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Unformatted text preview: Stat 402A, HW 6 Answers 1. Drugs in mice The filled in ANOVA table is: Source d.f. SS MS F pvalue litter 4 6.403 drug 3 1.845 0.615 11.58 < 0.001 error 12 0.637 0.053 c.total 19 8.885 (a) 3 pt. See d.f. column above. Notes: remember blocks d.f. = # blocks  1, trts d.f. = # trts  1, error = product (or error = c.total d.f.  other terms), c.total = # obs  1. (b) 2 pt. See MS, F, and p columns above. Notes: remember MS = SS / d.f., F = MS treatments / MS error . The observed F is com pared to quantiles of the d.f.(treatments), d.f.(error) F distribution. That is the F 3 , 12 distribution. Different books have different quantiles. Most have the 0.95 and 0.99; some have more. The 0.95 quantile = 3.49, 0.99 quantile = 5.59 and 0.999 quantile = 10.8. This observed pvalue is larger than the 0.999 quantile, so the pvalue is < 10.999, i.e. < 0.001. While saying p < 0.05 is technically correct, it is useful to be as significant as possible. If you have the exact pvalue (e.g. from computer output), the best approach is to report that. (c) 2 pt. s.e. of the diff = q MSE * 2 /n = q . 053 * 2 / 5 = 0 . 15. note: n is the number of blocks....
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This note was uploaded on 01/27/2010 for the course STAT 402 taught by Professor Staff during the Spring '08 term at Iowa State.
 Spring '08
 Staff

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