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ch17 - Chapter 17 Solutions 17.1 s n = 21.88 20 = 4.8925...

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Chapter 17 Solutions 17.1. s / n = 21 . 88 / 20 . = 4 . 8925 minutes. 17.2. The mean is x = 251 ms. Because s / n = s / 12 = 45 ms, we have s . = 155 . 88 ms. 17.3. (a) t = 2 . 015 for df = 5 and α = 0 . 05. (b) t = 2 . 518 for df = 21 and α = 0 . 01. 17.4. Use df = 24. (a) t = 2 . 064 for α = 0 . 025. (b) t = 0 . 685 for α = 0 . 25. 17.5. (a) Use df = 9 and α = 0 . 025: t = 2 . 262. (b) Use df = 19 and α = 0 . 005: t = 2 . 861. (c) Use df = 6 and α = 0 . 05: t = 1 . 943. 17.6. (a) The stemplot shows a slight skew to the right, but not so strong that it would invalidate the t procedures. (b) With x = 18 . 66 and s . = 10 . 2768, and t = 2 . 093 (df = 19), the 95% confidence interval for µ is 18 . 66 ± 2 . 093 10 . 2768 20 = 18 . 66 ± 4 . 8096 = 13.8504 to 23.4696. 0 67888 1 0004 1 67 2 012 2 57 3 00 3 78 17.7. State: What is the mean percent µ of nitrogen in ancient air? Plan: We will estimate µ with a 90% confidence interval. Solve: We are told to view the observations as an SRS. A stemplot shows some left-skewness; however, for such a small sample, the data are not unrea- sonably skewed. There are no outliers. With x = 59 . 5889% and s = 6 . 2553% nitrogen, and t = 1 . 860 (df = 8), the 90% confidence interval for µ is 59 . 5889 ± 1 . 860 6 . 2553 9 = 59 . 5889 ± 3 . 8783 = 55.71% to 63.47%. Conclude: We are 90% confident that the mean percent of nitrogen in ancient air is be- tween 55.71% and 63.47%. 4 9 5 1 5 5 4 5 5 6 0 6 33 6 445 17.8. (a) df = 14. (b) t = 1 . 82 is between 1.761 and 2.145, for which the one-sided P -values are 0.05 and 0.025, respectively. (Software reports that P = 0 . 0451.) (c) t = 1 . 82 is significant at α = 0 . 05 but not at α = 0 . 01. 17.9. (a) df = 24. (b) t = 1 . 12 is between 1.059 and 1.318, so 0 . 20 < P < 0 . 30. (Software reports that P = 0 . 2738.) (c) t = 1 . 12 is not significant at either α = 0 . 10 or α = 0 . 05. 17.10. State: Is there evidence that the percent of nitrogen in ancient air was different from the present 78.1%? Plan: We test H 0 : µ = 78 . 1% vs. H a : µ = 78 . 1%. We use a two-sided alternative because, prior to seeing the data, we had no reason to believe that the percent of nitrogen in ancient air would be higher or lower. Solve: We addressed the conditions for inference in Exercise 17.7. In that solution, we found x = 59 . 5889% and s = 6 . 2553% nitrogen, so t = 59 . 5889 78 . 1 6 . 2553 / 9 . = − 8 . 88. For df = 8, 195
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196 Chapter 17 Inference about a Population Mean this is beyond anything shown in Table C, so P < 0 . 001 (software gives P = 0 . 00002). Conclude: We have very strong evidence ( P < 0 . 001) that Cretaceous air contained less nitrogen than modern air. 17.11. Plan: Take µ to be the mean difference (monkey call minus pure tone) in firing rate. We test H 0 : µ = 0 vs. H a : µ > 0, using a one-sided alternative because the researchers suspect a stronger response to the monkey calls. Solve: We must assume that the monkeys can be regarded as an SRS. For each monkey, we compute the call minus pure tone differ- ences; a stemplot of these differences shows no outliers or deviations from Normality. The mean and standard deviation are x . = 70 . 378 and s . = 88 . 447 spikes / second, so t = x 0 s / 37 . = 4 . 84 with df = 36. This has a very small P -value: P < 0 . 0001. Conclude: We have very strong evidence that macaque neural response to monkey calls is stronger than the response to pure tones.
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