Unformatted text preview: Conclusion: there is no evi dence that captive Humboldt penguins have a bill length any different from wild penguins (6.4cm). 5. (16 pts) Consider H : p = 0 . 32 vs the twosided alternative H A : p 6 = 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 , . 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 chisquare test) so that the pvalue is p = 2 * . 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 , . 72), skewed left. (b) (5 pts) 0 . 180 (c) (5 pts) 0 . 18 + 0 . 18 . 18 * . 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 * . 18) because these events are independent. (d) (10 pts) We may argue for a onesided test, H A : p r < . 72 because of the scientist’s suspicion. We are not told, but this suspicion could be based upon prior data. H : p r = 0 . 72. The number of plots where oriental bittersweet is present has a binomial distribution X r ∼ B (10 , . . 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 pvalue 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 < . 32. Frequency 20 40 60 80 100 10 20 30 40 Summary of grades ● ● ● ● ● 63 75 84 1...
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 Fall '08
 Staff
 ChiSquare Test, Normal Distribution, Null hypothesis, Probability theory, Penguin

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