Solution Manual For Biostatistics for the Biological and Health Sciences, 3rd Edition by Marc M. Tri

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

Chapter 1: Introduction to Statistics Section 1-1: Statistical and Critical Thinking ............................................................................ 1 Section 1-2: Types of Data ......................................................................................................... 2 Section 1-3: Collecting Sample Data ......................................................................................... 3 Quick Quiz ................................................................................................................................. 5 Review Exercises ....................................................................................................................... 5 Cumulative Review Exercises .................................................................................................... 6

Chapter 2: Exploring Data with Tables and Graphs Section 2-1: Frequency Distributions for Organizing and Summarizing Data .......................... 7 Section 2-2: Histograms ........................................................................................................... 11 Section 2-3: Graphs That Enlighten and Graphs That Deceive ............................................... 14 Section 2-4: Scatterplots, Correlation, and Regression ............................................................ 17 Quick Quiz ............................................................................................................................... 19 Review Exercises ..................................................................................................................... 19 Cumulative Review Exercises .................................................................................................. 21

Chapter 3: Describing, Exploring, and Comparing Data Section 3-1: Measures of Center .............................................................................................. 23 Section 3-2: Measures of Variation .......................................................................................... 27 Section 3-3: Measures of Relative Standing and Boxplots ...................................................... 31 Quick Quiz ............................................................................................................................... 35 Review Exercises ..................................................................................................................... 35 Cumulative Review Exercises .................................................................................................. 36

Chapter 4: Probability Section 4-1: Basic Concepts of Probability .............................................................................. 39 Section 4-2: Addition Rule and Multiplication Rule ............................................................... 40 Section 4-3: Complements, Conditional Probability, and Bayes’ Theorem ............................ 42 Section 4-4: Risks and Odds .................................................................................................... 43 Section 4-5: Rates of Mortality, Fertility, and Morbidity ........................................................ 45 Section 4-6: Counting............................................................................................................... 47 Quick Quiz ............................................................................................................................... 49 Review Exercises ..................................................................................................................... 50 Cumulative Review Exercises .................................................................................................. 51

Chapter 5: Discrete Probability Distributions Section 5-1: Probability Distributions ...................................................................................... 53 Section 5-2: Binomial Probability Distributions ...................................................................... 54 Section 5-3: Poisson Probability Distributions......................................................................... 57 Quick Quiz ............................................................................................................................... 59

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

Review Exercises ..................................................................................................................... 59 Cumulative Review Exercises .................................................................................................. 60

Chapter 6: Normal Probability Distributions Section 6-1: The Standard Normal Distribution ....................................................................... 63 Section 6-2: Real Applications of Normal Distributions ......................................................... 64 Section 6-3: Sampling Distributions and Estimators................................................................ 66 Section 6-4: The Central Limit Theorem ................................................................................. 70 Section 6-5: Assessing Normality ............................................................................................ 74 Section 6-6: Normal as Approximation to Binomial................................................................ 77 Quick Quiz ............................................................................................................................... 79 Review Exercises ..................................................................................................................... 80 Cumulative Review Exercises .................................................................................................. 81

Chapter 7: Estimating Parameters and Determining Sample Sizes Section 7-1: Estimating a Population Proportion ..................................................................... 83 Section 7-2: Estimating a Population Mean ............................................................................. 88 Section 7-3: Estimating a Population Standard Deviation or Variance ................................... 91 Section 7-4: Bootstrapping: Using Technology for Estimates ................................................. 95 Quick Quiz ............................................................................................................................... 96 Review Exercises ..................................................................................................................... 97 Cumulative Review Exercises .................................................................................................. 98

Chapter 8: Hypothesis Testing Section 8-1: Basics of Hypothesis Testing ............................................................................. 101 Section 8-2: Testing a Claim About a Proportion .................................................................. 102 Section 8-3: Testing a Claim About a Mean .......................................................................... 108 Section 8-4: Testing a Claim About a Standard Deviation or Variance................................. 112 Section 8-5: Resampling: Using Technology for Hypothesis Testing ................................... 115 Quick Quiz ............................................................................................................................. 117 Review Exercises ................................................................................................................... 117 Cumulative Review Exercises ................................................................................................ 119

Chapter 9: Chapter 9: Inferences from Two Samples Section 9-1: Two Proportions................................................................................................. 121 Section 9-2: Two Means: Independent Samples .................................................................... 129 Section 9-3: Matched Pairs..................................................................................................... 136 Section 9-4: Two Variances or Standard Deviations ............................................................. 141 Section 9-5: Resampling: Using Technology for Inferences ................................................. 143 Quick Quiz ............................................................................................................................. 145 Review Exercises ................................................................................................................... 145 Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

Cumulative Review Exercises ................................................................................................ 147

Chapter 10: Correlation and Regression Section 10-1: Correlation ....................................................................................................... 151 Section 10-2: Regression ........................................................................................................ 157 Section 10-3: Prediction Intervals and Variation ................................................................... 165 Section 10-4: Multiple Regression ......................................................................................... 167 Section 10-5: Dummy Variables and Logistic Regression ..................................................... 170 Quick Quiz ............................................................................................................................. 171 Review Exercises ................................................................................................................... 172 Cumulative Review Exercises ................................................................................................ 173

Chapter 11: Goodness-of-Fit and Contingency Tables Section 11-1: Goodness-of-Fit ............................................................................................... 177 Section 11-2: Contingency Tables ......................................................................................... 182 Quick Quiz ............................................................................................................................. 186 Review Exercises ................................................................................................................... 186 Cumulative Review Exercises ................................................................................................ 187

Chapter 12: Analysis of Variance Section 12-1: One-Way ANOVA .......................................................................................... 191 Section 12-2: Two-Way ANOVA .......................................................................................... 193 Quick Quiz ............................................................................................................................. 194 Review Exercises ................................................................................................................... 195 Cumulative Review Exercises ................................................................................................ 196

Chapter 13: Nonparametric Tests Section 13-2: Sign Test .......................................................................................................... 199 Section 13-3: Wilcoxon Signed-Ranks Test for Matched Pairs ............................................. 200 Section 13-4: Wilcoxon Rank-Sum Test for Two Independent Samples ............................... 201 Section 13-5: Kruskal-Wallis Test for Three or More Samples ............................................. 205 Section 13-6: Rank Correlation .............................................................................................. 207 Quick Quiz ............................................................................................................................. 208 Review Exercises ................................................................................................................... 209 Cumulative Review Exercises ................................................................................................ 210

Chapter 14: Survival Analysis Section 14-1: Life Tables ....................................................................................................... 213 Section 14-2: Kaplan-Meier Survival Analysis...................................................................... 215 Quick Quiz ............................................................................................................................. 217 Review Exercises ................................................................................................................... 217 Cumulative Review Exercises ................................................................................................ 218

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

Chapter 1: Introduction to Statistics Section 1-1: Statistical and Critical Thinking 1. The respondents are a voluntary response sample or a self-selected sample. Because those with strong interests in the topic are more likely to respond, it is very possible that their responses do not reflect the opinions or behavior of the general population.

2. a. The sample consists of the 1046 adults who were surveyed. The population consists of all adults. b. When asked, respondents might be inclined to avoid the shame of the unhealthy habit of not washing their hands, so the reported rate of 70% might well be much higher than it is in reality. It is generally better to observe or measure human behavior than to ask subjects about it.

3. Statistical significance is indicated when methods of statistics are used to reach a conclusion that a treatment is effective, but common sense might suggest that the treatment does not make enough of a difference to justify its use or to be practical. It is possible for a study to have statistical significance, but not practical significance.

4. No. Correlation does not imply causation. The example illustrates a correlation that is clearly not the result of any interaction or cause effect relationship between per capita consumption of margarine and the divorce rate in Maine.

5. Yes, there does appear to be a potential to create a bias.

6. No, there does not appear to be a potential to create a bias.

7. No, there does not appear to be a potential to create a bias.

8. Yes, there does appear to be a potential to create a bias.

9. The sample is a voluntary response sample and has strong potential to be flawed.

10. The samples are voluntary response samples and have potential for being flawed, but this approach might be necessary due to ethical considerations involved in randomly selecting subjects and somehow imposing treatments on them.

11. The sampling method appears to be sound.

12. The sampling method appears to be sound.

13. The Ornish weight loss program has statistical significance, because the results are so unlikely (3 chances in 1000) to occur by chance. It does not have practical significance because the amount of lost weight (3.3 lb) is so small.

14. Because there is only one chance in a thousand of getting such success rates by chance, the difference does appear to have statistical significance. The 92% success rate for surgery appears to be substantially better than the 72% success rate for splints, so the difference does appear to have practical significance.

15. The difference between Mendel’s 25% rate and the result of 26% is not statistically significant. According to Mendel’s theory, 145 of the 580 peas would have yellow pods, but the results consisted of 152 peas with yellow pods. The difference of 7 peas with yellow pods among the 580 offspring does not appear to be statistically significant. The difference does not appear to have practical significance.

16. Because there is a 25% chance of getting such results with a program that has no effect, the program does not appear to have statistical significance. Because the average increase is only 3 IQ points, the program does not appear to have practical significance.

17. The sample percentage of males is 49.5%, and it appears to be very close to the percentage expected under normal circumstances. It does not appear to have statistical significance, nor does it appear to have practical significance.

18. Because there is a 15% chance of getting such results with a medication that has no effect, the medication does not appear to have statistical significance. Because the average decrease is only 2 mmHg, the medication does not appear to have practical significance.

19. Because there is an 8% chance of getting such nausea rates by chance, the results do not appear to have statistical significance. Also, they do not appear to have practical significance.

20. Because there is less than a 1% chance of getting the results obtained in this study, the results have statistical significance. They also appear to have practical significance.

21. Yes. Each column of 8 AM and 12 AM temperatures is recorded from the same subject, so each pair is matched.

22. No. The source is from university researchers who do not appear to gain from distorting the data.

23. The data can be used to address the issue of whether there is a correlation between body temperatures at 8 AM and at 12 AM. Also, the data can be used to determine whether there are differences between body temperatures at 8 AM and at 12 AM. 24. Because the differences could easily occur by chance (with a 64% chance), the differences do not appear to have statistical significance.

25. No. The white blood cell counts measure a different quantity than the red blood cell counts, so their differences are meaningless.

26. The issue that can be addressed is whether there is a correlation, or association, between white blood cell counts and red blood cell counts.

27. No. The National Center for Health Statistics has no reason to collect or present the data in a way that is biased.

28. No. Correlation does not imply causation, so a statistical correlation between white blood cell counts and red blood cell counts should not be used to conclude that higher white blood cell counts are the cause of higher red blood cell counts.

29. It is questionable that the sponsor is the Idaho Potato Commission and the favorite vegetable is potatoes.

30. The sample is a voluntary response sample, so there is a good chance that the results do not reflect the larger population of people who have a water preference.

31. The correlation, or association, between two variables does not mean that one of the variables is the cause of the other. Correlation does not imply causation. Clearly, sour cream consumption is not directly related in any way to motorcycle fatalities.

32. The sponsor of the poll is an electronic cigarette maker, so the sponsor does have an interest in the poll results.

The source is questionable.

33. The correlation, or association, between two variables does not mean that one of the variables is the cause of the other. Correlation does not imply causation. Common sense suggests that cheese consumption is not directly related in any way to fatalities from bedsheet entanglements.

34. The correlation, or association, between two variables does not mean that one of the variables is the cause of the other. Correlation does not imply causation.

35. The survey results are from subjects who chose to respond, so the results constitute a voluntary response sample. Consequently, the results are questionable.

36. Because the nutritionists are paid such large amounts of money, they might be more inclined to find favorable results. It is very possible that the results represent desired outcomes instead of actual outcomes.

37. a. 0.45(3014) = 1356.3 b. No. The actual count of survey subjects who have at least one chronic condition must be a whole number. c. 1356 adults (Any value from 1342 to 1371 would yield a percentage that rounds to 45%.) d. 1206 / (1206 +1808) = 0.40, or 40% 38. a. 0.828(198) = 163.944 patients b. No. Because the result is a count of patients, the result must be a whole number. c. 164 patients d. 198 / (199 +198) = 0.499, or 49.9%, or 50% rounded 39. The wording of the question is biased and tends to encourage negative responses. The sample size of 20 is too small. Survey respondents are self-selected instead of being randomly selected by the newspaper. If 20 readers respond, the percentages should be multiples of 5, so 87% and 13% are not possible results.

40. All percentages of success should be multiples of 5. The given percentages cannot be correct.

Section 1-2: Types of Data 1. The population consists of all drug tests of adults in the United States, and the sample is the 10 million drug tests that were analyzed. The value of 4.2% is a statistic because it is obtained from the sample.

2. a. quantitative d. quantitative b. categorical e. quantitative c. categorical

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

3.

4.

Only part (a) describes discrete data. a. The sample is the 36,000 adults who were surveyed. The population is all adults in the United States. b. statistic c. ratio d. discrete 5. statistic 17. discrete 6. statistic 18. continuous 7. parameter 19. discrete 8. parameter 20. continuous 9. statistic 21. nominal 10. parameter 22. ordinal 11. parameter 23. ordinal 12. parameter 24. ratio 13. continuous 25. interval 14. discrete 26. nominal 15. continuous 27. ratio 16. discrete 28. interval 29. The numbers are not counts or measures of anything. They are at the nominal level of measurement, and it makes no sense to compute the average (mean) of them.

30. The ranks are at the ordinal level of measurement. Differences between medical schools cannot be interpreted, so there is no way to know whether the difference between Harvard and New York University is the same as the difference between New York University and Duke.

31. The numbers on the pain scale are at the ordinal level of measurement. Differences cannot be determined, so there is no way to know whether the first patient has twice as much pain as the second patient. Ratios such as “twice” make no sense with ordinal data.

32. The temperatures are at the interval level of measurement. Because there is no natural starting point with 0ºF representing no heat, it is wrong to state that the second patient is “5% cooler than the first patient.”

33. a. Continuous, because the number of possible values is infinite and not countable. b. Discrete, because the number of possible values is finite. c. Discrete, because the number of possible values is finite. d. Discrete, because the number of possible values is infinite and countable.

34. Interval level of measurement. The direction of north represented by 0° is arbitrary, and 0° does not represent “no direction.” Differences between degrees are meaningful; the difference between 30° and 60° is the same as the difference between 150° and 180°. But ratios are not meaningful; the ratio of 60° to 30° does not result in twice some direction. (These degree measurements are directions, not amounts of rotation.)

Section 1-3: Collecting Sample Data 1. The study is an experiment because subjects were given treatments.

2. The subjects in the study did not know whether they were given the magnet treatment or the sham treatment, and those who administered the treatments also did not know.

3. The group sample sizes are large enough so that the researchers could see the effects of the two treatments, but it would have been better to have larger samples.

4. The sample appears to be a convenience sample. Given that the subjects were all patients at a Veterans Affairs hospital, it is not likely that the sample is representative of the population, so it is questionable whether the results can be generalized for the population of subjects with chronic low back pain.

5. The sample appears to be a convenience sample. By e-mailing the survey to a readily available group of Internet users, it was easy to obtain results. Although there is a real potential for getting a sample group that is not representative of the population, indications of which ear is used for cell phone calls and which hand is dominant do not appear to be factors that would be distorted much by a sample bias.

6.

7.

The study is an observational study because the subjects were not given any treatment.

With 717 responses, the response rate is 14%, which does appear to be quite low. In general, a very low response rate creates a serious potential for getting a biased sample that consists of those with a special interest in the topic.

8. Answers vary, but the following are good possibilities. a. Obtain a printed copy of the class roster, assign consecutive numbers (integers), then use a computer to randomly generate six of those numbers. b. Select every third student leaving class until six students are chosen. c. Randomly select three males and three females. d. Randomly select a row, and then select the students in that row. (Use only the first six to meet the requirement of a sample of size six.) e. Select the first six students who enter the class.

9. systematic 15. stratified 10. convenience 16. systematic 11. random 17. random 12. stratified 18. cluster 13. cluster 19. convenience 14. random 20. systematic 21. Observational study. The sample is a convenience sample consisting of subjects who decided to respond. Such voluntary response samples have a high chance of not being representative of the larger population, so the sample may well be biased, as it was in this case.

22. Experiment. The sample subjects consist of male physicians only. It would have been better to include females.

Also, it would be better to include male and females who are not physicians.

23. Experiment. This experiment would create an extremely dangerous and illegal situation that has a real potential to result in injury or death. It’s difficult enough to drive in New York City while being completely sober.

24. Observational study. The sample of eight respondents is too small.

25. Experiment. The biased sample created by using a small sample of college students cannot be fixed by using a larger sample. The larger sample will still be a biased sample that is not representative of the population of all adults.

26. Experiment. Calling the subjects and asking them to report their weights has a high risk of getting results that do not reflect the actual weights. It would have been much better to somehow measure the weights instead of asking the subjects to report them.

27. Observational study. Respondents are not likely to respond honestly because there is a “social desirability bias” causing respondents to reply in ways that will be viewed favorably.

28. Observational study. The number of responses is very small, and the response rate of only 1.52% is far too small. With such a low response rate, there is a real possibility that the sample of respondents is biased and consists only of those with special interests in the survey topic.

29. prospective study 33. matched pairs design 30. retrospective study 34. randomized block design 31. cross-sectional study 35. completely randomized design 32. prospective study 36. matched pairs design 37. Prospective: The experiment was begun and results were followed forward in time.

Randomized: Subjects were assigned to the different groups through a process of random selection, whereby they had the same chance of belonging to each group.

Double-blind: The subjects did not know which of the three groups they were in, and the people who evaluated results did not know either.

Placebo-controlled: There was a group of subjects who were given a placebo; by comparing the placebo group to the two treatment groups, the effects of the treatments might be better understood.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

38. a. Not a simple random sample, but it is a random sample. b. Simple random sample and also a random sample. c. Not a simple random sample and not a random sample.

Quick Quiz 1. No. The numbers do not measure or count anything.

2. nominal 6. statistic 3. continuous 7. no 4. quantitative data 8. observational study 5. ratio 9. The subjects did not know whether they were getting aspirin or the placebo.

10. simple random sample Review Exercises 1. The respondents are a voluntary response sample or a self-selected sample. Because those with strong interests in the topic are more likely to respond, it is very possible that their responses do not reflect the opinions or behavior of the general population.

2. a. The survey uses a voluntary response sample, and those with special interests are more likely to respond, so it is very possible that the sample is not representative of the population. b. Because the statement refers to 72% of all Americans, it is a parameter (but it is probably based on a 72% rate from the sample, and the sample percentage is a statistic). c. observational study 3. Randomized: Subjects were assigned to the different groups through a process of random selection, whereby they had the same chance of belonging to each group. Double-blind: The subjects did not know which of the three groups they were in, and the people who evaluated results did not know either.

4. No. Correlation does not imply causality.

5. a. systematic d. convenience b. stratified e. cluster c. simple random sample 6. The survey was sponsored by the American Laser Centers, and 24% said that the favorite body part is the face, which happens to be a body part often chosen for some type of laser treatment. The source is therefore questionable.

7. a. discrete b. ratio c. The mailed responses would be a voluntary response sample, so those with strong opinions or greater interest in the topics are more likely to respond. It is very possible that the results do not reflect the true opinions of the population of all state residents. d. stratified e. cluster 8.

a. 0.628(199) = 125 patients b. 164 / 198 = 82.8%

9.

a. interval data; systematic sample b. nominal data; stratified sample c. ordinal data; convenience sample 10. Because there is less than a 1% chance of getting the results by chance, the method does appear to have statistical significance. The result of 239 males in 291 births is a rate of 82% so it is above the 50% rate expected by chance, and it does appear to be high enough to have practical significance. The procedure appears to have both statistical significance and practical significance.

Cumulative Review Exercises 135 +149 +145 +129 +118 +119 +115 +133 +107 +188 +127 +131 = 133.0. The IQ score of 188 12 appears to be substantially higher than the other IQ scores.

1.

The mean is

2.

0.513 = 0.000122 203 - 176 = 4.50, which is an unusually high value.

6

3.

4.

5.

98.2 - 98.6 =- 6.64 0.62 106

6.

1.959962 ×0.25

7.

0.032

= 1068

188 - 107

= 20.25 4 (135 - 133.0)2 = 0.364 11

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

8.

9.

(98.4 - 98.6)2 + (98.6 - 98.6)2 + (98.8 - 98.6)2 3- 1

= 0.04 = 0.20

0.36 = 0.000729

10. 812 = 68,719,476,736 (or about 68,719,477,000) 11. 856 = 377,149,515,625 (or about 377,149,520,000) 12. 0.212 = 0.000000004096

Chapter 2: Exploring Data with Tables and Graphs Section 2-1: Frequency Distributions for Organizing and Summarizing Data 1. No. Instead of frequencies that start low, reach a maximum, and then decrease, the given distribution starts with a maximum frequency of 14 and the frequencies then decrease.

2.

Nicotine (mg) 1.0–1.1 1.2–1.3 1.4–1.5 1.6–1.7 1.8–1.9 3.

Relative Frequency 56% 16% 12% 12% 4%

There is overlap in the class limits of 1.0–1.2, 1.2–1.4, and so on. For an amount such as 1.2 mg, it would not be clear which class contains the value. The classes should not overlap.

4. The sum of the relative frequencies is 125%, but it should be 100%, with a small round off error. All of the relative frequencies appear to be roughly the same, but if they are from a normal distribution, they should start low, reach a maximum, and then decrease.

5. Class width: 2.0 Class midpoints: 2.95, 4.95, 6.95, 8.95, 10.95, 12.95, 14.95 Class boundaries: 1.95, 3.95, 5.95, 7.95, 9.95, 11.95, 13.95, 15.95 Number: 147 6. Class width: 2.0 Class midpoints: 2.95, 4.95, 6.95, 8.95, 10.95, 12.95 Class boundaries: 1.95, 3.95, 5.95, 7.95, 9.95, 11.95, 13.95 Number: 153 7. Class width: 100 Class midpoints: 49.5, 149.5, 249.5, 349.5, 449.5, 549.5, 649.5 Class boundaries: –0.5, 99.5, 199.5, 299.5, 399.5, 499.5, 599.5, 699.5 Number: 153 8. Class width: 100 Class midpoints: 149.5, 249.5, 349.5, 449.5, 549.5 Class boundaries: 99.5, 199.5, 299.5, 399.5, 499.5, 599.5 Number: 147 9. Yes. The frequencies start low, reach a maximum, and then decrease.

10. Not normal. The frequencies are not approximately symmetric. The maximum frequency is in the second class.

11. Yes. Except for the single value that lies between 600 and 699, the frequencies start low, reach a maximum of 90, and then decrease. The values below the maximum are very roughly a mirror image of those above it. (That single value between 600 and 699 is an outlier that makes the determination of a normal distribution somewhat questionable, but using a loose interpretation of the criteria for normality, it is reasonable to conclude that the distribution is normal.)

12. Yes. Except for two values that lie between 500 and 599, there is a low frequency of 25, then a maximum frequency of 92, and then a low frequency of 28. The values below and above the maximum are roughly a mirror image. (Those two values between 500 and 599 are outliers that make the determination of a normal distribution somewhat questionable, but using a loose interpretation of the criteria for normality, it is reasonable to conclude that the distribution is normal.)

13. The pulse rates do appear to be from a normal distribution.

Pulse Rate 40–49 50–59 60–69 70–79 80–89 90–99

Frequency 1 10 13 9 5 2

14. The pulse rates do appear to be from a normal distribution.

Pulse Rate 30–39 40–49 50–59 60–69 70–79 80–89 90–99 100–109

Frequency 1 0 4 7 10 11 6 1

15. The verbal IQ scores do appear to be from a population having a normal distribution.

Verbal IQ 50–59 60–69 70–79 80–89 90–99 100–109 110–119 120–129

Frequency 1 2 6 9 4 4 2 2

16. The verbal IQ scores do appear to be from a population having a normal distribution.

Verbal IQ

Frequency

60–69

1

70–79

7

80–89

8

90–99

4

100–109

1

17. No, the pulse rates are not dramatically different. The pulse rates appear to be from a normal distribution.

Pulse Rates 40–49 50–59 60–69 70–79 80–89 90–99

Frequency 2 23 53 43 25 5

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy Section 2-1: Frequency Distributions for Organizing and Summarizing Data 100–109

2

7

18. No, the pulse rates are not dramatically different. The pulse rates appear to be from a normal distribution.

Pulse Rates 30–39 40–49 50–59 60–69 70–79 80–89 90–99 100–109

Frequency 1 1 17 33 41 37 13 4

19. The distribution does appear to be a normal distribution.

Weight (kg) in September 50–59 60–69 70–79 80–89 90–99

Frequency 2 12 11 3 4

20. No. There does not appear to be such a dramatic weight gain.

Weight (kg) in April 50–59 60–69 70–79 80–89 90–99 100–109

Frequency 3 12 8 7 1 1

21. Because there are disproportionately more 0s and 5s, it appears that the heights were reported instead of measured. Consequently, it is likely that the results are not very accurate.

Last Digit 0 1 2 3 4 5 6 7 8 9

Frequency 9 2 1 3 1 15 2 0 3 1

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy Section 2-1: Frequency Distributions for Organizing and Summarizing Data 22. Because there are disproportionately more 0s and 5s, it appears that the heights were reported instead of measured. There does appear to be a gap due to the tendency of respondents to round their heights to values ending in 0 or 5. Because the results appear to be reported instead of measured, it is likely that the results are not very accurate.

Last Digit 0 1 2 3 4 5 6 7 8 9

Frequency 26 1 1 2 2 12 1 0 4 1

23. The two distributions appear to be very similar.

White Blood Cell Count 2.0–3.9 4.0–5.9 6.0–7.9 8.0–9.9 10.0–11.9 12.0–13.9 14.0–15.9

Females 4.8% 38.1% 31.3% 19.7% 5.4% 0.0% 0.7%

Males 5.9% 39.2% 32.7% 19.0% 2.0% 1.3% 0.0%

24. There do appear to be differences, but overall, they are not very substantial differences.

Blood Platelet Count 0–99 100–199 200–299 300–399 400–499 500–599 600–699

Males 0.7% 33.3% 58.8% 6.5% 0.0% 0.0% 0.7%

Females 0.0% 17.0% 62.6% 19.0% 0.0% 1.4% 0.0%

25.

White Blood Cell Count of Females Less than 4.0 Less than 6.0 Less than 8.0 Less than 10.0 Less than 12.0 Less than 14.0 Less than 16.0

Cumulative Frequency 7 63 109 138 146 146 147

9

26.

White Blood Cell Count of Males Less than 4.0 Less than 6.0 Less than 8.0 Less than 10.0 Less than 12.0 Less than 14.0

Cumulative Frequency 9 69 119 148 151 153

27. Because only the five leading causes of death are listed, we know only that any other cause of death must have fewer than 150,005 deaths.

Cause of Death Heart Disease Cancer Accidents Chronic Lower Respiratory Disease Stroke

Relative Frequency 37.9% 34.5% 10.0% 9.0% 8.6%

28. Yes, it appears that births occur on the days of the week with frequencies that are about the same.

Day Monday Tuesday Wednesday Thursday Friday Saturday Sunday

Relative Frequency 13.0% 16.5% 18.0% 14.3% 14.3% 10.8% 13.3%

29. The frequencies aren’t very symmetric, but using a very loose interpretation, the measurements do appear to be from a normal distribution.

Systolic Blood Pressure (mmHg) of Females 80–99 100–119 120–139 140–159 160–179 180–199

Frequency 8 62 53 21 2 1

Section 2-2: Histograms 1. The histogram should be bell-shaped.

2. Not necessarily. Because the sample subjects themselves chose to be included, the voluntary response sample might not be representative of the population.

3. With a data set that is so small, the true nature of the distribution cannot be seen with a histogram.

4. The outlier will result in a single bar that is far away from all of the other bars in the histogram, and the height of that bar will correspond to a frequency of 1.

5. Approximately 50 6. Approximate values: Class width: 0.5 mm, lower limit of first class: 2.0 mm, upper limit of first class: 2.5 mm 7. The largest possible value is approximately 4.5 mm, which is not an outlier.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy Section 2-2: Histograms 8.

9.

11

The histogram very roughly approximates a bell shape, so it appears that the sample is from a population having a normal distribution.

The distribution appears to be normal, and there are no outliers.

10. The distribution appears to be normal, and there are no outliers.

11. The distribution appears to be normal, and there are no outliers.

12. The distribution appears to be approximately normal (or perhaps skewed to the right), and there are no outliers.

13. The distribution does not appear to be normal. It appears to be skewed to the right.

14. The distribution does not appear to be normal. It appears to be skewed to the right.

15. The distribution does not appear to be normal. It appears to be skewed to the right.

16. The distribution is dramatically far from normal. It is skewed to the right.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy Section 2-2: Histograms

13

17. The digits 0 and 5 appear to occur more often than the other digits, so it appears that the heights were reported and not actually measured. This suggests that the data might not be very useful.

18. The digits 0 and 5 appear to occur more often than the other digits, so it appears that the weights were reported and not actually measured. This suggests that the data might not be very useful.

19. Only part (c) appears to represent data from a normal distribution. Part (a) has a systematic pattern that is not that of a straight line, part (b) has points that are not close to a straight-line pattern, and part (d) is really bad because it shows a systematic pattern and points that are not close to a straight-line pattern.

20. The histogram for the movie lengths suggests that the data have a distribution that is approximately normal with an outlier of 120 minutes. The histogram for the time of tobacco use appears to be similar to the histogram for the times of alcohol use, but both distributions appear to be skewed to the right, with each histogram having an outlier.

Section 2-3: Graphs That Enlighten and Graphs That Deceive 1. The data set is too small for a dotplot to reveal important characteristics of the data. Because the data are listed in order for each of the last several years, a time-series graph would be most effective for these data.

2. Yes, the original data values can be found from the stemplot.

5 4 6 8 7 0369 8 123 9 8 3.

4.

5.

No. Graphs should be constructed in a way that is fair and objective. The readers should be allowed to make their own judgments, instead of being manipulated by misleading graphs.

No. If the sample is a bad sample, such as one obtained from voluntary responses, there are no graphs or statistical methods that can be used to salvage the data.

The pulse rate of 36 beats per minute appears to be an outlier.

6.

There do not appear to be any outliers.

7.

The data are arranged in order from lowest to highest, as 36, 56, 56, and so on.

3 6 4 5 668 6 044666 7 6888 8 02468 9 4

8.

The two values closest to the middle are 72 mmHg and 74 mmHg. 6 0022468 7 0000246688 8 22468 9 00

9.

There was a steep jump in the first four years, but the numbers of triplets have shown a downward trend in the past several years.

10. The first five years don’t show much change, but the trend starts to climb dramatically in 1999.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

11. Misconduct includes fraud, duplication, and plagiarism, and it does appear to be a major factor.

12.

13.

14.

15. The distribution appears to be roughly bell-shaped, so the distribution is approximately normal.

16. The distribution appears to be roughly bell-shaped, so the distribution is approximately normal.

17. The two costs are one-dimensional in nature, but the baby bottles are three-dimensional objects. The $4500 cost isn’t even twice the $2600 cost, but the baby bottles make it appear that the larger cost is about five times the smaller cost.

18. The graph is misleading because it depicts one-dimensional data with three-dimensional boxes. See the first and last boxes in the graph. Workers with advanced degrees have annual incomes that are roughly 3 times the incomes of those with no high school diplomas, but the graph exaggerates this difference by making it appear that workers with advanced degrees have incomes that are roughly 27 times the amounts for workers with no high school diploma.

19. 96 96 59 97 0001112333444 97 55666666788888999 98 00000000000002222233444444444444 98 5555666666666666666777777888888899 99 001244 99 56 20. Because the data are listed in order according to day, a time-series graph would be most revealing about the nature of the data. The time-series graph shows two distinct major peaks: in around the beginning of 2021 and in September 2021.

Section 2-4: Scatterplots, Correlation, and Regression 1. The term linear refers to a straight line, and r measures how well a scatterplot of the sample paired data fits a straight-line pattern.

2. No. Finding the presence of a statistical correlation between two variables does not justify any conclusion that one of the variables is a cause of the other.

3.

A scatterplot is a graph of paired (x, y) quantitative data. It helps us by providing a visual image of the data plotted as points, and such an image is helpful in enabling us to see patterns in the data and to recognize that there may be a correlation between the two variables.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

4.

a. 1 b. 0

5.

c. 0 d. - 1 There does not appear to be a linear correlation between brain volume and IQ score.

6.

There does appear to be a linear correlation between the chest sizes and weights of bears.

7.

There does not appear to be a linear correlation between the heights of fathers and the heights of their first sons.

8.

There does not appear to be a correlation between pulse rates of females and males. The major flaw with this exercise is that the data are not paired as required. The results are therefore meaningless.

9.

With n = 5 pairs of data, the critical values are ±0.878. Because r = 0.127 is between –0.878 and 0.878, there is not sufficient evidence to conclude that there is a linear correlation.

10. With n = 7 pairs of data, the critical values are ±0.754. Because r = 0.980 is in the right tail region beyond 0.754, there is sufficient evidence to conclude that there is a linear correlation.

11. With n = 10 pairs of data, the critical values are ±0.632. Because r =- 0.017 is between –0.632 and 0.632, there is not sufficient evidence to conclude that there is a linear correlation.

12. With n = 10 pairs of data, the critical values are ±0.632. Because r =- 0.076 is between –0.632 and 0.632, there is not sufficient evidence to conclude that there is a linear correlation. The data are not paired, so the results are meaningless.

13. Because the P-value of 0.839 is not small (such as 0.05 or less), there is a high chance of getting the sample results when there is no correlation. There is not sufficient evidence to conclude that there is a linear correlation.

14. Because the P-value of 0.0001 is small (such as 0.05 or less), there is a small chance of getting the sample results when there is no correlation. There is sufficient evidence to conclude that there is a linear correlation.

15. Because the P-value of 0.963 is not small (such as 0.05 or less), there is a high chance of getting the sample results when there is no correlation. There is not sufficient evidence to conclude that there is a linear correlation.

16. Because the P-value of 0.835 is not small (such as 0.05 or less), there is a high chance of getting the sample results when there is no correlation. There is not sufficient evidence to conclude that there is a linear correlation.

17. The scatterplot shows a pattern suggesting, not too surprisingly, that as the numbers of new cases of COVID-19 increase, the numbers of deaths also increase. There appears to be a correlation. Among the 800 points, there appears to be a somewhat strong correlation except for the 45 points farthest to the right. A flaw in the paired data is that vaccinations began roughly a year into the pandemic, so the paired data of new cases and deaths consists of a population that is changing instead of being constant.

Quick Quiz 1. The class width is 20.

2. The class boundaries are 19.5 and 39.5.

3. No, it is impossible to determine the original values.

4. 153, 154, 158 7. time-series graph 5. The histogram will be bell-shaped.

8. scatterplot 6. variation 9. Pareto chart 10. A frequency distribution is in the format of a table; a histogram is a graph.

Review Exercises 1.

Reported Weight (lb) 135–159 160–184 185–209 210–234 235–259

Frequency 2 7 5 5 1

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

2.

The data appear to be from a population with a distribution that is approximately normal. The bars start low, reach a maximum, and then decrease, and the left half of the histogram is approximately a mirror image of the right half. The graph is approximately bell-shaped.

3.

By using fewer classes, the histogram does a better job of illustrating the distribution.

4.

There are no outliers.

13

8

14 15

0

16

0079

17

00

18

2

19

0038

20 5 21 26 22 355 23 24 7 5.

Yes. There is a pattern suggesting that there is a relationship.

6.

a. time-series graph c. Pareto chart

b. scatterplot

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy By using a vertical scale that starts at 49.0% instead of 0%, the difference is greatly exaggerated. The graph creates the false impression that female enrollees outnumber male enrollees by a ratio of roughly 2.5 to 1, but the actual percentages of 50.5% and 49.5% are very much closer than that. If the graph had been created with a vertical axis starting at 0%, the difference between the two bars would be almost imperceptible.

Cumulative Review Exercises 7.

1.

Grooming Time (min) 0–9 10–19 20–29 30–39 40–49

Frequency 2 3 9 4 2

2.

The histogram is approximately bell-shaped. The frequencies increase to a maximum and then decrease, and the left half of the histogram is roughly a mirror image of the right half. The data do appear to be from a population with a normal distribution.

3.

0 05 1 255 2 024555778 3 0055 4 05

4.

There are disproportionately more last digits of 0 and 5. Fourteen of the 20 times have last digits of 0 or 5.

It appears that the subjects reported their own results and they tended to round the results. The data do not appear to be very accurate.

Last Digit 0 1 2 3 4 5 6 7 8 9

Frequency 5 0 2 0 1 9 0 2 1 0

5.

6.

a. ratio b. continuous c. No. The grooming times are quantitative data. d. statistic The scatterplot helps address the issue of whether there is a correlation between the heights of mothers and the heights of their first daughters. The scatterplot does not reveal a clear pattern suggesting that there is a correlation.

Chapter 3: Describing, Exploring, and Comparing Data Section 3-1: Measures of Center 1. The term average is not used in statistics. The term mean should be used for the result obtained by adding all of the sample values and dividing the total by the number of sample values.

2. No. The 50 amounts are all weighed equally in the calculation that yields the mean of 17.5%, but states have different populations, so the mean should be calculated using a weighted mean that takes into account the populations in the different states. (The CDC reported a value of 17.1% for all of the states.)

80 + 94 + 58 + 66 + 56 3. Using the five values, the mean is x = = 70.8 bpm and the median is 66.0 bpm. Using the 5 six values that include the outlier, x = 182.3 bpm and the median is 73.0 bpm. The outlier caused the mean to change by a substantial amount, but the median did not change by very much. The median is resistant to the effect of the outlier; the mean is not resistant.

4. They all use different approaches for providing a value (or values) of the center or middle of the sorted list of data.

1.8 +1.7 +1.7 +1.6 +1.5 +1.4 +1.4 +1.3 +1.3 +1.2 +1.2 +1.1 = 1.43 mg.

12 1.4 +1.4 The median is = 1.40 mg.

2 The modes are 1.7 mg, 1.4 mg, 1.3 mg, and 1.2 mg.

1.1+1.8 The midrange is = 1.45 mg.

2 Apart from the fact that the other nicotine amounts are lower than those given, nothing meaningful can be known about the sample of all such amounts.

5.

The mean is x =

6.

The mean is x =

1+14 + 4 +16 + 2 +15 + 3 +15 +19 + 5 +11+13 +14 + 9 = 10.1 g.

14 11+13 The median is = 12.0 g.

2 The modes are 14 g and 15 g.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

The midrange is

1+19

= 10.0 g.

2 Americans consume some brands of cereal much more often than others, but the 14 brands are all weighted equally in the calculations, so the statistics are not necessarily representative of the population of all servings of cereal consumed by Americans.

7.

The mean is x =

88.3 + 86.5 + 71.3 + 81.6 + 75.6 +

+ 83.1+ 90.4 + 78.6 + 75.1+ 69.2 15

= 79.50 kg.

The median is 82.40 kg.

There is no mode.

59.9 + 90.4 The midrange is = 75.15 kg.

2 Because the measurements were made in 1988, they are not necessarily representative of the current population of all males in the Army.

8.

The mean is x = The median is

1560 +1665 +1711+1660 +1572 +

1660 +1665

+1668 +1654 +1666 +1599 +1673 = 1641.8 mm.

16

= 1662.5 mm.

2 The mode is 1707 mm.

1521+1711 The midrange is = 1616.0 mm.

2 The Army has height restrictions, so females in the Army are not necessarily representative of the population of all adult females.

9.

The mean is x = The median is

524 + 607 + 266 + 485 + 405 + 723 +

524 + 540

+ 540 + 249 + 692 + 675 + 545 + 373 = 491.6.

18

= 532.0.

2 There is no mode.

249 + 723 = 486.0.

The midrange is 2 The identification numbers are nominal data that are just replacements for names. They do not measure or count anything, so the resulting statistics are meaningless.

10. The mean is x = The median is

69 + 70 + 67 + 68 + 70 + 73 +

70 + 70

+ 68 + 69 + 70 + 73 + 72 + 72 = 70.5 in.

20

= 70.0 in.

2 The mode is 72 in.

66 + 81 = 73.5 in.

The midrange is 2 Because the heights were reported instead of being measured, it is very possible that the results are not likely to represent the population.

11. The mean is x = The median is

41, 945 + 42,196 + 43, 005 +

37, 423 + 37, 473 2

+ 37, 473 + 36, 560 + 36,120 = 38, 000.0 deaths 20

= 37, 448 deaths.

There is no mode.

3 2, 479 + 43, 510 The midrange is = 37, 994.5 deaths.

2 The data are time-series data, but the measures of center do not reveal anything about a trend consisting of a pattern of change over time.

2 +1+1+1+1+1+1+ 4 +1+ 2 + 2 +1+ 2 + 3 + 3 + 2 + 3 +1+ 3 +1+ 3 +1+ 3 + 2 + 2 = 1.9.

25 The median is 2.0.

The mode is 1.

1+ 4 The midrange is = 2.5.

2 The mode of 1 correctly indicates that the smooth-yellow peas occur more than any other phenotype, but the other measures of center do not make sense with these data at the nominal level of measurement.

12. The mean is x =

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

13. The mean is x = The median is

0+0+0+

38 + 41

+ 34 + 36 + 38 + 41+ 41+ 41+ 20

+ 53 + 54 + 55

= 32.6 mg.

= 39.5 mg.

2 The mode is 0 mg.

0 + 55 The midrange is = 27.5 mg.

2 Americans consume some brands much more often than others, but the 20 brands are all weighted equally in the calculations, so the statistics are not necessarily representative of the population of all cans of the same 20 brands consumed by Americans.

7.9 + 7.7 + 8 + 7.9 + 7.9 + 7.8 + 7.7 + 7.8 + 7.4 + 6.8 + 7.2 + 7 + 6.8 = 7.53%.

13 The median is 7.70%.

The mode is 7.9%.

6.8 + 8 = 7.40%.

The midrange is 2 The data are time-series data, but the measures of center do not reveal anything about a trend consisting of a pattern of change over time.

9 +10 +10 + 20 + 40 + 50 + + 0 + 0 + 0 + 0 + 0 + 0 15. The mean is x = = 2.8 cigarettes.

50 The median is 0.0 cigarettes.

The mode is 0 cigarettes.

0 + 50 The midrange is = 25.0 cigarettes.

2 Because the selected subjects report the number of cigarettes smoked, it is very possible that the data are not at all accurate. And what about that person who smokes 50 cigarettes (or 2.5 packs) a day? What are they thinking?

14. The mean is x =

1.18 +1.41+1.49 +1.04 +1.45 + 0.74 + 0.89 +1.42 +1.45 + 0.51+1.38 = 1.178 W/kg.

11 The median is 1.380 W/kg.

The mode is 1.45 W/kg.

0.51+1.49 = 1.000 W/kg.

The midrange is 2 If concerned about radiation absorption, you might purchase the cell phone with the lowest radiation level. All of the cell phones in the sample have radiation levels below the FCC maximum of 1.6 W/kg.

17. Systolic: 96 +116 +118 +120 +122 +126 +128 +136 +156 +158 The mean is x = = 127.6 mmHg, 10 122 +126 The median is = 124.0 mmHg. 2 Diastolic: 52 + 58 + 64 + 72 + 74 + 76 + 80 + 82 + 88 + 90 The mean is x = = 73.6 mmHg. 10 74 + 76 The median is = 75.0 mmHg. 2 Given that systolic and diastolic blood pressures measure different characteristics, a comparison of the measures of center doesn’t make sense. Because the data are matched, it would make more sense to investigate whether there is an association or correlation between systolic blood pressure measurements and diastolic blood pressure measurements. (FYI: A helpful measure is the mean arterial pressure, which is (1 / 3)(systolic) + (2 / 3)diastolic.)

16. The mean is x =

18. White blood cells: 8.7 + 4.9 + 6.9 + 7.5 + 6.1+ 5.7 + 4.1+ 8.1+ 8 + 5.6 + 8.3 + 6.9 = 6.73 (1000 cells mL).

12 6.9 + 6.9 = 6.90 (1000 cells mL).

The median is 2 Red blood cells: 4.8 + 4.7 + 4.5 + 4.3 + 5 + 4 + 4.7 + 4.6 + 4.1+ 5.5 + 4.4 + 4.2 The mean is x = = 4.57 ( million cells mL).

12 4.5 + 4.6 The median is = 4.55 ( million cells mL).

2 Given that the white and red blood cell counts measure different characteristics and use different units of measurement, a comparison of the measures of center doesn’t make much sense. Because the data are matched, it would make more sense to investigate whether there is an association or correlation between white blood cell counts and red blood cell counts.

19. Males: 4.9 + 7.5 + 6.1+ 5.7 + 4.1+ 5.6 + 8.3 + 5.1+ 9.5 + 6.1+ 5.7 + 5.4 The mean is x = = 6.17 (1000 cells mL).

12 5.7 + 5.7 The median is = 5.70 (1000 cells mL).

2 Females: 8.7 + 6.9 + 8.1+ 8 + 6.9 + 8.1+ 6.4 + 6.3 +10.9 + 4.8 + 5.9 + 7.2 The mean is x = = 7.35 (1000 cells mL).

12 6.9 + 7.2 = 7.05 (1000 cells mL).

The median is 2 Females appear to have higher white blood cell counts than males.

20. Single Line: 390 + 396 + 402 + 408 + 426 + 438 + 444 + 462 + 462 + 462 The mean is x = = 429.0 seconds.

10 426 + 438 The median is = 432.0 seconds.

2 Individual Lines: 252 + 324 + 348 + 372 + 402 + 462 + 462 + 510 + 558 + 600 The mean is x = = 429.0 seconds.

10 402+462 The median is = 432.0 seconds.

2 Although the measures of center are the same, the times with individual lines are much more varied than those with a single line.

The mean is x =

21. Using all values, the mean is x = 53.7 mg/dL and the median is 52.0 mg/dL. The highest value of 138 mg/dL appears to be an outlier. Excluding 138 mg/dL, the mean is x = 53.4 mg/dL and the median is 52.0 mg/dL. Excluding the outlier does not cause much of a change in the mean, and the median remains the same.

22. The mean is x = 99, 342.2 new cases and the median is 56,515.5 new cases. Because of vaccinations, the numbers of new cases were changing over time, so the population was not constant.

23. The mean is x = 98.20°F and the median is 98.40°F. These results suggest that the mean is less than 98.6°F. 24. The mean is x = 3152.0 g and the median is 3300.0 g. All of the weights end in 00, so they are all rounded to the nearest 100 grams. This suggests that the results should be rounded as follows: x = 3150.0 g and the median is 3300 g.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

1(49.5)+ 51(149.5)+ 90(249.5)+10(349.5)+ 0(449.5)+ 0(549.5)+1(649.5) 1+ 51+ 90 +10 + 0 + 0 +1 = 224.0 (1000 cells / mL). The mean from the frequency distribution is quite close to the mean of

25. The mean is x =

224.3 (1000 cells / mL)obtained by using the original list of values.

25(149.5)+ 92(249.5)+ 28(349.5)+ 0(449.5)+ 2(549.5) = 255.6 (1000 cells / mL). The mean 25 + 92 + 28 + 0 + 2 from the frequency distribution is quite close to the mean of 225.1 (1000 cells / mL) obtained by using the original list of values. æ63 + 91+ 88 + 84 + 79 ö + 0.10(86)+ 0.15(90)+ 0.15(70) = 81.2, 27. The mean is x = 0.60 çè ÷ ø 5 26. The mean is x =

so the student earned a B. 59, 470(64,966) + 89,310(7,924)+ 77, 000(55,176)+ 60, 780(35, 430)+106,950(337, 738)

= $93,957 64,966 + 7,924 + 55,176 + 35, 430 + 337, 738 The weighted mean salary of the 501,234 nurses is $93,957.

59, 470 + 89,310 + 77, 000 + 60, 780 +106, 950 = $78, 702. b. The mean of the five listed salaries is 5 c. The results differ by a considerable amount. The mean salary of $93,957 is better because it is a weighted mean that takes into account the different numbers of nurses in the five different states.

28. a.

29. a. The missing value is 5(78.0)- 82 - 78 - 56 - 84 = 90beats per minute. b. n - 1 30. If we include the censored values, the mean is 15.9 years or greater. The results of 15.5 years and 15.9 years do not differ by a considerable amount.

31. Mean: 113.7 mg/dL; 10% trimmed mean: 112.6 mg/dL; 20% trimmed mean: 112.2 mg/dL. The 10% trimmed mean and 20% trimmed mean are fairly close, but the untrimmed mean of 113.7 mg/dL differs from them because it is more strongly affected by the outliers.

Section 3-2: Measures of Variation 193.9 - 155.0 1. s » = 9.58 cm, which is in the general ballpark of the standard deviation of 7.10 cm calculated 4 using the 153 heights. The range rule of thumb does not necessarily give an estimate of s that is very accurate 2.

Significantly low values are less than or equal to 174.12 - 2(7.10) = 159.92 cm and significantly high values are greater than or equal to 174.12 + 2(7.10) = 188.32 cm. A height of 190 cm is significantly high.

3.

(7.10 cm)2 = 50.41 cm2

4.

(a) s, (b) s , (c) s2, (d) s 2; If sample data consist of weights measured in grams, s and s would also be in units of grams (g), but s2 and s 2 would be in units of grams squared

5.

(g ).

2

The range is 1.8 - 1.1 = 0.70 mg.

2

2

2

(1.8 - 1.43) +(1.7 - 1.43) + The variance is s =

+(1.2 - 1.43) 2 +(1.1 - 1.43) 2 = 0.05 mg.

12

The standard deviation is 0.05 = 0.23 mg.

Because the top 12 sample amounts are used, the variation is not at all typical for the entire sample.

6.

The range is 19 - 1 = 18.0 g.

2 2 2 2 2 2 = (1- 10.1) + (14 - 10.1) + + (14 - 10.1) + (9 - 10.1) = 35.8 g .

The variance is s 14 - 1 The standard deviation is s = 35.8 = 6.0 g.

Because the top 12 sample amounts are used, the variation is not at all typical for the entire sample. Americans consume some brands of cereal much more often than others, but the 14 brands are all weighted equally in the calculations, so the statistics are not necessarily representative of the population of all servings of cereal consumed by Americans.

7.

The range is 90.4 - 59.9 = 30.50 kg.

2 2 2 2 2 2 = (88.3 - 79.5) + (86.5 - 79.5) + + (75.1- 79.5) + (69.2 - 79.5) = 65.32 kg .

The variance is s 15 - 1 The standard deviation is s = 65.32 = 8.08 kg.

Because the measurements were made in 1988, they are not necessarily representative of the current population of all males in the Army.

The range is 1711- 1521 =190.0 mm.

2 The variance is s 2 = 16(43,173,572) - (26, 268) = 3205.5 mm2 .

16(16 - 1)

8.

9.

The standard deviation is s = 3205.5 = 56.6 mm.

The Army has height restrictions, so females in the Army are not necessarily representative of the population of all adult females.

The range is 723- 249 = 474.0.

2 The variance is s 2 = 18(4, 784,320) - (8848) = 25,590.4.

18(18 - 1) The standard deviation is s = 25,590.4 = 160.6.

The identification numbers are nominal data that are just replacements for names. They do not measure or count anything, so the resulting statistics are meaningless.

10. The range is 81- 66 =15.0 in.

2 The variance is s 2 = 20(99, 602) - (1410) = 10.4 in.2.

20(20 - 1) The standard deviation is s = 10.4 = 3.2 in.

Because the heights were reported instead of being measured, it is very possible that the results are not likely to represent the population.

11. The range is 43,510 - 32, 479 = 11,031.0 deaths.

2 The variance is s 2 = 20(29, 204,367, 422) - (759,990) = 17,111,969.3 deaths 2.

20(20 - 1) The standard deviation is s = 17,111, 969.3 = 4136.7 deaths.

The data are time-series data, but the measures of variation do not reveal anything about a trend consisting of a pattern of change over time.

12. The range is 4 - 1 = 3.0.

2 25(109) - (47) 2 = = 0.9.

The variance is s 25(25 - 1) The standard deviation is s = 0.9 = 0.9.

Because the data are at the nominal level of measurement, the measures of variation are meaningless.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

13. The range is 55 - 0 = 55.0 mg.

2

The variance is s 2 = 20(29,045)- (651) = 413.4 mg2.

20(20 - 1) The standard deviation is s2 = 413.4 = 20.3 mg.

Americans consume some brands much more often than others, but the 20 brands are all weighted equally in the calculations, so the statistics are not necessarily representative of the population of all cans of the same 20 brands consumed by Americans.

14. The range is 8 – 6.8 =1.20%.

2 The variance is s 2 = 13(739.57) - (97.9) = 0.19%2.

13(13 - 1) The standard deviation is s = 0.19%2 = 0.44%.

The data are time-series data, but the measures of variation do not reveal anything about a trend consisting of a pattern of change over time.

15. The range is 50 - 0 = 50.0 cigarettes.

50(4781) - (139)2 The variance is s2 = = 89.7 (cigarettes)2 50 (50 - 1) The standard deviation is s = 89.7 = 9.5 cigarettes.

Because the selected subjects report the number of cigarettes they smoke, it is very possible that the data are not at all accurate, so the results might not reflect the actual smoking behavior of California adults.

16. The range is 1.49 - 0.51 = 0.980 W/kg.

2

The variance is s =

11(16.4078) - (12.96)2 = 0.114 (W/kg)2 11(11- 1)

The standard deviation is s =

0.114 = 0.337 W/kg.

If concerned about radiation absorption, you might purchase the cell phone with the lowest absorption rate.

18.6 mmHg ×100% = 14.6%.

17. Systolic: x = 127.6 mmHg, s = 18.6 mmHg; The coefficient of variation is 127.6 mmHg 12.5 mmHg ×100% = 16.9%.

Diastolic: x = 73.6 mmHg, s = 12.5 mmHg; The coefficient of variation is 73.6 mmHg The variation is roughly about the same.

18. All units are (1000 cells mL).

White Blood Cell Counts: x = 6.73, s = 1.46; The coefficient of variation is

1.46

×100% = 21.7%.

6.73

All units are (million cells mL).

Red Blood Cell Counts: x = 4.57, s = 0.42; The coefficient of variation is White blood cell counts appear to vary more than red blood cell counts.

0.42

×100% = 9.2%.

4.57

19. All units are (1000 cells mL).

Male: x = 6.17, s = 1.53; The coefficient of variation is

1.53

×100% = 24.8%.

6.17 1.57 ×100% = 21.3%.

Female: x = 7.35, s = 1.57; The coefficient of variation is 7.35 The variation is roughly about the same for females and males.

28.6 sec ×100% = 6.7%.

20. Single Line: x = 429.0 sec, s = 28.6 sec; The coefficient of variation is 429.0 sec 109.3 sec ×100% = 25.5%.

Individual Lines: x = 429.0 sec, s = 109.3 sec; The coefficient of variation is 429.0 sec The single line has much less variation than with individual lines.

21. Using all data: Range = 112.0 mg/dL, s2 = 238.3 (mg/dL)2 , s = 15.4 mg/dL Excluding possible outlier of 138: Range = 87.0 mg/dL, s2 = 215.2 (mg/dL)2 , s = 14.7 mg/dL The measures of variation change, but not by substantial amounts.

22. Using all data: Range = 212.0 mg/dL, s2 = 1238.3 (mg/dL)2 , s = 35.2 mg/dL Excluding possible outlier of 251: Range = 184.0 mg/dL, s2 = 1179.0 (mg/dL)2 , s = 34.3 mg/dL Only the range changes by a fairly large amount.

23. Range = 3.10 F, s2 = 0.39 ( F)2, s = 0.62 F 24. Range = 4600.0 g, s2 = 480,848.1 g2, s = 693.4 g; All of the weights end in 00, so they are all rounded to the nearest 100 g. This suggests that the results should be rounded as follows: Range = 4600.0 g, s2 = 480,850 g2, s = 690 g.

138 - 26 25. The rule of thumb standard deviation is s » = 28.0 mg/dL, which is far from s = 15.4 mg/dL found 4 by using all of the data.

251- 39 26. The rule of thumb standard deviation is s » = 53.0 mg/dL, which is far from s = 35.2 mg/dL found by 4 using all of the data.

99.6 - 96.5 27. The rule of thumb standard deviation is s » = 0.78 F, which is not substantially different from 4 s = 0.62 F found by using all of the data.

4900 - 300 28. The rule of thumb standard deviation is s » = 1150.0 g, which differs from s = 693.4 g found by 4 using all of the data by a considerable amount. Several of the lowest weights correspond to premature births, and they cause the range to be larger, with the resulting estimate being larger.

29. Significantly low values are less than or equal to 74.0 - 2(12.5) = 49.0 beats per minute, and significantly high values are greater than or equal to 74.0 + 2(12.5) = 99.0 beats per minute. A pulse rate of 44 beats per minute is significantly low.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

30. Significantly low values are less than or equal to 69.6 - 2(11.3) = 47.0 beats per minute, and significantly high values are greater than or equal to 69.6 + 2(11.3) = 92.2 beats per minute. A pulse rate of 50 beats per minute is neither significantly low or high.

31. Significantly low values are less than or equal to 77.32 - 2(1.29) = 24.74 cm, and significantly high values are greater than or equal to 77.32 + 2(1.29) = 29.90 cm. A foot length of 30 cm is significantly high.

32. Significantly low values are less than or equal to 98.20 - 2(0.62) = 96.96 F, and significantly high values are greater than or equal to 98.20 + 2(0.62) = 99.44 F. A body temperature of 100 F is significantly high.

(

)

103 1×49.52 + +1×649.52 - (1×49.5 + 33. s =

+1×649.5)2 = 68.4, which is somewhat far from the exact

103(103 - 1)

value of 59.5.

34.

s=

(

147 25×149.52 +

)

+ 2×549.52 - (25×149.5 + 147 (147 - 1)

+ 2×549.5)2

= 69.5, which is not very far from the exact

value of 65.4.

9 +10 + 20

2

(9 - 13.0)2 +(10 - 13.0)2 +(20 - 13.0)2

2

= 13 cigarettes and s = = 24.7 cigarettes 3 3 b. The nine possible samples of two values are the following: {(9, 9), (9, 10), (9, 20), (10, 9), (10, 10), (10, 20), (20, 9), (20, 10), (20,20)} which have the following corresponding sample variances:{0, 0.5, 60.5, 0.5, 0, 50,

35. a. m =

60.5, 50, 0}, that have a mean of s2 = 2.47 cigarettes2. c. The population variances of the nine samples above are {0, 0.25, 30.25, 0.25, 0, 25, 30.25, 25, 0} that have a mean of s2 = 12.3 cigarettes2. d. Part (b), because repeated samples result in variances that target the same value (24.7 cigarettes2 ) as the population variance. Use division by n - 1. e. No. The mean of the sample variances (24.7 cigarettes2 ) equals the population variance (24.7 cigarettes2 ), but the mean of the sample standard deviations (3.5 cigarettes) does not equal the population standard deviation (5.0 cigarettes).

36. The mean absolute deviation of the population is 4.7 cigarettes. With repeated samplings of size 2, the nine different possible samples have mean absolute deviations of 0, 0, 0, 0.5, 0.5, 5, 5, 5.5, 5.5. With many such samples, the mean of those nine results is 2.4 cigarettes, showing that the sample mean absolute deviations tend to center about the value of 2.4 cigarettes instead of the mean absolute deviation of the population, which is 4.7 cigarettes. The sample mean deviations do not target the mean deviation of the population. This is not good.

This indicates that a sample mean absolute deviation is not a good estimator of the mean absolute deviation of a population.

Section 3-3: Measures of Relative Standing and Boxplots 1. Brady’s height is 2.66 standard deviations above the mean.

2. The waiting line represented by the bottom boxplot is better because the times have much less variation, so all customers have wait times that are closer together.

3. It appears that weights of U.S. Army males increased from 1988 to 2012. All values of the 5-number survey increased, except for the minimum.

4. 2.00 should be preferred, because it is 2.00 standard deviations above the mean and would correspond to the highest of the five different possible scores.

5.

a. The difference is 98 - 70.2 = 27.8 mmHg. 27.8 b. = 2.48 standard deviations 11.2 c. z = 2.48 d. The diastolic blood pressure of 98 mmHg is significantly high.

6.

a. The difference is 71.3- 40 = 31.3 mmHg. 31.3 b. = 2.61 standard deviations 12.0 c. z =- 2.61 d. The diastolic blood pressure of 40 mmHg is significantly low.

7.

a. The difference is 17.2 - 5 = 12.2.

12.2 b. = 0.91 standard deviation 13.4 c. z =- 0.91 d. The measurement of 5 is neither significantly low nor significantly high. a. The difference is 75 - 14.4 = 60.6.

60.6 b. = 5.05 standard deviations 12 c. z = 5.05 d. The measurement of 75 is significantly high.

8.

9.

Significantly low scores are less than or equal to 21.1- 2(5.2) = 10.7, and significantly high scores are greater than or equal to 21.1+ 2(5.2) = 31.5. Scores that are not significant are between 10.7 and 31.5.

10. Significantly low scores are less than or equal to 100 - 2(15) = 70, and significantly high scores are greater than or equal to 100 + 2(15) = 130. Scores that are not significant are between 70 and 130.

11. Significantly low knee heights are less than or equal to 21.4 - 2(1.2) = 19.0 in., and significantly high knee heights are greater than or equal to 21.4 + 2(1.2) = 23.8 in. Values that are not significant are between 19.0 in. and 23.8 in.

12. Significantly low hip breadths are less than or equal to 36.6 - 2(2.5) = 31.6 cm, and significantly high hip breadths are greater than or equal to 36.6 + 2(2.5) = 41.6 cm. Hip breadths that are not significant are between 31.6 cm and 41.6 cm.

272 - 174.12 54.6 - 174.12 13. The tallest man’s z score is z = = 13.79 and the shortest man’s z score is z = 7.10 7.10 =- 16.83. Chandra Bahadur Dangi has the more extreme height because his z score of –16.83 is farther from the mean than the z score of 13.79 for Robert Wadlow.

99 - 74.0 14. The female has a higher pulse rate because her z score is z = = 2.00, which is a higher number than 12.5 50 - 69.6 the z score of z = = - 1.73 for the male.

11.3 1500 - 3272.8 15. The male has a more extreme birth weight because his z score is z = = - 2.69, which is a lower 660.2 1500 - 3037.1 number than the z score of z = = - 2.18 for the female.

706.3

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

16. The egg put in the wren’s nest is more extreme since its z score is z = than the z score of z =

24.080 - 22.575

22.916 - 21.130

= 2.40, which is larger

0.744 = 2.20 for the robin’s nest.

0.685 17. For 0.48 W/kg,

3

×100 = 6, so it is the 6th percentile.

50 46 18. For 1.47 W/kg, ×100 = 92, so it is the 92nd percentile.

50 20 19. For 1.10 W/kg, ×100 = 40, so it is the 40th percentile.

50 16 20. For 98 W/kg, ×100 = 32, so it is the 32nd percentile.

50 30 ×50 0.93 + 0.97 = 15, so P = = 0.95 W/kg (Tech: Minitab: 0.942 W/kg; Excel: 0.958 W/kg) 21. L = 30 2 100 25 ×50 = 12.5, so Q = P = 0.91 W/kg 22. L = 1 25 100 75×50 = 37.5, so Q = P = 1.28 W/kg (Tech: Minitab: 1.285 W/kg) 23. L = 3 75 100 1.09 +1.10 40 ×50 = 1.095 W/kg (Tech: Minitab: 1.094 W/kg; Excel: 1.096 W/kg) = 20, so P = 24. L = 40 2 100 50 ×50 1.15 +1.16 = 25, so P = = 1.155 W/kg 25. L = 50 2 100 75×50 L = = 37.5, so Q = P = 1.28 W/kg (Tech: Minitab: 1.285 W/kg) 26.

3 75 100 25 ×50 = 12.5, so Q = P = 0.91 W/kg 27. L = 1 25 100 85 ×50 28. L = = 42.5, so P = 1.38 W/kg (Tech: Minitab: 1.387 W/kg; Excel: 1.3765 W/kg) 85 100 29. The 5-number summary is 1.5 mg, , 5.35 mg, 6.60 mg, 7.25 mg, 11 mg.

30. The 5-number summary is 4.8 mg, 5.60 mg, 6.30 mg, 7.70 mg, 8.8 mg.

31. The 5-number summary is 128 mBq, 140.0 mBq, 150.0 mBq, 158.5 mBq, 172 mBq. (Tech: Minitab yields Q1 = 139.0 mBq and Q3 = 159.75 mBq. Excel yields Q1 = 141.0 mBq and Q3 = 157.25 mBq.)

32. The 5-number summary is 120 mmHg, 130.0 mmHg, 132.5 mmHg, 140.0 mmHg, 150 mmHg. (Tech: Minitab yields Q1 = 128.75 mmHg and Q3 = 140.75 mmHg.)

33. The top boxplot represents males. Males appear to have slightly lower pulse rates than females.

(Tech: For males, Minitab yields Q3 = 77.)

Male Pulse

Female Pulse 34. The top boxplot represents the tar in king-size cigarettes, and the bottom boxplot represents the tar in menthol cigarettes. From the boxplots, it appears that there is substantially more tar in king-size cigarettes than in menthol cigarettes. (Tech: For the menthol cigarettes, Excel yields Q1 = 14.5 and Q3 = 15.5.)

King-size

Menthol 35. The top boxplot represents the weights from ANSUR I 1988 and the bottom boxplot represents weights from ANSUR II 2012. It appears that the weights of male Army personnel increased somewhat from 1988 to 2012.

(Tech: For ANSUR II 2012, Minitab yields Q1 = 75.575 and Q3 = 94.425. TI data: The values of the fivenumber summary for ANSUR I are 49.8, 70.7, 77.7, 86.0, 116.9, and for ANSUR II those values are 47.8, 75.2, 84.0, 93.2, 137.1.)

ANSUR I ANSUR II 36. The low lead level group represented in the top boxplot has much more variation and the IQ scores tend to be higher than the IQ scores from the high lead level group. (Tech: For the low lead level group, Minitab yields Q1 = 84.75 and Q3 = 101.25. For the high lead level group, Minitab yields Q3 = 93.50. The results suggest that greater exposure to lead corresponds to lower full IQ scores (although a direct cause/effect link is not established).

Low Lead

High Lead

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

37. The top boxplot represents males. Males appear to have slightly lower pulse rates than females. The outliers for males are 40 beats per minute, 102 beats per minute, and 104 beats per minute. The outlier for females is 36 beats per minute.

Males

Females Quick Quiz 8 + 7 + 5 + 7 + 4 + 7 + 6 + 7 +8 +8 +8 + 6

1.

The sample mean is x =

2.

7+7 The median is = 7.0 hr.

2

= 6.8 hr.

12

5.

6.

7.

3.

The modes are 7 hr and 8 hr.

4. The variance is (1.3 hr)2 =1.7 hr2.

The sleep time of 0 hr appears to be an outlier because it is substantially less than all the other sleep times.

5 - 6.3 z= = - 0.93; No, the sleep time of 5 hr is not significantly low, so it is not an outlier.

1.4 About 75% or 0.75(80) = 60 sleep times are less than Q3.

8.

9.

minimum, first quartile Q1, second quartile Q2 (or median), third quartile Q3, maximum 10 - 4 10. s» = 1.5 hr x, m, s, s , s2, s 2 4 Review Exercises 1. The differences are given in the table below.

Reported Measured

68 67.9

71 69.9

63 64.9

70 68.3

71 70.3

60 60.6

65 64.5

64 67

54 55.6

63 74.2

66 65

72 70.8

Difference

0.1

1.1

–1.9

1.7

0.7

–0.6

0.5

–3

–1.6

–11.2

1

1.2

0.1+1.1+ (- 1.9) +1.7 + 0.7 + (- 0.6) + 0.5 + (–3) + (- 1.6) + (- 11.2) +1+1.2 = - 1.00 in.

12 0.1+ 0.5 b. The median is = 0.30 in.

2 c. There is no mode.

- 11.2 +1.7 =- 4.75 in. d. The midrange is 2 e. The range is 1.7 - (- 11.2) = 12.90 in. a. The mean is x =

f.

s=

(0.1 - (- 1.0))2 + (1.1 - (- 1.0))2 +

2

2

+ (1- (- 1.0)) + (1.2 - (- 1.0))

12 - 1

= 3.52 in.

g. s2 = 3.522 = 12.39 in.2 (- 1.9)+(- 1.6) 25 ×12 h. L = = 3, so Q = = - 1.75 in. (Tech: Minitab: Q = - 1.83 in. Excel: Q = - 1.675 in.)

1 1 1 100 2 75 ×12 1+1.1 i. L = = 9, so Q = = 1.05 in. (Tech: Minitab: Q = 1.07 in. Excel: Q = 1.025 in.)

3 3 3 100 2

2.

3.

4.

The difference of –11.2 in. appears to be an outlier. If that outlier is excluded, the mean changes from –1.00 in. to –0.07 in., the median changes from 0.30 in. to 0.50 in., and the standard deviation changes from 3.52 in. to 1.51 in. The outlier has a strong effect on the mean and standard deviation, but very little effect on the median.

- 11.2 - (- 1.00) z= = - 2.90; The difference of –11.2 in. is significantly low (because its z score is less than or 3.52 equal to –2).

The 5-number summary is –11.2 in., –1.75 in., 0.30 in., 1.05 in., 1.70 in. (Tech: Minitab yields Q1 = - 1.83 in. and Q3 = 1.07 in. Excel yields Q1 = - 1.675 in. and Q3 = 1.025 in.)

12 +14 + 22 + 27 + 40

5.

The mean is x =

= 23.0. The numbers don’t measure or count anything. They are used as 5 replacements for the names of the categories, so the numbers are at the nominal level of measurement. In this case the mean is a meaningless statistic.

6.

Significantly low scores are less than or equal to 504.7 - 2(9.4) = 485.9, and significantly high scores are greater than or equal to 504.7 + 2(9.4) = 523.5.

7.

8.

The minimum value is 119 mm, the first quartile is 128 mm, the second quartile (or median) is 131 mm, the third quartile is 135 mm, and the maximum value is 141 mm.

The outlier is 646. The mean and standard deviation with the outlier included are x = 267.8 and s = 131.6.

Those statistics with the outlier excluded are x = 230.0 and s = 42.0. Both statistics changed by a substantial amount, so here the outlier has a very strong effect on the mean and standard deviation.

9.

Significantly low heights are 97.5 - 2(6.9) = 83.7 cm or less; significantly high heights are 97.5 + 2(6.9) = 111.3 cm or greater. The height of 87.8 cm is not significant, so the physician should not be concerned.

10. The male z score is z =

3400 - 3272.8

= 0.19 . The female z score is z =

3200 - 3037.1

660.2 the larger relative birth weight because the female has the larger z score.

= 0.23 . The female has

706.3

Cumulative Review Exercises 1. a. quantitative b. ratio level of measurement c. continuous d. sample e. statistic 2.

a. The mean is x = b. The median is

0.72 + 0.9 + 0.84 + 0.68 + 0.84 +

0.84 + 0.84

+ 0.95 + 0.86 + 0.88 + 0.85 + 0.87 = 0.830 mm.

20

= 0.840 mm.

2

c. The standard deviation is s =

20(13.8901)- (16.59)2 20(20 - 1)

d. The variance is s2 = (0.082)2 = 0.007 mm2.

3.

e. The range is 0.95 - 0.64 = 0.310 mm.

4 For 0.76 mm, ×100 = 20, so it is the 20th percentile.

20

= 0.082 mm.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

4.

Thorax Length (mm) 0.60–0.65 0.66–0.71 0.72–0.77 0.78–0.83 0.84–0.89 0.90–0.95

Frequency 1 1 3 1 9 5

5.

6.

7.

8.

The vertical scale does not begin at 0, so the differences among the outcomes are exaggerated.

Because the distribution is roughly bell-shaped, it does appear that the sample data are from a population with a normal distribution.

Based on the scatterplot, there does appear to be a correlation between heights of fathers and heights of their first sons. Because the points are not very close to a straight-line pattern, the correlation does not appear to be very strong.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

Chapter 4: Probability Section 4-1: Basic Concepts of Probability 1. The probability of randomly selecting someone with blue eyes is 0.35.

2. The probability of a baby being male is 1/2 or 0.5.

3.

4.

P( A) = 1- P( A) = 1- 0.512 = 0.488 The answers vary, but a high answer in the neighborhood of 0.999 is reasonable.

5.

0, 3 5, 1, 0.135

11. 1 4 , or 0.25

6.

1 5 , or 0.2

12. 0.292

7.

1 10 , or 0.1

13. 1 10 , or 0.1

8.

{bb, bg, gb, gg}

14. 1 2 , or 0.5

9.

1 2 , or 0.5

15. 0 16. 1

10. 1 5 , or 0.2

239 = 239 , or 0.821; Yes, the technique appears to be effective.

239 + 52 291 117 117 = , or 0.159; Headaches do appear to be a substantial adverse reaction.

18.

117 + 617 734 428 19. , or 0.738; Yes, it is reasonable.

580 c. He already knew.

20. a. 1 365 d. 0 b. yes 311 3732 = , or 0.730; Yes, it is likely for someone to use a social networking site.

21.

1380 + 3732 426 418 83, 600 , or 0.0160; Yes, in a passenger car crash, a rollover is unlikely.

22. = 83, 600 + 5,127, 400 26, 055 17.

23. a. brown/brown, brown/blue, blue/brown, blue/blue b. 1 4 c. 3 4 24. a. brown/blue, brown/blue, blue/blue, blue/blue b. 1 2 c. 1 2 25. 3 8, or 0.375

26. 3 8, or 0.375

27. {mmmm, fmmm, mfmm, mmfm, mmmf, ffmm, fmfm, fmmf, mffm, mfmf, mmff, mfff, fmff, ffmf, fffm, ffff}; 4 16 = 1 4 , or 0.25 28. 2 16 = 1 8, or 0.125 29. 53 females is neither significantly low nor significantly high.

30. 35 females is significantly low.

31. 75 females is significantly high.

32. 48 females is neither significantly low nor significantly high.

33. Because the probability of 0.364 is not small (less than or equal to 0.05), it appears that getting a result of 91 or higher can easily occur. The result of 91 or higher is not significantly high (nor is it significantly low).

34. Because the probability of 0.000 is small, and because 856 is higher than expected, it appears that the result of 856 is significantly high.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

35. The probability of 0.873 is not small, suggesting that 6062 deaths is not significantly low (nor is it significantly high). (Because there were more deaths in the week before Thanksgiving than in the week after, the claim of postponing death does not appear to be valid.)

36. The probability of 0.265 is not small (less than or equal to 0.05), so the result of 152 yellow peas could easily occur by chance. The result of 152 yellow peas is not significantly high (nor is it significantly low).

37. If pregnant women have no ability to predict the sex of their babies, then among 104 predictions, we expect about half of them (or 52) to be correct. The 57 correct predictions is greater than 52. The high probability of 0.189 is greater than 0.05, so 57 correct predictions is not significantly high (and it is not significantly low). It does not appear that pregnant women can correctly predict the sex of their babies.

38. Because the probability of 0.00000000978 is so low (less than or equal to 0.05), the result of 148 is significantly low. This suggests that the claim that the rate is less than 50% appears to be supported.

39. Because the probability of getting 604 or more respondents who have made new friends online is 0.00000306 (less than or equal to 0.05), it appears that 604 is significantly high. This suggests that the true rate is not 50%.

40. Because 0.000235 is so low (less than or equal to 0.05), and because 37 deaths is greater than half of the 49 deaths, it appears that 37 male deaths is significantly high. It appears that male selfie deaths are more likely than female selfie deaths.

41. In the following, the first letter represents the chromosome contributed by the father and the second letter represents the chromosome contributed by the mother. Let X1 and X2 represent the possible X chromosomes contributed by the mother. a. 0; The possible outcomes are {xX1, xX2, YX1, YX2}, neither son will have the disease. b. 0; The possible outcomes are {xX1, xX2, YX1, YX2}, neither daughter will have the disease. c. 1 2, or 0.5; The possible outcomes are {Xx1, XX2, Yx1, YX2}, one of the two sons will have the disease. d. 0; The possible outcomes are {Xx1, XX2, Yx1, YX2}, neither daughter will have the disease.

Section 4-2: Addition Rule and Multiplication Rule 1. P(B) represents the probability that when an adult is randomly selected, the person selected has blue eyes.

P(B) represents the probability that when an adult is randomly selected, the person selected does not have blue eyes.

2.

P(M | B) represents the probability of getting a male, given that someone with blue eyes has been selected.

Because P(B | M ) is the probability of selecting someone with blue eyes given that the selected person is a male, P(M | B) is not the same as P(B | M ).

3.

6.

Because the selections are made without replacement, the events are dependent. Because the sample size of 1068 is less than 5% of the population size of 30,488,983, the selections can be treated as being independent (based on the 5% guideline for cumbersome calculations).

Because the probability of getting a result of 47 or lower is only 0.000000987, it appears that 47 is significantly low. It does appear the claim is supported by the data.

580 - 428 152 617 617 = , or 0.262 = , or 0.841 7.

580 580 617 +117 734 1- 0.0025 = 0.9975, or 99.75%

8.

P(I ) denotes the probability of screening a driver and finding that he or she is not intoxicated, and

4.

5.

P(I ) = 0.99112, or 0.991 when rounded.

Use the following table for Exercises 9–20

Texted While Driving No Texting While Driving Total

Drove When Drinking Alcohol?

Yes No 731 3054 156 4564 887 7618

Total 3785 4720 8505

Use the table on the previous page for Exercises 9–20 887 4720 9. , or 0.104 10. , or 0.555 8505 8505 731 3941 3785 887 = , or 0.463; The two events are not disjoint.

11. + 8505 8505 8505 8505 4720 887 156 = 5451 , or 0.641; The two events are not disjoint.

12. + 8505 8505 8505 8505 887 887 × = 0.0109; Yes, the events are independent.

8505 8505 887 886 b. × = 0.0109; The events are dependent, not independent.

8505 8504 c. In this case, the results are the same when rounded to three significant digits, but with more significant digits, the results of 0.0108767364 and 0.0108657516 are different.

3785 3785 × = 0.198; Yes, the events are independent.

14. a.

8505 8505 3785 3784 b. × = 0.198; The events are dependent, not independent.

8505 8504 c. In this case, the results are the same when rounded to three significant digits, but with more significant digits, the results of 0.1980537782 and 0.1980247356 are different.

156 156 × = 0.0309; Yes, the events are independent.

15. a.

887 887 156 155 × = 0.0308; The events are dependent, not independent. b.

887 886 156 156 156 16. a. × × = 0.0000361; Yes, the events are independent.

4720 4720 4720 156 155 154 b. × × = 0.0000354; The events are dependent, not independent.

4720 4719 4718 731 887 886 885 884 17.

19. × × × = 0.0859 = 0.000118 8505 8505 8504 8503 8502 4564 3785 3784 3783 3782 = 0.537 18.

20. × × × = 0.0392 8505 8504 8503 8502 8505 21. Use the following table for parts (a) and (b).

13. a.

Positive Test Result Subject Used True Positive Marijuana 119 Subject Did Not Use False Positive Marijuana 24 Total 143 a. There was a total of 300 subjects in the study. b. 154 subjects had a true negative result.

119 + 3 +154 23 22. = = 0.920 300 25 24.

Negative Test Result Total False Negative 122 3 True Negative 178 154 157 300 154 77 c. = = 0.513 300 150 119 + 24 +154 99 = 0.990 23. = 300 100

178 89 = = 0.593; No, in the general population, the rate of subjects who do not use marijuana is probably 300 150 much greater than 0.593 or 59.3%.

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25. a. 0.22 ×0.22 = 0.0484, or 4.84% b. 1- 0.0484 = 0.9516, or 0.952 when rounded; This probability seems high, but both generators fail about 5% of the time that they are needed. Given the importance of the hospital’s needs, the reliability should be improved.

26. a. 1- 0.005 = 0.995 b. 0.005×0.052 = 0.000260; 1- 0.000260 = 0.99974 c. 0.005×0.052 ×0.001 = 0.00000026; 1- 0.000000260 = 0.99999974 d. Use all three of the alarm clocks for redundancy and improved reliability.

8330 8329 8328 × × = 0.838; The probability of 0.838 is high, so it is likely that the entire 27. 8834 - 504 = 8330, 8834 8833 8832 batch will be accepted, even though it includes many firmware defects.

392 391 390 × × = 0.941; It is likely that the entire lot would be accepted.

28. 400 - 8 = 392, 400 399 398 47, 637 47, 637 = = 0.299 29. a.

47, 637 +111,874 159, 511 b. Using the 5% guideline for cumbersome calculations, (0.299)5 = 0.00239. Using exact probabilities, 47, 637 ×47, 636 47, 635 47, 634 47, 633 = 0.00238. × × × 159, 511 159, 510 159, 509 159, 508 159,507 æ47, 637 - 188 ö40 30. Using the 5% guideline for cumbersome calculations, ç = 0.854. è 47, 637 ø÷ 31. a. P( A or B) = P( A) + P(B) - 2P( A and B) 3785 887 731 = 3210 , or 0.377 b. + - 2× 8505 8505 8505 8505 4564 7774 , or 0.537 and P( A or B) = , or 0.914; The results are different. In general, P( A or B) 32. P( A or B) = 8505 8505 is not the same as P( A or B).

Section 4-3: Complements, Conditional Probability, and Bayes’ Theorem 1.

A is event of not getting at least 1 defect among the 4 sphygmomanometers, which means that all 4

2.

3.

sphygmomanometers are good.

Parts (a) and (b) are correct.

The probability that the polygraph indicated a lie given that the subject did tell a lie.

4.

5.

7.

Confusion of the inverse is to think that P(L | Y ) = P(Y | L) or to switch one of those values for the other. That is, confusion of the inverse is to think that the following two probabilities are equal or to incorrectly use one of them for the other: (1) the probability that the subject told a lie given that the polygraph indicated a lie; (2) the probability that the polygraph indicated a lie given that the subject told a lie. æ1 ö4 15 6. 1 2 , or 0.5 1 - ç ÷ = , or 0.938 è2 ø 16 a. 1- 0.512 = 0.488

8.

a. 1- 0.526 = 0.474

b. (0.512)3 = 0.134

b. (0.474)4 = 0.0505

c. (0.488)3 = 0.116

c. (0.526)4 = 0.0765

d. 1- (0.512)3 = 0.866

d. 1- (0.526)4 = 0.923

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9.

æ3 ö5 1- ç ÷ = 0.763; The chance of getting at least one offspring with vestigial wings isn’t high enough to say that è4 ø the researchers can be “almost sure” of getting one with vestigial wings.

10. 1- (1- 0.20)10 = 0.893; There is a good chance of continuing.

11. 1- (1- 0.398)10 = 0.994; The probability is high enough so that she can be reasonably sure of getting a defective transducer for her work.

12. 1- (1- 0.31)5 = 0.844; With a probability of 0.844, the researcher can be reasonably confident that the five road crash deaths will include at least one that involved alcohol. Don’t drink and drive!

Use the following table for Exercises 13–16.

Subject Did Not Lie 15 32 47

Subject Lied Total Polygraph indicated that the subject lied.

42 57 Polygraph indicated that the subject did not lie.

9 41 Total 51 98 15 13. P (positive result | did not lie) = , or 0.319; This is the probability of the polygraph making it appear that the 47 subject lied when the subject did not lie, so the subject would be unfairly characterized as a liar.

9 14. P (negative result | lied) = , or 0.176; The subject appears to be truthful when telling a lie, so the subject 51 would receive undeserved credibility.

42 15. P (lied | positive result) = , or 0.737; To be truly effective, the probability should be much higher. There is 57 not sufficient evidence to conclude that the polygraph is effective.

32 16. P (did not lie | negative result) = , or 0.780; To be truly effective, the probability should be much higher.

41 There is not sufficient evidence to conclude that the polygraph is effective.

17. a. 1- (0.0122)2 = 0.999851 b. 1- (0.0122)3 = 0.999998; The usual round-off rule for probabilities would result in a probability of 1.00, which would incorrectly indicate that we are certain to have at least one working hard drive.

18. 1- (0.22)3 = 0.989; The result is quite high, but there is roughly a 1% chance that in the event of a power

failure, none of the backup generators will work. With the possibility of lives at risk, it would be wise to improve the reliability of the backup generators so that the 22% failure rate is lowered and the probability of 0.989 is increased.

19. 1- (1- 0.126)5 = 0.490; The probability is not low, so further testing of the individual samples will be

necessary for about 49% of the combined samples.

20. 1- (1- 0.005)10 = 0.0489; The probability is quite low, indicating that further testing of the individual samples

will be necessary for about 5% of the combined samples.

365 364 363 341 21. 1× × × × = 1- 0.431 = 0.569 365 365 365 365 Section 4-4: Risks and Odds 1.

pt is the proportion of the characteristic in the treatment group, and pc is the proportion of the characteristic

in the control group.

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Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

2.

3.

4.

5.

6.

7.

Relative risk may be misleading by suggesting that a treatment is superior or inferior when there is only a relatively small absolute difference between the rates in the treatment and control groups. Another important disadvantage is that it can be very misleading when used with a retrospective study.

We need to treat 37 subjects with the influenza vaccine in order to prevent one case of influenza. The result applies to a large number of subjects, not every particular group of 37 subjects.

The efficacy of the vaccine in a clinical trial is the percentage reduction of the disease in a treatment group compared to a control group not given the treatment. The effectiveness is a measure of the success of the vaccine when it is used in the general population. prospective 117 117 = = 0.159 P (headache | viagra treatment) = 117 + 617 734 117 117 29 1 = 0.040; P (headache | viagra treatment) = = = 0.159; P (headache | placebo) = = 117 + 617 734 29 + 696 25 The risk of a headache appears to be higher in the treatment group.

117 29 = 0.119 117 + 617 29 + 696

8.

Absolute risk reduction =

9.

The number needed to treat is

1 = 8.4; or 9 when rounded up.

117 29 117 + 617 29 + 696

117 117 10. Odds in favor of headache: 117 + 617 = , or 117:617 (roughly 1:5.2) 617 617 117 + 617 617 117 + 617 = 617 , or 617:117 (roughly 5.2:1) Odds against headache: 117 117 117 + 617 117 pt 117 + 617 2925 = = = 3.99, 3.99 or roughly 4. The risk of headaches among Viagra users is roughly 4 11. RR: 29 pc 734 29 + 696 times the risk of headaches among those who take a placebo. ad 117 ×696 = = 4.55; The odds of a headache for Viagra users are 4.55 times the odds for those using a 12. OR: bc 617 ×29 placebo. It does appear that substantially more headaches occur among Viagra users than among those using a placebo, so there is a basis for concern.

52 13. a. P (nausea | OxyContin) = = 0.229 52 +175 5 b. P (nausea | placebo) = = 0.111 5 + 40 52 5 c. The absolute risk reduction is = 0.118. The chance of nausea in the OxyContin treatment 52 +175 5 + 40 group is higher than in the placebo group. For those in the placebo group, there is a 11.8% reduced chance of nausea when compared to the OxyContin treatment group.

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1

14.

= 8.48, or 9 when rounded up. On average, giving 9 subjects a placebo instead of OxyContin 52 - 5 52 +175 5 + 40 will prevent nausea in one subject.

15. OxyContin: 52:175 or roughly 0.297:1 or 3:10 or 1:1.34. Placebo: 5:40 or 1:8 or 0.125:1. There is not a dramatic difference between these two results.

52 pt ad 52 ×40 468 52 +175 = = 2.38; RR: = = = 2.06; The relative risk of 2.06 shows that the risk with 16. OR: bc 5×175 5 pc 227 5 + 40 OxyContin is roughly twice the risk with a placebo, so OxyContin does appear to be associated with an increased risk of nausea.

17. ad = 26×(1671- 22) = 0.938; Because the odds ratio is slightly less than 1, the risk of a headache with bc (2103 - 26)×22 Nasonex treatment is slightly less than the risk of a headache with a placebo.

26 22 = 0.000802. There is a very small difference in the risk of 2103 1671 headache from Nasonex or from a placebo.

18. The absolute risk reduction is

1 = 1246.14, or 1247 when rounded up. On average, giving 1247 26 - 22 2103 1671 subjects a Nasonex treatment instead of a placebo will prevent headache in one subject.

26 26 20. Odds in favor of headache (Nasonex): 2103 = , or 26:2077 (roughly 0.0125:1) 2077 2077 2103 22 22 , or 22:1694 (roughly 0.0133:1) 1671 Odds in favor of headache (Placebo): = 1649 1649 1671 The odds in favor of a headache for the Nasonex group are slightly lower than those for the placebo group, but the results are not dramatically different. The odds in favor of a headache are roughly the same for the Nasonex treatment group and the placebo group.

19. The number needed to treat is

5 13,934 1537 = 0.0554; Efficacy: [1- RR]´ 100% = [1- 0.0554]´ 100% = 94.5% = = 90 pc 27,868 13,833

21. RR: pt

84 7154 = 2136 = = 0.846; Efficacy: [1- RR]´ 100% = [1- 0.846]´ 100% = 15.4% 22. RR: 95 pc 8455 2044 23. a. It is unethical to ask some drivers to not wear seat belts. b. Answers will vary. One possible answer: It could take years, and it could result in samples that are too small. c. Relative risk should not be used with retrospective studies because researchers may have used a sample with incidence rates that are very different from the actual incidence rates.

Section 4-5: Rates of Mortality, Fertility, and Morbidity 1. During a year in China, there are 11.4 births for every 1000 people in the population. pt

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Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

2.

3.

4.

5.

6.

For most people, the birth rate of “11.4 per 1000” is easier to understand than the rate expressed as the decimal value of 0.0114.

About 0.0114 ×1, 439,323, 776 = 16, 408, 291 A disease incidence rate is the ratio of the number of reported new cases of the disease to the population size.

For a particular point in time, the prevalence rate of a disease is the ratio of the number of people with the disease to the population size. Incidence refers to reported new cases, whereas prevalence is a measure of how widespread the disease is. Incidence is therefore a measure of the risk of contracting the disease, whereas prevalence in a measure of how widespread the disease is. æ 14, 642 ö The neonatal mortality rate is or 3.7 per 1000. = 3.7, ø1000 ç ÷ æ è3,9 5 7 , 5 4 2 ö The fetal mortality rate is 27, 2 5 or 6.8 per 1000.

1000 = 6.8, çè 3, 957, 542 + 27, 255 ÷ ø

æ 27, 255 +14, 642 ö The perinatal mortality rate is ç 1000 = 10.5, or 10.5 per 1000. è 3, 957, 542 + 27, 255 ÷ø æ 3,957,542 ö 8. The crude birth rate is ç 1000 = 12.0, or 12.0 per 1000. è 331, 002, 651÷ø æ3,957,542 ö 9. The general fertility rate is ç 1000 = 61.7, or 61.7 per 1000 females aged 15–44. è64,158, 629 ÷ø æ 42, 060 ö 10. The motor vehicle death incidence rate is ç 100, 000 = 12.7 or 12.7 deaths per 100,000. è331, 002, 651÷ø æ 1, 208,331 ö 11. The HIV infection prevalence rate is ç 1000 = 3.7, or 3.7 per 1000. è331, 002, 651÷ø æ 7825 ö 12. The HIV infection mortality rate is ç 1000 = 6.5, or 6.5 per 1000. è1, 208, 331÷ø 10.1 13. = 0.0101; The rate uses fewer decimal places and is easier to understand.

1000 14.

7.4 a. = 0.0074 1000 b. (0.0074)(0.0074) = 0.0000548 7.

c. (1- 0.0074)(1- 0.0074) = 0.985 15.

a.

9.4

= 0.0094 1000 b. (0.0094)(0.0094) = 0.0000884 c. 1- (0.0094)(0.0094) = 0.999912; Using three significant digits would result in a probability of 1.00, which

would be misleading because it would incorrectly suggest that it is certain that at least one survives the year. æ 726, 220 ö 16. a. ç 1000 = 2.2, or 2.2 per 1000 è331, 002, 651÷ø 726, 220 b. 331, 002, 651 = 0.00219 c. (1- 0.00219)(1- 0.00219)(1- 0.00219) = 0.993

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æ 726, 220 ö çè 3, 358,814 ÷100 = 21.6 ø æ 726, 220 öæ 726, 220 öæ 726, 220 ö b. ç1 1 1÷ç ÷ç ÷ = 0.481 è 3, 358,814 øè 3, 358,814 øè 3, 358,814 ø æ ö 5060 + 55,579 +180, 601 18. a. The crude mortality rate for Florida is or 11.1 per 1000. çè 6, 562, 007 +11, 297, 017 + 3,874, 287 ÷1000 = 11.1, ø æ ö 84,399 + 827,843 + 2, 446, 571 The crude mortality rate for the United States is ç 1000 = 10.1, or ÷ è109, 568,195 +173, 501, 714 + 47, 932, 741 ø 10.1 per 1000.

It appears that Florida has a higher mortality rate than the United States. æ 180, 601 ö b. The crude mortality rate for Florida (65 and older) is ç 1000 = 46.6, or 46.6 per 1000. è3, 874, 287 ÷ø æ 2, 446, 571 ö The crude mortality rate for the United States (65 and older) is ç 1000 = 51.0, or 51.0 per è47, 932, 741÷ ø 1000.

It appears that among people aged 65 and older, Florida has a lower mortality rate than the United States.

3,874, 287 c. 6,562, 007 +11, 297, 017 + 3,874, 287 = 0.178, or 17.8% of Florida’s is 65 and older.

17. a.

47,932, 741

= 0.145, or 14.5% of the United States is 65 and older.

109, 568,195 +173, 501, 714 + 47, 932, 741 With a higher percentage of people aged 65 and older, Florida is likely to have a higher crude mortality rate, but the results from part (b) do not support that conclusion.

19. No, the health of the nation is not necessarily declining. The increasing number of deaths each year is probably due to the growing and aging population.

20. The two populations have different compositions of factors such as age and race that are likely to have a significant effect on the mortality rate, so a comparison of the crude mortality rates is likely to be misleading.

Adjusted rates should be used for such comparisons.

21. The United States has a population distribution of 33.1019%, 52.4170%, and 14.4811% for the three age categories. If the 21,733,311 Florida residents have that same distribution, the three age groups would have these numbers of people: 7,194,139, 11,391,950, and 3,147,222, respectively. Using the same Florida mortality rates for the three individual age groups and using the new adjusted population sizes for the three Florida age categories, we get these numbers of Florida deaths: 5547, 56,046, and 146,709, respectively. Using the adjusted numbers of deaths and the adjusted population sizes for the different categories, the crude mortality rate for Florida becomes 9.6 per 1000, which is closer to the U.S. mortality rate of 10.1 per 1000 than the mortality rate of 11.1 per 1000 found for Florida before the adjustments.

Section 4-6: Counting 1. The symbol ! is the factorial symbol that represents the product of decreasing whole numbers, as in 5! = 5×4×3×2 ×1 = 120. Five people can be scheduled for X-ray films in 120 different ways.

2.

3.

4.

In permutations, different orderings of the same items are counted separately. In combinations, order is irrelevant.

9!

9C4 = = 126; The result of 126 is the number of different combinations that are possible when 4 (9 - 4)!4! items are selected without replacement from 9 different items that are available.

9!

9 P4 = = 3024; The result of 3024 is the number of different permutations that are possible when 4 (9 - 4)! items are selected without replacement from 9 different items that are available.

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Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

5.

1 1 1 1 × × × = 10 10 10 10 10, 000

7.

There are 19 C2 =

8.

1

11C3 =

11!

(11- 3)!3!

6.

1

1 1 1 1 1 × × × × = 10 10 10 10 10 100, 000

19! = 171 ways to choose two physicians. The probability is 1 171, or 0.00585.

(19 - 2)!2!

9.

= 165

8!= 40,320 The probability is 1 40, 320.

10. There are 43 = 64 possible linear triplets. The probability is 1 64.

11. There are 50 P5 = 12.

13.

14.

12!

50! = 254, 251, 200 possible routes. The probability is 1 254, 251, 200.

(50 - 5)!

= 3,326, 400

3!4!

1

1 = ; No, there are too many different possibilities.

100×100×100×100 100, 000, 000 5! = 10 (5 - 2)!2!

5C2 =

15. The groups can be formed in (15 C3)×(12 C3)×(9 C3)×(6 C3)×(3 C3) = 455×220 ×84 ×20×1 = 168,168, 000 ways.

16. 7! = 5040 1 1 17. = 5 100, 000 10 18.

14 C4 =

14! = 1001; The probability is 1 1001.

(14 - 4)!4!

19. 7! = 5040 The probability is 1 5040.

20.

11! = 34, 650 4!4!2!

21. a.

10 P4 =

b.

10! = 5040 (10 - 4)!

C = 10

4

10! = 210 (10 - 4)!4!

c. The probability is 1 210.

3 1 3 × = , or 0.188 4 4 16 c. Trick question. There is no finite number of attempts, because you could continue to get the wrong position every time.

22. a. 1 4 , or 0.25

23.

8 P5 =

24. a.

b.

8! = 6720 (8 - 5)!

220 = 1, 048,576

20! = 184, 756 (20 - 10)!10!

184, 756 c. 1, 048,576 = 0.176 b.

20 C10 =

d. With a probability of 0.176, the result is common, but it should not happen consistently.

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25.

16! = 653,837,184,000 2!2!2!2!2!

16! = 10, 461, 394,944, 000; (Most calculators will give a result in scientific notation, so an (16 - 14)! answer such as 10,461,395,000,000 is OK.) 16! = 120 b. 16 C14 = (16 - 14)!14!

26. a.

16 P14 =

c. The probability is 1 120.

27. a. 214 = 16,384 b.

14!

14 P1 = = 14 (14 - 1)! c. The probability is 14 16, 384 , or 0.000854 d. Yes, because the probability is so small and is so far away from the 7 females and 7 males that is typical.

2 1 12! = 924; The probability is = . Yes, if everyone treated is of one sex while everyone 28.

C = 12 6 (12 - 6)!6!

924 462 in the placebo group is of the opposite sex, you would not know if different reactions are due to the treatment or to sex.

C 29. 2 1 = 2 = 1 70 35 8C4 30. 9×20×30 ×10×20 ×30 ×10×20×20 ×10 ×10 = 12, 960, 000, 000, 000; The number of different possible MBIs is substantially larger than the United States population, so there is no danger of running out of MBIs. The number of different possible MBIs is substantially greater than 10 9 = 1, 000, 000, 000, which is the number of different possible Social Security Numbers. A major advantage of the new MBIs is that Social Security Numbers become much more secure because they are no longer being used for the widely circulated Medicare cards.

31. The number of arrangements is

45! = 5.74989777 ´ 1039, so there are 40 digits in the 2!6!1!6!3!2!4!9!2!2!4!1!2!1!

answer.

32. a.

c. 4! = 24 d. (n - 1)!

5C2 = 10

n(n - 1) 2 Quick Quiz 4 1. , or 0.8 5 2. 1- 0.30 = 0.7 b.

5.

3.

659,041 659,041 or 0.231 = , 659,041+ 2,195,797 2,854,838

4.

0.677×0.677 = 0.458

Answers will vary. The probability should be low, such as 0.01.

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Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

Use the following table for Exercises 6–10.

Vaccine Treatment Placebo Total 6.

109

Developed Flu 14 95 109

Did Not Develop Flu 1056 437 1493

Total 1070 532 1602

, or 0.0680

1602 14 1165 14 +1056 + 95 1165 109 1070 = , or 0.727 = , or + 1602 1602 1602 1602 1602 1602 109 108 14 = 7 = 0.00847 × = 0.00459 9.

8.

1602 1601 1602 801 14 7 10. P (developed flu | given vaccine) = = , or 0.0131 1070 535 Review Exercises Use the following table for Exercises 1–10.

7.

Female Male Total 1.

2.

3.

4.

8.

Right 1773 3577 5350

Writing Hand Left 190 466 656

Both 23 39 62

Total 1986 4082 6068

1986

= 0.327, which is not reasonably close to the proportion of females in the general population. It does not 6068 seem that the study subjects were randomly selected from the general population.

190 656 4082 466 = 4272 = 0.704 P (left hand | female) = = 0.0957 5. + 1986 6068 6068 6068 6068 190 62 61 = 0.000103 P (female | left hand) = = 0.290 6. × 656 6068 6067 2452 656 1986 190 = 62 62 = 0.000104 = 0.404 + 7. × 6068 6068 6068 6068 6068 6068 L is the event of randomly selecting one of the study subjects and getting someone who does not write with

5 412 = 0.892. = 6068 6068 9. M is the event of randomly selecting one of the study subjects and getting someone who is not male, so 4082 1 986 P(M ) = 1= = 0.327.

6068 6068 656 655 654 10. × × = 0.00126; Yes, because the probability of getting three lefties is so small.

6068 6067 6066 11. a. 1- 0.75 = 0.25, or 25% b. 0.75×0.75×0.75×0.75 = 0.316 A result of x successes among n trials is a significantly high number of successes if the probability of x or c. more successes is unlikely with a probability of 0.05 or less. That is, x is a significantly high number of successes if P( x or more) £ 0.05. (The value 0.05 is not absolutely rigid. Other values, such as 0.01, could be used to distinguish between results that are significant and those that are not significant.) their left hand only, so P(L) = 1-

656

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11. (continued) No, the probability of getting all four people using vision correction is 0.316, which is not unlikely with a d. small probability such as 0.05 or less. Because the probability of four people using vision correction is so high, that event can easily occur 10 and it is not a significant event. æ 43 ö = 0.0422; No, it is not likely.

12. æ 4.3 ö10 1- ç1÷ = 1- ç1÷ è 1000 ø è 10, 000 ø Cumulative Review Exercises 186 + 247 + 395 + 253 + 218 + 217 + 238 +198 + 201 + 204 1. a. The mean is x = = 235.7 (1000 cells mL ).

10 217 + 218 = 217.5 (1000 cells mL). b. The median is 2 186 + 395 c. The midrange is = 290.5 (1000 cells mL).

2 d. The range is 395 –186 = 209.0 (1000 cells mL).

(186 - 235.7)2 + (247 - 235.7)2 + e. s = f.

2.

+ (201- 235.7)2 + (204 - 235.7)2

10 - 1

= 60.2 (1000 cells mL)

s2 = (60.159)2 = 3619.1 (1000 cells mL)2

a. The 5-number summary is 186, 201, 217.5, 247, 395 (all in units of 1000 cells mL). (Minitab yields Q1 = 200.3 and Q3 = 248.5. XLSTAT yields Q1 = 201.750 and Q3 = 244.750.)

b.

3.

4.

5.

c. The value of 395 (1000 cells/µ L) appears to be an outlier.

2346 a. = 0.46 = 46% 5100 b. 0.460 c. stratified sample a. convenience sample b. If the students at the college are mostly from a surrounding region that includes a large proportion of one ethnic group, the results will not reflect the general population of the United States. c. 0.35 + 0.4 = 0.75 d. 1- (0.6)2 = 0.64 Based on the scatterplot, it is reasonable to conclude that there is no association between heights of presidents and the heights of the presidential candidates who were runners-up. It is also reasonable to conclude that there is a very weak association with increasing heights of winners corresponding to decreasing heights of runners-up.

(More objective criteria will be introduced in Chapter 10.)

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Chapter 5: Discrete Probability Distributions Section 5-1: Probability Distributions 1. The random variable is x, which is the number of green peas among five offspring. The possible values of x are 0, 1, 2, 3, 4, and 5. The values of the random variable x are numerical.

2. A random variable is discrete if it has a finite number of values or a countable number of values. Here, the random variable is discrete because it has six possible values, and 6 is a finite number.

3.

4.

5.

7.

SP(x) = 0.001+ 0.015 + 0.088 + 0.264 + 0.396 + 0.237 = 1.001; The sum is not exactly 1 because of a round-off

error. The sum is close enough to 1 to satisfy the requirement. Also, the variable x is a numerical random variable and its values are associated with probabilities. Each of the probabilities is between 0 and 1 inclusive, as required. The table does describe a probability distribution.

The probability of 0.136 is relevant; 56 is not significantly high because the probability of 56 or more females is 0.136, which is not small, such as 0.05 or less. With random chance, it is likely that the outcome could be 56 or more females. a. discrete random variable 6. a. discrete random variable b. continuous random variable b. not a random variable c. discrete random variable c. discrete random variable d. not a random variable d. not a random variable e. discrete random variable e. continuous random variable Probability distribution m = 0×0.031+1×156 + 2×0.313 + 3×313 + 4×0.156 + 5×0.031 = 2.5 s=

8.

(0 - 2.5)2 ×0.031+ (1- 2.5)2 ×0.156 +

+ (4 - 2.5)2 ×0.156 + (5 - 2.5)2 ×0.031 = 1.1

Probability distribution (The sum of the probabilities is 1.001, but that is due to rounding.) m = 0×0.659 +1×0.287 + 2×0.050 + 3×004 + 4×0.001+ 5×0 = 0.4 s = (0 - 0.4)2 ×0.659 + (1- 0.4)2 ×0.287 +

+ (4 - 0.4)2 ×0.001+ (5 - 0.4)2 ×0 = 0.6

9. Not a probability distribution because the sum of the probabilities is 0.94, which is not 1 as required.

10. Probability distribution m = 0×0.049 +1×0.220 + 2×0.372 + 3×0.280 + 4×0.079 = 2.1 s=

(0 - 2.1)2 ×0.049 + (1- 2.1)2 ×0.220 + (2 - 2.1)2 ×0.372 + (3 - 2.1)2 ×0.280 + (4 - 2.1)2 ×0.079 = 1.0

11. Probability distribution (The sum of the probabilities is 0.999, but that is due to rounding errors.) m = 0×0.4219 +1×0.4219 + 2×0.1406 + 3×0.0156 = 0.7 s = (0 - 0.7)2 ×0.4219 + (1- 0.7)2 ×0.4219 + (2 - 0.7)2 ×0.1406 + (3 - 0.7)2 ×0.0156 = 0.7 12. Not a probability distribution because the sum of the probabilities is 0.986, which is not 1 as required.

13. m = 0×0.000 +1×0.004 + 2×0.033+ 3×0.132 + 4 ×0.297 + 5×0.356 + 6 ×0.178 = 4.5 green peas s =

(0 - 4.5)2 ×0.000 + (1- 4.5)2 ×0.004 +

+ (5 - 4.5)2 ×0.356 + (6 - 4.5)2 ×0.178 = 1.1 green peas

14. The lower limit is m - 2s = 4.5 - 2(1.1) = 2.3 green peas. Because 0 green peas is less than or equal to 2.3 green peas, it is a significantly low number of green peas.

15. The upper limit is m + 2s = 4.5 + 2(1.1) = 6.7 green peas. Because 6 green peas is not greater than or equal to 6.7 green peas, it is not a significantly high number of green peas.

16. a. P(5 green peas) = 0.356 b. P(5 or more green peas) = 0.356 + 0.178 = 0.534 c. The probability from part (b), since it is the probability of the given or more extreme result. d. No, because the probability of 5 or more green peas is 0.534, which is not low (less than or equal to 0.05).

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Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

17. a. P(2 green peas) = 0.033 b. P(2 or fewer green peas) = 0.000 + 0.004 + 0.033 = 0.037 c. The result from part (b), since it is the probability of the given or more extreme result. d. Yes, because the probability of 2 or fewer green peas is 0.037, which is low (less than or equal to 0.05).

18. a. P(3 green peas) = 0.132 b. P(3 or fewer green peas) = 0.000 + 0.004 + 0.033 + 0.132 = 0.169 c. The result from part (b), since it is the probability of the given or more extreme result. d. No, because the probability of 3 or fewer green peas is 0.169, which is not low (less than or equal to 0.05).

19. m = 0×0.066 +1×0.238 + 2×0.344 + 3×0.249 + 4×0.090 + 5×0.013 = 2.1 drivers s = (0 - 2.1)2 ×0.066 + (1- 2.1)2 ×0.238 +

+ (4 - 2.1)2 ×0.090 + (5 - 2.1)2 ×0.013 = 1.1 drivers

20. Significantly high numbers of drivers who say that they text while driving is greater than or equal to m + 2s = 2.1+ 2(1.1) = 4.3. Because 4 drivers is not greater than or equal to 4.3, 4 is not a significantly high number of drivers who say that they text while driving.

21. Significantly low numbers of drivers who say that they text while driving is less than or equal to m - 2s = 2.1- 2(1.1) = - 0.1. Because 1 driver is not less than or equal to –0.1, 1 is not a significantly low number of drivers who say that they text while driving.

22. a. P(3 drivers) = 0.249 b. P(3 or more drivers) = 0.249 + 0.090 + 0.013 = 0.352 c. The probability from part (b), since it is the probability of the given or more extreme result. d. No, because the probability of 3 or more drivers who say that they text while driving is not low (less than or equal to 0.05).

23. a. P(2 drivers) = 0.344 b. P(2 or fewer drivers) = 0.066 + 0.238 + 0.344 = 0.648 c. The probability from part (b), since it is the probability of the given or more extreme result. d. No, because the probability of 2 or fewer drivers who say that they text while driving is not low (less than or equal to 0.05).

24. a. P(4 drivers) = 0.090 b. P(4 or more drivers) = 0.090 + 0.013 = 0.103 c. The probability from part (b), since it is the probability of the given or more extreme result. d. No, because the probability of 4 or more drivers who say that they text while driving is not low (less than or equal to 0.05).

25. Because the probability of 270 or more saying that we should use biometrics is 0.0995, which is not low (less than or equal to 0.05), 270 is not significantly high. Given that 270 is not significantly greater than 50%, there is not sufficient evidence to conclude that the majority of the population says that we should replace passwords with biometric security.

26. Because the probability of 481 or more saying that we should use federal tax money is 0.00389, which is low (less than or equal to 0.05), 481 is significantly high. Given that 481 is significantly greater than 50%, there is sufficient evidence to reject the politician’s claim.

27. The company will pay $0 for a female that survives the year and –$1,000,000 for a female that does not survive the year. The expected payout per policy would be $0 ×0.99963 - $1, 000, 000 ×(1- 0.99963) = - $370. The company should charge $400 + $370 = $770 per policy.

Section 5-2: Binomial Probability Distributions 1. The given calculation assumes that the first two peas have green pods and the last three peas have yellow pods, but there are other arrangements consisting of two peas with green pods and three peas with yellow pods. The probabilities corresponding to those other arrangements should also be included in the result.

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2.

a. n = 5, x = 2, p = 0.75, q = 0.25 b. m = np = 5×0.75 = 3.75, or 3.8 peas; s =

np(1- p) = 5×0.75×0.25 = 0.96, or 1.0 pea; s 2 = 1.0 pea2

3.

Because the 30 selections are made without replacement, they are dependent, not independent. Based on the 5% guideline for cumbersome calculations, the 30 selections can be treated as being independent. (The 30 selections constitute 3% of the population of 1020 responses, and 3% is not more than 5% of the population.)

The probability can be found by using the binomial probability formula.

4. The 0+ indicates that the probability is a very small positive value. The notation of 0+ does not indicate that the event is impossible; it indicates that the event is possible, but very unlikely.

5. Not binomial; each of the weights has more than two possible outcomes.

6. binomial 7. binomial 8. Not binomial; there are more than two possible outcomes.

9. Not binomial; there are more than two possible outcomes.

10. binomial 11. binomial 12. Not binomial; there are more than two possible outcomes.

4 4 1 13. a. P(WWC) = × × = 0.128 5 5 5 b. {WWC, WCW, CWW}; Each has a probability of 0.128. c. 0.128×3 = 0.384 14. a. P(EEEN) = 0.53×0.53×0.53×0.47 = 0.0700 b. {EEEN, EENE, ENEE, NEEE }; Each has a probability of 0.0700. c. 0.0700×4 = 0.280 15. 8C2 ×0.052 ×0.956 = 0.0515 (Table: 0.051) 16. 8C0 ×0.050 ×0.958 + 8C1 ×0.051 ×0.957 = 0.943 (Table: 0.942) 17. 1-

(8C ×0.05 ×0.95 + C ×0.05 ×0.95 + C ×0.05 ×0.95 ) = 0.00579 (Table: 0.005) 0

0

1

8 1

7

2

8 2

6

3

×0.955 = 0.00542 (Table: 0.005)

20. 1- 8C0 ×0.050 ×0.958 = 0.337 (Table: 0.335)

0

×0.958 = 0.663

21.

18.

8C3 ×0.05

19.

8C0 ×0.05

22. 1-

8

(20C ×0.168 ×0.832 + C ×0.168 ×0.832 + 0

0

20

20 1

1

19

5 15 = 0.131 20C5 ×0.168 0.832

+ 20C3 ×0.1683 ×0.83217 + 20C4 ×0.1684 ×0.83216

)

= 0.236 23.

20 C0 ×0.168

0

×0.83220 = 0.0253

24.

20C20 ×0.168

20

×0.8320 = 0 +

25. a. m = np = 36×0.5 = 18.0 females, s = np(1- p) = 36×0.5×0.5 = 3.0 females b. Values of 18.0 - 2(3.0) = 12.0 females or fewer are significantly low, values of 18.0 + 2(3.0) = 24.0 females or more are significantly high, and values between 12.0 females and 24.0 females are not significant. c. The result is significantly high because the result of 26 females is greater than or equal to 24.0 females. A result of 26 females would suggest that the XSORT method is effective.

26. a. m = np = 16×0.5 = 8.0 females, s = np(1- p) = 16×0.5×0.5 = 2.0 females b. Values of 8.0 - 2(2.0) = 4.0 females or fewer are significantly low, values of 8.0 + 2(2.0) = 12.0 females or more are significantly high, and values between 4.0 females and 12.0 females are not significant. c. The result is not significant because the result of 11 females is not greater than or equal to 12.0 females. A result of 11 females would not suggest that the XSORT method is effective.

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Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

27. a. m = np = 10×0.75 = 7.5 peas, s = np(1- p) = 10×0.75×0.25 = 1.4 peas b. Values of 7.5 - 2(1.4) = 4.7 peas or fewer are significantly low, values of 7.5 + 2(1.4) = 10.3 peas or more are significantly high, and values between 4.7 peas and 10.3 peas are not significant. c. The result is not significant because the result of 9 peas with green pods is not greater than or equal to 10.3 peas.

28. a. m = np = 16×0.75 = 12.0 peas, s = np(1- p) = 16×0.75×0.25 = 1.7 peas b. Values of 12.0 - 2(1.7) = 8.6 peas or fewer are significantly low, values of 12.0 + 2(1.7) = 15.4 peas or more are significantly high, and values between 4.7 peas and 10.3 peas are not significant. c. The result is significant because the result of 7 peas with green pods is less than or equal to 8.6 peas.

29. 1- 36C0 ×0.010 ×0.9936 = 0.304; It is not unlikely for such a combined sample to test positive.

30. 1- 16C0 ×0.0140 ×0.98616 = 0.202; It is somewhat likely for such a combined sample to test positive.

31. 40C0 ×0.030 ×0.9740 + 40 C1 ×0.031 ×0.9739 = 0.662; The probability shows that about 2/3 of all shipments will be accepted. With about 1/3 of the shipments rejected, the supplier would be wise to improve quality.

32. 50C0 ×0.020 ×0.9850 + 50C1 ×0.021 ×0.9849 + 50C2 ×0.022 ×0.9848 = 0.922; About 92% of all such shipments will be accepted. Almost all shipments will be accepted, and only 8% of the shipments will be rejected.

33. a. m = np = 14×0.5 = 7.0 females, s = np(1- p) = 14×0.5×0.5 = 1.9 females; Values between 7.0 - 2(1.9) = 3.2 females and 7.0 + 2(1.9) = 10.8 females are not significant (3.3 and 10.7 if using unrounded values). The result of 13 females is greater than 10.7, so 13 is a significantly high number of females. b. The probability of exactly 13 females is 14C13 ×0.513 ×0.51 = 0.000854. c. The probability of 13 or more females is 14C13 ×0.513 ×0.51 + 14C14 ×0.514 ×0.50 = 0.000916. d. The probability from part (c) is relevant. The result of 13 females is significantly high. e. The results suggest that the XSORT method is effective in increasing the likelihood that a baby is a female.

34. a. m = np = 40×0.6 = 24.0 successes, s = np(1- p) = 40×0.6×0.4 = 3.1 successes; Values between 24.0 - 2(3.1) = 17.8 successes and 24.0 + 2(3.1) = 30.2 successes are not significant. The result of 29 successes lies between these boundaries, so 29 is neither significantly low or high. b. The probability of exactly 29 successes is 40C29 ×0.629 ×0.411 = 0.0357. c. The probability of 29 or more successes is 40C29 ×0.629 ×0.411 + + 40C40 ×0.640 ×0.40 = 0.0709. d. The probability from part (c) is relevant. The result of 29 successes is not significantly high. e. The results suggest that the new treatment is not better than the old treatment.

35. a. m = np = 47 ×0.75 = 35.3 long stems, s = np(1- p) = 14×0.75×0.25 = 3.0 long stems; Values between 35.3 - 2(3.0) = 29.3 long stems and 35.3 + 2(3.0) = 14.3 long stems are not significant (29.3 and 41.2 if using unrounded values). Because 34 falls between those limits, it is neither significantly low nor significantly high. b. The probability of exactly 34 peas with long stems is 47C34 ×0.7547 ×0.2513 = 0.118. c. The probability of 34 or more peas with long stems is 47C34 ×0.7534 ×0.2513 + + 47C47 ×0.7547 ×0.250 = 0.390. d. The probability from part (c) is relevant. The value of 34 peas with long stems is not significantly low. e. The results do not provide strong evidence against Mendel’s claim of 75% for peas with long stems.

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36. a. m = np = 80×0.05 = 4.0 influenza cases, s = np(1- p) = 80×0.05×0.95 = 1.9 influenza cases; Values between 4.0 - 2(1.9) = 0.2 influenza cases and 4.0 + 2(1.9) = 7.8 influenza cases are not significant (0.1 and 7.9 if using unrounded values). Because 1 falls between those limits, the result of 1 subject experiencing influenza is neither significantly low nor significantly high. b. The probability of exactly 1 influenza case is 80C1 ×0.051 ×0.9579 = 0.0695. c. The probability of 1 or fewer influenza cases is 80C0 ×0.050 ×0.9570 + 80C1 ×0.051 ×0.9579 = 0.0861. d. The probability from part (c) is relevant. The value of influenza case is not significantly high. e. The results do not provide strong evidence for the vaccine’s effectiveness.

37. P(4) = 0.066(1- 0.066)3 = 0.0538 38.

78 ö5 æ22 ö2 æ21 ö3 10! ×æ ç ÷ ×ç ÷ ×ç ÷ = 0.0485; If we sample without replacement, the events are not independent and 5!2!3! è121ø è121ø è121ø the given expression cannot be used to calculate the probability

39. Without replacement: P(3) =

4!

16!

(4 +16)! = 0.139 × ¸ (4 - 3)!3! (16 - 20 + 3)!(20 - 3)! (4 +16 - 20)!20!

( )3 ( )5

With replacement: C8 ×3 4 20 × 1620 = 0.147 Section 5-3: Poisson Probability Distributions 1. m = 26,271 365 = 71.975342, which is the mean number of patient admissions per day. x = 85 because we want the probability that a randomly selected day has exactly 85 admissions, and e » 2.71828 which is a constant used in all applications of Formula 5-9.

2.

The mean is m = 26,271 365 = 71.975342, admissions, the standard deviation is s =

71.975342 = 8.5

admissions (rounded), and the variance is s 2 = 72.0 admissions2.

3.

The possible values of x are 0, 1, 2, … (with no upper bound), so x is a discrete random variable. It is not possible to have x = 90.3 patient admissions in a day.

4.

P(0) represents the probability of no occurrences of an event during the relevant interval. If x = 0, P(0) = e- m.

5.

P(12) =

16.474012 e- 16.4740

= 0.0584

6.

P(10) =

12!

16.474010 e- 16.4740

= 0.0284

10!

16.47400 e- 16.4740 = 0 + (or 0.0000000701) 0!

7.

P(0) =

8.

P(10 or 11) = 468

16.474010 e- 16.4740 16.474011e- 16.4740 + = 0.0710 10!

11!

0

- 1.282192

= 1.282192; murders, P(0) = 1.282192 ×e = 0.277; The probability of no murders in a day is 365 0!

0.277. There should be many days (roughly 28%) with no murders.

196 = 0.7 10. m = 280 0.70 ×e- 0.7 0.72 ×e- 0.7 P(0) = = 0.497 P(2) = = 0.122 a. c.

0!

2!

0.71 ×e- 0.7 0.73 ×e- 0.7 = 0.348 = 0.0284 b. P(1) = d. P(3) = 1!

3!

9.

m=

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Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

10. (continued) 4 - 0.7 0.7 ×e P(4) = = 0.00497; The expected frequencies of 139, 97, 34, 8 , and 1.4 compare reasonably well e.

4! to the actual frequencies, so the Poisson distribution does provide good results.

0.9292 ×e- 0.929 P(2) = = 0.170 11. a.

2! b. The expected number of regions with exactly 2 hits is between 97.9 and 98.2, depending on rounding. c. The expected number of regions with 2 hits is close to 93, which is the actual number of regions with 2 hits.

12. a. m = 12, 429×0.000011 = 0.136719 (or 0.137) 0 - 0.136719 1 - 0.136719 b. P(0 or 1) = 0.136719 ×e + 0.136719 ×e = 0.991464 (or 0.991 rounded) 0!

1! c. 1- 0.991464 = 0.00854 or 1- 0.991 = 0.009 d. No, the probability of more than one case is extremely small, so the probability of getting as many as four cases is even smaller.

0 - 4.4493153 1624 = 4.449315; P(no C-section births) = 4.449315 e = 0.0117 13. a. m = 0!

365 Yes, a day with 0 C-section births is a significantly low number of such births.

4.4493150 e- 4.4493153 b. P(at least one C-section birth) = 1 = 0.988 0!

c.

14. a.

P(at least two C-section births) = 1 P(no dandelions) =

7.00 e- 7.0

= 0.000912

0!

b. P(at least one dandelion) = 1c.

P(at most two dandelions) =

4.4493150 e- 4.4493153 4.4493151e- 4.4493153 = 0.936 0!

1!

7.00 e- 7.0

= 0.999

0!

7.00 e- 7.0

+

7.01 e- 7.0

0!

15. a. m = b.

1!

+

7.02 e- 7.0

= 0.0296

2!

53

= 5.3 10 5.30 e- 5.3 P(no cases of rubella) = = 0.00499

0!

5.34 e- 5.3 = 0.164 c. P(four cases of rubella) = 4!

With 2 of the 10 years having exactly 4 cases of rubella, the probability appears to be 2/10 or 0.2. The Poisson distribution yields a probability of 0.164, which is close to the actual result.

37 m= = 1.681818 16. a.

22 1.6818180 e- 1.681818 = 0.186 b. P(no cases of diphtheria) = 0!

5 - 1.681818 1.6818180 e- 1.681818 + + 1.681818 e = 0.992 c. P(five or fewer cases of diphtheria) = 0!

5!

Yes, because the probability of a result at least as extreme as 6 or more diphtheria cases is 1- 0.99239 = 0.00761 , which is small.

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17. The mean number of births per year is P(120 births) =

135 +130 +127 +126 +124 +125 +124 +122 +118 +116

= 124.7 and

10

124.7120 e- 124.7

= 0.0333. The data are time-series data. Examination of the data shows a 120! downward trend, indicating that the population is changing over time, so the Poisson distribution may not be appropriate.

Quick Quiz 1. 0+ indicates that the probability is a very small positive number. It does not indicate that it is impossible for event A to occur.

2. No, the sum of the probabilities is 1.5, which is greater than 1.

3. Yes. The random variable x has numerical values, the sum of the probabilities is 1, and each probability is between 0 and 1.

4. m = 1×0.2 + 2×0.2 + 3×0.2 + 4×0.2 + 5×0.2 = 3.0 5.

6.

The mean is a parameter.

This is probability distribution because the three requirements are satisfied. First, the variable x is a numerical random variable and its values are associated with probabilities. Second, SP(x) = 0.172 + 0.363 + 0.306 + 0.129 + 0.027 + 0.002 = 0.999, which is not exactly 1 due to rounding, but is close enough to satisfy the requirement. Third, each of the probabilities is between 0 and 1 inclusive, as required.

7.

P( X ³ 1) = 1- P( X = 0) = 1 - 0.172 = 0.828 or P( X ³ 1) = 0.363 + 0.306 + 0.129 + 0.027 + 0.002 = 0.827

8.

m = 0×0.172 +1×0.363+ 2×0.306 + 3×0.129 + 4×0.027 + 5×0.002 = 1.5 sleepwalkers

9.

s 2 = (1.019041635)2 = 1.0 sleepwalker2

10. Yes. If using the range rule of thumb, significantly high numbers of sleepwalkers are greater than or equal to m + 2s = 1.5 + 2(1.0) = 3.5 sleepwalkers, and 4 sleepwalkers is greater than or equal to 3.5 sleepwalkers. If using probabilities, the probability of 4 or more sleepwalkers is 0.027 + 0.002 = 0.029, which is small (less than 0.05).

Review Exercises 2

×0.9588 = 0.056

1.

10C2 ×0.042

2.

1- 10C0 ×0.0420 ×0.95810 = 0.349

3.

m = 10×0.042 = 0.4 workers and s =

4.

No, 0 is not significantly low. If using the range rule of thumb with m = 0.4 workers and s = 0.6 workers,

10 ×0.042×0.958 = 0.6 workers

values are significantly low if they are less than or equal to m - 2s = 0.4 - 2(0.6) = - 0.8, but 0 is not less than or equal to –0.8. If using probabilities, the probability of 0 workers testing positive is 10C0 ×0.0420 ×0.95810 = 0.651, which is not low (less than or equal to 0.05).

5.

Yes, 4 is significantly high. If using the range rule of thumb with m = 0.4 workers and s = 0.6 workers, values are significantly high if they are greater than or equal to m + 2s = 0.4 + 2(0.6) = 1.6, and 4 is greater than or equal to 1.6. If using probabilities, the probability of 4 or more workers testing positive is

(

6.

7.

)

1- 10C0 ×0.0420 ×0.95810 + 10C1 ×0.0421 ×0.9589 + 10C2 ×0.0422 ×0.9588 + 10C3 ×0.0423 ×0.9587 = 0.000533, which is low (less than or equal to 0.05).

This is not a probability distribution because the responses are not values of a numerical random variable.

This is not a probability distribution because the sum of the probabilities is 0.9686, which is not 1 as required.

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Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

8.

Probability distribution (The sum of the probabilities is 0.999, but that is due to rounding errors.) m = 0×0 +1×0.003+ 2×0.025+ 3×0.111+ 4×0.279+ 5×0.373+ 6×0.208 = 4.6 people s =

9.

(0 - 4.6)2 ×0 + (1- 4.6)2 ×0.003 +

+ (5 - 4.6)2 ×0.373 + (6 - 4.6)2 ×0.208 = 1.0 people

a. This is probability distribution because the three requirements are satisfied.

First, the variable x is a numerical random variable and its values are associated with probabilities.

Second, SP(x) = 0.304 + 0.400 + 0.220 + 0.064 + 0.011+ 0.001+ 0 = 1 Third, each of the probabilities is between 0 and 1 inclusive, as required. b. m = 0×0.304 +1×0.400 + 2×0.220 + 3×0.064 + 4 ×0.011+5×0.001+ 6 ×0 = 1.1 condoms

c. s = (0 - 1.1)2 ×0.304 + (1- 1.1)2 ×0.400 + (2 - 1.1)2 ×0.220 + + (5 - 1.1)2 ×0.001+ (6 - 1.1)2 ×0 = 0.9 condom d. Yes, 5 failures is significantly high. A number is significantly high if it is equal to or greater than m + 2s = 1.1+ 2(0.9) = 2.9, and 5 does exceed 2.9. Also, the probability of 5 or more failures is 0.001, which is a low probability. e. Here, the symbol 0+ represents a positive probability that is so small that it is 0.000 when rounded.

7 0.01920 e- 0.0192 10. a. m = = 0.0192 b. = 0.981 365 0! æ ç 0.01920 e- 0.0192 0.01921e- 0.0192 ö÷ 0!

1! c. 1+ = 0.00182 è ø d. No, because the event is so rare. (But it is possible that more than one death occurs in a car crash or some other such event, so it might be wise to consider a contingency plan.)

Cumulative Review Exercises 1.

6.8 + 7.1+ 7.0 + 8.7 + 5.6 + 5.6 + 7.0 +10.1+ 3.5 + 5.1+ 8.4 + 8.0 = 6.91 (1000 cells mL).

12 7.0 + 7.0 = 7.00 (1000 cells mL). b. The median is 2 3.5 +10.1 c. The midrange is = 6.80 (1000 cells mL).

2 d. The range is 10.1- 3.5 = 6.60 (1000 cells mL).

(6.8 - 6.9)2 + + (8.0 - 6.9)2 e. The standard deviation is s = = 1.79 (1000 cells mL).

12 - 1 a. The mean is x =

f. The variance is s2 = 1.792 = 3.20 (1000 cells mL)2 . g. Significantly low values are 6.9 - 2×1.79 = 3.32 (1000 cells mL) or lower.

(3.33 (1000 cells mL) using unrounded values.)

Significantly high values are 6.9 + 2 ×1.79 = 10.48 (1000 cells mL) or higher.

(10.49 (1000 cells mL) using unrounded values.) h. None of the listed values are significantly low or significantly high. i. ratio j. continuous

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2.

a.

x= s=

3.

4.

0(11)+1(11)+ 2(16)+ 3(14)+ 4(16)+ 5(24)+ 6(13)+ 7 (14)+ 8(19)+ 9(9) = 4.6 and 11+11+16 +14 +16 + 24 +13 +18 +19 + 9

(

147 11×02 +

)

+ 9×92 - (11×0 +

+ 3×3)2

147(147 - 1)

= 2.7; They are statistics.

b. The last digits appear to be random. None of the frequencies appears to be substantially different from the others. c. No. The values of x are numerical, but the frequencies are not probabilities, as required.

No. The causes are not numerical as required. (The sum of the probabilities is 1.001; that sum is not 1 because of rounding.) a. 1- 0.115 = 0.885 b. 0.115×0.115 = 0.0132 c. 1- 3C0 ×0.1150 ×0.8853 = 0.307 (Table: 0.256) d. m = 40×0.115 = 4.6 adults, s =

40 ×0.115 ×0.885 = 2.0 adults; These results are parameters.

5.

e. Significantly low numbers are m - 2s = 4.6 - 2(2.0) = 0.6 or lower, and significantly high numbers are m + 2s = 4.6 + 2(2.0) = 8.6 and higher. Because 10 is greater than 8.6, it is a significantly high number of adults with diabetes (among 40).

No vertical scale is shown, but a comparison of the numbers shows that 7,066,000 is roughly 1.2 times the number 6,000,000. However, the graph makes it appear that the goal of 7,066,000 people is roughly 3 times the number of people enrolled. The graph is misleading in the sense that it creates the false impression that actual enrollments are far below the goal, which is not the case. Fox News apologized for their graph and provided a corrected graph.

6.

a.

8 12 20C8 ×0.45 ×0.55 = 0.162

b. m = 20×0.45 = 9.0 adults c. s = 20 ×0.45 ×0.55 = 2.2 adults d. Yes. Using the range rule of thumb, significantly low numbers are m - 2s = 9.0 - 2(2.2) = 4.6 and lower, and 3 is less than 4.6. If probabilities are used, the probability of 3 or fewer adults with hypertension is 0 20 1 19 2 18 3 17 20C0 ×0.45 ×0.55 + 20C1 ×0.45 ×0.55 + 20C2 ×0.45 ×0.55 + 20C3 ×0.45 ×0.55 = 0.00493, which is small (such as less than 0.05).

7.

a.

8C5 ×0.7

5

×0.33 = 0.254

b.

8C7 ×0.7

7

×0.31 + 8C8 ×0.78 ×0.30 = 0.255 (Table: 0.256)

c.

m = 8×0.7 = 5.6 adults, s =

8×0.7×0.3 = 1.3 adults

d. Yes. (Using the range rule of thumb, the limit separating significantly low values is m - 2s = 5.6 - 2(1.3) = 3.0, and 1 is less than 3. Using probabilities, the probability of 1 or fewer people washing their hands is 8C0 ×0.7

8.

0

×0.38 + 8C1 ×0.71 ×0.37 = 0.00129, (Table: 0.001) which is low, such as less than 0.05.

a. false b. true c. false d. false e. measure of variation

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Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

Chapter 6: Normal Probability Distributions Section 6-1: The Standard Normal Distribution 1. The word “normal” has a special meaning in statistics. It refers to a specific bell-shaped distribution that can be described by Formula 6-1. The lottery digits do not have a normal distribution.

2.

3.

The mean is m = 0 and the standard deviation is s = 1.

4.

The notation za represents the z score with the property that a vertical boundary line at that z score has an area of a to its right.

5.

P(x > 3.00) = 0.2(5 - 3) = 0.4

8.

P(2.5 < x < 4.5) = 0.2(4.5 - 2.5) = 0.4

6.

P(x < 4.00) = 0.2(4 - 0) = 0.8

9.

P(z < 0.44) = 0.6700

7.

P(2 < x < 3) = 0.2(3 - 2) = 0.2

10.

P(z > - 1.04) = 0.8508

11. P(- 0.84 < z < 1.28) = P(z < 1.28)- P(z < - 0.84) = 0.8997 - 0.2005 = 0.6992 (Tech: 0.6993) 12. P(- 1.07 < z < 0.67) = P(z < 0.67)- P(z < - 1.07) = 0.7486 - 0.1423 = 0.6063 13. z = 1.23

15. z =- 1.45

14. z =- 0.51

16. z = 0.82

17. P(z < - 2.00) = 0.0228

19. P(z < 1.33) = 0.9082

18. P(z < - 0.50) = 0.3085

20. P(z < 2.33) = 0.9901

21. P(z > 1.00) = 1- P(z < 1.00) = 1- 0.8413 = 0.1587 22. P(z > 2.33) = 1- P(z < 2.33) = 1- 0.9901 = 0.0099 23. P(z > - 1.75) = 1- P(z < - 1.75) = 1- 0.0401 = 0.9599 24. P(z > - 2.09) = 1- P(z < - 2.09) = 1- 0.0183 = 0.9817 25. P(1.50 < z < 2.00) = P(z < 2.00)- P(z < 1.50) = 0.9772 - 0.9332 = 0.0440 (Tech: 0.0441) 26. P(1.37 < z < 2.25) = P(z < 2.25)- P(z < 1.37) = 0.9878 - 0.9147 = 0.0731 27. P(- 2.36 < z < - 1.22) = P(z < - 1.22)- P(z < - 2.36) = 0.1112 - 0.0091 = 0.1021 28. P(- 2.08 < z < - 0.45) = P(z < - 0.45)- P(z < - 2.08) = 0.3264 - 0.0188 = 0.3076 29. P(- 1.55 < z < 1.55) = P(z < 1.55)- P(z < - 1.55) = 0.9394 - 0.0606 = 0.8788 (Tech: 0.8789) 30. P(- 0.77 < z < 1.42) = P(z < 1.42)- P(z < - 0.77) = 0.9222 - 0.2206 = 0.7016 (Tech: 0.7015) 31. P(- 2.00 < z < 3.50) = P(z < 3.50)- P(z < - 2.00) = 0.9999 - 0.0228 = 0.9771 (Tech: 0.9770) 32. P(- 3.52 < z < 2.53) = P(z < 2.53)- P(z < - 3.52) = 0.9943- 0.0001 = 0.9942 (Tech: 0.9941) 33. P(z > - 3.77) = 1- P(z < - 3.77) = 1- 0.0001 = 0.9999 34. P(z < - 3.93) = 0.0000425 or 0.0000 rounded (Table: 0.0001) Copyright © 2024 Pearson Education, Inc.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

35. P(- 4.00 < z < 4.00) = P(z < 4.00)- P(z < - 4.00) = 0.9999 - 0.0001 = 0.9998 (Tech: 0.9999) 36. P(- 3.67 < z < 4.25) = P(z < 4.25)- P(z < - 3.67) = 0.9999 - 0.0001 = 0.9998 (Tech: 0.9999) 37. P99 = 2.33

41. z0.25 = 0.67

38. P15 = - 1.04

42. z0.90 = - 1.28

39. P1 = - 2.33, P99 = 2.33

43. z0.02 = 2.05

40. Q1 = P25 = - 0.67, Q2 = P50 = 0.00,

44. z0.05 = 1.64 (Table: 1.645)

Q3 = P75 = 0.67

45. P(- 1 < z < 1) = P(z < 1)- P(z < - 1) = 0.8413- 0.1587 = 0.6826 = 68.26% (Tech: 68.27%) 46. P(- 2 < z < 2) = P(z < 2)- P(z < - 2) = 0.9772 - 0.0228 = 0.9544 = 95.44% (Tech: 95.45%) 47. P(- 3 < z < 3) = P(z < 3)- P(z < - 3) = 0.9987 - 0.0013 = 0.9974 = 99.74% (Tech: 99.73%) 48. P(- 3.5 < z < 3.5) = P(z < 3.5)- P(z < - 3.5) = 0.9999 - 0.0001 = 0.9998 = 99.98% (Tech: 99.95%) 49. a.

P( z > 2) = 1- P(z < 2) = 1- 0.9772 = 0.0228 = 2.28%

b. P( z < - 2) = 0.0228 = 2.28% c. P(- 2 < z < 2) = P(z < 2)- P(z < - 2) = 0.9772 - 0.0228 = 0.9544 = 95.44% (Tech: 95.45%) 50. a. m = 2.5 min. and s = 5

12 = 1.4 min

b. The probability is1 3 or 0.5774, and it is very different from the probability of 0.6827 that would be obtained by incorrectly using the standard normal distribution. The distribution does affect the results very much.

Section 6-2: Real Applications of Normal Distributions 1. a. m = 0 and s = 1 2.

3.

4.

5.

6.

7.

8.

9.

b. The z scores are numbers without units of measurements. a. The area equals the maximum probability value of 1. b. The median is the middle value and for normally distributed scores that is also the mean, which is 3152.0 g. c. The mode is also 3152.0 g. d. The variance is the square of the standard deviation which is 480,803.6 g2.

The standard normal distribution has a mean of 0 and a standard deviation of 1, but a nonstandard normal distribution has a different value for one or both of those parameters.

No. Randomly generated digits have a uniform distribution, but not a normal distribution. The probability of a digit less than 3 is 3 / 10 = 0.3.

118 - 100 = 1.2; which has an area of 0.8849 to the left. z x=118 = 15 91- 100 = - 0.6; which has an area of 0.7257 to the right. z x=91 = 15 133 - 100 79 - 100 = = 2.2; which has an area of 0.9861 to the left. z = = - 1.4; which has an area of z x=133 x=79 15 15 0.0808 to the left. The area between the two scores is 0.9861- 0.0808 = 0.9053.

124 - 100 112 - 100 = = 1.6; which has an area of 0.9452 to the left. z = = 0.8; which has an area z x=124 x=112 15 15 of 0.7881 to the left. The area between the two scores is 0.9452 - 0.7881 = 0.1571. z = 2.44; so x = 2.44×15 +100 = 136

10. z = 1; so x = 1×15+100 = 115

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11. z = - 2.07; so x = - 2.07 ×15 +100 = 69 12. z = 1.33; so x = 1.33×15 +100 =120 60 - 69.6 = - 0.85, which has an area of 0.1977 to the left. (Tech: 0.1978) 13. zx=60 = 11.3 60 - 74.0 = - 1.12, which has an area of 0.1314 to the left.

14. zx=60 = 12.5 80 - 74.0 = 0.48, which has an area of 1- 0.6844 = 0.3156 to the right.

15. zx=80 = 12.5 80 - 69.6 = 0.92, which has an area of 0.1787 to the right. (Table: 1- 0.8212 = 0.1788) 16. zx=80 = 11.3 70 - 74.0 90 - 74.0 = = - 0.32, which has an area of 0.3745 to the left and z = = 1.28, which has an 17. zx=70 x=90 12.5 12.5 area of 0.8997 to the left. The area between the two scores is 0.8997 - 0.3745 = 0.5252.

65 - 69.6 85 - 69.6 = = - 0.41, which has an area of 0.3409 to the left and z = = 1.36, which has an 18. zx=65 x=85 11.3 11.3 area of 0.9131 to the left. The area between the two scores is 0.9131- 0.3409 = 0.5722. (Tech: 0.5716) 70 - 69.6 90 - 69.6 = = 0.04, which has an area of 0.5160 to the left and z = = 1.81, which has an 19. zx=70 x=90 11.3 11.3 area of 0.9649 to the left. The area between the two scores is 0.9649 - 0.5160 = 0.4489. (Tech: 0.4505) 60 - 74.0 70 - 74.0 = = - 1.12, which has an area of 0.1314 to the left and z = = - 0.32, which has an 20. zx=60 x=70 12.5 12.5 area of 0.3745 to the left. The area between the two scores is 0.3745 - 0.1314 = 0.2431.

21. The z score for P90 is 1.28, so P90 = 69.6 +1.28 ×11.3 = 84.1 beats per minute.

22. The z score for Q1 = P25 is - 0.67, so Q1 = 74.0 - 0.67 ×12.5 = 65.6 beats per minute.

23. The z score for P1 is - 2.326, so the lower male pulse rate is 69.6 - 2.326×11.3 = 43.3 beats per minute. The z score for P99 is 2.326, so the upper male pulse rate is 69.6 + 2.326×11.3 = 95.9 beats per minute. No, 90 beats per minute is not significantly high.

24. The z score for P2.5 is - 1.96, so the lower female pulse rate is 74.0 - 1.96 ×12.5 = 49.5 beats per minute. The z score for P97.5 is 1.96, so the upper female pulse rate is 74.0 +1.96×12.5 = 98.5 beats per minute. Yes, 48 beats per minute is significantly low.

78 - 63.7 62 - 63.7 = = 4.93; which has an area of 0.9999 to the left. z = = - 0.59; which has an 25. a. zx=78 x=62 2.9 2.9 area of 0.2776 to the left. Therefore, the percentage of qualified females is 0.9999 - 0.2776 = 0.7223, or 72.23% (Tech: 72.11%). Yes, about 28% of females are not qualified because of their heights. b. The z score with 3% of values to the left is –1.88, which corresponds to a height of - 1.88×2.9 + 63.7 = 58.2 in. The z score with 3% of values to the right is 1.88 which corresponds to a height of 1.88×2.9 + 63.7 = 69.2 in.

77 - 68.6 64 - 68.6 = = 3.00; which has an area of 0.9987 to the left. z = = - 1.64; which has an 26. a. zx=77 x=64 2.8 2.8 area of 0.0505 to the left. Therefore, the percentage of qualified males is 0.9987 - 0.0505 = 0.9482, or 94.82% (Tech: 94.84%). b. The z score with 2.5% of values to the left is –1.96, which corresponds to a height of - 1.96×2.8+ 68.6 = 63.1 in. The z score with 2.5% of values to the right is 1.96 which corresponds to a height of 1.96×2.8 + 68.6 = 74.1in.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

27. The z score with 5% of values to the left is –1.645, which corresponds to a head circumference of 570.0 - 1.645×18.3 = 539.9 mm and the z score with 5% of values to the right is 1.645 which corresponds to a head circumference of 570.0 +1.645×18.3 = 600.1 mm.

28. a. The z score that has 95% of the area to the left is 1.67 which corresponds to a height of 1.67 ×1.2 + 21.4 = 23.4 in. If there is clearance for 95% of males, there will certainly be clearance for all females in the bottom 5%.

23.5 - 21.4 b. The males’ z score is = 1.75; which has an area to the left of 0.9599, or 95.99%. The females’ z 1.2 23.5 - 19.6 score is = 3.55; which has an area to the left of 0.9999, or 99.99% (Tech: 99.98%). The desk 1.1 will fit almost everyone except about 4% of the males with the largest sitting knee heights.

15 - 2.5 = 1.16, which has an area of 0.8770 to the left. The probability that a male college student 29. zx=15 = 10.8 gains 15 pounds or more during their freshman year is 1- 0.8770 = 0.1230 (Tech: 0.1236). It appears that only about 12% of males gain 15 pounds or more, so the “Freshman 15” appears to be an exaggeration instead of an accurate description of reality.

308 - 268 = 2.67, which has an area of 0.9962 to the left. The probability of a pregnancy lasting 30. a. zx=308 = 15 308 days or longer is 1- 0.9962 = 0.0038. This is either a very rare event, or the husband is not the father. b. The z score with 3% of values to the left is - 1.88, which corresponds to a duration of 268 - 1.88×15 = 240 days.

1 00.6 - 98.2 31. a. The z score for a temperature of 100.6 is = 3.87; which corresponds to an area of 0.62 1- 0.9999 = 0.0001 = 0.01% (Tech: 0.02%) to the right, which suggests a cutoff of 100.4°F is appropriate. b. The z score for a probability of 2% is 2.05 which corresponds to a temperature of 2.05×0.62 + 98.2 = 99.47°F. 174 - 188.6 32. a. The z score for 174 lb. is =- 0.38; which has an area to the left of 0.3520 (Tech: 0.3537).

38.9 b. 3500 140 = 25 male passengers c. 3500 188.6 = 18.6, or 18 male passengers d. The mean weight is increasing over time, so safety limits must be periodically updated to avoid an unsafe condition.

33. The z score for Q1 is –0.67, and the z score for Q3 is 0.67. The IQR is 0.67 - (- 0.67) = 1.34. 1.5×IQR = 2.01, so Q1 - 1.5 ×IQR = - 0.67 - 2.01 = - 2.68 and Q3 +1.5 ×IQR = 0.67 + 2.01 = 2.68.

The percentage to the left of –2.68 is 0.0037 and the percentage to the right of 2.68 is 0.0037. Therefore, the percentage of an outlier is 0.0074 (Tech: 0.0070).

Section 6-3: Sampling Distributions and Estimators 1. a. In the long run, the sample proportions will have a mean of 0.00559. b. The sample proportions will tend to have a distribution that is approximately normal.

2. a. without replacement b. (1) When selecting a relatively small sample from a large population, it makes no significant difference whether we sample with replacement or without replacement.

(2) Sampling with replacement results in independent events that are unaffected by previous outcomes, and independent events are easier to analyze and they result in simpler calculations and formulas.

3. sample mean, sample variance, sample proportion 4. No, the data set is only one sample, but the sampling distribution of the mean is the distribution of the means from all samples, not the one sample mean obtained from the one sample in Data Set 1.

5.

6.

7.

No, the sample is not a simple random sample from the population of all births worldwide. The proportion of boys born in China is substantially higher than in other countries. a. The distribution of the sample means is approximately normal. b. The sample means target the mean annual income of all physicians.

4+5+9 a. The population mean is m = = 6.

3 2 2 2 The population variance is s 2 = (4 - 6) + (5 - 6) + (9 - 6) = 4.7.

3 b. The possible sample of size 2 are {(4, 4), (4, 5), (4, 9), (5, 4), (5, 5), (5, 9), (9, 4), (9, 5), (9, 9)} which have the following variances {0, 0.5, 12.5, 0.5, 0, 8, 12.5, 8, 0} respectively.

Sample Variance 0.0 0.5 8.0 12.5

Probability 3/9 2/9 2/9 2/9

3×0 + 2 ×0.5 + 2 ×8 + 2 ×12.5 = 4.7.

9 d. Yes. The mean of the sampling distribution of the sample variances (4.7) is equal to the value of the population variance of 4.7, so the sample variances target the value of the population variance. c. The mean is

8.

a. The population standard deviation (using the result from the previous problem) is s = b. By taking the square root of the sample variances from the previous problem we get Sample Standard Deviation 0.000 0.707 2.828 3.536

4.67 = 2.160.

Probability 3/9 2/9 2/9 2/9

3×0 + 2 ×0.707 + 2 ×2.828 + 2 ×3.536 = 1.571.

9 d. No, the mean of the sampling distribution of the sample standard deviations is 1.571, and it is not equal to the value of the population standard deviation (2.160), so the sample standard deviations do not target the value of the population standard deviation. a. The population median is 5. b. The possible sample of size 2 are {(4, 4), (4, 5), (4, 9), (5, 4), (5, 5), (5, 9), (9, 4), (9, 5), (9, 9)}, which have the following medians {4, 4.5, 6.5, 4.5, 5, 7, 6.5, 7, 9} with the following associated probabilities. c. The mean is

9.

Sample Median 4.0 4.5 5.0 6.5 7.0 9.0

Probability 1/9 2/9 1/9 2/9 2/9 1/9

4 + 2 ×4.5 + 5 + 2 ×6.52 ×7 + 9 = 6.0.

9 d. No, the mean of the sampling distribution of the sample medians is 6.0, and it is not equal to the value of the population median of 5.0, so the sample medians do not target the value of the population median. c. The mean is

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

10. a. The proportion of odd numbers is 2/3, or 0.7 (there are two odd numbers from the population of 4, 5, and 9). b. The possible sample of size 2 are {(4, 4), (4, 5), (4, 9), (5, 4), (5, 5), (5, 9), (9, 4), (9, 5), (9, 9)}, which have the following proportion of odd numbers {0, 0.5, 0.5, 0.5, 1, 1, 0.5, 1, 1}.

Sample Proportion 0.0 0.5 1.0 c. The mean is

0 + 4 ×0.5 + 4 ×1

=

Probability 1/9 4/9 4/9

2

, or 0.7.

9 3 d. Yes. The mean of the sampling distribution of the sample proportion of odd numbers is 2/3, and it is equal to the value of the population proportion of odd numbers of 2/3, so the sample proportions target the value of the population proportion 11. a. The possible samples of size 2 are {(2, 2), (2, 3), (2, 5), (2, 9), (3, 2), (3, 3), (3, 5), (3, 9), (5, 2), (5, 3), (5, 5), (5, 9), (9, 2), (9, 3), (9, 5), (9, 9)}, which have the following means and associated probabilities.

b. The mean of the population is

x 2.0 2.5 3.0 3.5 4.0 5.0 5.5 6.0 7.0 9.0 2+3+5+9

Probability 1/16 2/16 1/16 2/16 2/16 1/16 2/16 2/16 2/16 1/16 = 4.75 and the mean of the sample means is also

4 2.0 ×1+ 2.5×2 + 3.0 ×1+3.5×2 + 4.0 ×2+ 5.0 ×1+ 5.5×2 + 6.0 ×2+ 7.0 ×2+ 9.0 ×1 = 4.75.

16 c. The sample means target the population mean. Sample means make good estimators of population means because they target the value of the population mean instead of systematically underestimating or overestimating it.

12. a. The possible samples of size 2 are {(2, 2), (2, 3), (2, 5), (2, 9), (3, 2), (3, 3), (3, 5), (3, 9), (5, 2), (5, 3), (5, 5), (5, 9), (9, 2), (9, 3), (9, 5), (9, 9)}, which have the following medians and associated probabilities.

Sample Median 2.0 2.5 3.0 3.5 4.0 5.0 5.5 6.0 7.0 9.0

Probability 1/16 2/16 1/16 2/16 2/16 1/16 2/16 2/16 2/16 1/16

12. (continued) b. The median of the population is

3 +5

= 4.0, but the mean of the sample medians is 2 2.0 ×1+ 2.5×2 + 3.0 ×1+3.5×2 + 4.0 ×2+ 5.0 ×1+5.5×2 + 6.0 ×2+ 7.0 ×2+ 9.0 ×1 = 4.75, so the values are not 16 equal. c. The sample medians do not target the population median of 4.0, so the sample medians do not make good estimators of the population medians.

13. a. The possible samples of size 2 are {(2, 2), (2, 3), (2, 5), (2, 9), (3, 2), (3, 3), (3, 5), (3, 9), (5, 2), (5, 3), (5, 5), (5, 9), (9, 2), (9, 3), (9, 5), (9, 9)}, which have the following ranges and associated probabilities.

Sample Range 0.0 1.0 2.0 3.0 4.0 6.0 7.0

Probability 4/16 2/16 2/16 2/16 2/16 2/16 2/16

b. The range of the population is 9 - 2 = 7.0, but the mean of the sample ranges is 0 ×4 +1×2 + 2 ×2 + 3×2 + 4 ×2 + 6 ×2 + 7 ×2 = 2.875, so the values are not equal.

16 c. The sample ranges do not target the population range of 7.0, so sample ranges do not make good estimators of the population range.

14. a. The possible samples of size 2 are {(2, 2), (2, 3), (2, 5), (2, 9), (3, 2), (3, 3), (3, 5), (3, 9), (5, 2), (5, 3), (5, 5), (5, 9), (9, 2), (9, 3), (9, 5), (9, 9)}, which have the following variances and associated probabilities.

Sample Variance (s2) 0.0 0.5 2.0 4.5 8.0 18.0 24.5

Probability 4/16 2/16 2/16 2/16 2/16 2/16 2/16

The variance of the population is 7.2 The mean of the sample variances is 4 ×0+ 2 ×0.5 + 2 ×2.0 + 2 ×4.5 + 2 ×8.0 + 2 ×18.0 + 2 ×24.5 = 7.2. The two values are equal.

16 c. The sample variances do target the population variance of 7.2, so sample variances do make good estimators of the population variance.

15. The possible birth samples are {(m, m), (m, f), (f, m), (f, f)}.

Proportion of Females 0 0.5 1

Probability 0.25 0.50 0.25

Yes. The proportion of females in 2 births is 0.5, and the mean of the sample proportions is 0.5. The result suggests that a sample proportion is an unbiased estimator of the population proportion.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

16. The possible birth samples are {mmm, fmm, mfm, mmf, ffm, fmf, mff, fff}.

Proportion of Females 0 1/3 2/3 1

Probability 1/8 3/8 3/8 1/8

Yes. The proportion of females in 3 births is 0.5 and the mean of the sample proportions is 0.5. The result suggests that a sample proportion is an unbiased estimator of the population proportion.

17. The possibilities are: both questions incorrect, one question correct (two choices), both questions correct. a.

Proportion Correct 0 =0 2 1 = 0.5 2 2 =1 2 b. The mean is

Probability 4 4 16 × = 5 5 25 æ1×4 ÷ö 8 2×ç = è5 ø 25 5 æ1 1 ö 1 ç ×÷ = è5 ø 25 5

16 ×0 + 8×0.5 +1×1

= 0.2.

25 c. Yes, the sampling distribution of the sample proportions has a mean of 0.2 and the population proportion is also 0.2 (because there is 1 correct answer among 5 choices). Yes, the mean of the sampling distribution of the sample proportions is always equal to the population proportion.

18. a. The proportions of 0, 0.5, and 1 have the following probabilities.

Proportion with Yellow Pods Probability 0 1/25 0.5 8/25 1 16/25 1×0 + 8×0.5 +16 ×1 b. The mean is = 0.8.

25 c. Yes, the population proportion is 0.8 and the mean of the sampling proportions is also 0.8. The mean of the sampling distribution of proportions is always equal to the population proportion.

19.

1 1 The formula yields P(0) = = 0.25 , P(0.5) = = 0.5, and 2(2 - 2 ×0)!(2 ×0)!

2(2 - 2 ×0.5)!(2 ×0.5)!

1 P(1) = = 0.25, which describes the sampling distribution of the sample proportions. The 2(2 - 2×1)!(2×1)! formula is just a different way of presenting the same information in the table that describes the sampling distribution.

20. Sample values of the mean absolute deviation (MAD) do not usually target the value of the population MAD, so a MAD statistic is not good for estimating a population MAD. If the population of {4, 5, 9} from Example 5 is used, the sample MAD values of 0, 0.5, 2, and 2.5 have corresponding probabilities of 3/9, 2/9, 2/9, and 2/9. For these values, the population MAD is 2, but the sample MAD values have a mean of 1.1, so the mean of the sample MAD values (1.1) is not equal to the population MAD of 2.0.

Section 6-4: The Central Limit Theorem 1. The sample must have more than 30 values, or there must be evidence that the population of systolic blood pressures has a normal distribution.

2.

No, because the original population is normally distributed, the sample means will be normally distributed for any sample size, not just for n > 30.

3.

mx represents the mean of all sample means and s x represents the standard deviation of all sample means. For

samples of 36 IQ scores, mx = 100, and s x = 15 4.

5.

6.

7.

8.

9.

36 = 2.5.

No. The sample of 50 annual incomes will tend to have a distribution that is skewed to the right, no matter how large the sample is. If we compute the sample mean, we can consider that value to be one value in a normally distributed population.

0 - 1.2 = - 0.24; which has an area of 0.4052 to the left. (Tech: 0.4033) a. zx=0 = 4.9 0 - 1.2 b. z x=0 = = - 1.22; which has an area of 0.1112 to the left. (Tech: 0.1104) 4.9 / 25 c. Because the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

2 - 1.2 = = 0.16; which has an area of 1- 0.5636 = 0.4364 to the right. (Tech: 0.4352) a. zx=2 4.9 2 - 1.2 = 0.65; which has an area of 1- 0.7422 = 0.2578 to the right. (Tech: 0.2569) b. z x=2 = 4.9 / 16 c. Because the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

0 - 1.2 3 - 1.2 = = - 0.24 and z = = 0.37, which has an area of 0.6443- 0.4052 = 0.2391 between a. zx=0 x=3 4.9 4.9 them. (Tech: 0.2401) 0 - 1.2 3 - 1.2 = - 0.73 and z = 1.10, which has an area of 0.8643- 0.2327 = 0.6316 b. z x=0 = x=3 = 4.9 / 9 4.9 / 9 between them (Tech: 0.6335). c. Because the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

0.5 - 1.2 2.5 - 1.2 = = - 0.14 and z = = 0.27, which has an area of 0.6064 - 0.4443 = 0.1621 a. zx=0.5 x=2.5 4.9 4.9 between them. (Tech: 0.1614) 0.5 - 1.2 2.5 - 1.2 = = - 0.29 and z = 0.53, which has an area of 0.7019 - 0.3859 = 0.3160 b. z x=0.5 x=2.5 = 4.9 / 4 4.9 / 4 between them (Tech: 0.3146). c. Because the original population has a normal distribution, the distribution of sample means is a normal distribution for any sample size.

17.7 - 14.7 13.6 - 14.7 = = 2.31 and z = = - 0.85 which have 98.96%- 19.77% = 79.19% a. zx=17.7 x=13.6 1.3 1.3 (Tech: 79.08%) between them.

17.7 - 14.7 13.6 - 14.7 = = 20.8 and z b. z x=17.7 = - 7.62 which have 99.99%- 0.01% = 99.98% x=13.6 = 1.3 9 1.3 9 (Tech: 99.44%) between them.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

15.1- 13.0

12.1- 13.0 = 1.62 and z = - 0.69 which have 94.74%- 24.51% = 70.23% x=12.1 = 1.3 1.3 (Tech: 70.25%) between them.

15.1- 13.0 12.1- 13.0 = 4.85 and z b. z x=15.1 = = - 2.08 which have 99.99%- 1.88% = 98.11% x=12.1 = 1.3 9 1.3 9 between them.

90 - 70.2 = 1.77; which has 100%- 96.16% = 3.84% (Tech: 3.85%) to the right.

11. a. zx=90 = 11.2 90 - 70.2 = 7.07; which has 100%- 99.99% = 0.01% (Tech: 0.02%) to the right. b. zx=90 = 11.2 4 90 - 71.3 = 1.56; which has 100%- 94.06% = 5.94% (Tech: 5.96%) to the right.

12. a. zx=90 = 12.0 90 - 71.3 = 3.12; which has 100%- 99.91% = 0.09% to the right. b. zx=90 = 12.0 4 148 - 189 = - 1.05, which has an area of 1- 0.1469 = 0.8531 to the right (Tech: 0.8534).

13. a. zx=148 = 39 148 - 189 = - 5.46 which has an area of 1- 0.0001 = 0.9999 to the right (Tech: 0.99999998). b. zx=148 = 39 / 27 c. It appears that the elevator would be overloaded with 27 adult males, which is probably unlikely but possible. Also, it is likely that a safety factor is built in so that the elevator can safely take a load greater than the 4000 lb capacity on the placard. But to be safe, instead of boarding the elevator full of adult men, it would be wiser to wait for another elevator.

22 - 18.2 = 3.8; which has a probability of 0.9999. So the percentage is 99.99%.

14. a. zx=22 = 1 18.5 - 18.2 = = 1.8 which has a probability of 0.9641. No, when considering the diameters of b. z 10. a. zx=15.1

=

x=18.5

1 36 manholes, we should use a design based on individual males, not samples of 36 males.

15. a. The mean weight of adult male passengers is 3500 25 = 140 lb.

140 - 189 b. zx=140 = = - 6.28; which has a probability of 1- 0.0001 = 0.9999 (Tech: 0.999999998) to the 39 25 right.

175 - 189 = = - 1.61; which has a probability of 1- 0.0537 = 0.9463 (Tech: 0.9458) to the right. c. zx=175 39 20 d. The new capacity of 20 passengers does not appear to be safe enough because the probability of overloading is too high.

140 - 189 = - 8.88; which has a probability of 1- 0.0001 = 0.9999 to the right (Tech: 1.0000 when 16. a. zx=140 = 39 50 rounded to four decimal places).

174 - 189 = - 1.44; which has a probability of 1- 0.0449 = 0.9251 to the right (Tech: 0.9249). b. z x=174 = 39 14 Because there is a high probability of overloading, the new ratings do not appear to be safe when the boat is loaded with 14 male passengers.

140 - 171 211- 171 = - 0.67; which have a probability of 0.8078 - 0.2514 = 0.87 and z x=140 = 46 46 = 0.5564 between them (Tech: 0.5575).

211- 171 140 - 171 = 4.35 and z b. z x=211 = = - 3.37; which have a probability of 0.9999 - 0.0004 x=140 = 46 25 46 25 = 0.9995 between them (Tech: 0.9996). c. Part (a) because the ejection seats will be occupied by individual females, not groups of females.

167.6 - 189.0 = - 3.34; which has a probability of 1- 0.0004 = 0.9996 to the right. There is a 0.9996 18. zx=167.6 = 39 37 probability that the aircraft is overloaded. Because that probability is so high, the pilot should take action, such as removing excess fuel and/or requiring that some passengers disembark and take a later flight.

72 - 68.6 = 1.21; which has a probability of 0.8869 (Tech: 0.8877).

19. a. zx=72 = 2.8 72 - 68.6 = 12.1; which has a probability of 0.9999. (Tech: 1.0000 when rounded to four decimal b. zx=72 = 2.8 100 places.) c. The probability from part (a) is more relevant because it shows that 89% of male passengers will not need to bend. The result from part (b) gives us information about the mean for a group of 100 males, but it doesn’t give us useful information about the comfort and safety of individual male passengers. d. Because males are generally taller than females, a design that accommodates a suitable proportion of males will necessarily accommodate a greater proportion of females.

53 - 51.6 = 0.64, which has an area of 0.7389 to the left. The percentage is 73.89% (Tech: 20. a. zx=53 = 2.2 73.77%), which is too high. b. The z score with 95% of values to the left is 1.645, which corresponds to a distance of 51.6 - 1.645×2.2 = 48.0 in. c. The z score with 95% of values to the left is 1.645, which corresponds to a distance of 17. a.

zx=211 =

(

)

51.6 - 1.645× 2.2 / 40 = 51.0 in. Because the cockpits are occupied by individual pilots, it would be a big mistake to plan a design around the mean for 40 pilots. With a design of 51.0 in., about 39% of individual pilots would not have sufficient overhead grip reach, so far too many pilots would be disqualified.

2.6 - 15 = - 11.8, which has an area of 0.0001 to the left. If we assume that the true mean weight 21. zx=2.6 = 8.6 / 67 gain is 15 lb, the probability of getting a sample of 67 college students having a mean weight gain of 2.6 lb or less is 0.0001 (Tech: 0+). Because it is so unlikely to get a sample such as the one obtained, it appears that the assumption of a mean weight gain of 15 lb is an incorrect assumption. The Freshman 15 claim of a mean weight gain of 15 lb appears to be an incorrect claim.

22. zx=3.0 =

3.0 - 0

= 3.88, which has an area of 1- 0.9999 = 0.0001 to the right (Tech: 0.0000540). If the mean 4.9 / 40 really is 0 lb as assumed, the probability of getting a sample with a mean of 3.0 lb or higher is very small, so the mean of 3.0 lb is significantly high. There is strong evidence suggesting that the mean is actually higher than 0 lb, so the diet does appear to be effective. However, the mean weight loss of only 3.0 lb is not very much, so even though the diet appears to be effective, it doesn’t appear to be worth using this diet. The amount of weight loss appears to have statistical significance but not practical significance.

6.8 - 7 = - 0.35, which has an area of 0.3632 to the left (Tech: 0.3645). If the mean really is 7 hours 23. zx=6.8 = 2.0 / 12 as assumed, the probability of getting a sample with a mean of 6.8 hours or less is not low (such as 0.05 or less), so the mean of 6.8 hours is not significantly low. There isn’t enough evidence to support the claim that the mean sleep time is less than 7 hours.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

24. The sample mean is 97.78°F. zx=97.78 =

97.78 - 98.6

= - 4.18, which has an area of 1- 0.9999 = 0.0001 to the 0.62 / 10 left (Tech: 0.0000). The sample mean is significantly low. We can conclude that the population mean is lower than 98.6°F.

25. a. Yes, the sampling is without replacement and the sample size of 50 is greater than 5% of the finite 16 275 - 50 = 2.0504584 population size of 275. s = x 50 275 - 1 105 - 95.5 95 - 95.5 = 4.63 and z = b. z x=105 = = - 0.24; which have a probability of x=95 2.0504584 2.0504584 0.9999 - 0.4052 = 0.5947 between them (Tech: 0.5963).

Section 6-5: Assessing Normality 1. The requirement is satisfied because the sample size of 147 is greater than 30.

2. The histogram should be approximately bell-shaped, and the normal quantile plot should have points that approximate a straight-line pattern.

3. Either the points are not reasonably close to a straight-line pattern, or there is some systematic pattern that is not a straight-line pattern.

4. Because the histogram is roughly bell-shaped, we can conclude that the data are from a population having a normal distribution.

5. Normal, the points are reasonably close to a straight-line pattern, and there is no other pattern that is not a straight-line pattern.

6. Not normal, the points are not reasonably close to a straight-line pattern.

7. Not normal, the points are not reasonably close to a straight-line pattern.

8. Normal, the points are reasonably close to a straight-line pattern, and there is no other pattern that is not a straight-line pattern.

9. not normal 10 not normal

11. not normal

12. normal

13. Not normal, there is a systematic pattern that is not a straight-line pattern. The data is strongly right-skewed.

Graph for Exercise 13 Graph for Exercise 14

14. Not normal, there is a systematic pattern that is not a straight-line pattern. The data has a short tail at the left and a long tail at the right, relative to a normal distribution.

15. Normal, the points are reasonably close to a straight-line pattern.

Graph for Exercise 15 Graph for Exercise 16

16. Normal, the points are reasonably close to a straight-line pattern.

17. Normal, the points are reasonably close to a straight-line pattern and there is no other pattern that is not a straight-line pattern. The points have coordinates (97.9, - 1.28), (98.0, - 0.52), (98.2, 0), (98.4, 0.52), (98.7,1.28).

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

Graph for Exercise 17

18. Not normal. The points have coordinates of (4, - 1.38), (19,1.38).

Graph for Exercise 18

(17, - 0.67), (18, - 0.21), (18, 0.21), (19, 0.67),

19. Not normal, the points are not reasonably close to a straight-line pattern and there appears to be a pattern that is not a straight-line pattern. The points have coordinates (963, - 1.53), (1027, - 0.89), (1029, - 0.49), (1034, - 0.16), (1070, 0.16), (1079, 0.49), (1079, 0.89), (1439,1.53).

Graph for Exercise 19

Graph for Exercise 20

(166.1,- 0.97), (166.1,- 0.59), (170.2, - 0.28),

20. Normal. The points have coordinates of (163.9, - 1.59), (170.7, 0), (176.7, 0.28), (177.8, 0.59) , (179.4, 0.97), (183.8,1.59).

21.

a. Yes, the histogram or normal quantile plot will remain unchanged. b. Yes, the histogram or normal quantile plot will remain unchanged. c. No, the histogram and normal quantile plot will not indicate a normal distribution.

22. The original values are not from a normally distributed population.

After taking the logarithm of each value, the values appear to be from a normally distributed population. The original values are from a population with a lognormal distribution.

Section 6-6: Normal as Approximation to Binomial (Download Only) 1. a. the area below (to the left of) 502.5 b. the area between 501.5 and 502.5 c. the area above (to the right of) 502.5 2. Yes, the circumstances correspond to 100 independent trials of a binomial experiment in which the probability of success is 0.2. Also, with n = 100, p = 0.2, and q = 0.8, the requirements of np = 100×0.2 = 20 ³ 5 and nq = 100×0.8 = 80 ³ 5 are both satisfied.

3.

4.

5.

p = 0.2, q = 0.8, m = 20, s = 4; The value of m = 20 shows that for people who make random guesses for the 100 questions, the mean number of correct answers is 20. For people who make 100 random guesses, the standard deviation of s = 4 is a measure of how much the numbers of correct responses vary.

The histogram should be approximately normal or bell-shaped, because sample proportions tend to approximate a normal distribution.

The requirements for the normal approximation are satisfied with np = 20×0.512 = 10.2 ³ 5 and nq = 20×0.488 = 9.8 ³ 5. zx=7.5 =

7.5 - 20 ×0.512 = - 1.23; which has a probability of 0.1093 (Tech: 20×0.512×0.488

6.

0.1102) to the left.

The requirement of np = 8×0.512 = 4.1 ³ 5 is not satisfied. The normal approximation should not be used.

7.

The requirement of np = 20×0.2 = 4 ³ 5 is not satisfied. The normal approximation should not be used.

8.

The requirements for the normal approximation are satisfied with np = 50×0.2 = 10 ³ 5 and 12.5 - 50 ×0.2 11.5 - 50 ×0.2 = 0.88; which have a = 0.53 and z x=12.5 = nq = 50×0.8 = 40 ³ 5. zx=11.5 = 50×0.2×0.8 50×0.2×0.8 probability of 0.8106 - 0.7019 =0.1087 (Tech: 0.1096) between them.

9.

The requirements for the normal approximation are satisfied with np = 100×0.23 = 23 ³ 5 and 19.5 - 100 ×0.23 = - 0.83; which has a probability of 0.2033 (Tech: 0.2028) 100 ×0.23×0.77 to the left. No, 20 is not a significantly low number of white cars. (Tech: Using the binomial distribution: 0.2047.) nq = 100×0.77 = 77 ³ 5. zx=19.5 =

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

10. The requirements for the normal approximation are satisfied with np = 100×0.18 = 18 ³ 5 and nq = 100×0.82 24.5 - 100 ×0.18 = 1.69; which has a probability of 1- 0.9545 = 0.0455 (Tech: 0.2028) to = 82 ³ 5. zx=24.5 = 100×0.18×0.82 the right. Yes, 25 is a significantly high number of black cars. (Tech: Using the binomial distribution: 0.0496.)

11. The requirements for the normal approximation are satisfied with np = 100×0.10 = 10 ³ 5 and 13.5 - 100 ×0.10 14.5 - 100 ×0.10 = 1.17 and z x=14.5 = = 1.50; which have a 100×0.10×0.90 100×0.10 ×0.90 probability of 0.9524 - 0.9332 = 0.0542 (Tech: 0.0549) between them. Determination of whether 14 red cars is significantly high should be based on the probability of 14 or more red cars, not the probability of exactly 14 red cars. (Tech: Using the binomial distribution: 0.0513.)

12. The requirements for the normal approximation are satisfied with np = 100×0.16 = 16 ³ 5 and nq = 100×0.84 nq = 100×0.90 = 90 ³ 5. z x=13.5 =

9.5 - 100 ×0.16 10.5 - 100 ×0.16 = - 1.77 and z x=10.5 = = - 1.50; which have a probability 100×0.16 ×0.84 100 ×0.16 ×0.84 of 0.0668 - 0.0384 = 0.0284 (Tech: 0.0287) between them. Determination of whether 10 gray cars is significantly high should be based on the probability of 10 or fewer red cars, not the probability of exactly 10 red cars. (Tech: Using the binomial distribution: 0.0292.)

13. a. The requirements for the normal approximation are satisfied with np = 879 ×0.25 = 219.75 ³ 5 and = 84 ³ 5. z x=9.5 =

230.5 - 879 ×0.25 231.5 - 879 ×0.25 = 0.84 and z = = 0.92; x=231.5 879×0.25×0.75 879 ×0.25×0.75 which have a probability of 0.8212 - 0.7995 = 0.0217 (Tech: 0.0212) between them. (Tech: Using the binomial distribution: 0.0209.) b. The requirements for the normal approximation are satisfied with np = 879×0.25 = 219.75 ³ 5 and 230.5 - 879 ×0.25 = 0.84; which has a probability of nq = 879×0.75 = 659.25 ³ 5. zx=230.5 = 879×0.25×0.75 1- 0.7995 = 0.2005 (Tech: 0.2012) to the right. The result of 231 overturned calls is not significantly high.

(Tech: Using the binomial distribution: 0.2006.) nq = 879×0.75 = 659.25 ³ 5. z x=230.5 =

14. a. The requirements for the normal approximation are satisfied with np = 879×0.22 = 193.38 ³ 5 and 230.5 - 879 ×0.22 231.5 - 879 ×0.22 = 3.02 and z x=231.5 = = 3.10; 879×0.22×0.78 879 ×0.22 ×0.78 which have a probability of 0.9990 - 0.9987 = 0.0003 between them. b. The requirements for the normal approximation are satisfied with np = 879×0.22 = 193.38 ³ 5 and 230.5 - 879 ×0.22 = 3.02; which has a probability of nq = 879×0.78 = 685.62 ³ 5. zx=230.5 = 879×0.22 ×0.78 1- 0.9987 = 0.0013 to the right. The result of 231 overturned calls is significantly high. (Tech: Using the binomial distribution: 0.0015.)

15. a. The requirements for the normal approximation are satisfied with np = 250×0.51 = 127.5 ³ 5 and 109.5 - 250 ×0.51 = - 2.28; which has a probability of 0.0113 (Tech: nq = 250×0.49 = 122.5 ³ 5. zx=109.5 = 250×0.51×0.49 0.0114) to the left. (Tech: Using the binomial distribution: 0.0113.) b. The result of 109 is significantly low.

16. a. The requirements for the normal approximation are satisfied with np = 650×0.12 = 78 ³ 5 and 85.5 - 650 ×0.12 = 0.91; which has a probability of nq = 650×0.88 = 572 ³ 5. zx=85.5 = 650×0.12 ×0.88 1- 0.8186 = 0.1814 (Tech: 0.1827) to the right. (Tech: Using the binomial distribution: 0.1819.) b. The result of 86 people with green eyes is not significantly high. nq = 879×0.78 = 685.62 ³ 5. zx=230.5 =

17. a. The requirements for the normal approximation are satisfied with np = 929×0.75 = 696.75 ³ 5 and 704.5 - 929 ×0.75 = 0.59; which has a probability of nq = 929×0.25 = 232.25 ³ 5. zx=704.5 = 929 ×0.75 ×0.25 1- 0.7224 = 0.2776 (Tech: 0.2785) to the right. (Tech: Using the binomial distribution: 0.2799.) b. The result of 705 peas with red flowers is not significantly high. c. The result of 705 peas with red flowers is not strong evidence against Mendel’s assumption that 3 4 of peas will have red flowers.

18. a. The requirements for the normal approximation are satisfied with np = 1480×0.292 = 432.16 ³ 5 and 454.5 - 1480 ×0.292 = 1.28; which has a probability of nq = 1480×0.708 = 1047.84 ³ 5. zx=454.5 = 1480 ×0.292 ×0.708 1- 0.8997 = 0.1003 (Tech: 0.1008) to the right. (Tech: Using the binomial distribution: 0.1012.) b. The result of 455 who have sleepwalked is not significantly high. c. The result of 455 does not provide strong evidence against the rate of 29.2%.

19. a. Using the normal approximation: m = 1002×0.61 = 611.22, s =

1002×0.61×0.39 = 15.4394, and

700.5 - 611.22 = 5.78; which has a probability of 0.0001 (Tech: 0.0000) to the right.

1002 ×0.61×0.39 b. The result suggests that the surveyed people did not respond accurately. zx=700.5 =

20. a. Using normal approximation: m = 420, 095×0.00034 = 142.83, s = = 11.9492, and z x=135.5 =

420,095×0.000344×0.999656

135.5 - 142.83

= - 0.61; which has a probability of 0.2709 420, 095×0.000344 ×0.999656 (Tech: 0.2697) to the left. (Tech: Using the binomial distribution: 0.2726.) b. Media reports appear to be wrong.

21. (1) The requirements for the normal approximation are satisfied with np = 11×0.512 = 5.632 ³ 5 and 6.5 - 11×0.512 7.5 - 11×0.512 = 1.13; = 0.52 and z x=7.5 = 11×0.512 ×0.488 11×0.512×0.488 which have a probability of 0.8708 - 0.6985 = 0.1723 between them.

(2) 0.1704 (3) 0.1726 No, the approximations are not off by very much.

22. The z score that corresponds to a 0.95 probability is 1.645. This means that we have to solve the equation 1.645 × n ×0.9005 ×0.0995 + n ×0.9005 = 213 for n. This has a solution of 229 (Tech: 230) reservations. nq = 11×0.488 = 5.368 ³ 5. zx=6.5 =

Quick Quiz 1.

2.

z = P90 = 1.28

3.

P(z > 1.55) = 1- P(z < 1.55) = 1- 0.9394 = 0.0606

4.

P(- 1.00 < z < 2.00) = P(z < 2.00)- P(z < - 1.00) = 0.9772 - 0.1587 = 0.8185 (Tech: 0.8186)

5.

a. m = 0 and s = 1 b. mx represents the mean of all sample means, and s x represents the standard deviation of all sample means.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

6.

z x=80 =

7.

z x=60

8.

9.

=

80.0 - 70.2 11.2 60.0 - 70.2

= 0.88; which has a probability of 0.8106 (Tech: 0.8092) to the left. = - 0.91 and z

x=80

11.2 = 0.6292 (Tech: 0.6280) between them.

=

80.0 - 70.2

= 0.88; which have a probability of 0.8106 - 0.1814

11.2

The z score for P90 is 1.28, which corresponds to a diastolic blood pressure of 1.28×11.2 + 70.2 = 84.5 mmHg (Tech: 84.6 mmHg).

75.0 - 70.2 = 1.71; which has a probability of 0.9564 (Tech: 0.9568) to the left. zx=75 = 11.2 16

10. The normal quantile plot suggests that diastolic blood pressure levels of women are normally distributed.

Review Exercises 1.

a. P(z > - 1.37) = 1- P(z > - 1.37) = 1- 0.0853 = 0.9147 b. P(z < 2.34) = 0.9904 c. P(- 0.67 < z < 1.29) = P(z < 1.29)- P(z < - 0.67) = 0.9015 - 0.2514 = 0.6501 (Tech: 0.6500) d.

Q1 = - 0.67

e.

æ 0.23 - 0.0 ö P (z > 0.23) = P çz > = P (z > 0.69) = 1- 0.7549 = 0.2451 ÷ è 1 9 ø

2.

a. An unbiased estimator is a statistic that targets the value of the population parameter in the sense that the sampling distribution of the statistic has a mean that is equal to the mean of the corresponding parameter. b. mean, variance, and proportion c. true

3.

a. z0.99 = - 2.33 b. z0.01 = 2.33 c. z0.025 = 1.96

4.

a. The distribution of samples means is normal. b. mx = 33.64 cm c. s x = 4.14

5.

25 = 0.83

a. The area under the curve is 1. b. The median is 3037.1 g. c. The mode is 3037.1 g. d. The variance is (706.3 g)2 = 498,859.7 g2.

6.

7.

a. The z score with 2% of values to the right is 2.054, which corresponds to an IQ score of 100 + 2.054×15 = 131.

131- 100 = 4.13, which has an area of 0.0001 to the right. (Tech: 0.0000179) b. z x =131 = 15 / 4 c. No. It is very possible that the 4 subjects have a mean of 131 while some of them have scores below 131.

70.0 - 63.7 = = 2.17; which has a probability of 1- 0.9850 = 0.0150 , or 1.50% (Tech: 1.49%) to the a. zx=70 2.9 right. b. The z score for the upper 2.5% is 1.96 which corresponds to a doorway height of 1.96×2.9 + 63.7 = 69.4 in.

zx=54 =

54 - 59.7

= - 2.28, which has an area of 0.0113, or 1.13%, to the left.

2.5 b. The z score with 95% of values to the left is 1.645, which corresponds to a height of 59.7 +1.645×2.5 = 63.8 in.

8.

a.

9.

The z score with 1% of values to the left is - 2.326, which corresponds to a height of 59.7 - 2.326×2.5 = 53.9 in. The z score with 1% of values to the right is 2.326, which corresponds to a height of 59.7 + 2.326×2.5 = 65.5 in. Yes, 67 in. is significantly high.

10. Yes. The points are reasonably close to a straight-line pattern, and there is no other pattern that is not a straightline pattern.

Cumulative Review Exercises 1.

15.9 +18.7 + 24.2 + 28.7 + 28.8 + 28.9 + 28.9 + 28.9 + 29.0 + 29.1+ 29.3 + 31.4 = 26.82.

12 28.9 + 28.9 b. The median is = 29.90.

2 a. The mean is x =

c. The standard deviation is s = d. z x=31.4 =

2.

(15.9 - 26.82)2 + + (31.4 - 26.82)2 = 4.76.

12 - 1

31.4 - 26.82 = 0.96 4.76

e. ratio f. continuous 25×12 = 3, so Q 24.2 + 28.7 a. Q : L = = 26.45 1 1 = 100 2 50 ×12 Q : L= = 6, so Q = 28.9 + 28.9 = 28.90 2 2 100 2 75×12 Q : L= = 9, so Q 29.0 + 29.1 = = 29.05 3 3 100 2 b.

c. The sample does not appear to be from a population having a normal distribution.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

3.

a.

zx=221.5 =

b. z x=220

=

221.5 - 246.3

12.4 220 - 246.3

= - 2.00, which has an area of 0.0228 to the left.

= - 2.12 and z

x=250

=

250 - 246.3

= 0.30, which has an area of 0.6179 - 0.0170

12.4 12.4 = 0.6009 between them. (Tech: 0.6003) c. The z score with 95% of values to the left is 1.645, so P95 = 246.3 + 1.645×12.4 = 266.7 mm.

250 - 246.3 = 1.19, which has an area of 1- 0.8830 = 0.1170 to the right (Tech: 0.1163). d. z x =250 = 12.4 / 16 e. The result from part (c) is more helpful because it provides a foot length that is potentially the maximum to be used for shoe sizes. The result from part (d) is based on the distribution of means from groups of 16 women, which is irrelevant for planning shoe sizes.

4.

a. B is the event of selecting someone who does not have blue eyes. b. P(B) = 1- P(B) = 1- 0.35 = 0.65 c. 0.35×0.35×0.35 = 0.0429

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

×0.6560 + + 100C100 ×0.35100 ×0.650 = 0.1724 (Table: 0.1736) Using normal approximation (see Section 6-6.): 0.1727 e. No, a result of 40 people with blue eyes among 100 randomly selected people is not significantly high. d.

5.

100C40 ×0.35

40

97.6 + 98.2 + 99.6 + 98.7 + 99.4 + 98.2 + 980 + 98.6 + 98.6 = 196.54°F, which is far too 9 high to be reasonable as the mean body temperature of a group of adult males. b. The value of 980°F is an outlier and an error because it is not a possible body temperature of an adult male. c. If the value of 980°F is discarded as an error, the mean appears to be 98.61°F. It is also reasonable to believe that the value of 980°F was incorrectly entered without the decimal point, and if 980°F is changed to 98.0°F, the mean appears to be 98.54°F. a. The mean is x =

Chapter 7: Estimating Parameters and Determining Sample Sizes Section 7-1: Estimating a Population Proportion 1. The confidence level (such as 95%) was not provided.

2. When using 10.5% to estimate the value of the population percentage for middle school students who use e-cigarettes, the maximum likely difference between 10.5% and the true population percentage is one percentage point, so the interval from 10.5%- 1.0% = 9.5% to 10.5% +1.0% = 11.5% is likely to contain the true population percentage.

3.

pˆ = 0.105 is the sample proportion; qˆ = 1 - pˆ = 1 - 0.105 = 0.895; n = 8837 is the sample size; E = 0.010 is the margin of error; p is the population proportion, which is unknown. The value of a is 0.05.

4.

The 95% confidence interval is wider than the 80% confidence interval. A confidence interval must be wider in order for us to be more confident that it captures the true value of the population proportion. (Think of estimating the age of a classmate. You might be 90% confident that they are between 20 and 30, but you might be 99.9% confident that they are between 10 and 40.)

5.

z0.05 = 1.645

7.

z0.0025 = 2.81

6.

z0.005 = 2.576 (Table: 2.575)

8.

z0.01 = 2.33

0.375 + 0.425 = 0.400 2 0.275 + 0.425 = 0.350 10. pˆ = 2 9.

pˆ =

and E = and E =

0.425 - 0.375 2 0.425 - 0.275

= 0.025; 0.400 ± 0.025

= 0.075; 0.350 ± 0.075 2 11. pˆ = 0.0780 + 0.162 = 0.120 and E = 0.162 - 0.0780 = 0.0420; 0.120 - 0.042 < p < 0.120 + 0.042, or 2 2 0.0780 < p < 0.162 12. 0.070 - 0.021 < p < 0.070 + 0.021 , or 0.049 < p < 0.091 13. a.

pˆ = 0.105

b. E = za /2 c.

p̂q̂ = 1.96 n

(0.105)(0.895) = 0.006 8837

pˆ - E < p < pˆ + E 0.105 - 0.006 < x < 0.105 + 0.006 0.099 < x < 0.111

d. We have 95% confidence that the interval from 0.099 to 0.111 contains the true value of the population proportion of middle school students who use e-cigarettes.

14. a.

pˆ = 153 5924 = 0.0258

c.

( 153 )( 771 )

p̂q̂ = 2.576 5924 5924 = 0.00531 n 5924 pˆ - E < p < pˆ - E 0.0258 - 0.00531 < p < 0.0258 + 0.00531

b. E = za /2

0.0205 < p < 0.0311

d. We have 99% confidence that the interval from 0.0205 to 0.0311 actually does contain the true value of the population proportion of all Eliquis users who experience nausea.

15. a.

pˆ = 717 5000 = 0.143

b. E = za /2

p̂q̂ = 1.645 n

5000

5000

5000

= 0.00815

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

15. (continued) c.

pˆ - E < p < pˆ - E 0.143 - 0.00815 < p < 0.143 + 0.00815 0.135 < p < 0.152

d. We have 90% confidence that the interval from 0.135 to 0.152 actually does contain the true value of the population proportion of returned surveys.

16. a.

pˆ = 856 1228 = 0.697

p̂q̂ b. E = z a /2 c.

= 1.96

856 372 (1228 )(1228 ) = 0.0257

1228 n pˆ - E < p < pˆ - E

0.697 - 0.0257 < p < 0.697 + 0.0257 0.671 < p < 0.723 d. We have 95% confidence that the interval from 0.671 to 0.723 actually does contain the true value of the population proportion of medical malpractice lawsuits that are dropped or dismissed.

( )( )

426 434 426 p̂q̂ = ___ ±1.96 860 860 Þ 0.462 < p < 0.529; Because 0.512 is contained within the 860 n 860 confidence interval, there is not strong evidence against 0.512 as the value of the proportion of males in all births.

17. 95% CI: pˆ ± za /2

( )( )

428 152 428 p̂q̂ = ___ ± 2.576 580 580 Þ 0.691 < p < 0.785, or 69.1% < p < 78.5% n 580 580 b. No, the confidence interval includes 75%, so the true percentage could easily equal 75%.

18. a. 99% CI: pˆ ± za /2

(0.000321)(0.999679) p̂q̂ = 0.000321±1.645 Þ 0.000276 < p < 0.000366, or n 420, 095 0.0276% < p < 0.0366%; (Using x = 135: 0.0276% < p < 0.0367%)

19. a. 90% CI: pˆ ± za /2

b. No, because 0.0340% is included in the confidence interval.

( )( )

52 175 52 p̂q̂ = ___ 227 227 ±1.96 Þ 0.174 < p < 0.284, or 17.4% < p < 28.4% n 227 227 b. Because the two confidence intervals overlap, it is possible that the OxyContin treatment group and the placebo group have the same rate of nausea. Nausea does not appear to be an adverse reaction made worse with OxyContin.

21. a. 0.5 b. pˆ = 123 280 = 0.439

20. a. 95% CI: pˆ ± za /2

( )( )

123 157 123 p̂q̂ = ___ ± 2.576 280 280 Þ 0.363 < p < 0.516 n 280 280 d. If the touch therapists really had an ability to select the correct hand by sensing an energy field, their success rate would be significantly greater than 0.5, but the sample success rate of 0.439 and the confidence interval suggest that they do not have the ability to select the correct hand by sensing an energy field.

c. 99% CI: pˆ ± za /2

( )( )

751 4254 751 p̂q̂ = ____ ±1.96 5005 5005 Þ 0.140 < p < 0.160, or 14.0% < p < 16.0%; Because n 5005 5005 48% is not contained within the confidence interval, it appears that the percentage of U.S. adults who do not use the Internet is different from 48%. It appears that the percentage of those who do not use the Internet is now significantly less than 48%.

22. 95% CI: pˆ ± za /2

23. Placebo group: 95% CI: pˆ ± za /2

7 p̂q̂ = ±1.96 n 270

270

270

270

Treatment group:

Þ 0.00697 < p < 0.0449, or 0.697% < p < 4.49%

( 7 )(855 )

8 p̂q̂ = ___ ±1.96 863 863 Þ 0.00288 < p < 0.0157, or 0.288% < p < 1.57% 95% CI: pˆ ± za /2 n 863 863 Because the two confidence intervals overlap, there does not appear to be a significant difference between the rates of allergic reactions. Allergic reactions do not appear to be a concern for Lipitor users.

( )( )

901 111 901 p̂q̂ = ____ ± 2.576 1012 1012 Þ 0.865 < p < 0.916 or 86.5% < p < 91.6% n 1012 1012 Yes, because the proportion that oppose cloning appears to be greater than 0.5.

24. 99% CI: pˆ ± za /2

25. 95% CI: pˆ ± za /2

p̂q̂ = 0.042 ±1.96 n

(0.042)(0.958) Þ 10, 000, 000

0.0419 < p < 0.0421

(0.042)(0.958) Þ p̂q̂ = 0.042 ± 2.576 0.0418 < p < 0.0422 10, 000, 000 n For both confidence levels, the upper and lower confidence interval limits are very close to each other, suggesting that the estimates are very accurate. Also, there is not much difference between the 95% CI and the 99% CI. The extremely large sample size is giving us confidence interval estimates with a very narrow range, so we are getting very precise estimates.

99% CI: pˆ ± za /2

26. XSORT: 95% CI: pˆ ± za /2

879 p̂q̂ = ±1.96 n 945

945

945

945

Þ 0.914 < p < 0.946, or 91.4% < p < 94.6%

( )( )

239 52 p̂q̂ = 239 ±1.96 291 291 Þ 0.777 < p < 0.865, or 77.7% < p < 86.5% n 291 291 The two confidence intervals do not overlap. It appears that the success rate for the XSORT method is higher than the success rate for the YSORT method, but the big story here is that the XSORT method and the YSORT method both appear to be very effective because the success rates are well above the 50% rates expected with no treatments.

27. Sustained care:

YSORT: 95% CI: pˆ ± za /2

p̂q̂ = 0.828 ±1.96 n (Using x = 164: 77.6% < p < 88.1%) 95% CI: pˆ ± za /2

(0.828)(0.172) 198

Þ 0.775 < p < 0.881, or 77.5% < p < 88.1%;

Standard care: p̂q̂ (0.628)(0.372) Þ = 0.628 ±1.96 0.561 < p < 0.695, or 56.1% < p < 69.5% n 199 The two confidence intervals do not overlap. It appears that the success rate is higher with sustained care.

28. Biochemically confirmed results: 95% CI: pˆ ± za /2

p̂q̂ = 0.258 ±1.96 n (Using x = 51: 19.7% < p < 31.8%) 95% CI: pˆ ± za /2

(0.258)(0.742) 198

Þ 0.197 < p < 0.319, or 19.7% < p < 31.9%;

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

28. (continued) Reported results:

(0.409)(0.591) p̂q̂ = 0.409 ±1.96 Þ 0.341 < p < 0.477, or 34.1% < p < 47.7%; n 198 (Using x = 81: 34.1% < p < 47.8%) The two confidence intervals do not overlap. It appears that the success rate is higher with sustained care.

95% CI: pˆ ± za /2

( )( )

147 153 p̂q̂ = 147 ___ ±1.96 300 300 Þ 0.433 < p < 0.547, or 300 n 300 43.3% < p < 54.7%; Yes, the confidence interval includes the proportion of females in the general population.

29. pˆ = 147 / 300 = 0.49; 95% CI: pˆ ± za /2

(1986 )(4082 )

p̂q̂ 1986 = 6068 ±1.96 6068 6068 Þ 0.315 < p < 0.339, or n 6068 31.5% < p < 33.9%; No, the confidence interval does not appear to include the proportion of females in the general population. Data Set 3 includes measurements from 6068 U. S. Army personnel, not subjects randomly selected from the general population.

30. pˆ = 1986 / 6068 = 0.327; 95% CI: pˆ ± za /2

(5350)( 718 )

p̂q̂ 5350 = 6068 ±1.96 6068 6068 Þ 0.874 < p < 0.890, or 6068 n 87.4% < p < 89.0%; No, the confidence interval does not appear to include the proportion of 0.90 for the general population. Data Set 3 includes measurements from 6068 U. S. Army personnel, not subjects randomly selected from the general population.

32. The total number of new cases is 79,473,733 and the total number of deaths is 962,580; pˆ = 962, 580 79, 473, 733 = 0.0121.

31. pˆ = 5350 / 6068 = 0.882; 95% CI: pˆ ± za /2

(

962,580

)(

78,511,153

)

p̂q̂ 79, 473, 733 79, 473, 733 Þ 0.012089 < p < 0.012140, or = 962,580 ±1.96 a /2 79, 473,733 79, 473, 733 n 1.211% < p < 1.214%; The confidence interval suggests that the probability of dying from COVID-19 is close to 0.0121, but the introduction of vaccines well into the pandemic changed the probability of death. Because the original data are from a population that was changing over time due to vaccines, the result is not necessarily meaningful. Also, some of the deaths occurred with individuals who were not categorized as new cases, so the sample proportion is not exactly accurate.

95% CI: pˆ ± z

33. a. n =

[za /2 ]2 p̂q̂ [2.575]2 ×0.25

= 1842 (Tech: 1844) 0.032 2 [za /2 ] p̂q̂ [2.575] (0.22)(0.78) b. n = = 1265 (Tech: 1266) = E2 0.032 [za /2 ]2 p̂ qˆ [1.645]2 ×0.25 34. a. n = = = 1692 (Tech: 1691) 2 E2 0.02 [za /2 ]2 p̂ qˆ [1.645]2 (0.1)(0.9) b. n = = = 609 E2 0.022 c. Yes, there is a substantial decrease in the sample size.

E2 2

=

35. a. n =

[za /2 ]2 p̂ qˆ [1.96]2 ×0.25

= 4269 2 0.015 [za /2 ] p̂ qˆ [1.96]2 (0.037)(0.963) b. n = = 609 = 0.0152 E2 c. Yes, there is a substantial decrease in the sample size.

36. a. n =

E2 2

=

[za /2 ]2 p̂ qˆ [2.575]2 ×0.25

= 1037 2 0.042 [za /2 ] p̂ qˆ [2.575] (0.26)(0.74) b. n = = = 798 2 E2 0.04

37. a. n =

E2 2

=

[za /2 ]2 p̂ qˆ [2.575]2 ×0.25

= 1842 (Tech: 1844) 0.032 2 [za /2 ] p̂ qˆ [2.575] ×(0.10)(0.90) b. n = = = 664 2 E2 0.03 c. The required sample size is reduced substantially.

38. a. n =

E2

=

2

[za /2 ]2 p̂ qˆ [1.645]2 ×0.25

= 1692 (Tech: 1691) 2 0.02 2 [za /2 ] p̂ qˆ [1.645] ×(0.95)(0.05) b. n = = = 332 2 E2 0.02 c. Yes, the added knowledge results in a very substantial decrease in the required sample size.

E2

=

2

[za /2 ]2 p̂ qˆ [2.575]2 ×0.25

= 4145 (Tech: 4147) 0.022 [za /2 ]2 p̂ qˆ [2.575]2 ×(0.45)(0.55) b. n = = = 4103 (Tech: 4106) 2 E2 0.02 c. No, the sample size doesn’t change much.

39. a. n =

40. a. n =

41.

=

E2

[za /2 ]2 p̂ qˆ [2.575]2 (0.5)(0.5)

= 4145 (Tech: 4147) 2 0.02 2 [za /2 ] p̂ qˆ [2.575] (0.82)(0.18) b. n = = = 2447 (Tech: 4147) 2 E2 0.02 c. Randomly selecting adult females would result in an underestimate, because some females will give birth to their first child after the survey was conducted. It will be important to survey females who have completed the time during which they can give birth.

2 2 Np̂q̂ z 2500 (0.82)(0.18)[2.575] [ ] a /2 = 1237 (Tech: 1238) n= = E2 2

=

2

p̂ q̂ [ za /2 ] + (N - 1)E 2

(0.82)(0.18)[2.575]2 + (2500 - 1)0.022

42. Because we have 95% confidence that p is less than 0.0984, we can safely conclude that rate of headaches among OxyContin users is less than 10%. pˆ + za

p̂q̂ = 16 +1.645 227 n

227

227

227

Þ p < 0.0984

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

43. a. The requirement of at least 5 successes and at least 5 failures is not satisfied, so the normal distribution cannot be used. b. 3 40 = 0.075 Section 7-2: Estimating a Population Mean 1. a. 15.123 g/dL < m < 15.459 g/dL 15.123 +15.459

= 15.291 g/dL. 2 15.459 - 15.123 = 0.168 g/dL. The margin of error is E = 2 c. Because the sample size of 100 is greater than 30, we can consider the sample mean to be from a population with a normal distribution. b. The best point estimate of m is x =

2.

3.

4.

5.

6.

7.

8.

9.

a. df = 99 b. ta /2 = 1.984 c. In general, the number of degrees of freedom for a collection of sample data is the number of sample values that can vary after certain restrictions have been imposed on all data values.

We have 95% confidence that the limits of 15.123 g/dL and 15.459 g/dL contain the true value of the mean hemoglobin level of the population of all adult males. a. The sample is a simple random sample, and either or both of these conditions are satisfied: The population is normally distributed or n > 30. b. When we say that the confidence interval methods of this section are robust against departures from normality, we mean that these methods work reasonably well with distributions that are not normal, provided that departures from normality are not too extreme. c. Yes, the sample is a simple random sample and the sample size is 36, which is greater than 30.

The sample size is greater than 30 and the data appear to be from a population that is normally distributed. s = 98.20 ±1.983×0.62 Þ 98.08°F < m < 98.32°F; Because the confidence interval does 95% CI: x ± ta /2 n 106 not contain 98.6°F, it appears that the mean body temperature is not 98.6°F, as is commonly believed. s = 2.1±1.685 × 4.8 Þ 0.8 lb < m < 3.4 lb; Because the The sample size is greater than 30. 90% CI: x ± ta /2 n 40 confidence interval does not include 0 or negative values, it does appear that the weight loss program is effective, with a positive loss of weight. Because the amount of weight lost is relatively small, the weight loss program does not appear to be very practical.

It is assumed that the 16 sample values appear to be from a normally distributed population.

42.3 x ±t s = 98.9 ± 2.602 × Þ 71.4 min < m < 126.4 min; The confidence interval includes the 98% CI: a /2 n 16 mean of 102.8 min that was measured before the treatment, so the mean could be the same after the treatment.

This result suggests that the zopiclone treatment does not have a significant effect.

21.0 Þ - 6.8 mg/dL < m < 7.6 mg/dL; The sample size is greater than 30. 98% CI: x ± t s = 0.4 ± 2.407 × a /2 n 49 Because the confidence interval includes the value of 0, it is very possible that the mean of the changes in LDL cholesterol is equal to 0, suggesting that the garlic treatment did not affect LDL cholesterol levels. It does not appear that garlic has a significant effect in reducing LDL cholesterol. s = 2.6 ± 2.262 ×1.07 The sample appears to have a normal distribution. 95% CI: x ± ta /2 Þ 1.8 < m < 3.4; n 10 The given numbers are just substitutes for the four DNA base names, so the numbers don’t measure or count anything, and they are at the nominal level of measurement. The confidence interval has no practical use.

10. The sample appears to have a normal distribution. 90% CI:

x ±t a /2

s

= 6.35 ±1.833×

2.325

n 10 Þ 5.00 mg < m < 7.70 mg; No, the samples obtained from California might be very different from those obtained in Arkansas. s = 0.719 ± 3.143 0.366 11. The sample appears to have a normal distribution. 98% CI: x ± ta /2 × n 7 Þ 0.284 ppm < m < 1.153 ppm; Using the FDA guideline, the confidence interval suggests that there could be too much mercury in fish because it is possible that the mean is greater than 1 ppm. Also, one of the sample values exceeds the FDA guideline of 1 ppm, so at least some of the fish have too much mercury.

12. The presence of five zeros suggests that the sample is not from a normally distributed population, so the normality requirement is violated and the confidence interval might not be a good estimate of the population mean.

20.33 x ±t s Þ 19.5 mg < m < 45.6 mg; People consume some brands much 99% CI: = 32.6 ± 2.861× a /2 n 20 more often than others, but the 20 brands are all weighted equally in the calculations. This, along with the violation of the normality requirement, means the confidence interval might not be a good estimate of the population mean. s 342.7 1611.2 lb/acre < m < 2071.7 lb/acre 13. 95% CI: x ± t × Þ = 1841.5 ± 2.228 × a /2 n 11 s 14. 95% CI: x ± t × 332.9 1651.6 lb/acre < m < 2098.8 lb/acre; Because this confidence Þ = 1875.2 ± 2.228 × a /2 n 11 interval and the one obtained in the preceding exercise overlap, there does not appear to be a substantial difference between them.

15. The sample data meset the loose requirement of having a normal distribution.

0.423 0.70 W/kg < m < 1.169 W/kg; Because the confidence interval 90% CI: x ± t × Þ = 0.938 ±1.812 × a /2 n 11 is entirely below the standard of 1.6 W/kg, it appears that the mean amount of cell phone radiation is less than the FCC standard, but individual cell phones that might exceed the standard.

16. The sample data meset the loose requirement of having a normal distribution.

6.461 6.43 mg/g < m < 15.67 mg/g; We cannot conclude that the 95% CI: x ± t × Þ = 11.05 ± 2.262 × a /2 n 10 population mean is less than 7 because the confidence interval shows that the mean might be higher than that level. é za /2 ×s ù2 é1.96 ×15ù2 ú =ê 17. The sample size is n = ê û = 97. It does appear to be very practical.

3 ë E û2 ë 2 é za /2 ×s ù é2.33×15ù ú =ê 18. The sample size is n = ê ú = 305. It does appear to be very practical. û ë E 2û ë 2 2 é za /2s ù é1.645 ×17.65 ù 1.5 ú =ê 19. The sample size is n = ê úû = 375. Yes, the assumption seems reasonable.

2 2 E û ë éz ë s ù é1.96×17.7 ù a /2 ú = ê 0.5 ú = 4815 (Tech: 4814). Yes, the assumption seems 20. The required sample size is n = ê û ë ë E û reasonable.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

2

éza /2 ×s ù2 é2.575 ×16.0 ù 21. a. s = ú =ê = 16.0; The sample size is n = ê ú = 425.

2 ë û 4 E û ë 2 é za /2 ×s ù2 é2.575 ×11.3ù = ú ê ú = 212. b. The sample size is n = ê û 2 ë ë E û c. The result from part (a) is substantially larger than the result from part (b). The result from part (b) is likely to be better because it uses s instead of the estimated s obtained from the range rule of thumb.

2 éza /2 ×s ù2 é2.575 ×17.0 ù 104 - 36 22. a. s = ú =ê = 17.0; The sample size is n = ê ú = 480.

2 ë û 4 E û ë 2 é za /2 ×s ù2 é2.575 ×12.5 ù = ú ê ú = 260. b. The sample size is n = ê û 2 ë ë E û c. The result from part (a) is substantially larger than the result from part (b). The result from part (b) is likely to be better2 because it uses s instead of the estimated s obtained from the range rule of thumb. é za /2 ×s ù é1.96 ×1 2 ù = 38, 416 (Tech: 38,415); The sample appears to be too large to be practical for a 23. n = ê E ú =ê ú ë ë 0.01 û û telephone survey.

2 éza /2 ×s ù2 é2.33×0.775ù 99.6 - 96.5 24. a. s = ú =ê = 0.775; The sample size is n = ê 0.1 ú = 327 (Tech: 326).

4 E û ë ë2 û éza /2 ×s ù2 é2.33×0.62ù = 209. ú = ê 0.1 b. The sample size is n = ê ûú ë ë E û c. The result from part (a) is substantially larger than the result from part (b). The result from part (b) is likely to be better because it uses s instead of the estimated s obtained from the range rule of thumb.

25. Both samples appear to have a normal distribution.

0.916 x ±t s Þ 67.8 bpm < m < 71.4 bpm Males: 95% CI: = 69.58 ±1.976 × a /2 n 153 s = 74.0 ±1.976 ×1.03 Þ 72.0 bpm < m < 76.1 bpm Females: 95% CI: x ± ta /2 n 147 Although final conclusions about means of populations should not be based on the overlapping of confidence intervals, the intervals do not overlap, so adult females appear to have a mean pulse rate that is higher than the mean pulse rate of adult males.

26. SMOKER: 95% CI: x ± t ×s 140.5448 257.208 ng/mL < m < 275.577 ng/mL Þ = 266.3922 ±1.963× sa /2 n 902 ETS: 95% CI: x ± t × 106.1758 18.5419 ng/mL < m < 38.5995 ng/mL Þ = 28.5707 ±1.965 × a /2 n 2483.39488 s Þ - 0.1692 ng/mL < m < 5.8486 ng/mL NO ETS: 95% CI: x ± t × = 2.8396 ±1.967 × a /2 358 n None of the confidence intervals overlap, so the three groups appear to have different means. Increased exposure to smoke appears to result in higher levels of cotinine.

104 - 40

27. Males: 95% CI: x ± t

×

s

68.551 Þ 1754.1 mm < m < 1758.3 mm = 1756.2 ±1.96 × a /2 4082 n s = 1628.5 ±1.96 × 64.2 Þ Females: 95% CI: x ± t 1625.6 mm < m < 1631.3 mm × a /2 1986 n Because the two confidence intervals do not overlap, it appears that there is a significant difference between the heights of males and females.

14,1128.7 28. New Cases: 95% CI: 99, 342.2 ±1.96 × Þ 89, 547.8 new cases < m < 109,136.5 new cases 800 906.8 Deaths: 95% CI: 1203.2 ±1.96 × Þ 1140.3 deaths < m < 1266.2 deaths 800 Because of the increased use of vaccinations, both samples are from populations that are not stable. The confidence interval estimates are not very reliable. A comparison of the two confidence intervals makes no sense because they are estimates of different quantities; nothing can be concluded by comparing the two confidence intervals.

65.4 x ±t s 29. a. Large population: 95% CI: = 255.1± 2.023× a /2 n 40 Þ 234.2 (1000 cells mL) < m < 276.0 (1000 cells mL) 65.4 500 - 40 b. Finite population: 95% CI: x ± t s N- n = 255.1± 2.26× a /2 n N- 1 40 500 - 1 Þ 235.0 (1000 cells mL) < m < 275.2 (1000 cells mL) c. The second confidence interval is narrower, indicating that we have a more accurate estimate when the relatively large sample is selected without replacement from a relatively small finite population.

Section 7-3: Estimating a Population Standard Deviation or Variance 1.

a. Unlike confidence interval estimates of p or m, confidence interval estimates of s or s 2 are not created using a distribution that is symmetric, so there is no ±E as in confidence interval estimates of p or m. b. The format of 10.9 bpm ± 3.3 bpm implies that s = 10.9 bpm, which is not the case. Because the chi-square distribution is not symmetric, confidence interval estimates of s or s 2 cannot be expressed in the format of s ± E.

2.

a. The sample must be a simple random sample and the sample must be from a normally distributed population. b. For confidence interval estimates of s or s 2, the requirement of a normal distribution is much stricter, and there is no sample size (such as n > 30) suitable for waiving the normality requirement. c. No. The selected numbers are equally likely and have a uniform distribution, not a normal distribution as required.

3.

(0.360 (million cells/mL))2 < (s )2 < (0.454 (million cells/mL))2 0.130 (million cells/mL)2 < s 2 < 0.206 (million cells/mL)2

4.

df = 20 - 1 = 19; c 2L = 8.907 and c 2R = 32.852

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

5.

df = 25 - 1 = 24; c 2L = 12.401, c 2R = 39.364 (n - 1)s2 <s < c R2 (25 - 1)0.242

6.

(n - 1)s2 c L2

(n - 1)s2 <s < c R2 (20 - 1)65.42

(25 - 1)0.242 12.401 39.364 95% CI: 0.19 mg < s < 0.33 mg

7.

8.

<s <

df = 153 - 1 = 152; c L2 = 110.846,

c R2 = 193.761

cR2 = 200.657

(Table: c L2 = 67.328, c R2 = 140.169)

(Table: c L2 = 67.328, c 2R = 140.169)

(n - 1)s2 <s < c R2 (147 - 1)1.962

(n - 1)s2 c L2

(147 - 1)1.962 105.741 193.761 99% CI: 1.70 < s < 2.30 in (1000 cells/mL) Table: 2.00 < s < 2.89 in (1000 cells/mL) 9.

(n - 1)s2 c L2

(20 - 1)65.42 8.907 32.852 95% CI: 49.7 < s < 95.5 in (1000 cells mL)

<s <

df = 147 - 1 = 146; c L2 = 105.741,

df = 20 - 1 = 19; c 2L = 8.907, c 2R = 32.852

<s <

df = 106 - 1 = 105; c 2L = 74.222, c 2R = 129.561

(n - 1)s2 c R2

(106 - 1)0.622 129.561

<s <

<s <

c R2

(16 - 1)42.32

<s <

10. df = 40 - 1 = 39; c 2L = 26.509, c 2R = 55.758 c R2

74.222

(40 - 1)4.82

<s <

<s <

c L2

(40 - 1)4.82

Tech: 4.1 lb < s < 5.8 lb The confidence interval gives us information about the variation among the amounts of lost weight, but it does not give us information about the effectiveness of the diet.

12. df = 49 - 1 = 48; c 2L = 29.707, c 2R = 76.154

(n - 1)s2

c L2

c R2

5.229 30.578 98% CI: 29.6 min < s < 71.6 min No, the confidence interval does not indicate whether the treatment is effective.

(n - 1)s2

26.509 55.758 90% CI: 4.0 lb < s < 5.9 lb

(n - 1)s2 (16 - 1)42.32

<s <

Table: 7.39 cm < s < 10.67 cm

c L2

(106 - 1)0.622

(n - 1)s2 c L2

(153 - 1)7.102 110.846 200.657 99% CI: 6.18 cm < s < 8.31 cm

(n - 1)s2

11. df = 16 - 1 = 15; c L2 = 5.229, c 2R = 30.578 <s <

(153 - 1)7.102

(n - 1)s2

95% CI: 0.56 F < s < 0.74 F Tech: 0.55 F < s < 0.72 F

(n - 1)s2

(n - 1)s2 <s < c R2

(49 - 1)21.02

<s <

<s <

(n - 1)s2 c L2

(49 - 1)21.02

29.707 76.154 98% CI: 16.7 mg/dL < s < 26.7 mg/dL Tech: 16.9 mg/dL < s < 27.4 mg/dL No, the confidence interval does not indicate whether the treatment is effective.

13. Because the confidence interval includes the value of 2.9 in., it is very possible that the standard deviation of heights of female who are professional soccer players is the same as the standard deviation of heights of female in the general population. There is not sufficient evidence to support the claim that female who are professional soccer players have heights that vary less than female in the general population. df = 23 - 1 = 22; c L2 = 10.982, c R2 = 36.781 (n - 1)s2 <s < c R2 (23 - 1)2.41132

(n - 1)s2 c L2

(23 - 1)2.41132 10.982 36.781 95% CI: 1.9 in. < s < 3.4 in. <s <

14. The sample data include the value of 1.6 g repeated six times, so the data do not appear to be from a normally distributed population. Consequently, the confidence interval might not be a good estimate of s . df = 15 - 1 = 14, c L2 = 5.629, and c R2 = 26.119

(n - 1)s2 c R2

(15 - 1)0.258752

<s <

<s <

(n - 1)s2 c L2

(15 - 1)0.258752

5.629 26.119 95% CI: 0.019 g < s < 0.041 g 15. Because the two confidence intervals overlap and are reasonably close, there does not appear to be a difference in variation. df = 10 - 1 = 9; c L2 = 2.700, c 2R= 19.023 F:

(n - 1)s2 <s < c R2 (10 - 1)0.298332

(n - 1)s2 c L2

M:

(n - 1)s2 <s < c R2 (10 - 1)0.295152

(10 - 1)0.298332 2.700 19.023 95% CI: 0.21°F < s < 0.54°F

(n - 1)s2 c L2

(10 - 1)0.295152 2.700 19.023 95% CI: 0.20°F < s < 0.54°F

<s <

<s <

16. df = 9, c 2L = 2.700, c 2R = 19.023 a.

(n - 1)s2 <s < c R2 (10 - 1)0.476682

(n - 1)s2 c L2

b.

(n - 1)s2 <s < c R2

(n - 1)s2 c L2

(10 - 1)1.821632 2 (10 - 1)0.476682 < s < (10 - 1)1.82163 2.700 2.700 19.023 19.023 95% CI: 0.33 min < s < 0.87 min 95% CI: 1.25 min < s < 3.33 min c. The variation appears to be significantly lower with a single line. The single line appears to be better because patients are more satisfied if their waiting times are about the same.

17. The sample size is 85, which is very practical.

18. The sample size of 19,205 is too large to be practical. The United States has only 15,000 air traffic controllers.

19. The sample size is 192, which is practical.

20. The sample size is 14, which is very practical, but an estimate of s that is within 50% of the true value is not very accurate and may have limited value. <s <

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

21. a. Females: The sample appears to be from a population with a distribution that is far from normal, so the confidence interval estimate might not be very good. df = 204, c L2 = 166.338, and c 2R = 245.448

(n - 1)s2 c R2

(205 - 1)9.72

<s <

9 <s <

(n - 1)s2 c L2

(205 - 1)9.72

245.448 166.338 95% CI: 8.8 days < s < 10.7 days b. Males: The sample appears to be from a population with a distribution that is far from normal, so the confidence interval estimate might not be very good. df = 79, c L2 = 0.000, and c 2R = 0.000

(n - 1)s2 c R2

(195 - 1)8.22

<s <

<s <

(n - 1)s2 c L2

(195 - 1)8.22

157.321 234.465 95% CI: 7.5 days < s < 9.2 days c. The amounts of variation appear to be about the same. df = 204, c 2L = 166.338, c 2R = 245.448 (Using technology.)

22. a.

(n - 1)s2 c R2

(205 - 1)706.32

<s <

<s <

(n - 1)s2 c L2

(205 - 1)706.32

166.338 245.448 95% CI: 643.9 g < s < 782.1 g Table: 886.2 g < s < 1170.9 g b. df = 194, c 2L= 157.321, and c 2R= 234.465 (Using technology.)

(n - 1)s2 c R2

(195 - 1)660.22

<s <

<s <

(n - 1)s2 c L2

(195 - 1)660.22

157.321 234.465 95% CI: 600.5 g < s < 733.1 g Table: 807.8 g < s < 1076.3 g c. The amounts of variation are about the same.

2 2 1 1 23. c 2 = é- 1.96 + 2(106 - 1)- 1ù = 78.085 and c 2 = é+1.96 + 2(106 - 1)- 1ù = 134.756; The values L

R û û 2ë 2ë from the given approximation are quite close to the actual critical values.

1 æza /2 ö2 1 æ2.326347877 ö2 24. = 1083 (Using the given formula with Table A-2 yields 1086. Statdisk n= ç ø÷ ÷ = 2 çè 0.05 ø è 2 d yields 1087.)

Section 7-4: Bootstrapping: Using Technology for Estimates 1. (1) The sample must be collected in an appropriate way, such as a simple random sample.

(2) The sample means found from the generated bootstrap samples should have a distribution that is approximately symmetric.

2. For the given sample, a bootstrap sample is another sample of 15 of the listed times randomly selected with replacement.

3. There is no universal exact number, but there should be at least 1000 bootstrap samples, and the use of 10,000 or more is common.

4. The 1000 means will target the sample mean of 54.3 seconds. The bootstrap method will not improve upon a poor estimate from the sample.

9. Answers vary, but here are typical answers.

5. 0.000 < p < 0.500 a. - 0.8 kg < m < 7.8 kg 6. 0.125 < p < 0.750 7.

a.

0.1 kg < m < 8.6 kg

b. 1.9 kg < s < 6.3 kg 8.

a. 66.3 cW/kg < m < 98.3 cW/kg b.

16.7 cW/kg < s < 54.6 cW/kg

b. 1.2 kg < s < 7.0 kg 10. Answers vary, but here are typical answers. a. 50.0 cW/kg < m < 118.3 cW/kg b. 9.8 cW/kg < s < 57.3 cW/kg

11. a. Answers vary, but here is a typical answer: 1647.2 lb/acre < m < 2035.8 lb/acre. b. The bootstrap confidence interval does not differ by a substantial amount from the confidence interval of 1611.2 lb/acre < m < 2071.1 lb/acre found in Exercise 13 in Section 7-2.

12. a. Answers vary, but here is a typical answer: 1685.0 lb/acre < m < 2063.7 lb/acre. b. The bootstrap confidence interval does not differ by a substantial amount from the confidence interval of 1651.6 lb/acre < m < 2098.8 lb/acre found in Exercise 14 in Section 7-2.

13. a. Answers vary, but here is a typical answer: 0.709 W/kg < m < 1.176 W/kg. b. The bootstrap confidence interval does not differ by a substantial amount from the confidence interval of 0.707 W/kg < m < 1.169 W/kg found from Exercise 15 in Section 7-2.

14. a. Answers vary, but here is a typical answer: 7.43 mg/g < m < 14.83 mg/g. b. The bootstrap confidence interval does not differ by a substantial amount from the confidence interval of 6.43 mg/g < m < 15.67 mg/g found in Exercise 16 in Section 7-2.

15. a. Answers vary, but here is a typical answer: 1.5 in. < s < 3.1 in. b. The confidence interval obtained from the bootstrap method is close to this confidence interval of 1.9 in. < s < 3.4 in. found in Exercise 13 in Section 7-3.

16. a. Answers vary, but here is a typical answer: 0.10 g < s < 0.33 g. b. The bootstrap confidence interval does not differ by a substantial amount from the confidence interval of 0.19 g < s < 0.41 g found in Exercise 14 in Section 7-3. Because the sample data do not appear to be from a normally distributed population, the bootstrap confidence interval is likely to be a better result.

17. a. Answers vary, but here is a typical answer: 1080.0 cm3 < m < 1169.0 cm3. b. Answers vary, but here is a typical answer: 79.0 cm3 < s < 152.8 cm3.

18. a. Answers vary, but here is a typical answer: 19.8 mg < m < 42.9 mg. b. The confidence interval given in part (a) is likely to be better. Given that the sample data do not appear to be from a normally distributed population, the use of the t distribution is questionable, so the confidence interval found in Exercise 12 in Section 7-2 might not be a good estimate of m. c. The confidence interval obtained from the bootstrap method is very close to this confidence interval of 19.5 mg < m < 45.6 mg found in Exercise 12 in Section 7-2.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

19. Answers vary, but here is a typical result: 0.135 < p < 0.152. The result is essentially the same as the confidence interval of 0.135 < p < 0.152 found in Exercise 15 from Section 7-1.

20. Answers vary, but here is a typical result: 0.671 < p < 0.722. The result is very close to the confidence interval of 0.671 < p < 0.723 found in Exercise 16 from Section 7-1.

21. Answers vary, but here is a typical result: 0.0208 < p < 0.0317. This is quite close to the confidence interval of 0.0205 < p < 0.0311 found in Exercise 14 from Section 7-1.

22. Answers vary, but here is a typical result: 0.860 < p < 0.932.

23. a. Answers vary, but here is a typical result: 2.5 < s < 3.3. b. 2.4 < s < 3.7 c. The confidence interval from the bootstrap method is not very different from the confidence interval found using the methods of Section 7-3. Because a histogram or normal quantile plot shows that the sample appears to be from a population not having a normal distribution, the bootstrap confidence interval of 2.5 < s < 3.3 would be a better estimate of s .

24. a. Answers vary, but here is a typical result : 1.6 < m < 3.3. b. 1.6 < m < 3.3 c. The confidence interval found from the bootstrap method is essentially the same as the confidence interval found using the methods of Section 7-2.

25. Answers vary, but here is a typical result using 10,000 bootstrap samples: 1.7 < m < 3.3. This result is very close to the confidence interval of 1.6 < m < 3.3 found using 1000 bootstrap samples. In this case, increasing the number of bootstrap samples from 1000 to 10,000 does not have much of an effect on the confidence interval.

26. a. No. A histogram or normal quantile plot would show that the distribution of the sample data is far from normal. b. Yes. The sample means appear to have a distribution that is approximately normal. c. Yes. The sample standard deviations appear to have a distribution that is approximately normal.

27. The histogram of the 1000 bootstrap sample means is approximately symmetric as required.

28. The 1000 medians have a mean around 18 sec, and the 95% confidence interval of the 1000 medians is likely to be something like 0 sec < median < 37 sec. or 0 sec < median < 74 sec. The 95% confidence interval is good in the sense that it does capture the median of the population, but it is quite wide and does not give us a very accurate estimate of the population median. With this median, the bootstrap method does not work well because the 1000 medians are targeting a value near 18 seconds instead of targeting the median of 5.5 seconds or the sample median of 24.0 seconds.

Quick Quiz 1.

2.

0.110 + 0.150 = 0.130 2 We have 95% confidence that the limits of 0.110 and 0.150 contain the true value of the proportion of adults in the population who correct their vision by wearing contact lenses. pˆ =

3.

z0.01/2 = 2.576 (Table: 2.575)

4.

3%- 1.0% < p < 3% +1.0% Þ 2.0% < p < 4.0%, or 0.02 < p < 0.04

5.

n=

6.

7.

[za /2 ]2 p̂ qˆ [2.575]2 ×0.25

= = 1824 (Tech: 1844) E2 0.032 é za /2s ù2 é2.575 ×15ù2 ú =ê n=ê ú = 166; Yes, the sample size is small enough to be practical. û 3 ë E û ë There is a loose requirement that the sample values are from a normally distributed population.

8.

The degrees of freedom is the number of sample values that can vary after restrictions have been imposed on all of the values. For the sample data described in Exercise 7, df = 15 - 1 = 14.

9.

t0.05/2 = 2.145

10. No, the use of the c 2 distribution has a fairly strict requirement that the data must be from a normal distribution. The bootstrap method could be used to find a 95% confidence interval estimate of s .

Review Exercises 1.

2.

3.

4.

5.

6.

7.

8.

(

723

)(

222,277

)

p̂q̂ = 723 ±1.96 223,000 223,000 Þ 0.00301 < p < 0.00348; We have 95% confidence a /2 223,000 223, 000 n that the limits of 0.00301 and 0.00348 contain the value of the population proportion.

95% CI: pˆ ± z

n=

[za /2 ]2 pˆ qˆ [1.645]2 (0.25)

= 423 = 2 E2 0.04 a. x = 22.083 mm 0.077850 x ±t s Þ 22.034 mm < m < 22.133 mm b. 95% CI: = 22.083 ± 2.201× a /2 n 12 c. We have 95% confidence that the limits of 22.034 mm and 22.133 mm contain the value of the population mean m.

[za /2 ]2 pˆqˆ [1.96]2 (0.25)

= 1068 = 0.032 E2 é za /2s ù2 é1.96 ×39ù2 ú =ê b. n = ê ú = 234 ë 5 û ë E û c. You must take the larger sample of 1068. a. student t distribution b. None of the three distributions is appropriate, but a confidence interval could be constructed by using bootstrap methods. a.

n=

c.

c 2 (chi-square distribution)

d. normal distribution e. None of the three distributions is appropriate, but a confidence interval could be constructed by using bootstrap methods. p̂q̂ = 0.66 ± 2.575 (0.66)(0.34) Þ 99% CI: pˆ ± za /2 0.622 < p < 0.698, or 62.2% < p < 69.8%; Yes, the n 1028 confidence interval includes a range of values well above 0.5, so it safe to conclude that most U.S. adults say that it is morally acceptable to conduct medical research involving stem cells obtained from human embryos. a.

[za /2 ]2 p̂ qˆ [1.96]2 ×0.25

= 9604 2 0.01 2 [za /2 ] p̂ qˆ [1.96] (0.34)(0.66) b. n = = = 8621 2 E2 0.01 a. Yes, the requirement s are satisfied (n > 30). s 74.737 8.8 seconds < m < 59.3 seconds 95% CI: x ± t × Þ = 34.1± 2.030 × a /2 n 36 b. Construction of a confidence interval estimate of s has a fairly strict requirement that the sample should be from a population with a normal distribution, regardless of the sample size. That requirement is not satisfied.

19 of the 36 times are 0 seconds, so the sample does not appear to be from a population having a normal distribution. n=

E2 2

=

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

9.

a. The requirements are satisfied: 95% CI: x ± t

× a /2

s n

2

4.227 = 30.05 ± 2.262 ×

b. The requirements are satisfied. df = 10 - 1 = 9; c L= 2.700 c (n - 1)s2 <s < c R2 (10 - 1)4.2272

2

Þ 27.03 cm < m < 33.07 cm.

10

R= 19.023

(n - 1)s2 c L2

(10 - 1)4.2272 2.700 19.023 95% CI: 2.91 cm < s < 7.72 cm <s <

10. Answers vary, but here is a typical result: 27.55 cm < m < 32.49 cm. The bootstrap confidence interval does not differ by a substantial amount from the confidence interval of 27.03 cm < m < 33.07 cm found in part (a) of Exercise 9.

Cumulative Review Exercises 1.5 + 0.9 +1.3 +1.5 +1.3 +1.2 +1.4 +1.1+1.3 +1.4 +1.2 1. x = = 1.28 W/kg; min, Q 2= 1.30 W/kg, 11 2 2 2 (1.5 - 1.28) + (0.9 - 1.28) + + (1.3 - 1.28) + (1.4 + (1.2 - 1.28)2 s= = 0.18 W/kg, 11- 1 s2 = (0.18 W/kg)2 = 0.03 (W/kg)2 , range = 1.5 - 0.9 = 0.6 W/kg; These results are statistics.

2.

3.

4.

5.

Using the range rule of thumb, the limit separating significantly low values is m - 2s = 1.28 - 2(0.18) = 0.92 W/kg and the limit separating significantly high values is m + 2s = 1.28 + 2(0.18) = 1.64 W/kg. Because the lowest value of 0.9 W/kg is less than 0.92 W/kg, it is significantly low. ratio level of measurement; continuous data.

The graphs suggest that the sample appears to be from a population having a distribution that is approximately normal. s 0.18 1.16 W/kg < m < 1.40 W/kg; We have 95% confidence that the 95% CI: x ± t × Þ = 1.28 ± 2.228 × a /2 n 11 limits of 1.16 W/kg and 1.40 W/kg contain the actual value of the population mean m.

7.

é za /2 ×s ù2 é1.96 ×0.29 ù2 = 130 cell phones ú = ê 0.05 n=ê ûú E û ë ë 1.6 - 1.17 = 1.48; which has a probability of 1- 0.9306 = 0.0694 (Tech: 0.0691) to the right. a. z1.6 = 0.29 b. The z score for the lower 75% is 0.67, which correspond to a radiation amount of 1.17 + 0.67(0.29) = 1.36 W/kg (Tech: 1.37 W/kg).

8.

99% CI: pˆ ± za /2

9.

95% CI: pˆ ± za /2

6.

p̂q̂ 986 = 1043 ±1.96 n population is well informed.

1043

1043

1043

Þ 0.956 < p < 0.980, or 95.6% < p < 98.0%; Yes, the

p̂q̂ (0.66)(0.34) Þ = 0.66 ±1.96 0.639 < p < 0.681, or 63.9% < p < 68.1%; CVS n 2020 Pharmacy sells flu shots and could potentially benefit from the widespread belief that there will be high demand for flu shots, so there is a potential for bias (although the Harris Poll would likely conduct the survey without any bias).

9.44 = 128.5 ± 2.145 × Þ 123.2 mmHg < m < 133.7 mmHg n 15 b. Yes. Because the systolic blood pressure measurements were taken on consecutive days, those values could be investigated with a time-series plot. The time-series plot does not appear to reveal a pattern that is changing over time, so the blood pressure measurements appear to be stable. x ±t

10. a. 95% CI:

a /2

s

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

Chapter 8: Hypothesis Testing Section 8-1: Basics of Hypothesis Testing 1. a. H0: m = 325 mg of aspirin b. H1: m ¹ 325 mg of aspirin

2.

3.

4.

5.

6.

c. Reject the null hypothesis or fail to reject the null hypothesis. d. No, in this case, the original claim becomes the null hypothesis. For the claim that the mean is equal to 325 mg of aspirin, we can either reject that claim or fail to reject it, but we cannot state that there is sufficient evidence to support that claim.

Estimates and hypothesis tests are both methods of inferential statistics, but they have different objectives. We could use the sample data to construct a confidence interval estimate of the population mean, but hypothesis testing is used to test some claim made about the value of the mean amount of potassium chloride in the bottles.

The P-value of 0.001 is preferred because it corresponds to the sample evidence that most strongly supports the alternative hypothesis that the method is effective.

The most serious error is the first error (i) because it would result in distributing pills with lisinopril amounts that are very different from the intended amounts. Distributing pills with the wrong amount of a drug is more serious than distributing vitamin pills with the wrong amount of the vitamin. The error of failing to reject the false null hypothesis is a type II error. a. p < 0.80 13. a. Left-tailed b. P-value = P(z < - 0.75) = 0.2266 b. H0: p = 0.80; H1: p < 0.80 c.

a. p > 0.50

14. a. Right-tailed

b. H0: p = 0.50; H1: p > 0.50

b. P-value = P(z > 2.00) = 0.0228

7.

a. m < 123 mmHg

8.

b. H0: m = 123 mmHg; H1: m < 123 mmHg a. s > 5 mmHg

b. H0: s = 5 mmHg; H1: s > 5 mmHg 788 - 0.80 pˆ - p 1018 9. z = =z= = - 2.07 (0.80)(0.20) pq n 1018 2678 - 0.5 pˆ - p = z = 4464 = 13.35 10. z = (0.5)(0.5) pq n 4464 11. t = x - m = z = 122.96 - 123 = - 0.044

s n

0.2266 > 0.05; Fail to reject H 0.

15.85 300

c.

0.0228 < 0.05; Reject H 0.

15. a. Two-tailed b. P-value = 2×P(z > 1.80) = 0.0719 (Table: 0.0718) c. > 0.05; Fail to reject H .

0.0719 0

16. a. Two-tailed b. P-value = 2×P(z < - 1.60) = 0.1096 c. 0.1096 > 0.05; Fail to reject H .

0

17. a. z =- 1.645 b. - 0.75 > - 1.645; Fail to reject H 0.

18. a. z = 1.645 b. 2.00 > 1.645; Reject H 0.

2

(n - 1)s2 = (300 - 1)15.85 = 3004.621 12. c 2 = 52 s 2

19. a. z = ±1.96 b. - 1.96 < 1.80 < 1.96; Fail to reject H 0.

20.

a. z = ±1.96 b. - 1.96 < - 1.60 < 1.96; Fail to reject H 0.

21. a. 0.1830 > 0.05; Fail to reject H 0. b. There is not sufficient evidence to support the claim that more than 15% of adults think that the U.S. health care system is in a state of crisis.

22. a. 0.0000 < 0.05; Reject H 0. b. There is sufficient evidence to support the claim that most adults worry a fair amount or a great deal about the extinction of plant and animal species.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

23. a. 0.0095 < 0.05; Reject H 0. b. There is sufficient evidence to warrant rejection of the claim that the mean pulse rate (in beats per minute) of adult males is 72 bpm.

24. a. 0.3045 > 0.05; Fail to reject H 0. b. There is not sufficient evidence to support the claim that the standard deviation of pulse rates of adult males is more than 11 bpm.

25. Type I error: In reality p = 0.1, but we reject the claim that p = 0.1. Type II error: In reality p ¹ 0.1, but we fail to reject the claim that p = 0.1.

26. Type I error: In reality p = 0.35, but we reject the claim that p = 0.35. Type II error: In reality p ¹ 0.35, but we fail to reject the claim that p = 0.35.

27. Type I error: In reality p = 0.25, but we support the claim that p > 0.25. Type II error: In reality p > 0.25, but we fail to support that claim.

28. Type I error: In reality p = 1 / 2, but we support the claim that p < 1 / 2. Type II error: In reality p < 1 / 2, but we fail to support that conclusion.

29. The power of 0.96 shows that there is a 96% chance of rejecting the null hypothesis of p = 0.08 when the true proportion is actually 0.18. That is, if the proportion of Chantix users who experience abdominal pain is actually 0.18, then there is a 96% chance of supporting the claim that the proportion of Chantix users who experience abdominal pain is greater than 0.08.

(0.5)(0.5)

= 0.6028125 64 0.6028125 - 0.65 = - 0.791; Power = P(z > - 0.791) = 0.7852 (Tech: 0.7857) From p = 0.65, z = (0.65)(0.35)

30. a. From p = 0.5, pˆ = 0.5 +1.645

64

b. Assuming that p = 0.5, as in the null hypothesis, the critical value of z = 1.645 corresponds to pˆ = 0.6028125, so any sample proportion greater than 0.6028125 causes us to reject the null hypothesis, as shown in the shaded critical region of the top graph. If p is actually 0.65, then the null hypothesis of p = 0.5 is false, and the actual probability of rejecting the null hypothesis is found by finding the area greater than pˆ = 0.6028125 in the bottom graph, which is the shaded area. That is, the shaded area in the bottom graph represents the probability of rejecting the false null hypothesis.

31. From p = 0.5, pˆ = 0.5 +1.645

n

;

p = 0.55, since P(z > - 0.842) = 0.8000,

(0.55)(0.45)

pˆ = 0.55 - 0.842 0.5 +1.645

(0.5)(0.5) from

n

(0.5)(0.5) n

= 0.55 - 0.842

(0.55)(0.45) n

0.5 n +1.645 0.25 = 0.55 n - 0.842 0.2475 0.05 n = 1.645 0.25 + 0.842 0.2475 2 æ1.645 0.25 + 0.842 0.2475 ö n=ç 0.05 ÷ = 617 (or 618 or 619 depending on precision used.) è ø Section 8-2: Testing a Claim About a Proportion 1. a. 0.262×580 = 152 b. pˆ = 0.262; The symbol pˆ is used to represent a sample proportion. c.

p = 0.25

2.

a.

H0: p = 0.25; H1: p ¹ 0.25

b. z =

pˆ - p pq n

=

0.262 - 0.25 (0.25)(0.75) 580

= 0.67

3.

(1) Mendel’s experiment resulted in offspring peas that can be treated as a simple random sample. (2) There is a fixed number of trials (580), the trials are independent because any offspring pea is independent of the others, there are two categories of success (yellow or not yellow), and the probability remains the same for all offspring peas. (3) With n = 580 and p = 0.25, the conditions np = 0.25(580) = 145 ³ 5 and nq = 0.75(580) = 435 ³ 5 are both satisfied. The requirements are all satisfied.

4.

a. The method based on a confidence interval is not equivalent to the P-value method and the critical value method. b. If the P-value is very low (such as less than or equal to 0.05), “the null must go” means that we should reject the null hypothesis. c. The statement that “if the P is high, the null will fly” suggests that with a high P-value, the null hypothesis has been proved or is supported, but we should never make such a conclusion of proving or supporting the null hypothesis. d. Choosing a significance level with a number like 0.0483 would make it seem like you’re carefully choosing the significance level to reach a desired conclusion. a. left-tailed test c. P-value = 0.000004 b. z =- 4.46 d. H0: p = 0.10; Reject the null hypothesis.

5.

6.

7.

e. There is sufficient evidence to support the claim that less than 10% of treated subjects experience headaches. a. right-tailed test c. P-value < 0.0001 b. z = 8.04 d. H0: p = 0.5; Reject the null hypothesis. e. There is sufficient evidence to support the claim that most U.S. adults know that inserting a gene into a plant is an example of genetic engineering. a. right-tailed test c. P-value = 0.122 = 0.3; Fail to reject the null b. z = 1.16 d. hypothesis.

0

:

8.

p e. There is not sufficient evidence to support the claim that more than 30% of adults drink alcohol at least once a week. a. two-tailed test c. P-value = 0.1840 p = 0.5; Fail to reject the null b. z = 1.33 hypothesis. d.

0

9.

: e. There is not sufficient evidence to warrant rejection of the claim that half of us say that we should replace passwords with biometric security.

91 - 0.40 pˆ - p 220 H0: p = 0.40; H1: p > 0.40; Test statistic: z = = = 0.41 pq n

(0.40)(0.60) 220

P-value = P(z > 0.41) = 0.3409 (Tech: 0.3399); Critical value: z = 1.645; Fail to reject H 0. There is not sufficient evidence to support the claim that the sample of cast and crew members is from a population in which the rate of cancer is greater than 40%. There is not sufficient evidence to support a conclusion that the movie was cursed because of a significantly high number of cancer cases.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

19 - 0.50

pˆ - p 10. H0: p = 0.5; H1: p ¹ 0.5; z =

pq n

=

48

= - 1.44

(0.50)(0.50) 48

P-value = 2P(z < - 1.44) = 0.1498 (Tech: 0.1489); Critical values: z = ±1.96; Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim that dropped toast will land with the buttered side down 50% of the time. The real intent of the experiment was to test the common theory that when buttered toast is dropped, it will land with the buttered side down most of the time. That theory is not supported by the experimental data.

481 - 0.5 pˆ - p = 882 = 2.69 11. H0: p = 0.5; H1: p ¹ 0.5; Test statistic: z = pq (0.5)(0.5) n

882

P-value = 2×P(z > 2.69) = 0.0072 (Tech: 0.0071); Critical values: z = ±2.575 (Tech: z = ±2.576); Reject H 0. There is sufficient evidence to reject the claim that the proportion of those in favor is equal to 0.5. The

result suggests that the politician is wrong in claiming that the responses are random guesses equivalent to a coin toss.

72 - 0.06 pˆ - p 724 12. H0: p = 0.06; H1: p > 0.06; Test statistic: z = = = 4.47 pq (0.06)(0.94) n

724

P-value = P(z > 4.47) = 0.0001 (Tech: 0.00004); Critical value: z = 1.645; Reject H 0. There is sufficient evidence to support the claim that the rate of nausea is greater than the 6% rate experienced by flu patients given a placebo. The rate of nausea with the Tamiflu treatment appears to be greater than the rate for those given a placebo, and the sample proportion of 0.0994 suggests that about 10% of those treated experience nausea, so nausea is a concern for those given the treatment.

52 - 0.20 pˆ - p 227 13. H0: p = 0.20; H1: p > 0.20; z = = = 1.10 pq n

(0.20)(0.80) 227

P-value = P(z > 1.10) = 0.1357 (Tech: 0.1367); Critical value: z = 1.645; Fail to reject H 0. There is not sufficient evidence to support the claim that more than 20% of OxyContin users develop nausea. However, with pˆ = 0.229, we see that a large percentage of OxyContin users experience nausea, so that rate does appear to be very high.

856 - 0.50 pˆ - p 1228 = = 13.81 14. H : p = 0.5; H : p > 0.5; z = 0

1

pq n

(0.50)(0.50) 1228

P-value = P(z > 13.81) = 0.0001 (Tech: 0.0000); Critical value: z = 2.33; Reject H 0. There is sufficient evidence to support the claim that most medical malpractice lawsuits are dropped or dismissed. This should be comforting to physicians.

717 - 0.15 pˆ - p 5000 = = - 1.31 15. H0: p = 0.15; H1: p < 0.15; Test statistic: z = pq (0.15)(0.85) n

5000

P-value = P(z < - 1.31) = 0.0951 (Tech: 0.0956); Critical value: z = - 2.33; Fail to reject H 0. There is not sufficient evidence to support the claim that the return rate is less than 15%.

27 - 0.10 pˆ - p 300 16. H0: p = 0.10; H1: p < 0.10; Test statistic: z = = = - 0.58 pq (0.10)(0.90) n

300

P-value = P(z < - 0.58) = 0.2810 (Tech: 0.2819); Critical value: z = - 1.645; Fail to reject H 0. There is not sufficient evidence to support the claim that less than 10% of the test results are wrong. The sample results suggest that the test is wrong too often to be considered very reliable.

17. H : p = 0.512; H : p ¹ 0.512; z = 0

pˆ - p

1

=

pq n

860 - 0.512 426

= - 0.98

(0.512)(0.488) 860

P-value = 2P(z < - 0.98) = 0.3270 (Tech: 0.3286); Critical values: z = ±1.96; Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim that 51.2% of newborn babies are males. The results do not support the belief that 51.2% of newborn babies are males; the results merely show that there is not strong evidence against the rate of 51.2%.

153 - 0.03 pˆ - p 5924 = = - 1.88 18. H0: p = 0.03; H1: p ¹ 0.03; Test statistic: z = pq (0.03)(0.97) n

5924

P-value = 2×P(z < - 1.88) = 0.0602 (Tech: 0.0597); Critical values: z = ±1.96; Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim that 3% of Eliquis users develop nausea. The rate of nausea appears to be quite low, so it is not a problematic adverse reaction.

19. H 0: p = 0.5; H 1: p ¹ 0.5; Test statistic: z =

pˆ - p pq n

=

0.55 - 0.5 = 1.02 (z = 0.98 using x = 57) (0.5)(0.5) 104

P-value = 2×P(z > 1.02) = 0.3078; (0.3268, Table: 0.3270 using x = 57) Critical value: z = ±1.96; Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim that women who guess the gender of their babies have a success rate equal to 50%.

32 - 0.5 pˆ - p 20. H0: p = 0.5; H1: p ¹ 0.5; Test statistic: z = = 45 = 2.83 pq (0.5)(0.5) n

104

P-value = P(z > 2.83) = 0.0023; Critical value: z = 2.33; Reject H 0. There is sufficient evidence to support the claim that women with more than 12 years of education have a proportion of correct predictions that is greater than 0.5. It appears these women do have an ability to correctly predict the gender of their babies. pˆ - p 123 - 0.5 = 280 = - 2.03 21. H0: p = 0.5; H1: p ¹ 0.5; Test statistic: z = pq (0.5)(0.5) n

280

P-value = 2×P(z < - 2.03) = 0.0424 (Tech: 0.0422); Critical values: z = ±1.645; Reject H 0. There is sufficient evidence to warrant rejection of the claim that touch therapists use a method equivalent to random guesses. However, their success rate of 123 280, or 43.9%, indicates that they performed worse than random guesses, so they do not appear to be effective. pˆ - p 0.57 - 0.50 = = 4.56 (4.55 if using x = 0.57(1060) = 604) 22. H : p = 0.50; H : p ¹ 0.50; z = 0

1

pq n

(0.50)(0.50) 1060

P-value = 2P(z > 4.56) = 0.0002 (Tech: 0.000005); Critical values: z = ±2.575; Reject H 0. There is sufficient

evidence to warrant rejection of the claim that half of all teens have made new friends online. (The proportion appears to be greater than one half.) pˆ - p 0.33 - 0.35 = = - 2.18 (- 2.17 if using x = 0.33(2705) = 893) 23. H : p = 0.35; H : p ¹ 0.35; z = 0

1

pq n

(0.35)(0.65) 2705

P-value = 2P(z < - 2.17) = 0.0300 (Tech: 0.0303); Critical values: z = ±1.96; Reject H 0. There is sufficient evidence to warrant rejection of the claim that the percentage who make angry gestures while driving is equal to 35%. If the significance level is changed to 0.01, the conclusion does change to this: There is not sufficient evidence to warrant rejection of the claim that the percentage who make angry gestures while driving is equal to 35%.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

135

pˆ - p 24. H0: p = 0.00034; H0: p ¹ 0.00034; Test statistic: z =

pq n

=

- 0.000340

420,095

(0.000340)(0.99966)

= - 0.66

420,095

P-value = 2×P(z < - 0.66) = 0.5092 (Tech: 0.5122); Critical values: z = ±2.81; Fail to reject H 0. There is not sufficient evidence to support the claim that the rate is different from 0.0340%. Cell phone users should not be concerned about getting cancer of the brain or nervous system.

39 - 0.5 pˆ - p 71 25. H : p = 0.5; H : p > 0.5; Test statistic: z = = = 0.83 0

1

pq n

(0.5)(0.5) 71

P-value = P(z > 0.83) = 0.2033 (Tech: 0.2031); Critical value: z = 1.645; Fail to reject H 0. There is not sufficient evidence to support the claim that among smokers who try to quit with nicotine patch therapy, the majority are smoking one year after the treatment. The results show that about half of those who use nicotine patch therapy are successful in quitting smoking.

26. H 0: p = 0.75; H 1: p > 0.75; Test statistic: z =

pˆ - p pq n

=

0.817 - 0.75 = 8.48 (0.75)(0.25) 3005

P-value = P(z > 8.48) = 0.0001 (Tech: 0.0000); Critical value: z = - 2.33; Reject H 0. There is sufficient evidence to support the claim that more than 3 4 of adults use at least one prescription medication. With more than 3 4 of adults using at least one prescription medication, it appears that prescription use among adults is high.

239 - 0.50 pˆ - p 291 27. H0: p = 0.50; H1: p ¹ 0.50; z = = = 10.96 pq n

(0.50)(0.50) 291

P-value = 2P(z > 10.96) = 0.0001 (Tech: 0.0000); Critical values: z = ±2.575 (Tech: z = ±2.576); Reject H 0.

There is sufficient evidence to reject the claim that the YSORT method has no effect. pˆ - p 28. H0: p = 0.5; H1: p < 0.5; Test statistic: z =

6062 - 0.5

= 12,000

pq n

(0.5)(0.5) 12,000

= 1.13

P-value = P(z < 1.13) = 0.8708 (Tech: 0.8712); Critical value: z = - 1.645; Fail to reject H 0. There is not sufficient evidence to support the claim that less than 0.5 of the deaths occur the week before Thanksgiving.

Based on these results, there is no indication that people can temporarily postpone their death to survive Thanksgiving. (With 50.5% of the deaths occurring before Thanksgiving, there is no way that the claim of a proportion less than 0.5 could be supported.) pˆ - p 0.42 - 1 / 3 = = 8.72 (x = 0.42(2250) = 945) 29. H : p = 1 / 3; H : p > 1 / 3; z = 0

1

pq n

(1/3)(2/3) 2250

P-value = P(z > 8.72) = 0.0001 (Tech: 0.0000); Critical value: z = 2.33; Reject H 0. There is sufficient evidence to support the claim that more than 1/3 of adults believe in ghosts. pˆ - p 0.828 - 0.80 30. H : p = 0.80; H : p ¹ 0.80; Test statistic: z = = = 0.98 (0.99 using x = 0.828(198) = 164); 0

1

pq n

(0.80)(0.20) 198

P-value = 2×P(z > 0.98) = 0.3270 (Tech: 0.3246); Critical values: z = ±2.575 (Tech: z = ±2.576); Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim that 80% of patients stop smoking when given sustained care. With a success rate around 80%, it appears that sustained care is effective.

52 - 0.5

pˆ - p 31. H0: p = 0.5; H1: p < 0.5; Test statistic: z =

pq n

= 175

(0.5)(0.5) 175

= - 5.37

P-value = P(z < - 5.37) = 0.0001 (Tech: 0.0000); Critical value: z = - 2.33; Reject H 0. There is sufficient evidence to support the claim that the proportion of wrong indications is less than 0.5. In this experiment, the dogs were wrong about 30% of the time, so they should not be used for serious medical evaluations of malaria. pˆ - p 0.57 - 0.12 32. H : p = 0.12; H : p > 0.12; z = = = 47.99 (48.03 if using x = 0.57(1201) = 685) 0

1

pq n

(0.12)(0.88) 1201

P-value = P(z > 48.03) = 0.0001 (Tech: 0.0000); Critical value: z = 2.33; Reject H 0. There is sufficient

evidence to support the claim that the current rate of support for marijuana is greater than the 12% level in 1969.

33. H0: p = 0.5; H1: p ¹ 0.5; Normal approximation: 9 - 0.5 pˆ - p z= = 10 = 2.53; P-value = 2×P(z > 2.53) = 0.0114 pq (0.5)(0.5) n

10

Exact:

(

)

P-value = 2 × 10 C9 ×0.59 ×0.51 + 10C10 ×0.510 ×0.50 = 0.0215 Continuity Correæction: 9 1 10 0 1 P-value = 2 ×ç 10 C9 ×0.5 ×0.5 + 10C10 ×0.5 ×0.5 è H0: p = 0.4; H1: p ¹ 0.4;

9

1ö

(210 C9 ×0.5 ×0.5 ) ø÷ = 0.0117

Normal approximation: 9 - 0.4 pˆ - p 10 z= = = 3.23; P-value = 2×P(z > 3.23) = 0.0012 pq (0.4)(0.6) n

10

Exact:

(

)

P-value = 2 × 10 C9 ×0.49 ×0.61 + 10C10 ×0.410 ×0.60 = 0.0034 Continuity Correæction: 9 1 10 0 1

9

1ö

)

(

×0.6 ÷ = 0.0018 P-value = 2 ×ç 10 C9 ×0.4 ×0.6 + 10C10 ×0.4 ×0.6 10 C9 ×0.4 ø è 2 H0: p = 0.5; H1: p > 0.5; Normal approximation: pˆ - p 482 - 0.5 z= = 926 = 1.25; P-value = P(z > 1.25) = 0.1056 (Tech: 0.1059) pq (0.5)(0.5) n

926

Exact: P-value = 926C482 ×0.5482 ×0.5444 + Continuity Correction: P-value = 926 C482 ×0.5482 ×0.5444 +

+ 926C926 ×0.5926 ×0.50 = 0.1120 926

+ 926C926 ×0.5

×0.50 -

1

(

)

482 ×0.5444 = 0.1060 926 C482 ×0.5 2 The P-values agree reasonably well with the large sample size of n = 926. The normal approximation to the binomial distribution works better as the sample size increases.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

34. a. H 0: p = 0.10; H :1 p ¹ 0.10; . Test statistic: z =

pˆ - p pq n

=

0.119 - 0.1 = 2.00; Critical values: z = ±1.96; (0.1)(0.9) 1000

Reject H 0. There is sufficient evidence to warrant rejection of the claim that the proportion of zeros is 0.1. b. H 0: p = 0.10; H :1 p ¹ 0.10; Test statistic: z =

pˆ - p pq n

=

0.119 - 0.1 = 2.00; (0.1)(0.9) 1000

P-value = 2×P(z > 2.00) = 0.0456 (Tech: 0.0452); Reject H 0. There is sufficient evidence to warrant rejection of the claim that the proportion of zeros is 0.1. p̂q̂ (0.119)(0.881) Þ = 0.119 ±1.96 0.0989 < p < 0.139; Because 0.1 is contained n 1000 within the confidence interval, fail to reject H0: p = 0.10. There is not sufficient evidence to warrant rejection of the claim that the proportion of zeros is 0.1. d. The traditional and P-value methods both lead to rejection of the claim, but the confidence interval method does not lead to rejection of the claim.

(0.4)(0.6) = 0.286 35. a. From p = 0.40, pˆ = 0.4 - 1.645 50 pˆ - p 0.286 - 0.25 = = 0.588; Power = P(z < 0.588) = 0.7224 (Tech: 0.7219) From p = 0.25, z = pq (0.25)(0.75) c. 95% CI: pˆ ± za /2

n

50

b. 1- 0.7224 = 0.2776 (Tech: 0.2781) c. The power of 0.7219 shows that there is a reasonably good chance of making the correct decision of rejecting the false null hypothesis. It would be better if the power were even higher, such as greater than 0.8 or 0.9.

Section 8-3: Testing a Claim About a Mean 1. The requirements are (1) the sample must be a simple random sample, and (2) either or both of these conditions must be satisfied: The population is normally distributed or n > 30. Not enough information is given to determine whether the sample is a simple random sample. Because the sample size is not greater than 30, we must check for normality, but the value of 583 sec appears to be an outlier, and a normal quantile plot or histogram suggests that the sample does not appear to be from a normally distributed population. The requirements are not satisfied.

2.

df denotes the number of degrees of freedom. For the sample of 12 times, df = 12 - 1 = 11.

3.

A t test is a hypothesis test that uses the Student t distribution, such as the method of testing a claim about a population mean as presented in this section. The letter t is used in reference to the Student t distribution, which is used in a t test.

For a 0.05 significance level used in a one-tailed test, use a 90% confidence level. The given confidence interval does contain the value of 90 sec, so it is possible that the value of m is equal to 90 sec or some lower value.

There is not sufficient evidence to support the claim that the mean is greater than 90 sec.

4.

5.

P-value = 0.0437 (Table: 0.025 < P-value < 0.05); There is sufficient evidence to warrant rejection of the claim

that the mean body temperature is equal to 98.6°F. 6.

P-value = 0.0000 (Table: P-value < 0.01); There is sufficient evidence to warrant rejection of the claim that the

mean body temperature is equal to 98.6°F. 7.

P-value = 0.0000 (Table: P-value < 0.005); There is sufficient evidence to support the claim that the mean

cotinine level of smokers is greater than 2.84 ng/mL. 8.

P-value = 0.0001 (Table: P-value < 0.005); There is sufficient evidence to support the claim that the mean

body temperature of males is less than 98.6°F.

9.

H0: m = 98.6 F; H1: m ¹ 98.6 F; Test statistic: t = - 3.865; P-value = 0.0004; Critical values (a = 0.05): t = ±2.0262 (Table: ±2.028); Reject H 0. There is sufficient evidence to warrant rejection of the claim that the mean body temperature is equal to 98.6 F.

10. H0: m = 98.6°F; H1: m ¹ 98.6°F; Test statistic: t = - 13.272; P-value < 0.0001; Critical values (a = 0.05): t = ±1.995 (Table: ±1.994); Reject H 0. There is sufficient evidence to warrant rejection of the claim that the mean body temperature is equal to 98.6 F. 11. H0: m = 270 (1000 cells mL); H1: m < 270 (1000 cells mL); Test statistic: t = - 2.764; P-value = 0.0032; Critical value (with a = 0.05): t = - 1.665; (Table: - 1.660); Reject H 0. There is sufficient evidence to support the claim that the population of adult females has a mean platelet count less than 270 (1000 cells mL).

12. H0: m = 220 (1000 cells mL); H1: m > 220 (1000 cells mL); Test statistic: t = 0.89; P-value = 0.188; Critical value (with a = 0.05): t = 1.665; (Table: 1.660); Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim that the mean platelet count for adult males is greater than 220 (1000 cells mL).

13. n = 300 > 30; H0: m = 120 mmHg; H1: m > 120 mmHg; 122.96 - 120 x- m Test statistic: t = = = 3.234; Critical value: t = 2.339 P-value = 0.0007 s / n 15.85169 / 300 (Table: P-value < 0.005); Reject H 0. There is sufficient evidence to support the claim that the sample is from a population with a mean greater than 120 mmHg. 14. n = 300 > 30; H0: m = 60 mmHg; H1: m > 60 mmHg; x- m 70.75333 - 60 Test statistic: t = = = 16.034; Critical value: t = 2.339; P-value = 0.0000 s / n 11.61618 / 300 (Table: P-value < 0.005); Reject H 0. There is sufficient evidence to support the claim that the sample is from a population with a mean greater than 60 mmHg. 15. Since n = 21 < 30, assume the data follow a normal distribution. H0: m = 100; H1: m ¹ 100; x - m 86.90476 - 100 Test statistic: t = = = - 6.676; Critical values: t = ±2.086; P-value = 0.0000 s / n 8.988352 / 21 (Table: P-value < 0.01); Reject H 0. There is sufficient evidence to warrant rejection of the claim that the sample of children is from a population with mean IQ equal to 100. These results do not “prove” that exposure to lead has an adverse effect on IQ scores of children, but it strongly suggests that such an adverse effect is very possible.

16. The data appear to follow the loose definition for a normal distribution and n > 30.

H0: m = 3400 g; H1: m < 3400 g; x - m 3272.8 - 3400 Test statistic: t = = = - 2.690; Critical value: t = 2.345; P-value = 0.0039 s/ n 660.2 195 (Table: P-value < 0.005); Reject H 0. There is sufficient evidence to support the claim that the population mean of birth weights of males is less than 3400 g.

17. n = 40 > 30; H0: m = 0 lb; H1: m > 0 lb; x - m 3.0 - 0.0 = = 3.872; Critical value: t = 2.426; s / n 4.9 40 P-value = 0.0002 (Table: P-value < 0.005); Reject H 0. There is sufficient evidence to support the claim that the mean weight loss is greater than 0. Although the diet appears to have statistical significance, it does not appear to have practical significance, because the mean weight loss of only 3.0 lb does not seem to be worth the effort and cost.

Test statistic: t =

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

18. The data cannot be verified to follow a normal distribution, but n > 30.

H0: m = 0 mg/dL; H1: m > 0 mg/dL; x- m 0.4 - 0 Test statistic: t = = = - 0.133; Critical value: t = 1.677 (Table: t » 1.676); P-value = 0.4472 s / n 21.0 49 (Table: P-value > 0.10); Fail to reject H 0. There is not sufficient evidence to support the claim that with garlic treatment, the mean change in LDL cholesterol is greater than 0 mg/dL. There is not sufficient evidence to support a claim that the garlic treatment is effective in reducing LDL cholesterol levels.

19. The data cannot be verified to follow a normal distribution, but n > 30.

H0: m = 69.5 years; H1: m ¹ 69.5 years; x - m 73.4 - 69.5 Test statistic: t = = = 2.652; Critical values: t = ±2.032; P-value = 0.0121 s/ n 8.7 35 (Table: P-value < 0.02); Reject H 0. There is sufficient evidence to reject the claim that male symphony conductors have a mean life span equal to 69.5 years. Because males don’t become conductors until they have already lived for many years, the sample mean of 73.4 years should not be compared to the mean of 69.5 years.

20. The data cannot be verified to follow a normal distribution and n < 30, so proceed with caution.

H0: m = 102.8 min; H1: m < 102.8 min; x - m 98.9 - 102.8 Test statistic: t = = = - 0.369; Critical value (a = 0.05): t = - 1.753; P-value = 0.3587 s/ n 42.3 16 (Table: P-value > 0.10); Fail to reject H 0. There is not sufficient evidence to support the claim that after treatment with zopiclone, subjects have a mean wake time less than 102.8 min. These results suggest that the zoplicone treatment is not effective.

21. The sample data meet the loose requirement of having a normal distribution.

H0: m = 14 mg g; H1: m < 14 mg g; x - m 11.05 - 14.0 Test statistic: t = = = - 1.444; Critical value: t = - 1.883; P-value = 0.0913 s/ n 6.46 10 (Table: P-value > 0.05); Fail to reject H 0 . There is not sufficient evidence to support the claim that the mean lead concentration for all such medicines is less than 14 mg g.

22. The sample data meet the loose requirement of having a normal distribution.

H0: m = 60 sec; H1: m ¹ 60 sec; x - m 62.67 - 60.00 Test statistic: t = = = 0.530; Critical values: t = ±2.145; P-value = 0.6043 s/ n 9.48 15 (Table: P-value > 0.20); Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim that the times are from a population with a mean equal to 60 seconds. Although some of the individual estimates are off by a large amount, the group of 15 students yielded the mean of 62.7 sec, which is not significantly different from 60 sec, so as a group they appear to be reasonably good at estimating one minute.

23. n = 8 < 30, but the sample data meet the loose requirement of having a normal distribution.

H0: m = 1.6 W/kg; H1: m < 1.6 W/kg; x - m 1.04875 - 1.6 Test statistic: t = = = - 4.256; Critical value: t = - 2.998; P-value = 0.0019 s/ n 0.36635 / 8 (Table: P-value < 0.005); Reject H 0. There is sufficient evidence to support the claim that the sample is from a population of cell phones with a mean amount of radiation that is less than the FCC standard of 1.6 W/kg.

24. n = 15 < 30, but the sample data meet the loose requirement of having a normal distribution.

H0: m = 22.6 mm; H1: m < 22.6 mm; x- m 21.13 - 22.6 Test statistic: t = = = - 7.655; Critical value: t = - 2.624; P-value = 0.0000 s / n 0.743736 / 15 (Table: P-value < 0.01); Reject H 0. There is sufficient evidence to support the claim that the sample of wren egg lengths is from a population with a mean that is less than the population mean of 22.6 mm for robins.

25. The data appear to follow a normal distribution and n > 30.

H0: m = 75 bpm; H1: m < 75 bpm; Test statistic: t =

x - m 74.04 - 75.00 = = - 0.927; Critical value: t = - 1.655 (Table: t » - 1.660); s/ n 12.54 147

P-value = 0.1777 (Table: P-value > 0.10); Fail to reject H 0. There is not sufficient evidence to support the claim that the mean pulse rate of adult females is less than 75 bpm.

26. The data appear to follow a normal distribution and n > 30.

H0: m = 75 bpm; H1: m < 75 bpm;

Test statistic: t =

x - m 69.58 - 75 = = - 5.915; Critical value: t = - 1.655 (Table: t » - 1.660); s / n 11.33 153

P-value = 0.0000 (Table: P-value < 0.005); Reject H 0. There is sufficient evidence to support the claim that the mean pulse rate of adult males is less than 75 bpm.

27. n = 4082 > 30; H0: m = 1818 mm; H1: m < 1818 mm; x - m 1814.154 - 1818 = = - 2.903; Critical value: t =- 2.327 (Table: t = - 2.236); s / n 84.63228 / 4082 P-value = 0.0019 (Table: P-value < 0.005); Reject H 0. There is sufficient evidence to support the claim that the sample is from a population with a mean arm span less than 1818 mm. The sample mean is 1814.154 mm, and the claimed mean is 1818 mm. The difference of 3.846 mm (or 0.15 in.) is statistically significant, but does not appear to have practical significance. (TI data: Test statistic: t = 0.126, P-value = 0.550, critical value : t =- 2.334. Fail to reject H 0 . There is not sufficient evidence to support the claim that the sample is from a population with a mean arm span less than 1818 mm.)

Test statistic: t =

28. n = 4082 > 30; H0: m = 1755.8 mm; H1: m > 1755.8 mm; x - m 1756.215 - 1755.8 = 0.386; Critical value: t = 1.645; P-value = 0.3496 = s/ n 68.55079 / 4082 (Table: P-value > 0.10); Fail to reject H 0. There is not sufficient evidence to support the claim that the sample from 2012 is from a population with a mean height greater than 1755.8 mm. It does not appear that male army personnel were taller in 2012 than in 1988.

(TI data: Test statistic: t = 0.564, P-value = 0.2868, critical value: t = 1.648) Test statistic: t =

29. H0: m = 100; H1: m ¹ 100; 100.05 - 100 x- m Test statistic: t = = = 3.33; P-value = 0.0009; The difference between x = 100.05 and s / n 15 / 1, 000, 000 the claimed mean of 100 is statistically significant, but that difference is so small that it does not have practical significance.

30. The power of 0.9127 shows that the chance of recognizing that m < 1.6 W/kg is quite high when in reality m = 1.0 W/kg. This is a good value of power. b = 1- 0.9127 = 0.0873, so there is about a 9% chance of failing to recognize that m < 1.6 W/kg when in reality m = 1.0 W/kg.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

31.

A=

1.645(8×4081+ 3)

= 1.645100769 and t = = 1.645; The critical t score found 8 ×4081+1 using the given approximation is 1.645, which is the same value of 1.645 found from technology. The approximation appears to work quite well, and it provides us with a method for finding critical t scores when the number of degrees of freedom cannot be found from Table A-3 and suitable technology is unavailable.

x- m 6.83333333 - 7 = - 0.29; P-value = 0.3860 (Table: 0.3859); The P-value is smaller with 32. a. z = s / n = 1.99240984 / 12 the known value of s . The null and alternative hypotheses are the same and the conclusions are the same.

These results of this particular hypothesis test are not affected very much by the knowledge of s . b. Because the P-value is smaller with the knowledge of s , it is possible that the conclusion of a hypothesis test could change. It is possible to have a conclusion of fail to reject H 0 without knowledge of s , while the conclusion could be reject H 0 if s is known.

Section 8-4: Testing a Claim About a Standard Deviation or Variance 1. The sample must be a simple random sample, and the sample must be from a normally distributed population.

The normality requirement for a hypothesis test of a claim about a standard deviation is much stricter, which means that the distribution of the population must be much closer to a normal distribution.

2.

H0: s = 0.470 kg; H1: s ¹ 0.470 kg;

(16 - 1)0.6573182

2

c

3.

4.

5.

6.

7.

=

2

= 29.339; The sampling distribution of the test statistic is the c (chi-square)

0.4702 distribution. a. P-value = 0.0291 < 0.05; Reject H 0. b. Reject the claim that the sample is from a population with a standard deviation equal to 0.470 kg. c. It appears that the vitamin supplement does affect the variation among birth weights.

All of the values in the confidence interval are greater than the standard deviation of 0.470 kg, so it appears that the sample with the vitamin treatment has birth weights with a standard deviation different from 0.470 kg. It appears that the vitamin supplement does affect the variation among birth weights.

H0: s = 10 bpm; H1: s ¹ 10 bpm; Test statistic: c 2 = 195.172; P-value = 0.0208;

Reject H 0. There is sufficient evidence to warrant rejection of the claim that pulse rates of males have a standard deviation equal to 10 beats per minute. Using the range rule of thumb with the normal range of 60 to 100 beats per minute is not very good for estimating s in this case. (n - 1)s2 (147 - 1)12.52 = H0: s = 10 bpm; H1: s ¹ 10 bpm; Test statistic: c 2 = 229.718 or c 2 = = 228.125; s 2 102 P-value < 0.0001; Reject H 0. There is sufficient evidence to warrant rejection of the claim that pulse rates of females have a standard deviation equal to 10 beats per minute. Using the range rule of thumb with the normal range of 60 to 100 beats per minute is not very good for estimating s in this case.

2 (n - 1)s2 = (106 - 1)0.62 = 9.329; H0: s = 2.08°F; H1: s < 2.08°F; Test statistic: c 2 = s 2 2.082 P-value = 0.0000 (Table: P-value < 0.005); Critical value: c 2 = 74.252 (Table: c 2 » 70.065);

Reject H 0. There is sufficient evidence to support the claim that body temperatures have a standard deviation less than 2.08°F. It is very highly unlikely that the conclusion in the hypothesis test in Example 5 from Section 8-3 would change because of a standard deviation from a different sample.

(n - 1)s2 = (30 - 1)829.5

2

8.

H0: s = 660.2 hg; H1: s ¹ 660.2 hg; Test statistic: c 2 =

s 2

660.22

= 45.780;

P-value = 0.0493 (Table: P-value > 0.02); Critical values: c 2 = 13.121 and c 2 = 52.336;

9.

Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim that birth weights of females have the same standard deviation as the birth weights of males.

2 (n - 1)s2 = (18 - 1)9.6 H0: s = 15; H1: s < 15; Test statistic: c 2 = = 6.963; s 2 152 P-value = 0.0160 (Table: P-value < 0.025); Critical value: c 2 = 8.672;

Reject H 0. There is sufficient evidence to support the claim that IQ scores of physicians have a standard deviation less than 15.

2 (n - 1)s2 = (27 - 1)9.3 2 10. H0: s = 14.1; H1: s < 14.1; Test statistic: c = = 11.311; s 2 14.12 P-value = 0.0056 (Table: P-value < 0.01); Critical value: c 2 = 12.198;

Reject H 0. There is sufficient evidence to support the claim that the last class has less variation. The lower variation implies that the scores are closer together, but it does not imply that the scores are higher, so the lower standard deviation does not suggest that the class is doing better.

2 (n - 1)s2 = (71- 1)3.87 2 11. H0: s = 4.25 g; H1: s < 4.25 g; Test statistic: c = = 58.042; s 2 4.252 P-value = 0.1545 (Table: P-value > 0.10); Critical value: c 2 = 51.739;

Fail to reject H 0. There is not sufficient evidence to support the claim that the new process dispenses amounts with a standard deviation less than the standard deviation of 4.25 g for the old process. The new process does not appear to be better in the sense of dispensing amounts that are more consistent.

2 (n - 1)s2 = (40 - 1)4.9 2 = 26.011; 12. H0: s = 6.0 lb; H1: s ¹ 6.0 lb; Test statistic: c = s 2 6.02 P-value = 0.1101 (Table: P-value > 0.10); Critical values: c 2 = 19.996 and c 2 = 65.475 (Table: c 2 » 20.707

and c 2 » 66.766); Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim that the amounts of weight loss have a standard deviation equal to 6.0 lb.

2 (n - 1)s2 = (10 - 1)85.603 2 13. H0: s = 66.8 mm; H1: s ¹ 66.8 mm; Test statistic: c = = 14.780; s 2 66.82 P-value = 0.1943; (Table A-4 is not adequate here.) Critical values: c 2 = 2.700 and c 2 = 19.023; Fail to reject H 0. There is not sufficient evidence to support the claim that the listed sample is from a population with a standard deviation different from the standard deviation of 66.8 mm found from males in the ANSUR I survey in 1988. It appears that the variation of heights of male U.S. Army personnel did not change from 1988 to 2012.

2 (n - 1)s2 = (11- 1)9.65 = 413.878; 14. H0: s = 1.5 mmHg; H1: s > 1.5 mmHg; Test statistic: c 2 = s 2 1.52 P-value = 0.0000 (Table: P-value < 0.005); Critical value: c 2 = 18.307;

Reject H 0. There is sufficient evidence to support the claim that the sample of devices has a standard deviation greater than 1.5 mmHg. The results suggest that these aneroid sphygmomanometers are not accurate enough to be acceptable.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

(n - 1)s2 = (10 - 1)28.601

2

15. H0: s = 109.3 sec; H1: s < 109.3 sec; Test statistic: c 2 =

s 2

109.32

= 0.616;

P-value = 0.001 (Table: P-value < 0.005); Critical value: c 2 = 3.325;

Reject H 0. There is sufficient evidence to support the claim that with a single waiting line, the waiting times have a standard deviation less than 109.3 sec. Because the variation among waiting times appears to be reduced with the single waiting line, patients are happier because their waiting times are closer to being the same.

2 (n - 1)s2 = (15 - 1)0.25875 16. H0: s = 0.03 g; H1: s ¹ 0.03 g; Test statistic: c 2 = = 10.415; s 2 0.032 P-value = 0.5375 (Table: P-value > 0.20); Critical values: c 2 = 5.629 and c 2 = 26.119;

Reject H 0. There is sufficient evidence to warrant rejection of the claim that the weights of bumblebee bats have a standard deviation of 0.03 g. It appears the weights of these bats have a different variation than those from the region in Thailand.

17. H0: s = 59.6 (1000 cells mL); H1: s > 59.6 (1000 cells mL) 2 (n - 1)s2 (147 - 1)65.4168 = Test statistic: c 2 = = 175.899; s 2 59.62 P-value = 0.0464; (Table A-4 is not useful here because it shows that the critical value is greater than 135.807, but it doesn’t show whether it is greater than or less than the test statistic of 175.889.)

Critical value: c 2 = 188.666; (Again, Table A-4 is not adequate here.)

Fail to reject H 0 because the P-value of 0.0464 is not equal to or less than 0.01. There is not sufficient evidence to support the claim that the population of females has a standard deviation greater than the standard deviation of 59.6 (1000 cells mL) from the population of males. The conclusion does change with a significance level of 0.05 because the P-value of 0.0464 is less than 0.05.

18. H0: s = 20.9 kg; H1: s < 20.9 kg; Test statistic: c 2 =

(n - 1)s2

=

(153- 1)17.651442

= 108.420; s 2 20.92 P-value = 0.0030; (Table A-4 is not useful here because it shows only that the critical value is greater than

70.065.); Critical value: c 2 = 114.390; (Again, Table A-4 is not adequate here.); Reject H 0 because the P-value of 0.0030 is less than 0.01. There is sufficient evidence to support the claim the population of males has a standard deviation less than the standard deviation of 20.9 kg from the population of females.

2 (n - 1)s2 (4082 - 1)14.22070 2 = 19. H0: s = 11.2 kg; H1: s ¹ 11.2 kg; Test statistic: c = = 6579.190; s 2 11.22 P-value = 0.0000; Critical values: c 2 = 3905.835 and c 2 = 4259.952; Reject H 0. There is sufficient evidence to support the claim that the sample collected in 2012 is from a population with a standard deviation different from 11.2 kg. A histogram shows some skewness and a normal quantile plot shows a somewhat systematic pattern, so satisfaction of the requirements is somewhat questionable.

2 (n - 1)s2 (4082 - 1)68.55079 = 20. H0: s = 66.8 mm; H1: s ¹ 66.8 mm; Test statistic: c 2 = = 4297.725; s 2 66.82 P-value = 0.0180; Critical values: c 2 = 3905.835 and c 2 = 4259.952; Reject H 0. There is sufficient evidence to support the claim that the sample collected in 2012 is from a population with a standard deviation different from 66.8 mm. A histogram or normal quantile plot shows that the distribution is very close to being normal, so a requirement of the hypothesis test is satisfied.

(TI data: Test statistic: c 2 = 513.244; P-value = 0.0000; Critical values: c 2 = 270.492 and c 2 = 369.295.)

2 21. Critical c =

1

2

(2.33 + 2 ×147 - 1 ) = 187.961 (or 187.890 if using z = 2.326348 found from technology), 2

which is reasonably close to the value of 188.666 obtained from STATDISK. 3 æ 2 2 ö 2 22. Critical c = 146 ç1+ 2.33 ÷ = 18 .75 (or 188.678 if using z = 2.326348 found from 9×146 ø è 9×146 technology), which is very close to the value of 188.666 obtained from STATDISK. Section 8-5: Resampling: Using Technology for Hypothesis Testing 1. a. A sample of the same size (n = 5) is randomly selected from the given sample.

2.

3.

b. The sampling is done with replacement. If we were to sample without replacement, we would always obtain the same sample consisting of the same five waiting times, and those results would have no real value. c. A resampling method does not require that the sample data are from a population having a particular distribution (such as normal) or that the sample size meets some minimum requirement.

With bootstrapping, the original data values are resampled. With randomization, the original data values are somehow modified to conform to the assumed value of the population parameter. a.

pˆ = 426 / 860 = 0.495 and p = 0.512.

b. The sample proportions that are at least as extreme as pˆ = 0.495 are those that are 0.495 or lower and those that are 0.529 or higher. c. The result of 310 samples (among 1000) with a proportion at least as extreme as 426 / 860 shows that the event of getting such a sample proportion is quite common and can easily occur. It appears that there is not sufficient evidence to warrant rejection of the claim that the proportion of male births is equal to 0.512.

4.

a. The sample means at least as extreme as 98.13 F are those that are 98.13 F or lower and those that are 99.07 F or higher. b. Because none of the resampled data sets have means at least as extreme as the mean of 98.13 F that was obtained, it appears that 98.13 F is significantly different from the claimed mean of 98.6 F. It appears that a mean of 98.13 F is very unlikely to occur with a population having a mean of 98.6 F, so there is sufficient evidence to warrant rejection of the claim that the population has a mean body temperature equal

5.

6.

7.

to 98.6 F. Answers vary, but the following is typical. Resample 1000 times to find that 364 of the results have a proportion of 91 / 220 = 0.414 or greater. There is a likelihood of 0.364 of getting a sample proportion that is at least as extreme as the one obtained, so there is not sufficient evidence to support the claim that the sample of cast and crew members is from a population in which the rate of cancer is greater than 40%.

Answers vary, but the following is typical. The sample proportion is 0.396, so in this two-tailed case, the values at least as extreme as 0.396 are those that are 0.396 or lower plus those that are 0.604 or greater. (Note that 0.396 is below the assumed value of 0.5 by 0.104, so the upper bound becomes 0.104 above 0.5, which is 0.604.) Resample 1000 times to find that 184 of the results have a proportion at least as extreme as 0.396. There is a likelihood of 0.184 of getting a sample proportion that is at least as extreme as the one obtained, so there is not sufficient evidence to warrant rejection of the claim that dropped toast will land with the buttered side down 50% of the time.

Answers vary, but the following is typical: Create a column of 882 ones and zeros, with 0.5 of them being 1 (as assumed in the null hypothesis). The sample proportion is 481 882 = 0.545, so in this two-tailed case, the values “at least as extreme” as 0.545 are those that are 0.545 or higher or 0.455 or lower. Resample 1000 times to find that only 2 of the results have a sample proportion of 0.545 or higher, and only 4 of the results have a sample proportion of 0.455 or lower. There is a very small likelihood of getting a sample proportion that is at least as extreme as the one obtained, so there is sufficient evidence to reject the claim that the proportion of subjects who respond in favor is equal to 0.5.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

8.

Answers vary, but the following is typical: Create a column of 724 ones and zeros, with 0.06 of them being 1 (as assumed in the null hypothesis). The sample proportion is 72 724 = 0.0994, so in this right-tailed case, the values “at least as extreme” as 0.0994 are those that are 0.0994 or higher. Resample 1000 times to find that none of the results have a sample proportion of 0.0994 or higher. There is a very small likelihood of getting a sample proportion that is at least as extreme as the one obtained, so there is sufficient evidence to reject the claim that the rate of nausea is equal to the 6% rate experienced with flu patients given a placebo.

9.

Answers vary, but the following is typical. With x = 11.05 mg g. and m = 14 mg g. (as assumed in the null

hypothesis), add 2.95 mg g to each sample value (as Statdisk does) so that the sample has the claimed mean of 14 mg g. Resample the modified sample 1000 times to find that 53 of the results have a mean of 11.05 mg g or less. There appears to be a likelihood of 0.053 of getting a sample mean of 11.05 mg g or lower, so there is not sufficient evidence to support the claim that the mean lead concentration for all such medicines is less than 14 mg g. (Note: Because the likelihood of 0.053 is so close to the significance level of 0.05, it is very possible to find that the likelihood is less than 0.05, so the conclusion would be that there is sufficient evidence to support the claim.)

10. Answers vary, but the following is typical. With x = 62.66666667 sec and m = 60 sec (as assumed in the null hypothesis), subtract 2.66666667 sec from each sample value (as Statdisk does) so that the sample has the claimed mean of 60 sec. Resample the modified sample 1000 times. The sample means at least as extreme as 62.66666667 sec are those that are 62.66666667 sec or greater and those that are 57.33333333 sec or lower.

Among the 1000 generated samples, there are 579 that are at least as extreme as 62.66666667 sec, so there appears to be a likelihood of 0.579 of getting a sample mean at least as extreme as 62.66666667 sec. There is not sufficient evidence to warrant rejection of the claim that the times are from a population with a mean equal to 60 sec.

11. Answers vary, but the following is typical. With x = 1.04875 W/kg and m = 1.6 W/kg (as assumed in the null hypothesis), add 0.55125 W/kg to each sample value (as Statdisk does) so that the sample has the claimed mean of 1.6 W/kg. Resample the modified sample 1000 times. The sample means at least as extreme as 1.04875 W/kg are those that are 1.04875 W/kg or lower. Among the 1000 generated samples, there are none that are at least as extreme as 1.04875 W/kg, so there appears to be a likelihood of 0.000 of getting a sample mean at least as extreme as 1.04875 W/kg. There is sufficient evidence to support the claim that the sample is from a population of cell phones with a mean amount of radiation that is less than the FCC standard of 1.6 W/kg.

12. Answers vary, but the following is typical: With x = 21.130 mm and m = 22.6 mm (as assumed in the null hypothesis), add 1.47 mm to each sample value (as Statdisk does) so that the sample has the claimed mean of 22.6 mm. Resample the modified sample 1000 times. The sample means at least as extreme as 21.130 mm are those that are 21.130 mm or lower. Among the 1000 generated samples, none are 21.130 or lower, so it appears that the sample mean is unlikely to occur by chance, and it appears to be significantly low. There is sufficient evidence to support the claim that the sample of cuckoo eggs in wren nests is from a population with a mean that is less than the population mean of 22.6 mm for cuckoo eggs in robin nests.

13. For the data in Example 4, s = 10.09376 bpm. For testing the claim that s < 11.3 bpm, we assume in the null hypothesis that s = 11.3 bpm. To modify the data set so that s becomes 11.3 bpm, we multiply each of the data values by s / s = 11.3 /10.09376 = 1.119503535; then we resample 1000 times. A typical result is that among the 1000 resulting values of s, 265 of them are 10.097376 bpm or lower. That is, 265 of the 1000 resamples are at least as extreme as the value of s = 10.09376 bpm that was obtained. That result corresponds to an estimated P-value of 0.265, which is greater than the significance level of 0.05 used in Example 4. We therefore conclude that there is not sufficient evidence to support the claim that the listed pulse rates are from a population with a standard deviation that is less than 11.3 bpm. These results from randomization agree quite well with those from Example 4 in Section 8-4, and this suggests that randomization is very effective in this case.

14. Here are typical results: x = 62.1 mg/g, s = 6.8 mg/g, the range is 32.0 mg/g, the minimum is 47 mg/g and the maximum is 79 mg/g. The distribution appears to be approximately normal, and there are no outliers. The range of values from 47 mg/g to 79 mg/g is somewhat large. Given the way that the values were generated, it is not surprising that the distribution appears to be approximately normal.

Quick Quiz 1. H0: p = 0.5; H1: p > 0.5 pˆ - p

668 - 0.50

726

a. normal distribution b. right-tailed test

2.

z=

4.

a. 0.0000 < 0.05; Reject H 0. b. There is sufficient evidence to support the claim that with the XSORT method, the majority of births will be female.

5.

H0: m = 0.07; H1: m ¹ 0.07

7.

a. t distribution b. two-tailed test

pq n

=z=

3.

(0.50)(0.50) 726

= 22.64

6.

t=

x - m 0.50 - 0.07 = = 2.730 s / n 0.63 / 16

8.

a. 0.0155 < 0.05; Reject H 0. b. There is sufficient evidence to reject the claim that the sample is from a population with a mean test score equal to the value of 0.07 from those not given a treatment.

9. a. true d. false e. false b. false c. false 10. The t test requires that the sample is from a normally distributed population, and the test is robust in the sense that the test works reasonably well if the departure from normality is not too extreme. The c 2 (chi-square) test is not robust against a departure from normality, meaning that the test does not work well if the population has a distribution that is far from normal.

Review Exercises 1. a. H0: m = 1 mg; H1: m < 1 mg 2.

3.

c.

H0: p = 0.1; H1: p ¹ 0.1

b. H0: s = 10; H1: s > 10 a. There is not sufficient evidence to support the claim that the mean weight of fruit flies is less than 1 mg. b. There is sufficient evidence to support the claim that the standard deviation of IQ scores of biostatistics students is greater than 10. c. There is sufficient evidence to warrant rejection of the claim that the proportion of adults who are lefthanded is equal to 0.1.

H : p = 0.204; H : p ¹ 0.204; Test statistic: z = 0

1

124 - 0.204

572

= 0.76;

(0.204)(0.796) 572

P-value = 2×P(z > 0.76) = 0.4472 (Tech: 0.4480); Critical values: z = ±1.96; Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim that the rate of smoking by adult males is now the same as in 2008. The smoking rate appears to be about the same.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy Quick Quiz 23 4.

The data appear to follow a normal distribution.

H0: m = 25 mg; H1: m ¹ 25 mg; Test statistic: t =

24.89 - 25

= - 0.744; Critical values: t = ±2.145; 0.590 15 P-value = 0.4694 (Table: P-value > 0.10);

5.

Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim that the pills come from a population in which the mean amount of atorvastatin is equal to 25 mg.

The data cannot be verified to have a normal distribution, but n > 30.

H0: m = 5.4 million cells per microliter; H1: m < 5.4 million cells per microliter; Test statistic: t =

4.932 - 5.4

= - 5.873; Critical value: t = - 2.426; 0.504 40 P-value = 0.0000 (Table: P-value < 0.005);

6.

Reject H 0. There is sufficient evidence to support the claim that the sample is from a population with a mean less than 5.4 million cells per microliter. The test deals with the distribution of sample means, not individual values, so the result does not suggest that each of the 40 males has a red blood cell count below 5.4 million cells per microliter. n = 16 < 30; Must assume the sample data meet the requirement of having a normal distribution.

H0: m = 0.21; H1: m ¹ 0.21; 0.38 - 0.21

= 0.500; Critical values: t = ±2.131; 1.36 16 P-value = 0.6243 (Table: P-value < 0.10);

Test statistic: t =

7.

Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim that the sample is from a population with a mean test score equal to the value of 0.21 from those not given a treatment. It does not appear that the treatment has an effect.

Must assume the sample data meet the requirement of having a normal distribution.

2 (n - 1)s2 = (16 - 1)1.36 = 9.948; H0: s = 1.67; H1: s ¹ 1.67; Test statistic: c 2 = s 2 1.672 P-value = 0.3540 (Table: P-value > 0.20); Critical values: c 2 = 6.262 and c 2 = 27.488;

8.

9.

10.

Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim that the sample is from a population with a standard deviation equal to the value of 1.67 from those not given a treatment. It does not appear that the treatment has an effect on the variation of strategic test scores.

Answers vary, but the following is typical: Create a column of 572 ones and zeros, with 0.204 of them being 1 (as assumed in the null hypothesis). The sample proportion is 124 / 572 = 0.217, so in this two-tailed case, the values “at least as extreme” as 0.217 are those that are 0.217 or higher, or those that are 0.191 or lower.

Resample 1000 times to find that 197 of the results have a sample proportion of 0.217 or higher, and there are 231 that are 0.191 or lower. So among the 1000 results, there are 197 + 231 = 428 that are at least as extreme as 0.217. There is a high likelihood of getting a sample proportion that is at least as extreme as the one obtained, so there is not sufficient evidence to warrant rejection of the claim that the rate of smoking by adult males is now the same as in 2008. a. A type I error is the mistake of rejecting a null hypothesis when it is actually true. A type II error is the mistake of failing to reject a null hypothesis when in reality it is false. b. Type I error: In reality, the mean test score is equal to 0.21, but we reject the claim that it is 0.21. Type II error: In reality, the mean test score is different from 0.21, but we fail to reject the claim that it is equal to 0.21. a. false d. false b. true e. false c. false

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy Cumulative Review Exercises 1.

a.

x=

46 + 51+ 44 + 51+

b. Q2 = c. s =

2.

3.

4.

32 + 34

+ 45 + 27 + 34 + 29 + 20

= 33.0 deaths 2 (45 - 34.8)2 + (51- 34.8)2 +

+ 28 + 40 +16 + 20

+ (16 - 34.8)2 + (20 - 34.8)2

20 - 1

= 34.8 deaths

= 10.7 deaths

d. s2 = 10.7112 = 114.7 deaths2 e. range = 51- 16 = 35.0 deaths f. The pattern of the data over time is not revealed by the statistics. A time-series graph would be very helpful in understanding the pattern over time. The data appear to be trending downward. a. ratio b. discrete c. quantitative d. No, the data are from recent and consecutive years, so they are not randomly selected.

10.7 x ±t s = 34.8 ± 2.861× Þ 27.9 deaths < m < 41.6 deaths a. 99% CI: a /2 n 20 b. We have 99% confidence that the limits of 27.9 deaths and 41.6 deaths contain the value of the population mean. c. No. The data appear to be trending downward. Instead of having a stable population, the population characteristics appear to be changing over time.

H0: m = 72.6 deaths; H1: m < 72.6 deaths;

34.8 - 72.6

= - 15.804; Critical value: t = - 2.539; 10.7 20 P-value = 0.0000 (Table: P-value < 0.005);

Test statistic: t =

5.

6.

Reject H 0. There is sufficient evidence to support the claim that the mean number of annual lightning deaths is now less than the mean of 72.6 deaths from the 1980s. Possible factors: Shift in population from rural to urban areas; better lightning protection and grounding in electric and cable and phone lines; better medical treatment of people struck by lightning; fewer people use phones attached to cords; better weather predictions.

Because the vertical scale starts at 50 and not at 0, the difference between the number of males and the number of females is exaggerated, so the graph is deceptive by creating the false impression that males account for nearly all lightning strike deaths. A comparison of the numbers of deaths shows that the number of male deaths is roughly 4 times the number of female deaths, but the graph makes it appear that the number of male deaths is around 12 times the number of female deaths.

242 - 0.5 306 H0: p = 0.5; H1: p > 0.5; Test statistic: z = = 10.18; (0.5)(0.5) 306

P-value = P(z > 10.18) = 0.0001 (Tech: 0.0000); Critical value: z = 2.33; Reject H 0. There is sufficient evidence to support the claim that the proportion of male deaths is greater than 1 2. More males are involved in certain outdoor activities such as construction, fishing, and golf.

7.

( )( )

242 64 242 p̂q̂ = ___ ±1.96 306 306 Þ 0.745 < p < 0.836; Because the entire confidence n 306 306 interval is greater than 0.5, it does not seem feasible that males and females have equal chances of being killed by lightning. The confidence interval indicates that the proportion of male lightning deaths is substantially greater than 0.5.

95% CI: pˆ ± za /2

8.

a. 0.8×0.8×0.8 = 0.512 b. 0.2×0.2×0.2 = 0.008 c. 1- 0.2×0.2×0.2 = 0.992 d. 5C3 ×0.83 ×0.22 = 0.205 e.

m = np = 50×0.8 = 40 males; s =

npq = 50×0.8×0.2 = 2.8 males

f. Yes, using the range rule of thumb, values above m + 2s = 40.0 + 2(2.8) = 45.6 are considered significantly high. Since 46 is greater than 45.6, 46 males would be a significantly high number in a group of 50. Using probabilities, P (46 or more male victims among 50) = 50C46 ×0.846 ×0.24 + + 50C50 ×0.850 ×0.20 = 0.0185, which is very small, suggesting that 46 is significantly high.

331 4489 9. OR = = 1.14; The odds in favor of experiencing migraine headaches is about 1.14 times higher 358 5487 for overweight people than for people with normal weight.

10. Answers vary, but the following is typical: Create a column of 306 ones and zeros, with half of them being 1 (as assumed in the null hypothesis). The sample proportion is 242 / 306 = 0.791, so in this right-tailed case, the values “at least as extreme” as 0.791 are those that are 0.791 or higher. Resample 1000 times to find that none of the results have a sample proportion of 0.791 or higher. So among the 1000 results, there are none that are at least as extreme as 0.791. There is a very small likelihood of getting a sample proportion that is at least as extreme as the one obtained, so there is sufficient evidence to support the claim that the proportion of male deaths is greater than 1/2 or 0.5.

Chapter 9: Inferences from Two Samples Section 9-1: Two Proportions 1. The samples are simple random samples that are independent. For each of the two groups, the number of successes is at least 5 and the number of failures is at least 5. (Depending on what we call a success, the four numbers are 33, 115, 201,229 and 200,630 and all of those numbers are at least 5.) The requirements are satisfied.

33 = 0.000163992, qˆ = 1- 0.000163992 = 0.999836; 2. n = 201,229, pˆ = 1 1 1 201,299 n = 200,745, pˆ = 115 = 0.000572866, qˆ = 1- 0.000572866 = 0.999427; 2 2 2 200, 745 33 +115 p= = 0.000368183, q = 1- 0.000368183 = 0.999632 201, 299 + 200, 745 3.

a. H0: p1 = p2; H1: p1 < p2 b. There is sufficient evidence to support the claim that the rate of polio is less for children given the Salk vaccine than it is for children given a placebo. The Salk vaccine appears to be effective.

4.

a. The P-value method and the critical value method are equivalent in the sense that they will always lead to the same conclusion, but the confidence interval method is not equivalent to them. b. 0.90, or 90% c. Because the confidence interval limits do not contain 0, there appears to be a significant difference between the two proportions. Because the confidence interval consists of negative values only, it appears that the first proportion is less than the second proportion. There is sufficient evidence to support the claim that the rate of polio is less for children given the Salk vaccine than it is for children given a placebo.

5.

H0: p1 = p2; H1: p1 > p2; population1 = vinyl gloves, population2 = latex gloves;

Test statistic: z = 12.82; P-value = 0.0000; Critical value: z = 2.33; Reject H 0. There is sufficient evidence to support the claim that vinyl gloves have a greater virus leak rate than latex gloves.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

6.

H0: p1 = p2; H1: p1 > p2; population1 = surgery, population2 = splints;

Test statistic: z = 3.12; P-value = 0.0009; Critical value: z = 2.33; Reject H 0. There is sufficient evidence to support the claim that the success rate is better with surgery.

For Exercises 7–22, assume that the data fit the requirements for the statistical methods for two proportions unless otherwise indicated.

7.

a.

H0: p1 = p2; H1: p1 ¹ p2; population1 = vaccine, population2 = placebo;

Test statistic: z = - 11.79; P-value = 0.0000 (Table: 0.0001); Critical value: z = - 2.33; Reject H 0. There is sufficient evidence to support the claim that the rate of COVID-19 in the treatment group is less than the rate in the placebo group.

8 +162 170 ; q = 1- 170 36, 353 p= = = 18,198 +18, 325 36, 523 36, 523 36, 523 8 - 162 - 0 ( pˆ - pˆ ) - ( p - p ) 18,198 18,325 2 1 2 z= 1 = 36,353 36,353 170 170 pq pq + 36,523 36,523 36,523 36,523 + n1 n2 18,198 18, 325

(

(

)(

)

) (

)(

)

b. 98% CI: - 0.010 < p1 - p2 < - 0.007; Because the confidence interval limits do not include 0, there does appear to be a significant difference between COVID-19 rates for those treated with the vaccine and those given a placebo. There is sufficient evidence to support the claim that the rate of COVID-19 in the treatment group is less than the rate in the placebo group.

162 ö æ 8 8 8 162 1- 18,198 ( p̂ - pˆ )± z =ç 18,325 1÷ ± 2.326 + 1 2 a /2 è18,198 18, 325 ø 18,198 18,325

7.

(continued) c. The sample results suggest that the Pfizer-BioNTech vaccine appears to be effective in preventing COVID19.

8.

a. H0: p1 = p2; H1: p1 ¹ p2; population1 = inhaled anesthetics, population2 = intravenous anesthetics; Test statistic: z = - 0.37; P-value = 0.7121 (Table: 0.7114); Critical values: z = ±1.96; Fail to reject H 0.

There is not sufficient evidence to warrant rejection of the claim that the two treatments have the same rate of deaths. There does not appear to be a significant difference in the two rates of death.

75 + 79 77 ; q = 1- 77 2623 p= = = 2709 + 2691 2700 2700 2700 75 - 79 - 0 ( pˆ - pˆ ) - ( p - p ) 1 2 1 2 2709 2691 z= = 77 2623 77 2623 pq pq + 2700 2700 2700 2700 + n1 n2 2709 2691

( ) ( )( ) ( )( )

b. 95% CI: - 0.011 < p1 - p2 < 0.007; Because the confidence interval limits include 0, there is not sufficient evidence to warrant rejection of the claim that the two treatments have the same rate of deaths. There does not appear to be a significant difference in the two rates of death.

79 ö æ 75 ˆ pˆ ) ± z 75 2709 ( p1 - 2 a / 2 =ç - ____ ÷±1.96 2709 + 2691 1- 2691 è 2709 2691ø 2709 2691

) ( )(

9.

a.

H0: p1 = p2; H1: p1 < p2; population1 = used left ear, population2 = used right ear;

Test statistic: z = - 7.94; P-value = 0.0000 (Table: 0.0001); Critical value: z = - 2.33; Reject H 0. There is sufficient evidence to support the claim that the rate of right-handedness for those who prefer to use their left ear for cell phones is less than the rate of right-handedness for those who prefer to use their right ear for cell phones.

166 + 436 301 301 33 p= = ; q = 1= ; 216 + 452 334 334 334 166 - 436 - 0 ( pˆ - pˆ ) - ( p - p ) 1 2 1 2 216 452 z= = pq pq 301 33 301 33 + 334 334 + 334 334 n1 n2 216 452

(

)

( )( ) ( )( )

b. 98% CI: - 0.266 < p1 - p2 < - 0.126; Because the confidence interval limits do not contain 0, there is a significant difference between the two proportions. Because the interval consists of negative numbers only, it appears that the claim is supported. The difference between the populations does appear to have practical significance. æ166 436 ö ˆ pˆ )± z 166 436 436 1- 166 216 452 ( p1 - 2 a /2 =ç - ___ ÷± 2.33 216 + 452 è 216 452 ø 216 452 10. a.

H0: p1 = p2; H1: p1 < p2; population1 = single bill, population2 = multiple bills;

Test statistic: z = - 1.85; P-value = 0.0324 (Table: P-value = 0.0322); Critical value: z = - 1.645; Reject H 0. There is sufficient evidence to support the claim that when given a single large bill, a smaller

proportion of adults in China spend some or all of the money when compared to the proportion of adults in China given the same amount in smaller bills.

60 + 68 64 64 11 p= = ; q = 1= 75 + 75 75 75 75 60 - 68 - 0 ( pˆ - pˆ ) - ( p - p ) 2 1 2 75 75 z= 1 = 64 11 64 11 pq pq + 75 75 75 75 + n1 n2 75 75

(

)

( )( ) ( )( )

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

10. (continued) b. 90% CI: - 0.201 < p1 - p2 < - 0.0127; Because the confidence interval does not include 0 and it includes only negative values, it appears that the first proportion is less than the second proportion. There is sufficient evidence to support the claim that when given a single large bill, a smaller proportion of adults in China spend some or all of the money when compared to the proportion of adults in China given the same amount in smaller bills. æ60 68 ö ˆ pˆ ) ± z 68 60 60 68 ( p1 - 2 a / 2 pˆ1qˆ1 + pˆ2qˆ2 = ç - ___÷±1.645 75 1- 75 + 75 1- 75 è 75 75 ø 75 75 c. Because the P-value = 0.0324 (Table: 0.0322), the difference is significant at the 0.05 significance level, but not at the 0.01 significance level. The conclusion does change.

11. a.

H0: p1 = p2; H1: p1 > p2; population1 = over age 55, population2 = under age 25;

Test statistic: z = 6.44; P-value = 0.0000 (Table: 0.0001); Critical value: z = 2.33; Reject H 0. There is sufficient evidence to support the claim that the proportion of people over 55 who dream in black and white is greater than the proportion of those under 25.

68 +13 = 81 ; q = 1- 81 523 p= = 306 + 298 604 604 604 68 - 13 - 0 ( pˆ1 - pˆ2 ) - ( p1 - p2 ) 306 298 z= = 81 523 81 523 pq pq + 604 604 604 604 + n1 n2 306 298

( ) ( )( ) ( )( )

b. 98% CI: 0.117 < p1 - p2 < 0.240; Because the confidence interval limits do not include 0, it appears that the two proportions are not equal. Because the confidence interval limits include only positive values, it appears that the proportion of people over 55 who dream in black and white is greater than the proportion of those under 25.

13 ö æ 68 ˆ pˆ ) ± z 68 13 68 13 298 ( p1 - 2 a / 2 pˆ1qˆ1 + pˆ2qˆ2 = ç - ____÷± 2.33 306 1- 306 + 298 è306 298 ø 306 298 c. The results suggest that the proportion of people over 55 who dream in black and white is greater than the proportion of those under 25, but the results cannot be used to verify the cause of that difference.

12. a. H0: p1 = p2; H1: p1 ¹ p2; population1 = OxyContin, population2 = placebo; Test statistic: z = 1.78; P-value = 0.0757 (Table: P-value = 0.0750); Critical values: z = ±1.96; Fail to reject H 0. There is not sufficient evidence to support the claim that there is a difference between the rates of nausea for those treated with OxyContin and those given a placebo.

52 + 5 = 57 ; q = 1- 57 215 p= = 227 + 45 272 272 272 52 - 5 - 0 ( pˆ1 - pˆ2 ) - ( p1 - p2 ) 227 45 z= = pq pq 57 215 57 215 + 272 272 + 272 272 n1 n2 227 45 b. 95% CI: 0.0111 < p1 - p2 < 0.225; Because the confidence interval limits do not contain 0, there is a significant difference between the two proportions.

5 ö æ 52 ˆ pˆ ) ± z 5 52 52 5 1- 227 1- 45 pˆ qˆ 1 1 ( p1 - 2 a / 2 =ç - ___ ÷±1.96 227 + + 45 è 227 45 ø 45 227

(

)

( )( ) ( )( )

c. The conclusions from parts (a) and (b) are different. The conclusion from part (a) results from a hypothesis test instead of an estimate of the difference between the two rates of nausea, so there does not appear to be sufficient evidence to conclude that there is a difference between the rates of nausea for those treated with OxyContin and those given a placebo.

13. a. H0: p1 = p2; H1: p1 > p2; population1 = wearing seatbelt, population2 = not wearing seatbelt; Test statistic: z = 6.11; P-value = 0.0000 (Table: 0.0001); Critical value: z = 1.645; Reject H 0. There is sufficient evidence to support the claim that the fatality rate is higher for those not wearing seat belts.

31+16 47 ; q = 1- 47 10, 541 p= = = 2823 + 7765 10, 588 10, 588 10, 588 31 - 16 - 0 ( pˆ - pˆ ) - ( p - p ) 2 1 2 2823 7765 z= 1 = 10,541 10,541 47 47 pq pq + 10,588 10,588 10,588 10,588 + n1 n2 2823 7765 b. 90% CI: 0.00559 < p1 - p2 < 0.0123; Because the confidence interval limits do not include 0, it appears that the two fatality rates are not equal. Because the confidence interval limits include only positive values, it appears that the fatality rate is higher for those not wearing seat belts. æ 31 - 16 ö ±1.645 31 ( pˆ - p̂ ) ± z 1 pˆ1qˆ1 pˆ2qˆ2 = 2823 28 23 2 a /2 1 ç ÷ + + 7765 1- 7765 è 2823 7765 ø 2823 7765

(

(

)(

)

) (

)(

)

) ( )(

c. The results suggest that the use of seat belts is associated with fatality rates lower than those associated with not using seat belts.

14. a.

H0: p1 = p2; H1: p1 < p2; population1 = text warnings, population2 = picture warnings;

Test statistic: z = - 2.89; P-value = 0.0019; Critical value: z = - 2.33; Reject H 0 . There is sufficient evidence to support the claim that the proportion of smokers who tried to quit in the text warning group is less than the proportion in the picture warning group.

366 + 428 = 794 ; q = 1- 794 1355 p= = 1078 +1071 2149 2149 2149 366 428 ( pˆ1 - p̂ 2 )- ( p1 - p2 ) = 1078 - 1071 - 0 z= 794 1355 794 1355 pq pq + 2149 2149 + 2149 2149 n1 n2 1078 1071 b. 98% CI: - 0.108 < p1 - p2 < - 0.0118; Because the confidence interval limits do not contain 0, there is a significant difference between the two sample proportions. Because the confidence interval consists of negative values only, it appears that the proportion of smokers who tried to quit in the text warning group is less than the proportion in the picture warning group.

428 ö æ366 ˆ pˆ ) ± z 428 366 366 428 1- 1078 1071 ( p1 - 2 a / 2 =ç - ____ ÷± 2.326 1078 + 1071 è1078 1071ø 1071 1078

) ( ( )( ) ( )( )

15. a.

H0: p1 = p2; H1: p1 ¹ p2; population1 = malaria, population 2 = no malaria; Test statistic: z = - 4.41; P-value = 0.00001 (Table: P-value = 0.0002); Critical values: z = ±1.96; Reject H 0. There is sufficient evidence to reject the claim of no difference between the two rates of correct responses. It appears that the dogs do a better job of correctly identifying subjects without malaria.

123 +131 127 127 33 p= = ; q = 1= 175 +145 160 160 160 123 131 ( pˆ1 - pˆ2 )- ( p1 - p2 ) = 175 - 145 - 0 z= 127 33 127 33 pq pq + 160 160 160 160 + n1 n2 175 145

) ( ( )( ) ( )( )

b. 95% CI: - 0.284 < p1 - p2 < - 0.118; Because the confidence interval limits do not contain 0, there is a significant difference between the two sample proportions. Because the confidence interval consists of negative values only, it appears that the proportion of correct identifications made with subjects having malaria is less than the proportion of correct identifications made with subjects not having malaria.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

15. (continued) ˆ

( p1 -

pˆ )± z 2

a /2

pˆ1qˆ1

+

æ123 131ö =ç - ___ ÷±1.96 è175 145 ø

123 175

123 1- 175

175

131 145

131

+ 145

145

c. With correct identification rates of 70.3% and 90.3%, the dogs are doing quite well for subjects with malaria and subjects without malaria, but they do a much better job of correctly identifying patients without malaria.

16. a. H0: p1 = p2; H1: p1 < p2; population1 = used bednet, population2 = did not use bednet; Test statistic: z = - 2.44; P-value = 0.0074; Critical value: z = - 2.33; Reject H 0. There is sufficient evidence to support the claim that the incidence of malaria is lower for infants who use the bednets.

15 + 27 = 6 ; q = 1- 6 85 p= = 343 + 294 91 91 91 15 - 27 - 0 ( pˆ - pˆ )- ( p - p ) 1 2 1 2 343 294 z= = 6 85 6 85 pq pq + 91 91 91 91 + n1 n2 343 294 b. 98% CI: - 0.0950 < p1 - p2 < - 0.00125 (Table: - 0.0950 < p1 - p2 < - 0.00118); Because the confidence interval does not include 0 and it includes only negative values, it appears that the rate of malaria is lower for infants who use the bednets.

27 ö æ15 pˆ - pˆ ) ± z 15 27 15 27 294 ( 1 2 a /2 pˆ1qˆ1 + pˆ2qˆ2 = ç - ___ ÷± 2.33 343 1- 343 + 294 è343 294 ø 294 343

(

)

( )( ) ( )( )

c. The bednets appear to be effective.

17. a.

H0: p1 = p2; H1: p1 > p2; population1 = Music, population2 = Blank tape

Test statistic: z = 3.60; P-value = 0.0000 (Table: P-value = 0.0002); Critical value: z = 1.645; Reject H 0. There is sufficient evidence to support the claim that more of those given a treatment of music and soothing talk require no opioid use when compared to those who listen to a blank tape.

70 + 39 109 109 276 p= = ; q = 1= 191+194 385 385 385 70 - 39 - 0 ( pˆ - pˆ ) - ( p - p ) 1 2 1 2 191 194 z= = pq pq 109 276 109 276 + 385 385 + 385 385 n1 n2 191 194

(

)

( )( ) ( )( )

b. 90% CI: 0.091 < p1 - p2 < 0.240 Because the confidence interval consists of positive values only, it does appear that there is sufficient evidence to support the claim that more of those given a treatment of music and soothing talk require no opioid use when compared to those who listen to a blank tape. æ70 39 ö ˆ pˆ ) ± z 39 70 70 39 1- 191 194 ( p1 - 2 a / 2 =ç - ___ ÷± 1.645 191 + 194 è191 194 ø 194 191 18. a.

H0: p1 = p2; H1: p1 > p2; population1 = females, population2 = males; Test statistic: z = 11.35; P-value = 0.0000 (Table: P-value = 0.0002); Critical values: z = ±2.576 (Table: z = ±2.575); Reject H 0. There is sufficient evidence to support the claim that female survivors and male

survivors have different rates of thyroid diseases.

1397 + 436 1833 1833 2258 p= = ; q = 1= 2739 +1352 4091 4091 4091 1397 - 436 - 0 ( pˆ - pˆ ) - ( p - p ) 1 2 1 2 2739 1352 z= = pq pq 1833 2258 1833 2258 + 4091 4091 + 4091 4091 n1 n2 2739 1352

(

)

( )( ) ( )( )

18. (continued) b. 90% CI: 0.147 < p1 - p2 < 0.229 (Table: 0.147 < p1 - p2 < 0.228); Because the confidence interval limits do not contain 0, there is a significant difference between the two proportions. Because the confidence interval includes only positive values, it appears that the rate of thyroid diseases for females is greater than the rate for males. æ1397 436 ö pˆ - pˆ ) ± z 1397 436 436 1- 1397 2739 1352 ( 1 2 a / 2 pˆ1qˆ1 + pˆ2qˆ2 = ç - ____ ÷± 2.576 2739 + 1352 è 2739 1352 ø 2739 1352 19. a.

H0: p1 = p2; H1: p1 ¹ p2; population1 = women, population2 = men;

Test statistic: z = - 2.63; P-value = 0.0085 (Table: P-value = 0.0086); Critical values: z = ±2.575; Reject H 0. There is sufficient evidence to reject the claim that the proportions of blue eyes are the same for females and males.

729 370 + 359 ; q = 1- 729 = 1297 p= = 1107 + 919 2026 2026 2026 370 359 pˆ1 - p̂ 2 ) - ( p1 - p2 ) ( 1107 - 919 - 0 z= = 729 1297 729 1297 pq pq + 2026 2026 + 2026 2026 n1 n2 1107 919 b. 99% CI: - 0.112 < p1 - p2 < - 0.00116; Because the confidence interval limits do not contain 0, there appears to be a significant difference between the two sample proportions. Because the confidence interval consists of negative values only, it appears that the proportion of blue eyes among females is less than the proportion of blue eyes among males. æ370 359 ö ˆ pˆ )± z 359 370 370 359 1- 1107 919 ( p1 - 2 a / 2 =ç - ___ ÷± 2.575 1107 + 919 è1107 919 ø 919 1107

) ( ( )( ) ( )( )

c. The professors used a convenience sample of their students. Convenience samples are typically problematic for making inferences about population parameters, but in this case there isn’t anything about eye color that would seem to make someone more or less likely to be a student in a statistics class, so the data do not appear to have the typical bias of a convenience sample.

20. a.

H0: p1 = p2; H1: p1 ¹ p2; population1 = women, population2 = men;

Test statistic: z = 0.12; P-value = 0.9074 (Table: P-value = 0.9044); Critical values: z = ±2.575; Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim that the proportions of brown eyes are the same for females and males.

321 352 + 290 ; q = 1- 321 = 692 p= = 1107 + 919 1013 1013 1013 352 290 pˆ1 - pˆ2 )- ( p1 - p2 ) ( 1107 - 919 - 0 z= = 321 692 321 692 pq pq + 1013 1013 + 1013 1013 n1 n2 1107 919

) ( ( )( ) ( )( )

b. 99% CI: - 0.0511 < p1 - p2 < 0.0559; Because the confidence interval limits do contain 0, there is not a significant difference between the two sample proportions. It appears that proportions of brown eyes are the same for females and males. æ352 290 ö ˆ pˆ ) ± z 290 352 352 290 1- 1107 pˆ qˆ pˆ qˆ 1 1 2 2 919 ( p1 - 2 a / 2 =ç - ___ ÷± 2.575 1107 + + 919 è1107 919 ø 919 1107 c. The professors used a convenience sample of their students. Convenience samples are typically problematic for making inferences about population parameters, but in this case there isn’t anything about eye color that would seem to make someone more or less likely to be a student in a statistics class, so the data do not appear to have the typical bias of a convenience sample.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

21. a.

H0: p1 = p2; H1: p1 < p2; population1 = male, population2 = female;

Test statistic: z = - 1.17; P-value = 0.1214 (Table: 0.1210); Critical value: z = - 2.33; Fail to reject H 0.

There is not sufficient evidence to support the claim that the rate of left-handedness among males is less than that among females.

23 + 65 = 11 ; q = 1- 11 = 84 p= 240 + 520 95 95 95 23 - 65 - 0 ( pˆ - pˆ ) - ( p - p ) 1 2 1 2 240 520 z= = 11 84 11 84 pq pq + 95 95 95 95 + n1 n2 240 520

(

)

( )( ) ( )( )

b. 98% CI: - 0.0848 < p1 - p2 < - 0.0264 (Table: - 0.0849 < p1 - p2 < - 0.0265); Because the confidence interval limits include 0, there does not appear to be a significant difference between the rate of lefthandedness among males and the rate among females. There is not sufficient evidence to support the claim that the rate of left-handedness among males is less than that among females.

65 ö æ 23 pˆ - pˆ ) ± z 23 65 23 65 520 ( 1 2 a / 2 pˆ1qˆ1 + pˆ2qˆ2 = ç - ____ ÷± 2.33 240 1- 240 + 520 è 240 520 ø 520 240 c. The rate of left-handedness among males does not appear to be less than the rate of left-handedness among females.

22. a. H0: p1 = p2; H1: p1 > p2; population1 = helicopter, population2 = ground; Test statistic: z = 10.75; P-value = 0.0000 (Table: P-value = 0.0001); Critical value: z = 2.33; Reject H 0.

There is sufficient evidence to support the claim that the rate of fatalities is higher for patients transported by helicopter.

7813 +17, 775 25,588 25, 588 197,887 p= = ; q = 1= 61,909 +161,566 223, 475 223,1477,7575 223, 475 -0 7813 ( pˆ - pˆ ) - ( p - p ) z=

1

2

1

pq pq + n1 n2

2

=

( )(

61,909

(

25,588 223,475

161,566

) +(

197,887 223,475

61,909

)

)(

25,588 223,475

197,887 223,475

)

161,566

b. 98% CI: 0.0126 < p1 - p2 < 0.0198; Because the confidence interval limits do not contain 0, there is a significant difference between the two proportions. Because the entire range of values in the confidence interval consists of positive numbers, it appears that the rate of fatalities is higher for patients transported by helicopter.

17,775 17,775 7813 7813 ( p̂ - pˆ )± z æ 7813 17, 775 ö 1- 61,909 161,566 61,909 161,566 =ç ± 2.33 + ÷ 1 2 a /2 è61, 909 161, 566 ø 61,909 161,566

( )(

)(

)(

c. The fatality rate for the helicopter sample is 0.126, or 12.6%, and the fatality rate for the ground services sample is 0.110, or 11.0%. The large sample sizes result in a significant difference, but it does not appear that the difference has very much practical significance. Also, it is possible that the most serious of the serious traumatic injuries led to helicopter transportation, and that could partly explain the higher rate of fatalities with helicopter transportation. z2 1.962 23. n = a /2 = = 4082; The samples should include 4082 males and 4082 females.

2E2 2 ×0.022

)

24. a. The method of this section requires that both samples must have at least 5 successes and 5 failures, but the group not exposed to yawning includes a frequency of 4, which violates that requirement. b. P-value = 0.3729 (Table: 0.3745); which is not close to the P-value of 0.5128 from Fisher’s exact test.

H0: p1 = p2; H1: p1 > p2; population1 = yawning, population2 = not yawning; 7 10 + 4 ; q = 1- 7 = 18 p= = 34 +16 25 25 25 10 - 4 - 0 ( pˆ - pˆ ) - ( p - p ) 1 2 1 2 34 16 z= = 7 18 7 18 pq pq + 25 25 25 25 + n1 n2 34 16

( ) ( )( ) ( )( )

c. The P-value of 0.5128 shows that there is not sufficient evidence to support the claim that people are more likely to yawn when they are exposed to yawning. The results of the experiment do not support the common belief that yawning is contagious.

25. a. 95% CI: 0.0227 < p1 - p2 < 0.217; population1 = first sample, population2 = second sample; Because the confidence interval limits do not contain 0, it appears that p1 = p2 can be rejected. æ112 88 ö pˆ - pˆ ) ± z 112 88 88 1- 112 pˆ qˆ pˆ qˆ 1 1 2 2 200 200 ( 1 2 a /2 =ç - ____ ÷±1.96 200 + + 200 è 200 200 ø 200 200 112 ±1.96 p̂q̂ = Þ 0.491 < p1 < 0.629 b. First sample: 95% CI: pˆ ± za /2 n 200 200 88 88 88 p̂q̂ = ___ ±1.96 200 1- 200 Second sample: 95% CI: pˆ ± za /2 Þ 0.371 < p2 < 0.509 n 200 200 Because the confidence intervals do overlap, it appears that p1 = p2 cannot be rejected.

( )(

)

c. H0: p1 = p2; H1: p1 ¹ p2; population1 = first sample, population2 = second sample; Test statistic: z = 2.40; P-value = 0.0164; Critical values: z = ±1.96; Reject H 0. There is sufficient evidence to reject p1 = p2.

112 + 88 = 1 ; q = 1- 1 = 1 p= 200 + 200 2 2 2 112 - 88 - 0 ( pˆ - pˆ ) - ( p - p ) 1 2 1 2 200 200 z= = 1 1 1 1 pq pq + 2 2 + 2 2 n1 n2 200 200

( ) ( )( ) ( )( )

d. Reject p1 = p2. The least effective method is using the overlap between the individual confidence intervals.

26. Hypothesis test: H0: p1 = p2 ; H1: p1 ¹ p2 ; population1 = first sample, population2 = second sample; Test statistic: z = - 1.9615; P-value = 0.0498 (Table: P-value = 0.05); Critical values: z = ±1.96; Reject H 0.

There is sufficient evidence to reject p1 = p2.

7 10 +1404 7 3 p= = ; q = 1= 20 + 2000 10 10 10 10 - 1404 - 0 ( pˆ - pˆ ) - ( p - p ) 2 1 2 20 2000 z= 1 = pq pq 7 103 7 3 + 10 + 10 10 n1 n2 20 2000

(

)

( )( ) ( )( )

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

26. (continued) 95% CI: - 0.422 < p1 - p2 < 0.0180; population1 = first sample, population2 = second sample; which suggests that we should not reject p1 = p2 (because 0 is included). æ10 1404 ö ˆ pˆ ) ± z 1404 10 10 1404 2000 ( p1 - 2 a / 2 pˆ1qˆ1 + pˆ2qˆ2 = ç - ____ ÷±1.96 20 1- 20 + 2000 è 20 2000 ø 20 2000 The hypothesis test and confidence interval lead to different conclusions about the equality of p1 = p2.

27. H0: p1 = p2; H1: p1 ¹ p2; population1 = ANSUR I 1998, population2 = ANSUR I 2012; Test statistic: z = - 22.59; P-value = 0.0000 (Table: P-value = 0.0002); Critical values: z = ±2.575; Reject H 0. There is sufficient evidence to warrant rejection of the claim that the proportion of males in the sample in

Data Set 2 “ANSUR I 1988” is the same as the proportion of males in the sample in Data Set 3 “ANSUR II 2012.” The proportion of males in ANSUR II 2012 appears to be significantly higher than in ANSUR I 1988.

The 99% CI is - 0.253 < p1 - p2 < - 0.202. (TI data: With sample proportions of 227/500 and 319/500, test statistic is z = - 5.84, P-value = 0.0000, and the confidence interval is - 0.264 < p1 - p2 < - 0.104.)

1774 + 4082 976 976 699 p= = ; q = 1= 3982 + 6068 1675 1675 1675 z=

( pˆ1 - p̂ 2 ) - ( p1 - p2 ) = pq pq + n1 n2

pˆ - pˆ )± z

(1

2

a /2

4082 (1774 3982 - 6068 - )0 976 699 (1675 )(1675 ) +( 9761675)( 699 1675)

3982 6068 æ1774 4082 ö pˆ1qˆ1 pˆ2qˆ2 = - ____ ÷± 2.575 + ç è3982 6068 ø

1774 3982

1- 1774 3982 3982

4082 6068

4082

+ 6068

6068

Section 9-2: Two Means: Independent Samples 1. Parts (b) and (c) describe independent samples.

2. a. Because the confidence interval does not include 0, it appears that there is a significant difference between the mean pulse rate in females and the mean pulse rate in males. b. We have 95% confidence that the interval from 1.7 bpm to 7.2 bpm actually contains the value of the difference between the two population means (m1 - m2 ). c. - 7.2 bpm < m1 - m2 < - 1.7 bpm 3.

4.

5.

a. yes b. yes c. 98% The critical value of t =- 1.796 is more conservative than t =- 1.727 in the sense that rejection of the null hypothesis requires a greater difference between the sample means. The sample evidence must be stronger with t =- 1.796.

H0: m1 = m2; H1: m1 ¹ m2; population1 = Not Breastfed, population2 = Breastfed Test statistic: t = - 2.867; P-value = 0.0048 (Table: P-value < 0.01); Critical values: t = ±2.614 (Table: t = ±1.994); Reject H 0. There is sufficient evidence to warrant rejection of the claim that IQ scores of 11-

year-old children are about the same for those who were not breastfed and those who were breastfed. There does appear to be a difference. The given data and the hypothesis test result do not provide sufficient evidence to address the conclusion of the study. (x1 - x2 )- (m1 - m2 ) = (103.7 - 108.5)- 0 t= s12 s22 12.02 12.92 + + 72 207

6.

H0: m1 = m2; H1: m1 ¹ m2; population1 = Not Breastfed, population2 = Breastfed Test statistic: t = - 2.462; P-value = 0.0149 (Table: P-value > 0.01); Critical values: t = ±2.608 (Table: t = ±2.374); Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim that

breastfeeding has no effect on IQ scores. It appears that breastfeeding does not have an effect on IQ scores of 4-year-old children. The conclusion changes if the significance level is changed from 0.01 to 0.05. (x1 - x2 )- (m1 - m2 ) = (100.9 - 105.3)- 0 t= s12 s22 14.02 14.52 + + 85 237 7.

a. H0: m1 = m2; H1: m1 > m2; population1 = Washing with soap, population2 = Rubbing with alcohol Test statistic: t = 2.095; P-value = 0.0196 (Table: P-value > 0.01); Critical value: t = 2.372 (Table: t = 2.403); Fail to reject H 0. There is not sufficient evidence to support the claim that rubbing with alcohol results in a lower bacteria count. t=

(x1 - x2 )- (m1 - m2 ) = (69 - 35)- 0 s12

s2 + 2

1062 + 592 55 59

b. 98% CI: - 4 < m1 - m2 < 72 (Table: - 5 < m1 - m2 < 73)

(x1 - x2 )± ta /2

s12

s2 1062 592 + 2 = (69 - 35)± 2.403 + 55 59

(df = 55 - 1 = 54)

c. Yes, the null hypothesis is rejected with a 0.05 significance level, so the conclusion changes to this: There is sufficient evidence to support the claim that rubbing with alcohol results in a lower bacteria count.

8.

a. H0: m1 = m2; H1: m1 < m2; population1 = females, population2 = males Test statistic: t = - 3.450; P-value = 0.0003 (Table: P-value < 0.005); Critical value: t = - 2.336 (Table: t = - 2.345); Reject H 0. There is sufficient evidence to support the claim that at birth, females have a lower mean weight than males.

t=

(x1 - x2 )- (m1 - m2 ) = (3037.1- 3272.8)- 0 s12

s2 + 2

706.32 660.22 + 205 195 b. 98% CI: - 395.3 g < m1 - m2 < - 76.1 g (Table: - 395.5 g < m1 - m2 < - 75.5 g)

(x1 - x2 )± ta /2 9.

s12

s2 706.32 660.22 + 2 = (3037.1- 3272.8)± 2.345 + 205 195

(df = 195 - 1 = 194)

a. H0: m1 = m2; H1: m1 ¹ m2; population1 = red background, population2 = blue background Test statistic: t = 2.647; P-value = 0.0101 (Table: P-value < 0.02); Critical values: t = ±1.995 (Table: t = ±2.032); Reject H 0. There is sufficient evidence to warrant rejection of the claim that the samples are

from populations with the same mean. Color does appear to have an effect on word recall scores. Red appears to be associated with higher word memory recall scores. (x1 - x2 )- (m1 - m2 ) = (15.89 - 12.31)- 0 t= s12 s22 5.902 5.482 + + 35 36

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

9.

(continued) b. 95% CI: 0.88 < m1 - m2 < 6.28 (Table: 0.83 < m1 - m2 < 6.33); The confidence interval includes positive numbers only, so the mean score with a red background appears to be greater than the mean score with a blue background.

(x1 - x2 )± ta /2

s12

5.902 5.482 s2 + 2 = (15.89 - 12.31)± 2.032 + 35 36

(df = 35 - 1 = 34)

c. The background color does appear to have an effect on word recall scores. Red appears to be associated with higher word memory recall scores.

10. a. H0: m1 = m2; H1: m1 < m2; population1 = red background, population 2 = blue background Test statistic: t = - 2.979; P-value = 0.0021 (Table: P-value < 0.005); Critical value: t =- 2.392 (Table: t = - 2.441); Reject H 0. There is sufficient evidence to support the claim that blue enhances performance on a creative task. t=

(x1 - x2 )- (m1 - m2 ) = (3.39 - 3.97)- 0 s12

s2 + 2

0.972 0.632 + 35 36

b. 98% CI: - 1.05 < m1 - m2 < - 0.11 (Table: - 1.06 < m1 - m2 < - 0.10); The confidence interval consists of negative numbers only and does not include 0, so the mean creativity score with the red background appears to be less than the mean creativity score with the blue background. It appears that blue enhances performance on a creative task.

(x1 - x2 )± ta /2

s12

s2 0.972 0.632 + 2 = (3.39 - 3.97)± 2.032 + 35 36

(df = 35 - 1 = 34)

11. a. H0: m1 = m2; H1: m1 > m2; population1 = magnet treatment, population2 = sham treatment Test statistic: t = 0.132; P-value = 0.4480 (Table: P-value > 0.10); Critical value: t = 1.691 (Table: t = 1.729); Fail to reject H 0. There is not sufficient evidence to support the claim that the magnets are effective in reducing pain. t=

(x1 - x2 )- (m1 - m2 ) = (0.49 - 0.44)- 0 s12

s2 + 2

0.962 1.42 + 20 20 b. 90% CI: - 0.59 < m1 - m2 < 0.69 (Table: - 0.61 < m1 - m2 < 0.71); 0.962 1.42 s12 s22 + + = (0.49 - 0.44)±1.729 (df = 20 - 1 = 19) (x1 - x2 )± ta /2 20 20 c. Magnets do not appear to be effective in treating back pain. It is valid to argue that the magnets might appear to be effective if the sample sizes were larger.

12. a. H0: m1 = m2; H1: m1 > m2; population1 = exposed, population2 = not exposed Test statistic: t = 1.845; P-value = 0.0352 (Table: P-value < 0.05); Critical value: t = 1.673 (Table: t = 1.685); Reject H 0. There is sufficient evidence to support the claim that nonsmokers exposed to tobacco smoke have a higher mean cotinine level than nonsmokers not exposed to tobacco smoke. t=

(x1 - x2 )- (m1 - m2 ) = (60.58 - 16.35)- 0 s12

+

s22

2

138.082 + 62.53 40 40

12. (continued) b. 90% CI: 4.12 ng mL < m1 - m2 < 84.34 ng mL (Table: 3.85 ng mL < m1 - m2 < 84.61 ng mL); 138.082 62.532 s12 s22 + + (df = 40 - 1 = 39) = (60.58 - 16.35)±1.685 (x1 - x2 )± ta /2 40 40 c. Exposure to secondhand smoke appears to have the effect of being associated with greater amounts of nicotine than for those not exposed to secondhand smoke.

13. a. H0: m1 = m2; H1: m1 ¹ m2; population1 = Surgery, population2 = Bracing Test statistic: t = - 0.973; P-value = 0.3338 (Table: P-value > 0.20); Critical values: t = ±1.991 (Table: t = ±2.026; Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim of no difference between the two treatments. t=

(x1 - x2 )- (m1 - m2 ) = (8.9 - 12.0)- 0 s12

s2 + 2

2

14.32 + 14.5 38 44

b. 95% CI: - 9.4 < m1 - m2 < 3.2 (Table: - 9.6 < m1 - m2 < 3.4) 14.32 14.52 s12 s22 + + = (8.9 - 12.0)± 2.0262 (x1 - x2 )± ta /2 38 44

(df = 38 - 1 = 37)

c. Based on the DASH score results, there does not appear to be a difference between the two treatments.

14. a. H0: m1 = m2; H1: m1 > m2; population1 = low lead, population2 = high lead Test statistic: t = 2.282; P-value = 0.0132 (Table: P-value < 0.05); Critical value: t = 1.673 (Table: t = 1.725); Reject H 0. There is sufficient evidence to support the claim that the mean IQ score of people with low blood lead levels is higher than the mean IQ score of people with high blood lead levels. t=

(x1 - x2 )- (m1 - m2 ) = (92.88462 - 86.90476)- 0 s12

s2 + 2

15.344512 78

+

8.9883522 21

b. 90% CI: 1.6 < m1 - m2 < 10.4 (Table: 1.5 < m1 - m2 < 10.5);

(x1 - x2 )± ta /2

s12

s2 15.344512 8.9883522 + 2 = (92.88462 - 86.90476)±1.725 + 78 21

(df = 21- 1 = 20)

c. Yes, it does appear that exposure to lead has an effect on IQ scores.

15. a. H0: m1 = m2; H1: m1 > m2; population1 = alcohol solution, population2 = antiseptic soap Test statistic: t = 2.095; P-value = 0.0196 (Table: P-value < 0.05); Critical value: t = 1.663 (Table: t = 1.676); Reject H 0. There is sufficient evidence to support the claim that soap is more effective, with higher reductions in bacteria counts.

t=

(x1 - x2 )- (m1 - m2 ) = (69 - 35)- 0 s12

s2 + 2

2 1062 + 59 55 59

b. 90% CI: 7.0 < m1 - m2 < 61.0 (Table: 6.8 < m1 - m2 < 61.2) 1062 592 s12 s22 + + (df = 55 - 1 = 54) = (69 - 35)±1.674 (x1 - x2 )± ta /2 55 59 c. It appears that using the antiseptic soap is better because it does result in greater reductions in bacteria counts.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

16. a. H0: m1 = m2; H1: m1 ¹ m2; population1 = Disney movies, population 2 = other movies Test statistic: t = 0.462; P-value = 0.6465 (Table: P-value > 0.20); Critical values: t = ±2.012 (Table: t = ±2.120); Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim that Disney

animated children’s movies and other animated children’s movies have the same mean time showing tobacco use. (x1 - x2 )- (m1 - m2 ) = (61.6 - 49.3)- 0 t= s12 s22 118.82 69.32 + + 33 17 b. 95% CI: - 41.3 sec < m1 - m2 < 65.9 sec (Table: - 44.2 sec < m1 - m2 < 68.8 sec) 118.82 69.32 s12 s22 + + (df = 17 - 1 = 16) = (61.6 - 49.3)± 2.120 (x1 - x2 )± ta /2 33 17 c. The times appear to be from a population with a distribution that is not normal (the sample is right-skewed), but the methods in this section are robust against departures from normality. (Results obtained by using other methods confirm that the results obtained here are quite good, even though the non-Disney times appear to violate the normality requirement.)

17. a. H0: m1 = m2; H1: m1 < m2; population1 = ANSUR I, population2 = ANSUR II Test statistic: t = - 1.085; P-value = 0.1442 (Table: P-value > 0.10); Critical value: t =- 1.708 (Table: t = - 1.796); Fail to reject H 0. There is not sufficient evidence to support the claim that the mean weight of the 1988 male population is less than the mean weight of the 2012 male population. t=

(x1 - x2 )- (m1 - m2 ) = (77.67 - 82.25)- 0 s12

s2 + 2

9.432 12.462 + 12 15 b. 90% CI: - 11.77 kg < m1 - m2 < 2.63 kg (Table: - 12.14 kg < m1 - m2 < 3.00 kg)

(x1 - x2 )± ta /2

s12

s2 9.432 12.462 + 2 = (77.67 - 82.25)±1.796 + 12 15

(df = 12 - 1 = 11)

18. a. H0: m1 = m2; H1: m1 > m2; population1 = Pennsylvania, population2 = New York Test statistic: t = 3.265; P-value = 0.0024 (Table: P-value < 0.005); Critical value: t = 1.746 (Table: t = 1.796); Reject H 0. There is sufficient evidence to support the claim that the mean amount of strontium90 from Pennsylvania residents is greater than the mean amount from New York residents. t=

(x1 - x2 )- (m1 - m2 ) = (147.6 - 136.4)- 0 = 3.265 (df = 11) s12

s2 + 2

10.62 5.22 + 12 12

b. 95% CI: 5.2 mBq < m1 - m2 < 17.1 mBq (Table: 5.0 mBq < m1 - m2 < 17.3 mBq)

(x1 - x2 )± ta /2

s12

s2 10.62 5.22 + 2 = (147.6 - 136.4)±1.796 (df = 11) + 12 12

19. a. H0: m1 = m2; H1: m1 > m2; population1 = Low, population2 = High Test statistic: t = 0.043; P-value = 0.4833 (Table: P-value > 0.10); Critical value: t = 1.782 (Table: t = 1.833); Fail to reject H 0. There is not sufficient evidence to support the claim that the mean IQ score of people with low blood lead levels is higher than the mean IQ score of people with high blood lead levels.

t=

(x1 - x2 )- (m1 - m2 ) = (92.2 - 91.9)- 0 s12

s2 + 2

2

20.442 + 8.53 10 10

19. (continued) b. 90% CI: - 12.2 < m1 - m2 < 12.8 (Table: - 12.5 < m1 - m2 < 13.1) 20.442 8.532 s12 s22 + + (df = 10 - 1 = 9) = (92.2 - 91.9)±1.833 (x1 - x2 )± ta /2 10 10 c. Based on the listed data, it does not appear that exposure to lead has an effect on IQ scores.

20. H0: m1 = m2; H1: m1 ¹ m2; population1 = easy to difficult, population 2 = difficult to easy; Test statistic: t = - 2.657; P-value = 0.0114 (Table: P-value < 0.02); Critical values (a = 0.05): t = ±2.023 (Table: t = ±2.131); The conclusion depends on the choice of the significance level. There is a significant difference between the two population means at the 0.05 significance level, but not at the 0.01 significance level. (x1 - x2 )- (m1 - m2 ) = (27.115 - 31.728)- 0 = - 2.657 (df = 16 - 1 = 15) t= s12 s22 6.8572 4.2602 + + 25 16 21. a. H0: m1 = m2; H1: m1 < m2; population1 = ANSUR I, population2 = ANSUR II Test statistic: t = - 20.393; P-value = 0.0000 (Table: P-value < 0.005); Critical value: t = - 1.645; Reject H 0. There is sufficient evidence to support the claim that the mean weight of the 1988 male population is less than the mean weight of the 2012 male population. It does appear that the male population is getting heavier. (TI data: Test statistic is t = - 5.282, P-value = 0.0000, critical value is t = - 1.648.) t=

(x1 - x2 )- (m1 - m2 ) = (78.49 - 85.52)- 0 s12

s2 + 2

11.122 14.222 + 1774 4082

b. 90% CI: - 7.60 kg < m1 - m2 < - 6.47 kg (TI data: - 7.50 kg < m1 - m2 < - 3.93 kg)

(x1 - x2 )± ta /2

s12

s2 11.122 14.222 + 2 = (78.49 - 85.52)±1.645 + 1774 4082

(df = 1774 - 1 = 1773)

22. H0: m1 = m2; H1: m1 < m2; population1 = ANSUR I, population2 = ANSUR II Test statistic: t = - 0.212; P-value = 0.4159 (Table: P-value > 0.10); Critical value: t = - 1.645; Fail to reject H 0. There is not sufficient evidence to support the claim that the mean height of the 1988 male population is less than the mean height of the 2012 male population. The larger samples do not provide sufficient evidence to suggest that people are getting taller. (TI data: Test statistic is t =- 0.659 and P-value = 0.2552) t=

(x1 - x2 )- (m1 - m2 ) = (1755.8 - 1756.2)- 0 s12

s2 + 2

66.82 68.62 + 1774 4082

23. H0: m1 = m2; H1: m1 ¹ m2; population1 = female, population2 = male; Test statistic: t = - 0.863; P-value = 0.3887 (Table: P-value > 0.20); Critical values: t = ±1.968 (Table: t = ±1.984); Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim that women and men have the same mean diastolic blood pressure.

t=

(x1 - x2 )- (m1 - m2 ) = (70.16 - 71.32)- 0 = - 0.863 (df = 146) s12

s2 + 2

11.222 147

+

11.992 153

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

24. H0: m1 = m2; H1: m1 < m2; population1 = females, population2 = males; Test statistic: t = - 3.450; P-value = 0.0003 (Table: P-value < 0.005); Critical value (a = 0.05): t = - 1.649; (Table: t = - 1.653); Reject H 0. There is sufficient evidence to support the claim that at birth, females have a lower mean weight than males. (x - x )- (m1 - m2 ) = (3037.07 - 3272.82)- 0 = - 3.450 (df = 194) t= 1 2 706.2682 660.1542 s12 s22 + + 205 195 25. With pooling, df increases dramatically to 97, but the test statistic decreases from 2.282 to 1.705 (because the estimated standard deviation increases from 2.620268 to 3.507614), the P-value increases to 0.0457, and the 90% confidence interval becomes wider. With pooling, these results do not show greater significance. a.

H0: m1 = m2; H1: m1 > m2; population1 = low lead, population2 = high lead; Test statistic: t = 1.705; P-value = 0.0457 (Table: P-value < 0.05); Critical value: t = 1.661

(Table: t = 1.987); df = 78 + 21- 2 = 97; Reject H 0. There is sufficient evidence to support the claim that the mean IQ score of people with low blood lead levels is higher than the mean IQ score of people with high blood lead levels. (n1 - 1)s12 + (n2 - 1)s22 (78 - 1)15.344512 + (21- 1)8.9883522 s2 = = = 203.564 p (n1 - 1)+ (n2 - 1) (78 - 1)+ (21- 1) t=

(x1 - x2 )- (m1 - m2 ) = (92.88462 - 86.90476)- 0 s2p

+

sp2

203.564 203.564 + 78 21

b. 90% CI: 0.15 < m1 - m2 < 11.81;

(x1 - x2 )± ta /2

s12

s2 + 2 = (92.88462 - 86.90476)±1.662 203.564 + 203.564 (df = 97) 78 21

c. Yes, it does appear that exposure to lead has an effect on IQ scores.

26. df = 38.9884; Using df = smaller of n1 - 1 and n2 - 1 is a more conservative estimate of the number of degrees of freedom. Using Formula 9-1 results in a larger number of degrees of freedom, so critical t values will be smaller. With Formula 9-1, sample data will require smaller differences to be considered significant.

With Formula 9-1, confidence intervals will be narrower, so estimates of m1 - m2 will involve a smaller range of 2 possible values.

2 2 6.8572 25 + 4.2602 16 s2 n + s 1 1 2 n2 = df = 2 = 38.9884 6.8572 25 + 4.2602 16 s1 n1+ n1 - 1 n2 - 1 25 - 1 16 - 1

(

)

(

)

27. H0: m1 = m2; H1: m1 ¹ m2; population1 = treatment, population2 = placebo; Test statistic: t = 15.322; P-value = 0.0000 (Table: P-value < 0.01); Critical values: t = ±2.080; Reject H 0.

There is sufficient evidence to warrant rejection of the claim that the two populations have the same mean. (n1 - 1)s12 + (n2 - 1)s22 (22 - 1)0.0152 + (22 - 1)02 2 s = = = 0.000125; p (n1 - 1)+ (n2 - 1) (22 - 1)+ (22 - 1) t=

( x 1 - x 2 ) - ( m1 - m 2 ) = s2p

+

sp2

(0.049 - 0.000)- 0 0.000125 0.000125 = 15.322 + 22 22

Section 9-3: Matched Pairs 1.

H0: md = 0 admissions; H1: md < 0 admissions; difference = 6th - 13th;

2.

a. d = - 3.3 admissions; sd = 3.0 admissions Friday 6th 9 6 11 11 3 5 Friday 13th 13 12 14 10 4 12 Difference –4 –6 –3 1 –1 –7 b. md represents the mean of the differences in hospital admissions from the population of paired data.

3.

4.

5.

a. 90% b. t0.05 = 2.015 (df = n - 1 = 6 - 1 = 5) c. Because the confidence interval limits do not include 0 admissions and the range of values consists of negative values only, there is sufficient evidence to support the claim that fewer hospital admissions due to traffic accidents occur on Friday the 6th than on the following Friday the 13th. a. true b. False, the sample size is sufficiently large. c. False, the data are not matched pairs. d. True, the significance level is the same. e. False, the sample size is n = 100. a. H0: md = 0 lb; H1: md > 0 lb; difference = measured - reported Test statistic: t = 0.407; P-value = 0.3469 (Table: P-value > 0.10); Critical value: t = 1.833; Fail to reject H 0. There is not sufficient evidence to support the claim that for females, the measured weights tend to be higher than the reported weights. d - md 0.370 - 0 (df = 10 - 1 = 9) = t= sd / n 2.877 / 10 b. 90% CI: - 1.30 lb < md < 2.04 lb; Because the confidence interval includes 0, fail to reject H 0. There is not sufficient evidence to support the claim that for females, the measured weights tend to be higher than the reported wseights. d d ±t 2.877 = 0.370 ±1.833× (df = 10 - 1 = 9) a /2 n 10

6.

a. H0: md = 0 lb; H1: md > 0 lb; difference = measured - reported Test statistic: t = 0.926; P-value = 0.1893 (Table: P-value > 0.10); Critical value: t = 1.833; Fail to reject H 0. There is not sufficient evidence to support the claim that for males, the measured weights tend to be higher than the reported weights. d - md 3.25 - 0 (df = 10 - 1 = 9) t= = sd / n 11.097 / 10

(

)

b. 90% CI: - 3.2 lb < md < 9.7 lb; Because the confidence interval includes 0, fail to reject H 0. There is not sufficient evidence to support the claim that for males, the measured weights tend to be higher than the reported weights. s 11.097 d ± ta /2 d = 3.25 ±1.833 (df = 10 - 1 = 9) n 10 7.

a. H0: md = 0 mm; H1: md ¹ 0 mm; difference = Arm Span - Height Test statistic: t = - 3.116; P-value = 0.0124 (Table: P-value < 0.02); Critical values: t = ±2.262; Reject H 0. There is sufficient evidence to warrant rejection of the claim that for males, their height is the same as their arm span.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

7.

(continued)

(d - md )

- 41- 0

(df = 10 - 1 = 9) 41.606 / 10 sd / n b. 95% CI: - 70.8 mm < md < - 11.2 mm; Because the confidence interval does not include 0 mm, reject H 0.

There is sufficient evidence to warrant rejection of the claim that for males, their height is the same as their arm span. s 41.606 d ± ta /2 d = - 41- 2.262 (df = 10 - 1 = 9) n 10

t=

8.

=

a. H0: md = 0 cm3; H1: md ¹ 0 cm3; difference = First Born - Second Born Test statistic: t = - 0.474; P-value = 0.6466 (Table: P-value > 0.20); Critical values: t = ±3.250; Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim of no difference between brain volumes for first-born and second-born twins. There does not appear to be a significant difference. d - md - 8.5 - 0 (df = 10 - 1 = 9) t= = 56.679 / 10 sd / n

(

)

b. 99% CI: - 66.7 cm3 < md < 49.7 cm3 (Table: - 66.8 cm3 < md < 49.8 cm3); Because the confidence interval includes 0 cm3, the mean of the differences could be equal to 0 cm3, so there does not appear to be a significant difference.

56.679 sd d ± ta /2 n = - 8.5 ± 3.249 10 (df = 10 - 1 = 9) 9.

a. H0: md = 0; H1: md ¹ 0; difference = right ear - left ear Test statistic: t = 0.890; P-value = 0.3924 (Table: P-value > 0.20); Critical values: t = ±2.201; Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim that the population of differences has a mean equal to 0. d - md 2.08 - 0 (df = 12 - 1 = 11) = t= sd / n 8.11/ 12 b. 95% CI: - 3.1 < md < 7.2; Because the confidence interval includes 0, fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim that the population of differences has a mean equal to 0.

8.11 sd = 2.08 ± 2.201× (df = 12 - 1 = 11) d ± ta /2 n 12

10. a.

H0: md = 0; H1: md ¹ 0 in.; difference = right eye - left eye

Test statistic: t = 0.896; P-value = 0.3938 (Table: P-value > 0.20); Critical values: t = ±2.262; Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim that the population of differences has a mean equal to 0. d - md 1.50 - 0 (df = 10 - 1 = 9) = t= sd / n 5.30 / 10 b. 95% CI: - 2.3 < md < 5.3; Because the confidence interval includes 0, fail to reject H 0. There is not sufficient esvidence to warrant rejection of the claim that the population of differences has a mean equal to 0. d d ±t 5.30 (df = 10 - 1 = 9) = 1.50 ± 2.262 × a /2 n 10 11. a.

H0: md = 0; H1: md > 0; difference = before - after; Test statistic: t = 6.371; P-value = 0.0000 (Table: P-value < 0.005); Critical value: t = 2.718; Reject H 0.

There is sufficient evidence to support the claim that captopril is effective in lowering systolic blood pressure.

11. (continued) 18.58 - 0 d - md = = 6.371 (df = 11) sd n 10.103 12 b. 98% CI: 10s.7 mmHg < md < 26.5 mmHg d d ±t 10.103 = 18.58 ± 2.718 × (df = 10 - 1 = 9) a /2 n 12 t=

12. a.

H0: md = 0 cm; H1: md > 0 cm; difference = president - opponent;

Test statistic: t = 1.304; P-value = 0.1246 (Table: P-value > 0.10); Critical value: t = 2.015; Fail to reject H 0. There is not sufficient evidence to support the claim that for the population of heights of presidents and their main opponents, the differences have a mean greater than 0 cm (or that presidents tend to be taller than their opponents).

3.67 - 0 d - md = (df = 6 - 1 = 5) t= sd n 6.890 6 b. 90% CI: - 2s . 0 cm < md < 9.3 cm; The confidence interval includes 0, so it is possible that md = 0. d d ±t 6.890 (df = 6 - 1 = 5) = 3.67 ± 2.015 × a /2 n 6 13. H0: md = 0 in.; H1: md ¹ 0 in.; difference = mother - daughter; Test statistic: t = - 1.379; P-value = 0.2013 (Table: P-value > 0.20); Critical values: t = ±2.262; Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim that there is no difference in heights between mothers and their first daughters. d - md - 0.95 - 0 (df = 10 - 1 = 9) = t= sd n 2.179 10 14. H0: md = 0 in.; H1: md ¹ 0 in.; difference = father - son; Test statistic: t = 0.034; P-value = 0.9737 (Table: P-value > 0.20); Critical values: t = ±2.262; Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim that there is no difference in heights between fathers and their first sons. d - md 0.02 - 0 (df = 10 - 1 = 9) = t= sd n 1.863 10 15. difference = before - after; 95% CI: 0.69 < md < 5.66; Because the confidence interval limits do not contain 0 and they consist of positive values only, it appears that the “before” measurements are greater than the “after” measurementss, so hypnotism does appear to be effective in reducing pain. d d ±t 2.911 = 3.125 ± 2.365 × (df = 8 - 1 = 7) a /2 n 8 16. a.

H0: md = 0 hours; H1: md ¹ 0 hours; difference = Dextro - Laevo

Test statistic: t = - 4.672; P-value = 0.0012 (Table: P-value < 0.01); Critical value: t = ±3.250; Reject H 0. There is sufficient evidence to warrant rejection of the claim that both treatments have the same effect.

t=

- 1.670 - 0 d - md (df = 10 - 1 = 9) = sd / n 1.130 / 10

b. 99% CI: - 2.83 hours < md < - 0.51 hours; Because the confidence interval does not include 0, reject H 0.

There is susfficient evidence to warrant rejection of the claim that both treatments have the same effect. d d ±t 1.130 (df = 10 - 1 = 9) = - 1.670 ± 3.250 × a /2 n 10

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

17. The larger data set changed the results and conclusion. a.

H0: md = 0 lb; H1: md > 0 lb; difference = measured - reported Test statistic: t = 17.611; P-value = 0.0000 (Table: P-value < 0.005); Critical value: t = 1.645; Reject H 0.

There is sufficient evidence to support the claim that for females, the measured weights tend to be higher than the reported weights (TI data: Test statistic: t = 4.864, critical value: t = 1.651, P-value = 0.0000)

t=

3.245 - 0 d - md = (df = 2971- 1 = 2970) sd / n 10.045 / 2971

b. 90% CI: 2.94 lb < md < 3.55 lb; Because the confidence interval does not include 0, reject H 0. There is sufficient evidence to support the claim that for females, the measured weights tend to be higher than the reported wseights. (TI data: 1.02 lb < md < 2.08 lb) d d ±t 10.045 = 3.245 ±1.645 × (df = 2971- 1 = 2970) a /2 2971 n 18. The larger data set changed the results and conclusion.

H0: md = 0 lb; H1: md > 0 lb; difference = measured - reported Test statistic: t = 3.879; P-value = 0.0001 (Table: P-value < 0.005); Critical value: t = 1.645; Reject H 0.

There is sufficient evidence to support the claim that for males, the measured weights tend to be higher than the reported weights. (TI data: Test statistic: t = 0.773, critical value: t = 1.651, P-value = 0.2202; Using the 247 matched pairs did not have much of an effect on the conclusion from Example 1.)

0.811- 0 d - md = (df = 2784 - 1 = 2783) t= sd / n 11.034 / 2784 19. a. H0: md = 0 lb; H1: md > 0 lb; difference = measured - reported Test statistic: t = 17.611; P-value = 0.0000 (Table: P-value < 0.005); Critical value: t = 1.645; Reject H 0.

There is sufficient evidence to support the claim that for females, measured weights are greater than reported weights. d - md 3.245 - 0 (df = 2971- 1 = 2970) t= = sd / n 10.045 / 2971 b. 90% CI: 2.9 lb < md < 3.5 lb (Table: 2.9 lb < md < 3.6 lb); Because the confidence interval consists of positive values only, the mean of the differences appears to be greater than 0 lb, so there is sufficient evidence to support the claim that for females, measured weights are greater than reported weights. s 10.045 d ± ta /2 d = 3.245 ±1.645 (df = 2971- 1 = 2970) n 2971 20. difference = New cases - deaths; 95% CI: 88, 376.2 < md < 107, 901.7; Because it makes no sense to find differences bestween numbers of new cases and numbers of deaths, the confidence interval is meaningless. d d ±t 140, 672.74 = 98,138.9 ±1.963× (df = 800 - 1 = 799) a /2 800 n

(

)

21. H0: md = 0 cm; H1: md ¹ cm; difference = height - arm span Test statistic: t = - 77.006; P-value = 0.0000 (Table: P-value < 0.005); Critical values: t = ±1.961 Reject H 0.

There is sufficient evidence to warrant rejection of the claim that for males, their height is the same as their arm span. (TI data: Test statistic: t = - 22.16, critical values: t = ±1.967, P-value = 0.0000) t=

- 57.940 - 0 d - md = (df = 4082 - 1 = 4081) sd / n 48.072 / 4082

22. a. H0: md = 0°F; H1: md ¹ 0°F; difference = Day 1- Day 2 Test statistic: t = - 1.113; P-value = 0.2685 (Table: P-value > 0.20); Critical values: t = ±1.986 (Table: t = ±1.987); Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim of no difference between body temperatures measured at 12 AM on the two different days. There does not appear to be a difference.

- 0.089 - 0 d - md = (df = 92 - 1 = 91) t= sd n 0.76782 92 b. 95% CI: - 0.25°F < md < 0.07°F; Because the confidence interval includes 0°F, there is not sufficient evidence to warrant rejection of the claim of no difference between body temperatures measured at 12 AM on the two different days. There does not appear to be a difference. s 0.76782 d ± ta /2 d = - 0.089 ±1.987 (df = 92 - 1 = 91) n 92 23. H0: md = 0 in.; H1: md ¹ 0 in.; difference = mother - daughter Test statistic: t = - 4.090; P-value = 0.0001 (Table: P-value < 0.01); Critical values: t = ±1.978 (Table: t » ±1.974); Reject H 0. There is sufficient evidence to warrant rejection of the claim of no difference in heights between mothers and their first daughters. d - md - 0.93 - 0 (df = 134 - 1 = 133) = t= sd n 2.636 134 24. H0: md = 0 in.; H1: md ¹ 0 in.; difference = father - son Test statistic: t = - 6.347; P-value = 0.0000 (Table: P-value < 0.01); Critical values: t = ±1.978 (Table: t » ±1.984); Reject H 0. There is sufficient evidence to warrant rejection of the claim of no difference in heights between fathers and their first sons. d - md - 1.366 - 0 (df = 134 - 1 = 133) = t= sd n 2.491 134 25. a. difference = father - son; 95% CI: - 1.31 in. < md < 1.35 in.; Because the confidence interval includes the value of 0 in., fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim of no difference in heights between fathers and their first sons. There does not appear to be a significant difference. sd 1.863 d ±t = 0.02 ± 2.262 × (df = 10 - 1 = 9) a /2 n 10 b. Answers vary, but here is a typical 95% CI: - 0.98 in. < md < 1.20 in. Because the confidence interval includes the value of 0 in., fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim of no difference in heights between fathers and their first sons. There does not appear to be a significant difference. c. The two confidence intervals do not differ by amounts that are very substantial. The conclusions from the two confidence intervals are the same. It appears that the confidence interval constructed using the t distribution and the confidence interval constructed using the bootstrap method are reasonably consistent with each other.

26. a. Using the weights in kilograms instead of pounds, the test statistic changes from t = 17.611 to t = 17.623, but that is due to the precision of the data. If the sample weights were given with much greater precision, the test statistics would be the same. The critical value and P-value are the same. The results do not change if weights in kilograms are used instead of weights in pounds. b. The confidence interval based on pounds and the confidence interval based on kilograms are the same when the values are converted to the same units of measure.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

Section 9-4: Two Variances or Standard Deviations 1. a. No, the numerator will always be larger than the denominator in the fraction. b. No, both variances are nonnegative, so their quotient cannot be negative. c. The two samples have standard deviations (or variances) that are very close in value. d. skewed right 2. a. s2 = 4121.6 cm2 and s2 = 4045.0 cm2 1

b.

2

H0: s 2 = s 2; H : s 2 ¹ s 2 or H : s = s ; H : s 1

2

1

1

2

0

1

2

1

¹ s

1

2

population1 = ANSUR I females, population2 = ANSUR II females c. F = s2 s2 = 64.22 63.62 = 1.0189 1

3.

4.

5.

2

d. Fail to reject H 0. There is not sufficient evidence to support the claim that heights of female Army personnel in 1988 and 2012 have different amounts of variation.

No, unlike some other tests that have a requirement that samples must be from normally distributed populations or the samples must have more than 30 values, the F test has a requirement that the samples must be from normally distributed populations, regardless of how large the samples are.

The F test is very sensitive to departures from normality, which means that it works poorly by leading to wrong conclusions when either or both of the populations have a distribution that is not normal.

H0: s 1 = s 2; H1: s 1 ¹ s 2; population1 = Female, population2 = Male Test statistic: F = s2 s2 = 62.829982 46.462532 = 1.8286; P-value = 0.3740; Upper critical value: 1

2

F = 3.9121 (Table: 3.8682 < F < 3.9639); Fail to reject H 0 . There is not sufficient evidence to warrant rejection of the claim that variation among platelet measurements is the same for females and males.

6.

H0: s 1 = s 2; H1: s 1 > s 2; population1 = Pennsylvania, population2 = New York Test statistic: F = s2 s2 = 10.638342 5.2128922 = 4.1648; P-value = 0.0130; Critical value: F = 2.8179 1

2

(Table: 2.7876 < F < 2.8536); Reject H 0. There is sufficient evidence to support the claim that amounts of strontium-90 from Pennsylvania residents vary more than amounts from New York residents.

7.

H0: s 1 = s 2; H1: s 1 ¹ s 2; population1 = red background, population2 = blue background; Test statistic: F = s2 s2 = 0.972 0.632 = 2.3706; P-value = 0.0129; Upper critical value: F = 1.9678 1

2

(Table: 1.8752 < F < 2.0739); Reject H 0 . There is sufficient evidence to warrant rejection of the claim that creative task scores have the same variation with a red background and a blue background.

8.

H0: s 1 = s 2; H1: s 1 ¹ s 2; population1 = red background, population2 = blue background; Test statistic: F = s2 s2 = 5.902 5.482 = 1.1592; P-value = 0.6656; Upper critical value: F = 1.9678 1

2

(Table: 1.8752 < F < 2.0739); Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim that variation of scores is the same with the red background and blue background.

9.

H0: s 1 = s 2; H1: s 1 ¹ s 2; population1 = ethanol, population2 = placebo; Test statistic: F = s2 s2 = 2.202 0.722 = 9.3364; P-value = 0.0000; Critical value: F = 2.4086 1

2

(Table: 2.3675 < F < 2.4247); Reject H 0. There is sufficient evidence to reject the claim that the treatment and placebo groups have the same amount of variation among the errors.

10. a. H0: s 1 = s 2; H1: s 1 > s 2; population1 = exposed, population2 = not exposed; Test statistic: F = s2 s2 = 119.502 62.532 = 3.6522; P-value = 0.0000; Critical value: F = 1.7045 1

2

(Table: 1.6928 < F < 1.8409); Reject H 0. There is sufficient evidence to support the claim that the variation of cotinine in smokers is greater than the variation of cotinine in nonsmokers not exposed to tobacco smoke. b. The sample is not from a normally distributed population as required, so the results in part (a) are highly questionable.

11. H0: s 1 = s 2; H1: s 1 > s 2 ; population1 = Lighter bicycle, population2 = Heavier bicycle; Test statistic: F = s2 s2 = 6.32 4.92 = 1.6531; P-value = 0.0966; Critical value: F = 1.8915 1

2

(Table: 1.8543 < F < 1.9005); Fail to reject H 0. There is not sufficient evidence to support the claim that commuting times with the lighter bicycle have more variation than commuting times with the heavier bicycle.

12. H0: s 1 = s 2; H1: s 1 > s 2 ; population1 = Female babies, population2 = Male babies; Test statistic: F = s2 s2 = 706.32 660.22 = 1.1445; P-value = 0.1714; Critical value: F = 1.2639 1

2

(Table A-5 cannot be used with the given sample sizes). Fail to reject H 0. There is not sufficient evidence to support the claim that at birth, weights of female babies have more variation than weights of male babies.

13. H0: s 1 = s 2; H1: s 1 > s 2 ; population1 = Female pulse rates, population2 = Male pulse rates; Test statistic: F = s2 s2 = 13.962 7.152 = 3.8134; P-value = 0.0340; Critical value: F = 3.3129 1

2

(Table: 3.3472 < F < 3.2839); Reject H 0. There is sufficient evidence to support the claim that the variation among pulse rates of females is greater than the variation among males.

14. H0: s 1 = s 2; H1: s 1 ¹ s 2 ; population1 = Female weights in 2012 population2 = Female weights in 1988; Test statistic: F = s2 s2 = 12.962 5.742 = 5.0960; P-value = 0.0139; Critical value: F = 3.5879; Reject H 0 .

1

2

There is sufficient evidence to reject the claim that the variation among weights did not change from the ANSUR I study in 1988 to the ANSUR II study in 2012. It appears that the variation did change.

15. H0: s 1 = s 2; H1: s 1 > s 2 ; population1 = Medium lead exposure, population2 = High lead exposure; Test statistic: F = s2 s2 = 14.292 8.992 = 2.5285; P-value = 0.0213; Critical value: F = 2.1124 1

2

(Table: 2.0825 < F < 1.1242); Reject H 0. There is sufficient evidence to support the claim that IQ scores of subjects with medium lead levels vary more than IQ scores of subjects with high lead levels.

16. H0: s 1 = s 2; H1: s 1 ¹ s 2 ; population1 = Easy to difficult, population2 = Difficult to easy; Test statistic: F = s2 s2 = 6.8572 4.2602 = 2.5908; P-value = 0.0599; Critical value: F = 2.7006; Fail to 1

2

reject H 0. There is not sufficient evidence to support a claim that the two populations of scores have different amounts of variation.

17. a. Calculations not shown. b. c1 = 3, c2 = 0 log(0.05 / 2) c. Critical value = = 7.4569 25 ö log æ çè ÷ 25 +16 ø d. c1 = 3 < 7.4569; Fail to reject H 0. There is not sufficient evidence to support a claim that the two populations of scores have different amounts of variation.

18. H0: m1 = m2; H1: m1 ¹ m2; population1 = easy to difficult, population 2 = difficult to easy; Test statistic: t = 1.403; P-value = 0.1686 (Table: P-value > 0.10); Critical values: t = ±2.024 (Table: t = ±2.131); Fail to reject H 0 . There is not sufficient evidence to support a claim that the two populations of

scores have different amounts of variation. t=

19.

FL =

(x1 - x2 )- (m1 - m2 ) = (4.914 - 3.296)- 0 = 1.403 (df = 15) s12

1 2.4374

s2 + 2

2

4.7662 + 2.5998 16 25

= 0.4103, FR = 2.7006

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

Section 9-5: Resampling: Using Technology for Inferences 1. Bootstrapping is used for obtaining a confidence interval estimate, and it involves sampling with replacement of selected sample values. Randomization is used for testing hypotheses. When working with two independent samples, randomization involves sampling without replacement; when working with matched pairs, randomization involves sampling with replacement.

2. Both samples are convenience samples. Randomization will not overcome the flaws of those samples. It is likely that nothing can be done to overcome those flaws.

3. Only part (b) is a randomization because it has the same original sample sizes and the sample data selected without replacement.

4. Because the samples are small and the distributions are far from normal, the resampling method is more likely to yield better results.

5. a. Randomization: The difference between the two sample proportions is ⎼0.00840. Results vary but this is typical: Among 1000 resamples, there were no differences at least as extreme as ⎼0.00840. It appears that by chance, it is extremely difficult to get a difference like the one obtained, so there is sufficient evidence to support the claim that the rate of COVID-19 in the treatment group is less than the rate in the placebo group. b. Bootstrapping: Results vary but this is typical: 98% CI: - 0.0101 < p1 - p2 < - 0.00676. Because the confidence interval limits do not contain 0 and contain negative values only, there is sufficient evidence to support the claim that the rate of COVID-19 in the treatment group is less than the rate in the placebo group.

6. a. Randomization: The difference between the two sample proportions is 0.00167. Results vary but this is typical: Among 1000 resamples, there were 745 differences at least as extreme as 0.00167. It appears that by chance, it is easy to get a difference like the one obtained, so there is not sufficient evidence to warrant rejection of the claim that the two treatments have the same rate of deaths. b. Bootstrapping: Results vary but this is typical: 95% CI: - 0.0120 < p1 - p2 < 0.00679. Because the confidence interval limits do contain 0, there is not sufficient evidence to warrant rejection of the claim that the two treatments have the same rate of deaths.

7. a. Randomization: The difference between the two sample proportions is 0.196. Results vary but this is typical: Among 1000 resamples, differences at least as extreme as - 0.196 never occurred. It appears that by chance, it is very difficult to get a difference like the one obtained, so there is sufficient evidence to support the claim that the rate of right-handedness for those who prefer to use their left ear for cell phones is less than the rate of right-handedness for those who prefer to use their right ear for cell phones. b. Bootstrapping: Results vary but this is typical: 98% CI: - 0.266 < p1 - p2 < - 0.127. Because the confidence interval limits do not contain 0, there is a significant difference between the two sample proportions. There is sufficient evidence to support the claim that the rate of right-handedness for those who prefer to use their left ear for cell phones is less than the rate of right-handedness for those who prefer to use their right ear for cell phones.

8. a. Randomization: The difference between sample proportions is 0.107. Results vary but this is typical: Among 1000 resamples, differences at least as extreme as 0.107 occurred only 47 times. There is sufficient evidence to support the claim that when given a single large bill, a smaller proportion of adults spend some or all of the money when compared to the proportion of adults given the same amount in smaller bills. (Note: Because 47 is so close to 50, results could easily suggest that there is not sufficient evidence to support the claim.) b. Bootstrapping: Results vary but this is typical: 90% CI: - 0.200 < p1 - p2 < - 0.0133. Because the confidence interval does not include 0 and consists of negative values only, it appears that when given a single large bill, a smaller proportion of adults spend some or all of the money when compared to the proportion of adults given the same amount in smaller bills.

9.

a. Randomization: Using the sample data, we get x1 - x2 = - 4.57 kg. If we use randomization with the two sets of sample data to generate 1000 simulated differences, a typical result is that 150 of those differences will be - 4.57 kg or below, so it appears that such differences can easily occur. There is not sufficient evidence to support the claim that the mean weight of the 1988 male population is less than the mean weight of the 2012 male population.

9.

(continued) b. Bootstrapping: Results vary but this is typical: 90% CI: - 11.38 kg < m1 - m2 < 2.21 kg. Because the confidence interval does include 0, it appears that there is not a significant difference between the mean weight in 1988 and the mean weight in 2012. There is not sufficient evidence to support the claim that the mean weight of the 1988 male population is less than the mean weight of the 2012 male population.

10. a. Randomization: Using the sample data, we get x1 - x2 = 11.2 mBq. If we use randomization with the two sets of sample data to generate 1000 simulated differences, a typical result is that only 1 of those differences will be 11.2 or greater, so it appears that such differences are very rare. There is sufficient evidence to support the claim that the mean amount of strontium-90 from Pennsylvania residents is greater than the mean from New York residents. b. Bootstrapping: Results vary but this is typical: 90% CI: 5.9 mBq < m1 - m2 < 16.7 mBq. Because the confidence interval does not include 0 and consists of positive values only, it appears that the mean amount of strontium-90 from Pennsylvania residents is greater than the mean from New York residents.

11. a. Randomization: Using the sample data, we get x1 - x2 = 0.3. If we use randomization with the two sets of sample data to generate 1000 simulated differences, a typical result is that 480 of those differences will be 0.3 or greater, so it appears that it is very easy to get such differences. There is not sufficient evidence to support the claim that the mean IQ of the low lead population is greater than the mean IQ of the high lead population. b. Bootstrapping: Results vary but this is typical: 98% CI: - 10.3 < m1 - m2 < 10.9. The confidence interval does include 0. There is not sufficient evidence to support the claim that the mean IQ of the low lead population is greater than the mean IQ of the high lead population.

12. Randomization: Using the sample data, x1 - x2 = - 4.613. If we use randomization with the two sets of sample data to generate 1000 simulated differences, a typical result is that 18 of the differences are at least as extreme as 4.613. Whether using a 0.05 or 0.01 significance level, it appears that there is a significant difference between the two population means.

13. a. Randomization: Results vary but this is typical: Among 1000 resamples, 336 of the values of d are 0.37 lb or greater, so the value of d = 0.37 lb can easily occur by chance. There is not sufficient evidence to support the claim that for females, the measured weights tend to be higher than the reported weights. b. Bootstrapping: Results vary but this is typical: 90% CI: - 1.02 lb < md < 1.78 lb. Because the confidence interval does include 0, there is not sufficient evidence to support the claim that for females, the measured weights tend to be higher than the reported weights.

14. a. Randomization: Results vary but this is typical: With 1000 resamples, 193 of the values of d are 3.25 lb or less, so values at least as extreme as d = 3.25 lb occur at a rate of about 0.193. Because the significance level is 0.05, there is not sufficient evidence to support the claim that for males, the measured weights tend to be higher than the reported weights. b. Bootstrapping: Results vary but this is typical: 90% CI: - 2.2 lb < md < 9.5 lb. Because the confidence interval does include 0, there is not sufficient evidence to support the claim that for males, the measured weights tend to be higher than the reported weights.

15. a. Randomization: Results vary but this is typical: With 1000 resamples, 9 of the values of d are –41.0 mm or lower and 6 of the values of d are 41.0 mm or greater, so the values at least as extreme as d = - 41.0 mm occur at a rate of about 0.015. Because the significance level is 0.05, there is sufficient evidence to reject the claim that for males, their height is the same as their arm span. b. Bootstrapping: Results vary but this is typical: 95% CI: - 64.5 mm < md < - 13.1 mm. Because the confidence interval does not include 0, there is sufficient evidence to reject the claim that for males, their height is the same as their arm span.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

16. a. Randomization: Results vary but this is typical: with 1000 resamples, 323 of the values of d are - 8.5 cm3 or lower and 325 of the values d are 8.5 cm3 or greater, so values at least as extreme as d =- 8.5 cm3 occur at a rate of about 0.648. Because the significance level is 0.01, there is not sufficient evidence to reject the claim of no difference between brain volumes for first-born and second-born twins. b. Bootstrapping: Results vary but this is typical: 99% CI: - 49.0 cm3 < md < 37.3 cm3. Because the confidence interval includes 0 cm3, the mean of the differences could be equal to 0 cm3, so there does not appear to be a significant difference.

17. Results vary but this is typical from 1000 resamples: 90% CI: - 11.51 kg < m1 - m2 < 1.99 kg. Typical result from 10,000 resamples: - 11.20 kg < m1 - m2 < 2.11 kg. The differences are very small, so the larger number of resamples does not have much effect on the results.

18. Results vary but this is typical from 1000 resamples: 95% CI: 0.15 < s 21 / s 22< 1.4. Because the confidence interval includes 1, it appears that the weights from 1988 and the weights from 2012 have amounts of variation that are not significantly different.

Quick Quiz 1.

2.

H0: p1 = p2; H1: p1 ¹ p2; population1 = treatment, population2 = placebo

pˆ = 1

3.

102

= 0.354, p̂ 2

288

=

75 = 0.271, p = 102 + 75 = 0.313 277

288 + 277

a. P-value = 2 ×P(z > 2.14) = 0.0324 b. 0.0324 < 0.05; Reject H 0 and conclude that there is sufficient evidence to warrant rejection of the claim that patients treated with dexamethasone and patients given a placebo have the same rate of complete resolution. It appears that the rates of complete resolution are different.

4.

5.

6.

7.

8.

9.

a. 95% CI: 0.00732 < p1 - p2 < 0.159 b. The confidence interval does not include 0, so it appears that the two proportions are not equal. There appears to be a significant difference between the success rate in the treatment group and the success rate in the placebo group. a. Because the data consist of matched pairs, they are dependent.

- 3- 0 d - md = = - 0.647 b. t = sd / n 10.368 / 5 c. There is not sufficient evidence to warrant rejection of the claim that visual acuity is the same in the right eye and left eye. a. F = s2 s2 = 25.248762 17.175562 = 2.1610 or F = s2 s2 = 17.175562 25.248762 = 0.4627 1

2

1

2

b. There is not sufficient evidence to warrant rejection of the claim that the two samples are from populations having the same variation. a. Because each pair of data is matched by the same subject, the two sets of data are dependent. b. H0: md = 0°F; H1: md ¹ 0°F; difference = 8 AM - 12 AM Because the confidence interval includes the value of 0°F, it appears that there is not a significant difference between the temperatures at 8 AM and the temperatures at 12 AM. True, since n > 30.

10. False, the requirements are np ³ 5 and nq ³ 5.

Review Exercises 1.

H0: p1 = p2; H1: p1 < p2; population1 = $1 bill, population2 = 4 quarters;

Test statistic: z = - 3.49; P-value = 0.0002; Critical value: z = - 1.645; Reject H 0. There is sufficient evidence to support the claim that money in a large denomination is less likely to be spent relative to an equivalent amount in smaller denominations.

1.

(continued) 12 + 27 39 39 50 p= = ; q = 1= ; 46 + 43 89 89 89 12 - 27 - 0 ( pˆ - pˆ ) - ( p - p ) 1 2 1 2 46 43 z= = 39 50 39 50 pq pq + 89 89 89 89 + n1 n2 46 43

(

)

( )( ) ( )( )

2.

90% CI: - 0.528 < p1 - p2 < - 0.206; The confidence interval limits do not contain 0, so it appears that there is a significant difference between the two proportions. Because the confidence interval consists of negative values only, it appears that p1 is less than p2 , so it appears that money in a large denomination is less likely to be spent relative to an equivalent amount in smaller d en o m in a tions.

12 1 - 12 27 1 - 27 46 = 12 - 27 ± 1.645 × 46 43 ( pˆ1 - pˆ2 )± za /2 + 43 46 43 46 43

(

)

( )(

) ( )(

)

3.

H0: md = 0 kg; H1: md ¹ 0 kg; difference = pre - post; Test statistic: t = 2.301; P-value = 0.0469; Critical values: t = ±2.262; Reject H 0. There is sufficient evidence to conclude that there is a difference between pre-training and post-training weights. d - md = 2.0 - 0 = 2.301 (df = 9) t= sd n 2.749 10

4.

95% CI: 0.0 kg < md < 4.0 kg; The confidence interval does not include 0 kg and consists of positive values only. s 2.749 d ± ta /2 d = 2.0 ± 2.262 (df = 9) n 10

5.

H0: p1 = p2; H1: p1 > p2; population1 = sustained care, population2 = standard care Test statistic: z = 2.64; P-value = 0.0041; Critical value: z = 2.33; Reject H 0. There is sufficient evidence to support the claim that the rate of success for smoking cessation is greater with the sustained care program.

51+ 30 = 81 ; q = 1- 81 316 p= = 198 +199 397 397 397 51 30 ( pˆ1 - pˆ2 )- ( p1 - p2 ) = 198 - 199 - 0 z= 81 316 81 316 pq pq + 397 397 397 397 + n1 n2 198 199

) ( ( )( ) ( )( )

6.

a. 98% CI: 0.0135 < p1 - p2 < 0.200 (Table: 0.0134 < p1 - p2 < 0.200); Because the confidence interval limits do not contain 0, there is a significant difference between the two proportions. Because the interval consists of positive numbers only, it appears that the success rate for the sustained care program is greater than the success rate for the standard care program.

( pˆ1 - pˆ2 )± za /2

=

30 (19851 - 199)± 2.326 × 198 198 198 + 199 199 199

b. Based on the samples, the success rates of the programs are 25.8% (sustained care) and 15.1% (standard care). That difference does appear to be substantial, so the difference between the programs does appear to have practical significance. c. The more successful of the two programs has a success rate of only 25.8%, so there is a failure rate of about 74% for those who try to stop smoking with the sustained care program. The time required and cost of the sustained program relative to standard care must be considered when determining practical significance. If sustained care is much more time consuming and expensive, the practical significance will be lower.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy Cumulative Review Exercises 7.

33

H0: m1 = m2; H1: m1 < m2; population1 = seat belt, population2 = no seat belt Test statistic: t = - 2.330; P-value = 0.0102 (Table: P-value < 0.025); Critical value: t =- 1.649 (Table: t = - 1.660); Reject H 0. There is sufficient evidence to support the claim that children wearing seat belts have

a lower mean length of time in an ICU than the mean for children not wearing seat belts. Buckle up! t=

8.

(x1 - x2 )- (m1 - m2 ) = (0.83 - 1.39)- 0 (df = 123 - 1 = 122) s12

s2 + 2

90% CI: - 0.96 day < m1 - m2 < - 0.16 day; The confidence interval does not include 0 days, so there appears to be a significant difference. Because the confidence interval consists of negative values only, it appears that children wearing seat belts spend less time in intensive care units than children who don’t wear seat belts.

Children should wear seat belts (except for young children who should use properly installed car seats). population1 = seat belt, population2 = no seat belt s2 1.772 3.062 + 2 = (0.83 - 1.39)±1.660 (df = 123 - 1 = 122) + 123 290

s12

(x1 - x2 )± ta /2 9.

2 1.772 + 3.06 123 290

The waist circumferences from 1988 are measured from eight males, and the waist circumferences from 2012 are from eight different males, so the data are not paired or matched in any meaningful way. The stated claim makes no sense for these two independent samples. The results that would be obtained by blindly following the instruction to test the claim that the differences between the pairs of data are from a population with a mean of 0 mm are given below.

H0: md = 0 mm; H1: md ¹ 0 mm; difference = 1988 - 2012 Test statistic: t = - 0.834; P-value = 0.4320 (Table: P-value > 0.20); Critical values: t = ±2.365; Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim that the differences between the pairs of

data are from a population with a mean of 0 mm. However, these results are not valid and they make no sense with the two independent samples. d - md = - 38.4 - 0 t= sd / n 130.19 / 8 10. H0: s 1 = s 2; H1: s 1 ¹ s 2 ; population1 = no seat belt, population2 = seat belt Test statistic: F = s2 s2 = 3.062 1.772 = 2.9888; P-value = 0.0000; Upper critical value: F = 1.3630; Reject 1

2

H 0. There is sufficient evidence to warrant rejection of the claim that for children hospitalized after motor

vehicle crashes, the numbers of days in intensive care units for those wearing seat belts and for those not wearing seat belts have the same variation.

Cumulative Review Exercises 1. a. Key question: Is there a trend in the number of births in the United States? b. A time-series graph would be effective in identifying a trend. c. The time-series graph shows a distinct downward trend in the number of births.

2.

a. Key question: Is there a difference in braking reaction times of males and females? b. Use the methods for inferences from two independent means (Section 9-2). c. Males appear to have lower reaction times.

H0: m1 = m2; H1: m1 » m2; population1 = Male, population2 = Female Test statistic: t = - 3.259; P-value = 0.0019 (Table: P-value < 0.005); Critical values (a = 0.05): t = ±2.002 (Table: t = ±2.030); There is sufficient evidence to warrant rejection of the claim that males and females have the same mean braking reaction time.

(

)

1 1 1 1 95% CI: - 16.0 10,000 sec < m1 - m2 < - 3.8 10,000 sec Table: - 16.1 10,000 sec < m1 - m2 < - 3.7 10,000 sec ; Because the confidence interval consists of negative numbers and does not include 0, there appears to be a significant difference between the mean braking reaction times of males and females.

t=

(x1 - x2 )- (m1 - m2 ) = (44.36 - 54.28)- 0 s12

s2 + 2

(x1 - x2 )± ta /2 3.

s12

9.472 15.612 + 36 36

s2 + 2 = (44.36 - 54.28)- 2.030

9.472 36

+

15.612 36

a. Key question: Two good possible answers: (1) What is an estimate of the proportion of all males who are in favor the death penalty? (2) Do the majority of males favor the death penalty? b. Use a confidence interval to estimate the proportion of males who favor the death penalty (Section 7-1), or conduct a hypothesis test of the claim that most males favor the death penalty (Section 8-1). c. H0: p = 0.5; H1: p > 0.5; Test statistic: z = 2.30 (2.35 using x = 0.55(530) = 292); P-value = P(z > 2.35) = 0.0094 (Tech: 0.0095); Critical value (a = 0.05): z = 1.645; Reject H 0. There is sufficient evidence to support the claim that the majority of males are in favor of the death penalty.

95% CI: 0.509 < p1 - p2 < 0.593; Because the values contained within the confidence interval are all greater than 0.5, there is sufficient evidence to support the claim that the majority of males are in favor of the death penalty. pˆ - p 0.55 - 0.50 z= = (0.50)(0.50) pq 530 n

pˆqˆ (0.55)(0.45) = 0.55 ±1.96 530 n a. Key question: Does the bar graph correctly depict the data or is it somehow misleading? (Another reasonable question would be to ask whether significantly more males than females are in favor of the death penalty, but no sample sizes are provided to help address that question.) b. Carefully examine the graph to determine whether it is misleading. c. Because the vertical scale begins with 50% instead of 0%, the bottom portion of the graph is cut off, so the difference is visually exaggerated. The graph makes it appear that roughly twice as many males favor the death penalty when compared to females, but examination of the graph shows that the two percentages are about 55% and 52%, so they are actually quite close. a. Key question: Is there a difference between the proportions of males and females who are in favor of the death penalty? b. Use a hypothesis test for two proportions, or use a confidence interval estimate of the difference between two proportions (Section 9-1). c. H0: p1 = p2; H1: p1 ¹ p2; population1 = Males, population2 = Females Test statistic: z = 0.97; P-value = 0.3309 (Table: 0.3320); Critical values: (a = 0.05): z = ±1.96; Fail to reject H 0. There is not sufficient evidence to support the claim that there is a difference between the proportions of males and females who are in favor of the death penalty. pˆ ± za /2

4.

5.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

5.

(continued) 95% CI: - 0.0306 < p1 - p2 < 0.0909; Because the confidence interval limits include 0, there is not sufficient evidence to support the claim that there is a difference between the proportions of males and females who are in favor of the death penalty.

0.55(530) + 0.52(505) p= = 0.535; q = 1- 0.535 = 0.465 530 + 505 ( pˆ1 - pˆ2 )- ( p1 - p2 ) = (0.55 - 0.52)- 0 z= pq pq 0.535(0.465) 0.535(0.465) + + n1 n2 530 505

( pˆ - pˆ )± z 1

2

pˆ1qˆ1 a /2

+

pˆ2qˆ2 = (0.55 - 0.52)- 1.96 0.55(1- 0.55) 0.52 (1- 0.52) + 530 505

6.

a. Key question: Are the last digits selected in a way that appears to be random, with roughly the same frequency for each digit? b. Construct a histogram or dotplot to illustrate the frequencies c. The graph shows that the numbers do appear to be selected with roughly the same frequency. (Chapter 11 will introduce a procedure for testing “goodness-of-fit” with a uniform distribution, and that procedure is much more thorough and objective.)

7.

a. Key question: Based on the differences between the IQ scores of each matched pair in the sample, is there no difference between the IQ scores of pairs of twins in the population? b. Use the methods of Section 9-3 for matched pairs. c. There does not appear to be a difference between the IQ scores of first-born and second-born twins in the population.

H0: md = 0; H1: md ¹ 0; difference = first - second; Test statistic: t = 0.150; P-value = 0.8843 (Table: P-value > 0.20); Critical values (a = 0.05): t = ±2.262; Fail to reject H 0. 95% CI: - 5.6 md < 6.4. t=

8.

d - md

=

0.40 - 0 8.45 / 10

sd / n 8.45 s d ± ta /2 d = 0.40 ± 2.262 × n 10 a. Key question: Is eyewitness memory of police better (or the same as) with a non-stressful interrogation than with a stressful interrogation? b. Use the methods for inferences from two independent means (Section 9-2). c. It appears that eyewitness memory of police is better with a non-stressful interrogation than with a stressful interrogation.

H0: m1 = m2; H1: m1 > m2; Test statistic: t = 2.843; P-value = 0.0029 (Table: P-value < 0.005); Critical value: t = 1.665 (Table: t = 1.685); Reject H 0. There is sufficient evidence to support the claim that eyewitness memory of police is better with a non-stressful interrogation than with a stressful interrogation.

8.

(continued) 90% CI: 3.27 < m1 - m2 < 12.53 (Table: 3.22 < m1 - m2 < 12.58); The confidence interval limits do not contain 0 and the interval consists of positive numbers only. t=

(x1 - x2 )- (m1 - m2 ) = (53.3 - 45.4)- 0 s12

s2 + 2

(x1 - x2 )± ta /2 9.

s12

11.62 13.22 + 40 40

2 s2 11.62 + 13.2 + 2 = (53.3 - 45.4)±1.685 × 40 40

a. Key question: How do the different categories compare in terms of the numbers of deaths? b. Construct an effective graph such as a Pareto chart so that we can see which causes of death are most significant. c. A Pareto chart or bar chart or pie chart shows that pollution is the largest cause of deaths, followed by tobacco use. Deaths from the other causes are relatively much lower.

10. a. Key question: Is there a correlation between the numbers of pleasure boats and the numbers of manatee fatalities? b. A scatterplot (see Section 2-2) would be helpful in visualizing whether there is a correlation. (The following chapter will introduce much more thorough and objective criteria for analyzing correlations.) c. A scatterplot reveals that there does not appear to be a correlation between the numbers of pleasure boats and the numbers of manatee deaths.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

Chapter 10: Correlation and Regression Section 10-1: Correlation 1. a. r is a statistic that represents the value of the linear correlation coefficient computed from the paired sample data, and r is a parameter that represents the value of the linear correlation coefficient that would be computed by using all of the paired data in the population of all statistics students. b. The value of r is estimated to be 0, because it is likely that there is no correlation between heights of biostatistics students and their scores on the first statistics test. c. The value of r does not change if the heights are converted from centimeters to inches.

2.

3.

4.

5.

No, with r = 0, there is no linear correlation, but there might be some association with a scatterplot showing a distinct pattern that is not a straight-line pattern.

No, a correlation between two variables indicates that they are somehow associated, but that association does not necessarily imply that one of the variables has a direct effect on the other variable. Correlation does not imply causality. a. –1 d. 0.992 e. 1 b. 0.746 c. 0.268 r = 0.963; P-value = 0.000; Critical values: r = ±0.268 (Table: r » ±0.279); Yes, there is sufficient evidence

to support the claim that there is a linear correlation between the weights of bears and their chest sizes. It is easier to measure the chest size of a bear than the weight, which would require lifting the bear onto a scale. It does appear that chest size could be used to predict weight.

6.

r = 0.864; Critical values (n = 54): r = ±0.268 (Table: r » ±0.279); Yes, there is sufficient evidence to support

the claim that for bears there is a linear correlation between length and weight.

7.

r = 0.552; P-value < 0.0001; Critical values: r » ±0.196; Yes, there is sufficient evidence to support the claim

that there is a linear correlation between the heights of fathers and the heights of their sons.

8.

9.

r = 0.031; Critical values (n = 10): r = ±0.632; No, there is not sufficient evidence to support the claim that

there is a linear correlation between heights of mothers and heights of their first daughters. a. Answer will vary, but because there appears to be an upward pattern, it is reasonable to think that there is a linear correlation. b. H0: r = 0; H1: r ¹ 0; r = 0.906; Critical values (a = 0.05): r = ±0.632; P-value = 0.000 (Table: P-value < 0.01); There is sufficient evidence to support the claim of a linear correlation. c.

H0: r = 0; H1: r ¹ 0; r = 0; Critical values (a = 0.05): r = ±0.666; P-value = 1.000 (Table: P-value > 0.05); There is not sufficient evidence to support the claim of a linear correlation.

d. The effect from a single pair of values can be very substantial, and it can change the conclusion.

10. a. There does not appear to be a linear correlation. b. There does not appear to be a linear correlation. c. H0: r = 0; H1: r ¹ 0; r = 0; Critical values (a = 0.05): r = ±0.950; P-value = 1.000 (Table: P-value > 0.05); There does not appear to be a linear correlation. The same results are obtained with the four points in the upper right corner. d. H0: r = 0; H1: r ¹ 0; r = 0.985; Critical values (a = 0.05): r = ±0.707; P-value = 0.000 (Table: P-value < 0.01); There is sufficient evidence to support the claim of a linear correlation. e. There are two different populations that should be considered separately. It is misleading to use the combined data from females and males and conclude that there is a relationship between x and y.

11. a. See graph on next page. b. H0: r = 0; H1: r ¹ 0; r = 0.816; P-value = 0.002 (Table: P-value < 0.01); Critical values (a = 0.05): r = ±0.602; There is sufficient evidence to support the claim of a linear correlation between the variables. c. The scatterplot reveals a distinct pattern that is not a straight-line pattern.

Scatterplot for Exercise 11

Scatterplot for Exercise 12

12. a. See graph above. b. H0: r = 0; H1: r ¹ 0; r = 0.816; P-value = 0.002 (Table: P-value < 0.01); Critical values (a = 0.05): r = ±0.602; There is sufficient evidence to support the claim of a linear correlation between the variables. c. The scatterplot reveals a perfect straight-line pattern, except for the presence of one outlier.

13. H0: r = 0; H1: r ¹ 0; r = 0.441; P-value = 0.174 (Table: P-value > 0.05); Critical values: r = ±0.602; Fail to reject H 0. There is not sufficient evidence to support the claim that there is a linear correlation between IQ scores and brain volumes. It does not appear that people with larger brains have higher IQ scores.

Scatterplot for Exercise 13 Scatterplot for Exercise 14

14. H0: r = 0; H1: r ¹ 0; r = - 0.279; P-value = 0.407 (Table: P-value > 0.05); Critical values: r = ±0.602; Fail to reject H 0. There is not sufficient evidence to support the claim that there is a linear correlation between IQ scores and brain surface areas.

15. H0: r = 0; H1: r ¹ 0; r = 0.831; P-value = 0.001 (Table: P-value < 0.01); Critical values: r = ±0.576; Reject H 0. There is sufficient evidence to support the claim of a linear correlation between circumferences and heights of trees. Tall trees tend to have larger trunks.

16. H0: r = 0; H1: r ¹ 0; r = 0.930; P-value = 0.00001 (Table: P-value < 0.01); Critical values: r = ±0.576; Reject H 0. There is sufficient evidence to support the claim of a linear correlation between weights in September and weights in April.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

Scatterplot for Exercise 15

Scatterplot for Exercise 16

17. H0: r = 0; H1: r ¹ 0; r = 0.591; P-value = 0.294 (Table: P-value > 0.05); Critical values: r = ±0.878; Fail to reject H 0. There is not sufficient evidence to support the claim that there is a linear correlation between shoe print lengths and heights of males. The given results do not suggest that police can use a shoe print length to estimate the height of a male.

Scatterplot for Exercise 17 Scatterplot for Exercise 18

18. H0: r = 0; H1: r ¹ 0; r = 0.827; P-value = 0.084 (Table: P-value > 0.05); Critical values: r = ±0.878; Fail to reject H 0. There is not sufficient evidence to support the claim that there is a linear correlation between foot lengths and heights of males. The given results do not suggest that police can use a foot length to estimate the height of a male.

19. H0: r = 0; H1: r ¹ 0; r = 0.998; P-value = 0.000 (Table: P-value < 0.01); Critical values: r = ±0.514; Reject H 0. There is sufficient evidence to support the claim that there is a linear correlation between amounts of tar and nicotine in king size cigarettes.

Scatterplot for Exercise 19 Scatterplot for Exercise 20

20. H0: r = 0; H1: r ¹ 0; r = 0.003; P-value = 0.992 (Table: P-value > 0.05); Critical values: r = ±0.514; Fail to reject H 0. There is not sufficient evidence to support the claim that there is a linear correlation between amounts of tar and carbon monoxide in menthol cigarettes.

21. H0: r = 0; H1: r ¹ 0; r = - 0.231; P-value = 0.389 (Table: P-value > 0.05); Critical values: r = ±0.497; Fail to reject H 0. There is not sufficient evidence to support the claim that there is a linear correlation between the lengths of movies and the amounts of time spent showing the use of alcohol or tobacco.

Scatterplot for Exercise 21 Scatterplot for Exercise 22

22. H0: r = 0; H1: r ¹ 0; r = 0.941; P-value = 0.002 (Table: P-value < 0.01); Critical values: r = ±0.754; Reject H 0. There is sufficient evidence to support the claim that there is a linear correlation between the populations of storks and humans in Oldenburg, Germany.

23. H0: r = 0; H1: r ¹ 0; r = 0.857; P-value = 0.007 (Table: P-value < 0.01); Critical values: r = ±0.707; Reject H 0. There is sufficient evidence to support the claim that there is a linear correlation between footprint lengths and heights of males. The given results do suggest that police can use a footprint length to estimate the height of a male.

Scatterplot for Exercise 23 Scatterplot for Exercise 24

24. H0: r = 0; H1: r ¹ 0; r = 0.874; P-value = 0.005 (Table: P-value < 0.01); Critical values: r = ±0.707; Reject H 0. There is sufficient evidence to support the claim of a linear correlation between the number of cricket chirps and the temperature.

25. H0: r = 0; H1: r ¹ 0; r = 0.840; P-value = 0.005 (Table: P-value < 0.01); Critical values: r = ±0.666; Reject H 0. There is sufficient evidence to support the claim that there is a significant linear correlation between the numbers of registered pleasure boats and the numbers of manatee boat fatalities. The results show that there is an association between the numbers of pleasure boats and manatee deaths, but the results do not show that the numbers of pleasure boats are the cause of the manatee deaths.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

Scatterplot for Exercise 25

Scatterplot for Exercise 26

26. H0: r = 0; H1: r ¹ 0; r = 0.959; P-value = 0.000 (Table: P-value < 0.01); Critical values: r = ±0.632; Reject H 0. There is sufficient evidence to support the claim that there is a linear correlation between per capita consumption of mozzarella cheese and the numbers of civil engineering PhD degrees awarded each year.

Common sense suggests that there is no cause/effect relationship between consumption of mozzarella cheese and PhD degrees awarded in civil engineering.

27. H0: r = 0; H1: r ¹ 0; r = - 0.959; P-value = 0.010; Critical values: r = ±0.878; Reject H 0. There is sufficient evidence to support the claim that there is a linear correlation between weights of lemon imports from Mexico and U.S. car fatality rates. The results do not suggest any cause-effect relationship between the two variables.

Scatterplot for Exercise 27 Scatterplot for Exercise 28

28. H0: r = 0; H1: r ¹ 0; r = 0.948; P-value = 0.004 (Table: P-value < 0.01); Critical values: r = ±0.811; Reject H 0. There is sufficient evidence to support the claim of a linear correlation between the overhead width of a seal in a photograph and the weight of a seal.

29. H0: r = 0; H1: r ¹ 0; r = 0.895; P-value = 0.000 (Table: P-value < 0.01); Critical values (Tech): r = ±0.349; Reject H 0. There is sufficient evidence to support the claim of a linear correlation between weights in September and weights in April. In this case, the larger sample results in the same conclusion as the one reached with the smaller sample.

30. H0: r = 0; H1: r ¹ 0; r = - 0.132; P-value = 0.362 (Table: P-value > 0.05); Critical values (Tech): r = ±0.279; Fail to reject H 0. There is not sufficient evidence to support the claim that there is a linear correlation between the lengths of movies and the amounts of time spent showing the use of alcohol or tobacco.

In this case, the larger sample results in the same conclusion as the one reached with the smaller sample.

Scatterplot for Exercise 29

Scatterplot for Exercise 30

31. H0: r = 0; H1: r ¹ 0; r = 0.718; P-value = 0.000 (Table: P-value < 0.01); Critical values (Tech): r = ±0.031; Reject H 0. There is sufficient evidence to support the claim that there is a linear correlation between foot lengths and heights of males. In this case, the larger sample results in the same conclusion as the one reached with the smaller sample.

Scatterplot for Exercise 31 Scatterplot for Exercise 32

32. H0: r = 0; H1: r ¹ 0; r = 0.814; P-value = 0.000 (Table: P-value < 0.01); Critical values (Tech): r » ±0.374; Reject H 0. There is sufficient evidence to support the claim that there is a significant linear correlation between the numbers of registered pleasure boats and the numbers of manatee boat fatalities. In this case, the larger sample results in the same conclusion as the one reached with the smaller sample.

33. Answers vary, but the following is typical: Resampling 1000 times, there are no results that are at least as extreme as r = 0.930 found from the sample data. (The results “at least as extreme” are those with r = 0.930 or greater and those with r =- 0.930 or lower.) It appears that the likelihood of getting a result at least as extreme as the one obtained is 0.000, so there is sufficient evidence to support the claim that there is a linear correlation between weights in September and weights in April.

34. Answers vary, but the following is typical: Resampling 1000 times, there are 351 results that are at least as extreme as r =- 0.231 found from the sample data. (The results “at least as extreme” are those with r = 0.231 or greater and those with r =- 0.231 or lower.) It appears that the likelihood of getting a result at least as extreme as the one obtained is 0.351, so there is not sufficient evidence to support the claim that there is a linear correlation between the lengths of movies and the amounts of time spent showing the use of alcohol or tobacco.

In this case, the larger sample results in the same conclusion as the one reached with the smaller sample.

35. Answers vary, but the following is typical: Resampling 1000 times, there are 6 results that are at least as extreme as r = 0.857 found from the sample data. (The results “at least as extreme” are those with r = 0.857 or greater and those with r =- 0.857 or lower.) It appears that the likelihood of getting a result at least as extreme as the one obtained is 0.006, so there is sufficient evidence to support the claim that there is a linear correlation between foot lengths and heights of males. In this case, the larger sample results in the same conclusion as the one reached with the smaller sample.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

36. Answers vary, but the following is typical: Resampling 1000 times, there are 5 results that are at least as extreme as r = 0.840 found from the sample data. (The results “at least as extreme” are those with r = 0.840 or greater and those with r =- 0.840 or lower.) It appears that the likelihood of getting a result at least as extreme as the one obtained is 0.005, so there is sufficient evidence to support the claim that there is a significant linear correlation between the numbers of registered pleasure boats and the numbers of manatee boat fatalities. In this case, the larger sample results in the same conclusion as the one reached with the smaller sample.

37. With n = 4082, there are 4082 - 2 = 4080 degrees of freedom. From Table A-3 use the closest t value of 1.960 1.960 in the given formula to get the critical values of r = ± = ±0.031.

1.9602 + 4082 - 2 Using a more accurate value of t = 1.96054554 from technology leads to the same critical values of 1.96054554 r=± = ±00.31.

1.960545542 + 4082 - 2 38. r = 0.505; P-value = 0.000 (Table: P-value < 0.01); Critical values (a = 0.05): r = ±0.069; There is sufficient evidence to support the claim that there is a linear correlation between the numbers of new cases and numbers of deaths from COVID-19. Examination of the scatterplot shows that as the numbers of new cases of COVID19 increase, the numbers of deaths also increase. Among the 800 points, there appears to be a somewhat strong correlation except for the 45 points farthest to the right.

Section 10-2: Regression 1. a. yˆ represents the predicted value of the weight of a bear. b. The slope is 12.5 and the y-intercept is - 264. c. The predictor variable is chest size which is represented by x. d. yˆ = - 264 +12.5(40) = 236 lb 2.

The first equation represents the regression line that best fits sample data, and the second equation represents the regression line that best fits all paired data in a population. The values b0 and b1 are statistics representing the y-intercept and slope of the straight line that best fits the sample data. The values b 0 and b1 are parameters representing the y-intercept and slope of the straight line that best fits the data in the population of paired data.

3.

a. A residual is a value of y - yˆ, which is the difference between an observed value of y and a predicted value of y. b. The regression line has the property that the sum of squares of the residuals is the lowest possible sum.

4.

5.

The value of r and the value of b1 have the same sign. They are both positive or they are both negative or they are both 0. If r is positive, the regression line has a positive slope and rises from left to right. If r is negative, the slope of the regression line is negative and it falls from left to right.

With no significant linear correlation, the best predicted value is y = 161.69 cm.

6.

With a significant linear correlation, the best predicted value is yˆ = - 212 + 61.9(6.5) = 190 lb.

7.

With a significant linear correlation, the best predicted value is yˆ = - 106 +1.10(180) = 92.0 kg.

8.

With no significant linear correlation, the best predicted value is y = 1.26 mg of nicotine.

9.

yˆ = 3.00 + 0.500x; The data have a pattern that is not a straight line.

Scatterplot for Exercise 9

Scatterplot for Exercise 10

10. yˆ = 3.00 + 0.500x; There is an outlier.

11. a.

yˆ = 0.264 + 0.906x

b. yˆ = 2 + 0x (or yˆ = 2 ) c. The results are very different, indicating that one point can dramatically affect the regression equation.

12. a.

yˆ = 0.0846 + 0.985x

b. yˆ = 1.5 + 0x (or yˆ = 1.5 ) c.

yˆ = 9.5 + 0x (or yˆ = 9.5 )

d. The results are very different, indicating that combinations of clusters can produce results that differ dramatically from results within each cluster alone.

13. H0: r = 0; H1: r ¹ 0; r = 0.441; P-value = 0.174; yˆ = 713 + 3.99x; With no significant linear correlation, the best predicted value is y = 1108.5 cm3. The best predicted value isn’t very close to the actual brain volume of 1160 cm3.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

14. H0: r = 0; H1: r ¹ 0; r = - 0.279; P-value = 0.407; yˆ = 2441 - 5.49x; With no significant linear correlation, the best predicted value is y = 1896.6 cm2. The best predicted value isn’t very close to the actual brain surface area of 1914 cm2.

15. H0: r = 0; H1: r ¹ 0; r = 0.831; P-value = 0.001; yˆ = 21.6 + 5.40x; With a significant linear correlation, the best predicted value is yˆ = 21.6 + 5.40(6.0) = 54.0 ft.

16. H0: r = 0; H1: r ¹ 0; r = 0.930; P-value = 0.00001; yˆ = 1.33 + 0.941x; With a significant linear correlation, the best predicted value is yˆ = 1.33 + 0.941(70) = 67.2 kg. The best predicted value is close to the actual April weight of 68 kg.

17. H0: r = 0; H1: r ¹ 0; r = 0.591; P-value = 0.294; yˆ = 125 +1.73x; With no significant linear correlation, the best predicted value is y = 177.3 cm. The best predicted value is not close to the actual height of 190.5 cm.

Because the best predicted height is the mean height, it would not be helpful to police in trying to describe the male.

18. H0: r = 0; H1: r ¹ 0; r = 0.827; P-value = 0.084; yˆ = 85.2 + 3.52x; With no significant linear correlation, the best predicted value is y = 177.3 cm. The best predicted value is not close to the actual height of 185.4 cm.

Because the best predicted height is the mean height, it would not be helpful to police in trying to describe the male.

19. H0: r = 0; H1: r ¹ 0; r = 0.998; P-value = 0.000; yˆ = - 0.644 + 0.0866x; With a significant linear correlation, the best predicted value is yˆ = - 0.644 + 0.0866(25) = 1.5 mg. The best predicted value is the same as the actual amount of 1.5 mg of nicotine.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

20. H0: r = 0; H1: r ¹ 0; r = 0.003; P-value = 0.992; yˆ = 15.9 + 0.000727x; With no significant linear correlation, the best predicted value is y = 15.9 mg. The best predicted value is somewhat close to the actual amount of 18 mg carbon monoxide.

21. H0: r = 0; H1: r ¹ 0; r = - 0.231; P-value = 0.389; yˆ = 371- 3.06x; With no significant linear correlation, the best predicted value is y = 129.9 sec. The best predicted value is not at all close to the actual amount of 69 sec of time showing alcohol or tobacco.

22. H0: r = 0; H1: r ¹ 0; r = 0.941; P-value = 0.002; yˆ = 35.5 + 0.151x; With a significant linear correlation, the best predicted value is yˆ = 35.5 + 0.151(200) = 65.7 thousand (or 65,700).

23. H0: r = 0; H1: r ¹ 0; r = 0.857; P-value = 0.007; yˆ = 350 + 5.21x; With a significant linear correlation, the best predicted value is yˆ = 350 + 5.21(273) = 1772 mm. The best predicted height is close to the actual height.

24. H0: r = 0; H1: r ¹ 0; r = 0.874; P-value = 0.005; yˆ = 27.6 + 0.0523x; With a significant linear correlation, the best predicted value is yˆ = 27.6 + 0.0523(3000) = 185°F (or 184°F). The value of 3000 chirps in one minute is well beyond the scope of the listed sample data, so the extrapolation might be off by a considerable amount, especially if the cricket is dead from such a high temperature.

25. H0: r = 0; H1: r ¹ 0; r = 0.840; P-value = 0.005; yˆ = - 432 + 5.73x; With a significant linear correlation, the best predicted value is yˆ = - 432 + 5.73(97) = 123.8 fatalities. The predicted number of fatalities is not at all close to the actual number of 90 fatalities.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

26. H0: r = 0; H1: r ¹ 0; r = 0.959; P-value = 0.000; yˆ = - 988 +157x; With a significant linear correlation, the best predicted value is yˆ = - 988 +157(12.0) = 896 (or 897). Given the spurious nature of the data, it is not too likely that the predicted value will be accurate.

27. H0: r = 0; H1: r ¹ 0; r = - 0.959; P-value = 0.010; yˆ = 16.5 - 0.00282x; With a significant linear correlation, the best predicted value is yˆ = 16.5 - 0.00282(500) = 15.1 fatalities per 100,000 population. Common sense suggests that the prediction doesn’t make much sense.

28. H0: r = 0; H1: r ¹ 0; r = 0.948; P-value = 0.004; yˆ = - 157 + 40.2x; With a significant linear correlation, the best predicted value is yˆ = - 157 + 40.2(2) = - 76.6 kg (or - 76.5 kg). The prediction is a negative weight that cannot be correct. The overhead width of 2 cm is well beyond the scope of the sample widths, so the extrapolation might be off by a considerable amount. Clearly, the predicted negative weight makes no sense.

29. H0: r = 0; H1: r ¹ 0; r = 0.941; P-value = 0.000; yˆ = 5.05 + 0.941x; With a significant linear correlation, the best predicted value is yˆ = 5.05 + 0.941(70) = 70.9 kg. The best predicted value is somewhat close to the actual April weight.

30. H0: r = 0; H1: r ¹ 0; r = - 0.132; P-value = 0.362; yˆ = 248 - 2.00x; With no significant linear correlation, the best predicted value is y = 89.9 sec, which is not close to the actual value of 69 sec.

31. H0: r = 0; H1: r ¹ 0; r = 0.718; P-value = 0.000; yˆ = 737 + 3.76x; With a significant linear correlation, the best predicted value is yˆ = 737 + 3.76(273) = 1763 mm, which is close to the actual height of 1776 mm.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

32. H0: r = 0; H1: r ¹ 0; r = 0.814; P-value = 0.000; yˆ = - 65.5 +1.63x; With a significant linear correlation, the best predicted value is yˆ = - 65.5 +1.63(97) = 92.6 manatee fatalities, which is close to the actual value of 90 manatee fatalities.

33. a. See table below. b. For yˆ = - 30.33114785 +1.358246519x, the sum of squares of the residuals is 25.4765179 (or 25.715008 if using the rounded y-intercept of ⎼30.3 and the rounded slope of 1.36). c. For yˆ = - 30 + 2x, the sum of squares of the residuals is 20,686.34, which is larger than 25.4765179. yˆ = - 30.33114785 +1.358246519x

yˆ = - 30 + 2x

x

yˆ

y - yˆ

( y - yˆ)2

x

yˆ

y - yˆ

( y - yˆ)2

68.0 72.7 72.0

62.0 68.4 67.5

106.0 115.4 114.0

68.8 66.1 65.4 63.4 63.0 64.7 62.0

8.8231844 2.5173818 1.0761956 5.3863356 0.8021800 0.0055903 0.0125749 1.6393934 1.0943430 4.1193388

68.0 72.7 72.0

73.0 71.0 70.5 69.0 68.7 70.0 68.0

2.9703846 1.5866259 1.0373985 –2.3208480 0.8956450 0.0747683 0.1121380 –1.2803880 –1.0461085 –2.0296154

73.0 71.0 70.5 69.0 68.7 70.0 68.0

116.0 112.0 111.0 108.0 107.4 110.0 106.0

–41.0 –45.4 –45.5 –49.5 –45.0 –45.5 –44.5 –45.7 –46.3 –46.0

1681.00 2061.16 2070.25 2450.25 2025.00 2070.25 1980.25 2088.49 2143.69 2116.00

Sum:

25.4765179

Sum:

20,686.34

Section 10-3: Prediction Intervals and Variation 1.

2.

3.

The value of se = 16.27555 cm is the standard error of estimate, which is a measure of the differences between the observed weights and the weights predicted from the regression equation. It is a measure of the variation of the sample points about the regression line.

We have 95% confidence that the limits of 59.0 kg and 123.6 kg contain the value of the weight for a male with a height of 180 cm. The major advantage of using a prediction interval is that it provides us with a range of likely weights, so we have a sense of how accurate the predicted weight is likely to be. The terminology of prediction interval is used for an interval estimate of a variable, whereas the terminology of confidence interval is used for an interval estimate of a population parameter.

The coefficient of determination is r2 = 0.155. We know that 15.5% of the variation in weight is explained by the linear correlation between height and weight, and 84.5% of the variation in weight is explained by other factors and/or random variation.

4.

For the paired weights, se = 0 because there is an exact conversion formula with no error. For a student who weighs 100 lb, the predicted weight is 45.4 kg, and there is no prediction interval because the conversion yields an exact result.

5.

r2 = (0.874)2 = 0.764; 76.4% of the variation in temperature is explained by the linear correlation between chirps and temperature, and 23.6% of the variation in temperature is explained by other factors and/or random variation.

6.

r2 = (0.303)2 = 0.092; 9.2% of the variation in heights of sons is explained by the linear correlation between heights of mothers and heights of their sons, and 90.8% of the variation in heights of sons is explained by other factors and/or random variation.

7.

r2 = (0.934)2 = 0.872; 87.2% of the variation in weights of bears is explained by the linear correlation between neck size and weight, and 12.8% of the variation in weights is explained by other factors and/or random variation.

8.

r2 = (0.783)2 = 0.613; 61.3% of the variation in weight is explained by the linear correlation between head width and weight, and 38.7% of the variation in weight is explained by other factors and/or random variation.

9.

r = 0.726; Critical values (a = 0.05) : r = ±0.159, (Table: r » ±0.159); There is sufficient evidence to support a

claim of a linear correlation between waist circumference and arm circumference for males.

10. r2 = (0.726)2 = 0.527, or 52.7% 11. yˆ = 13.491069 + 0.20135359(100) = 33.6 cm 12. The 95% prediction interval estimate is 28.0 cm < y < 39.3 cm. We have 95% confidence that the limits of 28.0 cm and 39.3 cm contain the arm circumference.

13. 99% CI: yˆ ± E = 73.39 ± 34.99 Þ 38.4 manatee fatalities < y < 108.4 manatee fatalities yˆ = - 65.5 +1.634(85) = 73.39 manatee fatalities

E = ta /2se

1+

1

n

+

n (x 0 - x )2

= (2.779)(12.36941734) 1 +

28(85 - 86.4)2 1 = 34.99 + 28 28(211, 919)- (2419)2

14. 95% CI: yˆ ± E = 73.39 ± 26.44 Þ 68.2 manatee fatalities < y < 121.1 manatee fatalities yˆ = - 65.5 +1.634(98) = 94.63 manatee fatalities

E = ta /2se

1+

1

n

+

n (x 0 - x )2

= (2.0556)(12.36941734) 1+

28(98 - 86.4)2 1 = 26.44 + 28 28(211, 919)- (2419)2

15. 95% CI: yˆ ± E = 91.36 ± 26.27 Þ 65.1 manatee fatalities < y < 117.6 manatee fatalities yˆ = - 65.5 +1.634(96) = 91.36 manatee fatalities

E = ta /2se

1+ 1 + n

n (x 0 - x )2

= (2.0556)(12.36941734) 1+

28(96 - 86.4)2 1 + = 26.27 28 28(211, 919)- (2419)2

16. 99% CI: yˆ ± E = 76.66 ± 34.99 Þ 41.7 manatee fatalities < y < 111.6 manatee fatalities yˆ = - 65.5 +1.634(87) = 76.66 manatee fatalities E = ta /2se 1+

1

n

+

n (x 0 - x )2

= (2.779)(12.36941734) 1 +

28(87 - 86.4)2 1 = 34.99 + 28 28(211, 919)- (2419)2

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

17. a. 618.9541 b. 304.7126 c. 95% CI: 56.0 mmHg < y < 81.6 mmHg yˆ = 23.6 + 0.377 (120) = 68.8 1 + n

E = ta /2se

n (x 0 - x )2

= 2.2283(5.52008)

+

12 (120 - 121)2 1 = 12.8 + 12 12(180, 048)- (1452)2

18. a. 3714.141 b. 1659.426 c. 99% CI: 0.607 ft < y < 85.8 ft; The interval is wide enough to be of no practical use. yˆ = 21.6 + 5.401(4.0) = 43.2 E = ta /2se

1+

1 + n

n (x 0 - x )2

= 3.1693(12.88187) 1+

12 (4.0 - 4.91)2 1 = 42.6 + 12 12 (416.41)- (58.9)2

19. a. 352.7278 b. 109.3722 c. 90% CI: 71.09°F < y < 88.71°F yˆ = 27.628 + 0.05227(1000) = 79.90 E = ta /2 ×se 1+

1

n

2

+

n (x0 - x )

8(1000 - 1016.25)2 1 = 1.943(4.2695) 1+ + = 8.81.

8 8(8, 391, 204)- (8130)2

20. a. 8880.12 b. 991.1515 c. 99% CI: 125.0 kg < y < 284.5 kg yˆ = - 156.879 + 40.182 (9.0) = 204.76 E = ta /2 ×se 1+

1

n

2

+

n (x0 - x )

1

6(9.0 - 8.5)2

= 4.604 (15.7413) 1+ + = 79.79 6 6(439)- (51)2

21. 95% CI: yˆ ± E = 64.05 ±1.42 Þ 62.6 in. < y < 65.5 in. yˆ = - 30.33 +1.358(69.5) = 64.05 E = ta /2 ×se

1

n

+

n (x 0 - x )2

= (2.306)(1.784534879)

10 (69.5 - 70.29)2 1 + = 1.42 10 10 (49, 438.23)- (702.9)2

Section 10-4: Multiple Regression 1. The response variable is nicotine, and the predictor variables are tar and CO (carbon monoxide).

2. Only 9.9% of the variation in nicotine can be explained by the variables of tar and carbon monoxide, so 90.1% of the variation in nicotine can be explained by other factors and/or random variation. Given that the percentage of the explained variation is so small, it appears that the regression equation is not very useful for predicting nicotine when given amounts of tar and carbon monoxide.

3. The unadjusted R2 increases (or remains the same) as more variables are included, but the adjusted R2 is adjusted for the number of variables and sample size. The unadjusted R2 incorrectly suggests that the best multiple regression equation is obtained by including all of the available variables, but by taking into account the sample size and number of predictor variables, the adjusted R2 is much more helpful in weeding out variables that should not be included.

4.

No, it is not better to use the regression equation with the three predictor variables of length, chest size, and neck size. The adjusted R2 value of 0.925 is just a little less than 0.933, so in this case it is better to use two predictor variables instead of three.

5.

Son = 18.0 + 0.504 Father + 0.277 Mother

6.

a. P-value < 0.0001 b. R2 = 0.3649 c. adjusted R2 = 0.3552

7.

A P-value less than 0.0001 is low, but the values of R2 (0.3649) and adjusted R2 (0.3552) are not high.

Although the multiple regression equation fits the sample data best, it is not a good fit, so it should not be used for predicting the height of a son based on the height of his father and the height of his mother.

8.

Predicted height: Son = 18.0 + 0.504(70)+ 0.277(60) = 70 in.; This result is not likely to be a good predicted value because the multiple regression equation is not a good model (based on the results from Exercise 7).

Waist circumference, because it has the highest adjusted R2 (0.784) value and the P-values are the same (0.000) for the three individual variables.

10. Waist circumference and arm circumference, because they have the highest adjusted R2 value (0.878) and the P-values are the same (0.000) for the other combinations of two variables.

9.

11. Some subjective judgment is involved here. A good choice is to use all three variables because the adjusted R2 value (0.939) is considerably higher than any of the other values of adjusted R2. With this reasoning, the best regression equation is yˆ = - 147 + 0.632HT + 0.697WAIST +1.58ARM. 12. The best predicted weight is yˆ = - 147 + 0.632(186) + 0.697(107.8) +1.58(37.0) = 104 kg. Because the P-value is low and the adjusted R2 value (0.939) is fairly high, the predicted value is likely to be a fairly good estimate, but it might not be very accurate. (With technology, the 95% prediction interval shows that the actual weight is likely to be between 95.0 kg and 112.4 kg, so the predicted value of 104 kg should be in the general neighborhood of the actual weight.)

13. The best regression equation is yˆ = 0.127 + 0.0878x1 - 0.0250x2 , where x1 represents tar and x2 represents carbon monoxide. It is best because it has the highest adjusted R2 value of 0.927 and the lowest P-value of 0.000. It is a good regression equation for predicting nicotine content because it has a high value of adjusted R2 and a low P-value.

The possible models are shown below: yˆ = 0.127 + 0.0878x1 - 0.0250x2 ; adjusted R2 = 0.934 yˆ = 0.800 + 0.0633x1; adjusted R2 = 0.882 yˆ = 0.328 + 0.0397 x2; adjusted R2 = 0.460

14. The three different possible regression equations all have a P-value of 0.000. Using the two predictor variables of FootLength and ArmSpan yields an adjusted R2 of 0.827, using the single predictor variable of FootLength yields an adjusted R2 of 0.714, and using the single predictor variable of ArmSpan yields an adjusted R2 of 0.808. Because the highest adjusted R2 is 0.827, which is somewhat greater than the next highest adjusted R2 of 0.808, the best regression equation uses both predictor variables, so the best regression equation is HEIGHT = 393 + 0.540 ARMSPAN +1.40 FOOTLENGTH. Because the adjusted R2 of 0.827 isn’t close to 1, it is likely that predicted values will not be very accurate. (Comment: It would also be reasonable to conclude that the adjusted R2 of 0.808 is not dramatically less than 0.827, so it would be reasonable to conclude that the best regression equation is HEIGHT = 427 + 0.730 ARMSPAN.) (TI data: P-value of 0.000 for all three possible regression equations. Using FootLength and ArmSpan as predictor variables yields an adjusted R2 of 0.825, using only FootLength yields 0.702, and using only ArmSpan yields 0.815. Because 0.815 is only slightly less than 0.825, the best regression is HEIGHT = 451+ 0.716 ARMSPAN.)

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

14. (continued) The possible models are shown below: HEIGHT = 564 + 4.37FOOTLENGTH, adjusted R2 = 0.714 HEIGHT = 427 + 0.730ARMSPAN, adjusted R2 = 0.808 HEIGHT = 393+ 0.540ARMSPAN +1.40FOOTLENGTH, adjusted R2 = 0.827 15. The best regression equation is yˆ = 109 - 0.00670x1, where x1 represents volume. It is best because it has the highest adjusted R2 value of –0.0513 and the lowest P-value of 0.791. The three regression equations all have adjusted values of R2 that are very close to 0, so none of them are good for predicting IQ. It does not appear that people with larger brains have higher IQ scores.

The possible models are shown below: yˆ = 109 - 0.00670x1, adjusted R2 = - 0.0513 yˆ = 101- 0.00178x2, adjusted R2 = - 0.0555 yˆ = 108 - 0.00694x1 + 0.00722x2, adjusted R2 = - 0.113 16. The best regression equation is yˆ = - 10.0 + 0.567x1 + 0.532x2 , where x1 represents verbal IQ score and x2 represents performance IQ score. It is best because it has the highest adjusted R2 value of 0.999 and the lowest P-value of 0.000. Because the adjusted R2

is so close to 1, it is likely that predicted values will be very

accurate.

The possible models are shown below: yˆ = 11.5 + 0.940x1, adjusted R2 = 0.762 yˆ = 10.5 + 0.806x2, adjusted R2 = 0.814 yˆ = - 10.0 + 0.567x1 + 0.532x2, adjusted R2 = 0.999 0.769317 - 0 17. For H : b = 0, Test statistic: t = = 10.814; P-value < 0.0001; Reject H and conclude that the 0 1 0 0.0711414 regression coefficient of b1 = 0.769 should be kept.

1.009510 - 0 = 29.856; P-value < 0.0001; Reject H and conclude that the For H : b = 0, Test statistic: t = 0 2 0 0.0338123 regression coefficient of b2 = 1.01 should be kept.

It appears that the regression equation should include both independent variables of height and waist circumference.

18. 0.629 < b1 < 0.910; 0.943 < b 2 < 1.08; Neither confidence interval includes 0, so neither of the two variables should be eliminated from the regression equation.

CI for b1 b1 - E < b1 < b1 + E b1 - ta /2 sb < b1 < b1 + ta /2sb 1

1

0.7693 - 1.976(0.0711414) < b1 < 0.7693 +1.976(0.0711414) 0.629 < b1 < 0.910 CI for b 2 b2 - E < b 2 < b2 + E b2 - ta /2 sb < b 2 < b2 + ta /2 sb 2

2

1.0095 - 1.976(0.033812) < b 2 < 1.0095 - 1.976(0.033812) 0.943 < b 2 < 1.08

Section 10-5: Dummy Variables and Logistic Regression 1. A dummy variable is a variable that has only two values and those values (such as 0 and 1) are used to represent the different categories of a qualitative variable.

2.

3.

With logistic regression, the response variable ( y) is a dummy variable that assumes only the values of 0 and 1, whereas linear regression has a response variable that is continuous.

With simple logistic regression, there is only one predictor variable, but with multiple logistic regression, there are two or more predictor variables.

4.

False. If the only dummy variable is a predictor (x) variable, we can use the methods of multiple regression from Section 10-4.

5.

There is a probability of

6.

There is a probability of

7.

8.

9.

e- 40.6+0.242(170) 1+ e- 40.6+0.242(170) e- 40.6+0.242(152.4)

= 0.632 that the person is male.

= 0.0239 that the person is male, so there is a probability of 1+ e- 40.6+0.242(152.4) 1- 0.0237 = 0.976 that the person is female. æ p ö e- 1.63+0.0505(40) ln ç = 1.63 + 0.0505(Arm Circumference); There is a probability of = 0.596 that the è1 - p ÷ ø 1+ e- 1.63+0.0505(40) person is a male. æ p ö e- 1.63+0.0505(24.5) ln ç = 1.63 + 0.0505(Arm Circumference); There is a probability of = 0.403 that è1 - p ÷ ø 1+ e- 1.63+0.0505(24.5) the person is a male.

Length and weight are predictor variables. The response variable is sex, which is a dummy variable.

10. Because the regression equation has the high overall P-value of 0.218, the predicted value is not likely to be very accurate. e2.40- 0.0553(60)+0.00826(300) 11. The probability that the bear is a male is = 0.826. The probability that the bear is 2.40- 0.0553(60)+0.00826(300) 1+ e a female is 1- 0.826 = 0.174. e2.40- 0.0553(40)+0.00826(50) 12. The probability that the bear is a male is = 0.646. The probability that the bear is a 2.40- 0.0553(40)+0.00826(50) 1+ e female is 1- 0.646 = 0.354. The relatively small length and the very small weight suggest that the bear is very young and has not yet reached its adult length and weight.

13. Weight = 3.06 +82.4(Sex)+ 2.91(Age) a. Female: Weight = 3.06 +82.4(0)+ 2.91(20) = 61 lb b. Male: Weight = 3.06 +82.4(1)+ 2.91(20) =144 lb The sex of the bear does appear to have an effect on its weight. The regression equation indicates that the predicted weight of a male bear is about 82 lb more than the predicted weight of a female bear with other characteristics being the same.

14. Weight = - 56.2 + 0.828(Heights)- 2.37(Sex) a. Female: Weight = - 56.2 + 0.828(170)- 2.37(0) = 84.6 kg b. Male: Weight = - 56.2 + 0.828(170)- 2.37(1) = 82.2 kg The regression equation indicates that the predicted weight of a male is 2.4 kg less than the predicted weight of a female with the same height. This doesn’t make sense because we expect that for the same value of height, a male would likely weigh more than a female.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

æ p ö

= - 41.2 + 0.250(Height)- 0.00856(Weight) çè1 - p ÷ ø e- 41.2+0.250(170)- 0.00856(90) The probability of a male is = 0.629.

- 41.2+0.250(170)- 0.00856(90) 1+ e æ p ö = - 38.3 + 0.245(Height)- 0.0289(Pulse); The probability of a male is 16. ln çè1 - p ÷ ø 15. ln

e- 38.3+0.245(170)- 0.0389(60)

= 0.734. The negative coefficient for Pulse shows that higher pulse rates make - 38.3+0.245(170)- 0.0389(60) 1+ e males less likely æ p ö = - 101.5 + 0.191(Foot Length)+ 0.301(Height) 17. ln çè1 - p ÷ ø The probability of a male is

e- 101.5+0.191(28)+0.301(190) = 0.9999.

- 101.5+0.191(28)+0.301(190) 1+ e

æ p ö

e- 70.8+0.993(9)+0.350(170)

18. ln ç = - 70.8 + 0.993(Shoe Size)+ 0.350 (Height); The probability of a male is - 70.8+0.993(9)+0.350(170) è1- p ø÷ 1+ e = 0.086. Predictions made using foot length and height are likely to be more accurate than those made with shoe size and height because foot lengths have more precision than shoe sizes, which are basically rounded values of foot lengths. Foot lengths contain more and better information than shoe sizes. æ p ö 19. a. ln = - 0.905 + 0.000291(Birth Weight) + 0.00323(Length of Stay) çè1 - p ÷ ø b. The probability of a male is

e- 0.905+0.000291(4000)+0.00323(4)

= 0.568.

1+ e- 0.905+0.000291(4000)+0.00323(4) c. The P-values show that the birth weight and length of stay are good predictor variables, and the P-value of 0.000 for the regression equation shows that it is good for predictions. æ p ö 20. a. ln = 15.7 - 0.000781(Age) - 0.163(Body Temperature) çè1 - p ÷ ø e15.7- 0.000781(50)- 0.163(98.0) b. The probability of a male is = 0.422.

1+ e15.7- 0.000781(50)- 0.163(98.0) c. The P-values suggest that the predictor variable of age is not a good predictor variable, but the predictor variable of body temperature is a good predictor variable, and the regression equation is good for predictions.

21. a. The probability of a male is

e- 0.510- 0.00296(110)+0.0123(50)

= 0.445.

1+ e- 0.510- 0.00296(110)+0.0123(50) b. The P-values suggest that the regression equation is not good for predicting, and the variables of IQ and weight are not good predictor variables. c. The probability of 0.445 found in part (a) is not a good result.

Quick Quiz 1. 0.479 > 0.05 or - 0.632 < - 0.254 < 0.632; Conclude that there is not sufficient evidence to support the claim ofa linear correlation between systolic blood pressure measurements and diastolic blood pressure measurements 2. None of the given values change when the variables are switched.

3. The value of r does not change if all values of one of the variables are multiplied by the same constant.

4. Because r must be between –1 and 1 inclusive, the value of 1.500 is the result of an error in the calculations.

5.

- 0.632 < - 0.254 < 0.632; With no significant linear correlation, the best predicted value is

y = 77.8 mmHg,

the mean of the 10 measurements of diastolic blood pressure.

6.

0.997 > 0.632; With a significant linear correlation, the best predicted value is yˆ = 103 - 0.198(120)

= 79.2 mmHg. 7.

Because r2 = (- 0.254)2 = 0.065, it follows that 0.065 (or 6.5%) of the variation in diastolic blood pressure is

8.

9.

explained by the relationship between diastolic blood pressure and systolic blood pressure. false 10. r =- 1 false

Review Exercises 1. H0: r = 0; H1: r ¹ 0; r = 0.951; P-value = 0.000 (Table: < 0.01); Critical values (a = 0.05): ± 0.632; Reject H 0. There is sufficient evidence to support the claim that there is a linear correlation between verbal IQ scores and full IQ scores.

2.

a

yˆ = 16.6 + 0.857x

b. The best predicted value is of 101.

3.

yˆ = 16.6 + 0.857(95) = 98.0, which is reasonably close to the actual full IQ score

r2 = (0.9511)2 = 0.9046; 90.5% of the variation in full IQ scores can be explained by the correlation between verbal IQ scores and full IQ scores.

4.

a.

H0: r = 0; H1: r ¹ 0; r = 0.450; P-value = 0.192 (Table: P-value > 0.05); Critical values (a = 0.05): r = ±0.632; Fail to reject H 0. There is not sufficient evidence to support the claim that there is a linear

correlation between time and population size. b. Although there is no linear correlation between time and population size, the scatterplot shows a very distinct pattern revealing that time and population size are associated by a function that is not linear.

5.

6.

a. NICOTINE = - 0.443+ 0.0968TAR - 0.0262CO, or yˆ = - 0.443 + 0.0968x1 - 0.0262x2 , where x1 represents tar and x2 represents carbon monoxide. b. R2 = 0.936; adjusted R2 = 0.910; P-value = 0.001 c. With high values of R2 and adjusted R2 and a small P-value of 0.001, it appears that the regression equation can be used to predict the amount of nicotine given the amounts of tar and carbon monoxide. d. The predicted value is yˆ = - 0.443 + 0.0968(23) - 0.0262(15) =1.39 mg or 1.4 mg rounded, which is close to the actual value of 1.3 mg of nicotine. æ p ö ln = - 44.4 + 0.557 (8 AM Temp)- 0.098(12 AM Temp); The probability that the subject smokes is çè1 - p ÷ ø e- 44.4+0.557(98.25)- 0.098(98.25) = 0.667. Because the regression equation has coefficients with high P-values - 44.4+0.557(98.25)- 0.098(98.25) 1+ e of 0.271 and 0.871, the predicted value is not likely to be very accurate.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

Cumulative Review Exercises 1. a. Is there a difference between the mean IQ score of airline passengers and the mean IQ score of police officers? b. Test for a difference between the means of two independent populations using the methods of Section 9-2. c. H0: m1 = m2; H1: m1 ¹ m2; population1 = airline passengers, population2 = police officers Test statistic : t =- 1.557; P-value = 0.1516 (Table: P-value > 0.05); Critical values (a = 0.05): t = ±2.239 (Table: t ± 2.262); Fail to reject H 0. There is not sufficient evidence to support the claim that there is a difference between the mean IQ score of airline passengers and the mean IQ of police officers.

(This 95% confidence interval could also be used: - 20.0 < m1 - m2 < 3.59. Because the confidence interval includes 0, there is not sufficient evidence to support the claim that there is a difference between the mean IQ score of airline passengers and the mean IQ of police officers.) (x1 - x2 )- (m1 - m2 ) = (102.1- 110.3)- 0 t= s12 s22 16.362 3.092 + + 10 10

(x1 - x2 )± ta /2 2.

3.

4.

s12

2 2 s2 16.36 + 3.09 + 2 = (102.1- 110.3)± 2.262 × 10 10

(df = 10 - 1 = 9)

a. Was the training course effective in raising the IQ scores? That is, do the “before – after” differences have a mean that is less than 0, showing that the course is effective in raising IQ scores? b. Use the methods of Section 9-3 to test the claim that the mean of the “before – after” differences is less than 0, showing that the course is effective with larger after scores. c. H0: md = 0; H1: md < 0; difference = before - after Test statistic: t = - 1.541; P-value=0.0789 (Table: P-value > 0.05); Critical value (a = 0.05): t = - 1.833; Fail to reject H 0. There is not sufficient evidence to support the claim that the course is effective with higher after scores. (This 90% confidence interval could also be used: - 18.0 < md < 1.56). Because the confidence interval includes 0, there is not sufficient evidence to support the claim that the course is effective with higher after scores.

- 8.2 - 0 d - md t= = sd / sn 16.83 / 10 d 16.83 d ±t = - 8.2 ±1.833× (df = 10 - 1 = 9) a /2 n 10 a. For teenage females, is there a correlation between weights and post-exercise pulse rates? b. Use the methods of Section 10-1 to test for a linear correlation. c. H0: r = 0; H1: r ¹ 0; r = - 0.060; P-value = 0.869 (Table: P-value > 0.05); Critical values (a = 0.05): r = ±0.632; There is not sufficient evidence to support the claim that there is a linear correlation between weights and post-exercise pulse rates. a. Because the table lists time-series data, a key question is this: What is the trend of the data over time? b. Use the methods of Section 2-3 to construct a time-series graph that will reveal a trend of the data over time. c. A time-series graph clearly shows that there is a distinct trend of steadily increasing numbers of influenza cases until 2019, when the numbers of cases began to stabilize.

4.

(continued)

5.

a. What is an estimate of the proportion of all adults who say that they would not get the COVID-19 vaccine? b. Use the methods of Section 7-1 to construct a confidence interval estimate of the proportion of all adults who say that they would not get the COVID-19 vaccine. c. 95% CI: 0.233 < p < 0.287; With 95% confidence, it is estimated that between 23.3% and 28.7% of all adults say that they would not get the COVID-19 vaccine.

It would also be reasonable to conduct a hypothesis test, such as a test of the claim that fewer than 50% of adults have wireless earbuds. For the test: H0: p = 0.5; H1: p < 0.5, the test statistic is z =- 15.21 and P-value = 0.0000, so there is sufficient evidence to support the claim that fewer than 50% of adults say that they would not get the COVID-19 vaccine.

261 - 0.50 pˆ - p 1004 z= = (0.50)(0.50) pq 1004 n pˆ ± za /2

6.

pˆqˆ 261 = 1004 ± 1.96 n

261 261 (1004 ) )(1- 1004

1004 a. Is Anthony Fauci significantly short in the population of adult males? b. Using the methods of Section 3-3, convert Anthony Fauci’s height to a z score and use the range rule of thumb to determine whether his height is significantly high.

170.18 - 174.12 c. Converting Anthony Fauci’s height to a z score, we get z = = - 0.55. Anthony Fauci’s 7.10 height is 0.55 standard deviations below the mean, so his height is not significantly low.

170.18 - 174.12 Using the methods of Section 6-2, z = = - 0.55 has an area of 0.2895 to the left. (Table: 7.10 0.2912). Therefore, among adult males, about 29% of the population have heights equal to or less than the height of Anthony Fauci. His height is not significantly low.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

7.

a. Is the mean amount provided by the new device equal to 16 ounces? Is there anything else about the data suggesting that there is a problem with the new device? b. Explore the sample data to see if there are any undesirable characteristics. Use the methods of Section 8-3 to test the claim that the mean of the amounts is equal to 16 ounces. c. H0: m = 16 oz; H1: m ¹ 16 oz; Test statistic : t = 0.998; P-value = 0.3363 (Table: P-value > 0.20); Critical values (a = 0.05): t = ±2.160; Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim that the mean is equal to 16 oz. (This 95% confidence interval could also be used: 15.42 oz < m < 17.58 oz. Because the confidence interval includes 16 oz, there is not sufficient evidence to reject the claim that the mean is equal to 16 oz.)

Although the mean appears to be OK, exploring the data reveals that the variation appears to be too high.

The minimum sample value is 12.6 oz and the maximum is 19.2 oz, and they show that the amount of variation is far too high. Some containers would be dramatically underfilled while others would be overflowing. The filling device must be modified to correct the unacceptably high amount of variation. x - m 16.5 - 16.0 = s / n 1.874 / 14 s x ±t × 1.874 (df = 14 - 1 = 13) = 16.5 ± 2.160 × a /2 14 n a. What is an estimate of the proportion of wrong results in the population of all of the drug tests? b. Use the methods of Section 7-1 to construct a confidence interval estimate of the proportion p of wrong results in the population of all of the drug tests. Decide whether the proportion of wrong results is acceptable. c. 95% CI: 0.0576 < p < 0.122; This shows that we have 95% confidence that the percentage of wrong test results is between 5.76% and 12.2%. Because wrong test results could possibly have adverse implications, such as unfair rejection of job applicants, it appears that the proportion of wrong results is unacceptably high. t=

8.

pˆ ± za /2

27 p̂q̂ = ___ ±1.96 n 300

27 300

273 300

300

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

Chapter 11: Goodness-of-Fit and Contingency Tables Section 11-1: Goodness-of-Fit 1. a. Observed values are represented by O and expected values are represented by E. b. For the leading digit of 2, O = 62 and E = (317)(0.176) = 55.792. c. For the leading digit of 2, (O - E)2 E = 0.691.

2

H0: p1 = 0.301, p2 = 0.176, p3 = 0.125, , p9 = 0.046; H1: At least one of the proportions is not equal to the given claimed value.

3.

There is sufficient evidence to warrant rejection of the claim that the leading digits have a distribution that fits well with Benford’s law.

Because the leading digits of interarrival traffic times do not fit the distribution described by Benford’s law, it appears that those times are not typical. The anomaly could be due to other factors, but there is a good chance that the computer has been hacked. Computer experts should be used to correct the incursion and secure the computer against further incursions.

4.

5.

6.

7.

H0: p0 = p1 = p2 = p3 = p4 = p5 = p6 = p7 = p8 = p9 = 0.1; H1: At least one of the proportions is not equal to the given claimed value.

Test statistic: c 2 = 4.050; P-value = 0.908 (Table: P-value > 0.10); Critical value: c 2 = 16.919; Do not reject H 0. There is not sufficient evidence to warrant rejection of the claim that the last digits of the reported heights occur with about the same frequency. There is not sufficient evidence to conclude that the heights were reported instead of being measured.

(12 - 10.1)2 (9 - 10.1)2 (11- 10.1)2 (14 - 10.1)2 2 + + = 4.050 (df = 9) c = + + 0.1×101 0.1×101 0.1×101 0.1×101 H0: p0 = p1 = p2 = p3 = p4 = p5 = p6 = p7 = p8 = p9 = 0.1; H1: At least one of the proportions is not equal to the given claimed value.

Test statistic: c 2 = 96.762; P-value = 0.000 (Table: P-value < 0.005); Critical value: c 2 = 16.919; Reject H 0. There is sufficient evidence to reject the claim that the last digits of the reported heights occur with about the same frequency. The results suggest that the heights were reported and not measured. The larger data set in this exercise results in a different conclusion than the one obtained in the preceding exercise, so the larger data set does have a dramatic impact on the results.

(321- 278.4)2 (315 - 278.4)2 (297 - 278.4)2 (329 - 278.4)2 2 + = 96.762 (df = 9) c = + + + 0.1×2784 0.1×2784 0.1×2784 0.1×2784 H0: pMon = pTue = pWed = pThu = pFri = 1 5; H1: At least one of the proportions is not equal to the given claimed value.

Test statistic: c 2 = 0.523; P-value = 0.9712 (Table: P-value > 0.95); Critical value: c 2 = 9.488; Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim that injuries and illnesses occur with equal frequency on the different days of the week.

2 (23 - 21.4)2 (21- 21.4)2 (21 - 21.4)2 (19 - 21.4)2 2 (23 - 21.4) c = + + + + = 0.523 (df = 4) 107 5 107 5 107 5 107 5 107 5

8.

H0: p1 = p2 = p3 = p4 = 0.25; H1: At least one of the proportions is not equal to the given claimed value.

Test statistic: c 2 = 4.600; P-value = 0.204 (Table: P-value > 0.10); Critical value: c 2 = 7.815; Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim that the tires selected by the students are equally likely. It appears that when four students identify a tire, it is not likely that they would all select the same tire.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

8.

9.

(continued) 2 2 2 2 c 2 = (11- 10.25) + (15 - 10.25) + (18 - 10.25) + (6 - 10.25) = 4.600 (df = 3) 0.25×41 0.25×41 0.25×41 0.25×41 H0: p1 = 0.756, p2 = 0.091, p3 = 0.108, p4 = 0.038, p5 = 0.007; H1: At least one of the proportions is not equal to the given claimed value.

Test statistic: c 2 = 524.713; P-value = 0.0000 (Table: P-value < 0.005); Critical value: c 2 = 13.277; Reject H 0. There is sufficient evidence to warrant rejection of the claim that the distribution of clinical trial participants fits well with the population distribution. Hispanics have an observed frequency of 60 and an expected frequency of 391.027, so they are very underrepresented. Also, the Asian/Pacific Islander subjects have an observed frequency of 54 and an expected frequency of 163.286, so they are also underrepresented.

2 (60 - 391.027)2 (54 - 163.286)2 (12 - 30.079)2 2 (3855 - 3248.532) + + = 524.713 (df = 4) c = + + 0.756×4297 0.091×4297 0.038×4297 0.007 ×4297 10. H0: p1 = p2 = p3 = p4 = p5 = p6 = p7 = 1 7 ; H1: At least one of the proportions is not equal to the given claimed value.

Test statistic: c 2 = 30.017; P-value = 0.0000 (Table: P-value < 0.005); Critical value: c 2 = 12.592; Reject H 0. There is sufficient evidence to warrant rejection of the claim that the car crash fatalities occur with equal frequency on the different days of the week. Drinking on Friday night might be a cause of an exceptionally large number of deaths on early Saturday morning.

(132 - 117)2 (98 - 117)2 (133 - 117)2 (158 - 117)2 2 + c = + + + = 30.017 (df = 6) 819 7 819 7 819 7 819 7 11. H0: pYS = 9 16 , pGS = 3 16 , pYW = 3 16 , pGW = 1 16; H1: At least one of the proportions is not equal to the given claimed value.

Test statistic: c 2 = 11.161; P-value = 0.011 (Table: P-value < 0.025); Critical value: c 2 = 7.815; Reject H 0.

There is sufficient evidence to support the claim that the results contradict Mendel’s theory.

2 2 2 2 c 2 = (307 - 281.25) + (77 - 93.75) + (98 - 93.75) + (18 - 32.25) = 11.161 (df = 3) 0.5625×500 0.1875×500 0.1875×500 0.0625×500 12. H0: The frequency counts agree with the claimed distribution.

H1: The frequency counts do not agree with the claimed distribution.

Test statistic: c 2 = 0.976; P-value = 0.913 (Table: P-value > 0.10); Critical value: c 2 = 9.488; Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim that the actual frequencies fit a Poisson distribution. There is no way to prove that the frequencies fit a Poisson distribution even though it appears that they do.

2 2 2 2 2 c 2 = (229 - 227.5) + (211- 211.4) + (93 - 97.9) + (35 - 30.5) + (8 - 8.7) = 0.976 (df = 4) 227.5 211.4 97.9 30.5 8.7 13. H0: pJan = pFeb = pMar = pApr = pMay = pJun = pJul = pAug = pSep = pOct = pNov = pDec = 1 12; H1: At least one of the proportions is not equal to the given claimed value.

Test statistic: c 2 = 47.200; P-value = 0.0000 (Table: P-value < 0.005); Critical value: c 2 = 19.675; Reject H 0. There is sufficient evidence to warrant rejection of the claim that motorcycle fatalities occur with equal frequencies in the different months. Fatalities might be lower in winter months when colder weather is associated with substantially less use of motorcycles.

2 (8 - 16.67)2 (10 - 16.67)2 (8 - 16.67)2 2 (6 - 16.67) c = + + + + = 47.200 (df = 11) 200 12 200 12 200 12 200 12

14. H0: pAA = 0.25, pAa = 0.50, paa = 0.25; H1: At least one of the proportions is not equal to the given claimed value.

Test statistic: c 2 = 11.552; P-value = 0.0031 (Table: P-value < 0.005); Critical value: c 2 = 5.991; Reject H 0. There is sufficient evidence to warrant rejection of the claim that the actual frequencies correspond to the predicted distribution.

2 2 2 c 2 = (20 - 36.25) + (90 - 72.5) + (35 - 36.25) = 11.552 (df = 2) 0.25×145 0.50 ×145 0.25×145 15. H0: pbrown = 0.87, pblue = 0.08, pgreen = 0.05; H1: At least one of the proportions is not equal to the given claimed value.

Test statistic: c 2 = 9.658; P-value = 0.0080 (Table: P-value < 0.01); Critical value: c 2 = 5.991; Reject H 0.

There is sufficient evidence to warrant rejection of the claim that the actual frequencies correspond to the predicted distribution.

2 2 2 c 2 = (132 - 1129.63) + (17 - 11.92) + (0 - 7.45) = 9.658 (df = 2) 0.87×149 0.08×149 0.05×149 16. H0: pJan = pFeb = pMar = pApr = pMay = pJun = pJul = pAug = pSep = pOct = pNov = pDec = 1 12; H1: At least one of the proportions is not equal to the given claimed value.

Test statistic: c 2 = 93.072; P-value = 0.0000 (Table: P-value < 0.005); Critical value: c 2 = 19.675; Reject H 0. There is sufficient evidence to warrant rejection of the claim that American-born Major League Baseball players are born in different months with the same frequency. The sample data appear to support Gladwell’s claim.

2 (329 - 376.25)2 (398 - 376.25)2 (371- 376.25)2 2 (387 - 376.25) c = + + + + = 93.072 (df = 11) 4515 12 4515 12 4515 12 4515 12 17. H0: pSun = pMon = pTue = pWed = pThu = pFri = pSat = 1 7 ; H1: At least one of the proportions is not equal to the given claimed value.

Test statistic: c 2 = 9.500; P-value = 0.147 (Table: P-value > 0.10); Critical value: c 2 = 16.812; Fail to reject H 0. There is not sufficient evidence to support the claim that births do not occur on the seven different days of the week with equal frequency.

Day Sun Mon Tue Wed Thu Fri Sat 57.14 57.14 57.14 57.14 57.14 57.14 57.14 Expected 53 52 66 72 57 57 43 Observed 2 (52 - 57.14)2 (57 - 57.14)2 (43 - 57.14)2 2 (53 - 57.14) c = + + + + = 9.500 (df = 6) 400 7 400 7 400 7 400 7

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

18. H0: pSun = pMon = pTue = pWed = pThu = pFri = pSat = 1 7 ; H1: At least one of the proportions is not equal to the given claimed value.

Test statistic: c 2 = 10.760; P-value = 0.096 (Table: P-value > 0.05); Critical value: c 2 = 16.812; Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim that discharges occur on the seven different days of the week with equal frequency.

Day Expected Observed

Mon 57.14 50

(65 - 57.14)2 (50 - 57.14)2

2

c

Sun 57.14 65

=

400 7

+

400 7

+

+

Tue 57.14 47

Wed 57.14 48

Thu 57.14 73

(64 - 57.14)2 (53 - 57.14)2 400 7

+

400 7

Fri 57.14 64

Sat 57.14 53

= 10.760 (df = 6)

19. The frequencies of the digits 0 through 9 are 102, 0, 2, 1, 0, 25, 0, 4, 0, and 0, respectively. Subjective examination of the last digits shows that there are disproportionately more 0’s and 5’s than expected.

H0: p0 = p1 = p2 = p3 = p4 = p5 = p6 = p7 = p8 = p9 = 0.1; H1: At least one of the proportions is not equal to the given claimed value.

Test statistic: c 2 = 690.627; P-value = 0.0000 (Table: P-value < 0.005); Critical value: c 2 = 16.919; Reject H 0. There is sufficient evidence to warrant rejection of the claim that the digits 0 through 9 occur with about the same frequency. It appears that some strange rounding was used.

Last Digit 0 1 2 3 4 5 6 7 8 13.4 13.4 13.4 13.4 13.4 13.4 13.4 13.4 13.4 Expected Observed 102 0 2 1 0 25 0 4 0 2 2 2 2 2 (102 13.4) (0 13.4) (2 13.4) (113.4) (0 13.4) 2 c = + + + + 0.1×134 0.1×134 0.1×134 0.1×134 0.1×134 (25 - 13.4)2 (0 - 13.4)2 (4 - 13.4)2 (0 - 13.4)2 (0 - 13.4)2 + + + + + = 690.627 0.1×134 0.1×134 0.1×134 0.1×134 0.1×134 20.

9 13.4 0

H0: p0 = p1 = p2 = p3 = p4 = p5 = p6 = p7 = p8 = p9 = 0.1; H1: At least one of the proportions is not equal to the given claimed value.

Test statistic: c 2 = 11.400; P-value = 0.249 (Table: P-value > 0.10); Critical value: c 2 = 16.919; Fail to reject H 0. There is not sufficient evidence to support the claim that the sample is from a population of weights in which the last digits do not occur with the same frequency. That is, the last digits appear to occur with approximately the same frequencies. The results suggest that the weights were measured.

Last Digit 0 1 2 3 4 5 6 7 8 30 30 30 30 30 30 30 30 30 Expected 20 31 30 32 31 39 29 27 39 Observed 2 2 2 20 30 3130 ( ) ( ) 2 (39 - 30)2 (22 - 30) c = + + + + = 11.400 (df = 9) 30 30 30 30 21. H0: p1 = 0.301, p2 = 0.176, p3 = 0.125, , p7 = 0.058, p8 = 0.051, p9 = 0.046; H1: At least one of the proportions is not equal to the given claimed value.

9 30 22

Test statistic: c 2 = 3650.251; P-value = 0.000 (Table: P-value < 0.005); Critical value: c 2 = 20.090; Reject H 0. There is sufficient evidence to warrant rejection of the claim that the leading digits are from a population with a distribution that conforms to Benford’s law. It appears that the image has been corrupted.

2 (15 - 137.98)2 (23 - 39.98)2 (0 - 36.06)2 2 (0 - 235.98) + = 3650.251 (df = 8) c = + + + 0.301×784 0.176×784 0.051×784 0.046×784

22. H0: p1 = 0.301, p2 = 0.176, p3 = 0.125, , p7 = 0.058, p8 = 0.051, p9 = 0.046; H1: At least one of the proportions is not equal to the given claimed value.

Test statistic: c 2 = 18.955; P-value = 0.015 (Table: P-value > 0.01); Critical value: c 2 = 20.090; Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim that the leading digits are from a population with a distribution that conforms to Benford’s law. It does not appear that the image has been corrupted.

2 (58 - 44.00)2 (4 - 12.75)2 (9 - 11.50)2 2 (83 - 75.25) + = 18.955 (df = 8) c = + + + 0.051×250 0.301×250 0.176×250 0.046×250 23. H0: p1 = 0.301, p2 = 0.176, p3 = 0.125, , p7 = 0.058, p8 = 0.051, p9 = 0.046; H1: At least one of the proportions is not equal to the given claimed value.

Test statistic: c 2 = 10.327; P-value = 0.243 (Table: P-value > 0.10); Critical value: c 2 = 20.090; Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim that the leading digits are from a population with a distribution that conforms to Benford’s law. The author’s check amounts appear to be legitimate. He appears to be an honest person who is a true credit to society.

2 (45 - 52.8)2 (18 - 15.3)2 (12 - 13.8)2 2 (102 - 90.3) + + = 10.327 (df = 8) c = + + 0.051×300 0.046×300 0.301×300 0.176×300 24. H0: p1 = 0.301, p2 = 0.176, p3 = 0.125, , p7 = 0.058, p8 = 0.051, p9 = 0.046; H1: At least one of the proportions is not equal to the given claimed value.

Test statistic: c 2 = 45.618; P-value = 0.000 (Table: P-value < 0.005); Critical value: c 2 = 15.507; Reject H 0. There is sufficient evidence to warrant rejection of the claim that the leading digits are from a population with a distribution that conforms to Benford’s law.

2 (28 - 64.592)2 (11- 18.717)2 (11- 17.616)2 2 (112 - 110.467) + + = 45.618 (df = 8) c = + + 0.301×367 0.176×367 0.051×367 0.046×367 25. H0: Heights selected from a normal distribution.

H1: Heights not selected from a normal distribution. a. b. c.

Height (cm) Frequency Tech: Table: Tech:

Less than 155.45 26 0.2023 0.2033 0.2023(247)

155.45 – 162.05 46 0.3171 0.3166 0.3171(247)

162.05–168.65 49 0.3046 0.3039 0.3046(247)

Greater than 168.65 26 0.1761 0.1762 0.1761(247)

= 29.7381 0.2033(247)

= 46.6137 0.3166(247)

= 44.7762 0.3039(247)

= 25.8867 0.1762(247)

= 46.5402

= 44.6733

= 25.9014

Table:

= 29.8851 2

2

d. Test statistic: c = 0.877 (Table: c = 0.831); P-value = 0.831 (Table: P-value > 0.10); Critical value: c 2 = 11.345; Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim that heights were randomly selected from a normally distributed population. The test suggests that we cannot rule out the possibility that the data are from a normally distributed population.

2 2 2 2 c 2 = (26 - 29.7381) + (46 - 46.6137) + (49 - 44.7762) + (26 - 25.8867) = 0.877 (df = 3) 29.7381

46.6137

44.7762

25.8867

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

26. H0: p0 = p1 = p2 = p3 = p4 = p5 = p6 = p7 = p8 = p9 = 0.1; H1: At least one of the proportions is not equal to the given claimed value.

Test statistic: c 2 = 2940.582; P-value = 0.000 (Table: P-value < 0.005); Critical value: c 2 = 16.919; Reject H 0. There is sufficient evidence to warrant rejection of the claim that the last digits of the reported weights occur with about the same frequency, so it appears that the weights were reported and not measured. Also, examination of the frequencies shows that the frequency of 0 is 1069, which is far larger than any of the other frequencies, so it appears that many of the reported weights were rounded to a multiple of 10. Examination of the measured weights of females shows that they are all rounded to one decimal place, and it is highly unlikely that people would use that precision when reporting weights.

(TI data: Test statistic is c 2 = 235.656, critical value is c 2 = 16.919, and the P-value is 0.000.)

2 2 2 2 2 c 2 = (1069 - 297.1) + (78 - 297.1) + (189 - 297.1) + (155 - 297.1) + (165 - 297.1) 0.1×2971 0.1×2971 0.1×2971 0.1×2971 0.1×2971 (624 - 297.1)2 (160 - 297.1)2 (157 - 297.1)2 (261- 297.1)2 (113 - 297.1)2 + + + + + = 2940.58 0.1×2971 0.1×2971 0.1×2971 0.1×2971 0.1×2971 Section 11-2: Contingency Tables 1.

2.

3.

4.

5.

(123 +131)(123 + 52) = 138.906 123 +131+ 52 +14 H 0: Whether a dog makes a correct identification about malaria is independent of whether malaria is present.

H1: Whether a dog makes a correct identification and whether malaria is present are dependent.

E=

a. Test statistic: c 2 = 19.490; P-value = 0.000; Reject the null hypothesis of independence between whether the dog is correct and whether malaria is present. b. No. It is possible that the dogs are wrong significantly more than they are correct. However, with the given data, dogs were correct 70.3% of the time when malaria was present, and they were correct 90.3% of the time when malaria was not present, so they do appear to be effective in their identifications.

The test is right-tailed. The test statistic is based on differences between observed frequencies and the frequencies expected with the assumption of independence between the row and column variables. Only large values of the test statistic correspond to substantial differences between the observed and expected values, and such large values are located in the right tail of the distribution.

H0: Whether a subject lies is independent of polygraph indication.

H1: Subject lies depends on polygraph indication.

6.

Test statistic: c 2 = 25.571; P-value = 0.000 (Table: P-value < 0.005); Critical value: c 2 = 3.841; df = (2 - 1)(2 - 1) = 1; Reject H 0. There is sufficient evidence to warrant rejection of the claim that whether a subject lies is independent of the polygraph test indication. The results suggest that polygraphs are effective in distinguishing between truths and lies, but there are many false positives and false negatives, so they are not highly reliable.

2 2 2 2 c 2 = (15 - 27.3) + (42 - 29.7) + (32 - 19.7) + (9 - 21.3) = 25.571 27.3 29.7 19.7 21.3 H0: Whether testing positive for COVID-19 is independent of getting the vaccine.

H1: Testing positive for COVID-19 is depends on getting the vaccine.

Test statistic: c 2 = 136.086; P-value = 0.0000 (Table: P-value < 0.005); Critical value: c 2 = 6.635; df = (2 - 1)(2 - 1) = 1; Reject H 0. There is sufficient evidence to warrant rejection of the claim that getting the vaccine is independent of whether the subject tested positive for COVID-19. Given that the rates of positive results are 0.000440 (or 0.04%) for the vaccinated group and 0.00884 (or 0.9%) for the group not vaccinated, it appears that the vaccine is effective.

6.

(continued) 2

2

2

2

c 2 = (8 - 84.7) + (18,190 - 18,113.3) + (162 - 85.3) + (18,163 - 18, 239.7) = 136.086 84.7 18,113.3 85.3 18, 239.7 7.

H0: Whether texting while driving is independent of driving when drinking alcohol.

H1: Texting while driving depends on driving when drinking alcohol.

8.

Test statistic: c 2 = 576.224; P-value = 0.0000 (Table: P-value < 0.005); Critical value: c 2 = 3.841; df = (2 - 1)(2 - 1) = 1; Reject H 0. There is sufficient evidence to warrant rejection of the claim of independence between texting while driving and driving when drinking alcohol. Those two risky behaviors appear to be somehow related.

2 2 2 2 c 2 = (731- 394.7) + (3054 - 3390.3) + (156 - 492.3) + (4564 - 4227.7) = 576.224 394.7 3390.3 492.3 4227.7 H0: Whether correct identification is independent of whether evaluator is an expert or a novice.

H1: Correct identification depends on whether evaluator is an expert or a novice.

Test statistic: c 2 = 49.330; P-value = 0.0000 (Table: P-value < 0.005); Critical value: c 2 = 3.841; df = (2 - 1)(2 - 1) = 1; Reject H 0. There is sufficient evidence to warrant rejection of the claim of independence between whether the subject is a novice or expert and whether the fingerprint identification was correct or wrong. The experts were correct 92.1% of the time, compared to 74.5% for the novices, so the experts appear to be substantially better than the novices.

9.

2 2 2 2 c 2 = (331- 370) + (113- 74) + (409 - 370) + (35 - 74) = 49.330 370 74 370 74 H0: Adverse health condition is independent of restoration type.

H1: Adverse health condition depends on restoration type.

Test statistic: c 2 = 0.751; P-value = 0.3862 (Table: P-value > 0.10); Critical value: c 2 = 3.841; df = (2 - 1)(2 - 1) = 1; Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim of

10.

independence between the type of restoration and adverse health conditions. Amalgam restorations do not appear to affect health conditions.

2 2 2 2 c 2 = (135 - 140) + (145 - 140) + (132 - 127) + (122 - 127) = 0.751 140 140 127 127 H0: Sensory disorder is independent of restoration type.

H1: Sensory disorder depends on restoration type.

Test statistic: c 2 = 1.136; P-value = 0.2864 (Table: P-value > 0.10); Critical value: c 2 = 3.841; df = (2 - 1)(2 - 1) = 1; Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim of independence between the type of restoration and sensory disorder. Amalgam restorations do not appear to affect sensory disorder.

2 2 2 2 c 2 = (36 - 32) + (28 - 32) + (231- 235) + (239 - 235) = 1.136 32 32 235 235

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

11.

H0: Adverse effect is independent of treatment.

H1: Adverse effect depends on treatment.

Test statistic: c 2 = 42.568; P-value = 0.0000 (Table: P-value < 0.005); Critical value: c 2 = 9.210; df = (2 - 1)(3 - 1) = 2; Reject H 0. There is sufficient evidence to warrant rejection of the claim that experiencing an adverse reaction in the digestive system is independent of the treatment group. Treatments with 1332-mg doses of Campral appear to be associated with an increase in adverse effects of the digestive system.

2

2

2

c 2 = (344 - 284.1) + (89 - 143.7) + (8 - 13.2) 284.1 143.7 13.2 (1362 - 1421.9)2 (774 - 719.3)2 (71- 65.8)2 + + + = 42.568 1421.9 719.3 65.8 12. H0: Whether deaths on a shift is independent of whether Gilbert was working.

H1: Deaths on a shift depends on whether Gilbert was working.

Test statistic: c 2 = 86.481; P-value = 0.0000 (Table: P-value < 0.005); Critical value: c 2 = 6.635; df = (2 - 1)(2 - 1) = 1; Reject H 0. There is sufficient evidence to warrant rejection of the claim that deaths on shifts are independent of whether Gilbert was working. The results favor the guilt of Gilbert.

2 2 2 2 c 2 = (40 - 11.6) + (217 - 245.4) + (34 - 62.4) + (1350 - 1321.6) = 86.481 11.6 245.4 62.4 1321.6 13. H0: Whether eye color is independent of gender.

H1: Eye color depends on gender.

Test statistic: c 2 = 16.091; P-value = 0.001 (Table: P-value < 0.005); Critical value: c 2 = 11.345; df = (2 - 1)(4 - 1) = 3; Reject H 0 . There is sufficient evidence to warrant rejection of the claim of independence between gender and eye color. However, because the sample data were reported, it is possible that the sample is not representative of the population.

2 2 2 2 c 2 = (359 - 330.7) + (290 - 291.2) + (110 - 139.7) + (160 - 157.4) 330.7 291.2 139.7 157.4 (370 - 398.3)2 (352 - 350.8)2 (198 - 168.3)2 (187 - 189.6)2 + + + + = 16.091 398.3 350.8 168.3 189.6

14. H0: Whether amount of smoking is independent of seat belt use.

H1: Amount of smoking depends on seat belt use.

Test statistic: c 2 = 1.358; P-value = 0.715 (Table: P-value > 0.10); Critical value (a = 0.05): c 2 = 7.815; df = (2 - 1)(4 - 1) = 3; Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim that the amount of smoking is independent of seat belt use. The theory is not supported by the given data.

(175 - 171.5)2 (6 - 7.9)2 (149 - 152.5)2 (9 - 7.1)2 c2= + + + + + = 1.358 7.1 7.9 152.5 171.5 15. H0: Whether getting a cold is independent of treatment.

H1: Getting a cold depends on treatment.

Test statistic: c 2 = 2.925; P-value = 0.232 (Table: P-value > 0.10); Critical value: c 2 = 5.991; df = (2 - 1)(3 - 1) = 2; Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim that getting a cold is independent of the treatment group. The results suggest that echinacea is not effective for preventing colds.

2 2 2 2 2 2 c 2 = (88 - 88.6) + (48 - 44.7) + (42 - 44.7) + (15 - 14.4) + (4 - 7.3) + (10 - 7.3) = 2.925 88.6 44.7 44.7 14.4 7.3 7.3

16. H0: Whether injuries are independent of helmet color.

H1: Injuries depend on helmet color.

Test statistic: c 2 = 9.971; P-value = 0.041 (Table: P-value < 0.05); Critical value (a = 0.05): c 2 = 9.488; df = (2 - 1)(5 - 1) = 4; Reject H 0. There is sufficient evidence to warrant rejection of the claim that injuries are independent of helmet color. It appears that motorcycle drivers should use yellow or orange helmets.

(491- 509.5)2 (55 - 58.6)2 (213 - 194.5)2 (26 - 22.4)2 2 c = + + + + + = 9.971 22.4 58.6 194.5 509.5 17. H0: Whether cooperation of the subject is independent of age category.

H1: Cooperation of the subject depends on age category.

Test statistic: c 2 = 20.271; P-value = 0.0011 (Table: P-value < 0.005); Critical value: c 2 = 15.086; df = (2 - 1)(6 - 1) = 5; Reject H 0. There is sufficient evidence to warrant rejection of the claim that cooperation of the subject is independent of the age category. The age group of 60 and over appears to be particularly uncooperative.

2 2 2 2 c 2 = (73 - 73.1) + + (202 - 218.5) + (11- 10.9) + + (49 - 32.5) = 20.271 73.1 218.5 10.9 32.5 18. H0: Whether handedness of offspring is independent of parental handedness.

H1: Handedness of offspring depends on parental handedness.

Test statistic: c 2 = 784.647; P-value = 0.000 (Table: P-value < 0.005); Critical value: c 2 = 11.345; df = (4 - 1)(2 - 1) = 3; Reject H 0. There is sufficient evidence to warrant rejection of the claim that handedness of offspring is independent of parental handedness. It appears that handedness of the parents has an effect on handedness of the offspring, so handedness appears to be an inherited trait.

2 2 2 2 c 2 = (5360 - 6067.9) + (50, 928 - 50, 220.1) + (767 - 337.6) + (2736 - 3125.4) 6067.9 50, 220.1 337.6 3125.4 (741- 475.2)2 (3667 - 3932.8)2 (94 - 41.3)2 (289 - 341.7)2 + + + + = 784.647 475.2 3932.8 41.3 341.7 19.

H0: Infection is independent of treatment.

H1: Infection depends on treatment.

Test statistic: c 2 = 0.773; P-value = 0.8560 (Table: P-value > 0.10); Critical value: c 2 = 7.815; df = (2 - 1)(4 - 1) = 3; Fail to reject H 0 . There is not sufficient evidence to warrant rejection of the claim that infections are independent of treatment. Based on the given sample data, the atorvastatin treatment does not appear to have an effect on infections.

2 2 2 2 c 2 = (27 - 27.1) + (89 - 86.6) + (8 - 7.9) + (7 - 9.4) 27.1 86.6 7.9 9.4 (243 - 242.9)2 (774 - 776.4)2 (71- 71.1)2 (87 - 84.6)2 + + + + = 0.773 242.9 776.4 71.1 84.6

20. H0: Category of cause is independent of sex.

H1: Category of cause depends on sex.

Test statistic: c 2 = 122.446; P-value = 0.000 (Table: P-value < 0.005); Critical value (a = 0.05): c 2 = 7.815; df = (2 - 1)(4 - 1) = 3; Reject H 0 . There is sufficient evidence to warrant rejection of the claim that the category of cause of workplace fatalities is independent of sex.

(175 - 171.5)2 (6 - 7.9)2 (149 - 152.5)2 (9 - 7.1)2 2 c = + + + + + = 1.358 7.1 7.9 152.5 171.5

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

21. c 2 = 3.197517514 and z = 1.788160371, so z2 = c 2. The critical values are: c 2 = 3.841 and z2 = ±1.96, (or ±1.959963986 from technology) so z2 = c 2.

22. Without Yates’s correction, the test statistic is c 2 = 0.751. With Yates’s correction, the test statistic is c 2 = 0.608. Yates’s correction decreases the test statistic so that sample data must be more extreme in order to reject the null hypothesis of independence.

Without Yates’s correction: 2

2

2

2

c 2 = (135 - 140) + (145 - 140) + (132 - 127) + (122 - 127) = 0.751 140 140 127 127 With Yates’s Correction: c2=

(135 - 140 - 0.5)2 + (145 - 140 - 0.5)2 + (132 - 127 - 0.5)2 + (122 - 127 - 0.5)2 = 0.608 140

140

127

127

Quick Quiz 1.

H0: p0 = p1 = p2 = p3 = p4 = p5 = p6 = p7 = p8 = p9 = 0.1; H1: At least one of the probabilities is different from the others.

2. O = 31 and E = 0.1×134 = 13.4 4. df = n - 1 = 10 - 1 = 9 3. right-tailed 5. There is sufficient evidence to warrant rejection of the claim that the last digits are equally likely. Among the 134 last digits, 130 are either 0 or 5. It appears that either the weights were reported or some strange rounding occurred.

6.

H 0: Surviving the sinking is independent of whether the person is a man, woman, boy, or girl.

H1: Surviving the sinking and whether the person is a man, woman, boy, or girl are somehow related.

7. chi-squared distribution 9. df = (r - 1)(c - 1) = (2 - 1)(4 - 1) = 3 8. right-tailed 10. There is sufficient evidence to warrant rejection of the claim that surviving the sinking is independent of whether the person is a man, woman, boy, or girl. Most of the women survived, 45% of the boys survived, and most girls survived, but only about 20% of the men survived, so it appears that the rule was followed quite well.

Review Exercises 1.

H0: pJan = pFeb = pMar = pApr = pMay = pJun = pJul = pAug = pSep = pOct = pNov = pDec = 1 12; H1: At least one of the proportions is not equal to the given claimed value.

Test statistic: c 2 = 186.332; P-value = 0.0000 (Table: P-value < 0.005); Critical value: c 2 = 24.725; Reject H 0. There is sufficient evidence to warrant rejection of the claim that weather-related deaths occur in the different months with the same frequency. The warmer months appear to have disproportionately more weatherrelated deaths, and that is likely due to the fact that vacations and outdoor activities are much greater during those months.

2 (14 - 44.17)2 (96 - 44.17)2 (16 - 44.17)2 2 (61- 44.17) c = + + + + = 186.332 (df = 11) 530 12 530 12 530 12 530 12

2.

H0: p1 = 0.301, p2 = 0.176, p3 = 0.125, , p7 = 0.058, p8 = 0.051, p9 = 0.046; H1: At least one of the proportions is not equal to the given claimed value.

Test statistic: c 2 = 13.410; P-value = 0.0985 (Table: P-value > 0.05); Critical value (a = 0.05): c 2 = 15.507; Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim that the observed frequencies fit the distribution from Benford’s law. It appears that the data approximately fit the distribution from Benford’s law.

2.

(continued) 2 2 2 2 2 c 2 = (233 - 232.974) + (136 - 136.224) + (79 - 96.75) + (83 - 75.078) + (77 - 61.146) 0.301×774 0.176 ×774 0.125×774 0.097 ×774 0.079 ×774 (58 - 51.858)2 (50 - 44.892)2 (31- 39.474)2 (27 - 35.604)2 + + + + = 13.410 (df = 8) 0.067 ×774 0.058×774 0.051×774 0.046 ×774

3.

H0: Whether success is independent of type of treatment.

H1: Success depends on type of treatment.

Test statistic: c 2 = 9.750; P-value = 0.002 (Table: P-value < 0.005); Critical value: c 2 = 6.635; df = (2 - 1)(2 - 1) = 1; Reject H 0. There is sufficient evidence to warrant rejection of the claim that success is independent of the type of treatment. The results suggest that the surgery treatment is better.

2 2 2 2 c 2 = (60 - 67.6) + (23 - 15.4) + (67 - 59.4) + (6 - 13.6) = 9.750 67.6 4.

15.4

59.4

13.6

H0: Whether success is independent of type of treatment.

H1: Success depends on type of treatment.

Test statistic: c 2 = 58.393; P-value = 0.000 (Table: P-value < 0.005); Critical value: c 2 = 7.815; df = (4 - 1)(2 - 1) = 3; Reject H 0. There is sufficient evidence to warrant rejection of the claim that success is

independent of the treatment. Although these results of this test do not tell us which treatment is best, we can see from the table that the success rates of 81.8%, 44.6%, 95.9%, and 77.3% suggest that the best treatment is to use a non-weight-bearing cast for 6 weeks. These results suggest that the increasing use of surgery is a treatment strategy that is not supported by the evidence.

2 2 2 2 c 2 = (54 - 47.5) + (12 - 18.5) + (41- 66.2) + (51- 25.8) 47.5 18.5 66.2 25.8 (70 - 52.5)2 (3 - 20.5)2 (17 - 15.8)2 (5 - 6.2)2 + + + + = 58.393 52.5 20.5 15.8 6.2 Cumulative Review Exercises 1. a. Here are two different key questions that are appropriate for the data: (1) Is there a correlation between the scores of the females and the scores of their brothers?

(2) For the paired scores, is there a significant difference between the scores of the siblings? b. (1) Use the linear correlation coefficient, as in Section 10-1.

(2) Use the methods for inferences about differences from matched pairs, as in Section 9-3. c. (1) Using correlation: r = - 0.481; P-value = 0.412 (Table: P-value > 0.05); critical values: (a = 0.05): r = ±0.878; There is not sufficient evidence to support a claim of a correlation.

(2) The differences are - 6, 7,- 5,- 5,3. Test statistic: t =- 0.459 P-value = 0.670 (Table: P-value > 0.20); critical values: (a = 0.05): t = ±2.776; Fail to reject H 0 . There is

not sufficient evidence to support a claim of a difference between the scores of the siblings. x - md - 1.2 - 0 t= d = 5.8482 5 s2 n

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

2.

a. Are the last digits from a population in which they all occur with the same frequency? b. Use the goodness-of-fit test, as in Section 11-1. c. Test statistic: c 2 = 6.500; P-value = 0.689 (Table: P-value > 0.10); Critical value:

3.

(a = 0.05): c 2 = 16.919; Fail to reject the null hypothesis that the last digits are from a population in which they all occur with the same frequency.

2 2 2 2 2 c 2 = (14 - 16.0) + (22 - 16.0) + (11- 16.0) + (12 - 16.0) + (18 - 16.0) 0.1×160 0.1×160 0.1×160 0.1×160 0.1×160 (20 - 16.0)2 (15 - 16.0)2 (16 - 16.0)2 (17 - 16.0)2 (15 - 16.0)2 + + + + + = 6.500 0.1×160 0.1×160 0.1×160 0.1×160 0.1×160 a. Is there a significant difference between the distances run by males and the distances run by females? b. Use the test for a difference between the means of two independent populations, as in Section 9-2. c. H0: m1 = m2; H1: m1 ¹ m2; Test statistic: t = - 0.536; P-value = 0.613 (Table: P-value > 0.20). Critical values: (a = 0.05): t = ±2.492 (Table: ±2.776); Fail to reject the null hypothesis of no difference. There is not sufficient evidence to reject the claim of no significant difference between the distances run by males and females. (The same conclusion results from a 95% confidence interval: - 6.8 miles < m1 - m2 < 4.4 miles.) t=

(x1 - x2 )- (m1 - m2 ) = (15.4 - 16.6)- (0) s12

s2 + 2

(x1 - x2 )± ta /2 4.

s12

4.5612 2.0742 + 5 5 2

s2 4.561 + 2.074 + 2 = (15.4 - 16.6)± 2.776 × 5 5 2

(df = 4)

a. Are the variables of workday and shift independent? b. Use the c 2 test for a contingency table, as in Section 11-2.

5.

c. Test statistic: c 2 = 4.225; P-value = 0.376 (Table: 70.10). Critical value: (a = 0.05): c 2 = 9.488; df = (2 - 1)(5 - 1) = 4; Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim of independence between shift and day of the week.

2 2 2 2 2 c 2 = (14 - 16.4) + (22 - 17.8) + (11- 13) + (12 - 14) + (18 - 15.9) 16.4 17.8 13 14 15.9 (20 - 17.6)2 (15 - 19.2)2 (16 - 14)2 (17 - 15)2 (15 - 17.1)2 =+ + + + + = 4.225 17.6 19.2 14 15 17.1 a. Is there a correlation between diastolic blood pressure and height? b. Use the linear correlation coefficient, as in (Section 10-1). c. r = - 0.146; P-value = 0.782 (Table: P-value > 0.05); Critical values (a = 0.05): r = ±0.8114; There is not sufficient evidence to support the claim that for males, there is a linear correlation between diastolic blood pressure and height.

6.

38 + 64 + 35 + 67 + 42 + 29 + 68 + 62 + 74 + 58 = 53.7 years.

10 58 + 62 The median is = 60.0 years.

2 The mean is x =

+ (74 - 53.7)2 + (58 - 53.7)2 = 16.1 years.

10 - 1 2 2 2 +(74 - 53.7)2 +(58 - 53.7)2 2 (38 - 53.7) +(64 - 53.7) + = 258.9 years .

The variance is s = 10 - 1 Significantly low values are less than or equal to 53.7 - 2(16.1) = 21.5 years. An age of 16 years is significantly low.

16 - 53.7 z= = - 2.34; Because an age of 16 is more than 2 standard deviations below the mean of 53.7 years, it 16.1 is significantly low.

The sample data meet the loose requirement of having a normal distribution.

16.1 s = 53.7 ± 2.262 × Þ 42.2 years < m < 65.2 years; Yes, the confidence interval limits do 95% CI: x ± ta /2 n 10 contain the value of 65.0 years. a. The z score for the bottom 5% is –1.645, which correspond a forward grip reach of - 1.645×34 + 686 = 630 mm.

650 - 686 = - 1.06; which has a probability of 0.1448, or 14.48% (Table: 14.46%) to the left. That b. z x=650 = 34 percentage is too high, because too many women would not be accommodated.

680 - 686 = - 0.71; which has a probability of 1- 0.2401 = 0.7599 (Table: 0.7611) to the right. c. z x=680 = 34 16 Groups of 16 women do not occupy a driver’s seat or cockpit; because individual women occupy the driver’s seat/cockpit, this result has no effect on the design.

The standard deviation is s =

7.

8.

(38 - 53.7)2 + (64 - 53.7)2 +

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

Chapter 12: Analysis of Variance Section 12-1: One-Way ANOVA 1. a. The chest compression amounts are categorized according to the one characteristic of vehicle size category. b. The terminology of analysis of variance refers to the method used to test for equality of the four population means. That method is based on two different estimates of a common population variance.

2. As we increase the number of individual tests of significance, we increase the risk of finding a difference by chance alone (instead of a real difference in the means). The risk of a type I error (finding a difference in one of the pairs when no such difference actually exists) is too high. The method of analysis of variance helps us avoid that particular pitfall (rejecting a true null hypothesis) by using one test for equality of several means, instead of several tests that each compare two means at a time.

3. The test statistic is F = 3.815, and the F distribution applies.

4.

The P-value is 0.016. Because the P-value is less than the significance level of 0.05, we reject the null hypothesis of equal means. There is sufficient evidence to warrant rejection of the claim that small cars, midsize cars, large cars, and SUVs have the same mean chest compression in car crash tests. It appears that the four populations have means that are not all the same, but the ANOVA test does not enable us to identify which populations have means that are significantly different.

5.

Test statistic: F = 0.57; P-value = 0.638; Fail to reject H0: m1 = m2 = m3 = m4. There is not sufficient evidence to warrant rejection of the claim that the four vehicle size categories have the same mean force on the left femur in crash tests. Size of the car does not appear to have an effect on the force on the femur of the left leg in crash tests.

6.

Test statistic: F = 0.39; P-value: 0.677; Fail to reject H0: m1 = m2 = m3. There is not sufficient evidence to warrant rejection of the claim that the three categories of blood lead level have the same mean verbal IQ score.

Exposure to lead does not appear to have an effect on verbal IQ scores.

7.

Test statistic: F = 2.303; P-value: 0.104; Fail to reject H0: m1 = m2 = m3. There is not sufficient evidence to warrant rejection of the claim that the three categories of blood lead level have the same mean full IQ score.

There is not sufficient evidence to conclude that exposure to lead has an effect on full IQ scores.

8.

Test statistic: F = 969.744; P-value < 0.0001; Reject H0: m1 = m2 = m3. There is sufficient evidence to warrant rejection of the claim that the three samples are from populations with the same mean. It appears that exposure to tobacco smoke does affect the amount of nicotine absorbed by the body.

9.

Test statistic: F = 1.3037; P-value: 0.2746; Fail to reject H0: m1 = m2 = m3. There is not sufficient evidence to warrant rejection of the claim that males from the three age brackets have the same mean pulse rate. It appears that the pulse rates of males are not affected by age bracket.

10. Test statistic: F = 7.9338; P-value: 0.0005; Reject H0: m1 = m2 = m3. There is sufficient evidence to warrant rejection of the claim that females from the three age brackets have the same mean pulse rate. It appears that the pulse rates of females are affected by age bracket.

11. Test statistic: F = 27.2488; P-value = 0.000; Reject H0: m1 = m2 = m3. There is sufficient evidence to warrant rejection of the claim that the three different miles have the same mean time. These data suggest that the third mile appears to take longer, and a reasonable explanation is that the third lap has a hill.

Excel Output: ANOVA Source of Variation Between Groups Within Groups

SS 0.103444 0.022778

Total

0.126222

df 2 12 14

MS 0.051722 0.001898

F 27.24878

P-value 3.45E-05

F crit 3.885294

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

12. Test statistic: F = 22.9477; P-value = 0.000; Reject H0: m1 = m2 = m3. There is sufficient evidence to warrant rejection of the claim that the three different states have the same mean arsenic content in brown rice. Even though Texas has the highest mean, the results from ANOVA do not allow us to conclude that any one specific population mean is different from the others, so we cannot conclude that brown rice from Texas poses the greatest health problem.

Excel Output: ANOVA Source of Variation Between Groups Within Groups Total

SS 31.65167 22.75833

df 2 33

54.41

MS 15.82583 0.689646

F 22.94775

P-value 5.68E-07

F crit 3.284918

35

13. Test statistic: F = 18.9931; P-value = 0.0000; Reject H0: m1 = m2 = m3. There is sufficient evidence to warrant rejection of the claim that the three different types of cigarettes have the same mean amount of nicotine.

Given that the king size cigarettes have the largest mean of 1.26 mg per cigarette compared to the other means of 0.87 mg per cigarette and 0.92 mg per cigarette, it appears that the filters do make a difference, although this conclusion is not justified by the results from analysis of variance.

Excel Output: ANOVA Source of Variation Between Groups Within Groups Total

SS 2.20826667 4.1856 6.39386667

df 2 72 74

MS 1.10413333 0.05813333

F 18.9931193

P-value 2.3757E-07

F crit 3.12390745

14. Test statistic: F = 1.1810; P-value = 0.3167; Fail to reject H0: m1 = m2 = m3 = m4. There is not sufficient evidence to support the claim that the different hospitals have different mean birth weights. It appears that birth weights are about the same in urban and rural areas.

Excel Output: ANOVA Source of Variation Between Groups Within Groups Total

SS 1701400 190157000 191858400

df 3 396 399

MS 567133.333 480194.444

F 1.18104934

P-value 0.3166564

F crit 2.62744077

15. Test statistic: F = 5.3568; P-value: 0.0026; Reject H0: m1 = m2 = m3 = m4. There is sufficient evidence to warrant rejection of the claim that the four treatment categories yield poplar trees with the same mean weight.

Although not justified by the results from analysis of variance, the treatment of fertilizer and irrigation appears to be most effective.

Excel Output: ANOVA Source of Variation Between Groups Within Groups Total

SS 5.264327 18.34455 23.60887

df

3 56 59

MS 1.754776 0.327581

F 5.356765

P-value 0.002583

F crit 2.769431

16. Test statistic: F = 2.5423; P-value: 0.0929; Fail to reject H0: m1 = m2 = m3 = m4. There is not sufficient evidence to warrant rejection of the claim that the four treatment categories yield poplar trees with the same mean weight. No treatment appears to be more effective than any other in the sandy and dry region.

Excel Output: ANOVA Source of Variation Between Groups Within Groups Total

SS 3.036215 6.36944 9.405655

df

3 16 19

MS 1.012072 0.39809

F 2.542319

P-value 0.092852

F crit 3.238872

17. The Tukey test results show different P-values, but they are not dramatically different. The Tukey results suggest the same conclusions as the Bonferroni test.

18. a. Test statistic: F = 6.1413; P-value: 0.0056; Reject H0: m1 = m2 = m3. There is sufficient evidence to warrant rejection of the claim that the four treatment categories yield poplar trees with the same mean weight. b. The displayed Bonferroni results show that with a P-value of 0.039, there is a significant difference between the mean of the no treatment group (group 1) and the mean of the group treated with both fertilizer and irrigation (group 4). (Also, when comparing group 1 and group 2, there is no significant difference between means, and when comparing group 1 and group 3, there is no significant difference between means.) c. Test statistic: t = - 4.007; P-value: 6(0.001018) = 0.00611; Reject the null hypothesis that the mean weight from the irrigation treatment group is equal to the mean from the group treated with both fertilizer and irrigation.

Section 12-2: Two-Way ANOVA 1. The measurements are categorized using two different factors of (1) femur side (left, right) and (2) vehicle size (small, midsize, large, SUV).

2. No. To use two individual tests of one-way analysis of variance is to totally ignore the very important feature of the possible effect from an interaction between femur side (left, right) and vehicle size category. If there is an interaction, it doesn’t make sense to consider the effects of one factor without the other.

3. a. An interaction between two factors or variables occurs if the effect of one of the factors changes for different categories of the other factor. b. If there is an interaction effect, we should not proceed with individual tests for effects from the row factor and column factor. If there is an interaction, we should not consider the effects of one factor without considering the effects of the other factor. c. The lines are not too far from being parallel except for the mean for the left femur in small cars. That large difference does suggest that there is an interaction effect between femur side and size of vehicle.

4. Yes, the result is a balanced design because each cell has the same number (5) of values.

5.

6.

7.

For interaction, the test statistic is F = 4.4313 and the P-value is 0.0103, so there is sufficient evidence to warrant rejection of the null hypothesis of no interaction effect. Because there appears to be an interaction between femur side (left, right) and vehicle size category, we should not proceed with a test for an effect from the femur side (left/right) and a test for an effect from vehicle size category. It appears that an interaction between the femur side and vehicle size category has an effect on the force measurements. (Remember, these results are based on fabricated data used in one of the cells, so this conclusion does not necessarily apply to real data.)

In XLSTAT, the interaction is identified as “Sex*ANSUR.” For interaction, the test statistic is F = 0.085 and the P-value is 0.774, so there is not sufficient evidence to warrant rejection of the null hypothesis of no interaction effect. For the row variable of sex, the test statistic is F = 9.459 and the P-value is 0.006, so there does appear to be an effect from sex. For the column variable of ANSUR (ANSUR I or ANSUR II), the test statistic is F = 0.124 and the P-value is 0.728, so there does not appear to be an effect from ANSUR. It appears that weights are affected by sex.

For interaction, the test statistic is F = 3.6296 and the P-value is 0.0749, so there is not sufficient evidence to warrant rejection of the null hypothesis of no interaction effect. For the row variable of sex, the test statistic is F = 5.3519 and the P-value is 0.0343, so there does appear to be an effect from sex. For the column variable of handedness, the test statistic is F = 2.2407 and the P-value is 0.1539, so there does not appear to be an effect from handedness. It appears that the distance between pupils is affected by sex. For distances between pupils, we expect the following: (1) No interaction between sexes; (2) No effect from handedness (right, left); (3) There could be differences based on sex. The results are as expected.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

8.

For interaction, the test statistic is F = 0.9391 and the P-value is 0.3973, so there does not appear to be an interaction effect. For the row variable of age bracket, the test statistic is F = 1.8156 and the P-value is 0.1725, so there is not sufficient evidence to conclude that age bracket has an effect on pulse rate. For the column variable of sex, the test statistic is F = 13.5914 and the P-value is 0.0005, so there is sufficient evidence to support the claim that sex has an effect on pulse rate.

9.

For interaction, the test statistic is F = 2.1727 and the P-value is 0.1599, so there is not sufficient evidence to warrant rejection of the null hypothesis of no interaction effect. For the row variable of sex, the test statistic is F = 27.8569 and the P-value is 0.0001, so there does appear to be an effect from sex. For the column variable of handedness, the test statistic is F = 1.4991 and the P-value is 0.2385, so there does not appear to be an effect from handedness. It appears that sitting height is affected by sex. For sitting heights, it is reasonable to not expect an interaction between sexes and handedness, it is reasonable to not expect that handedness (right, left) would affect sitting heights, and it is reasonable to expect that there could be differences based on sex, so the results are as expected.

Excel Output: ANOVA Source of Variation Row Columns Interaction Within Total

SS 39427.2 2121.8 3075.2 22645.6 67269.8

df

1 1 1 16 19

MS 39427.2 2121.8 3075.2 1415.35

F 27.85686 1.499134 2.172749

P-value 7.5E-05 0.238527 0.159878

F crit 4.493998 4.493998 4.493998

10. For interaction, the test statistic is F = 0.3328 and the P-value is 0.5747, so there is not sufficient evidence to warrant rejection of no interaction effect. There does not appear to be an interaction between sex and smoking.

For the row variable of sex, the test statistic is F = 1.3313 and the P-value is 0.2710, so there is not sufficient evidence to warrant rejection of the claim of no effect from sex. For the column variable of smoking, the test statistic is F = 1.1186 and the P-value is 0.3110, so there is not sufficient evidence to conclude that smoking has an effect on body temperatures. It appears that body temperatures are not affected by an interaction between sex and smoking, they are not affected by sex, and they are not affected by smoking.

Excel Output: ANOVA Source of Variation Row Columns Interaction Within Total

11.

SS

0.36 0.3025 0.09 3.245 3.9975

df

1 1 1 12 15

MS

0.36 0.3025 0.09 0.270417

F 1.331279 1.118644 0.33282

P-value 0.271041 0.311038 0.574667

F crit 4.747225 4.747225 4.747225

a. Test statistics and P-values do not change. b. Test statistics and P-values do not change. c. Test statistics and P-values do not change. d. An outlier can dramatically affect and change test statistics and P-values.

Quick Quiz 1. The sample data are partitioned into the three different categories according to the one factor of epoch.

2.

H0: m1 = m2 = m3; H1: At least one of the three population means is different from the others.

3.

Test statistic: F = 4.0497; Larger test statistics result in smaller P-values.

4.

Reject H 0. There is sufficient evidence to support the claim that the different epochs have mean skull breadths that are not all the same.

Right-tailed. Yes, all one-way analysis of variance tests are right-tailed.

No. The method of analysis of variance does not justify a conclusion that any particular mean is different from the others.

With one-way analysis of variance, data from the different samples are categorized using only one factor, but with two-way analysis of variance, the sample data are categorized into different cells determined by two different factors.

5.

6.

7.

8.

Test statistic: F = 1.41; P-value = 0.281; Fail to reject the null hypothesis of no interaction effect. There is not sufficient evidence to warrant rejection of the claim that head injury measurements are not affected by an interaction between the type of car (foreign, domestic) and size of the car (small, medium, large). There does not appear to be an effect from an interaction between the type of car (foreign or domestic) and whether the car is small, medium, or large.

9.

Test statistic: F = 2.25 P-value = 0.159; Fail to reject the null hypothesis of no effect from the type of car.

There is not sufficient evidence to support the claim that whether the car is foreign or domestic has an effect on head injury measurements.

10. Test statistic: F = 0.44; P-value = 0.655; Fail to reject the null hypothesis of no effect from the size of the car.

There is not sufficient evidence to support the claim that whether the car is small, medium, or large has an effect on head injury measurements.

Review Exercises 1.

With test statistic F = 2.1436 and P-value = 0.096, fail to reject H0: m1 = m2 = m3 = m4. Conclude that there is not sufficient evidence to warrant rejection of the claim that subjects in the four age brackets have the same mean cholesterol level. It appears that for the given age brackets, age does not have a significant effect on cholesterol.

2.

Test statistic: F = 2.3831; P-value = 0.092; Fail to reject H0: m1 = m2 = m3 = m4. There is not sufficient evidence to warrant rejection of the claim that subjects in the four different age brackets have the same mean cholesterol level. It appears that for those four age brackets, age does not have an effect on LDL cholesterol.

Excel Output: ANOVA Source of Variation Between Groups Within Groups Total

SS 6069.163 22071.8 28140.97

df

3 26 29

MS 2023.054 848.9155

F 2.383104

P-value 0.092336

F crit 2.975154

3.

For interaction, the test statistic is F = 0.8476 and the P-value = 0.6413, so there is not sufficient evidence to warrant rejection of the null hypothesis of no interaction effect. For the row variable of hospital, the test statistic is F = 0.3549 and the P-value = 0.5426, so there does not appear to be an effect from hospital. For the column variable of day of weekday/weekend, the test statistic is F = 0.6423 and the P-value = 0.5267, so there does not appear to be an effect from whether the day is a weekday or weekend day. It appears that birth weights are not affected by an interaction between hospital and weekday/weekend, they are not affected by the hospital, and they are not affected by whether the day is a weekday or weekend day.

4.

For interaction, the test statistic is F = 1.0605 and the P-value = 0.3795, so there is not sufficient evidence to warrant rejection of the null hypothesis of no interaction effect. For the row variable of hospital, the test statistic is F = 0.6423 and the P-value = 0.5935, so there does not appear to be an effect from hospital. For the column variable of day of weekday/weekend, the test statistic is F = 0.4097 and the P-value = 0.5267, so there does not appear to be an effect from whether the day is a weekday or weekend day. It appears that birth weights are not affected by an interaction between hospital and weekday/weekend, they are not affected by the hospital, and they are not affected by whether the day is a weekday or weekend day.

Excel Output: ANOVA Source of Variation Sample Columns Interaction Within Total

SS

1.204 0.256 1.988 19.996 23.444

df

3 1 3 32 39

MS 0.401333 0.256 0.662667 0.624875

F 0.642262 0.409682 1.060479

P-value 0.593478 0.526688 0.379536

F crit 2.90112 4.149097 2.90112

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

Cumulative Review Exercises 1. a. Presidents: x = 15.9 years; Popes: x = 13.1 years; Monarchs: x = 22.7 years b. Presidents: s = 9.9 years; Popes: s = 9.0 years; Monarchs: s = 18.6 years c. Presidents: s2 = 97.4 years2; Popes: s2 = 80.3 years2; Monarchs: s2 = 346.1 years2 d. Visual inspection of the data shows that among the monarchs, the values of 59 years and 63 years appear to be possible outliers, but they are not outliers according to the criterion of being above the third quartile Q3 by more than 1.5 times the interquartile range. e. ratio level of measurement 2.

H0: m1 = m2; H1: m1 ¹ m2; population1 = presidents, population2 = popes; Test statistic: t = 1.136; P-value = 0.2610 (Table: P-value > 0.20); Critical values (a = 0.05): t = ±2.006 (Table: t = ±2.069); Fail to reject H 0. There is not sufficient evidence to support the claim that there is a difference between the mean longevity times of presidents and popes.

t=

3.

4.

5.

6.

(x1 - x2 )- (m1 - m2 ) = (15.872 - 13.125)- (0) s12

s2 + 2

2

9.706892 + 8.960360 24 39

Because the pattern of the points is reasonably close to a straight-line pattern, the longevity times of presidents do appear to be from a population with a normal distribution. s = 15.9 ± 2.024 × 9.9 Þ n = 39 > 30; 95% CI: x ± ta /2 12.7 years < m < 19.1 years; We have 95% n 39 confidence that the limits of 12.7 years and 19.1 years contain the value of the population mean for longevity times of presidents. a. H0: m1 = m2 = m3 b. Because the P-value of 0.054 is greater than the significance level of 0.05, fail to reject the null hypothesis of equal means. There is not sufficient evidence to warrant rejection of the claim that the three means are equal. The three populations do not appear to have means that are significantly different.

345 - 280 = = 1.00; which has a probability of 1- 0.8413 = 0.1587 to the right. a. zx=345 65 215 - 280 345 - 280 = = - 1.00 and z = = 1.00; which have a probability of b. z x=215 x=345 65 65 0.8413- 0.1587 = 0.6826 (Tech: 0.6827) between them.

319 - 280 = 3.00; which has a probability of 0.9987 to the left. c. z x=319 = 65 25 d. The z score for the bottom 80% is 0.84, which correspond to a blood platelet count of 0.84×65 + 280 = 334.6 (1000 cells mL). (Tech: 334.7)

7.

H0: Whether symptoms 28 days after concussion is independent of physical activity within 7 days of event.

H1: Symptoms 28 days after concussion depend on physical activity within 7 days of event.

Test statistic: c 2 = 85.945; P-value = 0.000 (Table: P-value < 0.005); Critical value: c 2 = 3.841; df = (2 - 1)(2 - 1) = 1; Reject H 0. There is sufficient evidence to warrant rejection of the claim of independence between physical activity within 7 days after an acute concussion and symptoms 28 days after the concussion. It appears that early physical activity during the week after a concussion does have an effect on symptoms 28 days after a concussion.

2 2 2 2 c 2 = (413 - 509.4) + (1264 - 1167.6) + (320 - 223.58) + (416 - 512.48) = 85.945 509.4 1167.6 223.58 512.48

8.

Using normal as approximation to binomial: P(19 out of 100) = 0.1020.

Exact result using binomial distribution: P(19 out of 100) = 0.0995.

Assuming that one-quarter of all offspring have blue eyes, the probability of getting 19 or fewer offspring with blue eyes is high, so there is not sufficient evidence to conclude that the one-quarter rate is wrong.

Normal approximation: np = 100×0.25 = 25 ³ 5 and nq = 100×0.75 = 75 ³ 5 zx=19.5 =

19.5 - 100 ×0.25 100 ×0.25×0.75

= - 1.27

Binomial distribution: 100C0 ×0.250 ×0.75100 +

+ 100C19 ×0.2519 ×0.7581

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

Chapter 13: Nonparametric Tests 13-2 : Sign Test

1.

2.

a. The only requirement for the matched pairs is that they constitute a simple random sample. b. There is no requirement of a normal distribution or any other specific distribution. c. The sign test is “distribution free” in the sense that it does not require a normal distribution or any other specific distribution.

There is 1 positive sign, 5 negative signs, 1 tie, n = 6, and the test statistic is x = 1 (the smaller of 1 and 5).

3.

H 0: There is no difference between hospital admissions due to traffic accidents that occur on Friday the

4.

5.

6.

7.

8.

6th and those that occur on the following Friday the 13th.

H1: There is a difference between hospital admissions due to traffic accidents that occur on Friday the 6th and those that occur on the following Friday the 13th.

The sample data do not contradict H1 because the numbers of positive signs (1) and negative signs (5) are not exactly the same.

The efficiency of 0.63 indicates that with all other things being equal, the sign test requires 100 sample observations to achieve the same results as 63 sample observations analyzed through a parametric test.

The test statistic of x = 2 is not less than or equal to the critical value of 0 (from Table A-7). There is not sufficient evidence to reject the claim of no difference in heights between mothers and their first daughters.

The test statistic of x = 3 is not less than or equal to the critical value of 0 (from Table A-7). There is not sufficient evidence to warrant rejection of the claim of no difference. There is not sufficient evidence to support the claim that there is a difference in heights between fathers and their first sons.

The test statistic of x = 2 is not less than or equal to the critical value of 1 (from Table A-7). There is not sufficient evidence to warrant rejection of the claim that for males, their height is the same as their arm span.

The test statistic of x = 0 is less than or equal to the critical value of 1 (from Table A-7). There is sufficient evidence to warrant rejection of the claim that Captopril affects blood pressure. It appears that Captopril lowers blood pressure. (x + 0.5) - n (401+ 0.5) - 882 2 2 The test statistic of z = = - 2.66 results in a P-value of 0.0078, and it is =

()

9.

n 2

( )

882 2

in the critical region bounded by z = - 2.575 and 2.575. There is sufficient evidence to warrant rejection of the claim that there is no difference between the proportions of those opposed and those in favor. (x + 0.5) - n (372 + 0.5) - 1228 2 2 10. The test statistic of z = = - 13.78 results in a P-value of 0.0000, and it =

()

n 2

( )

1228 2

is in the critical region bounded by z = - 2.575 and 2.575. There is sufficient evidence to support the claim that there is a difference between the rate of medical malpractice lawsuits that go to trial and the rate of such lawsuits that are dropped or dismissed.

(426 + 0.5)- 860 2 11. The test statistic of z = = - 0.24 results in a P-value of 0.8103, and it is not in the critical 860 2 region bounded by z = - 1.96 and 1.96. There is not sufficient evidence to reject the claim that boys and girls are equally likely.

(123 + 0.5)- 280 2 12. The test statistic of z = = - 1.97 results in a P-value of 0.0488, and it is not in the critical 280 2 region bounded by z = - 2.575 and 2.575. There is not sufficient evidence to warrant rejection of the claim that the touch therapists make their selections with a method equivalent to random guesses. The touch therapists do not appear to be able to select the correct hand.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

(x + 0.5) 13. The test statistic of z =

(n ) (19 + 0.5) - (84 ) 2

n 2

=

2

84 2

13-2: Sign Test vii = - 4.91 results in a P-value of 0.0000, and it is in

the critical region bounded by z = - 2.575 and 2.575. There is sufficient evidence to warrant rejection of the claim that the median is equal to 98.6°F. (x + 0.5) - n (117 + 0.5) - 273 2 2 14. The test statistic of z = = - 2.30 results in a P-value of 0.0214, and it is =

()

n 2

( )

273 2

in the critical region bounded by z = - 1.96 and 1.96. There is sufficient evidence to warrant rejection of the claim that the sample is from a population with a median diastolic blood pressure of 72 mmHg. (x + 0.5) - n (5 + 0.5) - 902 2 2 15. The test statistic of z = = - 29.67 results in a P-value of 0.0000, and it is =

()

n 2

( )

902 2

in the critical region bounded by z = - 2.575 and 2.575. There is sufficient evidence to warrant rejection of the claim that smokers have a median cotinine level equal to 2.84 ng/mL. (123 + 0.5)- 300 2 16. The test statistic of z = = - 3.06 results in a P-value of 0.0022, and it is in the critical 300 2 region bounded by z = - 1.96 and 1.96. There is sufficient evidence to warrant rejection of the claim that the systolic blood pressure of the population is equal to 125 mmHg. (30 + 0.5)- 105 2 = - 4.29 results in a P-value of 0.0000, and it is 17. Second approach: The test statistic of z = 105 2 in the critical region bounded by z =- 1.645 , so the conclusions are the same as in Example 4.

(38 + 0.5)- 1062 Third approach: The test statistic of z = = - 2.82 results in a P-value of 0.0024, and it is in 106 2 the critical region bounded by z =- 1.645 , so the conclusions are the same as in Example 4.

The different approaches can lead to very different results; as seen in the test statistics of –4.21, –4.29, and –2.82. The conclusions are the same in this case, but they could be different in other cases.

18. The column entries are *, *, *, *, *, 0, 0, 0.

13-3 : Wilcoxon Signed-Ranks Test for Matched Pairs

1.

2.

3.

4.

5.

6.

a. The only requirements are that the matched pairs be a simple random sample and the population of differences be approximately symmetric. b. There is no requirement of a normal distribution or any other specific distribution. c. The Wilcoxon signed-ranks test is “distribution free” in the sense that it does not require a normal distribution or any other specific distribution. a. –4, –6, –3, 1, –1, –7 d. 1.5 and 19.5 b. 4, 5, 3, 1.5, 1.5, 6 e. T = 1.5 c. –4, –5, –3, 1.5, –1.5, –6 f. The critical value of T is 1.

The sign test uses only the signs of the differences, but the Wilcoxon signed-ranks test uses ranks that are affected by the magnitudes of the differences.

The efficiency of 0.95 indicates that with all other things being equal, the Wilcoxon signed-ranks test requires 100 sample observations to achieve the same results as 95 sample observations analyzed through a parametric test.

Test statistic: T = 2.5 + 6 = 8.5; Critical value: T = 4; Fail to reject the null hypothesis that the population of differences has a median of 0. There is not sufficient evidence to reject the claim of no difference in heights between mothers and their first daughters. There does not appear to be a difference in the heights of mothers and their first daughters.

Test statistic: T = 2.5 + 8 + 6.5 = 17; Critical value: T = 4; Fail to reject the null hypothesis that the population of differences has a median of 0. There is not sufficient evidence to reject the claim of no difference in the

7.

8.

9.

heights of fathers and their first sons. There does not appear to be a difference in the heights of fathers and their first sons.

Test statistic: T = 1+ 4 = 5; Critical value: T = 8; Reject the null hypothesis that the population of differences has a median of 0. There is sufficient evidence to warrant rejection of the claim that for males, their height is the same as their arm span.

Test statistic: T = 0; Critical value: T = 7; Reject the null hypothesis that the population of differences has a median of 0. There is sufficient evidence to support the claim that captopril has an effect on systolic blood pressure.

Convert T = 491.5 to the test statistic z =- 5.77. P-value = 0.0000; Critical values: z = ±2.575; There is sufficient evidence to warrant rejection of the claim that the median is equal to 98.6°F. n(n +1) 84(84 +1) T491.5 4 4 z= = = - 5.77 84(84 +1)(2 ×84 +1) n(n +1)(2n +1) 24 24

10. Convert T = 16, 236 to the test statistic z =- 1.89. P-value = 0.0588; Critical values: z = ±1.96; There is not sufficient evidence to warrant rejection of the claim that the sample is from a population with a median diastolic blood pressure of 72 mmHg. n (n +1) 273(273 +1) T16, 236 4 4 z= = = - 1.89 n (n +1)(2n +1) 273(273 +1)(2 ×273 +1) 24 24 11. Convert T = 19 to the test statistic z =- 26.01. P-value = 0.0000; Critical values: z = ±2.575; There is sufficient evidence to warrant rejection of the claim that smokers have a median cotinine level equal to 2.84 ng/mL. n(n +1) 902(902 +1) T19 4 4 z= = = - 26.01 902(902 +1)(2 ×902 +1) n(n +1)(2n +1) 24 24 12. Convert T = 17,826 to the test statistic z =- 3.16. P-value = 0.0016; Critical values: z = ±1.96; There is sufficient evidence to warrant rejection of the claim that the times are from a population with a median systolic blood pressure of 125 mmHg. n (n +1) 300(300 +1) T17,826 4 4 z= = = - 3.16 n (n +1)(2n +1) 24

24

13. a. Smallest: T = 0; Largest: T = 1+ 2 + 3 + b.

+ 298 + 299 + 300 =

45,150

= 22,575 2 c. 45,150 - 12, 345 = 32,805

d.

300(300 +1)

= 45,150

2 n (n +1) -k 2

13-4 : Wilcoxon Rank-Sum Test for Two Independent Samples

1.

Yes, the two samples are independent. (The heights from ANSUR I and ANSUR II are not matched in any way.) Each sample has more than 10 values.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

2.

The ranks are given in the table below: R =12 + 25 + 3+ 27 + 9 + 21+15 + 7.5 +19 + 22 + 7.5 + 4 =172.

ANSUR I 1988

1620 (12) 1640 (19) 1672 (23) 1637 (18)

ANSUR II 2012

3.

1693 (25) 1660 (22) 1621 (13) 1718 (26)

1558 (3) 1597 (7.5) 1623 (14) 1588 (6)

1783 (27) 1569 (4) 1633 (17) 1520 (1)

1609 (9)

1649 (21)

1628 (15)

1597 (7.5)

1526 (2) 1618 (11)

1570 (5) 1631 (16)

1616 (10) 1642 (20)

1690 (24)

H 0: The sample of heights from ANSUR I and the sample of heights from ANSUR II are from populations

with the same median.

There are three different possible alternative hypotheses.

H1: The sample of heights from ANSUR I and the sample of heights from ANSUR II are from populations with different medians.

H1: The ANSUR I sample is from a population with a lower median than the median of the second population.

H1: The ANSUR I sample is from a population with a higher median than the median of the second population.

4.

The efficiency rating of 0.95 indicates that with all other factors being the same, the Wilcoxon rank-sum test requires 100 sample observations to achieve the same results as 95 observations with the parametric t test of Section 9-2, assuming that the stricter requirements of the parametric t-test are satisfied.

5.

R1 = 172; R2 = 206; mR = 168; s R = 20.4939; Test statistic: z = 0.20; P-value = 0.857; Critical values: z = ±1.96; Fail to reject the null hypothesis that the populations have the same median. There is not sufficient

evidence to warrant rejection of the claim that the sample of female heights from ANSUR I and the sample of female heights from ANSUR II are from populations with the same median. n (n + n +1) 12(12 +15 +1) m = 1 1 2 = = 168 R 2 2 n1n2(n1 + n2 +1) 12 ×15(12 +15 +1) sR = = = 20.4939 12 12 R - mR 172 - 168 z= = = 0.20 sR 20.4939 6.

R1 = 194.5; R2 = 105.5; mR = 150; s R = 17.321; Test statistic: z = 2.57; P-value = 0.0102; Critical values: z = ±1.96; Reject the null hypothesis that the populations have the same median. There is

sufficient evidence to reject the claim that the median amount of strontium-90 from Pennsylvania residents is the same as the median from New York residents. n1 (n1 + n2 +1) 12(12 +12 +1) m= = = 150 R 2 2 s

7.

R

=

n1n2 (n1 + n2 +1)

=

12 ×12(12 +12 +1)

12 R - mR 194.5 - 150 z= = = 2.57 sR 17.321

12

= 17.321

R1 = 253.5; R2 = 124.5; mR = 182; s R = 20.607; Test statistic: z = 3.47; P-value = 0.0005; Critical values: z = ±1.96; Reject the null hypothesis that the populations have the same median. There is sufficient evidence to reject the claim that for those treated with 20 mg of Lipitor and those treated with 80 mg of Lipitor, changes in LDL cholesterol have the same median. It appears that the dosage amount does have an effect on the change in LDL cholesterol.

7.

(continued) mR= s

8.

n1 (n1 + n2 +1) 13(13 +14 +1) = = 182 2 2 n1n2 (n1 + n2 +1)

13×14(13 +14 +1) = = 20.607 12 12 R - mR 253.5 - 182 z= = = 3.47 sR 20.607

R

=

R1 = 437; R2 = 424; mR = 525; s R = 37.4166; Test statistic: z = - 2.35; P-value = 0.0188; Critical values: z = ±2.575; Fail to reject the null hypothesis of no difference between the two population medians. There is not

sufficient evidence to support the claim that the arrangement of the test items has an effect on the score. mR =

n1 (n1 + n2 +1) 25(25 +16 +1) = = 525 2 2

n1n2 (n1 + n2 +1) 25×16(25 +16 +1) = = 37.4166 12 12 R - mR 437 - 525 z= = = - 2.35 sR 37.4166

sR =

9.

R1 = 36, 531.5; R2 = 43, 668.5; mR = 41,102.5; s R = 1155.782; Test statistic: z = - 3.95; P-value = 0.0001 (Table: 0.0002); Critical values: z = ±1.96; Reject the null hypothesis that the populations have the same

median. There is sufficient evidence to warrant rejection of the claim that girls and boys have the same median birth weight. n1 (n1 + n2 +1) 205(205 +195 +1) mR = = = 41,102.5 2 2 n1n2 (n1 + n2 +1) 205 ×195(205 +195 +1) = = 1155.782 12 12 R - mR 36, 531.5 - 43, 668.5 z= = = - 3.95 sR 1155.782

sR =

10. R1 = 863; R2 = 412; mR = 637.5; s R = 51.539; Test statistic: z = 4.38; P-value = 0.0000 (Table: 0.0001); Critical value: z = 2.33; Reject the null hypothesis that the populations have the same median. There is sufficient evidence to support the claim that the nonfiltered king size cigarettes have a median amount of nicotine that is greater than the median amount of nicotine in the 100-mm filtered cigarettes. n1 (n1 + n2 +1) 25(25 + 25 +1) mR = = = 637.5 2 2 n1n2 (n1 + n2 +1) 25×25(25 + 25 +1) = = 51.54 12 12 R - mR 863 - 637.5 z= = = 4.38 sR 51.54

sR =

11. R1 = 501; R2 = 445; mR = 484; s R = 41.15823; Test statistic: z = 0.41; P-value = 0.3409; Critical value: z = 1.645; Fail to reject the null hypothesis that the populations have the same median. There is not sufficient evidence to support the claim that subjects with medium lead levels have a higher median of the full IQ scores than subjects with high lead levels. Based on these data, it does not appear that lead level affects full IQ scores.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

11. (continued) mR = s

n1 (n1 + n2 +1) 22(22 + 21+1) = = 484 2 2 n1n2 (n1 + n2 +1)

22 ×21(22 + 21+1) = = 41.15823 12 12 R - mR 501- 484 z= = = 0.41 sR 41.15823

R

=

12. R1 = 4178; R2 = 772; mR = 3900; s R = 116.833; Test statistic: z = 2.38; P-value = 0.0087; Critical value: z = 1.645; Reject the null hypothesis that the populations have the same median. There is sufficient evidence to support the claim that subjects with low lead levels have performance IQ scores with a higher median than subjects with high lead levels. It appears that exposure to lead does have an adverse effect. n1 (n1 + n2 +1) 78(78 + 21+1) m= = = 3900 R 2 2 n1n2 (n1 + n2 +1) 78×21(78 + 21+1) = = 116.833 12 12 R - mR 4178 - 3900 z= = = 2.38 sR 116.833

sR =

13. The test statistic from the Mann-Whitney U test is the same test statistic from the Wilcoxon rank-sum test but with opposite sign. n n +1 12 12 +1) U = n n + 1 ( 1 ) - R = 12 ×15 + ( - 127 = 131 1 2 2 2 n1n2 12 ×15 U1312 2 z= = = 2.00 n1n2 (n1 + n2 +1) 12 12 14. a.

Rank Sum for Rank Treatment A 1 2 3 4 A A B B 3 A B A B 4 A B B A 5 B B A A 7 B A A B 5 B A B A 6 b. The R values of 3, 4, 5, 6, 7 have probabilities of 1 6 , 1 6 , 2 6 , 1 6 , and 1 6 , respectively c. No, none of the probabilities for the values of R would be less than 0.10.

13-5 : Kruskal-Wallis Test for Three or More Samples

1.

R1 = 103.5, R2 = 48.5, R3 = 90, R4 = 83

2.

Small Midsize Large SUV 29 (13) 32 (20) 27 (10.5) 24 (4) 31 (16.5) 28 (12) 32 (20) 31 (16.5) 35 (23) 26 (8) 39 (24.5) 31 (16.5) 33 (22) 23 (3) 27 (10.5) 25 (5.5) 26 (8) 25 (5.5) 31 (16.5) 30 (14) 32 (20) 26 (8) 39 (24.5) 21 (1) 22 (2) (The rank for each value is shown in parentheses.)

Yes, the samples are independent simple random samples, and each sample has at least five data values.

3.

n1 = 7, n2 = 5, n3 = 6, n4 = 7, N = 7 + 5 + 6 + 7 = 25

4.

The efficiency rating of 0.95 indicates that with all other factors being the same, the Kruskal-Wallis test requires 100 sample observations to achieve the same results as 95 observations with the parametric one-way analysis of variance test, assuming that the stricter requirements of the parametric test are satisfied.

5.

Test statistic: H = 4.9054; Critical value: c 2 = 5.991; (Tech: P-value = 0.0861); Fail to reject the null hypothesis of equal æ medians.

The data do not suggest that larger safer.

2 æ862cars97are 12 R2 2 2ö 12 482 ö 2 3 1 H = + R + R ÷- 3(N +1) = + + ç ÷- 3(21+1) ç 7 7 ø 21(21+1) è 7 N (N +1) è n1 n2 n3 ø = 4.9054

6.

Test statistic: H = 2.3503; Critical value: c 2 = 5.991; (Tech: P-value = 0.3088); Fail to reject the null hypothesis of equal medians.

These data do not provide sufficient evidence to suggest that larger cars are safer. æR2 R2 R2 ö æ1422 133.52 189.52 ö 12 12 1 H = + 2 + 3 ÷- 3(N +1) = + + ç ÷- 3(30 +1) ç 30(30 +1) è 10 10 10 ø N (N +1) è n1 n2 n3 ø = 2.3503

7.

Test statistic: H = 2.029; Critical value: c 2 = 7.815; (Tech: P-value = 0.566); Fail to reject the null hypothesis of equal medians. There is not sufficient evidence to warrant rejection of the claim that small, midsize, large, and SUV vehicles have the same median HIC measurement in car crash tests. æR2 æ103.52 48.52 902 832 ö 12 2 2 2ö 12 1 H = + R2 + R3 + R4 ÷- 3(N +1) = + + + ç ç ÷- 3(25 +1) 25(25 +1) è 7 5 6 7 ø N (N +1) è n1 n2 n3 n4 ø = 2.029

8.

Test statistic: H = 22.8157; Critical value: c 2 = 9.210; (Tech: P-value = 0.000); Reject the null hypothesis of equal medians. It appears the three states have median amounts arsenic that are not all the same. æR2 that æ1922 of 12 2 2ö 12 116.52 357.52 ö 2 3 1 H = + R + R ÷- 3(N +1) = + + ç ÷- 3(36 +1) ç 36(36 +1) è 12 12 12 ø N (N +1) è n1 n2 n3 ø = 22.8157

9.

Test statistic: H = 27.9098; Critical value: c 2 = 5.991; (Tech: P-value = 0.000); Reject the null hypothesis of equal medians. There is sufficient evidence to warrant rejection of the claim that the three different types of cigarettes have the same median amount of nicotine. It appears that the filters do make a difference.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

9.

(continued)

æ1413.52 650.52 7862 ö 2 2ö 12 æ 2 2 3 1 R R R H = + + + + ç ÷- 3(N +1) = ÷- 3(75 +1) ç 25 25 ø 75(75 +1) è 25 N (N +1) è n1 n2 n3 ø 12

= 27.9098 10. Test statistic: H = 43.3240; Critical value: c 2 = 5.991; (Tech: P-value = 0.000); Reject the null hypothesis of equal medians. There is sufficient evidence to warrant rejection of the claim that the three different types of cigarettes have the same amount of nicotine. It appears that2the filters do make a difference. æR2 median æ1534 12 2 2ö 12 6202 6962 ö 2 3 1 H = + + ç + R + R ÷- 3(N +1) = ÷- 3(75 +1) ç 25 25 ø 75(75 +1) è 25 N (N +1) è n1 n2 n3 ø = 43.3240 11. Test statistic: H = 2.5999; Critical value: c 2 = 7.815; (Tech: P-value = 0.458); Fail to reject the null hypothesis of equal æ medians.

It appears that the four hospitals have birth weights with the same median.

12 R2 2 2 2 ö 2 3 1 H = ç + R + R + 4R÷- 3(N +1) N (N +1) è n1 n2 n3 n4 ø æ 12 20,1892 21, 262.52 20,106.52 18, 6422 ö = ç 400(400 +1) + 100 + 100 ÷- 3(400 +1) 100 100 + è ø = 2.5999 12. Test statistic: H = 8.0115; Critical value: c 2 = 5.991; (Tech: P-value = 0.018); Reject the null hypothesis of equal medians. It appears that the three categories of blood lead level have different median performance IQ scores. æR2 æ5277.52 11122 991.52 ö 12 12 22 32 ö 1 R R H= + + + ÷- 3(N +1) = ç + ÷- 3(121+1) ç 22 21 ø 121(121+1) è 78 N (N +1) è n1 n2 n3 ø = 8.0115 13. Using the methods of this section, H = 0.694. æR2 R2 R2 ö æ862 50.52 53.52 ö 12 12 1 2 3 H = + + ÷- 3(N +1) = + + ç ç ÷- 3(19 +1) 19(19 +1) è 8 6 5 ø N (N +1) è n1 n2 n3 ø = 0.694 For the correction factor, the values of t are 2, 2, 2, 2, and 4 (See table below.), the values of T are 6, 6, 6, 6, and 60, and ST = 84.

84 ö æ Using ST = 84 and N = 8 + 6 + 5 = 19, the corrected value is H = 0.694 1= 0.703, which is not ç ÷ 3 è 19 - 19 ø substantially different from the value of 0.694.

In this case, the large numbers of ties do not appear to have a dramatic effect on the test statistic H. Lead Level 85 90 97 100 107

Rank 1.5 4.5 6.5 17.0 28.0

t

t3 - t

2 2 4 2 2 SUM:

6 6 60 6 6 84

13-6 : Rank Correlation

1.

5+6 2

= 5.5 and

8+9

= 8.5; Lower ranks correspond to lower costs.

2 Quality Rank Cost Rank

1 7

2 8.5

7 10

9 8.5

3 5.5

4 1

5 2

6 4

8 3

10 5.5

2.

rs = 0.018; Critical values: ±0.648; There is not sufficient evidence to support the claim of a correlation

3.

between the quality rank and cost of thermometers. r represents the linear correlation coefficient computed from sample paired data; r represents the parameterof the linear correlation coefficient computed from a population of paired data; rs denotes the rank correlation

4.

5.

coefficient computed from sample paired data; and r s represents the rank correlation coefficient computed from a population of paired data. The subscript s is used to distinguish the rank correlation coefficient from the linear correlation coefficient r. The subscript does not represent the standard deviation s. It is used in recognition of Charles Spearman, who introduced the rank correlation method.

The efficiency rating of 0.91 indicates that with all other factors being the same, rank correlation requires 100 pairs of sample observations to achieve the same results as 91 pairs of observations with the parametric test using linear correlation, assuming that the stricter requirements for using linear correlation are met. rs = 1; Critical values: rs = ±0.648. Reject the null hypothesis of r s = 0. There is sufficient evidence to

support a claim of a correlation between the ages and heights of trees.

Age (rank) Height (rank) 6.

1 9

2 8

Points from left to right 4 5 6 7 10 7 2.5 6

3 4

8 1

9 2.5

10 5

rs = - 1; Critical values: rs = ±0.648. Reject the null hypothesis of r s = 0. There is sufficient evidence to

support a claim of a correlation between the numbers of waiting patients and patient service times.

Age (rank) Height (rank) 7.

1 10

2 9

Points from left to right 4 5 6 7 7 6 5 4

3 8

8 3

9 2

10 1

rs = 0.857; Critical values: ±0.738; Reject the null hypothesis of r s = 0. There is sufficient evidence to

conclude that there is a correlation between the number of chirps in 1 min and the temperature.

Chirps (rank) Temperature (rank) 8.

2 1

7 8

6 6

1 3

8 7

5 5

4 2

3 4

rs = 1; Critical values: ±0.866; Reject the null hypothesis of r s = 0. There is sufficient evidence to conclude

that there is a correlation between overhead widths of seals from photographs and the weights of the seals.

Overhead width (rank) Weight (rank) 9.

1 1

2 2

6 6

5 5

4 4

3 3

rs = 0.624; Critical values: rs = ±0.587. Reject the null hypothesis of r s = 0. There is sufficient evidence to

support a conclusion that there is a correlation between the number of cigarettes smoked and the cotinine level.

Cigarettes/day (rank) Cotinine (rank)

1.5 1

3 4

1.5 2

4 6

9.5 13

5 3

12 5

9.5 8

13 10

14 14

11 12

8 11

6 9

7 7

10. rs = 0.778; Critical values: rs = ±0.538. Reject the null hypothesis of r s = 0. There is sufficient evidence to support the claim of a correlation between the circumference and height of a tree.

Circumference (rank) 1.5 Height (rank) 1

3 4

1.5 2

4 6

9.5 13

5 3

12 5

9.5 8

13 10

14 14

11 12

8 11

6 9

7 7

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

11. rs = 0.923; Critical values: ±0.648; Reject the null hypothesis of r s = 0. There is sufficient evidence to support the claim of a correlation between cheese consumption and numbers of civil engineering PhD degrees awarded. It should be noted that this is a spurious correlation. Common sense suggests that there is no real association between cheese consumption and numbers of civil engineering PhD degrees awarded.

Cheese Consumption (rank) Civil Engineering PhDs (rank)

1 1

3 2

3 3

3 5

5 4

6 6

7 7

10 8

8.5 10

8.5 9

12. rs = - 0.434; Critical values: ±0.700; Fail to reject the null hypothesis of r s = 0. There is not sufficient evidence to support the claim of a correlation between heights of winning presidential candidates and heights of their main opponents. Hopefully, we do not elect our presidents based on heights or any other physical appearance.

President (rank) Opponent (rank)

9 5.5 ±z

2 5.5

1 7

4 4

6 2

6 9

3 8

6 3

8 1

±1.96 = ±0.159; Reject the null hypothesis of r s = 0. There is 153 - 1 n- 1 sufficient evidence to support a conclusion that there is a correlation between the systolic and diastolic blood pressure in males. ±z ±1.96 = 14. rs = 0.106; Critical values: rs = = ±0.450; Fail to reject the null hypothesis of r s = 0. There 20 - 1 n- 1 is not sufficient evidence to support the claim of a correlation between brain volumes and IQ scores. ±z ±1.96 = 15. rs = 0.346; Critical values: rs = = ±0.044; Reject the null hypothesis of r s = 0. There is 1986 -1 n- 1 sufficient evidence to support the claim of a correlation between foot lengths and distances between pupils of females. ±z ±1.96 = 16. rs = 0.270; Critical values: rs = = ±0.031; Reject the null hypothesis of r s = 0. There is 4082 - 1 n- 1 sufficient evidence to support the claim of a correlation between foot lengths and distances between pupils of males.

13. rs = 0.360; Critical values: rs =

17. rs = ±

=

1.961162

= ±0.044 (df = 1986 - 2 = 1984); The critical values of ±0.044 are the same as 2 1.96116 +1986 - 2 those found by using Formula 13-1. (Using t = 1.960 for “large” n from Table A-3 results in the same critical values.)

Quick Quiz 1.

The ranks are 3, 8.5, 8.5, 3, 6, 1, 5, 10, 7, 3. The rank for 1.1 is found using found using

2.

3.

4.

8+9

2 +3+ 4

= 3 and the rank for 1.7 is

3

= 8.5.

2 The efficiency rating of 0.91 indicates that with all other factors being the same, rank correlation requires 100 pairs of sample observations to achieve the same results as 91 pairs of observations with the parametric test for linear correlation, assuming that the stricter requirements for using linear correlation are met. a. distribution-free test b. The term “distribution-free test” suggests correctly that the test does not require that a population must have a particular distribution, such as a normal distribution. The term “nonparametric test” incorrectly suggests that the test is not based on a parameter, but some nonparametric tests are based on the median, which is a parameter; the term “distribution-free test” is better because it does not make that incorrect suggestion.

Rank correlation should be used. The rank correlation test is used to investigate whether there is a correlation between red blood cell counts and platelets.

5.

No, the P-values are almost always different, and the conclusions may or may not be the same.

6.

Rank correlation can be used in a wider variety of circumstances than linear correlation. Rank correlation does not require a normal distribution for any population. Rank correlation can be used to detect some (not all) relationships that are not linear.

7. The sign test can be used to test claims involving matched pairs of sample data, it can be used to test claims involving nominal data with two categories, and it can be used to test claims about the median of a single population.

8. Because the sign test uses only signs of differences while the Wilcoxon signed-ranks test uses ranks of the differences, the Wilcoxon signed-ranks test uses more information about the data and tends to yield conclusions that better reflect the true nature of the data.

9. The Wilcoxon signed-ranks test is used to test a claim that a population of matched pairs has the property that they have differences with a median equal to zero or to test a claim that a single population of individual values has a median equal to some claimed value, whereas the Wilcoxon rank-sum test is used to test the null hypothesis that two independent samples are from populations having equal medians.

10. One-way analysis of variance can be used instead of the Kruskal-Wallis test. Like many other nonparametric tests, the Kruskal-Wallis test has no requirement that the populations have a normal distribution or any other particular distribution.

Review Exercises 1. The two sets of ranks appear to agree reasonably well.

2.

3.

4.

September (rank)

2

4

9

April (rank)

2

3

9

8 7 +8 2 = 7.5

1

6

1

6

4

4

7

4+5 2 = 4.5

4+5 2 = 4.5

7 +8 2 = 7.5

The test statistic of x = 0 is less than or equal to the critical value of 0 (from Table A-7). There is sufficient evidence to warrant rejection of the claim that for the population of freshman male college students, there is not a significant difference between the weights in September and the weights in the following April. Based on the given data, there does appear to be a significant difference.

The test statistic T = 0 is less than or equal to the critical value of T = 2. There is sufficient evidence to warrant rejection of the claim that the median of the differences is equal to 0. There does appear to be a difference. The test does not address the specific weight gain of 15 lb. rs = 0.983; Critical values: rs = ±0.700; Reject the null hypothesis of r s = 0. There is sufficient evidence to

support the claim of a correlation between the September weights and the April weights. The presence of a correlation tells us nothing about the belief that college students gain 15 lb (or 6.8 kg) during their freshman year.

5.

Test statistic: H = 0.465; (Tech: P-value = 0.792); Critical value: c 2 = 5.991; Fail to reject the null hypothesis of equal medians. There is not sufficient evidence to warrant rejection of the claim that the lengths of stay at the three hospitals have the same median. æR2 æ254.52 2822 243.52 ö 12 12 R2 R2 ö 1 H= + + ç + 2 + 3 ÷- 3(N +1) = ç ÷- 3(39 +1) 39(39 +1) è 13 13 13 ø N (N +1) è n1 n2 n3 ø = 0.465 (x + 0.5) -

6.

7.

Sign test: Test statistic: z =

(n ) (907 + 0.5) - (2015 ) 2

n 2

=

2

2015 2

= - 4.46; P-value = 0.0000 (Table: 0.0001);

Critical value: z = - 1.645; Reject the null hypothesis of p = 0.5. There is sufficient evidence to support the claim that the majority of adults obtain medical information more often from the Internet than from a doctor.

The test statistic of x = 3 is less than or equal to the critical value of 7 from Table A-7. There is sufficient evidence to warrant rejection of the claim that the sample is from a population with a median equal to 5 min.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

8.

Test statistic: H = 6.6305; (Tech: P-value = 0.0363); Critical value: c 2 = 5.991; Reject the null hypothesis of equal medians. Interbreeding of cultures is suggested by the data. æR2 æ912 112.52 174.52 ö 12 2 2ö 12 2 3 1 H = + R + R ÷- 3(N +1) = + + ç ÷- 3(27 +1) ç 9 9 ø 27 (27 +1) è 9 N (N +1) è n1 n2 n3 ø = 6.6305

9.

R1 = 60; R2 = 111; mR = 85.5; s R = 11.3248; Test statistic: z = - 2.25; (Tech: P-value = 0.0244); Critical values: z = ±1.96; Reject the null hypothesis that the populations have the same median. Skull breadths from

4000 B.C. appear to have a different median than those from A.D. 150. mR =

n1 (n1 + n2 +1) 9(9 + 9 +1) = = 85.5 2 2

n1n2 (n1 + n2 +1) 9 ×9(9 + 9 +1) = = 11.3248 12 12 R - mR 60 - 85.5 z= = = - 2.25 sR 11.3248 ±z ±1.96 = = ±0.741; Fail to reject the null hypothesis of r = 0. There is 10. r = 0.714. Critical values: r = s s s 8- 1 n- 1 not sufficient evidence to support the claim that there is a correlation between the student ranks and the magazine ranks. When ranking colleges, students and the magazine do not appear to agree. sR =

Cumulative Review Exercises 1.

2.4 + 3.6 + 2.2 + 3.1+ 2.8 + 3.7 + 3.4 + 3.1+ 3.2 + 2.4 = 2.99 kg.

10 3.1+ 3.1 = 3.10 kg.

The median is Q2 = 2 The range is 3.7 - 2.2 = 1.50 kg.

The mean is x =

The standard deviation is s =

(2.4 - 2.99)2 + (3.6 - 2.99)2 +

(3.2 - 2.99)2 + +(2.4 - 2.99)2

10 - 1

= 0.52 kg.

The variance is s2 = (0.52 kg)2 = 0.27 kg2.

2.

A histogram is not too helpful with such a small data set, but a normal quantile plot shows that the points approximate a straight-line pattern, so the birth weights do appear to be from a population that has a normal distribution.

3.

H0: m = 3.25 kg; H1: m ¹ 3.25 kg

Test statistic: t =

4.

5.

6.

7.

8.

9.

x- m

=

2.99 - 3.25

= - 1.570; Critical values: t = ±2.262; P-value = 0.1509 s / n 0.524 / 10 (Table: P-value > 0.10); Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim that the female babies are from a population with a mean of 3.25 kg.

The test statistic of x = 3 is not less than or equal to the critical value of 1 from Table A-7. There is not sufficient evidence to warrant rejection of the claim that the female babies are from a population with a median of 3.25 kg.

The test statistic of T = 5 + 6.5 + 3 = 14.5 is not less than or equal to the critical value of T = 8. Fail to reject the null hypothesis that the population has a median of 3.25 kg. There is not sufficient evidence to warrant rejection of the claim that the female babies are from a population with a median of 3.25 kg. s 95% CI: x ± t × 0.524 Þ 2.62 kg < m < 3.36 kg (df = 10 - 1 = 9); Because the confidence = 2.99 ± 2.262 × a /2 n 10 interval contains the claimed mean of 3.25 kg, there is not sufficient evidence to warrant rejection of the claim that the female babies are from a population with a mean of 3.25 kg.

Answers vary, but here is a typical answer: 2.70 kg < m < 3.29 kg. Because the confidence interval contains the claimed mean of 3.25 kg, there is not sufficient evidence to warrant rejection of the claim that the female babies are from a population with a mean of 3.25 kg. n=

[za /2 ]2 pˆqˆ [1.96]2 ×0.25

= 2401; The sample size does appear to be reasonable. = 2 E2 0.02 a. The scatterplot does not reveal a pattern that suggests a correlation. There does not appear to be a correlation.

b. H0: r = 0; H1: r ¹ 0; r = - 0.413; Critical values (a = 0.05): r = ±0.632; P-value = 0.235 (Table: P-value > 0.05); Fail to reject H 0. There is not sufficient evidence to support a claim that there is a linear correlation between systolic blood pressure and HDL cholesterol measurements. c. rs = - 0.343; Critical values (a = 0.05): rs = ±0.648; Fail to reject the null hypothesis of r s = 0. There does not appear to be a correlation between systolic blood pressure and HDL cholesterol measurements.

Systolic(rank) HDL(rank)

3.5 4

6 8.5

9 5

3.5 8.5

8 6

6 2

10 1

6 7

2 10

1 3

10. There must be an error, because the rates of 13.7% and 10.6% are not possible with samples of size 100.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

Chapter 14: Survival Analysis Section 14-1: Life Tables 1. A cohort life table is a record of the actual observed mortality experience for a particular group, whereas a period life table provides mortality and longevity data for a hypothetical group that lived with the same mortality conditions throughout their lives.

2. Period life tables are more practical because the data comes from one year. Cohort life tables require that an actual population of people is followed through life, and this would take many, many years. Also, there would probably be many missing or incomplete data values.

3. The values in columns 3 through 6 would be halved, but the other values would remain the same.

4. Possible answers: Table 14-1 assumes that the same conditions affecting mortality in the year 2018 remain in effect throughout the person's lifetime, but those conditions might change substantially. Also, Table 14-1 was designed for the total population, but the person might belong to a subgroup (such as white females) that has a very different mortality experience.

5. There were 99,586 white females alive on their first birthday, and there were 25 deaths during the interval.

The probability of dying in that interval was 25 99, 586 = 0.000251. (Using greater precision, the actual published value is 0.000252.)

6. There were 99,561 white females alive at their second birthday, and the probability of dying between their second and third birthdays was 0.000195. The number of deaths during the interval was 99,561(0.000195) = 19.

7.

There were 100,000 white females at birth, and there were 99,561 that survived until their second birthday.

The probability of surviving until the second birthday is 99,561 100, 000 = 0.99561.

8.

The values would be: 0–2

9.

414 + 25 100, 000 = 0.00439

100,000

414 + 25

99,637 + 99,574

= 439

= 199, 211

8,109,660

81.1

From column 2, the probability of dying between the 20th birthday and the 21st birthday is 0.000750, so the probability of surviving that same period is 1- 0.000750 = 0.999250. If 5000 people live to their 20th birthday, we expect that 5000(0.999250) = 4996.25 will survive to their 21st birthday.

74 Using the values from columns 3 and 4, 1= 0.999252 and 5000(0.999252) = 4996.26.

98,964

10. a. The expected remaining lifetime for someone who has just reached their 60th birthday is 23.3 years. For someone reaching their 61st birthday, the expected remaining lifetime is 22.6 years. b. The expected age of death for someone who has just reached their 60th birthday is 60 + 23.2 = 83.3 years. For someone reaching their 61st birthday, the expected age of death is 61+ 22.6 = 83.6 years. c. The results would be equal only if the person has no chance of dying between the 60th birthday and 61st birthday. But the mortality rate for the age interval of 60–61 is the reason that someone alive at age 61 is expected to die at a later age than someone who is alive at age 60. The person who has reached their 61st birthday has already survived the age interval of 60–61, but the person who has reached their 60th birthday has a chance of dying during that age interval.

11. a. There were 99,175 people alive on their 16th birthday, and 82,763 people alive on their 66th birthday.

The probability of surviving is 82, 763 99,175 = 0.834515. b. 3554(0.834515) = 2965.87, or 2966 12. a. There were 98,890 people alive on their 21st birthday and 58,141 alive on their 80th birthday. The probability of surviving is 58,141 98,890 = 0.587936. b. 8847(0.587936) = 5201.47, or 5201

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

13. a. 0.005650 + 0.000367 = 0.006017 or

565 + 36

= 0.006010 100, 000 26 +19 = 0.000453 b. 0.000264 + 0.000187 = 0.000451 or 99, 399 The results are so different because of the exceptionally high mortality rate at or very near birth.

14. The values would be: 15. The values would be: 565 + 36

0–2 26 +19

2–4

= 0.006010 or 100, 000 0.005650 + 0.000367 = 0.006017 100,000 565 + 36 = 601 99,505 + 99,417 = 198, 922

= 0.000453 or 99, 399 0.000264 + 0.000187 = 0.000451 99,399 26 +19 = 45 99,385 + 99,363 = 198, 748

7,873,749 78.7

7,674,827 77.2

16. The values would be: 0–10 565 + 36 + 26 +19 +15 +14 +13 +11+10 + 9 = 0.007180 or 100, 000 0.00565 + 0.000367 + 0.000264 + 0.000187 + 0.000148 + 0.000142 + 0.000126 + 0.000114 +0.000103 + 0.000094 = 0.007195 100,000 565 + 36 + 26 +19 +15 +14 +13+11+10 + 9 = 718 99, 505 + 99, 417 + 99, 385 + 99, 363 + 99,346 + 99,332 + 99, 319 + 99, 307 + 99, 296 +99, 286 = 993, 556 7,873,749 78.7 17. H0: p = 0.000750; H1: p > 0.000750 (the death rate in column 2); The test statistic is z = 1.84, the P-value is 0.0330 (Table: 0.0329), and the critical value is z = 1.645. Reject H 0 and conclude that there is sufficient evidence to support the claim that the number of deaths is significantly high. (There are other ways to calculate the probability of dying in the time interval, and they produce similar results.)

15 - 0.000750 pˆ - p 12,500 z= = = 1.84 pq (0.000750)(1 - 0.000750) n 12, 500 18. H0: p = 0.003950; H1: p > 0.003950; The test statistic is z = 1.08, the P-value is 0.1407 (Table: 0.1401), and the critical value is z = 1.645. Fail to reject H 0 and conclude that there is not sufficient evidence to support the claim that the number of deaths is significantly high. (There are other ways to calculate the probability of dying in the time interval, and they produce similar results.)

29 - 0.003950 pˆ - p 6015 z= = = 1.08 pq (0.003950)(1- 0.003950) n 6015

19. H0: p = 1- 97,946 = 0.012392; H1: p > 0.012392; The test statistic is z = 3.62, the P-value is 0.0001, and the 99,175 critical value is z = 1.645. Reject H 0 and conclude that there is sufficient evidence to support the claim that the number of deaths is significantly high. (There are other ways to calculate the probability of dying in the time interval, and they produce similar results.)

147 - 0.012392 pˆ - p 8774 z= = = 3.69 pq (0.012392)(1- 0.012392) n 8774 20. H0: p = 1- 97,946 98,964= 0.010287; H1: p < 0.010287; The test statistic is z = - 2.24, the P-value is 0.0127 (Table: 0.0125), and the critical value is z =- 1.645. Reject H 0 and conclude that there is sufficient evidence to support the claim that the number of deaths is significantly low. (There are other ways to calculate the probability of dying in the time interval, and they produce similar results.)

27 - 0.010333 pˆ - p 3427 z= = = - 2.24 pq (0.010333)(1- 0.010333) n 3427 21. Using Table 14-1, 5-year-old students have an expected remaining lifetime of 74.3 years, so their typical total lifetime is 5 + 74.3 = 79.3 years. Fifty-year-old symphony conductors have an expected remaining lifetime of 31.7 years, so their typical total lifetime is 50 + 31.7 = 81.7 years. It appears that being a student is more dangerous than being a symphony conductor, but that is deceptive because a 50-year-old has already survived through ages 5 to 50. A fair comparison would involve 5-year-old students and 5-year-old people who eventually become conductors at age 50, but that approach has an obvious obstacle.

Section 14-2: Kaplan-Meier Survival Analysis 1. A survivor is a subject that successfully lasted throughout a particular time period without reaching some terminating event. A survivor could be a person or an object or some other entity such as a marriage.

2. Censored data are subjects that are removed from the study because they survived past the end of the study or were no longer available to be included in the study.

3. A life table is based on fixed intervals of time, but a table of survival data and Kaplan-Meier calculations is based on times that vary according to the terminating event.

4. It is a graphical representation of the information generated using Kaplan-Meier calculations.

5. The fast walkers are more likely to be healthier with greater longevity, so they correspond to the top (green) curve. The moderate group corresponds to the middle (black) curve. The slow group of walkers is more likely to have health issues with lower longevity, so they correspond to the bottom (red) curve.

6. The event is surviving for the given time period. The probability of survival decreases with the passage of time.

7. a. 4 years b. 1 year 8. The red graph reaches zero, so none of the slow walkers survived past the study.

9. The five-year survival rates are 0.2 for the group of slow walkers, 0.5 for the group with moderate walking speeds, and 0.9 for the group of fast walkers. The differences are substantial, and they suggest that after five years, those with faster walking speeds have much greater survival rates, and those with slow walking speeds have much lower survival rates. The data do not necessarily suggest that we can get older people to live longer by somehow getting them to walk faster. It’s very possible that walking speed is one manifestation of overall health status, so longevity and walking speed are likely both affected by one or more other extraneous variables.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

10.

Day 2

Status 0 = Censored 1 = Failed 0

Number of Patients

Patients Not Requiring Retreatment

Proportion of Patients Not Requiring Retreatment

6

1

9

8

8 / 9 = 0.889

10

1

8

7

7 / 8 = 0.875

11

1

7

6

6 / 7 = 0.857

0.67 (0.889 ´ 0.875 ´ 0.857)

13

1

6

5

5 / 6 = 0.833

0.56 (0.889 ´ 0.875´ 0.857 ´ 0.833)

14 14 14 14 14

0 0 0 0 0

Day 2

Status 0 = Censored 1 = Failed 0

Number of Patients

Patients Not Requiring Retreatment

Proportion of Patients Not Requiring Retreatment

Cumulative Proportion of Patients Not Requiring Retreatment

10

1

7

6

6 / 7 = 0.857

0.86

12

1

6

5

5 / 6 = 0.833

0.71 (0.857 ´ 0.833)

0 0 0 0 0

0 0 0 0 0

Cumulative Proportion of Patients Not Requiring Retreatment 0.89 0.78 (0.889 ´ 0.875)

11.

12.

13. The graph does not show any information about the six censored survival times. The graph shows information about the two survival times that were not censored.

14. Estimates will vary. Placebo: 0.50; Treatment: 0.80; It appears that the treatment is effective.

15. The graph shows that for the first few days, there is not much of a difference between the treatment and the placebo, but the treatment appears to kick in on Day 5 when the number of subjects experiencing dysphagia drops to zero. The treatment appears to be effective.

16.

Day 1 3 4 5 10 11 12 13 13 14

Status 0 = Censored 1 = Failed 1 1 0 0 0 0 0 0 0 1

Quick Quiz 1. life table 2. survival table with Kaplan-Meier calculations 3. A period life table describes mortality and longevity data for a hypothetical group that would have lived with the same mortality conditions throughout their lives.

4. A cohort life table is a record of the actual observed mortality experience for a particular group.

5. false 8. true 6. true 9. The entries are 1, 0, 0, and 0.

7. false 65 = 0.999337 or 1- 0.000611 = 0.999339 10. 198, 030 Review Exercises 99, 516 - 99, 490 1. = 0.000261 (Using more precise results, the published value is 0.000264.)

99, 516 2.

99, 516 - 99, 490 = 26

3.

99,490 100, 000

= 0.99490

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy

4.

The values would be: 484 + 26 100, 000 = 0.00510

0–2 5.

484 + 26

100,000

= 510

99,574 + 99,503 = 199, 077

8,431,083

84.3

84.3 years; 83.7 years; the second value is less than the first value. As we age, our expected remaining lifetime steadily decreases.

6.

Day

Status 0 = Censored 1 = Failed

Number of Patients

Patients Not Requiring Retreatment

Proportion of Patients Not Requiring Retreatment

Cumulative Proportion of Patients Not Requiring Retreatment

3

1

4

3

3 / 4 = 0.75

0.75

12

1

3

2

2 / 3 = 0.667

0.50 (0.75 ´ 0.667)

20

1

2

1

1 / 2 = 0.5

0.25 (0.75 ´ 0.667 ´ 0.5)

30

0

7.

8.

The treatment group and the placebo group appear to have approximately the same behavior. The treatment does not appear to be effective.

9. The graph shows that instead of fostering the lives of bacteria, the treatment causes them to die prematurely.

The treatment is not only ineffective but also doing the opposite of what it is supposed to do.

10. Because of the large increase in the number of deaths from COVID-19, the probabilities of dying during the different intervals will increase, the numbers surviving to the next time interval will decrease, the numbers of deaths in each interval will increase, and the entries in the last three columns of Table 14-1 will all decrease.

Cumulative Review Exercises 1.

The table shows that among 100,000 births, 99,490 survived to the second birthday, so the probability of dying is 1- 0.99490 = 0.00510. H0: p = 0.00510; H1: p < 0.00510, the test statistic is z = - 0.68, the P-value is 0.2473 (Table: 0.2483), and the critical value is z =- 1.645 (assuming a 0.05 significance level). Fail to reject the null hypothesis of p = 0.00510 and conclude that there is not sufficient evidence to support the claim that the proportion of deaths is less than 0.00510. There isn’t sufficient evidence to conclude that the program is effective in reducing the mortality rate.

5 - 0.00510 pˆ - p 1328 z= = = - 0.68 pq (0.00510)(1- 0.00510) n 1328

2.

95% CI: 0.99294 < p < 0.99953; The confidence interval limits contain 0.99490, so it appears that the mortality rate has not been lowered by a significant amount. z = pˆ ± za /2

p̂q̂ 1323 = 1328 ±1.96 n

1328

3.

(1- 0.004840)3 = 0.986

4.

1- (1- 0.004840)4 = 0.019220; No, because being in the same family causes the events to be dependent, instead of being independent as required by the multiplication rule. It is reasonable to expect that four Hispanic females in the same family are more likely to experience similar environmental and hereditary characteristics.

The graph is misleading. The vertical scale begins with a frequency of 800 instead of 0, so the difference between the “yes” and “no” responses is greatly exaggerated.

60.0 - 68.6 80.0 - 68.6 = = - 3.07 and z = = 4.07, which have a probability of 0.9999 - 0.0011 a. zx=60 x=80 2.8 2.8 = 0.9988, or 99.88% (Tech: 99.89%) between them.

70.0 - 68.6 = 1.00, which has a probability of 1- 0.8413 = 0.1587 to the right. b. z x=70 = 2.8 4

5.

6.

7.

8.

9.

Data Set 1 “Body Data” includes subjects selected by the National Center for Health Statistics, and those subjects are between 18 and 80 years of age, whereas Data Set 3 “ANSUR II 2012” includes subjects who were U.S. Army personnel with ages between 17 and 58 years. The samples are from different populations.

The most dramatic difference is that Group A has a standard deviation of 13.4, whereas Group B has a standard deviation of 6.1. Group B appears to be a sample from a population with IQ scores that have much less variation than the variation of IQ scores of the other population.

H0: m1 = m2; H1: m1 ¹ m2; population1 = Group A, population2 = Group B Test statistic: t = 0.300; P-value = 0.7694 (Table: P-value > 0.20); Critical values: t = ±2.168 (Table: ± 2.262); Fail to reject H 0. There is not sufficient evidence to warrant rejection of the claim that both

samples are from populations with the same mean. t=

(x1 - x2 )- (m1 - m2 ) = (98.1- 96.7)- 0 s12

s2 + 2

2

13.442 + 6.13 10 10

10. The scatterplot shows a very distinct pattern that is linear except for the ages of 80 and greater. As expected, the life expectancy decreases as age increases, so there is a negative correlation.

Instructors Solution Manual James Lapp for Biostatistics for The Biological and Health Science 3rd Edition by Marc M Triola, Mario F Triola, Jason Roy