APPENDIX D 789 7. H0: s = 2.08°F. H1: s 6 2.08°F. Test statistic: x2 = 9.329. P-value = 0.0000 (Table: 60.005). Critical value: x2 = 74.252 (Table: 70.065 approximately). Reject H0. 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. 9. H0: s = 0.0400 g. H1: s 6 0.0400 g. Test statistic: x2 = 13.486. P@value = 0.1872. Critical value: x2 = 7.633. Fail to reject H0. There is not sufficient evidence to support a claim that M&Ms are manufactured so that they have weights with a standard deviation less than 0.0400 g. It does not appear that the goal is being satisfied. 11. H0: s = 9.9 min. H1: s 7 9.9 min. Test statistic: x2 = 19.556. P@value = 0.0518 (Table: 0.05 6 P@value 6 0.10). Critical value: x2 = 19.675. Fail to reject H0. There is not sufficient evidence to support a claim that Friday afternoon ride times have greater variation than the Friday morning ride times. 13. H0: s = 32.2 ft. H1: s 7 32.2 ft. Test statistic: x2 = 29.176. P-value: 0.0021. Critical value: x2 = 19.675. Reject H0. There is sufficient evidence to support the claim that the new production method has errors with a standard deviation greater than 32.2 ft. The variation appears to be greater than in the past, so the new method appears to be worse because there will be more altimeters that have larger errors. The company should take immediate action to reduce the variation. 15. H0: s = 0.2000 g. H1: s ≠ 0.2000 g. Test statistic: x2 = 5.275. P@value = 0.7664 (Table: P@value 7 0.20). Critical values: 0.831, 12.833. Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim that the sample is from a population having weights with a standard deviation equal to 0.2000 g. 17. H0: s = 20 minutes. H1: s ≠ 20 minutes. Test statistic: x2 = 1153.950. P@value = 0.0009. Critical values: 887.620, 1117.889. Reject H0. There is sufficient evidence to warrant rejection of the claim that the sample is from a population with a standard deviation equal to 20 minutes. A histogram or normal quantile plot shows that the distribution is far from being normal, so a requirement of the hypothesis test is violated and the results of this hypothesis test are not necessarily valid. (TI data: Test statistic: x2 = 679.168, P@value = 0.0000. Critical values: 421.384, 584.125.) 19. Critical x2 = 81.540 (or 81.494 if using z = 2.326348 found from technology), which is close to the value of 82.292 obtained from Statdisk and Minitab. Section 8-5 1. a. A sample of the same size 1n = 52 is randomly selected from the given sample. 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. 23. H0: m = 1.6 W>kg. H1: m 6 1.6 W>kg. Test statistic: t = -4.256. P@value = 0.0019 (Table: P@value 6 0.005). Critical value: t = -2.998. Reject H0. 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. 25. H0: m = 35 minutes. H1: m 6 35 minutes. Test statistic: t = -5.820. P@value = 0.000 (Table: P@value 6 0.005). Critical value: t = -2.330. Reject H0. There is sufficient evidence to support the claim that the mean commute time is less than 35 minutes. The sample mean is 31.0 minutes, and the claimed mean is 35 minutes. That difference is statistically significant, but does not appear to have much practical significance. (TI data: Test statistic is t = -2.62 and P@value = 0.005.) 27. H0: m = 1818 mm. H1: m 6 1818 mm. Test statistic: t = -2.903. P@value = 0.0019 (Table: P@value 6 0.005). Critical value: t = -2.327 (Table: -2.326). Reject H0. 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 H0. 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.) 29. The hypothesis test yields a P-value of 0.0009, so the difference between x = 100.05 and the claimed mean of 100 is statistically significant, but that difference is so small that it does not have practical significance. 31. The critical t score found 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. Section 8-4 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 different in these ways: (1) It is much more strict, meaning that the distribution of the population must be much closer to a normal distribution; (2) the normality requirement applies for any sample size, not just for small samples with n … 30. 3. a. Reject H0. b. There is sufficient evidence to support the claim that the standard deviation is less than 0.04000 g. c. It appears that with the new minting process, the variation among weights has decreased, so the weights are more consistent. The new minting process appears to be better than the original minting process. 5. H0: s = 10 bpm. H1: s ≠ 10 bpm. Test statistic: x2 = 195.172. P-value = 0.0208. Reject H0. There is sufficient evidence to warrant rejection of the claim that pulse rates of men 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.
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