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Unformatted text preview: Stat 500 Homework 11 answers 2009 My notes are marked by 1. Power of ANOVA tests (a) The standard errors of each contrast are: 1) q 2 / 3 / n , 2) 2 / n , 3) / n , 4) 2 / n , The s.e. for contrast 4 (the interaction) is the largest. (b) When all else is equal, the estimate with the largest s.e. leads to the test with the smallest power and viceversa. The test of interaction is the least powerful; the test of the main effect of reinforcement (contrast 1) is the most powerful. (c) The general equation is = ( t 1 / 2 + t ) s.e. = ( t 1 / 2 + t )2 / n . Solving for n gives you n 2 2 ( t 1 / 2 + t ) 2 ( / ) 2 . Starting with t 1 / 2 = 2 and t = 0 . 85 gives n 4 * 2 . 85 2 * (16 / 25) = 20 . 7, i.e. use n = 21 replicates per treatment. The above design has 6*(211) = 120 error d.f. so better quantiles are t 1 / 2 = 1 . 98 and t = 0 . 8446. Using these gives n 4 * 2 . 825 2 * (16 / 25) = 20 . 4. Use n = 21 replicates per treatment. 2. Threefactor factorial My SAS code: data freeway; infile C:\500\Homework 11\freeway.txt firstobs=2; input rod segment lane location width; logwidth=log(width); run; proc glm data=freeway; class rod lane location; model width=rod lane location rod*lane rod*location lane*location rod*lane*location; run; proc glm data=freeway; class rod lane location; model logwidth=rod lane location rod*lane rod*location 1 lane*location rod*lane*location; estimate rod1rod2 rod 1 1 0; estimate rod1rod3 rod 1 0 1;...
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This note was uploaded on 02/11/2012 for the course STAT 500 taught by Professor Staff during the Fall '08 term at Iowa State.
 Fall '08
 Staff
 Standard Error

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