A weighted average of the treatment means and is

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a weighted average of the treatment means, and is equal to (12 * 8 + 6 * 6 + 15 . 5 * 4) / 18 = 10 . 78. Then SSTrt = 8 * (12 - 10 . 78) 2 +6 * (6 - 10 . 78) 2 +4 * (15 . 5 - 10 . 78) 2 = 238 . 1. SSTot is 238 . 1+277 = 515 . 1, MSTrt = 238 . 1 / 2 = 119 . 05, MSErr = 277 / 15 = 18 . 47, and F = 119 . 05 / 18 . 476 = 6 . 44. F 2 , 15 , 0 . 01 = 6 . 36 and F 2 , 15 , 0 . 005 = 7 . 70, so 0 . 005 < p < 0 . 01. We would reject the null at the 1% level or lower, and conclude that at least one of the treatments has a different average re-growth score than one other treatment. 4 (16 points) Since Levene’s test has a very low p-value, we would conclude that the variances probably dif- fer, and use the variances allowed to differ t-test. The t-statistic is (13 . 5 - 9 . 19) / (21 . 6 2 ) / 12 + (8 . 7 2 ) / 9 = 0 . 6267. The appropriate df is 15.3, round down to 15 to be conservative. t 15 , 0 . 1 = 1 . 341, so the p-value is larger than 2 * 0 . 1 = 0 . 2. We do not have suf- ficient evidence to reject the null at the 10% level, and conclude that the average wing spans of the two sub-species are the same. 5 (18 points) We must assume independence, no re- porting bias, and large sample size. Independence is somewhat questionable, since there could be infec- tions among family members, or a particularly bad strain could travel through a hospital, etc. There is also likely to be some reporting bias, since it is likely that many cases that did not result in deaths may not have been reported. This data would then probably over-estimate the probability of survival. We can use the normal approximation so long as n * p 5 and n * (1 - p ) 5. They are, since (37 / 42) * 42 = 37 5, and (5 / 42) * 42 = 5 5. Then the CI is 5 / 42 ± z 0 . 1 / 2 * ((5 / 42) * (37 / 42)) / 42, or 0 . 119 ± 0 . 082: ( . 037 , . 201). Frequency 30 40 50 60 70 80 90 100 0 10 20 30 40 50 60 Summary of grades 81 87 91 1
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  • Fall '08
  • Staff
  • 10%, 1%, researcher, 571 Second

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Christopher Reinemann
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