HW4_STA5166_complete

HW4_STA5166_complete - Jaime Frade Problem 4.1 (J ) a) What...

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Jaime Frade Problem 4.1 (J ) a) What kind of an experimental design is this? Randomized Block Design b) Make a graphical analysis and an ANOVA. Analyzing each of the paint suppliers, the mean is different for each of them as illustrated above.
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Jaime Frade ANOVA output > summary(fit1) Df Sum Sq Mean Sq F value Pr(>F) supplier 3 665.13 221.71 20.387 1.503e-05 *** site 5 568.71 113.74 10.459 0.0001808 *** Residuals 15 163.13 10.88 --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 55 60 65 70 75 -2 0 2 4 6 fit1$fitted.values fit1$residuals Analyzing the residuals, assume no effect since no pattern and wide spread. CODE >Prob4.1=read.table(file="E:/FSU-FALL07-School/Fall07-school/STA5166/BHH2- Data/prb0401.dat", header=TRUE) >Prob4.1 > par(mfrow=c(1,1)) >plot(Prob4.1$supplier, Prob4.1$y) >fit1= aov(y~supplier + factor(site), data=Prob4.1) >fit2= aov(y~supplier, data=Prob4.1) >summary(fit1) >attributes(fit1) >par(mfrow=c(1,1)) >plot(fit1$fitted.values, fit1$residuals) ##LOAD BBH2 package >par(mfrow = c(1, 1), cex = 0.7) >anovaPlot(fit1, main = "Anova plot: problem 4.1",labels = TRUE, cex.lab = 0.6 ) c) Obtain confidence limits for the supplier averages
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Jaime Frade Using R General Linear Hypotheses Multiple Comparisons of Means: Tukey Contrasts Linear Hypotheses: Estimate GS - FD == 0 9.8333 L - FD == 0 -2.0000 ZK - FD == 0 9.0000 L - GS == 0 -11.8333 ZK - GS == 0 -0.8333 ZK - L == 0 11.0000 Simultaneous Confidence Intervals for General Linear Hypotheses Multiple Comparisons of Means: Tukey Contrasts Fit: aov(formula = y ~ supplier, data = Prob4.1) Estimated Quantile = 2.799 Linear Hypotheses: Estimate lwr upr GS - FD == 0 9.83333 0.05797 19.60869 L - FD == 0 -2.00000 -11.77536 7.77536 ZK - FD == 0 9.00000 -0.77536 18.77536 L - GS == 0 -11.83333 -21.60869 -2.05797 ZK - GS == 0 -0.83333 -10.60869 8.94203 ZK - L == 0 11.00000 1.22464 20.77536 95% family-wise confidence level Simultaneous Tests for General Linear Hypotheses Multiple Comparisons of Means: Tukey Contrasts Fit: aov(formula = y ~ supplier, data = Prob4.1) Linear Hypotheses: Estimate Std. Error t value p value GS - FD == 0 9.8333 3.4925 2.816 0.0641 . L - FD == 0 -2.0000 3.4925 -0.573 1.0000 ZK - FD == 0 9.0000 3.4925 2.577 0.1080 L - GS == 0 -11.8333 3.4925 -3.388 0.0175 * ZK - GS == 0 -0.8333 3.4925 -0.239 1.0000 ZK - L == 0 11.0000 3.4925 3.150 0.0303 * --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 (Adjusted p values reported -- bonferroni method)
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Jaime Frade CODE: (R) ##load package mvtnorm ##load package multcomp ##load package multtest fit2= aov(y~supplier, data=Prob4.1) summary(fit2) int= glht(fit2, linfct = mcp(supplier = "Tukey")) int print(confint(int)) plot(confint(int)) summary(int, test = adjusted( "bonferroni"))
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Jaime Frade Using Splus fit1$fitted.values fit1$residuals 60 62 64 66 68 70 72 -10 -5 0 5 10
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HW4_STA5166_complete - Jaime Frade Problem 4.1 (J ) a) What...

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