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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 twosided Ha , we double the uppertail probabilities to ﬁnd the pvalue: 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 pvalue 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 pvalue is 0.7388. We can not reject H0 .
(c) The 95% CI is (−0.0117, 0.0087). 2 ...
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 Summer '08
 AndrewB.Nobel

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