For the US crime data of Example 13.4.2, carry out the PCA for the covariates and then perform the regression analysis on the PC scores. Investigate if the multicollinearity problem persists in the fitted regression model based on the PC scores.

Data from in Example 13.4.2

Consider the girder experiment, which has 9 blocks and 4 treatments. The aim of the experiment is to find if the shear strength depends on the blocks, varying from S1.1 to S4.2, or on the method of preparation: Aarau, Karisruhe, Lehigh, and Cardiff. A bit of data preparation is needed from the csv file, and post the preparation, the ANOVA table is obtained using the aov function in R. Recollect the analysis of this dataset in Example 4.5.3 of Section 4.5 through themedian polish algorithm. The data arrangement is slightly different in the block design model.

After reading the data, the names of the variables have been curtailed to ensure the output of model. matrix displays all the variables in a single line. The codes rep(girder$Girder,each=4) and rep(colnames(girder)[2:5],9) with as. character ensures that the block design is clearly told that these two inputs are factors. The vector ss now contains the shear strength for each combination of the method of preparation and the girder. Finally, aov(ss gf + mf, data=girdernew) sets up the required block design in R. The interpretation is as follows. The large p-value of 0.172 with girderf suggests that the blocks have insignificant effect on the output. However, the method of preparation is found to have significant effects on the strength of the material.

As seen in Models 13.2, 13.10, etc., these linear models need to be investigated for model assumptions and other regression diagnostics too. Can it be said without loss of generality that leverages are not a concern in DOE since all the covariate values are fixed? Now, we consider an example from Montgomery (2005), wherein we will investigate issues related to model adequacy.

Transcribed Image Text:

> data (girder) > names (girder) [2:5] <- c("Aar", "Kar", "Leh", "Car") > gf <- as. character > mf <- as.character (rep (girder$Girder, each=4) ) (rep (colnames (girder) [2:5],9)) SS <- NULL # Shear Strength > for (i in 1:nrow (girder)) ss <- c (ss, as.numeric (girder [i, 2:5])) > girdernew <- data.frame (mf, gf, ss) > ssaov <- aov (ss-gf + mf, data=girdernew) > summary (ssaov) gf mf th th Df Sum Sq Mean Sq F value Pr (>F) 8 0.0895 0.0112 1.619 0.172 3 1.5138 0.5046 73.027 3.3e-12 *** 0.0069 Residuals 24 0.1658 1 2 Signif. codes: 0 ** > model.matrix (ssaov) (Intercept) gfs1.2 1 1 0 0 0.001**' 0.01 '*' 0.05. 0.1 1 mf Car mfKar mfLeh 0 0 0 1 0 0