# ch 18 - Chapter 18 Solutions 18.1 s √ n = 21 88 √ 20 =...

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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 µ is x ± t ∗ µ s √ n ¶ = 1 . 1182 ± 2 . 228 µ . 04378 √ 11 ¶ = 1 . 1182 ± . 0294 = 1 . 089 to 1 . 148 . (c) Because this interval does not include 1, we can reject H : µ = 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 µ of nitrogen in ancient air? Formulate: 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 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 µ is 59 . 5889 ± 1 . 860 µ 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 one-sided P-values are 0.05 and 0.025, respectively. (Software reports that P = . 0451.) (c) t = 1 . 82 is significant at α = . 05 but not at α = . 01. 200 202 Chapter 18 Inference about a Population Mean 18.13. (a) The outlier (31 µ 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 µ > 0, because we computed control rate minus experimental rate, so we test H : µ = 0 versus H a : µ > 0. For all 12 newts, x . = 6 . 4167 and s . = 10 . 7065 µ m / hr, so t = 6 . 4167 − 10 . 7065 / √ 12 ....
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## This note was uploaded on 12/30/2009 for the course PSYC PSYC 60 taught by Professor ? during the Fall '09 term at UCSD.

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ch 18 - Chapter 18 Solutions 18.1 s √ n = 21 88 √ 20 =...

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