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Unformatted text preview: Stat 500 2009 Midterm I Solution Notes about my answers are marked by General things about my grading: If I wrote math error, you had the right idea/equation, but made a mistake in the calculation. I deducted one point for these. The exam had many places where results from early parts were used in subsequent parts. I deducted points for the mistake when (and only when) they occurred. I did not deduct subsequent points because your answer(s) in later parts didnt match mine. If you did the right things in subsequent parts, I gave you full credit. These solutions indicate some common mistakes. If you dont understand a comment on your exam or something in these solutions, please come and see me (or call/email if youre offcampus). If I misadded points, or you dont understand why I deducted points, please see / call / email me. 1. Making components  design based inference (a) experimental study. Treatments are randomly assigned to components (the e.u.s). (b) p = 15/1000 = 0.015. The observed ratio = 14.65/7.34 = 1.996. The pvalue comes from the randomization distribution. How many values are 1.996? I count 14. So p = (N+1)/(R+1) = 15/1000. (c) T = (1.996  1.5)/0.0946 = 5.24 I didnt ask you for a pvalue because the d.f. here are not trivial. The usual 2sample formula doesnt apply because the s.e. is not s p q 1 /n 1 + 1 /n 2 . (d) Not reasonable. The randomization distribution is not symmetric. Various folks said that the randomization distribution was not centered at 0. Thats not a problem. The rand. distribution looks centered at 1.5. If you calculated the rand. distribution of ratio1.5, that would be centered at 0. The T distribution is centered at 0 because the T formula includes a 1.5. Saying check assumptions got no credit because the usual assumptions are for appropri ate tests of differences, not ratios. You have to think about the randomization distribution. 2. Litter size in pigs (a) The mean of the experimental group is known more precisely because it has the smaller s.e. You are looking for the group with the smallest s.e. Since you do not assume equal variances, calculate the s.e. for each group using its own s.d. I get control = 0.54 and expt = 0.47. Because you are not pooling variances, larger sample size is not sufficient. Because the sample sizes are unequal, larger s.d. is not sufficient....
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 Fall '08
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