Lec20-21-ex

Lec20-21-ex - 7.15(a df = 14 t∗ = 2.145(b df = 24 t∗ =...

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Unformatted text preview: 7.15 (a) df = 14, t∗ = 2.145. (b) df = 24, t∗ = 2.064. (c) df = 24, t∗ = 1.711. (d) For a given confidence level C , t∗ (hence the margin of error m) decreases as sample size n increases. Given n, t∗ increases as the confidence level C increases. 7.21 (a) df = 23. (b) 2.177 < t < 2.500. (c) For the two-sided Ha , we double the upper-tail probabilities to find the p-value: 0.02 < p < 0.04. (d) t = 2.40 is significant at the level α = 0.05, but not at the level α = 0.01. 7.24 √ (b) x = 43.17, s = 4.4149, n = 20, s/ n = 0.9872. For df = 19, the critical value with α/2 = 0.025 ¯ is t∗ = 2.093, hence the margin of error is m = 2.0662. (c) The 95% CI is (41.1038, 45.2362). 7.32 (a) and (b) The mean for paired differences is x = 4.73125 with SD s = 1.7457. ¯ √ (c) SEX = s/ n = 0.4364, df = 15, t∗ = 2.131, the margin of error m = 0.9300, so the 95% CT ¯ is (3.8012, 5.6613). (d) The 95% CI is (8.3626, 12.4549). (e) For testing H0 : µ = 16, with df = 15 we have t = −5.823 which gives a p-value smaller than 0.0001 (highly significant). (f) The data suggest that the excess calories were not converted to weight ... 7.39 1 (a) For the paired differences, x = −0.0015, s = 0.0012. ¯ (b) t = −0.347 with df = 7, so the p-value is 0.7388. We can not reject H0 . (c) The 95% CI is (−0.0117, 0.0087). 2 ...
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This note was uploaded on 11/18/2011 for the course STOR 155 taught by Professor Andrewb.nobel during the Summer '08 term at UNC.

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Lec20-21-ex - 7.15(a df = 14 t∗ = 2.145(b df = 24 t∗ =...

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