  0.330  0.05 we do not reject H0. We do not have convincing evidence that fresh tortilla chips are louder than stale tortilla chips. 11.90 a

d = mean difference in systolic blood pressure between dental setting and medical

setting (dental − medical) 2. H0: d  0 3. Ha: d  0 4.   0.01 x  hypothesized value 5. t  d sd n 6. We need to assume that the subjects formed a random sample of patients. With this assumption, since n  60  30, we can proceed with the paired t test. 4.47  0  3.948 7. t  8.77 60 8. df = 59 P-value  P(t59  3.948)  0 9. Since P-value  0  0.01 we reject H0. We have strong evidence that the mean blood pressure is higher in a dental setting than in a medical setting. b

d = mean difference in pulse rate between dental setting and medical setting (dental −

medical) 2. H0: d  0 3. Ha: d  0 4.   0.05 x  hypothesized value 5. t  d sd n 6. We need to assume that the subjects formed a random sample of patients. With this assumption, since n  60  30, we can proceed with the paired t test. 1.33  0  1.165 7. t  8.84 60 8. df = 59 P-value  2  P(t59  1.165)  0.249 9. Since P-value  0.249  0.05 we do not reject H0. We do not have convincing evidence that the mean pulse rate in a dental setting is different from the mean pulse rate in a medical setting. 11.91 a Location Surface pH Subsoil pH Difference

1 6.55 6.78 -0.23

2 5.98 6.14 -0.16

3 5.59 5.8 -0.21

4 6.17 5.91 0.26

5 5.92 6.1 -0.18

6 6.18 6.01 0.17

7 6.43 6.18 0.25

8 5.68 5.88 -0.2

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Chapter 11: Comparing Two Populations or Treatments Check of Conditions

-0.3

-0.2

-0.1

0.0 Difference

0.1

0.2

0.3

The boxplot shows a sample distribution of differences that, when we look at the middle 50% of the data, is positively skewed. However, for a small sample this is not inconsistent with the assumption of normality in the population of differences (especially since there are no outliers in the sample of differences). Additionally, we are told that the agricultural locations were randomly selected. Calculation df = 7. The 90% confidence interval for d is

xd  (t critical value)

sd 0.221  0.0375  1.895  (0.186, 0.111) n 8

Interpretation We are 90% confident that the mean difference between surface and subsoil pH is between −0.186 and 0.111. b

As stated in Part (a), we must assume that the distribution of differences across all locations is normal.

Chapter 12 The Analysis of Categorical Data and Goodness-of-Fit Tests Note: In this chapter, numerical answers were found using values from a calculator. Students using statistical tables will find that their answers differ slightly from those given. 12.1

The number of degrees of freedom is 4 – 1 = 3. P( 32  6.4)  0.094. This is greater than 0.05. In other words, if the population of purchases were equally divided between the four colors it would not be particularly unlikely that you would get an X2 value as large as 6.4. So you conclude that there is not convincing evidence that the purchases are not equally divided between the four colors.

P( 32  15.3)  0.002 , which is less than 0.01. So you would conclude that there is convincing evidence that the purchases are not equally divided between the four colors.

12.2

df = 6 – 1 = 5. P( 52  13.7)  0.018 , which is less than 0.05. So you would conclude that there is convincing evidence that the purchases are not equally divided between the four colors.

P-value  P(  22  7.5)  0.024. H0 is not rejected.

P-value  P(  62  13.0)  0.043. H0 is not rejected.

P-value  P(  92  18.0)  0.035. H0 is not rejected.

P-value  P(  42  21.3)  0.0002. H0 is rejected.

P-value  P( 32  5.0)  0.172. H0 is not rejected.

12.3 Age Group 18 to 24

Observed Count 7

Expected Count

25 to 34

35 to 44

45 to 54

55 to 64

106

65 and older

136

375(0.14)  52.5 375(0.20)  75 375(0.19)  71.25 375(0.18)  67.5 375(0.14)  52.5 375(0.15)  56.25

Total

375

1. Let p1 , p2 , p3 , p4 , p5 , and p6 be the proportions of lottery ticket purchasers in the different age groups. 2. H 0 : p1  0.14, p2  0.20, p3  0.19, p4  0.18, p5  0.14, p6  0.15 3. Ha: H0 is not true

343

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Chapter 12: The Analysis of Categorical Data and Goodness-of-Fit Tests 4.

  0.05

(observed cell count  expected cell count)2 expected cell count all cells 6. We are told that these data resulted from a representative sample of 375 Texas Lottery players. All the expected counts are greater than 5 (see table above). 5.

X2 



 7  52.52  32  752 

52.5



136  56.25 2 56.25

 251.981

8. df = 5 P-value  P( X 52  251.981)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that one or more of the age groups buys a disproportionate share of Texas Lottery tickets. 12.4

Observed Proportion 0.25 0.21 0.16 0.11 0.10 0.17 1.00

Color White Black Grey Silver Red Other Total

Observed Count

1, 200(0.25)  300 1, 200(0.21)  252 1, 200(0.16)  192 1, 200(0.11)  132 1, 200(0.10)  120 1, 200(0.17)  204 1200

1. Let p1 , p2 , p3 , p4 , p5 , and p6 be the proportions of new cars sold for the six color categories. 2. H 0 : p1  p2  p3  p4  p5  p6  1 6 3. Ha: H0 is not true 4.   0.05 (observed cell count  expected cell count)2 5. X 2  expected cell count all cells 6. We are told to assume that this sample is representative of all male smokers who die of lung cancer. In addition, because the sample size was n = 1,200, all expected counts in each



1 6

category are all equal to 1, 200    200 , which is greater than 5, so the sample size is large enough. 7.

X  2

 300  200 2  252  200 2 200



200



 204  200 2 200

 119.04

8. df = 5 P-value  P( X 52  119.04)  0

9. Since P-value  0  0.05 we reject H0. We have convincing evidence that the proportions of new cars sold are not the same for all of the six color categories.

Chapter 12: The Analysis of Categorical Data and Goodness-of-Fit Tests

12.5

1. Let p1 , p2 , p3 , p4 , p5 ,

345

, and p12 be the proportions of births in each of the 12 lunar cycles.

2. H 0 : p1  p2  p3  p4  p5  p6  p7  p8  p9  p10  p11  p12  1 12 3. Ha: H0 is not true 4.   0.05 (observed cell count  expected cell count)2 5. X 2  expected cell count all cells 6. We are told that these data resulted from a random sample. The expected counts in each



1   83.9167 for each lunar cycle category, which are all greater than 5,  12 

category are 1, 007 

so the sample size is large enough. 7.

X  2

 90  83.9167 2 81  83.9167 2 

83.9167



85  83.9167 2 83.9167

 6.017

8. df = 11 2 P-value  P( X11  6.017)  0.872

9. Since P-value  0.872  0.05 , we fail to reject H0. We do not have convincing evidence that the proportion of births is not uniform. This is consistent with the researcher’s claim that the frequency of births is not related to lunar cycle. 12.6

1. Let p1 , p2 , p3 , and p4 be the proportions of severe trauma cases in Germany in each of the four seasons. 2. H 0 : p1  p2  p3  p4  1 4  0.25 3. Ha: H0 is not true 4.   0.05 (observed cell count  expected cell count)2 2 X  5. expected cell count all cells 6. We are told that this sample is representative of all severe trauma cases in Germany. In addition, because the sample size was n = 1,200, all expected counts in each category are all equal to 1, 200(0.25)  300 , which is greater than 5, so the sample size is large enough.



X  2

 228  300 2  332  300 2  352  300 2  288  300 2 300



300



300



300

 30.1867

8. df = 3 P-value  P( X 32  30.1867)  0

9. Since P-value  0  0.05 we reject H0. We have convincing evidence that the proportions of severe trauma are not the same for the four seasons. The data do support the theory that the proportion of severe trauma cases is not the same for the four seasons. 12.7 Twitter Type IS Observed Count 51 Expected Count 70

OC RT ME O 61 64 101 73 70

1. Let p1 , p2 , p3 , p4 , and p5 be the proportions of the five Twitter types among the population of Twitter users. 2. H0: p1  p2  p3  p4  p5  0.2

346

Chapter 12: The Analysis of Categorical Data and Goodness-of-Fit Tests 3. Ha: H0 is not true 4. α = 0.05 5. 6. 7. 8. 9.

12.8

(observed cell count  expected cell count)2 expected cell count all cells We are told that the 350 Twitter users included in the study formed a random sample from the population of Twitter users. All the expected counts are greater than 5. (51  70)2 (73  70)2 X2     20.686 70 70 df = 4 P-value  P(  42  20.686)  0.0004 Since P-value  0.0004  0.05 we reject H0. We have convincing evidence that the proportions of Twitter users falling into the five categories are not all the same. X2 



a Direction Left Field Left Center Center Right Center Right Field

Observed Expected Count Count 18 17.4 10 17.4 7 17.4 18 17.4 34 17.4

1. Let p1 , p2 , p3 , p4 and p5 be the proportions of home runs in the given five directions. 2. H0: p1  p2  p3  p4  p5  0.2 3. Ha: H0 is not true 4.   0.05 (observed cell count  expected cell count)2 5. X 2  expected cell count all cells 6. We are told to regard the 87 home runs as representative of home runs hit at Yankee Stadium, and therefore it is reasonable to assume that the sample is a random sample from that population. All the expected counts are greater than 5. (18  17.4)2 (34  17.4)2    25.241 7. X 2  17.4 17.4 8. df = 4 P-value  P(  42  25.241)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that the proportions of home runs hit are not the same for all five directions.



For home runs going to left center and center field the observed counts are significantly lower than the numbers that would have been expected if the proportion of home runs hit was the same for all five directions, while for right field the observed count is much high than the number that would have been expected.

12.9 Ethnicity African-American 57 Observed Count

Asian 11

Caucasian 330

Hispanic 6

Chapter 12: The Analysis of Categorical Data and Goodness-of-Fit Tests

Expected Count

71.508

12.928

296.536

347

23.028

1. Let p1 , p2 , p3 , and p4 be the proportions of appearances of the four ethnicities across all commercials. 2. H0: p1  0.177, p2  0.032, p3  0.734, p4  0.057 3. Ha: H0 is not true 4.   0.01 5. 6. 7. 8. 9.

(observed cell count  expected cell count)2 expected cell count all cells We need to assume that the set of commercials included in the study form a random sample from the population of commercials. All the expected counts are greater than 5. (57  71.508)2 (6  23.028)2 X2     19.599 71.508 23.028 df = 3 P-value  P(  32  19.599)  0 Since P-value  0  0.01 we reject H0. We have convincing evidence that the proportions of appearances in commercials are not the same as the census proportions. X2 



12.10 a

4 2 p(0)     9  3 2!  1  2  4 p(1)   1!1!  3   3  9 2

1 1 p(2)     9  3 b Number Correct Observed Count Expected Count

0 21 19.556

1 10 19.556

2 13 4.889

1. Let p0 , p1 , and p2 be the proportions of the given numbers of correct identifications 2. H0: p0  4 9, p1  4 9, p2  1 9 3. Ha: H0 is not true 4.   0.05

(observed cell count  expected cell count)2 expected cell count all cells 6. One of the expected counts is a little less than 5, but, as instructed in the question, we will proceed. (Note that no assumption of randomness is necessary, since we are simply asking whether observed counts as far from the expected counts as those obtained would be feasible if the students were purely guessing.) (21  19.556)2 (13  4.889)2    18.233 7. X 2  19.556 4.889 8. df = 2 P-value  P(  22  18.233)  0 5.

X2 



348

Chapter 12: The Analysis of Categorical Data and Goodness-of-Fit Tests 9. Since P-value  0  0.05 we reject H0. The numbers of correct identifications are significantly different from those that would have been expected by guessing.

12.11 Since the P-value is small, there is convincing evidence that the population proportions of people who respond “Monday” and “Friday” are not equal. 12.12 State California Virginia Washington Florida Maryland

Observed Expected Count Count 250 200.970 56 41.818 34 34.510 33 97.440 33 31.262

1. Let p1 , p2 , p3 , p4 , and p5 be the actual proportions of hybrid car sales for the five states in the following order: California, Virginia, Washington, Florida, Maryland. 2. H0: p1  0.495, p2  0.103, p3  0.085, p4  0.240, p5  0.077 3. Ha: H0 is not true 4.   0.01 (observed cell count  expected cell count)2 5. X 2  expected cell count all cells 6. We need to assume that the sample was a random sample of hybrid car sales. All the expected counts are greater than 5, so the sample size is large enough to use the chi-square test. (250  200.970)2 (33  31.262)2    59.492 7. X 2  200.970 31.262 8. df = 4 P-value  P(  42  59.492)  0 9. Since P-value  0  0.01 we reject H0. There is convincing evidence that hybrid sales are not proportional to population size for the five states listed.



12.13 a

1. Let p1 , p2 , and p3 be the proportions of the three phenotypes for this particular plant. 2. H0: p1  0.25, p2  0.5, p3  0.25 3. Ha: H0 is not true 4.   0.05 (observed cell count  expected cell count)2 5. X 2  expected cell count all cells 6. We are told that the 200 plants included in the study form a random sample from the population of this type of plant. Since 200(0.25)  50  5 and 200(0.5)  100  5, all the expected counts are greater than 5. 7. X 2  4.63 8. df = 2 P-value  P(  22  4.63)  0.099 9. Since P-value  0.099  0.05 we do not reject H0. We do not have convincing evidence that the theory is incorrect.



Chapter 12: The Analysis of Categorical Data and Goodness-of-Fit Tests b

The analysis and conclusion would be the same.

12.14 a

df  (6  1)(3  1)  10

df  (7  1)(3  1)  12

df  (6  1)(4  1)  15

12.15 a

2 df  (4  1)(5  1)  12. P-value  P( 12  7.2)  0.844. Since the P-value is greater than 0.1, we do not have convincing evidence that education level and preferred candidate are not independent.

df  (4  1)(4  1)  9. P-value  P(  92  14.5)  0.106. Since the P-value is greater than 0.05, we do not have convincing evidence that education level and preferred candidate are not independent.

12.16 a United States Spain Italy India b

349

Female 550 (378.333) 195 (189.167) 190 (189.167) 200 (378.333)

Male 450 (621.667) 305 (310.833) 310 (310.833) 800 (621.667)

H0: The gender proportions are the same for the four countries. Ha: The gender proportions are not the same for the four countries.   0.05 (observed cell count  expected cell count)2 X2  expected cell count all cells We are told that samples were representative of the populations of smartphone users in the four countries. All the expected counts are greater than 5. (550  378.333)2 (800  621.667)2 X2     260.809 378.333 621.667 df = 3 P-value  P( 32  260.809)  0 Since P-value  0  0.05 we reject H0. There is convincing evidence that the gender proportions are not the same for the four countries.



350

Chapter 12: The Analysis of Categorical Data and Goodness-of-Fit Tests

12.17 Accepts Credit Cards for Tuition Payment 92 (86.67) 68 (86.67) 100 (86.67)

Public Four-Year Colleges Private Four-Year Colleges Community Colleges

Does Not Accept Credit Cards for Tuition Payment 8 (13.33) 32 (13.33) 0 (13.33)

H0: The proportions falling into the two credit card response categories are the same for all three types of colleges Ha: The proportions falling into the two credit card response categories are the same for all three types of colleges   0.05 (observed cell count  expected cell count)2 X2  expected cell count all cells We are told to suppose that these three samples are representative of their respective populations in the U.S. The table above shows observed counts and, in parentheses, expected counts. Note that all expected counts are at least 5.



X  2

 92  86.67 2  68  86.67 2 86.67



86.67



 32  13.332  0  13.332 

13.33

 48.000

df = 2 P-value  P( X 22  48.000)  0 Since P-value  0  0.05 we reject H0. We have convincing evidence that the proportions in each of the two credit card categories is not the same for all three types of colleges. 12.18 a Freedom of Speech Students Teachers

650 (525) 400 (525)

Most Important First Amendment Right Freedom to Freedom of Freedom of Peacefully the Press Religion Assemble 30 (45) 250 (335) 20 (35) 60 (45) 420 (335) 50 (35)

Freedom to Petition the Government 50 (60) 70 (60)

H0: The proportions of the five First Amendment right categories are the same for teachers and students Ha: The proportions of the five First Amendment right categories are not the same for teachers and students   0.01 (observed cell count  expected cell count)2 X2  expected cell count all cells We are told that the samples of teachers and students are representative of their respective populations. The table above shows observed counts and, in parentheses, expected counts. Note that all expected counts are at least 5.



X  2

 650  5252  30  452 525



df = 4 P-value  P( X 42  128.849)  0



 50  352  70  60 2 35



 128.849

Chapter 12: The Analysis of Categorical Data and Goodness-of-Fit Tests

351

Since P-value  0  0.01 we reject H0. We have convincing evidence that the proportions falling into the five First Amendment right categories are not the same for teachers and students. b

The result of part (a) tells us that there is evidence that students and teachers have different beliefs regarding the most important First Amendment right categories. Analysis of the observed and expected counts indicate that the largest deviations between students and teachers occur in the freedom of speech and freedom of religion categories. Students rank freedom of speech higher than the teachers do, and teachers rank freedom of religion higher than the students do.

12.19

2004 2014

People should be allowed to burn or deface the American flag as a political statement Strongly Strongly Agree Mildly Agree Mildly Disagree Don’t know Disagree 80 (75) 80 (75) 110 (110) 630 (645) 100 (95) 70 (75) 70 (75) 110 (110) 660 (645) 90 (95) H0: The proportions falling into each of the response categories were the same for high school student in 2004 and 2014 Ha: The proportions falling into each of the response categories were not the same for high school student in 2004 and 2014   0.05 (observed cell count  expected cell count)2 X2  expected cell count all cells We are told that the samples of high school students in 2004 and 2014 are representative of their respective populations. The table above shows observed counts and, in parentheses, expected counts. Note that all expected counts are at least 5.



X  2

80  752 80  752 75



 660  645 2  90  95 2 645



 2.557

df = 4 P-value  P( X 42  2.557)  0.634 Since P-value  0.634  0.05 we fail to reject H0. We do not have convincing evidence that the proportions falling into each of the response categories were not the same for high school student in 2004 and 2014. 12.20 Never Rarely Sometimes Frequently

Age 18 to 20 72 (54.94) 36 (34.38) 30 (37.75) 12 (22.92)

Age 21 to 22 62 (59.34) 34 (37.13) 42 (40.77) 24 (24.76)

Age 23 to 24 29 (48.72) 32 (30.49) 40 (33.47) 32 (20.32)

H0: There is no association between age group and the response to the question Ha: There is an association between age group and the response to the question   0.01 (observed cell count  expected cell count)2 X2  expected cell count all cells



352

Chapter 12: The Analysis of Categorical Data and Goodness-of-Fit Tests We are told that the sample is representative of the population of college students. The table above shows observed counts and, in parentheses, expected counts. Note that all expected counts are at least 5.

X  2

 72  54.94 2  62  59.342 54.94



59.34



 24  24.762  32  20.322 24.76



20.32

 28.646

df = 6 P-value  P( X 62  28.646)  0 Since P-value  0  0.01 we reject H0. We have convincing evidence that there is an association between age group and the response to the question. 12.21

High School Diploma Associate Degree Bachelor’s Degree

Less than $20,000

Income Level $20,000 to $40,000 to $39,999 $59,999

8 (28.02)

68 (120.58)

106 (111.66)

243 (164.74)

11 (12.26)

56 (52.77)

56 (48.87)

63 (72.10)

47 (25.71)

160 (110.65)

101 (102.47)

82 (151.17)

$60,000 or more

H0: There is no association between income category and education level Ha: There is an association between income category and education level   0.05 (observed cell count  expected cell count)2 X2  expected cell count all cells We are told to assume that the sample in this study is representative of adult Americans who are employed full-time. The table above shows observed counts and, in parentheses, expected counts. Note that all expected counts are at least 5.



X  2

8  28.02 2  68  120.582 28.02



120.58



101  102.47 2  82  151.17 2 102.47



151.17

 148.520

df = 6 P-value  P( X 62  148.520)  0 Since P-value  0  0.05 we reject H0. We have convincing evidence that there is an association between income category and education level. 12.22 a

A chi-square test of independence would be the appropriate test because the purpose of the study is to compare two categorical variables (pay full balance or not each month and age category) from one population (adult Americans age 20 to 39 who have at least one credit card).

Age 20 to 24 years Age 25 to 29 years Age 30 to 34 years Age 35 to 39 years

Pay Full Balance Each Month 75 (61.84) 74 (72.75) 75 (80.03) 67 (76.39)

Carry Balance from Month to Month 95 (108.16) 126 (127.25) 145 (139.97) 143 (133.61)

Chapter 12: The Analysis of Categorical Data and Goodness-of-Fit Tests

353

H0: There is no association between whether or not people pay their balance in full each month and age Ha: There is an association between whether or not people pay their balance in full each month and age   0.05 (observed cell count  expected cell count)2 X2  expected cell count all cells We are told to suppose that this data resulted from a random sample of 800 adult Americans age 20 to 39 years old who have at least one credit card. The table above shows observed counts and, in parentheses, expected counts. Note that all expected counts are at least 5.



X2 

 75  61.84 2  95  108.16 2 61.84



108.16



 67  76.392 143  133.612 76.39



133.61

 6.746

df = 3 P-value  P( X 32  6.746)  0.080 Since P-value  0.080  0.05 we fail to reject H0. We do not that whether or not people pay their balance in full each month is related to age. c

It is reasonable to generalize this conclusion to the population of adult Americans age 20 to 39 years old who have at least one credit card because that is the population from which the sample was randomly selected.

12.23

Freshman Sophomore Junior Senior

Body Piercings Only 61 (49.714) 43 (37.878) 20 (23.378) 21 (34.031)

Tattoos Only 7 (15.086) 11 (11.494) 9 (7.094) 17 (10.327)

Both Body Piercing and Tattoos 14 (18.514) 10 (14.106) 7 (8.706) 23 (12.673)

No Body Art 86 (84.686) 64 (64.522) 43 (39.822) 54 (57.969)

H0: Class standing and body art response are independent Ha: Class standing and body art response are not independent   0.01 (observed cell count  expected cell count)2 X2  expected cell count all cells We are told to regard the sample as representative of the students at this university, so we are justified in treating it as a random sample from that population. All the expected counts are greater than 5. (61  49.714)2 (54  57.969)2 X2     29.507 49.714 57.969 df = 9 P-value  P(  92  29.507)  0.001 Since P-value  0.001  0.01 we reject H0. We have convincing evidence of an association between class standing and response to the body art question.



12.24 Character Type

354

Chapter 12: The Analysis of Categorical Data and Goodness-of-Fit Tests

Sex Male Female

Positive 255 (262.072) 85 (77.928)

Negative 106 (90.954) 12 (27.046)

Neutral 130 (137.973) 49 (41.027)

H0: Sex and character type are independent Ha: Sex and character type are not independent   0.05 (observed cell count  expected cell count)2 X2  expected cell count all cells We are told to regard the sample as representative of smoking movie characters, so we are justified in treating it as a random sample from that population. All the expected counts are greater than 5.



X2 

(255  262.072)2  262.072



(49  41.027)2  13.702 41.027

df = 2 P-value  P(  22  13.702)  0.001 Since P-value  0.001  0.05 we reject H0. We have convincing evidence of an association between sex and character type for movie characters who smoke. 12.25 a Usually Eat Rarely Eat 3 Meals a Day 3 Meals a Day 26 (21.755) 22 (26.245) Male 37 (41.245) 54 (49.755) Female H0: The proportions falling into the two response categories are the same for males and females. Ha: The proportions falling into the two response categories are not the same for males and females.   0.05 (observed cell count  expected cell count)2 X2  expected cell count all cells We are told to assume that the samples of male and female students were random samples from the populations. All the expected counts are greater than 5. (26  21.755)2 (54  49.755)2 X2     2.314 21.755 49.755 df = 1 P-value  P( 12  2.314)  0.128 Since P-value  0.128  0.05 we do not reject H0. We do not have convincing evidence that the proportions falling into the two response categories are not the same for males and females.



Yes.

Yes. Since P-value  0.127  0.05 we do not reject H0. We do not have convincing evidence that the proportions falling into the two response categories are not the same for males and females.

Chapter 12: The Analysis of Categorical Data and Goodness-of-Fit Tests d

355

The two P-values are almost equal, in fact the difference between them is only due to rounding errors in the MINITAB program. In other words, if complete accuracy had been maintained throughout, the two P-values would have been exactly equal. (Also, the chisquare statistic in Part (a) is the square of the z statistic in Part (c).) It should not be surprising that the P-values are at least similar, since both measure the probability of getting sample proportions at least as far from the expected proportions, given that the proportions who usually eat three meals per day are the same for the two populations.

12.26

Compulsive Buyer?

Yes No

Locus of Control Internal External 3 (7.421) 14 (9.579) 52 (47.579) 57 (61.421)

H0: Locus of control and compulsive buyer behavior are independent Ha: Locus of control and compulsive buyer behavior are not independent   0.01 (observed cell count  expected cell count)2 X2  expected cell count all cells We are told to regard the sample as representative of college students at mid-western public universities, so we are justified in treating it as a random sample from that population. All the expected counts are greater than 5. (3  7.421)2 (57  61.421)2 X2     5.402 7.421 61.421 df = 1 P-value  P( 12  5.402)  0.020 Since P-value  0.020  0.01 we do not reject H0. We do not have convincing evidence of an association between locus of control and compulsive buyer behavior.



12.27

School Performance

1 110 (79.25) Excellent 328 (316) Good Average/Poor 239 (281.75)

Alcohol Exposure Group 2 3 93 (79.25) 49 (79.25) 325 (316) 316 (316) 259 (281.75) 312 (281.75)

4 65 (79.25) 295 (316) 317 (281.75)

H0: Alcohol exposure and school performance are independent Ha: Alcohol exposure and school performance are not independent   0.05 (observed cell count  expected cell count)2 X2  expected cell count all cells We are told to regard the sample as a random sample of German adolescents. All the expected counts are greater than 5. (110  79.25)2 (317  281.75)2 X2     46.515 79.25 281.75 df = 6



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P-value  P(  62  46.515)  0 Since P-value  0  0.05 we reject H0. We have convincing evidence of an association between alcohol exposure and school performance. 12.28 View Sex ID Correct Incorrect

Front 23 (26) 17 (14)

Profile 26 (26) 14 (14)

ThreeQuarter 29 (26) 11 (14)

H0: The proportions of correct sex identifications are the same for all three nose views. Ha: The proportions of correct sex identifications are not the same for all three nose views.   0.05 (observed cell count  expected cell count)2 X2  expected cell count all cells We need to assume that the students were randomly assigned to the nose views. All the expected counts are greater than 5.



X2 

(23  26)2  26



(11  14)2  1.978 14

df = 2 P-value  P(  22  1.978)  0.372 Since P-value  0.372  0.05 we do not reject H0. We do not have convincing evidence that the proportions of correct sex identifications are not the same for all three nose views. 12.29 a

H0: Gender and number of servings of water consumed per day are independent Ha: Gender and number of servings of water consumed per day are not independent df  (5  1)(2  1)  4

The P-value for the test was 0.086, which is greater than the new significance level of 0.05. So, for a significance level of 0.05, we do not have convincing evidence of a difference between males and females with regard to water consumption.

H0: Gender and number of consumptions of fried potatoes are independent Ha: Gender and number of consumptions of fried potatoes are not independent   0.05 (observed cell count  expected cell count)2 X2  expected cell count all cells No assumption regarding the randomness of the sample is necessary since we are merely asking whether there is a significant association between the two variables for this set of students. All the expected counts (shown in the output) are greater than 5. X 2  15.153 df = 5 P-value  0.015 Since P-value  0.015  0.05 we reject H0. We have convincing evidence of an association between gender and number of consumptions of fried potatoes. This agrees with the authors’



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conclusion that there was a significant association between gender and consumption of fried potatoes. 12.30 a

The number of men in the sample who napped is 744(0.38) = 282.72, which we round to 283, since the number of men who napped must be a whole number. The number of men who did not nap is therefore 744  283  461. The observed frequencies for the women are calculated in a similar way. (The table below also shows the expected frequencies in parentheses.)

Men Women b

Napped 283 (257) 231 (257)

Did Not Nap 461 (487) 513 (487)

Row Total 744 744

H0: Gender and napping are independent Ha: Gender and napping are not independent   0.01 (observed cell count  expected cell count)2 X2  expected cell count all cells We are told that the sample was nationally representative, so we are justified in treating it as a random sample from the population of American adults. All the expected counts are greater than 5. (283  257)2 (513  487)2 X2     8.034 257 487 df = 1 P-value  P( 12  8.034)  0.005 Since P-value  0.005  0.01 we reject H0. We have convincing evidence of an association between gender and napping.



Yes. We have convincing evidence at the 0.01 significance level of an association between gender and napping in the population. This is equivalent to saying that we have convincing evidence at the 0.01 significance level that the proportions of men and women who nap are different (a two-tailed test of a difference of the proportions). Thus, converting this to a onetailed test, since in the sample the proportion of men who napped was greater than the proportion of women who napped, we have convincing evidence at the 0.005 level that a greater proportion of men nap than women.

12.31 It is not possible to decide which, if either, of the two conclusions is correct. Since the results were obtained from an observational study, no conclusion regarding causality can be reached. 12.32 Day Sunday Monday Tuesday Wednesday Thursday Friday Saturday

Observed Count 14 13 12 15 14 17 15

Expected Count 14.286 14.286 14.286 14.286 14.286 14.286 14.286

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1. Let p1 , , p7 be the proportions of all fatal bicycle accidents occurring on the seven days. 2. H0: p1  p2   p7  1 7 3. Ha: H0 is not true 4.   0.05 (observed cell count  expected cell count)2 5. X 2  expected cell count all cells 6. We are told that the 100 accidents formed a random sample from the population of fatal bicycle accidents. All the expected counts are greater than 5.



X2 

(14  14.286)2  14.286



(15  14.286)2  1.08 14.286

8. df = 6 P-value  P( 62  1.08)  0.982 9. Since P-value  0.982  0.05 we do not reject H0. We do not have convincing evidence that the proportion of accidents is not the same for all days of the week. 12.33 Color Blue Green Yellow Red

Observed Expected Count Count 16 8.25 8 8.25 6 8.25 3 8.25

1. Let p1 , p2 , p3 , and p4 be the proportions of first pecks for the colors given. 2. H0: p1  p2  p3  p4  0.25 3. Ha: H0 is not true 4.   0.01 (observed cell count  expected cell count)2 2 X  5. expected cell count all cells 6. We need to assume that the birds in the study formed a random sample of 1-day-old bobwhites. All the expected counts are greater than 5. (16  8.25)2 (3  8.25)2    11.242 7. X 2  8.25 8.25 8. df = 3 P-value  P( 32  11.242)  0.0105 9. Since P-value  0.0105  0.01 we do not reject H0. We do not have convincing evidence of a color preference.



12.34 Excellent Male Female

103 (83.61) 83 (102.39)

Perception of Money Management Skills Good Average Not Very Good 139 (148.78) 89 (100.69) 17 (17.53) 192 (182.22) 135 (123.31) 22 (21.47)

Poor 8 (5.39) 4 (6.61)

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H0: There is no association between sex and how students perceive their money management skills Ha: There is an association between sex and how students perceive their money management skills   0.05 (observed cell count  expected cell count)2 X2  expected cell count all cells We are told that the sample is representative of college students. The table above shows observed counts and, in parentheses, expected counts. Note that all expected counts are at least 5.



X  2

103  83.612 139  148.782 83.61



148.78



 22  21.47 2  4  6.612 21.47



6.61

 14.121

df = 4 P-value  P( X 42  14.121)  0.007 Since P-value  0.007  0.05 we reject H0. We have convincing evidence that there is an association between sex and how students perceive their money management skills. 12.35 Perception of Money Management Skills Excellent Good Average Not Very Good/Poor 73 (81.70) 146 (147.06) 104 (99.37) 25 (19.87) 54 (59.87) 103 (107.76) 85 (72.81) 13 (14.56) 58 (43.43) 84 (78.18) 36 (52.82) 7 (10.56)

Age 18 to 20 Age 21 to 22 Age 23 to 24

H0: There is no association between age and how students perceive their money management skills Ha: There is an association between age and how students perceive their money management skills   0.05 (observed cell count  expected cell count)2 X2  expected cell count all cells We are told that the sample is representative of college students. The table above shows observed counts and, in parentheses, expected counts. Note that all expected counts are at least 5.



X  2

 73  81.70 2 146  147.06 2 81.70



147.06



 36  52.82 2  7  10.56 2 52.82



10.56

 17.346

df = 6 P-value  P( X 62  17.346)  0.008 Since P-value  0.008  0.05 we reject H0. We have convincing evidence that there is an association between age and how students perceive their money management skills.

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12.36

Risk Low Risk of Malnutrition Medium Risk of Malnutrition High Risk of Malnutrition

Survived 322 (281.50) 29 (32.10) 91 (128.40)

Did not survive 20 (60.50) 10 (6.90) 65 (27.60)

H0: There is no association between survival and risk of malnutrition Ha: There is an association between survival and risk of malnutrition   0.05 (observed cell count  expected cell count)2 X2  expected cell count all cells We are told to assume the sample is representative of the population of stroke patients. The table above shows observed counts and, in parentheses, expected counts. Note that all expected counts are at least 5.



X  2

 322  281.50 2  20  60.50 2 281.50



60.50



 91  128.40 2  65  27.60 2 128.40



27.60

 96.219

df = 2 P-value  P( X 22  96.219)  0 Since P-value  0  0.05 we reject H0. We have convincing evidence that there is an association between survival and the risk of malnutrition, which is consistent with what the authors concluded.

Chapter 13 Simple Linear Regression and Correlation: Inferential Methods Note: In this chapter, numerical answers to questions involving the normal, t, and chi square distributions were found using values from a calculator. Students using statistical tables will find that their answers differ slightly from those given. 13.1

y  5.0  0.017 x

When x  1000, y  5  0.017(1000)  12. When x  2000, y  5  0.017(2000)  29. y 45 40 35 30 25 20 15 10 5 x

0 -5 0

13.2

13.3

13.4

1000

2000

When x  2100, y  5  0.017(2100)  30.7. The mean gas usage for houses with 2100 square feet of space is 30.7 therms.

0.017 therms

100(0.017) = 1.7 therms

No. The given relationship only applies to houses whose sizes are between 1000 and 3000 square feet. The size of this house, 500 square feet, lies outside this range.

For x  10, y  0.12  0.095(10)  0.83. For x  15, y  0.12  0.095(15)  1.305.

Since the slope of the population regression line is 0.095, the average increase in flow rate associated with a 1-inch increase in pressure drop is 0.095.

When x  15,  y  0.135  0.003(15)  0.18 micrometers When x  17,  y  0.135  0.003(17)  0.186 micrometers

361

3000

When x  15,  y =0.18, so P( y  0.18)  0.5.

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When x  14,  y  0.135  0.003(14)  0.177,

0.175  0.177   so P( y  0.175)  P  z    P( z  0.4)  0.655. 0.005   0.178  0.177   P( y  0.178)  P  z    P( z  0.2)  0.579. 0.005   13.5

13.6

13.7

For each one-unit increase in horsepower the predicted fuel efficiency decreases by 0.150 mpg.

The number yˆ  29.0 is both an estimate of the mean fuel efficiency when the horsepower is 100 and a prediction for the horsepower of a car whose horsepower is 100.

When x  300, yˆ  44.0  0.15(300)  1.0. This result cannot be valid, since it is not possible to have a car whose fuel efficiency is negative. This has probably occurred because 300 is outside the range of horsepower ratings for the small cars used in the sample; the estimated regression equation is not valid for values outside this range.

This tells us that 68.0% of the variation in fuel efficiency is attributable to the approximate linear relationship between horsepower and fuel efficiency.

This tells us that 3.0 mpg is a typical deviation of the fuel efficiency of a car in the sample from the value predicted by the least-squares line.

Average change in price associated with one extra square foot of space = $47. Average change in price associated with 100 extra square feet of space = 100(47) = $4700.

When x  1800,  y  23000  47(1800)  107600.

110000  107600   So P( y  110000)  P  z    P( z  0.48)  0.316. 5000   100000  107600   P( y  100000)  P  z    P( z  1.52)  0.064. 5000   13.8

y     x is the equation of the population regression line, which relates the mean value of y to the value of x. ŷ  a  bx is the equation of an estimated regression line, which is an estimate of the population regression line obtained from a particular set of ( x , y ) observations.

 is the slope of the population regression line. b is an estimate of  obtained from a particular set of ( x , y ) observations.

   ( x*) is the mean value of y when x  x *. a  b( x*) is an estimate of the mean value of y (and also a predicted y value) when x  x *.

The simple linear regression model assumes that for any fixed value of x, the distribution of y is normal with mean    x and standard deviation  . Thus  is the shared standard

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deviation of these y distributions. The quantity se is an estimate of  obtained from a particular set of ( x , y ) observations. 13.9

r 2  23.48%

The point estimate of  e is se  5.36559 . The typical amount by which the depression score change value deviates from the value in the sample and what was predicted using the least squares regression line is 5.36599.

The estimated average change in depression score change associated with a 1 kg/m2 increase in BMI change is 5.08.

The estimated mean depression score change when x = 1.2 kg/m2 is yˆ  6.87  5.08(1.2)  12.966 .

13.10 a

The equation of the estimated regression line is yˆ  45.572  1.335x, where x = percent of women using HRT and y = breast cancer incidence.

1.335 cases per 100,000 women.

For x  40, yˆ  45.572  1.335(40)  98.990 cases per 100,000 women.

No. This is not advisable as 20 is not within the range of the x-values in the original data set, and we have no reason to believe that the relationship summarized by the estimated regression line applies outside this range.

13.11 a

r 2  0.830. Eighty-three percent of the variability in breast cancer incidence can be explained by the approximate linear relationship between breast cancer incidence and percent of women using HRT.

13.12 a

se  4.154. This is a typical deviation of breast cancer incidence value in this data set from the value predicted by the estimated regression line. The point estimate of  is se 

SSResid 1235.470   9.749. n2 13

df = 13. b

r2  1 

SSResid 1235.470  1  0.951. So the required percentage is 95.1%. SSTo 25321.368

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13.13 a y 0.14

0.12

0.10

0.08

0.06

0.04 0.04

0.05

0.06

0.07

0.08

0.09

0.10

0.11

yˆ  0.00227  1.247 x When x  0.09, yˆ  0.00227  1.247(0.09)  0.110.

r 2  0.436. This tells us that 43.6% of the variation in market share can be explained by the linear regression model relating market share and advertising share.

A point estimate of  is se 

SSResid 0.00551   0.026. The number of degrees of n2 8 freedom associated with this estimate is n  2  10  2  8.

13.14 The simple linear regression model assumes that for any fixed value of x, the distribution of y is normal with mean    x and standard deviation  . Thus  is the shared standard deviation of these y distributions. To understand the meaning of the symbol  b , consider the process, for a given set of x values, of randomly selecting a set of y values (one for each x value), and calculating the slope, b, of the least-squares regression line for the set of ( x , y ) observations obtained. The quantity  b is the standard deviation of the set of all possible such values of b. The quantity sb is an estimate, from a particular set of ( x , y ) observations, of  b . 13.15 a

Yes. The P-value for the hypothesis test of H0: β = 0 versus Ha: β ≠ 0 is 0.023. Since this value is less than 0.05, we have convincing evidence at the 0.05 significance level of a useful linear relationship between shrimp catch per tow and oxygen concentration density.

Yes. The value of the correlation coefficient for the sample is r  0.496  0.704 . So there is a moderate to strong linear relationship for the values in the sample.

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365

We must assume that the conditions for inference are met. df = 10 – 2 = 8. The 95% confidence interval for  is b  (t critical value)  sb  97.22  (2.306)(34.63)  (17.363,177.077). We are 95% confident that the slope of the population regression line relating mean catch per tow to O2 saturation is between 17.363 and 177.077.

13.16 Yes. We know that  b  

( x  x ) . Therefore, the larger the spread of the x values, the 2

smaller the variability of the statistic b. The set of x values given in the question has a greater spread than the set of x values given in the example, and therefore is likely to give a more reliable estimate for  than the one obtained in the example.

13.17 a

SSResid 1235.470   9.749. n2 13 s 9.749 sb  e   0.154. S xx 4024.2

se 

We must assume that the conditions for inference are met. df = 13. The 95% confidence interval for  is

b  (t critical value)  sb  2.5  (2.160)(0.154)  ( 2.168, 2.832). We are 95% confident that the slope of the population regression line relating hardness of molded plastic and time elapsed since the molding was completed is between 2.168 and 2.832. c

13.18 a

Yes. Since the confidence interval is relatively narrow it seems that  has been somewhat precisely estimated. The required proportion is r 2  1 

se 

SSResid 561.46  1  0.766. SSTo 2401.85

SSResid 561.46   6.572. n2 13

  x   13.92  14.1  0.666. x  2

S xx 



n se 6.572 sb    8.053. S xx 0.666 c

We must assume that the conditions for inference are met. df = 13. x y (14.1)(1438.5) S xy  xy   1387.20   35.01. n 15



    

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S xy

35.01  52.568 S xx 0.666 The 90% confidence interval for  is b  (t critical value)  sb  52.568  (1.771)(8.053)  ( 38.306, 66.829). We are 90% confident that the average change in revenue associated with a $1000 increase in advertising expenditure is between 38.306 and 66.829. b

13.19 a

S xy 





xy 

  x   y   44194  (50)(16705)  2431.5

x2 

  x   150  (50)  25.

S xx  x



50 16705  2.5, y   835.25 20 20

S xy

2431.5  97.26. S xx 25 The y intercept of the population regression line is estimated by a  y  bx  835.25  97.26(2.5)  592.1. The slope of the population regression line is estimated by b 



When x  2, yˆ  a  bx  592.1  97.26(2)  786.62. Residual  y  yˆ  757  786.62  29.62.

We require a 99% confidence interval for  . We must assume that the conditions for inference are met. SSResid  y 2  a y  b xy  14194231  592.1(16705)  97.26(44194)  4892.06.



SSResid 4892.06   16.486. n2 18 s 16.486 sb  e   3.297. S xx 25 se 

df = 18. The 99% confidence interval for  is b  (t critical value)  sb  97.26  (2.878)(3.297)  (87.769,106.751) We are 99% confident that the slope of the population regression line relating amount of oxygen consumed and time spent exercising is between 87.769 and 106.751. 13.20 a

1.  = slope of the population regression line relating typing speed to surface angle. 2. H0:   0 3. Ha:   0 4.   0.05 b  (hypothesized value) b  0  5. t  sb sb 6. We are told to assume that the basic assumptions of the simple linear regression model are reasonably met.

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7. t  0.09 8. df = 3 P-value  0.931 9. Since P-value  0.931  0.05 we do not reject H0. We do not have convincing evidence of a useful linear relationship between typing speed and surface angle. b

13.21 1.

Since r 2  0.003, virtually none of the variation in typing speed can be attributed to the approximate linear relationship between typing speed and surface angle as represented by the estimated regression line. This is consistent with our conclusion that we do not have convincing evidence of a useful linear relationship between the two variables. Now se  0.512. This value is relatively small when compared to typing speeds of around 60, and tells us that the y values are typically deviating by around only 0.5 from the values predicted by the virtually horizontal estimated regression line. This is not inconsistent with the conclusion from Part (a).

 = slope of the population regression line relating brain volume change with mean

childhood blood lead level. 2. H0:   0 3. Ha:   0 4.   0.05 b  (hypothesized value) b  0  5. t  sb sb 6. We are told to assume that the basic assumptions of the simple linear regression model are reasonably met. 7. t  3.66 8. P -value  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that the slope of the population regression line relating brain volume change with mean childhood blood lead level is not equal to zero, that is, that there is a useful linear relationship between these two variables. 13.22 1. 2. 3. 4. 5. 6. 7. 8. 9.

 = slope of the population regression line relating log(weekly gross earnings in dollars) to height. H0:   0 Ha:   0   0.05 b  (hypothesized value) b  0 t  sb sb We are told to assume that the basic assumptions of the simple linear regression model are reasonably met. 0.023  0 t  5.75 0.004 The number of degrees of freedom is large, and so the normal distribution can be used. P -value  2  P( Z  5.75)  0 Since P-value  0  0.05 we reject H0. We have convincing evidence of a useful linear relationship between height and the logarithm of weekly earnings.

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13.23 a

For the data given, b  0.140, se  0.402, sb  0.026 . The data are plotted in the scatterplot below. Pleasantness Rating 3.0 2.5 2.0 1.5 1.0 0.5 0.0 20

28 30 Firing Frequency

 = slope of the population regression line relating pleasantness rating to firing

frequency. 2. H0:   0 3. Ha:   0 4.   0.05 b  (hypothesized value) b  0  5. t  sb sb 6. The conditions for inference were checked in Part (a). 0.140  0  5.451 7. t  0.026 8. df = 8 P-value  2  P(t8  5.451)  0.001 9. Since P-value  0.001  0.05 we reject H0. We have convincing evidence of a useful linear relationship between firing frequency and pleasantness rating. 13.24 a

The estimated regression line is yˆ  51.305  0.16327 x , where ŷ is the predicted percentage of alumni who strongly agree that their education was worth the cost, and x is the 2015 university ranking.

Chapter 13: Simple Linear Regression and Correlation: Inferential Methods 1.

 = slope of the population regression line relating mean percentage of alumni who

strongly agree that their education was worth the cost and the 2015 university ranking. 2. H0:   0 3. Ha:   0 4.   0.05 b  (hypothesized value) b  0 5. t   sb sb 6. The data are plotted in the scatterplot below. 75 Percentage who Strongly Agree

369

50 20

50 60 70 University Ranking

100

The pattern in the scatterplot above, although weak, looks linear, and the spread does not seem to differ much for different values of x. The boxplot of residuals, shown below, is slightly left-skewed and there are no outliers, so it reasonable to conclude that the distribution of e is approximately normal.

-5



0.16327

 2.55 sb 0.06396 8. df = 15 P -value  2  P (t  2.55)  0.022 9. Since P-value=0.022  0.05 we reject H0. We have convincing evidence of a useful linear relationship between the percentage of alumni who strongly agree that their education was worth the cost and the 2015 university ranking. 7.

t

0 Residuals

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13.25 a

By definition, this is the slope of the estimated regression line, or b  0.359 g/kg.

When x = 250 s, the predicted acrylamide concentration is yˆ  87  0.359(250)  176.75 g/kg.

1. 2. 3. 4. 5. 6.

 = slope of the population regression line relating mean acrylamide concentration and frying time. H0:   0 Ha:   0   0.05 b  (hypothesized value) b  0 t  sb sb We must suppose that the basic assumptions of the simple linear regression model are met.

7. t  0.82 8. P-value  0.459 9. Since P-value=0.459  0.05 we fail to reject H0. We do not have convincing evidence of a useful relationship between acrylamide concentration and frying time. 13.26 a

The scatterplot suggests that the simple linear regression model might be appropriate because there is an increasing linear trend in the plot.

yˆ  6.8725681  5.077821x , where ŷ is the predicted depression score change and x is the BMI change.

 = slope of the population regression line relating mean depression score change and

BMI change. 2. H0:   0 3. Ha:   0 4.   0.05 b  (hypothesized value) b  0  5. t  sb sb 6. We must check the basic assumptions. We are not told how the data were collected or if the observations are independent. We shall proceed as if this is the case. In addition, the pattern in the given scatterplot looks linear, and the spread does not seem to differ much for different values of x. b 5.077821  1.75 7. t   sb 2.898557 8. P -value = 2  P(t  1.75)  0.1104 9. Since P-value=0.1104  0.05 we fail to reject H0. We do not have convincing evidence of a useful linear relationship between depression score change and BMI change. 13.27 a

 = average change in sales revenue associated with a 1-unit increase in advertising expenditure. 2. H0:   0 3. Ha:   0 1.

Chapter 13: Simple Linear Regression and Correlation: Inferential Methods 4. 5.

371

  0.05 b  (hypothesized value) b  0 t  sb sb

6. We must assume that the conditions for inference are met. 52.27  0  6.493 7. t  8.05 8. df = 13 P-value  2  P(t13  6.493)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that the slope of the population regression line relating sales revenue and advertising expenditure is not equal to zero. We conclude that there is a useful linear relationship between sales revenue and advertising expenditure. b

 = average change in sales revenue associated with a 1-unit increase in advertising

expenditure. 2. H0:   40 3. Ha:   40 4.   0.01 b  (hypothesized value) b  40  5. t  sb sb 6. We must assume that the conditions for inference are met. 52.27  40  1.524 7. t  8.05 8. df = 13 P-value  P(t13  1.524)  0.076 9. Since P-value  0.076  0.01 we do not reject H0. We do not have convincing evidence that the average change in sales revenue associated with a 1-unit (that is, $1000) increase in advertising expenditure is greater than $40,000. 13.28 a

1. 2. 3. 4. 5. 6.

 = slope of the population regression line relating growth rate to research and development expenditure. H0:   0 Ha:   0   0.05 b  (hypothesized value) b  0 t  sb sb The data are plotted in the scatterplot below.

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Growth Rate (% per year) 4

-1 0

1000

2000 3000 4000 R & D Expenditure (thousands of dollars)

5000

We require a 90% confidence interval for  , the slope of the population regression line. The conditions were checked in Part (a). df = 6. The 90% confidence interval for  is

b  (t critical value)  sb  0.000575  (1.943)(0.00025175)  (0.000086, 0.001064) We are 90% confident that the mean change in growth rate associated with a $1000 increase in research and development expenditure is between 0.000086 and 0.001064. 13.29 a

S xx 

( x  x )  (5  15)  2

 (25  15) 2  250. So  b 

Now S xx  2(250)  500. So  b  Part (a).

 S xx



 S xx



4  0.253. 250

4  0.179. No,  b is not half of what it was in 500

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373

Four observations should be taken at each of the x values, since then S xx would be multiplied by 4, and so  b would be divided by 2. To verify, S xx  4(250)  1000, so

b 

 S xx



4  0.126  1 2  (0.253). 1000

13.30 1.  = slope of the population regression line relating cranial capacity to chord length. 2. H0:   20 3. Ha:   20 4.   0.05 b  (hypothesized value) b  20  5. t  sb sb 6. The data are plotted in the scatterplot below. Cranial Capacity (cubic centimeters) 1050 1000 950 900 850 800 750 75.0

77.5

80.0 82.5 Chord Length (mm)

85.0

87.5

The plot shows a linear pattern, and the vertical spread of points does not appear to be changing over the range of x values in the sample. If we assume that the distribution of errors at any given x value is approximately normal, then the simple linear regression model seems appropriate. 7. b  22.257, se  55.575, S xx  123.429, sb  5.002 22.257  20 t  0.451 5.002 8. df = 5 P-value  2  P(t5  0.451)  0.671 9. Since P-value  0.671  0.05 we do not reject H0. We do not have convincing evidence that the increase in cranial capacity associated with a 1-mm increase in chord length is not 20 cm3. 13.31 There is a random scatter of points in the residual plot, implying that a linear model relating squirrel population density to percentage of logging is appropriate. The residual plot shows no tendency for the size (magnitude) of the residuals to either increase of decrease as percentage of logging increases. So it is justifiable to assume that the vertical deviations from the population regression line have equal standard deviations. The last condition is that the vertical deviations be

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Chapter 13: Simple Linear Regression and Correlation: Inferential Methods normally distributed. The fact that the boxplot of the residuals is roughly symmetrical and shows no outliers suggests that this condition is satisfied.

13.32 a Standardized Residual 1.5 1.0 0.5 0.0 -0.5 -1.0 -1.5 -2.0 20

28 30 Firing Frequency

There are no particularly unusual features in the standardized residual plot. The only slightly unusual feature is the point whose standardized residual is −1.83, which is relatively far from zero, but not particularly extreme. The plot supports the assumption that the simple linear regression model applies. b

Yes. Since the normal probability plot shows a roughly linear pattern we can conclude that is it reasonable to assume that the error distribution is approximately normal.

13.33 The standardized residual plot is shown below. Standardized Residual 1.5 1.0 0.5 0.0 -0.5 -1.0 -1.5 -2.0 0

20 Sunflower Meal (%)

There is clear indication of a curve in the standardized residual plot, suggesting that the simple linear regression model is not adequate.

Chapter 13: Simple Linear Regression and Correlation: Inferential Methods 13.34 a

Letting x = minimum width and y = maximum width, the least-squares regression line is yˆ  0.939  0.873x. The residuals and the standardized residuals are shown in the table below. Product 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27

Minimum Width 1.8 2.7 2 2.6 3.15 1.8 1.5 3.8 5 4.75 2.8 2.1 2.2 2.6 2.6 2.9 5.1 10.2 3.5 1.2 1.7 1.75 1.7 1.2 1.2 7.5 4.25

Maximum Width 2.5 2.9 2.15 2.9 3.2 2 1.6 4.8 5.9 5.8 2.9 2.45 2.6 2.6 2.7 3.1 5.1 10.2 3.5 2.7 3 2.7 2.5 2.4 4.4 7.5 4.25

The standardized residual plot is shown below.

Residual Standardized Residual -0.011 -0.016 -0.397 -0.601 -0.535 -0.816 -0.309 -0.469 -0.489 -0.742 -0.511 -0.780 -0.649 -0.996 0.543 0.825 0.595 0.918 0.714 1.096 -0.484 -0.734 -0.323 -0.491 -0.260 -0.396 -0.609 -0.924 -0.509 -0.773 -0.371 -0.563 -0.292 -0.451 0.355 0.748 -0.495 -0.751 0.713 1.100 0.577 0.882 0.233 0.356 0.077 0.117 0.413 0.637 2.413 3.721 0.013 0.021 -0.400 -0.610

375

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Standardized Residual 4

-1 0

6 Minimum Width

The standardized residual plot shows that there is one point that is a clear outlier (the point whose standardized residual is 3.721). This is the point for product 25. c

The equation of the least-squares regression line is now yˆ  0.703  0.918 x. A computer analysis gives sb  0.065 . Thus the change in slope from 0.873 to 0.918 expressed in standard deviations is (0.918  0.873) 0.065  0.692. Removal of the point resulted in a reasonably substantial change in the equation of the estimated regression line.

For every 1-cm increase in minimum width, the mean maximum width is estimated to increase by 0.918 cm. The intercept would be an estimate of the mean maximum width when the minimum width is zero. It is clearly impossible to have a container whose minimum width is zero.

e Standardized Residual 2.0 1.5 1.0 0.5 0.0 -0.5 -1.0 0

6 Minimum Width

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377

The standardized residual plot for the data with the Coke bottle removed is shown above. The pattern in this plot suggests that the variances of the y distributions decrease as x increases, and therefore that the assumption of constant variance is not valid. 13.35 There is no clear curve in the standardized residual plot, initially suggesting that the assumptions of the simple linear regression model might be satisfied. Furthermore, there is no point with an extreme x value (a point with an extreme x value would be potentially influential, and therefore undesirable), and there is no one point particularly far-removed from the other points in the standardized residual plot. However, there does seem to be a tendency for points with more central x values to have larger standardized residuals (positive or negative) than points with larger or smaller x values, and there is one point with a standardized residual quite a bit larger than 2. These two facts suggest that the assumptions of the simple linear regression model might not be satisfied. 13.36 a Residual 150

100

-50

-100 10

15 20 Traffic flow (thousands of cars per 24 hrs)

There is the smallest hint of a curve in the residual plot, but this is not strong enough to suggest that the simple linear regression model is inappropriate.

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b Standardized residual 1.5 1.0 0.5 0.0

-0.5 -1.0

15 20 Traffic flow (thousands of cars per 24 hrs)

This plot is very similar in appearance to the one in Part (a). Online Exercises 13.37 Suppose we constructed a 95% confidence interval for the mean value of y when x = x*. We would then be 95% confident that the mean value of y was within that interval. If we were to construct the 95% prediction interval at x = x* we would be 95% confident that an observed y value, y*, at that value of x will be within the interval. The 95% confidence level for the prediction interval is interpreted as follows. The prediction interval is constructed using a set of independent y values for a given set of x values. Imagine this being done a large number of times, with the prediction interval at x  x* being calculated for each set of ( x , y ) points. Imagine, also, a large number of y values being selected at x = x*. Then if one interval is chosen at random, and one y value is chosen at random, on average 95 times out of 100 the y value will be within the interval. 13.38 If a confidence interval for  is required, you are asked for a confidence interval either for the slope of the regression line or for the mean increase in y associated with a one-unit increase in x. If a confidence interval for    x* is required, you are given a value of x and asked to find a confidence interval for the mean value of y at that value of x.

13.39 a

We use sa bx*  se

1 ( x *  x )2 .  n S xx

Here sa  b(2.0)  16.486

1 (2  2.5) 2   4.038. 20 25

Since 3 is the same distance from 2.5 as is 2, sa b(3.0)  sa b(2.0)  4.038.

sa  b(2.8)  16.486

1 (2.8  2.5)2   3.817. 20 25

Chapter 13: Simple Linear Regression and Correlation: Inferential Methods

379

The estimated standard deviation of a  bx * is smallest for x*  2.5, since the distance of this value from the mean value of x is zero.

13.40 The scatterplot given in the example shows a linear pattern that is consistent with the assumptions of the simple linear regression model. The point estimate of    (0.6) is a  b(0.6)  0.6678  1.3957(0.6)  0.16962.

  x   3.1804  (5.82)  0.3577. x  2

S xx 



n 12 The estimated standard deviation of a  b(0.6) is

1 (0.6  x )2 1 (0.6  5.82 12) 2   0.285   0.034287. n S xx 12 0.3577 The critical value of the t distribution with 10 degrees of freedom for a 95% confidence interval is 2.228. So the required confidence interval is 0.16962  2.228(0.034287)  ( 0.093, 0.246). We are 95% confident that the mean vote-difference proportion for congressional races where 60% judge candidate A as more competent is between 0.093 and 0.246. se

13.41 a

The equation of the estimated regression line is yˆ  0.001790  0.0021007 x, where x = mean childhood blood lead level and y = brain volume change.

We need to assume that the conditions for inference are met. The point estimate of    (20) is a  b(20)  0.001790  0.0021007(20)  0.043804. The estimated standard deviation of a  b(20) is

1 (20  x )2 1 (20  11.5)2   0.031   0.0069979. n S xx 100 1764

The critical value of the t distribution with 98 degrees of freedom for a 90% confidence interval is 1.661. So the required confidence interval is 0.043804  1.661(0.0069979)  ( 0.055,  0.032). We are 90% confident that the mean brain volume change for people with a childhood blood lead level of 20 μg/dL is between −0.055 and −0.032. c

The estimated standard deviation of the amount by which a single y observation deviates from the value predicted by an estimated regression line is

se2  sa2bx*  0.0312  0.00699792  0.03178. The critical value of the t distribution with 98 degrees of freedom for a 90% confidence interval is 1.661. So the required confidence interval is 0.043804  1.661(0.03178)  ( 0.097, 0.009). We are 90% confident that the brain volume change for a person with a childhood blood lead level of 20 μg/dL will be between −0.097 and 0.009. d

The answer to Part (b) gives an interval in which we are 90% confident that the mean brain volume change for a person with a childhood blood lead level of 20 μg/dL lies. The answer to Part (c) states that if we were to find the brain volume change for one person with a childhood blood lead level of 20 μg/dL, we are 90% confident that this value will lie within

380

Chapter 13: Simple Linear Regression and Correlation: Inferential Methods the interval found.

13.42 a

The required proportion is r 2  0.630.

Yes. The P-value of 0.000 indicates that the data provide convincing evidence that the slope of the population regression line is not equal to zero.

We need to assume that the conditions for inference are met. The point estimate of    (200) is a  b(200)  4.7896  0.014388(200)  7.6672. We are given that sa  b(200)  0.347. The critical value of the t distribution with 15 degrees of freedom for a 95% confidence interval is 2.131. So the required confidence interval is 7.6672  2.131(0.347)  (6.928, 8.407). We are 95% confident that the mean time necessary when the depth is 200 feet is between 6.928 and 8.407 minutes.

We need to assume that the conditions for inference are met. The point estimate of    (200) is a  b(200)  4.7896  0.014388(200)  7.6672. We are given that sa  b(200)  0.347. Thus, the estimated standard deviation of the amount by which a single y observation deviates from the value predicted by an estimated regression line is se2  sa2bx*  1.4322  0.3472  1.47344. The critical value of the t distribution with 15 degrees of freedom for a 95% confidence interval is 2.131. So the required confidence interval is 7.6672  2.131(1.47344)  (4.527, 10.808). We are 95% confident that the time necessary when the depth is 200 feet will be between 4.527 and 10.808 minutes.

The critical value of the t distribution with 15 degrees of freedom for a 95% confidence interval is 2.131. The critical value of the t distribution with 15 degrees of freedom for a 99% confidence interval is 2.947. Thus the required interval will have the same center as the given interval and will have a width that is 2.947 2.131 times the width of the given interval. Therefore, the required interval will have center (8.147  10.065) 2  9.106 and the distance from this center to the top and the bottom of the required interval will be  2.947 2.13110.065  9.106  1.326. So the required interval is 9.106  1.326  (7.780,10.432)

13.43 a

The equation of the regression line is yˆ  133.02  5.919 x, where x = snout vent length and y = clutch size.

sb  1.127.

Yes. Since the estimated slope is positive and since the P-value is small (given as 0.000 in the output) we have convincing evidence that the slope of the population regression line is positive.

Chapter 13: Simple Linear Regression and Correlation: Inferential Methods d

381

We need to assume that the conditions for inference are met. The point estimate of    (65) is a  b(65)  133.02  5.919(65)  251.715.

S xx 

 x  nx  45958  14(56.5)  1266.5. 2

The estimated standard deviation of a  b(65) is

1 (65  x )2 1 (65  56.5)2   33.90   12.151. n S xx 14 1266.5 Therefore, the estimated standard deviation of the amount by which a single y observation deviates from the value predicted by an estimated regression line is se

se2  sa2bx*  33.902  12.1512  36.012. The critical value of the t distribution with 12 degrees of freedom for a 95% confidence interval is 2.179. So the required confidence interval is 251.715  2.179(36.012)  (173.252, 330.178). We are 95% confident that the clutch size for a salamander whose snout-vent length is 65 will be between 173.252 and 330.178. e

It would not be appropriate to use the estimated regression line to predict the clutch size for a salamander with a snout-vent length of 105, since 105 is a long way outside the range of the x values in the original data set.

13.44 a

1.  = slope of the population regression line relating maximum width to minimum width. 2. H0:   0 3. Ha:   0 4.   0.05 b  (hypothesized value) b  0  5. t  sb sb 6. The standardized residual plot given in the solution to the earlier exercise shows one point with a large standardized residual. We will nonetheless assume that the conditions for the simple linear regression model apply, and proceed with the hypothesis test. 7. b  0.87310, se  0.67246, S xx  108.64963, sb  0.06451 0.87310  0 t  13.534 0.06451 8. df = 25 P-value  P(t25  13.534)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence of a positive linear relationship between minimum width and maximum width.

se  0.672. This is a typical deviation of a maximum width in the sample from the value predicted by the least-squares regression line.

We proceed with calculation of the confidence interval even though the standardized residual plot shows that there is a point with a large standardized residual. The point estimate of    (6) is a  b(6)  0.939  0.873(6)  6.178. The estimated standard deviation of a  b(6) is

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Chapter 13: Simple Linear Regression and Correlation: Inferential Methods

1 (6  x )2 1 (6  3.096)2   0.672   0.228. n S xx 27 108.650 The critical value of the t distribution with 25 degrees of freedom for a 95% confidence interval is 2.060. So the required confidence interval is 6.178  2.060(0.228)  (5.709, 6.647). We are 95% confident that the mean maximum width when the minimum width is 6 cm is between 5.709 and 6.647. (Note that some values in this interval are not actually possible for this mean, since, when the minimum width is 6 cm, the maximum width can’t be less than 6 cm.) se

We proceed with calculation of the confidence interval even though the standardized residual plot shows that there is a point with a large standardized residual. As calculated in Part (c), the point estimate of    (6) is 6.178, and the estimated standard deviation of a  b(6) is 0.228. Therefore, the estimated standard deviation of the amount by which a single y observation deviates from the value predicted by an estimated regression line is

se2  sa2bx*  0.6722  0.2282  0.710. The critical value of the t distribution with 25 degrees of freedom for a 95% confidence interval is 2.060. So the required confidence interval is 6.178  2.060(0.710)  (4.716, 7.640). We are 95% confident that, when the minimum width is 6 cm, the maximum width will be between 4.716 and 7.640. (Note that some values in this interval are not actually possible for this predicted value, since, when the minimum width is 6 cm, the maximum width can’t be less than 6 cm.) 13.45 a

The equation of the estimated regression line is yˆ  2.78551  0.04462 x, where x = time on shelf and y = moisture content.

 = slope of the population regression line relating moisture content to shelf time. H0:   0 Ha:   0   0.05 b  (hypothesized value) b  0 t  sb sb A standardized residual plot is shown below.

Chapter 13: Simple Linear Regression and Correlation: Inferential Methods

Standardized Residual 2

-1

-2

-3 0

20 Shelf Time (days)

Apart from one outlier, the standardized residual plot shows a random pattern that is consistent with the simple regression model.

  x   7445  (269)  2276.357. x  2

S xx 



se  0.196246 s sb  e  0.00411 S xx 0.04462  0  10.848 0.00411 df = 12 P-value  2  P(t12  10.848)  0 Since P-value  0  0.05 we reject H0. We have convincing evidence that the simple regression model provides useful information for predicting moisture content from knowledge of shelf time. t

The conditions for inference were checked in Part (b). The point estimate of    (30) is a  b(30)  2.78551  0.04462(30)  4.124.

S xx 

 x  nx  7745  14(269 14)  2576.357. 2

se  0.196246. The estimated standard deviation of a  b(30) is

1 (30  x )2 1 (30  19.214286) 2   0.196246   0.067006. n S xx 14 2576.357

Therefore, the estimated standard deviation of the amount by which a single y observation deviates from the value predicted by an estimated regression line is

se2  sa2bx*  0.1962462  0.0670062  0.207. The critical value of the t distribution with 12 degrees of freedom for a 95% confidence interval is 2.179. So the required confidence interval is 4.124  2.179(0.207)  ( 3.672, 4.576).

383

384

Chapter 13: Simple Linear Regression and Correlation: Inferential Methods We are 95% confident that the moisture content for a box of cereal that has been on the shelf for 30 days will be between 3.672 and 4.576 percent. d

Since 4.1 is included in the confidence interval constructed in Part (c), a moisture content exceeding 4.1 percent is quite plausible when the shelf time is 30 days.

13.46 Since 17 is further from the mean than is 20, both the confidence interval and the prediction interval will be wider for x*  17 than for x*  20. 13.47 The data are displayed in the scatterplot below. Hours of chiller operation per day 17.5

15.0

12.5

10.0

7.5

5.0 70

80 85 Maximum outdoor temperature

The scatterplot shows a linear pattern that is consistent with the assumptions of the simple linear regression model. n  6, x  82.667, S xx  269.333 a  46.410, b  0.702, a  b(82)  11.132 se  0.757

sa b(82)  se

t

1 (82  x )2 1 (82  82.667)2   0.757   0.311 n S xx 6 269.333

a  b(82)  12 11.132  12   2.794 sa b(82) 0.311

P-value  P(t4  2.794)  0.025 Since this P-value is greater than 0.01, we do not reject H0. We do not have convincing evidence at the 0.01 level that, when the maximum output is 82°F, the mean number of hours of chiller operation is less than 12. 13.48 The statistic r is the correlation coefficient for a sample, while  denotes the correlation coefficient for the population. 13.49 The first statement is not correct. It is theoretically possible, by chance, to obtain a sample correlation coefficient of 1 when the population correlation coefficient is not 1. This occurs when the points representing the sample values just happen to lie in a perfect straight line. (Also, in the

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385

special case when the sample size is 2, the value of r will necessarily be 1, whatever the value of  .) The second statement is correct. If the population correlation coefficient is 1, then all the points in the population lie on a perfect straight line. Thus, when a sample is taken from that population, all the points in the sample will lie on that same straight line. As a result, the sample correlation coefficient is equal to 1. 13.50 1. 2. 3. 4. 5.

6. 7.

8. 9.

 = the correlation between teaching evaluation index and annual raise for the population from which the sample was selected. H0:   0 Ha:   0   0.05 r t 1  r2 n2 We must assume that the variables have a bivariate normal distribution and that the sample was a random sample from the population. 0.11 t  2.073 1  0.112 351 df = 351 P-value  2  P(t351  2.073)  0.039 Since P-value  0.039  0.05 we reject H0. We have convincing evidence of a linear association between teaching evaluation index and annual raise.

This result might be initially surprising, since 0.11 seems to be a relatively small value for the sample correlation coefficient. However, what the result shows is that for a sample size as large as 353, a sample correlation as large as 0.11 would be very unlikely if the population correlation were zero. 13.51 1. 2. 3. 4. 5.

6. 7.

 = the correlation between annual dollar rent per square foot and annual dollar sales per square foot. H0:   0 Ha:   0   0.05 r t 1  r2 n2 We must assume that the variables have a bivariate normal distribution. We are told that the sample was randomly selected. 0.37 t  2.844 1  0.372 51 df = 51 P-value  P(t51  2.844)  0.003

386

Chapter 13: Simple Linear Regression and Correlation: Inferential Methods 9. Since P-value  0.003  0.05 we reject H0. We have convincing evidence of a positive linear association between annual dollar rent per square foot and annual dollar sales per square foot for the population of all such stores.

13.52 a

1. 2. 3. 4. 5.

6. 7.

8. 9.

13.53 a

 = the correlation between time spent watching television and grade point average for the population from which the sample was selected. H0:   0 Ha:   0   0.01 r t 1  r2 n2 We must assume that the variables have a bivariate normal distribution. We are told that the sample was a random sample. 0.26 t  6.175 1  ( 0.26)2 526 df = 526 P-value  P(t526  6.175)  0 Since P-value  0  0.01 we reject H0. We have convincing evidence of a negative correlation between time spent watching television and grade point average.

Since r 2  ( 0.26)2  0.0676, only 6.76% of the observed variation in grade point average would be explained by the regression line. This is not a substantial percentage. We first calculate the value of r using r  This formula can be proved as follows:  x  x  y  y     zx z y  sx   s y  r   n 1 n 1

 



S xy S xx S yy

( x  x )( y  y )  (n  1) sx s y

According to the summary statistics given we have



  x   56700  (860)  7393.333 x 

S yy 



y2 

  y   8954  (348)  880.4

S xy 



xy 

    y   22265  (860)(348)  2313

S xx 

So r 

S xy S xx S yy

n x



2313  0.907. (7393.333)(880.4)

S xy S yy S xx (n  1) n 1 n 1



S xy S xx S yy

Chapter 13: Simple Linear Regression and Correlation: Inferential Methods 1. 2. 3. 4. 5.

6. 7.

8. 9.

387

 = the correlation between particulate pollution and luminance for the population from which the sample was selected. H0:   0 Ha:   0   0.05 r t 1  r2 n2 We must assume that the variables have a bivariate normal distribution. We are told that the sample was representative of the population, so we can treat it as a random sample. 0.907 t  7.746 1  0.9072 13 df = 13 P-value  P(t13  7.746)  0 Since P-value  0  0.05 we reject H0. We have convincing evidence of a positive correlation between particulate pollution and luminance for the population from which the sample was selected.

The required proportion is r 2  0.9072  0.822.

13.54 1.  = the correlation between surface and subsurface concentration. 2. H0:   0 3. Ha:   0 4.   0.05 r 5. t  1  r2 n2 6. We must assume that the sample was a random sample from the population under consideration.

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Surface 50

0 -1.5

-1.0

-0.5

0.0 Normal Score

0.5

1.0

1.5

-1.0

-0.5

0.0 Normal Score

0.5

1.0

1.5

Subsurface 7

3 -1.5

The curved pattern in the first normal probability plot tells us that it is unlikely that the variables have a bivariate normal distribution, but we will nevertheless proceed with the hypothesis test. 7. r  0.574 0.574 t  1.855 1  0.5742 7 8. df = 7 P-value  2  P(t7  1.855)  0.106 9. Since P-value  0.106  0.05 we do not reject H0. We do not have convincing evidence of a linear relationship between surface and subsurface concentration. 13.55 1.  = population correlation 2. H0:   0 3. Ha:   0

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389

  0.05 r t 1  r2 n2

6. We must assume that the variables have a bivariate normal distribution and that the sample was randomly selected from the population. 0.022 7. t   2.200 1  0.0222 9998 8. df = 9998 P-value  2  P(t9998  2.200)  0.028. 9. Since P-value  0.028  0.05 we reject H0. We have convincing evidence that   0 . This tells us that, although the sample correlation coefficient was apparently close to 0, the sample is large enough for the result to convince us nonetheless that the population correlation coefficient is not zero. 13.56 a

The slope of the estimated regression line for y = verbal language score against x = height gain from age 11 to 16 is 2.0. This tells us that for each extra inch of height gain the average verbal language score at age 11 increased by 2.0 percentage points. The equivalent results for nonverbal language scores and math scores were 2.3 and 3.0. Thus the reported slopes are consistent with the statement that each extra inch of height gain was associated with an increase in test scores of between 2 and 3 percentage points.

The slope of the estimated regression line for y = verbal language score against x = height gain from age 16 to 33 is −3.1. This tells us that for each extra inch of height gain the average verbal language score at age 11 decreased by 3.1 percentage points. The equivalent results for nonverbal language scores and math scores were both −3.8. Thus the reported slopes are consistent with the statement that each extra inch of height gain was associated with a decrease in test scores of between 3.1 and 3.8 percentage points.

Between the ages of 11 and 16 the first boy grew 5 inches more than the second boy. So the first boy’s age 11 math score is predicted to be 5  3  15 percentage points higher than that of the second boy. Between the ages of 16 and 33 the second boy grew 5 inches more than the first boy. According to this information the first boy’s age 11 math score is predicted to be 5  3.8  19 percentage points higher than that of the second boy. These two results are consistent with the conclusion that on the whole boys who did their growing early had higher cognitive scores at age 11 than those whose growth occurred later.

13.57 a

For a 1-day increase in elapsed time, the average biomass concentration is estimated to decrease by 0.640 g/cm3.

When x  40, yˆ  106.3  0.640(40)  80.7 g/cm3 .

 = slope of the population regression line relating biomass concentration to elapsed

time. 2. H0:   0

390

Chapter 13: Simple Linear Regression and Correlation: Inferential Methods 3. Ha:   0 4.   0.05 b  (hypothesized value) b  0 5. t   sb sb 6. We need to assume that the assumptions of the simple linear regression model apply. 7. Since the slope of the estimated regression line is negative, r   0.470  0.686. b0 r As explained at the end of Section 13.5, t   . sb 1  r2

n2 So here, t 

0.686  7.047 1  0.470 56

8. df = 56 P-value  2  P(t56  7.047)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence of a useful linear relationship between elapsed time and biomass concentration. 13.58 a

t



0.18

 3.399. 1 r 1  ( 0.18) 2 n2 345 Thus, for a two-tailed test, the P-value is 2  P(t345  3.399)  0.001. Since the P-value for a one-tailed test would be a half of this, it is indeed correct, whether this be a one- or two-tailed test, that P-value < 0.05. 2

Yes. One would expect, generally speaking, that those with greater coping humor ratings would have smaller depression ratings.

No. Since r 2  ( 0.18)2  0.0324, we know that only 3.2% of the variation in depression scale values is attributable to the approximate linear relationship with the coping humor scale. So the linear regression model will generally not give accurate predictions.

13.59 a

1. 2. 3. 4. 5.

 = the correlation between soil hardness and trail length for the population of penguin burrows. H0:   0 Ha:   0   0.05 r t 1  r2 n2 We must assume that the variables have a bivariate normal distribution and that the sample was a random sample of penguin burrows.

Chapter 13: Simple Linear Regression and Correlation: Inferential Methods

r   0.386  0.621. (We know that r  0 since the slope of the least-squares line is negative.) 0.621 t  6.090 1  ( 0.621)2 59 8. df = 59 P-value  P(t59  6.090)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence of a negative correlation between soil hardness and trail length. 7.

We need to assume that the conditions for inference are met. The point estimate of    (6.0) is a  b(6.0)  11.607  1.4187(6.0)  3.0948. The estimated standard deviation of a  b(6.0) is

1 (6.0  x )2 1 (6.0  4.5)2   2.35   0.374. n S xx 61 250 Therefore, the estimated standard deviation of the amount by which a single y observation deviates from the value predicted by an estimated regression line is se

se2  sa2bx*  2.352  0.3742  2.380. The critical value of the t distribution with 59 degrees of freedom for a 95% confidence interval is 2.001. So the required prediction interval is 3.0948  2.001(2.380)  ( 1.667, 7.856). We are 95% confident that the trail length when the soil hardness is 6.0 will be between −1.667 and 7.856. c

No. For x  10 the least-squares line predicts y  2.58. Since it is not possible to have a negative trail length, it is clear that the simple linear regression model does not apply at x  10. So the simple linear regression model is not suitable for this prediction.

13.60 a Peak photovoltage 1.4 1.2 1.0 0.8 0.6 0.4 0.2 0.0 5

15 20 Percentage of light absorption

The scatterplot suggests that the simple linear regression model might be appropriate.

391

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Chapter 13: Simple Linear Regression and Correlation: Inferential Methods

The equation of the estimated regression line is yˆ  0.0826  0.0446 x, where x = percentage of light absorption and y = peak photovoltage.

The required proportion is r 2  0.983.

When x  19.1, yˆ  0.0826  0.0446(19.1)  0.770. The peak photovoltage is predicted to be 0.770. The residual is y  yˆ  0.68  0.770  0.090.

1. 2. 3. 4. 5. 6. 7.

8. 9.

 = slope of the population regression line relating peak photovoltage to percentage of light absorption. H0:   0 Ha:   0   0.05 b  (hypothesized value) b  0 t  sb sb We are told in Part (b) to assume that the simple linear regression model is appropriate. b  0.0446, se  0.0611, S xx  746.4, sb  0.00224 0.0446  0 t  19.959 0.00224 df = 7 P-value  2  P(t7  19.959)  0 Since P-value  0  0.05 we reject H0. We have convincing evidence of a useful linear relationship between percent light absorption and peak photovoltage.

We are told in Part (b) to assume that the simple linear regression model is appropriate. df = 7. The 95% confidence interval for  is b  (t critical value)  sb  0.0446  (2.365)(0.00224)  (0.039, 0.050). We are 95% confident that the mean increase in peak photovoltage associated with a 1percentage point increase in light absorption is between 0.039 and 0.050.

We are told in Part (b) to assume that the simple linear regression model is appropriate. The point estimate of    (20) is a  b(20)  0.08259  0.04465(20)  0.810. The estimated standard deviation of a  b(20) is

1 (20  x )2 1 (20  19.967) 2   0.06117   0.0204. n S xx 9 746.4

The critical value of the t distribution with 7 degrees of freedom for a 95% confidence interval is 2.365. So the required confidence interval is 0.810  2.365(0.0204)  (0.762, 0.859). We are 95% confident that the mean peak photovoltage when the percentage of light absorption is 20 is between 0.762 and 0.859. 13.61 a

 = slope of the population regression line relating x = age and y = percentage of the

cribriform area of the lamina scleralis occupied by pores. 2. H0:   0.5 3. Ha:   0.5

Chapter 13: Simple Linear Regression and Correlation: Inferential Methods 4. 5.

393

  0.1 b  (hypothesized value) b  ( 0.5) t  sb sb

6. The data are plotted in the scatterplot below. y 80

30 20

50 x

sb 

se 6.75598   0.1096 S xx 3797.529

0.447  ( 0.5)  0.488 0.1096 8. df = 15 P-value  2  P(t15  0.488)  0.633 9. Since P-value  0.633  0.1 we do not reject H0. We do not have convincing evidence that the average decrease in percentage area associated with a 1-year age increase is not 0.5. t

As shown in the solution to Part (a), the scatterplot shows a linear pattern that is consistent with the assumptions of the simple linear regression model. The point estimate of    (50) is a  b(50)  72.918  0.447(50)  50.591. The estimated standard deviation of a  b(50) is

1 (50  x )2 1 (50  48.294)2   6.75598   1.649. n S xx 17 3797.529

The critical value of the t distribution with 15 degrees of freedom for a 95% confidence interval is 2.131.

394

Chapter 13: Simple Linear Regression and Correlation: Inferential Methods So the required confidence interval is 50.591  2.131(1.649)  (47.076, 54.106). We are 95% confident that the mean percentage area at age 50 is between 47.076 and 54.106.

13.62 a

The estimated regression line is yˆ  1.752  0.685x, where x = actual air temperature and y = temperature as measured by a satellite.

sa  se

1 (0  x )2 1 (0  2.5)2   0.804   0.337. n S xx 10 82.5

H0:   0 Ha:   0 We need to assume that the assumptions of the simple linear regression model apply. a  0 1.752 t   5.199 sa 0.337

P-value  2  P(t8  5.199)  0.001 Since P-value  0.001  0.05 we reject H0. We have convincing evidence that the y intercept of the population regression line differs from zero. c

We need to assume that the assumptions of the linear regression model apply. The 95% confidence interval for  is a  (t critical value)sa  1.752  (2.306)(0.337)  ( 2.529,  0.975). We are 95% confident that the y intercept of the population regression line lies between −2.529 and −0.975. The fact that zero does not lie within this interval tells us, exactly as concluded in Part (b), that we have convincing evidence at the 0.05 level that the y intercept of the population regression differs from zero.

13.63 For leptodactylus: SSResid  0.30989 Sample size = 9 b  0.31636 S xx  42.82 For bufa: SSResid  0.12792 Sample size = 8 b  0.35978 S xx  34.54875

s2 

0.30989  0.12792  0.03368 76

H 0:     Ha:       0.05 b  b t  s2 s2   S xx S xx

0.31636  0.35978  1.03457 0.03368 0.03368  42.82 34.54875

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df = 13 P-value  2  P(t13  1.03457)  0.320 Since P-value  0.319  0.05 we do not reject H0. We do not have convincing evidence that the slopes of the population regression lines for the two different frog populations are not equal. 13.64 y1 11 10 9 8 7 6 5 4 5.0

7.5

10.0

12.5

15.0

The scatterplot of the first data set (above) shows no apparent curve, no extreme residuals, no clear change in the variance of the y values as the x values change, and no influential points. Therefore the simple linear regression model seems appropriate. y2 10 9 8 7 6 5 4 3 5.0

7.5

10.0

12.5

15.0

The plot for the second data set (above) shows a clear curve, and so it seems that the assumptions of the simple linear regression model do not apply.

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Chapter 13: Simple Linear Regression and Correlation: Inferential Methods

y3 13 12 11 10 9 8 7 6 5 5.0

7.5

10.0

12.5

15.0

The scatterplot for the third data set (above) shows one point that has a very large residual, and therefore it seems that the assumptions of the simple linear regression model do not apply. y4 13 12 11 10 9 8 7 6 5 8

14 x4

The scatterplot for the fourth data set (above) shows one point that is clearly influential, and therefore it seems that the assumptions of the simple linear regression model do not apply. 13.65 If the point (20, 33000) is not included, then the slope of the least-squares line would be relatively small and negative (appearing close to horizontal when drawn to the scales of the scatterplot given in the question). If the point is included then the slope of the least-squares line would still be negative, but much further from zero. 13.66 a

The equation of the estimated regression line is yˆ  57.964  0.0357 x, where x = fermentation time and y = glucose concentration. 1.

 = slope of the population regression line relating glucose concentration to fermentation time.

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2. H0:   0 3. Ha:   0 4.   0.1 b  (hypothesized value) b  0 5. t   sb sb 6. We need to assume that the assumptions of the simple linear regression model apply. 7. b  0.03571, se  9.8896, S xx  42, sb  1.5260 0.03571  0 t  0.0234 1.5260 8. df = 6 P-value  2  P(t6  0.0234)  0.982 9. Since P-value  0.982  0.1 we do not reject H0. We do not have convincing evidence of a useful linear relationship between fermentation time and glucose concentration. c

The residuals are shown in the table below. x 1 2 3 4 5 6 7 8

y 74 54 52 51 52 53 58 71

Residual 16.000 -4.036 -6.071 -7.107 -6.143 -5.179 -0.214 12.750

The residual plot is shown below. Residual 15

-5

-10 0

4 5 Fermentation time

The residual plot shows a clear curve, and so the simple linear regression model seems not to be appropriate.

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Chapter 13: Simple Linear Regression and Correlation: Inferential Methods

13.67 The small P-value indicates that there is convincing evidence of a useful linear relationship between percentage raise and productivity. 13.68 The points appear to be varying further from the regression line as the values of x increase, and there are some points with comparatively very large residuals, and so it seems that the simple linear regression model is not appropriate. A plot of the standardized residuals will show points close to the horizontal axis for small values of x, and points spreading further from the horizontal axis for larger values of x. There will be three points with large magnitudes of standardized residuals. 13.69 a

The values e1 , , en are the vertical deviations of the y observations from the population regression line. The residuals are the vertical deviations from the sample regression line.

False. The simple linear regression model states that the mean value of y is equal to    x.

No. You only test hypotheses about population characteristics; b is a sample statistic.

Strictly speaking this statement is false, since a set of points lying exactly on a straight line will give a zero result for SSResid. However, it is certainly true to say that, since SSResid is a sum of squares, its value must be nonnegative.

This is not possible, since the sum of the residuals is always zero.

This is not possible, since SSResid (here said to be equal to 731) is always less than or equal to SSTo (here said to be 615).

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399

Cumulative Review Exercises CR13.1 Randomly assign the 400 students to two groups of equal size, Group A and Group B. (This can be done by writing the names of the students onto slips of paper, placing the slips into a hat, and picking 200 at random. These 200 people will go into Group A, and the remaining 200 people will go into Group B.) Have the 400 students take the same course, attending the same lectures and being given the same homework assignments. The only difference between the two groups should be that the students in Group A should be given daily quizzes and the students in Group B should not. (This could be done by having the students in Group A take their quizzes in class after the students in Group B have been dismissed.) After the final exam the exam scores for the students in Group A should be compared to the exam scores for the students in Group B. CR13.2 a

The two samples are paired, since the two sets of blood cholesterol measurements were conducted on the same set of subjects.

1. 2. 3. 4. 5.

d = mean difference in total cholesterol (regular diet − pistachio diet) H0: d  0 Ha: d  0   0.01 x  hypothesized value t d sd n

6. We are told to assume that the sample is representative of adults with high cholesterol, and therefore we are justified in treating it as a random sample from that population. We are also told to assume that total cholesterol differences are approximately normally distributed. Therefore we can proceed with the paired t test. 11  0  1.775 7. t  24 15 8. df = 14 P-value  P(t14  1.775)  0.049 9. Since P-value  0.049  0.01 we do not reject H0. We do not have convincing evidence that eating the pistachio diet for four weeks is effective in reducing total cholesterol level. CR13.3 a

Median = 2 Lower quartile = 1.5 Upper quartile = 6.5 IQR = 6.5 − 1.5 = 5

20 Number of Fines

Two of the observations, 23 and 36, are (extreme) outliers.

400

Chapter 13: Simple Linear Regression and Correlation: Inferential Methods b

The two airlines with the highest numbers of fines assessed may not be the worst in terms of maintenance violations since these airlines might have more flights than the other airlines.

We would expect 400(0.05) = 20 to be steroid users and 400(0.95) = 380 to be non-users.

We would expect 20(0.95) = 19 of the steroid users to test positive.

We would expect 380(0.05) = 19 of the non-users to test positive.

The proportion of those who test positive who are users is

The players who test positive consist of 95% of the 5% of players who use steroids and 5% of the 95% who do not. Thus one-half (not 0.95) of those who test positive use steroids.

Check of Conditions 1. Since npˆ  1003(0.68)  682  10 and n(1  pˆ )  1003(0.32)  321  10, the sample size is large enough. 2. The sample size of n = 1003 is much smaller than 10% of the population size (the number of adult Americans). 3. We are told that the survey was nationally representative, so it is reasonable to regard the sample as a random sample from the population of adult Americans. Calculation The 95% confidence interval for p is pˆ (1  pˆ ) (0.68)(0.32) pˆ  1.96  0.68  1.96  (0.651, 0.709). n 1003 Interpretation We are 95% confident that the proportion of all adult Americans who view a landline phone as a necessity is between 0.651 and 0.709.

1. p = proportion of all adult Americans who considered a TV set a necessity 2. H0: p = 0.5 3. Ha: p > 0.5 4.   0.05 pˆ  p pˆ  0.5  5. z  p(1  p ) (0.5)(0.5) n 1003 6. The sample was nationally representative, so it is reasonable to treat the sample as a random sample from the population. The sample size is much smaller than the population size (the number of adult Americans). Furthermore, np  1003(0.5)  501.5  10 and n(1  p )  1003(0.5)  501.5  10 , so the sample is large enough. Therefore the large sample test is appropriate. 0.52  0.5  1.267 7. z  (0.5)(0.5) 1003 8. P -value  P ( Z  1.267)  0.103

CR13.4

19 19   0.5. 19  19 38

CR13.5

Chapter 13: Simple Linear Regression and Correlation: Inferential Methods

401

9. Since P-value  0.103  0.05 we do not reject H0. We do not have convincing evidence that a majority of adult Americans consider a TV set a necessity. c

2. 3. 4. 5.

p1 = proportion of adult Americans in 2003 who regarded a microwave oven as a necessity p2 = proportion of adult Americans in 2009 who regarded a microwave oven as a necessity H0: p1  p2  0 Ha: p1  p2  0   0.01 pˆ1  pˆ 2 z pˆ c (1  pˆ c ) pˆ c (1  pˆ c )  n1 n2

6. We are told that the 2009 survey was nationally representative, so it is reasonable to treat the sample in 2009 as a random sample from the population. We need to assume that the sample in 2003 was a random sample from the population. Also, n1 pˆ1  1003(0.68)  682  10, n1 (1  pˆ1 )  1003(0.32)  321  10, n2 pˆ 2  1003(0.47)  471  10, and n2 (1  pˆ 2 )  1003(0.53)  532  10, so the samples are large enough. n pˆ  n pˆ 1003(0.68)  1003(0.47)  0.575 7. pˆ c  1 1 2 2  n1  n2 1003  1003

0.68  0.47  9.513 (0.575)(0.425) (0.575)(0.425)  1003 1003 8. P -value  P ( Z  9.513)  0 9. Since P-value  0  0.01 we reject H0. We have convincing evidence that the proportion of adult Americans who regarded a microwave oven as a necessity decreased between 2003 and 2009. z

CR13.6 The angles are distorted by the fact that the somewhat circular shape is viewed at a slant. Nonetheless, the display does an acceptable job of representing the proportions in this study. CR13.7 a

P ( x  0)  1  0.38  0.62.

P(2  x  5)  0.5(0.38)  0.19. P ( x  5)  0.18(0.38)  0.0684. So P ( x  1)  0.38  0.19  0.0684  0.1216.

0.19

0.0684

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CR13.8

Before Fees After Fees

Number of Checked Bags 0 1 2 7 (14.5) 70 (67) 23 (18.5) 22 (14.5) 64 (67) 14 (18.5)

H0: The proportions of passengers falling into the three “number of checked bags” categories were the same before and after fees were imposed. Ha: H0 is not true.   0.05 (observed cell count  expected cell count)2 X2  expected cell count all cells The samples were random samples from the populations of passengers before and after fees were imposed. All the expected counts are greater than 5.



X2 

(7  14.5)2  14.5



(14  18.5)2  10.216 18.5

df = 2 P-value  P(  22  10.216)  0.006 Since P-value  0.006  0.05 we reject H0. We have convincing evidence that the proportions falling into the three “number of checked bags” categories were not the same before and after fees were imposed.

CR13.9 a Number of Songs 800 700 600 500 400 300 200 100 0 0

20 Number of Months

Yes, the relationship looks approximately linear. b

The equation of the estimated regression line is yˆ  12.887  21.126 x, where x = number of months the user has owned the MP3 player and y = number of songs stored.

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403

c Standardized Residual 2

-1

-2 0

20 Number of Months

There is a random pattern in the standardized residual plot, and there is no suggestion that the variance of y is not the same at each x value. There are no outliers. The assumptions of the simple linear regression model would therefore seem to be reasonable. d

 = slope of the population regression line relating the number of songs to the number of

months. 2. H0:   0 3. Ha:   0 4.   0.05 b  (hypothesized value) b  0  5. t  sb sb 6. As explained in Part (c), the assumptions of the simple linear regression model seem to be reasonable. 7. sb  0.994 21.126  0 t  21.263 0.994 8. df = 13 P-value  2  P(t13  21.263)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence of a useful linear relationship between the number of songs stored and the number of months the MP3 player has been owned. CR13.10 1.

1 = mean score for those given ginkgo  2 = mean score for those given the placebo 2. H0: 1  2  0 3. Ha: 1  2  0 4.   0.05

404

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8. 9.

t

( x1  x2 )  (hypothesized value)



( x1  x2 )  0

s12 s22 s12 s22   n1 n2 n1 n2 We are told that the participants were randomly assigned to the treatment groups. Also n1  104  30 and n2  115  30 , so we can proceed with the two-sample t test. 5.6  5.5 t  1.232 0.62 0.62  104 115 df = 214.807 P-value  P(t214.807  1.232)  0.110 Since P-value  0.110  0.05 we do not reject H0. We do not have convincing evidence that taking 40 mg of ginkgo three times a day is effective in increasing mean performance on the Wechsler Memory Scale.

CR13.11 Political Affiliation Democrat Republican Independent/Unaffiliated Other

2005 397 (379.706) 301 (343.448) 458 (440.473) 60 (52.373)

Year 2004 409 (375.647) 349 (339.776) 397 (435.764) 48 (51.813)

2003 325 (375.647) 373 (339.776) 457 (435.764) 48 (51.813)

H0: The proportions falling in each of the response categories are the same for the three years. Ha: H0 is not true.   0.05 (observed cell count  expected cell count)2 X2  expected cell count all cells The samples were considered to be representative of the populations of undergraduates for the given years, and so it is reasonable to assume that they were random samples from those populations. All the expected counts are greater than 5. (397  379.706)2 (48  51.813)2 X2     26.175 379.706 51.813 df = 6 P-value  P(  62  26.175)  0 Since P-value  0  0.05 we reject H0. We have convincing evidence that the distribution of political affiliation is not the same for all three years for which the data are given.



CR13.12 Year Response All of the time Most of the time Some of the time Never

2005 132 (155.478) 337 (431.054) 554 (473.411) 169 (132.057)

2002 180 (156.522) 528 (433.946) 396 (476.589) 96 (132.943)

H0: The proportions falling into the four response categories are the same for the two years.

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Ha: The proportions falling into the four response categories are not all the same for the two years.   0.05 (observed cell count  expected cell count)2 X2  expected cell count all cells We are told that the samples were independently selected and representative of the populations, and therefore it is reasonable to treat them as independent random samples. All the expected counts are greater than 5.



X2 

(132  155.478)2  155.478



(96  132.943)2  95.921 132.943

df = 3 P-value  P(  32  95.921)  0 Since P-value  0  0.05 we reject H0. We have convincing evidence that the proportions falling into the four response categories are not all the same for the two years. CR13.13

Region Northeast Midwest South West

Credit Card? At Least One No Credit Credit Card Card 401 (429.848) 164 (135.152) 162 (150.637) 36 (47.363) 408 (397.895) 115 (125.105) 104 (96.621) 23 (30.379)

H0: Region of residence and having a credit card are independent Ha: Region of residence and having a credit card are not independent   0.05 (observed cell count  expected cell count)2 X2  expected cell count all cells We are told that the sample was a random sample of undergraduates in the US. All the expected counts are greater than 5. (401  429.848)2 (23  30.379)2 X2     15.106 429.848 30.379 df = 3 P-value  P(  32  15.106)  0.002 Since P-value  0.002  0.05 we reject H0. We have convincing evidence that region of residence and having a credit card are not independent.



CR13.14 Region Northeast Midwest South West

Balance Over $7000 No Yes 28 (87.341) 537 (477.659) 162 (53.178) 182 (290.822) 42 (80.849) 481 (442.151) 9 (19.632) 118 (107.368)

406

Chapter 13: Simple Linear Regression and Correlation: Inferential Methods H0: Region of residence and having or not having a credit card balance over $7000 are independent Ha: Region of residence and having or not having a credit card balance over $7000 are not independent   0.01 (observed cell count  expected cell count)2 X2  expected cell count all cells We are told that the sample was a random sample of undergraduates in the US. All the expected counts are greater than 5.



X2 

(28  87.341)2  87.341



(118  107.368)2  339.994 107.368

df = 3 P-value  P( 32  339.994)  0 Since P-value  0  0.01 we reject H0. We have convincing evidence of an association between region of residence and having or not having a credit card balance over $7000. CR13.15 1. 2. 3. 4. 5.

1 = mean alkalinity upstream  2 = mean alkalinity downstream H0: 1  2  50 Ha: 1  2  50   0.05 ( x  x )  (hypothesized value) ( x1  x2 )  ( 50) t 1 2  s12 s22 s12 s22   n1 n2 n1 n2

6. We need to assume that the water specimens were chosen randomly from the two locations. We are given that n1  24 and n2  24, so neither sample size was greater than or equal to 30. We therefore need to assume that the distributions of alkalinity at the two locations are approximately normal. (75.9  183.6)  ( 50)  113.169 7. t  1.832 1.702  24 24 8. df = 45.752 P-value  P(t45.752  113.169)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that the mean alkalinity is higher downstream than upstream by more than 50 mg/L.

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407

CR13.16 Flight 1 2 3 4 5 6 7 8

Method 1 Method 2 Difference 27.5 34.4 -6.9 41.3 38.6 2.7 3.5 3.5 0 24.3 21.9 2.4 27 24.4 2.6 17.7 21.4 -3.7 12 11.8 0.2 20.9 24.1 -3.2

1. d = mean difference between the methods (Method 1 − Method 2) 2. H0: d  0 3. Ha: d  0 4.   0.05 x  hypothesized value 5. t  d sd n 6.

-8

-6

-4

-2 Difference

The boxplot shows that the distribution of the differences is close enough to being symmetrical and has no outliers, so we are justified in assuming that the population distribution of differences is normal. Additionally, we need to assume that this set of flights was a random sample from the population of flights. 7. xd  0.7375, sd  3.526 0.7375  0 t  0.592 3.526 8 8. df = 7 P-value  2  P(t7  0.592)  0.573 9. Since P-value  0.573  0.05 we do not reject H0. We do not have convincing evidence of a difference in mean radiation measurement for the two methods.

408

Chapter 13: Simple Linear Regression and Correlation: Inferential Methods

CR13.17 Direction 0° to <45° 45° to < 90° 90° to <135° 135° to <180° 180° to <225° 225° to <270° 270° to <315° 315° to <360°

Observed Expected Count Count 12 15 16 15 17 15 15 15 13 15 20 15 17 15 10 15

1. Let p1 , , p8 be the proportions of homing pigeons choosing the twelve given directions. 2. H0: p1   p8  0.125 3. Ha: H0 is not true 4.   0.1 (observed cell count  expected cell count)2 5. X 2  expected cell count all cells 6. We need to assume that the study was performed using a random sample of homing pigeons. All the expected counts are greater than 5.



X2 

(12  15)2  15



(10  15)2  4.8 15

8. df = 7 P-value  P(  72  4.8)  0.684 9. Since P-value  0.684  0.1 we do not reject H0. We do not have convincing evidence that the birds exhibit a preference. CR13.18 a

Check of Conditions We cannot truly treat the samples as random samples from the populations, but we will nonetheless continue with construction of the confidence interval. We have n1 pˆ1  120(26 120)  26  10, n1 (1  pˆ1 )  120(94 120)  94  10, n2 pˆ 2  419(222 419)  222  10, and n2 (1  pˆ 2 )  419(197 419)  197  10, so the samples are large enough. Calculation The 95% confidence interval for p1  p2 is

( pˆ1  pˆ 2 )  ( z critical value)

pˆ1 (1  pˆ1 ) pˆ 2 (1  pˆ 2 )  n1 n2

(26 120)(94 120) (222 419)(197 419)  26 222     1.96   120 419  120 419   ( 0.401,  0.225) Interpretation of Interval We are 95% confident that p1  p2 lies between −0.401 and −0.225, where p1 is the proportion of cardiologists who did not know that carbohydrate was the diet component most likely to raise triglycerides and p2 is the proportion of internists who did not know that fact.

Chapter 13: Simple Linear Regression and Correlation: Inferential Methods b

Only 16% of those receiving the questionnaire responded, and it’s quite possible that those who did respond differed from those who did not in some way relevant to the study.

409

Chapter 14 Multiple Regression Analysis Note: In this chapter, numerical answers to questions involving the normal, t, chi square, and F distributions were found using values from a calculator. Students using statistical tables will find that their answers differ slightly from those given. 14.1

14.2

14.3

14.4

410

A deterministic model does not have the random deviation component e, while a probabilistic model does contain such a component.

y = taxi fare x1 = distance traveled x2 = time taken y    1 x1   2 x2

y = test score x1 = average for previous tests x2 = study time x3 = number of hours of sleep y    1 x1   2 x2  3 x3  e

0  1 x1  2 x2

When the measure of weight-bearing activity is fixed, a 1-kg increase in body weight is associated with a mean bone mineral density increase of 0.587 g/cm3.

The population regression function is 30  0.90 x1  0.08 x2  4.50 x3 .

The population regression coefficients are 0.90, 0.08, and −4.50.

When dynamic hand grip endurance and trunk extension ratio are fixed, the mean increase in rating of acceptable load associated with a 1-cm increase in extent of left lateral bending is 0.90 kg.

When extent of left lateral bending and dynamic hand grip endurance are fixed, the mean decrease in rating of acceptable load associated with a 1-N/kg increase in trunk extension ratio is 4.50 kg.

Mean of y  30  0.90(25)  0.08(200)  4.50(10)  23.5 kg.

For these values of the independent variables, the distribution of y is normal, with mean 23.5 and standard deviation 5. We require 33.5  23.5   13.5  23.5 P(13.5  y  33.5)  P  z   P( 2  z  2)  0.9545. 5 5  

411

14.5

14.6

Chapter 14: Multiple Regression Analysis

When x1  20 and x2  50, mean weight  21.658  0.828(20)  0.373(50)  13.552 g.

When length is fixed, the mean increase in weight associated with a 1-mm increase in width is 0.828 g. When width is fixed, the mean increase in weight associated with a 1-mm increase in length is 0.373 g.

The predicted ecology score is yˆ  3.6  0.01(250)  0.01(32)  0.07(0)  0.12(1)  0.02(16)  0.04(1)  0.01(3)

 0.04(0)  0.02(0)  1.79.

14.7

The variable x4 decreases by 1, so the predicted value of y decreases by 0.12. Therefore, the predicted value of y would be 1.79  0.12  1.67.

For women, the value of x3 is 0, and it is 1 for men. Thus, if all the other variables are unchanged, the estimated mean ecology score for women is 0.07 more than for men.

If all the other variables are fixed, then the increase in the predicted ecology score associated with a $1000 increase in income is 0.01.

Consider the ideology variable, x6 . According to the model as it is given, the effect on the predicted ecology score of a change from, for example, liberal to left of center is the same as the effect of a change from left of center to middle of the road. In reality this might not be the case. A better way of dealing with this variable would be to use in its place four variables. We could then say that if the person is a liberal then the values of all four variables are 0; if the person is left of center then the value of the first variable is 1, and all the other variables are 0; if the person is middle of the road, then the value of the second variable is 1, and all the other variables are 0; and so on. Then the values of the coefficients of these variables in the model would take care of possibly differing effects on the ecology score from the various changes in ideology. A similar argument applies to the social class variable.

Mean yield  yˆ  113,527  0.974(130,000)  0.057(400,000)  35,893 hundreds of millions of kwh

14.8

When the gross domestic product is fixed, the mean increase in population (in millions) associated with a 1-million increase in population is 0.974.

Mean value of y  yˆ  183.560  1.589(300)  0.069(2500)  465.64 mg/L

Mean value of y  yˆ  183.560  1.589(230)  0.069(2700)  368.21 mg/L

Chapter 14: Multiple Regression Analysis

14.9

412

a x Mean of y

20 0.848

30 0.833

40 0.832

50 0.845

60 0.872

70 0.913

The values calculated in Part (a) show us that the gait instability is higher for age 60 than for age 40.

As age increases from 20 to 30, the change in Mean gait instability is 0.833  0.848  0.015 . As age increases from 50 to 60, the change in Mean gait instability is 0.872  0.845  0.027 .

14.10 a b

The population regression function is 1.09  0.653x1  0.0022 x2  0.0206 x12  0.4 x22 .

x1  16, x2  4.118.

Mean y  1.09  0.653(16)  0.0022(4.118)  0.0206(16)2  0.4(4.118)2  33.057. c

Since 1  0, larger values of x1 are expected to lead to larger yields. Thus planting on May 22 is expected to bring about a greater yield.

No, since there is a further term involving x1 .

413

Chapter 14: Multiple Regression Analysis

14.11 a Mean y 14

x1=10 x1=20 x1=30

12 10 8 6 4 2 0

b Mean y 54

x2=50 x2=55 x2=60

52 50 48 46 44 42 0

15 x1

The fact that there is no interaction between x1 and x2 is reflected by the fact that in each of the graphs, the lines are parallel.

Chapter 14: Multiple Regression Analysis

414

d Mean y x1=10 x1=20 x1=30

0 0

x2 Mean y 110

x2=50 x2=55 x2=60

100 90 80 70 60 50 40 0

15 x1

The presence of an interaction term causes the lines in the graphs to be nonparallel. 14.12 a

To take account of the intake settings we could define the variables x1 and x2 as shown in the table below. Intake setting Low Medium High

x2 0 1 0

x3 0 0 1

Then the model equation will be

Mean y    1 x1   2 x2  3 x3  e .

415

Chapter 14: Multiple Regression Analysis

1 is the amount by which the predicted particulate matter concentration increases when the flue temperature increases by 1 degree (the intake setting remaining fixed).  2 is the amount by which the predicted particulate matter concentration increases when the intake setting is changed from low to medium (the flue temperature remaining fixed)  3 is the amount by which the predicted particulate matter concentration increases when the intake setting is changed from low to high (the flue temperature remaining fixed). b 14.13 a

We would need to add to the model terms involving x1 x2 and x1 x3 .

y    1 x1  2 x2  3 x3  e

y    1 x1   2 x2  3 x3   4 x12  5 x22  6 x32  e

y    1 x1  2 x2  3 x3  4 x2 x3  e y    1 x1  2 x2  3 x3  4 x1 x3  e y    1 x1  2 x2  3 x3  4 x1 x2  e

d 14.14 a

y    1 x1   2 x2  3 x3   4 x12  5 x22   6 x32  7 x2 x3  8 x1 x3  9 x1 x2  e Mean y value  86.8  0.123x1  5.09 x2  0.0709 x3  0.001x4

 86.8  0.123(3200)  5.09(57)  0.0709(57)2  0.001(3200)(57)  64.6241. b 14.15 a

No, since there are other terms involving x2 . We need additional variables x3 , x4 , and x5 . The values of these variables could be defined as shown in the table. Size Class Subcompact Compact Midsize Large

x3 x4 0 0 1 0 0 1 0 0

x5 0 0 0 1

The model equation is y    1 x1  2 x2  3 x3  4 x4  5 x5  e b 14.16 a

The additional predictors are x1 x3 , x1 x4 , and x1 x5 . When all the other variables are fixed, the predicted increase in percentage of body fat mass associated with a 1-unit increase in log(years) is 6.65. When all the other variables are fixed, the predicted increase in percentage of body fat mass would be 11.40 greater for females than for males. The required proportion is R2  1 

SSResid 180,000 1  0.75 SSTo 540,000  180,000

Chapter 14: Multiple Regression Analysis

The value of  is estimated using s 

SSResid 180,000   7.505 n  (k  1) 3,200  4

 n  1   SSResid  180,000  3,200  1   Adjusted R 2  1   1      0.7498   3,200  4  540,000  180,000   n  (k  1)   SSTo  The value of the adjusted R 2 is slightly smaller than that of R 2 , reflecting the fact that the number of predictors used in the model is three, rather than one.

14.17 a

P-value  P( F3,15  4.23)  0.024.

P-value  P( F4,18  1.95)  0.146.

P-value  P( F5,20  4.10)  0.010.

P-value  P( F4,35  4.58)  0.004.

14.18 a

P-value  P( F2,18  2.47)  0.113.

P-value  P( F8,16  5.98)  0.001.

P-value  P( F5,20  3.00)  0.035.

The model is y    1 x1   2 x2    3 x12   4 x22  5 x1 x2  e . Thus, k  5. So P-value  P( F5,14  8.25)  0.001.

P-value  P( F5,94  2.33)  0.048.

14.19 a

416

1. The model is y    1 x1  2 x2  3 x3  e where y = surface area, x1 = weight, x2 = width, and x3 = length. 2. H0: 1   2  3  0 3. Ha: At least one of the i ’s is not zero. 4.   0.05 R2 k 5. F  (1  R 2 )  n  (k  1)  6. Since we do not have the original data set we are unable to check the conditions. We need to assume that the variables are related according to the model given above, and that the random deviations, e, are normally distributed with mean zero and fixed standard deviation. 0.996 3  12118 7. F  (1  0.996) 146 8. P-value  P( F3,146  12118)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that the multiple regression model is useful.

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Since the P-value is small and r 2 is close to 1 there is strong evidence that the model is useful.

The model in Part (b) should be recommended, since adding the variables x1 and x2 to the model (to obtain the model in Part (a)) only increases the value of R 2 a small amount (from 0.994 to 0.996).

14.20 a

yˆ  26.43  0.3499 x1  0.3033x2 where y = home-range size, x1 = age, and x2 = weight.

b Age Weight 10.5 54 6.5 40 28.5 62 6.5 55 7.5 56 6.5 62 5.5 42 7.5 40 11.5 59 9.5 51 5.5 50

Home-Range Size 43.1 46.6 57.4 35.6 62.1 33.9 39.6 32.2 57.2 24.4 68.7

Standardized Residual -0.23602 0.46077 0.52091 -0.70921 1.16163 -1.16291 -0.11325 -0.72486 0.64420 -1.44993 1.79165

Standardized residual 2.0 1.5 1.0 0.5 0.0 -0.5 -1.0 -1.5 -2

-1

0 Normal Score

The plot is reasonably straight, indicating that the assumption of normality of the random deviation distribution is justified. c

1. The model is y    1 x1  2 x2  e where y = home-range size, x1 = age, and x2 = weight. 2. H0: 1   2  0

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3. Ha: At least one of 1 ,  2 is not zero. 4.   0.05 5. 6. 7. 8. 9.

R2 k (1  R 2 )  n  (k  1)  The conditions were checked in Part (b). The computer output gives F  0.38. The computer output gives P-value  0.697. Since P-value  0.697  0.05 we do not reject H0. We do not have convincing evidence that the predictors age and weight are useful for predicting home-range size. F

14.21 1. The model is y    1 x1   2 x2  3 x3   4 x4  5 x5  6 x6  e , where y = species richness, x1 = watershed area, x2 = shore width, x3 = drainage, x4 = water color, x5 = sand percentage, and x6  alkalinity . 2. H0: 1   2  3   4  5  6  0 3. Ha: At least one of the i ’s is not zero. 4.   0.01

R2 k (1  R 2 )  n  (k  1)  6. Since we do not have the original data set we are unable to check the conditions. We need to assume that the variables are related according to the model given above, and that the random deviations, e, are normally distributed with mean zero and fixed standard deviation. 0.83 6  24.412 7. F  (1  0.83) 30 5.

F

8. P-value  P( F6,30  24.412)  0 9. Since P-value  0  0.01 we reject H0. We have convincing evidence that the chosen model is useful. 14.22 1. The model is y    1 x1   2 x2  3 x3   4 x4  5 x5  6 x6  7 x7  e , where y = spring math comprehension, x1 = previous fall test score, x2 = previous fall academic motivation, x3 = age, x4 = number of credit hours, x5 = residence, x6 = hours worked on campus, and x7 = hours worked off campus. 2. H0: 1  2  3  4  5  6  7  0 3. Ha: At least one of the i ’s is not zero. 4.   0.05 R2 k 5. F  (1  R 2 )  n  (k  1)  6. Since we do not have the original data set we are unable to check the conditions. We need to assume that the variables are related according to the model given above, and that the random deviations, e, are normally distributed with mean zero and fixed standard deviation. 0.543 7  34.288 7. F  (1  0.543) 202 8.

P-value  P( F7,202  34.288)  0

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Chapter 14: Multiple Regression Analysis 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that there is a useful linear relationship between y and at least one of the predictors.

14.23 1. The model is y    1 x1  2 x2  3 x3  e , where y = percentage of body fat mass, x1 = log of body mass index, x2 = log of age, and x3 = sex. 2. H0: 1  2  3  0 3. Ha: At least one of the i ’s is not zero. 4.   0.1 SSRegr k 5. F  SSResid  n  (k  1)  6. Since we do not have the original data set we are unable to check the conditions. We need to assume that the variables are related according to the model given above, and that the random deviations, e, are normally distributed with mean zero and fixed standard deviation. 540,000 3  3,196 7. F  180,000 (3, 200  4) 8.

P-value  P( F3,3196  2.996)  0

9. Since P-value  0  0.1 we reject H0. We have convincing evidence that the model is useful. 14.24 a

The estimated regression equation is yˆ  86.85  0.12297 x1  5.090 x2  0.07092 x3  0.0015380 x4 , where y = tar content, x1 = rotor speed, x2 = gas inlet temperature, x3  x22 , and x4  x1 x2 .

1. The model is y    1 x1  2 x2  3 x3  4 x4  e , where the variables are defined as above. 2. H0: 1  2  3  4  0 3. Ha: At least one of the i ’s is not zero. 4.   0.01 SSRegr k 5. F  SSResid  n  (k  1)  6. Since we do not have the original data set we are unable to check the conditions. We need to assume that the variables are related according to the model given above, and that the random deviations, e, are normally distributed with mean zero and fixed standard deviation. SSRegr 4 5896.6 4   64.4 7. F  SSResid 26 595.1 26 8. P-value  P( F4,26  64.4)  0 9. Since P-value  0  0.01 we reject H0. We have convincing evidence that the multiple regression model is useful.

R 2  0.908 . This is the proportion of the observed variation in tar content that can be explained by the fitted model. se  4.784 . This is a typical deviation of a tar content value in the sample from the value predicted by the estimated regression equation.

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14.25 1. The model is y    1 x1  2 x2  3 x3  4 x4  5 x5  6 x6  7 x7  8 x8  9 x9  e , where y = ecology score, x1 = age times 10, x2 = income, x3 = gender, x4 = race, x5 = number of years of education, x6 = ideology, x7 = social class, x8 = postmaterialist (0 or 1), and x9 = materialist (0 or 1). 2. H0: 1  2  3  4  5  6  7  8  9  0 3. Ha: At least one of the i ’s is not zero. 4.   0.05

F

8. P-value  P( F9,1126  7.986)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that the multiple regression model is useful. 14.26 a

The MINITAB output is shown below. The regression equation is Weight = - 511 + 3.06 Length - 1.11 Age Predictor Constant Length Age

Coef -510.9 3.0633 -1.113

S = 94.2379

SE Coef 286.1 0.8254 9.040

R-Sq = 59.3%

T -1.79 3.71 -0.12

P 0.096 0.002 0.904

R-Sq(adj) = 53.5%

Analysis of Variance Source Regression Residual Error Total

DF 2 14 16

SS 181364 124331 305695

MS 90682 8881

F 10.21

P 0.002

The estimated regression equation is yˆ  510.9  3.0633x1  1.113x2 , where y = weight, x1 = length, and x2 = age. b

1. The model is y    1 x1  2 x2  e , with the variables as defined above. 2. H0: 1   2  0 3. Ha: At least one of the i ’s is not zero. 4.   0.05 R2 k 5. F  2 (1  R )  n  (k  1) 

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Chapter 14: Multiple Regression Analysis 6. The normal probability plot of the standardized residuals is shown below. Standardized Residual 3

-1

-2 -2

-1

0 Normal Score

The plot at first seems to show a slight curve, but this curvature can be attributed to just one point. We are justified in assuming that the random deviations are normally distributed. 7. F  10.21 8. P-value  0.002. 9. Since P-value  0.002  0.05 we reject H0. We have convincing evidence that the predictors length and age, together, are useful for predicting weight. 14.27 a

The MINITAB output is shown below. The regression equation is Catch Time = 1.44 - 0.0523 Prey Length + 0.00397 Prey Speed Predictor Constant Prey Length Prey Speed

Coef 1.43958 -0.05227 0.0039700

S = 0.0930752

SE Coef 0.08325 0.01459 0.0006194

R-Sq = 75.0%

T 17.29 -3.58 6.41

P 0.000 0.002 0.000

R-Sq(adj) = 71.9%

Analysis of Variance Source Regression Residual Error Total

DF 2 16 18

SS 0.41617 0.13861 0.55478

MS 0.20809 0.00866

F 24.02

P 0.000

The estimated regression equation is yˆ  1.43958  0.05227 x1  0.0039700 x2 , where y = catch time, x1 = prey length, and x2 = prey speed. When x1  6 and x2  50, yˆ  1.43958  0.05227(6)  0.0039700(50)  1.324 seconds.

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1. The model is y    1 x1  2 x2  e , with the variables as defined above. 2. H0: 1   2  0 3. Ha: At least one of the i ’s is not zero. 4.   0.05

R2 k (1  R 2 )  n  (k  1)  6. The normal probability plot of the standardized residuals is shown below. 5.

F

Standardized Residual 2

-1

-2

-3 -2

-1

0 Normal Score

There is a linear pattern in the plot, so we are justified in assuming that the random deviations are normally distributed. 7. F  24.02 8. P-value  0.000. 9. Since P-value  0.000  0.05 we reject H0. We have convincing evidence that the multiple regression model is useful for predicting catch time.

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The values of the new variable are shown in the table below. Prey Prey Catch Length Speed Time 7 20 1.10 6 20 1.20 5 20 1.23 4 20 1.40 3 20 1.50 3 40 1.40 4 40 1.36 6 40 1.30 7 40 1.28 7 80 1.40 6 60 1.38 5 80 1.40 7 100 1.43 6 100 1.43 7 120 1.70 5 80 1.50 3 80 1.40 6 100 1.50 3 120 1.90

x 0.3500 0.3000 0.2500 0.2000 0.1500 0.0750 0.1000 0.1500 0.1750 0.0875 0.1000 0.0625 0.0700 0.0600 0.0583 0.0625 0.0375 0.0600 0.0250

The MINITAB output is shown below. The regression equation is Catch Time = 1.59 - 1.40 x Predictor Constant x

Coef 1.58648 -1.4044

S = 0.122096

SE Coef 0.04803 0.3124

R-Sq = 54.3%

T 33.03 -4.50

P 0.000 0.000

R-Sq(adj) = 51.6%

Analysis of Variance Source Regression Residual Error Total

DF 1 17 18

SS 0.30135 0.25342 0.55478

MS 0.30135 0.01491

F 20.22

P 0.000

The estimated regression equation is yˆ  1.58648  1.4044 x . e

Since both the R 2 and the adjusted R 2 values shown in the computer outputs are greater for the first model than for the second, the first model is preferable to the second. The first model is the one that accounts for the greater proportion of the observed variation in catch time.

Chapter 14: Multiple Regression Analysis 14.28 a

424

The MINITAB output is shown below. The regression equation is Volume = - 859 + 23.7 Minimum Width + 226 Maximum Width + 225 Elongation Predictor Constant Minimum Width Maximum Width Elongation

Coef -859.2 23.72 225.81 225.24

SE Coef 272.9 85.66 85.76 90.65

S = 286.974

R-Sq = 67.6%

T -3.15 0.28 2.63 2.48

P 0.005 0.784 0.015 0.021

R-Sq(adj) = 63.4%

Analysis of Variance Source Regression Residual Error Total

DF 3 23 26

SS 3960700 1894141 5854841

MS 1320233 82354

F 16.03

P 0.000

The estimated regression equation is yˆ  859.2  23.72 x1  225.81x2  225.24 x3 , where y = volume, x1 = minimum width, x2 = maximum width, and x3 = elongation score. b

We should use adjusted R 2 because it takes into account the number of predictors used in the model.

1. The model is y    1 x1  2 x2  3 x3  e , with the variables as defined above. 2. H0: 1   2  3  0 3. Ha: At least one of the i ’s is not zero. 4.   0.05 R2 k 5. F  (1  R 2 )  n  (k  1)  6. The normal probability plot of the standardized residuals is shown below.

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Standardized Residual 5 4 3 2 1 0 -1 -2 -3 -4 -2

-1

0 Normal Score

There seems to be a nonlinear pattern at the ends of the plot, but we will nonetheless proceed with the assumption that the random deviations are normally distributed. 7. F  16.03 8. P-value  0.000. 9. Since P-value  0.000  0.05 we reject H0. We have convincing evidence that the multiple regression model is useful. 14.29 a b

SSResid  390.435, SSTo  1618.209, SSRegr  1618.209  390.435  1227.775. SSResid 390.435  1  0.759. SSTo 1618.209 This tells us that 75.9% of the observed variation in shear strength can be explained by the fitted model. R2  1 

1. The model is y    1 x1  2 x2  3 x3  4 x4  5 x5  e , where y = shear strength, x1 = depth, x2 = water content, x3  x12 , x4  x22 , and x5  x1 x2 . 2. H0: 1   2  3   4  5  0 3. Ha: At least one of the i ’s is not zero. 4.   0.05 R2 k 5. F  (1  R 2 )  n  (k  1)  6. The normal probability plot of the standardized residuals is shown below.

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Standardized Residual 2

-1

-2 -2

-1

0 Normal Score

The plot shows a linear pattern, so we are justified in assuming that the random deviations are normally distributed. 0.759 5  5.031 7. F  0.241 8 8. P-value  P( F5,8  5.031)  0.022 9. Since P-value  0.022  0.05 we reject H0. We have convincing evidence that the multiple regression model is useful. 2 14.30 R  0.15 2 1. The model is y  0.92  0.005 x  0.00007 x  e where y = gait instability, x = age. 2. H0: 1   2  0 3. Ha: At least one of the i ’s is not zero. 4.   0.05 R2 k 5. F  (1  R 2 )  n  (k  1)  6. Since we do not have the original data set we are unable to check the conditions. We need to assume that the variables are related according to the model given above, and that the random deviations, e, are normally distributed with mean zero and fixed standard deviation. 0.15 2  8.559 7. F  (1  0.15) (100  3) 8.

P-value  P( F2,97  8.559)  0.00038

9. Since P-value  0.00038  0.05 we reject H0. We have convincing evidence that the quadratic regression model is useful. 14.31 1. The model is y    1 x1  2 x2  e , where y = yield, x1 = defoliation level, and x2  x12 . 2. H0: 1   2  0 3. Ha: At least one of the i ’s is not zero. 4.   0.01

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F

8. P-value  P( F2,21  96.643)  0 9. Since P-value  0  0.01 we reject H0. We have convincing evidence that the quadratic model specifies a useful relationship between y and x. 14.32

R2 k 0.9 k 9(14  k )   . 2 k (1  R )  n  (k  1)  0.1 15  (k  1)  Using the above formula and Appendix Table 6 the following table is obtained. (The critical values shown are for a 0.05 significance level.) F

df1

df2

Critical Value

1 2 3 4 5 6 7 8 9 10

117 1 54 2 33 3 22.5 4 16.2 5 12 6 9 7 6.75 8 5 9 3.6 10

13 12 11 10 9 8 7 6 5 4

4.67 3.89 3.59 3.48 3.48 3.58 3.79 4.15 4.77 5.96

For k  9 the value of the F statistic is greater than the critical value, and so there is convincing evidence that the model is useful. For k  10 the value of the F statistic is less than the critical value. Moreover, for k  10 the value of F continues to decrease and we presume that the critical value continues to increase, and thus for k  10 there is not convincing evidence that the model is useful. So the model would be judged to be useful for k  9 , and this large value of R 2 does not necessarily imply a useful model. 14.33 The MINITAB output is shown below. The regression equation is y = - 151 - 16.2 x1 + 13.5 x2 + 0.0935 x3 - 0.253 x4 + 0.492 x5 Predictor Constant x1 x2 x3 x4 x5

Coef -151.4 -16.216 13.476 0.09353 -0.2528 0.4922

SE Coef 134.1 8.831 8.187 0.07093 0.1271 0.2281

T -1.13 -1.84 1.65 1.32 -1.99 2.16

P 0.292 0.104 0.138 0.224 0.082 0.063

Chapter 14: Multiple Regression Analysis

S = 6.98783

R-Sq = 75.9%

R-Sq(adj) = 60.8%

Analysis of Variance Source Regression Residual Error Total

DF 5 8 13

SS 1227.57 390.64 1618.21

MS 245.51 48.83

F 5.03

P 0.022

This verifies the estimated regression equation given in the question. 14.34 a

The MINITAB output is shown below. The regression equation is y = 76.4 - 7.3 x1 + 9.6 x2 - 0.91 x3 + 0.0963 x4 - 13.5 x1^2 + 2.80 x2^2 + 0.0280 x3^2 - 0.000320 x4^2 + 3.75 x1x2 - 0.750 x1x3 + 0.142 x1x4 + 2.00 x2x3 - 0.125 x2x4 + 0.00333 x3x4 Predictor Constant x1 x2 x3 x4 x1^2 x2^2 x3^2 x4^2 x1x2 x1x3 x1x4 x2x3 x2x4 x3x4

Coef 76.437 -7.35 9.61 -0.915 0.09632 -13.452 2.798 0.02798 -0.0003201 3.750 -0.7500 0.14167 2.0000 -0.12500 0.003333

S = 0.352900

SE Coef 9.082 10.80 10.80 1.068 0.09834 6.599 6.599 0.06599 0.0002933 8.823 0.8823 0.05882 0.8823 0.05882 0.005882

R-Sq = 88.5%

T 8.42 -0.68 0.89 -0.86 0.98 -2.04 0.42 0.42 -1.09 0.43 -0.85 2.41 2.27 -2.13 0.57

P 0.000 0.506 0.387 0.404 0.342 0.058 0.677 0.677 0.291 0.676 0.408 0.028 0.038 0.049 0.579

R-Sq(adj) = 78.3%

Analysis of Variance Source Regression Residual Error Total

DF 14 16 30

SS 15.2642 1.9926 17.2568

MS 1.0903 0.1245

F 8.75

P 0.000

The estimated regression equation is yˆ  76.437  7.35x1  9.61x2  0.915x3  0.09632 x4  13.452 x12  2.798x22

0.02798 x32  0.0003201x42  3.750 x1 x2  0.7500 x1 x3  0.14167 x1 x4 2.0000 x2 x3  0.125x2 x4  0.003333x3 x4 .

428

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1. The model is y    1 x1   2 x2  3 x3   4 x4  5 x12  6 x22  7 x32  8 x42  9 x1 x2  10 x1 x3 2. 3. 4. 5. 6.

 11 x1 x4  12 x2 x3  13 x2 x4  14 x3 x4  e. H0: 1   14  0 Ha: At least one of the i ’s is not zero.   0.05 R2 k F (1  R 2 )  n  (k  1)  The normal probability plot of the standardized residuals is shown below. Standardized Residual 4 3 2 1 0 -1 -2 -3 -2

-1

0 Normal Score

The plot shows a linear pattern, so we are justified in assuming that the random deviations are normally distributed. 7. F  8.75 8. P-value  0.000 9. Since P-value  0.000  0.05 we reject H0. We have convincing evidence that the multiple regression model is useful. c

SSResid  1.9926. This is the sum of the squares of the deviations of the actual brightness values from the values predicted by the fitted model. se  0.352900. This is a typical deviation of a brightness value in the sample from the value predicted by the fitted model. R 2  0.885. This is the proportion of the observed variation in brightness that can be explained by the fitted model.

14.35 The MINITAB output is shown below. The regression equation is y = 35.8 - 0.68 x1 + 1.28 x2 Predictor Constant

Coef 35.83

SE Coef 53.54

T 0.67

P 0.508

Chapter 14: Multiple Regression Analysis x1 x2

-0.676 1.2811

S = 22.9789

1.436 0.4243

-0.47 3.02

R-Sq = 55.0%

430

0.641 0.005

R-Sq(adj) = 52.1%

Analysis of Variance Source Regression Residual Error Total

DF 2 31 33

SS 20008 16369 36377

MS 10004 528

F 18.95

P 0.000

The estimated regression equation is yˆ  35.83  0.676 x1  1.2811x2 , where y = infestation rate, x1 = mean temperature, and x2 = mean relative humidity. 1. The model is y    1 x1  2 x2  e , with the variables defined as above. 2. H0: 1   2  0 3. Ha: At least one of the i ’s is not zero. 4.   0.05

R2 k (1  R 2 )  n  (k  1)  6. The normal probability plot of the standardized residuals is shown below. 5.

F

Standardized Residual 2

-1

-2 -2

-1

0 Normal Score

The plot shows a linear pattern, so we are justified in assuming that the random deviations are normally distributed. 7. F  18.95 8. P-value  0.000 9. Since P-value  0.000  0.05 we reject H0. We have convincing evidence that the multiple regression model is useful.

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Online Exercises 14.36 It is possible that the proposed predictors, when used in linear combination, actually have virtually no relationship to the response variable – in other words, that just using the mean value of the response variable in the sample would be as good as any attempt to fit a model using the proposed predictors. If this is the case, then there is little point in using those variables as predictors. However, the model utility test can establish (beyond all reasonable doubt) that there is a useful relationship between the given set of variables and the response variable, and therefore that it is worthwhile to use them as predictors. 14.37 a

We need to assume that the mean value of a vacant lot is related to the predictors according to the multiple regression model, and that the random deviations from the model are normally distributed with mean zero and fixed standard deviation. df  n  ( k  1)  100  8  92. The required confidence interval is b2  (t critical value) sb2  0.489  (1.986)(0.1044)  ( 0.696, 0.282). We are 95% confident that, when all the other predictors are fixed, the average decrease in value of a vacant lot associated with a one-unit increase in the distance from the city’s major east-west thoroughfare is between 0.282 and 0.696.

1. 2. 3. 4. 5.

1 is the mean increase in value of a vacant lot for a one unit increase in the “residential use” variable when all the other predictors are held fixed. H0: 1  0 Ha: 1  0   0.05 b t 1 sb1

6. We need to assume that the mean value of a vacant lot is related to the predictors according to the multiple regression model, and that the random deviations from the values given by the population regression function are normally distributed with mean zero and fixed standard deviation. b 0.183  0.599 7. t  1  sb1 0.3055 8.

df  n  ( k  1)  100  8  92

P-value  2  P(t92  0.599)  0.551 9. Since P-value  0.551  0.05 we do not reject H0. We do not have convincing evidence that the predictor that indicates whether the lot was zoned for residential use provides useful information about the value of a vacant lot, over and above the information contained in the other predictors. 14.38 a

We need to assume that the number of fish at intake is related to the predictors according to the multiple regression model, and that the random deviations from the model are normally distributed with mean zero and fixed standard deviation. df  n  ( k  1)  26  5  21. The required confidence interval is b3  (t critical value) sb3  9.378  (2.080)(4.356)  ( 18.437, 0.319).

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We are 95% confident that, when all the other predictors are fixed, the average decrease in the number of fish at intake associated with a one-unit increase in sea state is between 0.319 and 18.437. b

We require a 90% confidence interval for 1 . We need to assume that the number of fish at intake is related to the predictors according to the multiple regression model, and that the random deviations from the model are normally distributed with mean zero and fixed standard deviation. df  n  ( k  1)  26  5  21. The required confidence interval is b1  (t critical value) sb1  2.179  (1.721)(1.087)  ( 4.049, 0.309). We are 90% confident that, when all the other predictors are fixed, the average decrease in the number of fish at intake associated with a 1° increase in temperature is between 0.309 and 4.049.

14.39 a

1. The model is y    1 x   2 x 2  e , where y = MDH activity and x = electrical conductivity. 2. H0: 1   2  0 3. Ha: At least one of the i ’s is not zero. 4.   0.05

F

8. P-value  P( F2,47  144.357)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that the quadratic model is useful. b

1.  2 is defined as in Part (a). 2. H0: 2  0 3. Ha: 2  0 4.   0.01 b 5. t  2 sb2 6. We need to assume that the mean MDH activity is related to electrical conductivity according to the quadratic model given in Part (a), and that the random deviations from the values given by the quadratic regression function are normally distributed with mean zero and fixed standard deviation. b 0.0446  4.330 7. t  2  sb2 0.0103 8.

df  n  ( k  1)  50  3  47

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P-value  2  P(t47  4.330)  0 9. Since P-value  0  0.01 we reject H0. We have convincing evidence that the quadratic predictor x 2 is important. c

We need to make the same assumptions as in Parts (a) and (b). When x  40, yˆ  0.1838  0.0272(40)  0.0446(40) 2  72.2642. df  n  ( k  1)  50  3  47. The required confidence interval is yˆ  (t critical value)s yˆ  72.2642  (1.678)(0.120)  (72.063, 72.466). We are 90% confident that the mean MDH activity when the conductivity is 40 is between 72.063 and 72.466.

14.40 a

The MINITAB output is shown below. The regression equation is Volume = - 859 + 23.7 Minimum Width + 226 Maximum Width + 225 Elongation Predictor Constant Minimum Width Maximum Width Elongation

Coef -859.2 23.72 225.81 225.24

SE Coef 272.9 85.66 85.76 90.65

S = 286.974

R-Sq = 67.6%

T -3.15 0.28 2.63 2.48

P 0.005 0.784 0.015 0.021

R-Sq(adj) = 63.4%

Analysis of Variance Source Regression Residual Error Total

DF 3 23 26

SS 3960700 1894141 5854841

MS 1320233 82354

F 16.03

P 0.000

Since the P-value associated with the coefficient of minimum width is as large as 0.784, this variable could be eliminated from the regression. b

When x1  2.5, x2  3.0, and x3  1.55, yˆ  859.2  23.72(2.5)  225.81(3.0)  225.24(1.55)  226.652.

The standardized residual plot shown in the solution to the earlier exercise shows a curved shape at its ends, but we will nonetheless proceed with construction of the prediction interval. df  n  ( k  1)  27  4  23. The standard error of ŷ is 73.8. The required prediction interval is

yˆ  (t critical value) se2  s 2yˆ  226.652  (2.069) 286.974 2  73.82 )  ( 386.315, 839.619).

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However, since negative volumes are not possible, the interval is adjusted to (0, 839.619). We are 95% confident that, when the minimum width is 2.5, the maximum width is 3.0, and the elongation is 1.55, the volume will be between 0 and 839.619 ml. 14.41 a

The value 0.469 is an estimate of the increase in mean exam score associated with a one-point increase in expected score on the exam, when time spent studying and GPA are fixed. 1. The model is y    1 x1  2 x2  3 x3  e , with variables as defined in the question. 2. H0: 1   2  3  0 3. Ha: At least one of the i ’s is not zero. 4.   0.05

F

8. P-value  P( F3,103  75.008)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that there is a useful linear relationship between exam score and at least one of the three predictors. c

We need to assume that the exam score is related to the predictors according to the multiple regression model, and that the random deviations from the model are normally distributed with mean zero and fixed standard deviation. df  n  ( k  1)  107  4  103. The required confidence interval is b2  (t critical value) sb2  3.369  (1.983)(0.456)  ( 2.464, 4.273). We are 95% confident that, when all the other predictors are fixed, the average increase in exam score associated with a 1-hour increase in study time is between 2.464 and 4.273.

When x1  75, x2  8, and x3  2.8, yˆ  2.178  0.469(75)  3.369(8)  3.054(2.8)  72.856.

We need to make the same assumptions as in Part (c). df  n  ( k  1)  107  4  103. To calculate se : SSResid SSResid R2  1  . So here 0.686  1  , giving SSTo 10200 SSResid  10200(1  0.686)  3202.8.

SSResid 3202.8   5.576. n  ( k  1) 103 The required prediction interval is So se 

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yˆ  (t critical value) se2  s 2yˆ  72.856  (1.983) 5.5762  1.2 2 )  (61.544, 84.169). Our estimate of the interval containing 95% of scores of students with the given predictor values is 61.544 to 84.169. 14.42 a

 n  1   SSResid  Adjusted R 2  1   .   n  (k  1)   SSTo  SSResid  296   299  SSResid  So here, 0.774  1   , which gives     (1  0.774)  0.224. SSTo  296  SSTo   299  SSResid  1  0.224  0.776. From this we get R 2  1  SSTo R2 k 0.776 3 Thus, for the model utility test we get F    342.336. 2 (1  R )  n  (k  1)  (1  0.776) 296 Therefore, P-value  P( F3,296  342.336)  0 , telling us that at least one of the three predictors provides useful information about y.

 3 is the mean increase in benevolence payments for a one unit increase in the

urban/nonurban variable when all the other predictors are held fixed. 2. H0: 3  0 3. Ha: 3  0 4.   0.05 b 5. t  3 sb3 6. We need to assume that the mean of the benevolence payments is related to the predictors according to the multiple regression model, and that the random deviations from the values given by the population regression function are normally distributed with mean zero and fixed standard deviation. b 101.1  0.162 7. t  3  sb3 625.8

df  n  ( k  1)  300  4  296 P-value  2  P(t296  0.162)  0.872 9. Since P-value  0.872  0.05 we do not reject H0. We do not have convincing evidence that the indicator should be retained in the model. 8.

14.43 1. 2. 3. 4. 5.

 3 is the coefficient of x1 x2 in the population regression function relating y to x1 , x2 , and x1 x2 , with variables as defined in the question. H0: 3  0 Ha: 3  0   0.05 b t 3 sb3

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6. We need to assume that the mean number of indictments disposed of in a given month is related to the predictors according to the multiple regression model, and that the random deviations from the values given by the population regression function are normally distributed with mean zero and fixed standard deviation. b 0.00002 7. t  3   2.222 sb3 0.000009 8.

df  n  ( k  1)  367  4  363

P-value  2  P(t363  2.222)  0.027 9. Since P-value  0.027  0.05 we reject H0. We have convincing evidence that the interaction predictor is important. 14.44 1.

 2 is the coefficient of x 2 in the population regression function relating y to x and x 2 ,

with variables as defined in the question. 2. H0: 2  0 3. Ha: 2  0 4.   0.05 b 5. t  2 sb2 6. A normal probability plot of the standardized residuals is shown below. Standardized residual 1.5 1.0 0.5 0.0 -0.5 -1.0

-1.0

-0.5

0.0 0.5 Normal Score

1.0

1.5

The plot is quite straight, indicating that normality of the random error distribution is plausible. b 1.7155  8.426 7. t  2  sb2 0.2036

df  n  ( k  1)  7  3  4 P-value  2  P(t4  8.426)  0.001 9. Since P-value  0.001  0.05 we reject H0. We have convincing evidence that the quadratic predictor is important. 8.

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14.45 a

1. The model is y    1 x1  2 x2  3 x3  e , with variables as defined in the question. 2. H0: 1   2  3  0 3. Ha: At least one of the i ’s is not zero. 4.   0.05

F

8. P-value  P( F3,6  5.463)  0.038 9. Since P-value  0.038  0.05 we reject H0. We have convincing evidence that the multiple regression model is useful. b

1. 2. 3. 4. 5.

6. We need to assume that the mean value of y is related to the predictors according to the multiple regression model, and that the random deviations from the values given by the population regression function are normally distributed with mean zero and fixed standard deviation. b 7. t  3  0.04 sb3

df  n  ( k  1)  10  4  6 P-value  2  P(t6  0.04)  0.969 9. Since P-value  0.969  0.05 we do not reject H0. We do not have convincing evidence that the interaction predictor is important. 8.

14.46 a

No. The model utility test indicates that all the predictors together provide a useful model. The fact that the t ratio values are small indicates that any one of the predictors could be dropped from the model as long as the other predictors are retained. 1. The model is y    1 x1  2 x2  e , with variables as defined in the question. 2. H0: 1   2  0 3. Ha: At least one of the i ’s is not zero. 4.   0.05 R2 k 5. F  (1  R 2 )  n  (k  1) 

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6. Since we do not have the original data set we are unable to check the conditions. We need to assume that the variables are related according to the model given above, and that the random deviations, e, are normally distributed with mean zero and fixed standard deviation. 0.964 2 7. F   334.722 (1  0.964) 25 8. P-value  P( F2,25  334.722)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that the multiple regression model is useful. b

Since we do not have the original data set we are unable to check the conditions. We need to assume that the variables are related according to the model given in Part (a), and that the random deviations, e, are normally distributed with mean zero and fixed standard deviation. df  n  ( k  1)  28  3  25. The required confidence interval is b2  (t critical value) sb2  0.33563  (2.060)(0.01814)  (0.298, 0.373). We are 95% confident that, when all the other predictors are fixed, the average decrease in y associated with a one-unit increase in x2 is between 0.298 and 0.373.

1 is the mean increase in y for a one unit increase x1 when all the other predictors

are held fixed. 2. H0: 1  0 3. Ha: 1  0 4.   0.05 b 5. t  1 sb1 6. We need to assume that the mean value of y is related to the predictors according to the multiple regression model, and that the random deviations from the values given by the population regression function are normally distributed with mean zero and fixed standard deviation. 7. t  6.95 8. P-value  0.000 9. Since P-value  0.000  0.05 we reject H0. We have convincing evidence that 1  0 . ii

 2 is the mean increase in y for a one unit increase x2 when all the other predictors

are held fixed. 2. H0: 2  0 3. Ha: 2  0 4.   0.05 b 5. t  2 sb2 6. We need to assume that the mean value of y is related to the predictors according to the multiple regression model, and that the random deviations from the values given

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Chapter 14: Multiple Regression Analysis by the population regression function are normally distributed with mean zero and fixed standard deviation. 7. t  18.51 8. P-value  0.000 9. Since P-value  0.000  0.05 we reject H0. We have convincing evidence that 2  0 . d

Yes. Since we have convincing evidence that 1  0 and that 2  0 , we conclude that both independent variables are important.

We need to assume that the mean value of y is related to the predictors according to the multiple regression model, and that the random deviations from the values given by the population regression function are normally distributed with mean zero and fixed standard deviation. When x1  11.7 and x2  57, yˆ  19.440  1.4423(11.7)  0.33563(57)  55.446. df  n  ( k  1)  28  3  25. The required confidence interval is yˆ  (t critical value) s yˆ  55.446  (1.708)(0.522)  (54.554, 56.337). We are 90% confident that the mean water absorption for wheat with 11.7% protein and a starch damage of 57 is between 54.554 and 56.337.

yˆ  (t critical value) se2  s 2yˆ  55.446  (1.708) 1.0942  0.522 2 )  (53.375, 57.516). Our estimate of the interval containing 90% of water absorption values when the predictors have the given values is 53.375 to 57.516. 14.47 a

Referring to the solution to the earlier question, the P-value associated with the coefficient of length is 0.002, which is small, and so length should not be eliminated. The P-value associated with the coefficient of age is 0.904, which is large, and so age could be eliminated from the model. Let the indicator variable for year caught take the value 0 for Year 1 and 1 for Year 2. The MINITAB output for the required multiple regression model is shown below. The regression equation is Weight = - 1516 + 5.71 Length + 2.48 Age - 238 Year Predictor Constant Length Age

Coef -1515.8 5.7149 2.479

SE Coef 290.8 0.7972 5.925

T -5.21 7.17 0.42

P 0.000 0.000 0.683

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-237.95

S = 61.2044

52.96

R-Sq = 84.1%

-4.49

440

0.001

R-Sq(adj) = 80.4%

Analysis of Variance Source Regression Residual Error Total

DF 3 13 16

SS 256997 48698 305695

MS 85666 3746

F 22.87

P 0.000

 3 is the increase in weight associated with a one-unit increase in year, when all the other predictors are fixed. 2. H0: 3  0 1.

3. Ha: 3  0 4.   0.05 b 5. t  3 sb3 6. A normal probability plot of the standardized residuals is shown below. Standardized Residual 4 3 2 1 0 -1 -2 -2

-1

0 Normal Score

Apart from the first and last points, the plot is quite straight, and so the assumption of normality of the random error distribution is acceptable. 7. t  4.49 8. P-value  0.001 9. Since P-value  0.001  0.05 we reject H0. We have convincing evidence that year is a useful predictor given that length and age are included in the model. 14.48 a

1. The model is y    1 x1  2 x2  e , with variables as defined in the question. 2. H0: 1   2  0 3. Ha: At least one of the i ’s is not zero. 4.   0.1

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SSRegr k SSResid  n  (k  1)  6. Since we do not have the original data set we are unable to check the conditions. We need to assume that the variables are related according to the model given above, and that the random deviations, e, are normally distributed with mean zero and fixed standard deviation. 649.75 2 7. F   9.057 538.03 15 5.

F

8. P-value  P( F2,15  9.057)  0.003 9. Since P-value  0.003  0.1 we reject H0. We have convincing evidence that the multiple regression model is useful. b

We need to assume that maximum heart rate is related to the predictors according to the multiple regression model, and that the random deviations from the values given by the regression function are normally distributed with mean zero and fixed standard deviation. df  n  ( k  1)  18  3  15. The required confidence interval is b1  (t critical value) sb1  0.8  (2.131)(0.280)  ( 1.396, 0.203). We are 95% confident that, when all the other predictors are fixed, the average decrease in maximum heart rate associated with a one-year increase in age is between 0.203 and 1.396.

SSResid 538.03   5.989. n  ( k  1) 15 The required prediction interval is se 

yˆ  (t critical value) se2  s 2yˆ  177.1  (2.947) 5.989 2  3.52 2  (156.630, 197.570). Our estimate of the interval containing 99% of maximum heart rates for marathon runners with the given predictor values is 156.630 to 197.570. d

We need to assume that maximum heart rate is related to the predictors according to the multiple regression model, and that the random deviations from the values given by the regression function are normally distributed with mean zero and fixed standard deviation. When x1  30 and x2  77.2, yˆ  179  0.8(30)  0.5(77.2)  193.6. df  n  ( k  1)  18  3  15. The required confidence interval is yˆ  (t critical value)s yˆ  193.6  (1.753)(2.97)  (188.393, 198.807). We are 90% confident that the mean maximum heart rate for marathon runners with the given predictor values is between 188.393 and 198.807

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14.49 a

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The prediction interval will be wider than the interval computed in Part (d), since the prediction interval is for the actual value (rather than the mean value) of the maximum heart rate for a runner who has the given predictor values, and therefore the prediction interval has to take into account the fact that the actual maximum heart rate value will have some deviation from the mean maximum heart rate value. For the model utility test, F 

SSRegr k 237.520 2   30.812 . SSResid  n  (k  1)  26.980 7

So P-value  P( F2,7  30.812)  0 . Since this P-value is small, we have convincing evidence that the quadratic model is useful. b

The t ratios associated with the coefficients of both of these predictors are large, and so neither could be eliminated from the model.

We need to assume that plant height is related to the predictors according to the multiple regression model, and that the random deviations from the values given by the regression function are normally distributed with mean zero and fixed standard deviation. When x  2, yˆ  41.7422  6.581(2)  2.3621(4)  45.4558. df  n  ( k  1)  7. The required confidence interval is yˆ  (t critical value)s yˆ  45.4558  (2.365)(1.037)  (43.004, 47.908). We are 95% confident that the mean plant height when x  2 is between 43.004 and 47.908.

We need to assume that plant height is related to the predictors according to the multiple regression model, and that the random deviations from the values given by the regression function are normally distributed with mean zero and fixed standard deviation. When x  1, yˆ  41.7422  6.581(1)  2.3621(1)  45.9611. df  n  ( k  1)  7. The required confidence interval is yˆ  (t critical value) s yˆ  45.9611  (1.895)(1.031)  (44.008, 47.914). We are 90% confident that the mean plant height when the wheat is treated with 10 M of Mn is between 44.008 and 47.914.

14.50 a

The MINITAB output is shown below. The regression equation is y = 110 - 2.29 x + 0.0329 x^2 Predictor Constant x x^2

Coef 109.771 -2.2943 0.032857

S = 1.35225

SE Coef 1.273 0.1508 0.003614

R-Sq = 99.7%

T 86.26 -15.22 9.09

P 0.000 0.004 0.012

R-Sq(adj) = 99.3%

Analysis of Variance Source

443

Chapter 14: Multiple Regression Analysis Regression Residual Error Total

2 2 4

1111.54 3.66 1115.20

555.77 1.83

303.94

0.003

The estimated regression equation is yˆ  109.771  2.2943x  0.032857 x2 , where x = distance from fertilizer band and y = plant height. b

1. The model is y    1 x   2 x 2  e , with variables as defined in the question. 2. H0: 1   2  0 3. Ha: At least one of the i ’s is not zero. 4.   0.05 SSRegr k 5. F  SSResid  n  (k  1)  6. Since we do not have the original data set we are unable to check the conditions. We need to assume that the variables are related according to the model given above, and that the random deviations, e, are normally distributed with mean zero and fixed standard deviation. 7. F  303.94 8. P-value  P( F2,2  303.94)  0.003 9. Since P-value  0.003  0.05 we reject H0. We have convincing evidence that the quadratic regression model is useful.

R 2  0.997 . This is the proportion of the observed variation in plant Mn that can be explained by the fitted model. se  1.352. This is a typical deviation of a plant Mn value in the sample from the value predicted by the estimated regression equation.

1 is the coefficient of x in the population regression function relating y to x and x 2 ,

with variables as defined in the question. 2. H0: 1  0 3. Ha: 1  0 4.   0.05 b 5. t  1 sb1 6. A normal probability plot of the standardized residuals is shown below.

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Standardized residual 1.5 1.0

0.5 0.0 -0.5

-1.0

-1.5

-1.0

-0.5

0.0 Normal Score

0.5

1.0

The plot is quite straight, indicating that normality of the random error distribution is plausible. 7. t  15.22 8. P-value  0.004 9. Since P-value  0.004  0.05 we reject H0. We have convincing evidence that the linear predictor is important. 1.

 2 is the coefficient of x 2 in the population regression function relating y to x and x 2 ,

with variables as defined in the question. 2. H0: 2  0 3. Ha: 2  0 4.   0.05 b 5. t  2 sb2 6. A normal probability plot of the standardized residuals is shown below.

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Chapter 14: Multiple Regression Analysis

Standardized residual 1.5 1.0

0.5 0.0 -0.5

-1.0

-1.5

-1.0

-0.5

0.0 Normal Score

0.5

1.0

The plot is quite straight, indicating that normality of the random error distribution is plausible. 7. t  9.09 8. P-value  0.012 9. Since P-value  0.012  0.05 we reject H0. We have convincing evidence that the quadratic predictor is important. We have convincing evidence that both of the predictors are important. e

When x  30, yˆ  109.771  2.2943(30)  0.032857(30) 2  70.5133. MINITAB tells us that the standard error associated with this estimate is 0.824. df  n  ( k  1)  5  3  2. The required confidence interval is yˆ  (t critical value)s yˆ  70.5133  (2.920)(0.824)  (68.107, 72.919). We are 90% confident that the mean plant Mn for plants that are 30 cm from the fertilizer band is between 68.107 and 72.919.

14.51 It is possible that some sort of stepwise procedure was used. For example, the authors might have started with the model that includes all independent variables, all quadratic terms, and all interaction terms (twenty predictors in all), and then used backward elimination to arrive at the given estimated regression equation. 14.52 The model with the largest value of adjusted R2 is the one with predictors x1 , x2 , x3 , x5 , x6 , x7 , and x8 . So we should consider that model and any others whose adjusted

R2 values are nearly as large. The model with predictors x3 and x5 has an adjusted R2 value that is almost as large as the one mentioned above, but has only two predictors. Therefore, the model with predictors x3 and x5 could be considered the best choice. 14.53 The model with the largest value of adjusted R2 is the one with predictors x3 , x5 , x6 , x9 , and x10 . So we should consider that model and any others whose adjusted R2 values are nearly as large. The model with predictors x3 , x9 , and x10 has an adjusted R2 value that is almost as large as the

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one mentioned above, but has only three predictors. Therefore, the model with predictors x3 , x9 , and x10 could be considered the best choice. 14.54 The first candidate for elimination would be x1 , since the t ratio associated with that variable is the one closest to zero. Since the t ratio for that variable, −0.155, is between −2 and 2, use of the criterion 2  t ratio  2 would lead to elimination of the variable x1 . 14.55 a

1. The model is y    1 x1  2 x2  3 x3  4 x4  5 x5  6 x6  7 x7  8 x8  9 x9  e . 2. H0: 1  2  3  4  5  6  7  8  9  0 3. Ha: At least one of the i ’s is not zero. 4.   0.05

F

8. P-value  P( F9,1846  91.807)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that the multiple regression model is useful. b

14.56 a

The single predictor that resulted in the largest r 2 value was smoking habits. When all of the two-predictor models consisting of smoking habits and one other variable were considered, the one with the largest R2 value was the one consisting of smoking habits and alcohol consumption. Then, when each three-predictor model consisting of these two predictors and one other predictor was considered, it was found that none of these three-predictor models produced an R2 value that was much larger than the one for the model that used smoking habits and alcohol consumption as predictors. The t-ratio (given by bi sbi ) closest to zero is the one for x7 (t  0.333) . Since

2  0.333  2, the predictor would be eliminated. b

14.57 a

No. Since we don’t know the t ratios for the new (eight-predictor) model, it is not possible to tell what the second candidate for elimination would be. Yes. The two regressions show similar patterns, with the intercept and each of the coefficients having the same order of value in the two cases. Each t ratio is given by ti  bi sbi . So, for log of sales in 1992, 6.56  0.372 sb2 , giving

sb2  0.372 6.56  0.057. c

The variable that would be considered first is return on equity, since this is the variable whose t ratio, 0.33, is closest to zero. It would be eliminated because its t ratio is between −2 and 2.

447

Chapter 14: Multiple Regression Analysis d

No. For the 1991 regression the first variable to be considered for elimination would be CEO tenure.

df  n  ( k  1)  160  7  153. P-value  P(t153  1.73)  0.043.

14.58 Yes. Multicollinearity might well be a problem, since there is likely to be a strong correlation between any two of the three predictors given. 14.59 At Step 1, x1 was eliminated, since this variable has the t value closest to zero, and that t value is between −2 and 2. At Step 2, x2 was eliminated, since this variable has the t value closest to zero, and that t value is between −2 and 2. At Step 3, no variable was eliminated, since both of the remaining variables have t values whose magnitudes are greater than 2. 14.60 A relevant MINITAB output is shown below. Response is y Vars 1 1 2 2 3 3 4

R-Sq 82.4 69.8 87.2 85.2 87.9 87.5 87.9

R-Sq(adj) 81.9 69.0 86.5 84.3 86.8 86.4 86.5

Mallows Cp 14.0 48.4 2.9 8.5 3.1 4.1 5.0

S 3.0573 4.0059 2.6450 2.8490 2.6139 2.6522 2.6474

x x x x 1 2 3 4 X X X X X X X X X X X X X X X X

Looking at the adjusted R 2 values we see that the best two-predictor model offers significant improvement over the best one-predictor model. However, the best three-predictor model offers little improvement over the best two-predictor model. Thus we would choose the best twopredictor model, which is the one using x2 and x4 as predictors. This is a different model from the one chosen by the backward elimination method. 14.61 A relevant MINITAB output is shown below. Response is y Vars 1 2 3 4 5

R-Sq 6.7 11.1 22.1 29.3 34.0

R-Sq(adj) 2.6 3.1 11.0 15.1 16.6

Mallows Cp 5.8 6.6 5.4 5.4 6.0

S 12.144 12.116 11.609 11.337 11.239

x x x x x 1 2 3 4 5 X X X X X X X X X X X X X X X

Looking at the adjusted R 2 values we see that the best three-predictor model offers significant improvement over the best one- and two-predictor models. The best four-predictor model offers some improvement over the best three-predictor model, but it is arguable as to whether this improvement is worthwhile when considering the increase in the number of predictors. The best

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five-predictor model does not offer significant improvement over the best four-predictor model. Thus, we would choose either the model with predictors x1 , x3 , and x4 , or the model with predictors x1 , x3 , x4 and x5 . 14.62 a

Since the predictors included in both of the models whose R 2 values we know are included in the model introduced in this part of the question, the R 2 value for the model introduced in this part of the question must be greater than or equal to both of the R 2 values for the models whose R 2 values we know. Thus the required R 2 value is greater than or equal to 0.723.

Since both of the predictors included in the model introduced in this part of the question are included in the model whose R 2 value is 0.723, we know that the required R 2 value is less than or equal to 0.723.

14.63 a

F

R2 / k

1  R  /  n  (k  1)  2



0.64 / 11  12.28 and 1  0.64  / 88  (11  1) 

P-value  P( F11,76  12.28)  0  0.01 . Therefore, H0 is rejected, and we conclude that there is

b c

a useful linear relationship.  n 1   88  1  2 Radj  1  1  R2    1  1  0.64     0.5879  n  (k  1)   88  (11  1) 





To calculate the 95% confidence interval for 1, use b1   t critical value  sb1 . The t critical

value for 95% confidence with df  n  ( k  1)  88  12  76 is 1.99. The t-ratio can be used b 0.458  0.149 . Therefore, the confidence interval is to find sb1  1  t1 3.08

b1   t critical value  sb1 0.458  (1.99)(0.149) 0.458  0.297

 0.161, 0.755 d e

14.64 a

The predictor with a t-ratio closest to zero would be considered first. In this case, we would consider eliminating x9. The t test would fail to reject H 0 : 9  0 . H 0 : 3   4  5   6  0 H a : at least one of the i 's is not zero None of the procedures presented in this chapter could be used. The procedures presented allow us to test for model utility or significance of individual predictors. This null hypothesis has us testing a subset containing four predictors.

F

R2 / k

1  R  /  n  (k  1)  2



0.50 / 6  11.83 and 1  0.50  /  78  (6  1) 

P-value  P( F6,71  11.83)  0, which is smaller than any reasonable significance level.

Therefore, H0 is rejected, and we conclude that there is a useful linear relationship.  n 1   78  1  2 Radj  1  1  R2    1  1  0.50     0.458  n  (k  1)   78  (6  1) 





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Chapter 14: Multiple Regression Analysis

Carry out the test H 0 : 6  0 versus H a : 6  0 . The t-ratio is given as t = 0.92, and df  n  ( k  1)  78  7  71 . The P-value  2  P(t71  0.92)  0.36 , which results in a failure to reject the null hypothesis. We conclude that x6 is not useful when the other five predictors remain in the model.

14.65 a

Glucose Concentration (g/L)

50 0

Fermentation Time (days)

Based on the scatterplot, a quadratic model seems appropriate. R2 / k 0.895 / 2 F   21.3 and 2 1  0.895  / 8  (2  1)  1  R /  n  (k  1)  





P-value  P( F2,5  21.3)  0.004 which is smaller than any reasonable significance level.

Therefore, H0 is rejected, and we conclude that the quadratic model specifies a useful relationship between x and y. Carry out the test H 0 :  x2  0 versus H a :  x2  0 . The t-ratio is given as t = 6.52, and

df  n  (k  1)  8  3  5 . The P-value  2  P(t5  21.3)  0 , which results in rejecting the null hypothesis. We conclude that the quadratic term should not be eliminated from the model. 14.66 a

F

R2 / k

1  R  /  n  (k  1)  2



0.69 / 4  15.58 and 1  0.69  /  33  (4  1) 

P-value  P( F4,28  15.58)  0, therefore H0 is rejected. We conclude that there is a useful b

linear relationship with that model. The predicted base salary is yˆ  2.60  0.125(50)  0.893(1)  0.057(8)  0.014(12)  10.031 .

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450

The table below shows the coefficients, standard deviations, and t-ratios for the four b predictors. The t-ratios are calculated using t  i . sbi x1 x2 x3 x4

sbi

t-ratio

0.125 0.893 0.057 -0.014

0.064 0.141 0.014 0.005

1.95 6.33 4.07 -2.80

The t-ratio closest to zero is 1.95, which indicates that the first variable to consider for elimination is x1. With df  n  ( k  1)  33  5  28 , the P-value  2  P(t28  1.95)  0.06 ,

which results in failing to reject the null hypothesis of H 0 : 1  0 at the 0.05 significance level. We conclude that this variable can be eliminated from the model. To calculate the 95% confidence interval for 2, use b2   t critical value  sb2 . The t critical value for 95% confidence with df  n  ( k  1)  33  5  28 is 2.05. Therefore, the confidence interval is b2   t critical value  sb2

0.893  (2.05)(0.141) 0.893  0.289

 0.604, 1.182  2 14.67 Given that n = 21 and k = 10, use the relationship between R 2 and Radj :

 n 1  2 Radj  1  1  R2    n  (k  1)   21  1   1  1  R2    21  (10  1) 









 20   1  1  R2    10 

   1  1  R   2  2

2 2  0 and solving for R 2 yields a cutoff between positive and negative values of Radj Set Radj .

  0  1  1  R   2  0  1   2  2R 

2 Radj  1  1  R2  2 2

0  1  2  2R2 0  1  2 R 2 1  2R2 0.5  R 2 2 Values of R 2 less than 0.5 will yield negative values of Radj .

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14.68 a

First, the model using all five variables was fit. The variable body mass (x3) was deleted because it had a t-ratio closest to zero and it was between –2 and 2. Then, the model with the remaining four variables was fit. The variable body fat (x4) was deleted because its t-ratio was closest to zero and was between –2 and 2. Finally, the model using the three variables age (x1), sex (x2), and lean body mass (x5) was fit. None of these variables could be eliminated because their t-ratios were greater than 2 in absolute magnitude. The final model includes age, sex, and lean body mass. R2  0.786; Approximately 78.6% of the observed variation in total body electrical conductivity can be explained by the fitted model. se  4.914; This is a typical deviation of a total body electrical conductivity value in the sample from the value predicted by the fitted model. For women, the value of x2 is 1 and it is 0 for men. Thus, if all the other variables are unchanged, the estimated mean total body electrical conductivity for men is 6.988 more than for women. To calculate the 95% confidence interval for the mean total body electrical conductivity, use yˆ   t critical value  s yˆ . The t critical value for 95% confidence with

df  n  ( k  1)  19  4  15 is 2.13. The predicted total body electrical conductivity of all 31-year-old females whose lean body mass is 52.7 kg is yˆ  15.175  0.3771(31)  6.988(1)  0.37779(52.7)  9.43663 . Therefore, the confidence interval is yˆ   t critical value  s yˆ

9.43663  (2.13)(1.42) 9.43663  3.0246

 6.412, 12.461 14.69

First, the model using all four variables was fit. The variable age at loading (x3) was deleted because it had the t-ratio closest to zero and it was between –2 and 2. Then, the model using the three variables x1, x2, and x4 was fit. The variable time (x4) was deleted because its t-ratio was closest to zero and was between –2 and 2. Finally, the model using the two variables x1 and x2 was fit. Neither of these variables could be eliminated since their t-ratios were greater than 2 in absolute magnitude. The final model, then, includes slab thickness (x1) and load (x2). The predicted tensile strength for a slab that is 25 cm thick, 150 days old, and is subjected to a load of 200 kg for 50 days is yˆ  13  0.487(25)  0.0116(200)  3.145 MPa.

14.70 a

b c d

From the Minitab output, the F statistic is 8.63, and the associated P-value is 0.001. Therefore, the null hypothesis is rejected. The usefulness of the multiple regression model is confirmed. The variable x3 (Fetus weight) can be eliminated because the t-ratio of 0.03 is close to zero, and also lies between –2 and 2. The variable x1 (Sex) can be eliminated because the t-ratio of –1.12 is the closest to zero, and also lies between –2 and 2. R2  0.527; Approximately 52.7% of the observed variation in fetus progesterone level can be explained by the fitted model.

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se  4.116; This is a typical deviation of a fetal progesterone level value in the sample from the value predicted by the fitted model. For each 1-cm increase in fetal length, the fetal progesterone level is predicted to increase by 0.231 mg.

14.71 a 60 50

Percentage Hatch

30 20

0 -10 10

Water Temperature

b 14.72 a

The claim is reasonable because 14 is close to where the curve has its maximum value.

F

R2 / k

1  R  /  n  (k  1)  2



0.9 / 15  2.4 . This F statistic value is less than 1  0.9   /  20  (15  1) 

the critical value of 5.86 that captures the upper-tail area 0.05 under the F curve. Therefore, the model does not appear to specify a useful relationship between y and the predictor variables. Alternatively, the P-value  P( F15,4  2.4)  0.21  0.05, therefore H0 is not

b c

rejected. We conclude that there is not a useful relationship between y and the predictor variables. The high R2 value does not indicate that the model is useful. Having a large number of predictor variables might make one suspicious of a model with a high R2 value. R2 must be at least 0.9565. To algebraically determine how large R2 must be for the model to be judged useful at the 0.05 significance level, F must be at least 5.86. Setting F equal to 5.86 and solving the formula in part (a) for R2, we can calculate how large R2 must be. The algebraic details are worked out below.

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Chapter 14: Multiple Regression Analysis

F

R2 / k

R 2 / 15



1  R  /  n  (k  1)  1  R  / 4 2

5.86  5.86 

R 2 / 15

1  R  / 4 2

R2 4  15 1  R 2





15(5.86) R2  4 1  R2 4 1  R2  15(5.86) R2 1 0.045506  2  1 R 1 1.045506  2 R 1  R2 1.045506 0.956474  R 2

d e

yˆ  1.5645  0.2372 x1  0.000249 x2 Using a statistical software package, F = 70.66, with an associated P-value of 0.000. Reject H0 and conclude that the profit margin model using net revenues and number of branch offices as predictors is useful. R2  0.865; Approximately 86.5% of the total variation in the observed values for profit margin has been explained by the fitted regression equation. se  0.0533; This is a typical deviation of a profit margin value in the sample from the value predicted by the fitted model. No. Both variables have associated t-ratios that exceed 2 in absolute magnitude and so neither can be eliminated from the model. There do not appear to be any influential observations. However, there is substantial evidence of multicollinearity. The plot shows a strong relationship between x1 and x2. This is evidence of multicollinearity between x1 and x2. 9500

9000

Number of Branch Offices (x2)

14.73 a b

8500

8000

7500 7000

6500

6000 3.0

3.5

4.0

Net revenues (x1)

4.5

5.0

Chapter 15 Analysis of Variance Note: In this chapter, numerical answers to questions involving the normal, t, chi square, and F distributions were found using values from a calculator. Students using statistical tables will find that their answers differ slightly from those given. 15.1

15.2

15.3

P-value  P( F4,15  5.37)  0.007.

P-value  P( F4,15  1.90)  0.163.

P-value  P( F4,15  4.89)  0.010.

P-value  P( F3,20  14.48)  0.000.

P-value  P( F3,20  2.69)  0.074.

P-value  P( F4,50  3.24)  0.019.

df1  k  1  4, df2  N  k  20  5  15. P-value  P( F4,15  5.37)  0.007.

df1  k  1  4, df 2  N  k  23  5  18. P-value  P( F4,18  2.83)  0.055.

df1  k  1  2, df 2  N  k  15  3  12. P-value  P( F2,12  5.02)  0.026.

df1  k  1  2, df 2  N  k  14  3  11. P-value  P( F2,11  15.90)  0.001.

df1  k  1  3, df 2  N  k  52  4  48. P-value  P( F3,48  1.75)  0.169.

Let 1 , 2 , 3 , 4 be the mean lengths of stay for people participating in the four health plans. H0: 1  2  3  4 Ha: At least two among 1 , 2 , 3 , 4 are different.

df1  k  1  3, df 2  N  k  32  4  28. P-value  P( F3,28  4.37)  0.012. Since P-value  0.012  0.01 we do not reject H0. We do not have convincing evidence that mean length of stay is related to health plan.

454

Chapter 15: Analysis of Variance

15.4

455

1. Let 1 , 2 , 3 , 4 be the mean social marginality scores for the four given populations. 2. H0: 1  2  3  4 3. Ha: At least two among 1 , 2 , 3 , 4 are different. 4.   0.01 MSTr 5. F  MSE 6. We are told to regard the four samples as representative of their respective populations, and so it is reasonable to treat the samples as random samples from those populations. Also, we are told to assume that the population social marginality distributions are approximately normal with the same standard deviation. 7. N  106  255  314  36  711 Grand total  106(2.00)  255(3.4)  314(3.07)  36(2.84)  2145.22

2145.22  3.017 711 SSTr  n1 ( x1  x ) 2  n2 ( x2  x ) 2  n3 ( x3  x ) 2  n4 ( x4  x ) 2 x

 106(2.00  3.017)2  255(3.40  3.017) 2  314(3.07  3.017)2  36(2.84  3.017)2  149.050 Treatment df  k  1  3 SSE  ( n1  1) s12  (n2  1) s22  (n3  1) s32  (n4  1) s42  105(1.56)2  254(1.68)2  313(1.66)2  35(1.89)2  1959.9439 Error df  N  k  711  4  707 MSTr SSTr treatment df 149.050 3 F    17.922 MSE SSE error df 1959.9439 707 8. P-value  P( F3,707  17.922)  0 9. Since P-value  0  0.01 we reject H0. We have convincing evidence that the mean social marginality score is not the same for all four age groups. 15.5

Summary statistics are given in the table below.

n x s s2

7+ label 10 4.8 2.098 4.4

12+ label 16+ label 10 10 6.8 7.1 1.619 1.524 2.622 2.322

18+ label 10 8.1 1.449 2.1

1. Let 1 , 2 , 3 , 4 be the mean ratings for the four restrictive rating labels. 2. H0: 1  2  3  4 3. Ha: At least two among 1 , 2 , 3 , 4 are different.

456

Chapter 15: Analysis of Variance 4. 5.

  0.05 MSTr F MSE

6. Boxplots for the four groups are shown below.

7+ label

12+ label

16+ label

18+ label 0

Rating

The boxplots are close enough to being symmetrical, and there are no outliers. The largest standard deviation (2.098) is not more than twice the smallest (1.449). We are told to assume that the boys were randomly assigned to the four age label ratings. 7. N  10  10  10  10  40 Grand total  10(4.8)  10(6.8)  10(7.1)  10(8.1)  268

268  6.7 40 SSTr  n1 ( x1  x ) 2  n2 ( x2  x ) 2  n3 ( x3  x ) 2  n4 ( x4  x ) 2 x

 10(4.8  6.7) 2  10(6.8  6.7) 2  10(7.1  6.7) 2  10(8.1  6.7) 2  57.4 Treatment df  k  1  3

SSE  (n1  1) s12  ( n2  1) s22  ( n3  1) s32  ( n4  1) s42  9(4.4)  9(2.622)  9(2.322)  9(2.1)  103 Error df  N  k  40  4  36 MSTr SSTr treatment df 57.4 3 F    6.687 MSE SSE error df 103 36 8. P-value  P( F3,36  6.687)  0.001 9. Since P-value  0.001  0.05 we reject H0. We have convincing evidence that the mean ratings for the four restrictive rating labels are not all equal. 15.6

Summary statistics are shown in the table below.

n x s s2

7+ label 10 5.5 1.958 3.833

12+ label 16+ label 10 10 5.2 6.7 1.989 1.767 3.956 3.122

18+ label 10 6.7 1.703 2.900

Chapter 15: Analysis of Variance

457

1. Let 1 , 2 , 3 , 4 be the mean ratings for the four restrictive rating labels. 2. H0: 1  2  3  4 3. Ha: At least two among 1 , 2 , 3 , 4 are different. 4.   0.05 MSTr 5. F  MSE 6. Boxplots for the four groups are shown below.

7+ label

12+ label

16+ label

18+ label 3

Rating

There are hints at skewness in two of the boxplots; however the distributions shown are consistent with normal population distributions for group sizes this small. There is one outlier, but we will nonetheless proceed with the test. The largest standard deviation (1.989) is not more than twice the smallest (1.703). We need to assume that the girls were randomly assigned to the four age label ratings. 7. N  10  10  10  10  40 Grand total  10(5.5)  10(5.2)  10(6.7)  10(6.7)  241

241  6.025 40 SSTr  n1 ( x1  x )2  n2 ( x2  x ) 2  n3 ( x3  x ) 2  n4 ( x4  x ) 2 x

 18.675 Treatment df  k  1  3 SSE  ( n1  1) s12  ( n2  1) s22  ( n3  1) s32  ( n4  1) s42  124.3 Error df  N  k  40  4  36 MSTr SSTr treatment df 18.675 3 F    1.803 MSE SSE error df 124.3 36 8. P-value  P( F3,36  1.803)  0.164 9. Since P-value  0.164  0.05 we do not reject H0. We do not have convincing evidence that the mean ratings for the four restrictive rating labels are not all equal. 15.7

1. Let 1 , 2 , 3 be the mean hunger rating for the three different treatments (“healthy” snack, “tasty” snack, no snack) 2. H0: 1  2  3 3. Ha: At least two among 1 , 2 , 3 are different.

458

Chapter 15: Analysis of Variance 4. 5.

  0.05 MSTr F MSE

6. We are told that the volunteers were randomly assigned to the one of the three treatments. The largest sample standard deviation ( s2  1.581) is not more than twice as large as the smallest sample standard deviation ( s3  1.167) , so the equal population standard deviations condition is reasonably met. Boxplots (shown below) are roughly symmetric and contain no outliers, so we can assume that the normality condition is also reasonably met.

Healthy

Tasty

No Snack

4 Hunger Rating

N  9  9  9  27 Grand total  9(5.2)  9(3.3)  9(4.1)  113.4 113.4 x  4.2 27

SSTr  n1  x1  x   n2  x2  x   n3  x3  x  2

 9(5.2  4.2) 2  9(3.3  4.2) 2  9(4.1  4.2) 2  16.38 Treatment df  k  1  2

SSE  (n1  1) s12  (n2  1) s22  (n3  1) s32  (9  1)(1.562 )  (9  1)(1.582 )  (9  1)(1.17 2 )

 50.3912 Error df  N  k  27  3  24 MSTr SSTr / treatment df 16.38 / 2 F    3.86 MSE SSE / error df 50.3912 / 24 P-value  P( F2,24  3.86)  0.035

9. Since P-value  0.035  0.05 we reject H0. We have convincing evidence that the mean hunger rating is not the same for all three treatments (“healthy” snack, “tasty” snack, no snack).

Chapter 15: Analysis of Variance b

459

It is not reasonable to conclude that the mean hunger rating is greater for people who do not get a snack because the sample mean hunger rating is highest for the healthy snack group.

15.8

1. Let 1 , 2 , 3 , 4 be the mean test score for the four different treatments. 2. H0: 1  2  3  4 3. Ha: At least two among 1 , 2 , 3 , 4 are different. 4.   0.05 MSTr 5. F  MSE 6. We are told to assume that the boys were randomly assigned to the one of the four treatments. Given that we don’t have sufficient detail to check the remaining conditions, we will assume all the conditions have been satisfied. 7. Test Statistic: F  2.39 8. P-value = 0.077 9. Since P-value  0.077  0.05 we do not reject H0. We do not have convincing evidence to conclude that the mean test score is not the same for the four different treatments.

15.9

Summary statistics are shown in the table below.

n x s s2

Treatment 1 18 5.778 4.081 16.654

Treatment 2 25 6.480 3.441 11.843

Treatment 3 17 3.529 2.401 5.765

Treatment 4 14 2.929 2.129 4.533

1. Let 1 , 2 , 3 , 4 be the mean numbers of pretzels consumed for the four treatments. 2. H0: 1  2  3  4 3. Ha: At least two among 1 , 2 , 3 , 4 are different. 4.   0.05 MSTr 5. F  MSE 6. Boxplots for the four groups are shown below.

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Chapter 15: Analysis of Variance

Treatment 1

Treatment 2

Treatment 3

Treatment 4 0

4 6 8 10 Number of Pretzels Consumed

The boxplots are roughly symmetric, and there are no outliers. The largest standard deviation (4.081) is not more than twice the smallest (2.129). We are told that the men were randomly assigned to the four treatments. 7. N  n1  n2  n3  n4  74 Grand total  n1 x1  n2 x2  n3 x3  n4 x4  367

grand total  4.959 N SSTr  n1 ( x1  x )2  n2 ( x2  x ) 2  n3 ( x3  x ) 2  n4 ( x4  x ) 2 x

 162.363 Treatment df  k  1  3 SSE  ( n1  1) s12  ( n2  1) s22  ( n3  1) s32  ( n4  1) s42  718.515 Error df  N  k  70 MSTr SSTr treatment df F   5.273 MSE SSE error df 8. P-value  P( F3,70  5.273)  0.002 9. Since P-value  0.002  0.05 we reject H0. We have convincing evidence that the mean numbers of pretzels consumed for the four treatments are not all equal. 15.10 The summary statistics are given in the table below.

n x s s2

1 39 6.31 3.75 14.0625

2 42 3.31 3.06 9.3636

Semester 3 32 1.79 3.25 10.5625

4 32 1.83 3.13 9.7969

5 34 1.5 2.37 5.6169

Chapter 15: Analysis of Variance

461

1. Let 1 , 2 , 3 , 4 , 5 be the mean percentages of words that are plagiarized for the five semesters. 2. H0: 1  2  3  4  5 3. Ha: At least two among 1 , 2 , 3 , 4 , 5 are different. 4.   0.05 MSTr 5. F  MSE 6. We are told to assume that the conditions necessary for the ANOVA F test are reasonable. 7. N  n1  n2  n3  n4  n5  179 Grand total  n1 x1  n2 x2  n3 x3  n4 x4  n5 x5  551.95 grand total x  3.084 N SSTr  n1 ( x1  x )2  n2 ( x2  x ) 2  n3 ( x3  x ) 2  n4 ( x4  x ) 2

 597.231 Treatment df  k  1  4 SSE  ( n1  1) s12  ( n2  1) s22  ( n3  1) s32  ( n4  1) s42  1734.782 Error df  N  k  174 MSTr SSTr treatment df F   14.976 MSE SSE error df 8. P-value  P( F4,174  14.976)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that the mean percentages of words that are plagiarized are not the same for all five semesters. 15.11 a

Random assignment ensures that our experiment does not systematically favor one experimental condition over any other and attempts to create experimental groups that are as much alike as possible.

1. Let 1 , 2 , 3 be the mean average speed for the three different treatment groups. 2. H0: 1  2  3 3. Ha: At least two among 1 , 2 , 3 are different. 4.   0.05 MSTr 5. F  MSE 6. The subjects were randomly assigned to the three treatment groups. Additionally, the largest standard deviation (17.06) is not more than twice the smallest standard deviation (11.42). It appears as if the necessary conditions have been satisfied. 7. Test Statistic: F  2.44 8. P-value = 0.094 9. Since P-value  0.094  0.05 we fail to reject H0. We do not have convincing evidence to support the claim that the mean average speed is not the same for all three treatments (Stretching Method 1, Stretching Method 2, and No Stretching).

The authors were likely surprised by the results of this study because they did not find evidence that supports a difference in mean average speed for the three treatments (the null

462

Chapter 15: Analysis of Variance hypothesis was not rejected). This means that there is not sufficient evidence that stretching improves performance. These results contradict the results of other studies, which concluded that stretching can improve performance.

15.12 1. Let 1 , 2 , 3 be the mean Hopkins scores for the three populations. 2. H0: 1  2  3 3. Ha: At least two among 1 , 2 , 3 are different. 4.   0.05 MSTr 5. F  MSE 6. We are told to treat the samples as random samples from their respective populations. We have to assume that the population Hopkins score distributions are approximately normal with the same standard deviation. 7. N  n1  n2  n3  234

Grand total  n1 x1  n2 x2  n3 x3  7064.66 grand total x  30.191 N SSTr  n1 ( x1  x )2  n2 ( x2  x )2  n3 ( x3  x ) 2  100.786 Treatment df  k  1  2 SSE  ( n1  1) s12  ( n2  1) s22  ( n3  1) s32  ( n4  1) s42  4409.036 Error df  N  k  231 MSTr SSTr treatment df F   2.640 MSE SSE error df 8. P-value  P( F2,231  2.640)  0.074 9. Since P-value  0.074  0.05 we do not reject H0. We do not have convincing evidence that the mean Hopkins scores are not the same for all three student populations. 15.13

k  4, N  20. Treatment df  k  1  3. Error df  N  k  16. SSTr  SSTo  SSE  310500.76  235419.04  75081.72. The completed table is shown below. Source of Variation Treatments Error Total

df 3 16 19

Sum of Squares 75081.72 235419.04 310500.76

Mean Square 25027.24 14713.69

F 1.701

1. Let 1 , 2 , 3 , 4 be the mean number of miles until failure for the four given brands of spark plug. 2. H0: 1  2  3  4 3. Ha: At least two among 1 , 2 , 3 , 4 are different. 4.   0.05

Chapter 15: Analysis of Variance

5. 6.

7. 8. 9.

463

MSTr MSE We need to treat the samples as random samples from their respective populations, and assume that the population distributions are approximately normal with the same standard deviation. F  1.701 P-value  P( F3,16  1.701)  0.207 Since P-value  0.207  0.05 we do not reject H0. We do not have convincing evidence that the mean number of miles to failure is not the same for all four brands of spark plug. F

15.14 Since the interval for 2  3 is the only one that contains zero, we have evidence of a difference between 1 and 2 , and between 1 and 3 , but not between 2 and 3 . Thus, statement c is the correct choice. 15.15 Since there is a significant difference in all three of the pairs we need a set of intervals none of which includes zero. Set 3 is therefore the required set. 15.16 In increasing order of the resulting mean numbers of pretzels eaten the treatments were: slides with related text, slides with no text, no slides, and slides with unrelated text. There were no significant differences between the results for slides with related text and slides with no text, or for no slides and slides with unrelated text. However, there was a significant difference between the mean numbers of pretzels eaten for no slides and slides with no text (and also between the results for no slides and slides with related text). Likewise, there was a significant difference between the mean numbers of pretzels eaten for slides with unrelated text and slides with no text (and also between the results for slides with unrelated text and slides with related text). 15.17 a

In decreasing order of the resulting mean numbers of pretzels eaten the treatments were: slides with related text, slides with no text, slides with unrelated text, and no slides. There were no significant differences between the results for slides with no text and slides with unrated text, and for slides with unrelated text and no slides. However there was a significant difference between the results for slides with related text and each one of the other treatments, and between the results for no slides and for slides with no text (and for slides with related text).

The results for the women and men are almost exactly the reverse of one another, with, for example, slides with related text (treatment 2) resulting in the smallest mean number of pretzels eaten for the women and the largest mean number of pretzels eaten for the men. For the men, treatment 2 was significantly different from all the other treatments; however for women treatment 2 was not significantly different from treatment 1. For both women and men there was a significant difference between treatments 1 and 4 and no significant difference between treatments 3 and 4. However, between treatments 1 and 3 there was a significant difference for the women but no significant difference for the men.

15.18 a

The sample sizes and means are given in the table below.

n x

7+ label 10 4.8

12+ label 16+ label 18+ label 10 10 10 6.8 7.1 8.1

464

Chapter 15: Analysis of Variance As calculated in Exercise 15.5, MSE = 103/36 = 2.861 and Error df = 36. Thus the T-K interval for i   j (using Error df = 40 in Appendix Table 7) is

xi  x j  q

MSE  1 1  MSE 2.861 .  xi  x j  3.79     xi  x j  q 2  n1 n2  n 10

So the intervals are as shown below.

Sample mean

Difference 1  2

Interval (−4.027, 0.027)

Includes 0? Yes

1  3 1  4  2  3 2  4 3   4

(−4.327, −0.273)

(−5.327, −1.273) (−2.327, 1.727)

No Yes

(−3.327, 0.727)

Yes

(−3.027, 1.027)

Yes

7+ label 4.8

12+ label 6.8

16+ label 7.1

18+ label 8.1

Certainly the more restrictive the age label on the video game the higher the sample mean rating given by the boys used in the experiment. However, according to the T-K intervals the only significant differences were between the means for the 7+ label and the 16+ label, and between the means for the 7+ label and the 18+ label.

15.19 a

1. Let 1 , 2 , 3 be the mean gender stereotyping score for the three different treatment groups 2. H0: 1  2  3 3. Ha: At least two among 1 , 2 , 3 are different. 4.   0.05 MSTr 5. F  MSE 6. We are told that the study subjects were randomly assigned to the three treatment groups. The largest sample standard deviation ( s2  0.800) is not more than twice as large as the smallest sample standard deviation ( s3  0.580) , so the equal population standard deviations condition is reasonably met. We will assume that the distributions of gender stereotyping scores for the three different treatment groups are approximately normal. Therefore, the necessary conditions have been satisfied. 7. Treatment df = k – 1 = 3 – 1 = 2 Error df = N – k = 234 – 3 = 231 Test Statistic: F  3.34 8. P-value = 0.041 9. Since P-value  0.041  0.05 we reject H0. We have convincing evidence that the mean gender stereotyping scores for the three different treatment groups differ.

Chapter 15: Analysis of Variance

465

The mean measure of gender stereotyping is significantly different for the Control group and the Spider Man group. The mean for X-Men was not significantly different from either of the other two groups. Because lower values indicate attitudes more accepting of equality, you can conclude that those who did not watch a video were more accepting of equality than those who watched Spider Man.

15.20 The sample means for the five different fabrics are: 16.350, 11.633, 10.500, 14.960, 12.300. Therefore, the fabric types, listed in order of their sample means, are: 3, 2, 5, 4, 1. The significant differences are between 1 and 2 , 1 and 3 , 1 and 5 , 2 and 4 , 3 and 4 , 4 and 5 . The underscoring pattern is shown below.

Sample mean 15.21 a

Fabric 3 10.5

Fabric 2 11.633

Fabric 5 12.3

Fabric 4 14.96

Fabric 1 16.35

N  n1  n2  n3  n4  80 Grand total  n1 x1  n2 x2  n3 x3  n4 x4  158 grand total x  1.975 N SSTr  n1 ( x1  x )2  n2 ( x2  x ) 2  n3 ( x3  x ) 2  n4 ( x4  x ) 2  13.450 Treatment df  k  1  3 SSE  ( n1  1) s12  ( n2  1) s22  ( n3  1) s32  ( n4  1) s42  7.465 Error df  N  k  76 MSTr SSTr treatment df F   45.644 MSE SSE error df The ANOVA table is shown below. Source of Variation Treatments Error Total

df 3 76 79

Sum of Squares 13.450 7.465 20.915

Mean Square 4.483 0.098

F 45.644

1. Let 1 , 2 , 3 , 4 be the mean numbers of seeds germinating for the four treatments. 2. H0: 1  2  3  4 3. Ha: At least two among 1 , 2 , 3 , 4 are different. 4.   0.05 MSTr 5. F  MSE

466

Chapter 15: Analysis of Variance 6. We need to assume that the samples of 100 seeds collected from each treatment were random samples from those populations, and that the population distributions of numbers of seeds germinating are approximately normal with the same standard deviation. 7. F  45.644 8. P-value  P( F3,76  45.644)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that the mean number of seeds germinating is not the same for all four treatments. b

We will construct the T-K interval for 2  3 . Appendix Table 7 gives the 95% Studentized range critical value q  3.74 (using k  4 and error df = 60, the closest tabled value to df  n  k  76 ). The T-K interval for 2  3 is

0.098  1 1     (0.388, 0.912).  2  20 20  Since this interval does not contain zero, we have convincing evidence that seeds eaten and then excreted by lizards germinate at a different rate from those eaten and then excreted by birds. Therefore, since the sample mean was higher for the lizard dung treatment than for the bird dung treatment, we have convincing evidence that seeds eaten and then excreted by lizards germinate at a higher rate from those eaten and then excreted by birds. (2.35  1.70)  3.74

15.22 a

Summary statistics are shown in the table below.

n x s s2

Imperial 4 14.1 0.356 0.127

Parkay 5 12.8 0.430 0.185

Blue Bonnet Chiffon 4 4 13.825 13.1 0.443 0.594 0.196 0.353

Mazola 5 17.14 0.598 0.358

Fleischmann’s 4 18.1 0.648 0.420

1. Let 1 , 2 , 3 , 4 , 5 , 6 be the mean PAPUFA percentages for the six brands. 2. H0: 1  2  3  4  5  6 3. Ha: At least two among 1 , 2 , 3 , 4 , 5 , 6 are different. 4.   0.05 MSTr 5. F  MSE 6. Boxplots for the four groups are shown below.

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467

Imperial Parkay Blue Bonnet Chiffon Mazola Fleischmann’s 12

15 16 PAPUFA percentage

The boxplots are roughly symmetric, and there are no outliers. The largest standard deviation (0.420) is more than twice the smallest (0.127), but we will nonetheless proceed. We need to assume that the specimens of margarine used formed random samples from their respective populations. 7. N  n1  n2  n3  n4  n5  n6  26 Grand total  n1 x1  n2 x2  n3 x3  n4 x4  n5 x5  n6 x6  386.2 grand total x  14.854 N

SSTr  n1 ( x1  x )2  n2 ( x2  x ) 2  n3 ( x3  x )2  n4 ( x4  x ) 2  n5 ( x5  x ) 2  n6 ( x6  x ) 2  108.185 Treatment df  k  1  5 SSE  ( n1  1) s12  ( n2  1) s22  ( n3  1) s32  ( n4  1) s42  ( n5  1) s52  ( n6  1) s62  5.459 Error df  N  k  20 MSTr SSTr treatment df F   79.264 MSE SSE error df 8. P-value  P( F5,20  79.264)  0 9. Since P-value  0  0.05 we reject H0. We have convincing evidence that the mean PAPUFA percentages for the six brands are not all equal. b

The confidence intervals are as follows:

1  2 : (14.1  12.8)  4.45

0.272975  1 1   4  5   (0.197, 2.403) 2  

1  3: (14.1  13.825)  4.45

1  4 : (14.1  13.1)  4.45

0.272975  1 1   4  4   ( 0.887, 1.437) 2  

0.272975  1 1   4  4   ( 0.162, 2.162) 2  

468

Chapter 15: Analysis of Variance

1  5: (14.1  17.14)  4.45

0.272975  1 1   4  5   ( 4.143, 1.937) 2  

1  6 : (14.1  18.1)  4.45

0.272975  1 1   4  4   (5.162, 2.838) 2   0.272975  1 1   5  4   ( 2.128, 0.078) 2  

2  3: (12.8  13.825)  4.45

2  4 : (12.8  13.1)  4.45

0.272975  1 1   5  4   ( 1.403, 0.803) 2  

2  5: (12.8  17.14)  4.45

0.272975  1 1   5  5   (5.380, 3.300) 2  

2  6 : (12.8  18.1)  4.45

0.272975  1 1   5  4   ( 6.403,  4.197) 2  

3  4 : (13.825  13.1)  4.45

0.272975  1 1   4  4   ( 0.437, 1.887) 2  

3  5: (13.825  17.14)  4.45

0.272975  1 1   4  5   ( 4.418, 2.212) 2  

3  6 : (13.825  18.1)  4.45

0.272975  1 1   4  4   ( 5.437, 3.113) 2  

4  5: (13.1  17.14)  4.45

0.272975  1 1   4  5   ( 5.143, 2.937) 2  

4  6 : (13.1  18.1)  4.45

0.272975  1 1   4  4   ( 6.162, 3.838) 2  

5  6 : (17.14  18.1)  4.45

0.272975  1 1   5  4   ( 2.063, 0.143) 2  

The underscoring pattern is shown below.

Sample mean

Parkay 12.8

Chiffon 13.1

Blue Bonnet 13.825

Imperial 14.1

Mazola 17.14

Fleischmann’s 18.1

Chapter 15: Analysis of Variance

469

Online Exercises 15.23 A randomized block experiment was used to control the factor value of house, which definitely affects the assessors’ appraisals. If a completely randomized experiment had been done, then there would have been danger of having the assessors appraising houses which were not of similar values. Therefore, differences between assessors would be partly due to the fact that the homes were dissimilar, as well as to systematical differences in the appraisals made. 15.24 a

Source of Variation

Degrees of Freedom

Sum of Squares

Mean Square

Treatments

11.7

5.85

0.37

Blocks

113.5

28.375

Error

125.6

15.7

Total

250.8

Ho: The mean appraised value does not depend on which assessor is doing the appraisal. Ha: The mean appraised value does depend on which assessor is doing the appraisal.  = 0.05 Test statistic: F 

MSTr MSE

df1 = k  1 = 2 df2 = (k 1)(l  1) = 8. From the ANOVA table, F = 0.37. From Appendix Table 6, P-value > 0.10. Since the P-value exceeds , the null hypothesis is not rejected. The mean appraised value does not seem to depend on which assessor is doing the appraisal. 15.25 Ho: The mean lead concentration does not depend on proportion of ash. Ha: The mean lead concentration does depend on proportion of ash.  = 0.01 Test statistic: F =

MSTr MSE

470

Chapter 15: Analysis of Variance Computations: Source of Variation

Degrees of Freedom

Sum of Squares

Mean Square

Treatment

1395.61

697.806

117.36

Block

7.08

2.361

Error

35.68

5.946

Total

1438.37

df1 = 2 , df2 = 6 and F = 117.36. From Appendix Table 6, P-value < 0.001. Since the P-value is less than , the null hypothesis is rejected. There is convincing evidence that the mean lead concentration differs for the three ash concentrations. 15.26 Ho: 1   2  3 Ha: The three means are not all equal  = 0.05 Test statistic: F =

MSTr MSE

Computations: Source of Variation

Degrees of Freedom

Sum of Squares

Mean Square

Treatment

3199140

1599570

641.49

Block

15028

1670

Error

44883

2494

Total

3259052

df1 = 2 , df2 = 18 and F = 641.49. From Appendix Table 6, P-value < 0.001. Since the P-value is less than , there is evidence that the null hypothesis should be rejected. 15.27 Ho: The mean soil moisture is the same for the three treatments. Ha: The mean soil moisture is not the same for all three treatments.  = 0.05 Test statistic: F =

MSTr MSE

Chapter 15: Analysis of Variance

471

Computations: Source of Variation

Degrees of Freedom

Sum of Squares

Mean Square

Treatment

7.541

3.77033

0.42

Block

47.112

5.23467

Error

162.246

9.01367

Total

216.899

df1 = 2 , df2 = 18 and F = 0.42. From Appendix Table 6, P-value > 0.1. Since the P-value exceeds , the null hypothesis is not rejected. We do not have convincing evidence that the mean soil moisture is not the same for all three treatments. 15.28 Ho: The mean soil moisture is the same for the three treatments. Ha: The mean soil moisture is not the same for all three treatments.  = 0.05 Test statistic: F =

MSTr MSE

Computations: Source of Variation

Degrees of Freedom

Sum of Squares

Mean Square

Treatment

7.046

3.523

0.95

Block

37.52

4.16893

1.13

Error

66.641

3.70226

Total

111.207

df1 = 2 , df2 = 18 and F = 0.95. From Appendix Table 6, P-value > 0.1. Since the P-value exceeds , the null hypothesis is not rejected. We do not have convincing evidence that the mean soil temperature is not the same for all three treatments. 15.29 The statement indicates that there is interaction between caffeine consumption and exercise. The protective effect of increasing exercise by a particular amount is greater for higher levels of caffeine consumption.

472

Chapter 15: Analysis of Variance

15.30 a

The graph is shown below.

The graph does not seem to support the conclusion of no interaction. It appears that for students who did not study the increase from 4 to 6 choices resulted in an increase in mean percentage correct, this not being the case for students who studied (with or without review). b

Ho: There are no prior study main effects. Ha: There are prior study main effects.  = 0.05 Test statistic: FA =

MSA MSE

df1 = k − 1 = 3 − 1 = 2, df2 = 69, and FA = 66.25. From Appendix Table 6, P-value < 0.001. Since the P-value is less than , the null hypothesis of no prior study main effects is rejected. The conclusion of significant main effects for prior study is justified. Ho: There are no number of choices main effects. Ha: There are number of choices main effects.  = 0.05 Test statistic: F B =

MSB MSE

df1 = l − 1 = 3 − 1 = 2, df2 = 69, and FB = 73.76. From Appendix Table 6, P-value < 0.001. Since the P-value is less than , the null hypothesis of no number of choices main effects is rejected. The conclusion of significant main effects for number of choices is justified.

Chapter 15: Analysis of Variance

473

The conclusion of the authors mentioned in Part (a) tells us that the effect of changing the number of answer choices from 2 to 4 and from 4 to 6 is not significantly changed according to the level of prior study. From the graph we can see that increasing the number of answer choices resulted in a decrease, on average, of the percentage correct on the exam, and the authors’ conclusion implies that this effect is roughly equal for all levels of prior study. The conclusion of the first hypothesis test in Part (b) is that this effect is significant. We also see from the graph that whatever the number of answer choices, an increase in the level of study is associated with an increase in the percentage correct, and that this applies for all numbers of answer choices. The conclusion of the second hypothesis test in Part (b) tells us that this effect is significant.

15.31 a

Yes. The roughly horizontal lines for “low approach” show that the level of thoughts of passion has very little effect on the satisfaction level for people in that category. The small positive slope for “average approach” shows that an increase in thoughts of passion results in a relatively small increase in satisfaction for people in that category. However, the greater slope for “high approach” shows that an increase in thoughts of passion results in a larger increase in satisfaction for people in that category.

Yes. The small positive slope for “low avoidance” shows that an increase in thoughts of insecurity results in a small increase in satisfaction for people in that category. The small negative slope for “average avoidance” shows that an increase in thoughts of insecurity results in a small decrease in satisfaction for people in that category. The larger negative slope for “high avoidance” shows that an increase in thoughts of insecurity results in a larger decrease in satisfaction for people in that category.

Yes. In the top graph the slope for “high approach” is more positive than that of either of the other two categories, and in the bottom graph the slope for “high avoidance” is more negative than that of either of the other two categories.

15.32 a

474

Chapter 15: Analysis of Variance b

The graphs for males and females are very nearly parallel. There does not appear to be an interaction between gender and type of odor.

15.33 When an AB interaction is present, the change in mean response, when going from one level of factor A to another, depends upon which level of factor B is being used. Since these effects are different for different levels of factor B, the individual effects of factor A or factor B cannot be interpreted. 15.34 a

The completed ANOVA table is shown below. Source of Variation

Degrees of Freedom

Sum of Squares

Mean Square

Status (A)

0.34776

0.17388

14.49

Gender (B)

0.00180

0.15

Status by Gender

0.02280

0.01140

0.95

Error

0.79200

0.01200

Total

1.16436

Ho: There is no interaction between status (A) and gender (B). Ha: There is interaction between status (A) and gender (B).  = 0.05 (for illustration) The test statistic is: F AB =

MSAB . MSE

df1 = 2 , df2 = 66, FAB = 0.95. From Appendix Table 6, P-value > 0.10. Since the P-value is greater than , the null hypothesis is not rejected. The data do not suggest existence of an interaction between status and gender. Hence tests about main effects may be performed. c

Ho: There are no gender main effects. Ha: There are gender main effects.  = 0.05 (for illustration) The test statistic is: F B =

MSB MSE

df1 = 1, df2 = 66, and FB = 0.15 From Appendix Table 6, P-value > 0.10.

Chapter 15: Analysis of Variance

475

Since the P-value exceeds , the null hypothesis of no gender main effects is not rejected. The data do not suggest the existence of a difference between mean “Rate of Talk” scores for girls and boys. d

Ho: There are no status main effects. Ha: There are status main effects.  = 0.05 (for illustration) The test statistic is: FA =

MSA MSE

df1 = 2, df2 = 66, and FA = 14.49 From Appendix Table 6, 0.001 > P-value. Since the P-value is smaller than , the null hypothesis of no status main effects is rejected. The data provide evidence to conclude that there is a difference among the mean “Rate of Talk” scores across the three status groups. 15.35 Source of Variation

Degrees of Freedom

Sum of Squares

Mean Square

Size (A)

0.088

0.044

4.00

Species (B)

0.048

4.364

Size by Species

0.048

0.024

2.18

Error

0.132

0.011

Total

0.316

Ho: There is no interaction between Size (A) and Species (B). Ha: There is interaction between Size and Species.  = 0.01 The test statistic is: F AB =

MSAB . MSE

df1 = 2 , df2 = 12, and FAB = 2.18 From Appendix Table 6, P-value > 0.10.

476

Chapter 15: Analysis of Variance Since the P-value exceeds , the null hypothesis is not rejected. The data are consistent with the hypothesis of no interaction between Size and Species. Hence, hypothesis tests on main effects will be done. Ho: There are no size main effects. Ha: There are size main effects.  = 0.01 The test statistic is: F A =

MSA . MSE

df1 = 2, df2 = 12, and FA = 4.00. From Appendix Table 6, 0.05 > P-value > 0.01. Since the P-value exceeds , the null hypothesis of no size main effects is not rejected. When using =0.01, the data do not support the conclusion that there are differences between the mean preference indices for the three sizes of bass. Ho: There are no species main effects. Ha: There are species main effects.  = 0.01 The test statistic is: F B =

MSB . MSE

df1 = 1, df2 = 12, and FB = 4.364 From Appendix Table 6, P-value > 0.05. Since the P-value exceeds , the null hypothesis of no species main effects is not rejected. At a significance level of  = 0.01, the data are consistent with the hypothesis that there are no differences between the mean preference indices for the three species of bass. 15.36 a b

There are two age classes. There were twenty-one observations made for each age-sex combination. Source of Variation

Degrees of Freedom

Sum of Squares

Mean Square

Age (A)

0.614

0.6140

8.73

Sex (B)

1.754

1.7540

24.95

Age by Sex (AB)

0.146

0.1460

2.08

Chapter 15: Analysis of Variance

477

c Error

5.624

Total

8.138

0.0703

Ho: There is no interaction between age (A) and sex (B). Ha: There is interaction between age and sex.  = 0.01 (for illustration) Test statistic: F AB =

MSAB MSE

df1 = 1 , df2 = 80 , and FAB = 2.08. From Appendix Table 6, P-value > 0.10. Since the P-value exceeds , the null hypothesis is not rejected. The data are consistent with the hypothesis of no interaction between age and sex. Hence, hypothesis tests on main effects will be done. Ho: There are no age main effects. Ha: There are age main effects.  = 0.01 Test statistic: F A =

MSA MSE

df1 = 1 , df2 = 80 , and FA = 8.73. From Appendix Table 6, 0.01 > P-value > 0.001. Since the P-value is less than , the null hypothesis of no age main effects is rejected. The data supports the conclusion that there is a difference between the mean values of the preference index for the two age groups. Ho: There are no sex main effects. Ha: There are sex main effects.  = 0.01 Test statistic: F B =

MSB MSE

df1 = 1 , df2 = 80 , and FB = 24.95. From Appendix Table 6, 0.001 > P-value. Since the P-value is less than , the null hypothesis of no sex main effects is rejected. The data supports the conclusion that there is a difference between the mean territory size for the two sexes.

478

Chapter 15: Analysis of Variance

15.37 a Source of Variation

Degrees of Freedom

Sum of Squares

Mean Square

Race

857

5.57

Sex

291

1.89

Race by Sex

0.21

Error

5541

153.92

HTotal 39 between 6721 race and sex. o: There is no interaction Ha: There is interaction between race and sex.  = 0.01 Test statistic: F AB =

MSAB MSE

df1 = 1 , df2 = 36 , and FAB = 0.21. From Appendix Table 6, P-value > 0.10. Since the P-value exceeds , the null hypothesis of no interaction between race and sex is not rejected. Thus, hypothesis tests for main effects are appropriate. b

Ho: There are no race main effects. Ha: There are race main effects.  = 0.01 Test statistic: F A =

MSA MSE

df1 = 1 , df2 = 36 , and FA = 5.57. From Appendix Table 6, 0.05 > P-value > 0.01 Since the P-value exceeds , the null hypothesis of no race main effects is not rejected. The data are consistent with the hypothesis that the true average lengths of sacra do not differ for the two races. c

Ho: There are no sex main effects. Ha: There are sex main effects.  = 0.01 Test statistic: F B =

MSB MSE

df1 = 1 , df2 = 36 , and FB = 1.89. From Appendix Table 6, P-value > 0.10.

Chapter 15: Analysis of Variance

479

Since the P-value exceeds , the null hypothesis of no sex main effects is not rejected. The data are consistent with the hypothesis that the true average lengths of sacra do not differ for males and females. 15.38 a

The required boxplot obtained using MINITAB is shown below. Price per acre values appear to be similar for 1996 and 1997 but 1998 values are higher. The mean price per acre values for each year are also plotted as a solid square within each box plot.

Year

1998

1997

1996

30000

40000

50000

Price

Let i denote the mean price per acre for vineyards in year i (i = 1, 2, 3). Ho: 1 = 2 = 3 Ha: At least two of the three i's are different.  = 0.05 Test statistic: F =

MSTr MSE

df1 = k  1 = 2 df2 = N  k = 15-3 = 12.

x1  35600, x2  36000, x3  43600 x=

[5(35600) + 5(36000) + 5(43600) ] = 38400 (15)

MSTr = [5(35600  38400)2 + 5(36000  38400)2 + 5(43600  38400)2 ]/2 = 101600000

s1  3847.077, s2  3807.887, s3  3911.521 MSE = [(5-1)(3847.077)2 + (5-1)(3807.887)2 + (5-1)(3911.521)2 ]/12 = 14866667

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Chapter 15: Analysis of Variance

MSTr 101600000 = = 6.83 MSE 14866667

From Appendix Table 6, 0.05 > P-value > 0.01. Since the P-value is less than , the null hypothesis is rejected. At a significance level of  = 0.05, the data support the claim that the true mean price per acre for the three years under consideration are not all the same. 15.39 a

From the given information:

n1  6, x1  10.125, s12  2.093750

n2  6, x2  11.375, s22  3.59375

n3  6, x3  11.708333, s32  2.310417

n4  5, x4  12.35, s42  .925

x

T 261   11.347826 N 23

SSTr = 6(10.125  11.347826)2 + 6(11.375  11.347826)2 + 6(11.708333  11.347826)2 + 5(12.35  11.347826)2 = 8.971822 + 0.004431 + 0.779791 + 5.021763 = 14.7778 SSE = 5(2.093750) + 5(3.59375) + 5(2.310417) + 4(0.925) = 43.689585. df1 = k  1 = 4  1 = 3 df2 = (6 + 6 + 6 + 5)  4 = 19

MSTr 

MSE 

F

SSTr 14.7778   4.925933 k 1 3

SSE 43.689585   2.299452 N k 19

MSTr 4.925933   2.142 MSE 2.299452

From Appendix Table 6, P-value > 0.10. b

Ho: 1 = 2 = 3 = 4 Ha: At least two of the four i's are different.  = 0.05 Test statistic: F =

MSTr  2.142 MSE

df1 = k  1 = 3 df2 = N  k = 19. From the ANOVA table, F = 2.142. From Appendix Table 6, P-value > 0.10.

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481

Since the P-value exceeds , the null hypothesis is not rejected. The data do not suggest that there are differences in true average responses among the treatments. 15.40 The interval for 1  2 contains zero, and hence 1 and 2 are judged not different. The intervals for 1  3 and 2  3 do not contain zero, so 1 and 3 are judged to be different and 2 and 3 are judged to be different. Hence there is evidence that 3 is different from the other two means. 15.41 a

Ho: 1 = 2 = 3 = 4 = 5 Ha: At least two of the five i's are different.  = 0.05 Test statistic: F =

MSTr MSE

df1 = k  1 = 4 df2 = N  k = 15. From the data, the following statistics were calculated. Hormone n mean Total variance

x

1 4 12.75 51 17.58

2 4 17.75 71 12.917 3

3 4 17.50 70 8.333

4 4 11.50 46 21.667

5 4 10.00 40 15.333

T 278   13.9 N 20

SSTr = 4(12.75  13.9)2 + 4(17.75  13.9)2 + 4(17.50  13.9)2 + 4(11.5  13.9)2 + 4(10  13.9)2 = 5.29 + 59.29 + 51.84 + 23.04 + 60.840 = 200.3 SSE = 3(17.583) + 3(12.917) + 3(8.333) + 3(21.667) + 3(15.333) = 227.499 df1 = k  1 = 5  1 = 4 df2 = (4 + 4 + 4 + 4 + 4)  5 = 15

MSTr 

MSE 

F

SSTr 200.3   50.075 k 1 4

SSE 227.499   15.1666 N k 15

MSTr 50.075   3.30 MSE 15.1666

From Appendix Table 6, 0.05 > P-value > 0.01.

482

Chapter 15: Analysis of Variance Since the P-value is less than , Ho is rejected. The data support the conclusion that the mean plant growth is not the same for all five growth hormones. b

k=5

Error df = 15

From Appendix Table 7, using a 95% confidence level, q = 4.37. Since the sample sizes are the same, the  factor is the same for each comparison.

4.37

15.1666  1 1   +   8.51 2 4 4

1  2 : (12.75  17.75)  8.51  5  8.51  (13.51, 3.51) 1  3: (12.75  17.50)  8.51  4.75  8.51  (13.26, 3.76) 1  4: (12.75  11.5)  8.51  1.25  8.51  (7.26, 9.76) 1  5: (12.75  10)  8.51  2.75  8.51  (5.76, 11.26) 2  3: (17.75  17.50)  8.51  0.25  8.51  (8.26, 8.76) 2  4: (17.75  11.5)  8.51  6.25  8.51  (2.26, 14.76) 2  5: (17.75  10)  8.51  7.75  8.51  (0.76, 16.26) 3  4: (17.5  11.5)  8.51  6.0  8.51  (2.51, 14.51) 3  5: (17.5  10)  8.51  7.5  8.51  (1.01, 16.01) 4  5: (11.5  10)  8.51  1.5  8.51  (7.01, 10.01) No significant differences are determined using the T-K method. 15.42 The test for no interaction would have df1 = 2 , df2 = 120, and FAB = 1. From Appendix Table 6, P-value > 0.10. Since the P-value exceeds the  of 0.05, the null hypothesis of no interaction is not rejected. Since there appears to be no interaction, hypothesis tests on main effects are appropriate. The test for no A main effects would have df 1 = 2 , df2 = 120, and FA = 4.99. From Appendix Table 6, 0.01 > P-value > 0.001. Since the P-value is less than , the null hypothesis of no A main effects is rejected. The data suggest that the expectation of opportunity to cheat affects the mean test score. The test for no B main effects would have df 1 = 1 , df2 =120, and FB = 4.81. From Appendix Table 6, 0.05 > P-value > 0.01. Since the P-value is less than , the null hypothesis of no B main effects is rejected. The data suggests that perceived payoff affects the mean test score.

Chapter 15: Analysis of Variance

483

15.43 Ho: 1 = 2 = 3 Ha: at least two among 1, 2, 3 are not equal.

MSTr MSE

N = 90, df1 = 2 df2 = 87 T = 30(9.40 + 11.63 + 11.00) = 960 .9

x

T 960.9   10.677 N 90

SSTr = 30[(9.40  10.677)2 + (11.63  10.677)2 + (11.00  10.677)2 ] = 30[1.630729 + 0.908209 + 0.104329] = 79.298

MSTr 

79.298  39.649 2

749.85  8.619 87 MSTr 39.649 F=   4.60 MSE 8.619 MSE 

From Appendix Table 6, 0.05 > P-value > 0.01. If  = 0.05 is used, then the P-value is less than  and the null hypothesis would be rejected. The conclusion would be that the true population mean score is not the same for the three types of students. If  = 0.01 is used, then the P-value exceeds  and the null hypothesis would not be rejected. The conclusion would be that the data are consistent with the hypothesis that the mean score is the same for the three types of students. 15.44 a

Ho: 1 = 2 = 3 = 4 Ha: at least two among 1, 2, 3, 4 are not equal.  = 0.05 MSTr F= MSE

n1  6, x1  4.923, s12  .000107

n2  6, x2  4.923, s22  .000067

n3  6, x3  4.917, s32  .000147

n4  6, x4  4.902, s42  .000057

T = 6(4.923) + 6(4.923) + 6(4.917) + 6(4.902) = 117.99

484

Chapter 15: Analysis of Variance

x

117.99  4.916 24

SSTr = 6(4.923  4.916)2 + 6(4.923  4.916)2 + 6(4.917  4.916)2 + 6(4.902  4.916)2 = 0.000294 + 0.000294 + 0.000006 + 0.001176 = 0.001770 SSE = 5(0.000107) + 5(0.000067) + 5(0.000147) + 5(0.000057) = 0.001890

MSTr 

SSTr .001770   .00059 k 1 3

MSE 

SSE .001890   .0000945 N k 24  4

MSTr .00059   6.24 MSE .0000945

df1 = 3 , df2 = 20. From Appendix Table 6, 0.01 > P-value > 0.001. Since the P-value is less than , the null hypothesis is rejected. The sample data support the conclusion that there are differences in the true average iron content for the four storage periods. b

k = 4 , Error df = 20. From Appendix Table 7, for a 95% confidence level, q = 3.96. Since the sample sizes are equal, the  factor is the same for each comparison.

 factor  3.96

.0000945  .0157 6

Storage Period Mean

4 4.902

2 4.917

1 4.923

0 4.923

The mean for the 4-month storage period differs significantly from the means for the 1month and 0 storage periods. No other significant differences are present. 15.45 Source of Variation

Degrees of Freedom

Sum of Squares

Mean Square

Locations

0.6

0.04286

1.89

Months

2.3

0.20909

9.20

Error

154

3.5

0.02273

Total

179

6.4

Chapter 15: Analysis of Variance

485

Test for locations effects:  = 0.05 df1 = 14 and df2 = 154. The F ratio to test for locations is F = 1.89. From Appendix Table 6, 0.05 > P-value > 0.01. Since the P-value is less than , the null hypothesis of no location main effects is rejected. The data suggest that the true concentration differs by location. Test for months effects:  = 0.05 df1 = 11 and df2 = 154. The F ratio to test for months is F = 9.20. From Appendix Table 6, 0.001 > P-value. Since the P-value is less than , the null hypothesis of no month main effects is rejected. The data suggest that the true mean concentration differs by month of year. 15.46 Multiplying each observation in a single-factor ANOVA will change x i , x , and si by a factor of c. Hence, MSTr and MSE will be also changed, but by a factor of c 2. However, the F ratio remains unchanged because c 2MSTr/c2MSE = MSTr/MSE. That is, the c 2 in the numerator and denominator cancel. It is reasonable to expect a test statistic not to depend on the unit of measurement. 15.47 Let 1, 2, and 3 denote the mean lifetime for brands 1, 2 and 3 respectively. Ho: 1 = 2 = 3 Ha: At least two of the three i's are different.  = 0.05 Test statistic: F =

MSTr MSE

Computations:  x 2 = 45171, x  46.1429, x1 = 44.571, x2  45.857, x3  48 SSTo = 45,171  21(46.1429)2 = 45171  44712.43 = 458.57 SSTr = 7(44.571)2 + 7(45.857)2 + 7(48)2  44712.43 = 44754.43  44712.43 = 42 SSE = SSTo  SSTr = 458.57  42 = 416.57

Source of Variation

Degrees of Freedom

Sum of Squares

Mean Square

Treatments

0.907

Error

416.57

23.14

Total

458.57

df1 = 2 and df2 = 18. The F ratio is 0.907. From Appendix Table 6, P-value > 0.10. Since the P-value exceeds , Ho is not rejected at level of significance 0.05. The data are consistent with the hypothesis that there are no differences in true mean lifetimes of the three brands of batteries.

486

15.48

Chapter 15: Analysis of Variance

c1  1, c2  .5, c3  .5 c1 x1  c2 x2  c3 x3  44.571 .5(45.857)  .5(48)  2.3575

 c2 c2 c2   1 (.5) 2 (.5) 2  MSE  1  2  3   23.14      23.14(.2143)  4.9586 7 7  7  n1 n2 n3  t critical = 2.10 The desired confidence interval is: 2.3575  2.10 4.9586  2.3575  2.10(2.2268)  2.3575  4.6762  (7.0337, 2.3187). 15.49 The transformed data are:

Brand 1 Brand 2 Brand 3 Brand 4

3.162 4.123 3.606 3.742

3.742 3.742 4.243 4.690

2.236 2.828 3.873 3.464

3.464 3.000 4.243 4.000

2.828 3.464 3.162 4.123

mean 3.0864 3.4314 3.8254 4.0038

SSTo = 264.001  20(3.58675)2 = 264.001  257.2955 = 6.7055 SSTr = [5(3.0864)2 + 5(3.4314)2 + 5(3.8254)2 + 5(4.0038)2]  257.2955 = 259.8409  257.2955 = 2.5454 SSE = 6.7055  2.5454 = 4.1601 Let 1, 2, 3, 4 denote the mean of the square root of the number of flaws for brand 1, 2, 3 and 4 of tape, respectively. Ho: 1 = 2 = 3 = 4 Ha: At least two of the four i's are different.  = 0.01

MSTr MSE

df1 = 3 and df2 = 16.

F

2.5454 / 3 = 3.26 4.1601 / 16

From Appendix Table 6, P-value > 0.01.

Chapter 15: Analysis of Variance

487

Since the P-value exceeds , H0 is not rejected. The data are consistent with the hypothesis that there are no differences in true mean square root of the number of flaws for the four brands of tape.

Chapter 16 Nonparametric (Distribution-Free) Statistical Methods Online Exercises 16.1

Let 1 denote the mean fluoride concentration for livestock grazing in the polluted region and  2 denote the mean fluoride concentration for livestock grazing in the unpolluted regions.

H 0 : 1   2 = 0

H a : 1   2 > 0

 = 0.05 The test statistic is: rank sum for polluted area (sample 1). Sample 2 1 1 2 2 2 1 1 2 1 1 1

Ordered Data 14.2 16.8 17.1 17.2 18.3 18.4 18.7 19.7 20.0 20.9 21.3 23.0

Rank 1 2 3 4 5 6 7 8 9 10 11 12

Rank sum = (2 + 3 + 7 + 8 + 10 + 11 + 12) = 53 P-value: This is an upper-tail test. With n1 = 7 and n 2 = 5, Chapter 16 Appendix Table 1 tells us that the P-value > 0.05. Since the P-value exceeds , Ho is not rejected. The data do not support the conclusion that there is a larger average fluoride concentration for the polluted area than for the unpolluted area. 16.2

Let 1 denote the true average developing time without modification and  2 the true average developing time with modification.

H 0 : 1   2 = 1

H a : 1   2 > 1

 = 0.05 The test statistic is: rank sum for (unmodified times - 1). (A) original process  1 (B) modified process

488

7.6 4.1 3.5 4.4 5.3 5.6 4.7 7.5 5.5 4.0 3.8 6.0 5.8 4.9 7.0 5.7

Chapter 16: Nonparametric (Distribution-Free) Statistical Methods

Sample A B B A A A B A B A B B B B A A

Ordered Data 3.5 3.8 4.0 4.1 4.4 4.7 4.9 5.3 5.5 5.6 5.7 5.8 6.0 7.0 7.5 7.6

489

Rank 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

Rank sum = 1 + 4 + 5 + 6 + 8 + 10 + 15 + 16 = 65 P-value: This is an upper-tail test. With n1 = 8 and n2 = 8, Chapter 16 Appendix Table 1 tells us that the P-value is greater than 0.05. Since the P-value exceeds  , H 0 is not rejected. There is not enough evidence suggesting that the mean reduction in development time resulting from the modification exceeds 1 second. 16.3

Let 1 denote the true average ascent time using the lateral gait and  2 denote the true average ascent time using the four-beat diagonal gait H 0 :  1   2 = 0 H a :  1   2 =/ 0 A value for  was not specified in the problem, so a value of 0.05 was chosen for illustration. The test statistic is: Rank sum for diagonal gait. Gait D L L D D L L D L L L D D

Ordered Data 0.85 0.86 1.09 1.24 1.27 1.31 1.39 1.45 1.51 1.53 1.64 1.66 1.82

Rank sum = 2 + 3 + 6 + 7 + 9 + 10 + 11 = 48

Rank 1 2 3 4 5 6 7 8 9 10 11 12 13

490

Chapter 16: Nonparametric (Distribution-Free) Statistical Methods P-value: This is a two-tailed test. With n1 = 7 and n 2 = 6, Chapter 16 Appendix Table 1 tells us that the P-value > 0.05. Since the P-value exceeds , Ho is not rejected. The data do not provide sufficient evidence that there is a difference in mean ascent time for the diagonal and lateral gaits. b

16.4

We can be at least 95% confident (actually 96.2% confident) that the difference in the mean ascent time using lateral gait and the mean ascent time using diagonal gait may be as small as .43 to as large as 0.3697.

Let 1 and  2 denote the mean conduction velocity of workers who are exposed to lead and not exposed to lead, respectively. H 0 :  1   2 = 0 H a :  1   2 =/ 0  = 0.05 The test statistic is: Rank sum of the exposed group. Group Ordered Data E 31 E 36 E 38 E 41 N 41 N 42 E 43 N 44 N 45 N 46 E 46 E 46.5 N 50.5 N 54

Rank 1 2 3 4.5 4.5 6 7 8 9 10.5 10.5 12 13 14

Rank sum = 1 + 2 + 3 + 4.5 + 7 + 10.5 + 12 = 40 P-value: This is a two-sided test. With n1 = 7 and n 2 = 7, Chapter 16 Appendix Table 1 tells us that P-value > 0.05. Since the P-value exceeds , H 0 is not rejected. There is not sufficient evidence to conclude that mean conduction velocity for those exposed to lead differs from the mean conduction velocity for those not exposed to lead. 16.5

Let 1 denote the mean number of binges per week for people who use a placebo and  2 denote the mean number of binges per week for people who use Imipramine. H 0 : 1   2 = 0 H a : 1   2 > 0  = 0.05

Chapter 16: Nonparametric (Distribution-Free) Statistical Methods

491

The test statistic is: Rank sum for the Imipramine group. Group I I I I I P P P

Ordered Data 1 1 2 2 3 3 3 4

Rank

Group

1.5 1.5 3.5 3.5 6 6 6 8.5

P I P I P P I P

Ordered Data 4 5 6 7 8 10 12 15

Rank 8.5 10 11 12 13 14 15 16

Rank sum = 6 + 6 + 8.5 + 8.5 + 11 + 13 + 14 + 16 = 83 P-value: This is an upper-tail test. With n1 = 8 and n 2 = 8, Chapter 16 Appendix Table 1 tells us that the P-value > 0.05. Since the P-value exceeds , H 0 is not rejected. The data does not provide enough evidence to suggest that Imipramine is effective in reducing the mean number of binges per week. 16.6

The 90.5% C.I. for 1  2 is given as (16.00, 95.98). Based on the sample information, we estimate with 90.5% confidence that the difference between the bond strength of Adhesive 1 and Adhesive 2 is that Adhesive 1 is stronger by, on average, between 16.00 units and 95.98 units.

16.7

The 95.5% C.I. for 1  2 is given as (-0.4998, 0.5699). The confidence interval indicates that with 95.5% confidence the mean burning time of oak may be as much as 0. 5699 hours longer than pine; but also that the mean burning time of oak may be as much as 0.4998 hours shorter than pine.

16.8

Let d denote the true mean difference of Peptide secretion of children with autistic syndrome (Before – After). H 0 : d  0 H a : d  0 Test statistic is the signed-rank sum. With n = 10 and  = 0.01, reject H 0 if the signed rank sum is greater than or equal to 45. (Critical Value from Chapter 16 Appendix Table 2.) The differences and signed ranks are: Differences: 5, 7, 13, 15, 16, 22, 31, 55, 58, 77 Signed-ranks: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 Signed-rank sum = 55 Since the signed-rank sum of 55 exceeds the critical value of 45, the null hypothesis is rejected. There is sufficient evidence to conclude that the restricted diet is successful in reducing Peptide secretion.

16.9

Let d denote the true mean difference of in peak force on the hand (Post – Pre).

492

Chapter 16: Nonparametric (Distribution-Free) Statistical Methods

H 0 : d  6

H a : d  6

Test statistic is the signed-rank sum. With n = 6 and  = 0.109, reject H 0 if the signed rank sum is greater than or equal to 13. (Critical Value from Chapter 16 Appendix Table 2.) Differences = (Post – Pre) - 6 The differences and signed ranks are: Differences: 5.5, −3.1, 1.1, 1.9, 11.3, −4.9. Signed-ranks: 5, −3, 1, 2, 6, −4. Signed-rank sum = 7 Since the signed-rank sum of 7 is less than the critical value of 13, the null hypothesis is not rejected. There is not sufficient evidence to conclude that the mean postimpact force is greater than the mean preimpact force by more than 6 Newtons. 16.10 Let d denote the true mean difference in epinephrine concentration for the two anesthetics (isoflurane − halothane).

H 0 : d  0

H a : d  0

Test statistic is the signed-rank sum. With n = 10 and  = 0.048, reject H 0 if the signed rank sum is less than or equal to -39, or greater than or equal to 39. (Critical Value from Chapter 16 Appendix Table 2.) The differences and signed ranks are: Differences: 0.21, 0.61, -0.24, -0.09, 0.15, -0.56, 0.3, -0.34, 0.01, -0.14 Signed-ranks: 5, 10, -6, -2, 4, -9, 7, -8, 1, -3 Signed-rank sum = -1 Since the calculated value of the test statistic does not fall into the rejection region, the null hypothesis is not rejected. The data do not suggest that there is a difference in the mean epinephrine concentration for the two anesthetics. The assumption that must be made is that the population distribution of differences of epinephrine concentration (isoflurane − halothane) is symmetrical. 16.11 Let d denote the true mean difference in concentration of strontium-90 (nonfat minus 2% fat).

H 0 : d  0

H a : d  0

Test statistic is the signed-rank sum. With n = 5 and  = 0.031, reject H 0 if the signed rank sum is less than or equal to -15. (Critical Value from Chapter 16 Appendix Table 2.) The differences and signed ranks are:

Chapter 16: Nonparametric (Distribution-Free) Statistical Methods

493

Differences: -0.7, -4.1, -4.7, -2.8, -2.7. Signed-ranks: -1, -4, -5, -3, -2. Signed-rank sum = -15. Since the calculated signed-rank sum of -15 is less than or equal to the critical value of -15, the null hypothesis is rejected. The data provide convincing evidence that the true mean strontium-90 concentration is higher for 2% milk than for nonfat milk. 16.12 Let d denote the true mean phosphate content when using the gravimetric technique and  2 the true mean phosphate content using the spectrophotometric technique. The differences are: -0.3, 2.8, 3.9, 0.6, 1.2, -1.1. The pairwise averages for these differences are:

Difference

-1.1 -0.3 0.6 1.2 2.8 3.9

-1.1 -1.1

Difference -0.3 0.6 1.2 -0.7 -0.25 0.05 -0.3 0.15 0.45 0.6 0.9 1.2

2.8 0.85 1.25 1.7 2.0 2.8

3.9 1.4 1.8 2.25 2.55 3.35 3.9

Arranging the pairwise averages in order yields: -1.1, -0.7, -0.3, -0.25, 0.05, 0.15, 0.45, 0.6, 0.85, 0.9, 1.2, 1.25, 1.4, 1.7, 1.8, 2, 2.25, 2.55, 2.8, 3.35, 3.9. With n = 6, and a confidence level of 96.9%, Chapter 16 Appendix Table 3 yields a d value equal to 1. Counting in 1 value from each end of the ordered pairwise averages yields an approximate 96.9% confidence interval of (-1.1, 3.9) for d . Thus, with 96.9% confidence, the difference in mean phosphate content as determined by the two procedures is between -1.1 and 3.9 units. 16.13 a

Let d denote the true mean difference in height velocity (during minus before).

H 0 : d  0

H a : d  0

Test statistic is the signed-rank sum. With n = 14 and  = 0.05, reject H 0 if the signed rank sum is greater than or equal to 53. (Critical Value from Chapter 16 Appendix Table 2.) The differences and signed ranks are: Differences: 2.7, 7.6, 2.0, 4.9, 3.5, 7.7, 5.3, 5.3, 4.4, 2.0, 6.4, 4.3, 5.8, 2.6. Signed-ranks: 4, 13, 1.5, 8, 5, 14, 9.5, 9.5, 7, 1.5, 12, 6, 11, 3. Signed-rank sum = 105.

494

Chapter 16: Nonparametric (Distribution-Free) Statistical Methods

Since the calculated signed-rank sum of 105 exceeds the critical value of 53, the null hypothesis is rejected. The data provide evidence that the growth hormone therapy is successful in increasing the mean height velocity. b

16.14 a

We need to assume that the population distribution of height velocity differences is symmetrical. Let d denote the true mean difference in time from entry to first stroke (hole minus flat).

H 0 : d  0

H a : d  0

Test statistic is the signed-rank sum. With n = 10 and  = 0.02, reject H 0 if the signed rank sum is less than or equal to -45, or greater than or equal to 45. (Critical Value from Chapter 16 Appendix Table 2.) The differences and signed ranks are: Differences: 0.12, -0.13, 0.11, -0.07, -0.17, 0.14, 0.18, -0.25, -0.01, -0.11. Signed-ranks: 5, -6, 3.5, -2, -8, 7, 9, -10, -1, -3.5. Signed-rank sum = -6. Since the calculated value of -6 does not fall in the rejection region for  = 0.02, it does not fall in the rejection region for  = 0.01, and so H 0 is not rejected. Thus, the data do not provide convincing evidence that there is a difference in mean time from entry to first stroke for the two entry methods. b

Let d denote the true mean difference in initial velocity (hole minus flat).

H 0 : d  0

H a : d  0

Test statistic is the signed-rank sum. The differences and signed ranks are: Differences: -1.1, 0.1, -2.4, -1.0, -3.0. -0.9, -1.4, -1.7, 1.2. Signed-ranks: -4, 1, -8, -3, -9, -2, -6, -7, -5. Signed-rank sum = -33 Chapter 16 Appendix Table 2 tells us that the critical region for a two-tailed 0.054 test starts at 33 and at −33. Thus our signed-rank sum of −33 is significant at the 0.054 level. At the 0.054 level the data provide convincing evidence that there is a difference in the mean initial velocity for the two entry methods. 16.15 Let d denote the true mean difference in cholesterol synthesis rates (potato minus rice).

H 0 : d  0

H a : d  0

Chapter 16: Nonparametric (Distribution-Free) Statistical Methods

495

Test statistic is the signed-rank sum. With n = 8 and  = 0.054, reject H 0 if the signed rank sum is less than or equal to -28, or greater than or equal to 28. (Critical Value from Chapter 16 Appendix Table 2.) The differences and signed ranks are: Differences: 0.18, -1.24, 0.25, -0.56, 1.01, 0.96, 0.60, 0.16 Signed-ranks: 2, -8, 3, -4, 7, 6, 5, 1. Signed-rank sum = 12 Since the calculated value of 12 does not fall into the rejection region, H 0 is not rejected. The data do not provide evidence that there is a difference in the true mean cholesterol synthesis for the two sources of carbohydrates. 16.16 Let d denote the true mean difference in lung capacity (post-operative minus pre-operative).

H 0 : d  0

H a : d  0

Test statistic is z 

signed-rank sum . n(n+1)(2n+1) 6

For  = 0.05, reject H 0 if z > 1.645. The differences and signed ranks are: Differences: 80, 340, 350, 100, 640, -115, 545, 220, 630, 800, 120, 240, -20, 880, 570, 40, 20, 580, 130, 70, 450, 110. Signed-ranks: 5, 13, 14, 6, 20, -8, 16, 11, 19, 21, 9, 12, -1.5, 22, 17, 3, -1.5, 18, 10, 4, 15, 7. Signed-rank sum = 231

z

231  3.75 22(23)(45) 6

Since the calculated value of z exceeds the z critical value of 1.645, H 0 is rejected. The data provide convincing evidence that surgery increases the mean lung capacity.

496

Chapter 16: Nonparametric (Distribution-Free) Statistical Methods

16.17 Using d as defined in Exercise 16.13, and the differences computed there, the pairwise averages are:

2.0 2.0 2.6 2.7 3.5 4.3 4.4 4.9 5.3 5.3 5.8 6.4 7.6 7.7

2.0 2.0

2.0 2.0 2.0

2.6 2.3 2.3 2.6

2.7 2.35 2.35 2.65 2.7

3.5 2.75 2.75 3.05 3.1 3.5

4.3 3.15 3.15 3.45 3.5 3.9 4.3

4.4 3.2 3.2 3.5 3.55 3.95 4.35 4.4

4.9 3.45 3.45 3.75 3.8 4.2 4.6 4.65 4.9

5.3 3.65 3.65 3.95 4.0 4.4 4.8 4.85 5.1 5.3

5.3 3.65 3.65 3.95 4.0 4.4 4.8 4.85 5.1 5.3 5.3

5.8 3.9 3.9 4.2 4.25 4.65 5.05 5.1 5.35 5.55 5.55 5.8

6.4 4.2 4.2 4.5 4.55 4.95 5.35 5.4 5.65 5.85 5.85 6.1 6.4

7.6 4.8 4.8 5.1 5.15 5.55 5.95 6 6.25 6.45 6.45 6.7 7.0 7.6

7.7 4.85 4.85 5.15 5.20 5.60 6.0 6.05 6.3 6.5 6.5 6.75 7.05 7.65 7.7

With n = 14, and a confidence level of 89.6%, d = 27. Counting in 27 averages from each end of the ordered pairwise averages yields the confidence interval of (3.65, 5.55). Thus, with 89.6% confidence, the true mean difference in height velocity is estimated to be between 3.65 and 5.55 units. 16.18 Let  denote the true mean processing time.

H0 :   2

Ha :   2

The test statistic is the signed rank sum. With n = 10, and  = 0.053, H 0 is rejected if the calculated signed rank is greater than or equal to 33. Process Time: Process time minus 2: Signed Rank: Signed Rank Sum = 9

1.4, 2.1, 1.9, 1.7, 2.4, 2.9, 1.8, 1.9, 2.6, 2.2. -0.6, .0.1, -0.1, -0.3, 0.4, 0.9, -0.2, -0.1, 0.6, 0.2. -8.5, 2, -2, -6, 7, 10, -4.5, -2, 8.5, 4.5.

Since the signed-rank sum of 9 does not exceed the critical value of 33, the null hypothesis is not rejected. There is not sufficient evidence to conclude that the mean processing time exceeds two minutes. 16.19 Let 1 ,  2 , and 3 denote the mean importance scores for lower, middle, and upper social class, respectively.

H 0 : 1  2  3 H a : at least two of the three i ’s are different Test statistic: Kruskal Wallis

Chapter 16: Nonparametric (Distribution-Free) Statistical Methods

497

Rejection region: The number of df for the chi-squared approximation is k – 1 = 2. For  = 0.05, Chapter 16 Appendix Table 4 gives 5.99 as the critical value. H 0 will be rejected if KW > 5.99. KW = .17 as given. Since this computed value does not exceed the critical value of 5.99, the null hypothesis is not rejected. The data do not provide enough evidence for concluding that the mean importance scores for lower, middle and upper social classes are not all the same. 16.20 Let 1 ,  2 , and 3 denote the true mean protoporphyrin levels for the normal workers, alcoholics with sideroblasts and alcoholics without sideroblasts, respectively,

H 0 : 1  2  3 H a : at least two of the three i ’s are different Test statistic: Kruskal Wallis Rejection region: The number of df for the chi-squared approximation is k – 1 = 2. For  = 0.05, Chapter 16 Appendix Table 4 gives 5.99 as the critical value. H 0 will be rejected if KW > 5.99.

Normal Workers

Sum of Ranks ni

ri KW 

5 6 20.5 10.5 16.5 29.5 7.5 25 27 24 10.5 18 23 22 13 258 15 17.2

RANKS Alcoholics w/sideroblasts 29.5 35 37 31 38 39 36 41 32 34 40

392.5 11 35.68

Alcoholics w/o sideroblasts 14.5 7.5 16.5 19 20.5 9 12 4 26 3 14.5 1 28 33 2 210.5 15 14.03

12 15(17.2  21) 2  11(35.68  21) 2  15(14.03  21) 2   23.11  (41)(42) 

Since 23.11 > 5.99, the null hypothesis is rejected. The data strongly suggest that mean protoporphyrin levels are not the same for all three groups. 16.21. Let 1 ,  2 , 3 and  4 denote the true mean phosphorous concentrations for the four soil treatments.

498

Chapter 16: Nonparametric (Distribution-Free) Statistical Methods

H 0 : 1  2  3  4 H a : at least two of the four i ’s are different Test statistic: Kruskal Wallis Rejection region: The number of df for the chi-squared approximation is k – 1 = 3. For  = 0.01, Chapter 16 Appendix Table 4 gives 11.34 as the critical value. H 0 will be rejected if KW > 11.34.

Treatment 1 2 3 4

KW 

Ranks 4 8 11 16

1 7 15 20

2 10 14 19

3 6 12 17

5 9 13 18

ri 3 8 13 18

12 5(3  10.5)2  5(8  10.5)2  5(13  10.5)2  5(18  10.5)2   17.86  (20)(21) 

Since 17.86 > 11.34, the null hypothesis is rejected. The data strongly suggests that the mean phosphorous concentration is not the same for the four treatments. 16.22

H 0 : The mean skin potential does not depend on emotion.

H a : The mean skin potential differs for at least two of the four emotions. Test statistic: Friedman Rejection region: With  =0.05, and k-1 = 3, Chapter 16 Appendix Table 4 gives a chi-square critical value of 7.82. H 0 will be rejected if Fr > 7.82.

Fr 

Emotion

Ranks Subject (blocks) 2 3 4 5 6 7 8

Fear Happiness Depression Calmness

4 3 1 2

4 2 3 1

3 2 4 1

4 1 2 3

1 4 3 2

4 3 2 1

4 1 2 3

3 4 2 1

ri 3.375 2.5 2.375 1.75

(12)(8) (3.375  2.5) 2  (2.5  2.5) 2  (2.375  2.5) 2  (1.75  2.5) 2   6.45  (4)(5) 

Since 6.45 < 7.82, the null hypothesis is not rejected. The data do not provide sufficient evidence that the mean skin potential is not the same for all four emotions. 16.23

H 0 : The mean permeability is the same for the four treatments. H a : The mean permeability differs for at least two of the four treatments.

Chapter 16: Nonparametric (Distribution-Free) Statistical Methods

499

Test statistic: Friedman Rejection region: With  = 0.01, and k − 1 = 3, Chapter 16 Appendix Table 4 gives a chi-square critical value of 11.34. H 0 will be rejected if Fr > 11.34.

Fr 

Treatment

Ranks Subject (blocks) 3 4 5 6 7 8 9

1 2 3 4

2 1 4 3

2 1 3 4

2 1 4 3

10 2 1 4 3

ri 2 1 3.9 3.1

(12)(10) (2  2.5)2  (1  2.5)2  (3.9  2.5)2  (3.1  2.5)2   28.92  (4)(5) 

Since 28.92 > 11.34, the null hypothesis is rejected. The data provide convincing evidence that the mean permeability is not the same for all four treatments. 16.24

H 0 : The mean food consumption is the same for the three experimental conditions. H a : The mean food consumption differs for at least two of the three experimental conditions. Test statistic: Friedman Rejection region: With  = 0.01, and k − 1 = 2, Chapter 16 Appendix Table 4 gives a chi-square critical value of 9.21. H 0 will be rejected if Fr > 9.21.

Fr 

Hours

0 24 72

1 2 3

1 1 2 2.5 3 2.5

Ranks Rats (blocks) 4 5 6 7

1 3 2

1 2.5 2.5

1 2 3

1 1 3 2 2 3

ri 1 2.375 2.625

(12)(8) (1  2)2  (2.375  2)2  (2.625  2)2   12.25  (3)(4) 

Since 12.25 > 9.21, the null hypothesis is rejected. The data provide convincing evidence that the mean food consumption depends on the number of hours of food deprivation. 16.25

H 0 : The mean survival rate does not depend on storage temperature. H a : The mean survival rate does depend on storage temperature. Test statistic: Friedman

500

Chapter 16: Nonparametric (Distribution-Free) Statistical Methods Rejection region: With  = 0.05, and k − 1 = 3, Chapter 16 Appendix Table 4 gives a chi-square critical value of 7.82. H 0 will be rejected if Fr > 7.82.

Fr 

Temperature

Ranks Storage Time 24 48 120 168

15.6 21.1 26.7 32.2

3 4 2 1

ri 3 4 2 1

(12)(5) (3  2.5) 2  (4  2.5) 2  (2  2.5) 2  (1  2.5) 2   15  (4)(5) 

Since 15 > 7.82, the null hypothesis is rejected. The data provide convincing evidence that the mean survival rate differs for at least two of the different storage temperatures.