HW4_STA5166_complete

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

This preview shows pages 1–6. Sign up to view the full content.

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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
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)

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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"))
Jaime Frade Using Splus fit1\$fitted.values fit1\$residuals 60 62 64 66 68 70 72 -10 -5 0 5 10

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 12/14/2011 for the course STAT 5166 taught by Professor Staff during the Fall '11 term at FSU.

### Page1 / 17

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

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online