Opinion About Life After Death by Gender Opinion About Life After Death. Male Female

MATH 117 - Elements of Statistics Fall 2014 Final Exam Review for Chapters 1 – 11, 15 From the textbook: Statistics: The Art and Science of Learning f...
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MATH 117 - Elements of Statistics Fall 2014 Final Exam Review for Chapters 1 – 11, 15 From the textbook: Statistics: The Art and Science of Learning from Data (3rd ed.), by Alan Agresti and Christine Franklin

1.

UW Student Survey In a University of Wisconsin (UW) study about alcohol abuse among students, 100 of the 40,858 members of the student body in Madison were sampled and asked to complete a questionnaire. One question asked was, “On how many days in the past week did you consume at least one alcoholic drink?” a. Identify the population and the sample. b. For the 40,858 students at UW, one characteristic of interest was the percentage who would respond “zero” to this question. For the 100 students sampled, suppose 29% gave this response. Does this mean that 29% of the entire population of UW students would make this response? Explain. c. Is the numerical summary of 29% a sample statistic, or a population parameter?

2.

Median versus mean sales price of new homes In December 2010, the US Census Bureau reported that the median US sales price of new homes was $241,500. Would you expect the mean sales price to have been higher or lower? Explain.

3.

Female Heights According to a recent report from the US National Center for Health Statistics, females between 25 and 34 years of age have a bell-shaped distribution for height, with a mean of 65 inches and standard deviation of 3.5 inches. a. Give an interval within which about 95% of the heights fall. b. What is the height for a female who is 3 standard deviations below the mean? Would this be a rather unusual height? Why?

4.

High School Graduation Rates The distribution of high school graduation rates in the United States in 2004 had a minimum value of 78.3 (Texas), first quartile of 83.6, median of 87.2, third quartile of 88.8, and maximum value of 92.3 (Minnesota) (Statistical Abstract of the United States, 2006). a. Report the range and the interquartile range. b. Would a box plot show any potential outliers? Explain.

5.

Blood Pressure A World Health Organization study (the MONICA project) of health in various countries reported that in Canada, systolic blood pressure readings have a mean 121 and a standard deviation of 16. A reading above 140 is considered to be high blood pressure. a. What is the z-score for a blood pressure reading of 140? How is this z-score interpreted? b. The systolic blood pressure values have a bellshaped distribution. Report an interval within which about 95% of the systolic blood pressure values fall.

6.

Life After Death for Males and Females In a recent General Social Survey, respondents answered the question, “Do you believe in a life after death?” The table shows the responses cross-tabulated with gender. Opinion About Life After Death by Gender Gender Opinion About Life After Death Yes No Male 621 187 Female 834 145 a. b.

7.

Construct a table of conditional proportions. Summarize results. Is there much difference between responses of males and females?

Women in Government and Economic Life The OECD (Organization for Economic Cooperation and Development) consists of advanced, industrialized countries that accept the principles of representative democracy and a free market economy. For the nations outside of Europe that are in the OECD, the table shows UN data from 2007 on the percentage of seats in parliament held by women and female economic activity as a percentage of the male rate. a. Treating women in parliament as the response variable, prepare a scatterplot and find the correlation. Explain how the correlation relates to the trend shown in the scatterplot. b. Use software or a calculator to find the regression equation. Explain why the y-intercept is not meaningful. c. Find the predicted value and residual for the United States. Interpret the residual. d. With UN data for all 23 OECD nations, the correlation between these variables is 0.56. For women in parliament, the mean is 26.5% and the standard deviation is 9.8%. For female economic activity, the mean is 76.8 and the standard deviation is 7.7. Find the prediction equation, treating women in parliament as the response variable. Nation Iceland Australia Canada Japan United States New Zealand

Women in Parliament (%) 33.3 28.3 24.3 10.7 15.0 32.2

Female Economic Activity 87 79 83 65 81 81

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8.

9.

Predict Crime Using Poverty A recent analyses of data for the 50 US states on y = violent crime rate (measured as number of violent crimes per 100,000 people in the state) and x = poverty rate (percent of people in the state living at or below the poverty level) yielded the regression equation 𝑦̂ = 209.9 + 25.5𝑥. a. Interpret the slope. b. The state poverty rates ranged from 8.0 (for Hawaii) to 24.7 (for Mississippi). Over this range, find the range of predicted values for the violent crime rate. c. Would the correlation between these variables be positive or negative? Why? Football Discipline A large southern university had problems with 17 football players being disciplined for team rule violations, arrest charges, and possible NCAA violations. The online Atlanta Journal Constitution ran a poll with the question, “Has the football coach lost control over his players?” having possible responses, “Yes, he’s been too lenient,” and “No, he can’t control everything teenagers do.” a. Was there potential for bias in this study? If so, what types of bias? b. The poll results after two days were Yes No

6012 487

93% 7%

Does this large sample size guarantee that the results are unbiased? Explain. 10. Video games mindless? “Playing video games not so mindless.” This was the headline of a CNN news report about a study that concluded that young adults who regularly play video games demonstrated better visual skills than young adults who do not play regularly. Sixteen young men volunteered to take a series of tests that measured their visual skills; those who had played video games in the previous six months performed better on the test than those who hadn’t played. a. What are the explanatory and response variables? b. Was this an observational study or an experiment? Explain. c. Specify a potential lurking variable. Explain. 11. Peyton Manning Completions As of the end of the 2010 NFL season, Indianapolis Colts quarterback Peyton Manning, throughout his 13-year career, completed 65% of all of his pass attempts. Suppose the probability each pass attempted in the next season has probability 0.65 of being completed. a. Does this mean that if we watch Manning throw 100 times in the upcoming season, he would complete exactly 65 passes? Explain.

b. Explain what this probability means in terms of observing him over a longer period, say for 1000 passes over the course of the next two seasons assuming Manning is still at his typical playing level. Would it be surprising if his completion percentage over a large number of passes differed significantly from 65%? 12. Driver’s Exam Three 15-year-old friends with no particular background in driver’s education decide to take the written part of the Georgia Driver’s Exam. Each exam was graded as a pass (P) or a failure (F). a. How many outcomes are possible for the grades received by the three friends together? Using a tree diagram, list the sample space. b. If the outcomes in the sample space in part a are equally likely, find the probability that all three pass the exam. c. In practice, the outcomes in part a are not equally likely. Suppose that statewide 70% of 15-year-olds pass the exam. If these three friends were a random sample of their age group, find the probability that all three pass. d. In practice, explain why probabilities that apply to a random sample are not likely to be valid for a sample of three friends. 13. Grandparents Let X = the number of living grandparents that a randomly selected adult American has. According to recent General Social Surveys, its probability distribution is approximately P(0) = 0.71, P(1) = 0.15, P(2) = 0.09, P(3) = 0.03, P(4) = 0.02. a. Does this refer to a discrete or a continuous random variable? Why? b. Show that the probabilities satisfy the two conditions for a probability distribution. c. Find the mean of this probability distribution. 14. Z-score and Tail Probability a. Find the z-score for the number that is less than only 1% of the values of a normal distribution. Sketch a graph to show where this value is. th b. Find the z-scores corresponding to the (i) 90 and (ii) th 99 percentiles of a normal distribution. 15. Cloning Butterflies The wingspans of recently cloned monarch butterflies follow a normal distribution with mean 9 inches and standard deviation 0.75 inches. What proportion of the butterflies has a wingspan a. less than 8 inches? b. wider than 10 inches? c. between 8 and 10 inches? d. ten percent of the butterflies have a wingspan wider than how many inches?

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16. Exam Performance An exam consists of 50 multiplechoice questions. Based on how much you studied, for any given question you think you have a probability of p = 0.70 of getting the correct answer. Consider the sampling distribution of the sample proportion of the 50 questions on which you get the correct answer. a. Find the mean and standard deviation of the sampling distribution of this proportion. b. What do you expect for the shape of the sampling distribution? Why? c. If truly p = 0.70, would it be very surprising if you got correct answers on only 60% of the questions? Justify your answer by using the normal distribution to approximate the probability of a sample proportion of 0.60 or less.

20. Grandpas Using Email When the GSS asked in 2004, “About how many hours per week do you spend sending and answering email?” the eight males in the sample of age at least 75 responded: 0, 1, 2, 2, 7, 10, 14, 15 a. The TI-83+/84 screen shot shows results of a statistical analysis for finding a 90% confidence interval. Identify the results shown and explain how to interpret them. b. Find and interpret a 90% confidence interval for the population mean. Explain why the population distribution may be skewed right. If this is the case, is the interval you obtained in part b useless, or is it still valid? Explain. TInterval (2.34, 10.41) 𝑥 = 6.38 𝑆𝑥 = 6.02 𝑛 = 8.00

17. Aunt Erma’s Restaurant In Example 5 about Aunt Erma’s Restaurant, the daily sales follow a probability distribution that has a mean of µ = $900 and a standard deviation of σ 21. US Popularity In 2007, a poll conducted for the BBC of = $300. This past week the daily sales for the seven days 28,389 adults in 27 countries found that the United States had a mean of $980 and a standard deviation of $276. had fallen sharply in world esteem since 2001 a. Identify the mean and standard deviation of the (www.globescan.com). The United States was rated third population distribution. most negatively (after Israel and Iran), with 30% of those b. Identify the mean and standard deviation of the data polled saying they had a positive image of the United States. distribution. What does the standard deviation a. In Canada, for a random sample of 1008 adults, 56% describe? said the United States is mainly a negative influence in c. Identify the mean and the standard deviation of the the world. True or false: The 99% confidence interval of sampling distribution of the sample mean for samples (0.52, 0.60) means that we can be 99% confident that of seven daily sales. What does this standard between 52% and 60% of the population of all Canadian deviation describe? adults have a negative image of the United States. b. In Australia, for a random sample of 1004 people, 66% 18. Approval Rating for President Obama A July 2011 Gallup said the United States is mainly a negative influence in poll based on the responses of 1500 adults indicated that the world. True or false: The 95% confidence interval of 46% of Americans approve of the job Barack Obama is (0.63, 0.69) means that for a random sample of 100 doing as president. One way to summarize the findings of people, we can be 95% confident that between 63 and the poll is by saying, “It is estimated that 46% of American 69 people in the sample have a negative image of the approve of the job Barack Obama is doing as president. United States. This estimate has a margin of error of plus or minus 3%.” How could you explain the meaning of this to someone 22. Driving After Drinking In December 2004, a report based on who has not taken a statistics course? the National Survey on Drug Use and Health estimated that 20% of all Americans of ages 16 to 20 drove under the 19. Vegetarianism Time magazine (July 15, 2002) quoted a influence of drugs or alcohol in the previous year (AP, poll of 10,000 Americans in which only 4% said they were December 30, 2004). A public health unit in Wellington, New vegetarian. Zealand, plans a similar survey for young people of that age a. What has to be assumed about this sample to in New Zealand. They want a 95% confidence interval to construct a confidence interval for the population have a margin of error of 0.04. proportion of vegetarians? a. Find the necessary sample size if they expect results b. Construct a 99% confidence interval for the similar to those in the United States. population proportion. Explain why the interval is so b. Suppose that in determining the sample size, they use narrow, even though the confidence level is high. the safe approach that sets 𝑝̂ = 0.50 in the formula for c. In interpreting this confidence interval, can you n. Then, how many records need to be sampled? conclude that fewer than 10% of Americans are Compare this to the answer in part a. Explain why it is vegetarians? Explain your reasoning. better to make an educated guess about what to expect for 𝑝̂ , when possible.

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23. Mean property tax. A tax assessor wants to estimate the mean property tax bill for all homeowners in Madison, Wisconsin. A survey 10 years ago got a sample mean and standard deviation of $1400 and $1000. a. How many tax records should the tax assessor randomly sample for 95% confidence interval for the mean to have a margin of error equal to $100? What assumption does your solution make? b. In reality, suppose that they’d now get a standard deviation equal to $1500. Using the sample size you derived in part a, without doing any calculation, explain whether the margin of error for a 95% confidence interval would be less than $100, equal to $100 or more than $100. c. Refer to part b. Would the probability that the sample mean falls with $100 of the population mean be less than 0.95, equal to 0.95 or greater than 0.95? Explain. 24. H0 or Ha? For each of the following hypothesis explain whether it is a null hypothesis or alternative hypothesis: a. For females, the population mean on the political ideology scale is equal to 4.0. b. For males, the population proportion who support the death penalty is larger than 0.50. c. The diet has an effect; the population mean can change in weight being less than 0. d. For all subway sandwich stores worldwide, the difference between sales this month and in the corresponding month last year has been a mean of 0. 25. ESP A person who claims to possess extrasensory perception (ESP) says she can guess more often than not the outcome of a flip of a balanced coin. Out of 20 flips, she guesses correctly 12 times. Would you conclude that she truly has ESP? Answer by reporting all five steps of a significance test of the hypothesis that each of her guesses has probability 0.50 of being correct against the alternative that corresponds to her having ESP. 26. Jurors and gender A jury list contains the names of all individuals who may be called for jury duty. The proportion of the available jurors on the list who are women is 0.53. If 40 people are selected to serve as candidates for being picked on the jury, show all steps of significance test of the hypothesis that the selections are random with respect to gender. a. Set up notation and hypotheses, and specify assumptions. b. 5 of the 40 selected were women. Find the test statistic. c. Report the P-value, and interpret. d. Explain how to make a decision using a significance level or 0.01.

27. Type I and Type II errors Refer to the previous exercise. a. Explain what type I and Type II errors mean in the context of that exercise. b. If you made an error with the decisions in part d, is it a Type I or Type II error? 28. Tennis balls in control? When it is operating correctly a machine for manufacturing tennis balls produces balls with a mean weight of 57.6 grams. The last eight balls manufactured had weights 57.3, 57.4, 57.2, 57.5, 57.4, 57.1, 57.3, 57.0 a.

Using a calculator or software, find the test statistic and P-value for a test of whether the process is in control against the alternative that the true mean of the process now differs from 57.6. b. For significance level of 0.05, explain what you would conclude. Express your conclusion so it would be understood by someone who never studied statistics. c. If your decision in part b is in error, what type of error have you made? 29. Wage claim false? Management claims that the mean income for all senior-level assembly-line workers in a large company equals $500 per week. An employee decides to test the claim, believing that it is actually less than $500. For a random sample of nine employees, the incomes are: 430, 450, 450, 440, 460, 420, 430, 450, 440. Conduct a significance test of whether the population mean income equals $500 per week against the alternative that is less. Include all assumptions, the hypotheses, test statistics, P-value, and interpret the results in context. 30. Legal trial errors Consider the analogy discussed in Section 9.4 between making a decision about null hypothesis in a significance test and making a decision about the innocence or guilt of a defendant in a criminal trial. a. Explain the difference between Type I and Type II errors in the trial setting. b. In this context, explain intuitively why decreasing the chance of Type I error increases the chance of Type II error.

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31. Gender and belief in afterlife. This table shows results from the 2008 General Social Survey on gender and whether or not on believes in an afterlife. Belief in Afterlife Yes No 599 111 425 168

Gender Female Male

Total 710 593

a.

Denote the population proportion who believe in an afterlife by p1 for females and by p2 for males. Estimate p1, p2 and (p1 – p2 ). b. Find the standard error for the estimate of (p1 – p2 ). Interpret. c. Construct a 95% confidence interval for (p1 – p2 ). Can you conclude which of 𝑝1 and 𝑝2 is larger? Explain. d. Suppose that, unknown to us, p1 = 0.81 and p2 = 0.72. Does the confidence interval in part c contain the parameter it is designed to estimate? Explain. 32. Belief depend on gender? Refer to the previous exercise. a. Find the standard error of 𝑝̂1 − 𝑝̂2 for a test of 𝐻0 : 𝑝1 = 𝑝2 . b. For two-sided test, find the test statistic and P-value, and make a decision using significance level 0.05. Interpret. c. Suppose that actually p1 = 0.81 and p2 = 0.72. Was the decision in part b in error? d. State the assumption on which the methods in this exercise are based. 33. Heavier horseshoe crabs more likely to mate? A study of a sample of horseshoe crabs on a Florida island (J. Brockman, Ethology, vol. 102, 1996, pp. 1-21) investigated the factors that were associated with whether or not female crabs had a male crab mate. Basic statistics, including five-number summary on weight (kg) for the 111 female crabs who had a male crab nearby and for the 62 female crabs who did not have a male nearby, are given in the table. Assume that these horseshoe crabs have the properties of random sample of all such crabs. Summary Statistics for Weights of Horseshoe Crabs Mate No Mate

a.

b. c. d.

# 111 62

Mean 2.6 2.1

Std. Dev. 0.6 0.4

Min 1.5 1.2

Q1 2.2 1.8

Med 2.6 2.1

Q3 3.0 2.4

Max 5.2 3.2

Sketch box plots for the weight distributions of the two groups. Interpret by comparing the groups with respect to the shape, center, and variability. Estimate the difference between the mean weights of the female crabs who have mates and those who don't. Find the standard error for the estimate in part b. Construct a 90% confidence interval for the difference between the population mean weights, and interpret.

34. Sex roles A study of the effect of the gender of the tester on 16 the sex-role differentiation scores in Manhattan gave a random sample of preschool children the Occupational Preference Test. Children were asked to give three choices of what they wanted to be when they grew up. Each occupation was rated on a scale from 1 (traditionally feminine) to 5 (traditionally masculine), and a child’s score was the mean of the three selections. When the tester was male, the 50 girls had 𝑥 = 2.9 and s = 1.4, whereas when the tester was female, the 90 girls had 𝑥 = 3.2 and s = 1.2. Show all steps of a test of the hypothesis that the population mean is the same for the female and male testers, against the alternative that they differ. Report the P-value and interpret. 35. Internet book prices Anna’s project for her introductory statistics course was to compare the selling prices of textbooks at two Internet bookstores. She first took a random sample of 10 textbooks used that term in courses at her college, based on the list of texts compiled by the college bookstore. The prices of those textbooks at two Internet sites were Site A: $115, $79, $43, $140, $99, $30, $80, $99, $119, $69 Site B: $110, $79, $40, $129, $99, $30, $69, $99, $109, $66 a.

Are these independent samples or dependent samples? Justify your answer. b. Find the mean for each sample. Find the mean of the difference scores. Compare, and interpret. c. Using software or a calculator, construct a 90% confidence interval comparing the population mean prices of all the textbooks used that term at her college. Interpret. 36. Comparing book prices 2 For the data in the previous exercise, use software or a calculator to perform a significance test comparing the population mean prices. Show all steps of the test, and indicate whether you would conclude that the mean price is lower at one or the two Internet bookstores.

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37. Down syndrome diagnostic test The table shown, from Example 8 in Chapter 5, cross-tabulates whether a fetus has Down syndrome by whether or not the triple blood diagnostic test for Down syndrome is positive (that is, indicates that the fetus has Down syndrome). a.

Tabulate the conditional distributions for the blood test result given the true Down syndrome status.

b.

For the Down cases, what percentage was diagnosed as positive by the diagnostic test? For the unaffected cases, what percentage got a negative test result? Does the diagnostic test appear to be a good one?

c.

Construct the conditional distribution on Down syndrome status, for those who have a positive test result. (Hint: You condition on the first column total and find proportions in that column.) Of those cases, what percentage truly have Down syndrome? Is the result surprising? Explain why this probability is small.

Down Syndrome Status D (Down) 0 D (unaffected) Total

Blood Test Result Positive Negative Total 48 6 54 1307 3921 5228 1355 3927 5282

38. Down and chi-squared For the data in the previous exercise, 𝑥 2 = 114.4. Show all steps of the chi-squared test of independence. 39. Gender gap? Exercise 11.1 showed a 2 x 3 table relating gender and political party identification, shown again here. The chi-squared statistic for these data equals 8.294. Conduct all five steps of chi-squared test. Sex

Dem

F M

422 299

Political Party Indep Repub 381 365

273 232

40. Tanning experiment Suppose the tanning experiment described in Examples 1 and 2 used only four participants, two for each treatment. a. Show the six possible ways the four ranks could be allocated, two to each treatment, with no ties. b. For each possible sample, find the mean rank for each treatment and the difference between the mean ranks. c. Presuming H0 is true of identical treatment effects, construct the sampling distribution of the difference between the sample mean ranks for the two treatments. 41. Test for tanning experiment Refer to the previous exercise. For the actual experiment, suppose the participants using the tanning studio got ranks 1 and 2 and the participants using the tanning lotion got ranks 3 and 4. a. Find and interpret the P-value for the alternative hypothesis that the tanning studio tends to give better tans than the tanning lotion. b. Find and interpret the P-value for the alternative hypothesis that the treatments have different effects. c. Explain why it is a waste of time to conduct this experiment if you plan to use a 0.05 significance level to make a decision.

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Answers 1.

UW Student Survey

a. The population is the entire UW student body of 40,858. The sample is the 100 students who were asked to complete the questionnaire.

b. This value would not necessarily equal the value of the entire population of UW students. It is quite possible that the sample of 100 is not exactly c.

representative of the whole student body. This percentage is only an estimate of the percentage of all students who would respond this way. It is unlikely that any single sample of 100 would have a percentage that was exactly the percentage of the entire population. The numerical summery is the sample statistics because it only summarizes for a sample, not for a population.

2.

Median versus mean sales price of new homes We would expect the mean sales price to have been higher due to the distribution being skewed to the right. A few very expensive homes will greatly affect the mean, but not the median sales price. 3. Female Heights a. According to the Empirical Rule, 95% of the scores in a bell-shaped distribution fall within two standard deviation of the mean. 𝑥̅ − 2𝑆 = 65 − 2(3.5) = 58 𝑥̅ + 2𝑆 = 65 + 2(3.5) = 72 Thus, 95% of the heights likely fall between 58 and 72 inches. b. The height for a woman who is three standard deviations below the mean is 54.5. 𝑥̅ − 3𝑆 = 65 − 3(3.5) = 54.5 This is on the cusp of what would be considered an outlier according to the z-score criterion. Scores that are beyond three standard deviations from the mean are considered to be potential outliers. So, yes, this height is bordering on unusual. 4. a.

High School Graduation Rates The range is the difference between the lowest and the highest scores: 92.3 − 78.3 = 14. th th The interquartile range (IQR) is the difference between scores at the 25 and 75 percentiles: IQR=Q3−Q1= 88.8 − 83.6 = 5.2. b. 1.5(IQR)= 7.8 from Q1 or Q3; this criterion suggests that potential outliers would be scores less than 75.8 and greater than 96.6. There are no scores beyond these values, so it would not indicate any potential outliers. 5. Blood Pressure 𝑥−𝑥̅ 140−121 a. 𝑧 = = = 1.19 𝑠 16 A 𝑧 −score of 1.19 indicates that a person with a blood pressure of 140, the cutoff having a high blood pressure, falls 1.19 standard deviations above the mean. b. About 95% of all the values in a bell-shaped distribution fall within two standard deviation of the mean−in this case 32. Subtracting two times the standard deviation from the mean and adding two times the standard deviation to the mean tells us about 95% of systolic blood pressures fall between 89 and 153. 6.

Life After Death for Males and Females Using 2008 data:

a.

Opinion about life after death Gender Yes No Total Male 0.77 0.23 808 Female 0.85 0.15 979 Overall, both male and women are more likely to believe in life after death than not, but women are somewhat more likely to do so.

b.

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7.

Women in Government and Economic Life

a.

The correlation between women in parliament and female economic activity is 0.745. This correlation is supported by the positive linear trend evident in the scatterplot, but note this is largely driven by the point (for Japan) having female economic activity very low (65). b. The regression equation is 𝑦̂ = −48.91 + 0.1986𝑥. Since the 𝑦 –intercept correspond to an 𝑥 value of 0, the 𝑦 –intercept is not meaningful in this case. (Female economic activity=0 is outside the range of the observed data). c. The predicted value for U.S. is −48.91 + 0.1986(81) = 25.5 with 15.0 − 25.5 = −10.5 as the corresponding residual. The regression equation underestimates the percentage of women in parliament by 10.5% for the U.S. 9.8

d. 𝑏 = 0.56 ( ) = 0.7127 and 𝑎 = 26.5 − 0.7127(76.8) = −28.24. Thus the prediction equation is 𝑦̂ = −28.24 + 0.7127𝑥. 7.7

8. a.

Predict Crime Using Poverty The slope of 25.5 indicates that for each percentage increase in the poverty, the predicted violent crime rate increases by 25.5 crimes per 100, 000people statewide. b. The range predicted values runs from 413.9 to 839.8 crimes per 100, 000 people. 𝑦̂ = 209.9 + 25.5(8.0) = 413.9 𝑦̂ = 209.9 + 25.5(24.7) = 839.75 (rounds to 839.8) c.

9. a.

The correlation would be positive we know this because the slope is positive and the slope and the correlation have always the same sign.

Football Discipline Because this is a volunteer sample, there is the potential for sampling bias, both because of the sample is not selected randomly (those who have responded might have been those felt the most strongly) and because of the undercover age (anyone without Internet access would not have been able to participate). There also is potential for response biases because of the statements are leading. b. If the sample is biased due to undercoverage and lack of random sampling, it does not matter how big the sample. It is almost always better to have a small random sample than a large volunteer sample.

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10. Video games mindless? a. The explanatory is the history playing video games, and the response variable is visual skills. b. This was an observation study because the men were not randomly assigned to treatment (played video games versus hadn’t played); those who already were in these groups were observed. c. One possible lurking variable is reaction time. Excellent reaction times might make it easier and therefore more fun, to play video games, leading young people to be more likely to play. Excellent reaction times also might lead young men to perform better on tasks measuring visual skills. These young men might have performed better on tasks measuring visual skills regardless of whether they played video games. 11. Peyton Manning Completions a. No. What it means that in 100 passes we expect to see about 65 completions, but actually the number may vary somewhat. b. If Manning is still at his typical paying level, it would be quite surprising if his completion percentage over a large number of passes differed significantly from 0.6. The more passes he throws, the closer the observed percentage should be to 0.65. 12. Driver’s Exam a. There are 2 × 2 × 2 = 8 possible outcomes. Construct the tree yourself. An alternative way of presenting the outcomes: let P=”pass” and F=”failure”, the outcomes are: PPP, PPF, PFP, FPP, PFF, FPF, FFP, FFF. b.

1

If the eight outcomes are equally likely, all three pass the exam is = 0.125. This also be calculated by multiplying the probability that 8

15.

the first would pass(0.5), by the probability that the second would pass(0.5), by the probability that the third would pass(0.5), 0.5 × 0.5 × 0.5 = 0.125. If the three friends were a random sample of their age group, the probability that all would pass 0.7 × 0.7 × 0.7 = 0.343. The probabilities that apply to a random sample are not likely to be valid for a sample of three friends because the three friends are likely to be similar on many characteristics that might affect the performance on such a test ( e.g., IQ) . In addition it is possible that they studied together. Grandparents This refers to a discrete random variable because there can only be whole numbers of grandparents. One can’t have 1.78 grandparents. The probabilities satisfy the two conditions for a probability distribution because they each fall between 0 and 1, and the sum of the probabilities of all possible values is 1. The mean of this probability is 0(0.71) + 1(0.15) + 2(0.09) + 3(0.03) + 4(0.02) ≈ 0.5. Z-score and Tail Probability The z- score that is less than only 1% of the values would be greater than 99% of the values. If we look up 0.99 on Table A, we see the z-score is 2.33. (i) The z-score that is above 0.90 is 1.28. (ii) The z-score that is above 0.99 is 2.33. Cloning Butterflies

a.

𝑧=

c. d.

13. a. b. c. 14. a. b.

b. 𝑧 =

8−9

= −1.33; according to Table A, 0.092 of the butterflies have a wingspan less than 8 inches.

0.75 10−9 0.75

= 1.33; according to Table A, 0.092 of the butterflies have a wingspan wider than 10 inches.

c. From parts a and b, 1 − 2(0.092) = 0.816 of the butterflies have wingspans between 8 and 10 inches. th d. From Table A, the 90 percentile is 1.28. 1.28 × 0.75 + 9 = 9.96 inches. Thus 10% of the butterflies have wingspan wider than 9.96 inches. 16. Exam Performance a.

𝑝(1−𝑝)

Mean= 𝑝 = 0.70, standard error= √

𝑛

=√

0.7(1−0.7) 50

= 0.0648.

b. Since 𝑛 = 50, by the Central Limit Theorem, we would expect the shape of the sampling distribution to be approximately normal with mean=0.70 and standard deviation≈ 0.0648. c.

The z-score for 0.60 is answers correct.

0.60−0.70 0.0648

= −1.54 giving a cumulative probability of 0.06. It would not be surprising to only get 60% of the

1 0

17. Aunt Erma’s Restaurant a. The population distribution has a mean $900 and a standard deviation of $300. b. The population distribution has a mean $980 and a standard deviation of $276. The standard deviation of the data distribution describes the spread of the daily sales values for this past week. c.

The mean of the sampling distribution of the sample mean = 𝜇 =$900; standard error=

𝜎 √𝑛

=

300 √7

≈ 113.4 dollars. The standard

error describes the spread of the sample means based on sample of seven days sales. 18. Approval Rating for President Obama We could tell to someone who hadn’t taken a statistics course that we do not know the exact percentage of the population who approve of the job Barack Obama is doing as president, but we are quite sure that it is within 3% of 46%, that is, between 43% and 49%. 19. Vegetarianism a. We must assume that the data were obtained randomly. b.

c.

𝑠𝑒 = √

𝑝̂(1−𝑝̂) 𝑛

0.04(1−0.04)

=√

10,000

≈ 0.002.

The confidence interval is 𝑝̂ ± 𝑧(𝑠𝑒) Lower limit: 0.04 − 2.58(0.002) = 0.035. Upper limit: 0.04 + 2.58(0.002) = 0.045. The interval is so narrow, even though the confidence level is high, mainly because of the very large sample size. The very large sample size contributes to a small standard error by providing a very large denominator for the standard error calculation. We can conclude that fewer than 10% of Americans are vegetarians because 10% falls above the highest believable value in the confidence interval.

20. Grandpas Using Email a. The first result is a 90% confidence interval for the mean hours spent per week sending and answering e-mail for males at least age 75. The sample mean, 𝑥̅ , is listed as 6.38 hours. Thus, the estimated mean spent per week sending and answering e-mail for males at least age 75 is 6.38 hours. The sample standard deviation is 6.02. This quantity estimates the population standard deviation which tells us how far we can expect a typical observation to vary from the mean. These estimates are based on a sample of size 8. b. The confidence interval is 2.34 to 10.41. We can be 90% confident that the populations mean numbers of hours spent per week sending and answering e-mail for males at least age 75 is 2.34 to 10.41 hours. c. Since there are likely to be a lot of men over the age of 75 who do not use email but also some who use e-mail regularly, this distribution is likely skewed right. Since the t-distribution is robust to violations of normal assumption, the interval is still valid. 21. US Popularity a. True b. False 22. Driving After Drinking a. b.

𝑛= 𝑛=

[𝑝̂(1−𝑝̂)]𝑧 2 𝑚2 [𝑝̂(1−𝑝̂)]𝑧 2 𝑚2

= =

[0.2(1−0.2)](1.96)2 (0.04)2 [0.5(1−0.5)](1.96)2 (0.04)2

= 385. = 601.

This is larger than the answer in a. If we can make an educated guess about what to expect for the proportion, we can use a smaller sample size, saving possibly unnecessary time and money. 23. Mean property tax a.

𝑛=

𝜎2𝑧 2 𝑚2

=

(1000)2 (1.96)2 (100)2

= 385.

The solution makes the assumption that the standard deviation will be similar now. b. The margin of error would be more than $100 because the standard error would be larger than predicted. c. With a larger margin of error, the 95% confidence interval is wider; thus, the probability the sample mean is within $100(which is less than the margin of error from b) of the population mean is less than 0.95.

1 1

24. H0 or Ha? a. null hypothesis b. alt alternative hypothesis c. alternative hypothesis d. null hypothesis 25. ESP 1. Assumptions: The data are categorical (correct vs. incorrect guesses) and are obtained randomly. The expected successes and failures are less than 15 under null hypothesis 𝐻0 : 𝑛𝑝 = (20)(0.5) = 10 < 15 and 𝑛(1 − 𝑝) = (20)(0.5) = 10 < 15, so this test is approximate. 2. Hypotheses: 𝐻0 : 𝑝 = 0.5; 𝐻𝑎 : 𝑝 > 0.5 3.

Test statistic: 𝑧 =

0.6−0.5 √0.5(1−0.5)/20

= 0.89

4. 5.

P-value:0.19 Conclusion: If the null hypothesis were true, the probability would be 0.19 of getting a test statistic at least as extreme as the value observed. There is no strong evidence that the population proportion correct guesses is higher than 0.50. 26. Jurors and gender a.

The proportion of jurors who are women Hypotheses: 𝐻0 : 𝑝 = 0.53; 𝐻𝑎 : 𝑝 ≠ 0.53

b.

Test statistic: 𝑧 =

0.125−0.53 √0.53(1−0.53)/40

= −5.1

c.

P-value is 0.000. If the null hypothesis were true, the probability would be almost 0 of getting a test statistic at least as extreme as the value observed. d. This P-value is more extreme than the significance level of 0.01. We can reject the null hypothesis; we have strong evidence that women are not being selected in numbers proportionate to their representation in their jury pool. 27. Type I and Type II errors a. In the previous exercises, a Type I error would have occurred if we had rejected the null hypothesis, concluding that the women were passed over jury duty, when they really were not. A Type II error would have occurred if we had failed to reject the null hypothesis, but women were picked disproportionate to their representation in their jury pool. b. If we made an error, it was a Type I error. 28. Tennis balls in control? a. Software indicates a test statistic of −5.5 and a P-value of 0.001. b. For a significance level of 0.05, we conclude that the process is not in control. The machine is producing tennis balls that weight less than they are supposed to. c. If we rejected the null hypothesis when in fact it is true, we have made a Type I error and conclude that the process is not in control when it actually is. d. 29. Wage claim false? 1. 2. 3.

Assumptions: The data are quantitative. The data seem to have been produced using randomization. We also assume an approximately normal population distribution. Hypotheses: 𝐻0 : 𝜇 = 500; 𝐻𝑎 : 𝜇 < 500 The sample mean is 441.11, and the standard deviation is 12.69. Test statistic: 𝑡 =

4. 5.

0441.11−500 12.69 √9

= −13.9

P-value:0.000 Conclusion: If the null hypothesis were true, the probability would be almost 0 of getting a test statistic at least as extreme as the value observed. There is extremely strong evidence that the population mean is less than 500; we can conclude that the mean income is less than $500 per week.

1 2

30. Legal trial errors a. A Type I error in a trail setting would occur if we convicted a defendant who was not guilty. A Type II error would occur if we failed to convict a guilty defendant. b. To decrease the chance of a Type I error, we would decrease the significance level. In doing this, it is more difficult to reject the null hypothesis (i.e., find someone guilty) . Thus, there will be more guilty people who are not found guilty, a Type II error. 31. Gender and belief in afterlife. a.

The sample proportions who report that they believe in afterlife for females is

599 710

= 0.8437, for males

425 593

= 0.7167, and for the

difference between females and males is 0.8437 − 0.7167 = 0.127. b. The standard error for the estimate of (𝑝1 − 𝑝2 ) is 𝑠𝑒 = √

𝑝̂1 (1−𝑝̂1 ) 𝑛1

+

𝑝̂2 (1−𝑝̂2 ) 𝑛2

=√

0.8437(1−0.8437) 710

+

0.7167(1−0.7167) 593

= 0.0230

This is the standard deviation for the difference between males and females for the sample of these sizes. The confidence interval is 𝑝̂1 − 𝑝̂2 ± 𝑧(𝑠𝑒); lower endpoint: 0.127−1.96(0.0230) = 0.0819; upper end point: 0.127+1.96(0.0230) = 0.172. The confidence interval is (0.0819, 0.172). Because 0 is less than all plausible values given in the confidence interval, we can conclude that the proportion of females who report that they believe in afterlife is higher than the proportion of males. d. The difference between these population proportions, 0.09, is in the confidence interval. The confidence interval in part c) contains the parameter it is designated to estimate. 32. Belief depend on gender c.

a.

𝑠𝑒 = √𝑝̂ (1 − 𝑝̂ ) (

b. 𝑧 =

(𝑝̂1 −𝑝̂2 )−0 𝑠𝑒

=

1 𝑛1

+

0.127 0.0228

1 𝑛2

) = √0.7859(1 − 0.7859) (

1 710

+

1 593

) = 0.0228.

= 5.57; P-value≈ 0.

If the null hypothesis were true, the probability would be approximately 0 of getting a test statistic at least as extreme as the value observed. Therefore, we reject the null hypothesis, and conclude that the population proportions believing in afterlife are different for females and males. c. If the population difference were 0.81−0.72=0.09, our decision would have been correct. d. The assumptions on which the methods in this exercises are based are independent random samples for the two groups and that we had at least 5 successes and 5 failures. 33. Heavier horseshoe crabs more likely to mate? a. Construct yourself two boxplots of weight for female carbs who have mates and who do not have mates on the same scale. Then conclude that the female crabs have a higher median and a bigger spread if they had a mate than if they did not have a mate. The distribution for female crabs with a mate is right-skewed, whereas the distribution for female crabs without a mate is symmetrical. b. The estimated difference between the mean weights of female crabs who have mates and do not have mates is 2.6−2.1=0.5. c.

𝑠𝑒 = √

𝑠12 𝑛1

+

𝑠22 𝑛2

0.36

=√

111

+

0.16 62

= 0.076

d. (𝑥̅1 − 𝑥̅2 ) ± 𝑡0.05 (𝑠𝑒) Because 𝑛1 and 𝑛2 are large, we will approximate t with the normal distribution using z=1.645. 0.5 − 1.645(0.076) = 0.375 0.5 + 1.645(0.076) = 0.625 (0.375, 0.625) We can be 90% confidence the difference between the population mean weights of female crabs with and without a mate is between 0.375 and 0.625. Because 0 does not fall in this interval, we can conclude that female crabs with a met weight more than do female crabs without a mate.

1 3

34. Sex roles 1. Assumptions: The data are quantitative (Child’s score); the samples are independent and we will assume that they were obtained randomly; we will assume that the populations scores distribution are approximately normal for each group. 2. Hypotheses: 𝐻0 : 𝜇1 = 𝜇2; 𝐻𝑎 : 𝜇1 ≠ 𝜇2 where group 1 represents with the group with male tester group 2 represents the group with female tester. 3.

𝑠𝑒 = √

1.4 2 50

+

1.22 90

= 0.2349, 𝑡 =

2.9−3.2 0.2349

= −1.28

4. 5.

P-value:0.205 If the null hypothesis were true, the probability would be 0.205 of getting a test statistic at least as extreme as the value observed. Since the P-value is quite large, there is not much evidence of a difference in the population mean of the children’s scores when the tester is male than female. 35. Internet book prices a. The samples are dependent because they are prices of the same ten books at two different internet sites b. Let group 1 be the prices from Site A and group 2 from Site B. Then, 𝑥̅1 = $87.30, 𝑥̅2 = $83.00, 𝑥̅𝑑 = $4.30. Sample mean price for the books from Site A is higher than the sample mean price for the book from Site B. Thus, the sample mean of the difference between the prices from the two sites is positive. c. A 90% confidence interval for 𝜇1 − 𝜇2 is given by (1.53, 7.03). Since 0 is less than the values in the confidence interval, we can conclude that the prices for textbooks used at her college are more expensive at Site A than at Site B. 36. Comparing book prices 2 1. Assumptions: the differences in prices are a random sample from a population that is approximately normal. 2. 𝐻0 : 𝜇𝑑 = 0; 𝐻𝑎 : 𝜇𝑑 ≠ 0 3.

𝑡=

𝑑̅ −0 𝑠𝑑/√𝑛

=

4.3 4.7152/√10

= 2.88

4. 5.

P-value:0.02 If the null hypothesis is true, the probability obtaining a difference in sample means extreme as the value observed is 0.02. We would reject the null hypothesis and conclude there is a significant difference in the prices of textbooks used at her college between the two sites for 𝛼 =0.05, or 0.10, but not for 𝛼 =0.01. 37. Down syndrome diagnostic test a. BLOOD TEST RESULT STATUS Positive Negative Total D(Down) 0.89 0.11 54 𝑐 𝐷 (Unafected) 0.25 0.75 5228 b. For the Down cases, 89% were correctly diagnosed. For the unaffected cases, 75% get a negative result. The test seems fairly good, but there are a good number of false positives and false negatives. c.

BLOOD TEST RESULT STATUS Positive Negative D(Down) 0.035 0.002 𝐷𝑐 (Unafected) 0.965 0.998 Total 1355 3927 Of the positive cases, only 0.035 truly have Down syndrome. This result is not surprising because there are so few cases overall. The fairly large numbers of false positives will overwhelm the much smaller number of actual cases.

38. Down and chi-squared 1. The assumptions are there are two categorical variables (Down syndrome status and blood test result), that randomization was used to obtain the data and that the expected count was at least five in all cells. 2. 𝐻0 : Down syndrome status and blood test result are independent 𝐻𝑎 : Down syndrome status and blood test result are dependent 3. 𝜒 2 = 114.4, df=1 4. P-value:0.000

1 4

5.

If the null hypothesis were true, the probability would be almost 0 of getting a test statistic at least as extreme as the value observed. There is very strong evidence of an association between test result and actual status.

39. Gender gap? 1.

The assumptions are there are two categorical variables (party identification and gender), that randomization was used to obtain the data and that the expected count was at least five in all cells. 2. 𝐻0 : Party identification and gender are independent 𝐻𝑎 : Party identification and gender are dependent 3. 𝜒 2 = 8.294, df=2 4. P-value:0.016 5. If the null hypothesis were true, the probability would be 0.016 of getting a test statistic at least as extreme as the value observed. There is very strong evidence that party identification depends on gender. 40. Tanning experiment a. Treatments Ranks Lotion (1, 2) (1, 3) (1, 4) (2, 3) (2, 4) (3, 4) Studio (3, 4) (2, 4) (2, 3) (1, 4) (1, 3) (1, 2) b. Lotion mean rank Studio mean rank Difference of mean ranks

1.5 3.5 -2.0

2.0 3.0 -1.0

2.5 2.5 0.0

2.5 2.5 0.0

3.0 2.0 1.0

3.5 1.5 2.0

c. Difference between mean ranks -2.0 -1.0 0.0 1.0 2.0

probability 1/6 1/6 2/6 1/6 1/6

41. Test for tanning experiment a. The P-value is 1/6 =0.17; if the treatments had identical effects, the probability would be 0.17 of getting sample like we observed, or even more extreme, in this direction. It is plausible that the null hypothesis is correct, and that the studio does not lead to better results than the lotion. b. The P-value is 2/6 =0.33; if the treatments had identical effects, the probability would be 0.33 of getting sample like we observed, or even more extreme, in this direction. It is plausible that the null hypothesis is correct, and that the treatments do not lead to different results. c. It is a waste of time to conduct this experiment if we plan to use a 0.05 significance level because the smallest possible P-value is 0.17.