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Therefore we will use a 2 sided paired t test on the

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right hand side). Therefore, we will use a 2-sided paired-t-test on the log-transformed values. (b) (12 pts) H 0 : μ 1 = μ 2 where μ i is the mean log- concentration of E. Coli in the water at beach i . H A : μ 1 = μ 2 . t = (7 . 1471 - 7 . 0708) / ( . 22579 / 12) = 1 . 14 on 11 df. Using the table from N & C, the closest we can get to the p-value is > 2 * 0 . 10. Using the table online, we can get slightly closer and would find p-value 2 * 0 . 147 using t = 1.1, or p-value 2 * 0 . 128 using t = 1.2. Either way, there is no evidence to reject H 0 . We conclude that the average E. Coli concentrations at the two beaches are not significantly different. 3. (15 pts) Interval: 10 ± 1 . 98 * 2 / 150. The t- multiplier 1 . 98 was obtained using 149 df, but it is very close to the z-multiplier 1 . 96. We get 10 ± 0 . 323 or (9 . 677 , 10 . 323). The first assumption is that nests were randomly sampled. The second assumption is that the number of acorns is nor- mally distributed (which we know is not true) or that the sample size is large enough for the CLT to apply. The sample size here (150) is deemed large enough, especially knowing that the skew is moderate. 4. (15 pts) Interval center: 0 . 20 - 0 . 35 = - 0 . 15. SE: . 20 * . 80 50 + . 35 * . 65 80 = . 0777. For 90% confidence, the z-multiplier is 1 . 64. We get: - 0 . 15 ± 1 . 64 * 0 . 0777 i.e. - 0 . 15 ± . 1275 or ( - . 277 - 0 . 022). Zero is not in this CI, so we reject the hypothesis that π 1 = π 2 and conclude that the two bird popula- tions, even though they are from the same species, have different proportions of individuals with the blunt beak trait. We are assuming that birds were
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