790 APPENDIX D s>s = 0.062>0.0480164 = 1.2912255, then resample 1000 times. A typical result is that among the 1000 resulting values of s, 69 of them are 0.0480164 or lower. That is, 69 of the 1000 resamples are at least as extreme as the value of s = 0.0480164 g that was obtained. That result corresponds to an estimated P-value of 0.069, 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 weights are from a population with a standard deviation that is less than 0.062 g. These results from randomization agree quite well with those from Example 1 in Section 8-4, and this suggests that randomization is very effective in this case. Chapter 8: Quick Quiz 1. H0: m = 2.000 pounds. H1: m 6 2.000 pounds. 2. t = -0.658 3. a. t distribution b. No. The relevant requirement is that the population is normally distributed or n 7 30. With n = 62, that requirement is satisfied and it is not necessary to test for normality. 4. a. Fail to reject H0. b. There is not sufficient evidence to support the claim that the sample is from a population with a mean less than 2.000 pounds. 5. H0: p = 0.45. H1: p 7 0.45. 6. z = 2.76 7. a. Reject H0. b. There is sufficient evidence to support the claim that for the population of all TV households, more than 45% have at least one stand-alone streaming device. 8. a. t distribution b. Normal distribution c. Chi-square distribution 9. a. True b. False c. False d. False e. 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 x2 (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. Chapter 8: Review Exercises 1. H0: p = 0.50. H1: p 7 0.50. Test statistic: z = 8.85. P@value = 0.0000 (Table: 0.0001). Critical value: z = 2.33. Reject H0. There is sufficient evidence to support the claim that the majority of employees are searching for new jobs. 2. H0: m = 27.2 km>h. H1: m ≠ 27.2 km>h. Test statistic: t = 92.116. P@value = 0.0000 (Table: 0.0002). Critical values: t = {2.093. Reject H0. There is sufficient evidence to reject the claim that current winning speeds are not significantly different from the 27.2 km>h overall speed of winners from the first few races. Current winning speeds appear to be significantly higher. 3. H0: m = 5.4 million cells per microliter. H1: m 6 5.4 million cells per microliter. Test statistic: t = -5.873. P-value = 0.0000 (Table: 60.005). Critical value: t = -2.426. Reject H0. 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 3. a. pn = 426>860 = 0.495 and p = 0.512. b. The sample proportions that are at least as extreme as pn = 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. 5. 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%. 7. Answers vary, but the following is typical. Create a column of 2005 ones and zeros, with 0.5 of them being 1 (as assumed in the null hypothesis). The sample proportion is 0.11, so in this lefttailed case, the values “at least as extreme” as 0.11 are those that are 0.11 or lower. Resample 1000 times to find that none of the results have a proportion of 0.11 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 support the claim that fewer than half of all Americans know what the symbol designates. 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.) 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. 13. For the data in Example 4, s = 0.0480164 g. For testing the claim that s 6 0.062 g, we assume in the null hypothesis that s = 0.062 g. To modify the data set so that s becomes 0.062 g, multiply each of the data values by
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