Unformatted text preview: n A = 13 and n B = 10. Consulting the MannWhitney table at 0.05 level, T would need to be less than or equal to 88 to reject. It is not, so we conclude that there is no evidence that genetic diversity affects abundance ( p > . 05). (b) (4 points) Since the deviations from respective medi ans are exactly the same in both groups, the pvalue would be exactly 1. Thus, p > . 20 is the correct choice. (c) (2 points) True. We only get as many df as number of pairs. (d) (2 points) True. If there is more variability between sample units than there is within a sample unit, pair ing is usually a good idea. (e) (4 points) Power = α when the true difference equals the difference under the null, which is zero in this example. Thus, the solid curve is not correct be cause the power is less than 0.05 at zero. Of the two other curves, the one that slopes up more steeply (dotdashed) has the higher sample size of 30. It has a value of 0.85 at a difference of 1, and thus the an swer is n = 30. 3 (22 points) df Trt = 3 1 = 2, df Err = 8 + 6 + 4 3 = 15, and df Tot = 2 + 15 = 17. The grand mean is 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 , . 01 = 6 . 36 and F 2 , 15 , .....
View
Full Document
 Fall '08
 Staff
 10%, 1%, researcher, 571 Second

Click to edit the document details