5.(16 pts) ConsiderH0:p= 0.32 vs the two-sidedalternativeHA:p= 0.32, wherepis the gerni-mation rate of oriental bittersweet seeds in tulippoplar stands. The number of seeds that germi-nate, out of 120 seeds, has a binomialB(120,0.32)distribution if the null hypothesis is true, which isapproximated by a normalN(38.4,5.112) distri-bution (np= 38.4 andnq= 81.6 are both>5).We observez= (58-38.4)/5.11 = 3.835 (orX2= 14.71 with a chi-square test) so that thep-value isp= 2*0.00006 = 0.0001. There is verystrong evidence that the germination rate in tulippoplar stands is greater than 0.32.6.(a) (6 pts) BinomialB(10,0.72), skewed left.(b) (5 pts) 0.180(c) (5 pts) 0.18 + 0.18-0.18*0.18 = 0.327 becausethe two events are not mutually exclusive.Wecan calculate the probability of finding orientalbittersweet in exactly 6 plots out of 10 inbothlocations (0.18*0.18) because these events areindependent.(d) (10 pts) We may argue for a one-sided test,HA:pr<0.72 because of the scientist’s suspicion. Weare not told, but this suspicion could be basedupon prior data.H0:pr= 0.72. The number ofplots where oriental bittersweet is present has abinomial distributionXr∼ B(10, .0.32) if the nullhypothesis is true, with an expected value of 3.2.The values that are as extreme as or more extremethan what was observed arex= 0 andx= 1, sothe p-value isP(Xr= 0) +P(Xr= 1).Usingthe 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 thatpr<0.32.Frequency20406080100010203040Summary of grades●●●●6375841
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