that New Yorkers sleep an amount different than 8 hours per night on average. (e) Yes, as we re- jected H 0 .
332 APPENDIX B. END OF CHAPTER EXERCISE SOLUTIONS 4.7 t ? 19 is 1.73 for a one-tail. We want the lower tail, so set -1.73 equal to the T score, then solve for ¯ x : 56.91. 4.9 (a) For each observation in one data set, there is exactly one specially-corresponding ob- servation in the other data set for the same geographic location. The data are paired. (b) H 0 : μ diff = 0 (There is no difference in av- erage daily high temperature between January 1, 1968 and January 1, 2008 in the continental US.) H A : μ diff 6 = 0 (Average daily high tem- perature in January 1, 1968 is different than the average daily high temperature in January, 2008 in the continental US.) (c) Independence: loca- tions are random and represent less than 10% of all possible locations in the US. We are not given the distribution to check the skew. In prac- tice, we would ask to see the data to check this condition, but here we will move forward under the assumption that it is not strongly skewed. (d) T = 1 . 60, df = 51 - 1 = 50. → p-value be- tween 0.1, 0.2 (two tails!). (e) Since the p-value > α (since not given use 0.05), fail to reject H 0 . The data do not provide strong evidence of a temperature change in the continental US on the two dates. (f) Type 2, since we may have incorrectly failed to reject H 0 . There may be an increase, but we were unable to detect it. (g) Yes, since we failed to reject H 0 , which had a null value of 0. 4.11 (a) (-0.28, 2.48). (b) We are 95% con- fident that the average daily high on January 1, 2008 in the continental US was 0.24 degrees lower to 2.44 degrees higher than the average daily high on January 1, 1968. (c) No, since 0 is included in the interval. 4.13 (a) Each of the 36 mothers is related to exactly one of the 36 fathers (and vice-versa), so there is a special correspondence between the mothers and fathers. (b) H 0 : μ diff = 0. H A : μ diff 6 = 0. Independence: random sample from less than 10% of population. The skew of the differences is, at worst, slight. T = 2 . 72, df = 36 - 1 = 35 → p-value ≈ 0 . 01. Since p- value < 0.05, reject H 0 . The data provide strong evidence that the average IQ scores of mothers and fathers of gifted children are different, and the data indicate that mothers’ scores are higher than fathers’ scores for the parents of gifted chil- dren. 4.15 Independence: Random samples that are less than 10% of the population. In practice, we’d ask for the data to check the skew (which is not provided), but here we will move forward under the assumption that the skew is not ex- treme (there is some leeway in the skew for such large samples). Use t ? 999 ≈ 1 . 65. 90% CI: (0.16, 5.84). We are 90% confident that the average score in 2008 was 0.16 to 5.84 points higher than the average score in 2004. 4.17 (a) H 0 : μ 2008 = μ 2004 → μ 2004 - μ 2008 = 0 (Average math score in 2008 is equal to aver- age math score in 2004.) H A : μ 2008 6 = μ 2004 → μ 2004 - μ 2008 6 = 0 (Average math score in 2008 is different than average math score in 2004.) Con- ditions necessary for inference were checked in Exercise 4.15 . T = - 1 . 74, df = 999 → p-value between 0.05, 0.10. Since the p-value < α , re- ject H 0 . The data provide strong evidence that
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