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lecture10solutions - ISYE 6414 Summer 2009 Solution Lecture...

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ISYE 6414 - Summer 2009 Solution Lecture 10. Problem 16.1 pag.722 Whit the fitted regression function is possible to infer the mean sales when the price is \$68. In the ANOVA model, there is not a continuous functional relation between the response and the predictors. Problem 16.2 pag.722 The ANOVA model is correct, given that is enough to know the means of the response for each of 5 different treatments. A functional relation is not necessary. Problem 16.4 pag.722 E ( MSE ) = σ 2 = 9. E ( MSTR ) = σ 2 + 1 2 ( 25(65 - 80) 2 + 25(95 - 80) 2 ) = 9 + 25 · 15 2 . Then, E ( MSTR ) >> E ( MSE ), that is, the means of the three-lenght-of-service groups are not equal. Problem 16.5 pag.723 1. E ( MSE ) = σ 2 = 2 . 8 2 . E ( MSTR ) = σ 2 + 100 3 ( (5 . 1 - 7 . 2) 2 + (6 . 3 - 7 . 2) 2 + (7 . 9 - 7 . 2) 2 + (9 . 5 - 7 . 2) 2 ) = 9+366 . 667. Then, E ( MSTR ) >> E ( MSE ), that is, the means of the four groups are not equal. 2. If μ 2 = 5 . 6 and μ 3 = 9, then, E ( MSTR ) is going to be larger than in part b. Even though the range of the means is the same, the dispersion of them increases, and then, the E ( MSTR ). Problem 16.6 pag.723 The F statistic is claculated with cuadratic expressions (dispersion esti- mates). In particualar: E ( MSE ) = σ 2 and E ( MSTR ) = σ 2 + i n i ( μ i - μ ) 2 r - 1 . Then, if the factor level means are not equal, E ( MSE ) < E ( MSTR ). Problem 16.8 pag.723 1. Plot: Figure 1. 2. Fitted values: > mod=aov(y ~ color, data=data) > model.tables(mod,type="means") Tables of means Grand mean: 29 color b g o 29.4 29.6 28.0 3. Residuals 1

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Figure 1: Data points foe each color > mod\$residuals 1 2 3 4 5 6 -1.40000e+00 -3.40000e+00 1.60000e+00 -2.40000e+00 5.60000e+00 4.40000e+00 7 8 9 10 11 12 -6.00000e-01 -4.60000e+00 1.40000e+00 -6.00000e-01 3.00000e+00 -3.00000e+00 13 14 15 -1.00000e+00 1.00000e+00 7.13134e-17 4. ANOVA table > summary(mod)
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