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We are assuming that birds were randomly sampled and

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blunt beak trait. We are assuming that birds were randomly sampled, and that the sample sizes were large enough for the CLT to apply. Specifically, we can tell this last assumption is met because 50 * 0 . 20 = 10, 50 - 10 = 40, 80 * 0 . 35 = 28 and 80 - 28 = 52 are all greater than 5. 5(a) (9 pts) H 0 : The distribution of the area colonized by X. perforans beetles is the same on trees from species 1 and on trees from species 2. In partic- ular, they have the same mean. H A : these two distributions differ. Samples have size n 1 = 7, n 2 = 8. The sum of ranks in sample 1 is 44, which is still smaller than 7 * (7+8+1) - 44 = 68 so T = 44. This is larger than 38, the criti- cal value to get p . 05. Therefore p > 0 . 05. There is no evidence that X. perforans beetles colonize trees from species 1 any differently than trees from species 2. (b) (9 pts) Source df SS MS F p-value Tree sp. 3 681 227 2.95 . 01 < p < . 05 Residual 36 2767 76.86 Total 39 3448 (c) (5 pts) H 0 : μ 1 = μ 2 = μ 3 = μ 4 where μ i is the mean area colonized by X. perforans beetles on timber samples from trees in species i . H A : not all μ i ’s are the same, which means that the average colonized area differs in at least one tree species versus another. There is moderate evidence to reject H 0 . We conclude that X. perforans beetles 1
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colonize a larger average area (or show preference) in at least one tree species versus another.
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