This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: STAT 4220 HW7 Solution 1 Problem 1 (a) Fitting the appropriate model yields the following: > summary(aov(g)) Df Sum Sq Mean Sq F value Pr(>F) factor(Day) 4 125.2 31.3 1.5343 0.2806 factor(Operator) 4 167.2 41.8 2.0490 0.1800 factor(Machine) 4 3424.8 856.2 41.9706 2.062e-05 *** factor(Method) 4 2857.6 714.4 35.0196 4.075e-05 *** Residuals 8 163.2 20.4--- Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 The Tukey multiple comparisons plot is given below. By inspection of the plot, we find that the pairs (B,A), (D,A), (E,A), (C,B), (D,B), (D,C) and (E,C) are all significantly different from one another.-20-10 10 20 30 40 E-D E-C D-C E-B D-B C-B E-A D-A C-A B-A 95% family-wise confidence level Differences in mean levels of factor(Method) (b) From the ANOVA table above, we see that the factors Day and Operator are not significant sources of vari- ability. While we would definitely need to keep the Machine blocking factor in the model, we could simplify things by removing the other two blocking variables. At the cost of removing these two factors, we will notthings by removing the other two blocking variables....
View Full Document
This note was uploaded on 06/06/2011 for the course STAT 4220 taught by Professor Smith during the Spring '08 term at University of Georgia Athens.
- Spring '08