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Unformatted text preview: Chapter 18 Solutions 18.1. s / n = 21 . 88 / 20 . = 4 . 8925 minutes. 18.2. Mean plus or minus the standard error of the mean is 89 . 01 5 . 36, so the mean is x = 89 . 01 mg / dl, and the standard error of the mean, s / n , is 5.36 mg / dl. If s / 6 = 5 . 36, then the standard deviation is s = 5 . 36 6 . = 13 . 13 mg / dl. 18.3. (a) t = 2 . 015. (b) t = 2 . 518. 18.4. Use df = 24. (a) t = 2 . 064. (b) t = . 685. 18.5. (a) Use df = 9: t = 2 . 262. (b) Use df = 19: t = 2 . 861. (c) Use df = 6: t = 1 . 440. 18.6. (a) A stemplot shows that the data are not skewed and have no outliers. (There is a gap between 1.12 and 1.18, but as 1.18 appears three times, we would not consider it an outlier.) (b) Calculate x . = 1 . 1182 and s . = . 04378. We have n = 11, so df = n 1 = 10. The critical value for a 95% confidence interval is t = 2 . 228, and the interval for is x t s n = 1 . 1182 2 . 228 . 04378 11 = 1 . 1182 . 0294 = 1 . 089 to 1 . 148 . (c) Because this interval does not include 1, we can reject H : = 1 in favor of the two sided alternative. 10 77 10 88 11 11 11 22 11 11 11 888 18.7. State: What is the mean percent of nitrogen in ancient air? Formulate: We will estimate with a 90% confidence interval. Solve: We are told to view the observations as an SRS. A stemplot shows some leftskewness; however, for such a small sample, the data are not unreasonably 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 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 between 55.71% and 63.47%. 4 9 5 1 5 5 4 5 5 6 6 33 6 445 18.8. (a) df = 14. (b) t = 1 . 82 is between 1.761 and 2.145, for which the onesided Pvalues are 0.05 and 0.025, respectively. (Software reports that P = . 0451.) (c) t = 1 . 82 is significant at = . 05 but not at = . 01. 200 202 Chapter 18 Inference about a Population Mean 18.13. (a) The outlier (31 m / hr, from newt #8) is the only significant departure from Normality visible in the stemplot. (b) The healing rate is higher in the control limbs means > 0, because we computed control rate minus experimental rate, so we test H : = 0 versus H a : > 0. For all 12 newts, x . = 6 . 4167 and s . = 10 . 7065 m / hr, so t = 6 . 4167 10 . 7065 / 12 ....
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 Fall '09
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