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Lecture

ANOVA; model checking.pdf

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Department
Statistics
Course
STAT368
Professor
Douglas Wiens
Semester
Winter

Description
110 15. ANOVA; model checking Summary. Theoretical ANOVA table for a two factor factorial experiment wobservations per cell: Source SS df MS F0 A 1 = = 1 0 B 1 = 0 = 1 AB ( ) = ( ) 0= Error ( ) = ( ) Total 1 ( ) = ( X1)( 1) ( X) = ( 1) = ˆ2 = ˆ2 X ³ ´ X ³ ´ = c 2 = ¯ 2 ˆ = ¯ ¯ ˆ = ¯ ¯ ³c ´ = ¯ ¯ ¯ + ¯ 111 In the battery experiment the various averages are > means for(i in 1:3) {for (j in 1:3) means[i,j] means [Temp=15] [Temp=70] [Temp=125] [Type1] 134.75 57.25 57.5 [Type2] 155.75 119.75 49.5 [Type3] 144.00 145.75 85.5                                   Fig. 5.2 112 > g anova(g) Analysis of Variance Table Response: y Df Sum Sq Mean Sq F value Pr(>F) type 2 10684 5342 7.9114 0.001976 temp 2 39119 19559 28.9677 1.909e-07 type:temp 4 9614 2403 3.5595 0.018611 Residuals 27 18231 675 As suspected, the interaction e ects are quite signi- cant. There is no battery type which is ‘best’ at all temperatures. If interactions were NOT signicant one could compare the + by seeing which of the di erences ˆ + ˆ = ¯ were signicaqtly di erent from each other (using (¯ ¯ ) = 2 ). 113 As it is, we can only make comparisons at xed levels of the other factor.For instance when = 70, ( = 2) we can compare the means 2 = + + 2+ ( )
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