hw44f08solu

hw44f08solu - STAT 514 Design of Experiments Homework 4...

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STAT 514 Design of Experiments Homework 4 Solution Problem 1 (a) SAS generates the following ANOVA table. Sum of Source DF Squares Mean Square F Value Pr > F Model 3 6026.83333 2008.94444 6.97 0.0022 Error 20 5767.00000 288.35000 Corrected Total 23 11793.83333 The p -value for the F -statistic is 0 . 0022, much less than 0.05. Hence we reject the null hypothesis and conclude that there exists difference between the treatments. (b) Since the experiment is a balanced design, two contrasts are orthogonal to each other if and only their inner product is 0. Let Γ 1 2 and Γ 3 be the contrasts of “Hormone I vs Hormone II”, “Low Level vs High Level” and “Equivalence of Level”, respectively. Then Γ t 1 Γ 2 =1 * 1+1 * ( - 1) + ( - 1) * 1+( - 1) * ( - 1) = 1 - 1 - 1+1=0 , Γ t 1 Γ 3 * * ( - 1) + ( - 1) * ( - 1) + ( - 1) * 1=1 - - 1=0 , Γ t 2 Γ 3 * - 1) * ( - 1) + 1 * ( - 1) + ( - 1) * 1=1+1 - 1 - . Hence the three contrasts are orthogonal to each other. (c) The SAS output for contrast sums of squares and contrasts testing is as follows. Contrast DF Contrast SS Mean Square F Value Pr > F C1 1 864.000000 864.000000 3.00 0.0989 C2 1 5162.666667 5162.666667 17.90 0.0004 C3 1 0.166667 0.166667 0.00 0.9811 Based the table above, Contrast Γ 1 (or C 1 ) is not significant because the p -value 0.0989 is greater than α . Therefore, the average effect of hormone I and the average effect of hormone II on are not different from each other; Contrast Γ 2 (or C 2 ) is significant ( p -value = 0 . 0004), this suggests that the average effect of the high levels of hormones and the average effect of the low levels of hormones are different. Contrast Γ (or C 3 ) is not significant at all ( p -value = 0 . 9811), so the difference between the high-level and low-level of hormone I is the same as that between the high-level and low-level of hormone II. Problem 2 (a) The ANOVA table is given below.
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hw44f08solu - STAT 514 Design of Experiments Homework 4...

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