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Unformatted text preview: Stat332 Short Solutions to Assignment 2 1. Look at the solution of Question 4 in the exercises of Chapter 3 in your course notes. 2. (1) s 2 3 = 0 . 500 and s 2 4 = 0 . 125. (2) Model Y ij = μ + τ i + R ij , i = 1 ,..., 4 and j = 1 , 2 with R ij ∼ G (0 ,σ ) and the constraint ∑ i τ i = 0. For a fixed j , μ is the overall average response cross all the treatments. τ i is the increase of the treatment 1 from the overall average response, and called treatment effect. R ij is the individual error term. (3) The ANOVA table for this experiment is Sum of Degrees of Mean Ratio to the Source Squares (SS) Freedom (df) Square (MS) residual MS Treatments 22.094 3 7.365 15.704 Residual 1.875 4 0.469 Total 23.969 7 (4) ˆ τ 1 = 2 . 4375, ˆ τ 2 = . 4375, ˆ τ 3 = 0 . 8125 and ˆ τ 4 = 2 . 0625. (5) The null hypothesis would be H o : τ 1 = τ 2 = τ 3 = τ 4 = 0. F = 15 . 704 with 3 and 4 df, which is nearly significant at the 1% significance level....
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 Spring '09
 Xu(Sunny)Wang
 Statistics, Rij G, Residual Total Squares, Source Treatments Residual, Treatments Residual Total, overall average response

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