5 16 pts consider h p 0 32 vs the two sided

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5. (16 pts) Consider H 0 : p = 0 . 32 vs the two-sided alternative H A : p = 0 . 32, where p is the gerni- mation rate of oriental bittersweet seeds in tulip poplar stands. The number of seeds that germi- nate, out of 120 seeds, has a binomial B (120 , 0 . 32) distribution if the null hypothesis is true, which is approximated by a normal N (38 . 4 , 5 . 11 2 ) distri- bution ( np = 38 . 4 and nq = 81 . 6 are both > 5). We observe z = (58 - 38 . 4) / 5 . 11 = 3 . 835 (or X 2 = 14 . 71 with a chi-square test) so that the p-value is p = 2 * 0 . 00006 = 0 . 0001. There is very strong evidence that the germination rate in tulip poplar stands is greater than 0 . 32. 6.(a) (6 pts) Binomial B (10 , 0 . 72), skewed left. (b) (5 pts) 0 . 180 (c) (5 pts) 0 . 18 + 0 . 18 - 0 . 18 * 0 . 18 = 0 . 327 because the two events are not mutually exclusive. We can calculate the probability of finding oriental bittersweet in exactly 6 plots out of 10 in both locations (0 . 18 * 0 . 18) because these events are independent. (d) (10 pts) We may argue for a one-sided test, H A : p r < 0 . 72 because of the scientist’s suspicion. We are not told, but this suspicion could be based upon prior data. H 0 : p r = 0 . 72. The number of plots where oriental bittersweet is present has a binomial distribution X r ∼ B (10 , . 0 . 32) if the null hypothesis is true, with an expected value of 3.2. The values that are as extreme as or more extreme than what was observed are x = 0 and x = 1, so the p-value is P ( X r = 0) + P ( X r = 1). Using the binomial formula, we get . 000003+ . 000076 = 8 . 10 - 5 . (we could double this by two for a two- sided test). There is very strong evidence that p r < 0 . 32. Frequency 20 40 60 80 100 0 10 20 30 40 Summary of grades 63 75 84 1
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