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Chapter 9 Review and QuestionsA significance test shows how strong the evidence is for some claim about a parameter. In significance testing, the null hypothesis often says “no effect” or “no difference.” The alternative hypothesis typically reflects what we hope to find evidence for.The P-value is the probability that a random sample or randomized experiment would produce a result at least as extreme as the observed result if the null hypothesis really were true. That is, the P-value tells us how surprising the observed outcome is. Very surprising outcomes (small P-values) are good evidence that the null hypothesis is not true. If the P-value is less than a specified significance level α, we reject H0 in favor of Ha. Otherwise, we fail to reject H0 and do not have convincing evidence that Ha is true.If we decide to reject H0 when H0 is true, we have committed a Type I error. The probability of making a Type I error is equal to the significance level α. If we fail to reject H0 when H0 is false, we have made a Type II error. The power of a significance test is the probability that the test will detect a specific alternative value of the parameter. Power = 1 ? P(Type II error). We can increase the power of a test by increasing the sample size or by increasing our willingness to make a Type I error.Section?9.1 presented the underlying logic and structure of significance tests. In Section?9.2, we looked at how to perform significance tests about a population proportion p. We also noted that confidence intervals give more information than a mere reject or fail to reject conclusion in a significance test. Section?9.3 discussed the one-sample t test for a population mean μ. We saw that paired data require a one-sample t test on the differences in each pair. Several important cautions for the wise use of significance tests were discussed at the end of Section?9.3.All the methods of this chapter work well when the three important conditions—Random, Normal, and Independent—are satisfied. Be sure to verify these conditions before you proceed to calculations.Chapter Review ExercisesThese exercises are designed to help you review the important ideas and methods of the chapter. Relevant learning objectives are provided in bulleted form before each exercise.State correct hypotheses for a significance test about a population proportion or mean.R9.1. Stating hypotheses State the appropriate null and alternative hypotheses in each of the following cases.(a) The average height of 18-year-old American women is 64.2 inches. You wonder whether the mean height of this year’s female graduates from a large local high school (over 3000 students) differs from the national average. You measure an SRS of 48 female graduates and find that X = 63.1 inches.(b) Mr. Starnes believes that less than 75% of the students at his school completed their math homework last night. The math teachers inspect the homework assignments from a random sample of students at the school to help Mr. Starnes test his claim.Correct Answer(a) H0: μ = 64.2; Ha: μ ≠ 64.2 (b) H0: p = 0.75; Ha: p < 0.75Check conditions for carrying out a test about a population proportion or mean.Interpret P-values in context.R9.2. Eye black Athletes performing in bright sunlight often smear black grease under their eyes to reduce glare. Does eye black work? In one experiment, 16 randomly selected student subjects took a test of sensitivity to contrast after 3 hours facing into bright sun, both with and without eye black. Here are the differences in sensitivity, with eye black minus without eye black: HYPERLINK "JavaScript:top.ShowFootnote('9_34')" 34We want to know whether eye black increases sensitivity on the average.(a) State hypotheses. Be sure to define the parameter.(b) Check conditions for carrying out a significance test.(c) The P-value of the test is 0.047. Interpret this value in context.Correct Answer(a) H0: μd = 0; Ha: μd > 0. μd = true mean difference (with eye black ? without eye black) in sensitivity among athletes. (b) Random: The students were chosen randomly. Normal: The histogram shows that the distribution is somewhat right-skewed with no outliers. We should be safe using t procedures. Independent: The sample size is 16, which is much less than 10% of all possible students. The conditions are met.(c) If eye black does not increase sensitivity, on average, the chance of finding a sample where the difference is as big as ours is 4.7%.Interpret a Type I error and a Type II error in context, and give the consequences of each.Understand the relationship between the significance level of a test, P(Type II error), and power.R9.3. Strong chairs? A company that manufactures classroom chairs for high school students claims that the mean breaking strength of the chairs that they make is 300 pounds. One of the chairs collapsed beneath a 220-pound student last week. You wonder whether the manufacturer is exaggerating the breaking strength of the chairs.(a) State null and alternative hypotheses in words and symbols.(b) Describe a Type I error and a Type II error in this situation, and give the consequences of each.(c) Would you recommend a significance level of 0.01, 0.05, or 0.10 for this test? Justify your choice.(d) The power of this test to detect μ = 294 is 0.71. Explain what this means to someone who knows little statistics.(e) Explain two ways that you could increase the power of the test from (d).Correct Answer(a) H0: μ = 300 (the company’s claim is true) versus Ha: μ < 300 (the mean breaking strength is less than the company’s claim). (b) Type I error: conclude that the company’s claim is incorrect (μ < 300) when, in fact, it is legitimate (μ = 300). We might falsely accuse the company of exaggerating its claim. Type II error: conclude that the company’s claim is legitimate when, in fact, it is exaggerated. This could result in more injuries from broken chairs. (c) Since a Type II error is more serious, we would like to reduce the likelihood of that error. This means increasing the likelihood of a Type I error. This suggests that we should use a significance level of 0.10. (d) If, in fact, the mean breaking strength is 294 lb, there is a 71% chance that we will reject the null hypothesis in our test. That is, there is a 71% chance that we will decide that the mean breaking strength is less than 300 lb. (e) Increase the sample size or the significance level.If conditions are met, conduct a significance test about a population proportion.R9.4. Flu vaccine A drug company has developed a new vaccine for preventing the flu. The company claims that fewer than 5% of adults who use its vaccine will get the flu. To test the claim, researchers give the vaccine to a random sample of 1000 adults. Of these, 43 get the flu.(a) Do these data provide convincing evidence to support the company’s claim? Perform an appropriate test to support your answer.(b) Which kind of mistake—a Type I error or a Type II error—could you have made in (a)? Explain.(c) From the company’s point of view, would a Type I error or Type II error be more serious? Why?Correct Answer(a) State: H0: p = 0.05 versus Ha: p < 0.05, where p is the actual proportion of adults who use the vaccine and who will get the flu if they use the vaccine. Plan: One-sample z test for p. Random: The sample was randomly selected. Normal: The expected number of successes np0 = 50 and failures n(1 ? p0) = 950 are both at least 10. Independent: There were 1000 adults, which is clearly less than 10% of the population. Do: The sample proportion is . The corresponding test statistic is z = ?1.02. Since this is a one-sided test, the P-value is P(z < ?1.02) = 0.1539. Conclude: Since the P-value is greater than 0.05, we fail to reject H0. We do not have enough evidence to conclude that fewer than 5% of adults who receive this vaccine will get the flu. (b) Type II error, since we failed to reject H0. (c) Type I error. This would mean concluding that the vaccine is more effective than it really is.R9.5. Roulette An American roulette wheel has 18 red slots among its 38 slots. In a random sample of 50 spins, the ball lands in a red slot 31 times.(a) Do the data give convincing evidence that the wheel is unfair? Carry out an appropriate test at the α = 0.05 significance level to help answer this question.(b) The casino manager uses your data to produce a 99% confidence interval for p and gets (0.44, 0.80). He says that this interval provides convincing evidence that the wheel is fair. How do you respond?Correct Answer(a) State: H0: p = 0.4737 versus Ha: p ≠ 0.4737, where p is the actual proportion of times the ball lands in a red slot. Plan: One-sample z test for p. Random: The sample was randomly selected. Normal: The expected number of successes np0 = 23.685 and failures n(1 ? p0) = 26.315 are both at least 10. Independent: There were 50 spins, which is clearly less than 10% of all possible spins. Do: The sample proportion is . The corresponding test statistic is z = 2.07. Since this is a two-sided test, the P-value is 0.0384. Conclude: Since P-value < 0.05, we reject H0. We have enough evidence to conclude that the wheel is unfair. (b) By using a 99% confidence level, the manager has obtained a very wide interval of plausible values that includes 0.4737. You might ask why he didn’t choose a 95% confidence level!If conditions are met, conduct a one-sample t test about a population mean μ.R9.6. Improving health A large company’s medical director launches a health promotion campaign to encourage employees to exercise more and eat better foods. One measure of the effectiveness of such a program is a drop in blood pressure. The director chooses a random sample of 50 employees and compares their blood pressures from physical exams given before the campaign and again a year later. The mean change (after ? before) in systolic blood pressure for these 50 employees is ?6 and the standard deviation is 19.8.(a) Do these data provide convincing evidence of an average decrease in blood pressure among all of the company’s employees during this year? Carry out a test at the α = 0.05 significance level.(b) Can we conclude that the health campaign caused a decrease in blood pressure? Why or why not?Correct Answer(a) State: H0: μd = 0 versus Ha: μd < 0, where μd is the actual mean difference (after ? before) in blood pressure among the company’s employees between after the campaign and before it. Plan: One-sample t test for μ. Random: The employees were chosen at random. Normal: The sample size was 50, which is at least 30. Independent: There were 50 people in the sample. Since the company is described as being a large company, there are very likely more than 500 employees, so this is less than 10% of all employees. Do: The sample mean and standard deviation are and sx = 19.8. The corresponding test statistic is t = ?2.143. The P-value from technology is 0.0186. Conclude: Since P-value < 0.05, we reject H0. We have enough evidence to conclude that the actual mean change in blood pressure among the company’s employees is negative. People’s blood pressure was reduced over the year, on average. (b) No. This was an observational study with no comparison group.R9.7. Fonts and reading ease Does the use of fancy type fonts slow down the reading of text on a computer screen? Adults can read four paragraphs of text in the common Times New Roman font in an average time of 22 seconds. Researchers asked a random sample of 24 adults to read this text in the ornate font named Gigi. Here are their times, in seconds:Do these data provide good evidence that the mean reading time for Gigi is greater than 22 seconds? Carry out an appropriate test to help you answer this question.Correct AnswerState: H0: μ = 22 versus Ha: μ > 22, where μ is the actual mean amount of time it takes adults to read the paragraph in Gigi font. Plan: One-sample t test for μ. Random: The adults were chosen at random. Normal: The sample size was only 24, so we look at the histogram below. It shows that the distribution is roughly symmetric with no outliers. Independent: There were 24 people in the sample. This is clearly less than 10% of all adults.Do: The sample mean and standard deviation are and sx = 4.728. The corresponding test statistic is t = 4.659. The P-value from technology using df = 23 is approximately 0. Conclude: Since P-value < 0.05, we reject H0. We have enough evidence to conclude that the mean time for adults to read the paragraph in Gigi font is more than 22 seconds.Use a confidence interval to draw a conclusion for a two-sided test about a population mean.R9.8. Radon detectors Radon is a colorless, odorless gas that is naturally released by rocks and soils and may concentrate in tightly closed houses. Because radon is slightly radioactive, there is some concern that it may be a health hazard. Radon detectors are sold to homeowners worried about this risk, but the detectors may be inaccurate. University researchers placed a random sample of 11 detectors in a chamber where they were exposed to 105 picocuries per liter of radon over 3 days. A graph of the radon readings from the 11 detectors shows no strong skewness or outliers. The Minitab output below shows the results of a one-sample t interval. Is there significant evidence at the 10% level that the mean reading μ differs from the true value 105? Give appropriate evidence to support your answer.Correct AnswerState: We want to perform a test of H0: μ = 105 versus Ha: μ ≠ 105, where μ is the actual mean reading from radon detectors. We will perform the test at the α = 0.10 significance level. Plan: If conditions are met, we should do a one-sample t test for the population mean μ. Random: The radon detectors were chosen at random. Normal: The sample size was only 11, but a graph of the data shows no strong skewness or outliers. Independent: There were 11 radon detectors in the sample. This is clearly less than 10% of all possible radon detectors. All conditions have been met. Do: The 90% confidence interval is (99.61, 110.03). Conclude: Since the 90% confidence interval contains 105, we do not reject the null hypothesis. We do not have enough evidence to conclude that the radon detectors are inaccurate.Recognize paired data and use one-sample t procedures to perform significance tests for such data.R9.9. Better barley Does drying barley seeds in a kiln increase the yield of barley? A famous experiment by William S. Gosset (who discovered the t distributions) investigated this question. Eleven pairs of adjacent plots were marked out in a large field. For each pair, regular barley seeds were planted in one plot and kiln-dried seeds were planted in the other. The following table displays the data on yield (lb/acre).35(a) How can the Random condition be satisfied in this study?(b) Perform an appropriate test to help answer the research question. Assume that the Random condition is met. What conclusion would you draw?Correct Answer(a) By randomly allocating which plot got the regular barley seeds and which one got the kiln-dried seeds. (b) State: We want to perform a test of H0: μd = 0 versus Ha: μd < 0, where μd is the actual mean difference (regular ? kiln) in yield between regular barley seeds and kiln-dried barley seeds. We will perform the test at the α = 0.05 significance level. Plan: If conditions are met, we should do a paired t test for the population mean μ. Random: We are told that this was a randomized experiment. Normal: The sample size was 11, so we check a graph. The histogram below shows no strong skewness or outliers in the data. Independent: There were 11 pairs of plots used in the sample. This is clearly less than 10% of all possible plots. All conditions have been met.Do: The sample mean and standard deviation are and sx = 66.2. The corresponding test statistic is t = ?1.688. The P-value from technology using df = 10 is 0.0612. Conclude: Since our P-value is greater than 0.05, we fail to reject the null hypothesis. We do not have enough evidence to conclude that kiln-dried barley seeds produce a larger yield. ................
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