hw11ans - Stat 500 Homework 11 answers 2009 My notes are...

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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 vice-versa. 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*(21-1) = 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. Three-factor 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 rod1-rod2 rod 1 -1 0; estimate rod1-rod3 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.

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hw11ans - Stat 500 Homework 11 answers 2009 My notes are...

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