Multiple Linear Regression

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-4.368 -1.670 -0.158 1.792 4.358Coefficients:Estimate Std. Error t value Pr(>|t|)(Intercept) -0.955101 63.013630 -0.015 0.988Girth -2.796569 1.468677 -1.904 0.068 .I(Girth^2) 0.265446 0.051689 5.135 2.35e-05 ***Height 0.119372 1.784588 0.067 0.947I(Height^2) 0.001717 0.011905 0.144 0.886---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1Residual standard error: 2.674 on 26 degrees of freedom<strong>Multiple</strong> R-squared: 0.9771, Adjusted R-squared: 0.9735F-statistic: 277 on 4 and 26 DF, p-value: < 2.2e-16In this ill-formed model nothing is significant except Girth and Girth^2. Let us continuedown this path and suppose that we would like to try a reduced model which contains nothing butGirth and Girth^2 (not even an Intercept). Our two models are nowthe full model: Y = β 0 + β 1 x 1 + β 2 x 2 1 + β 3x 2 + β 4 x 2 2 + ɛ,the reduced model: Y = β 1 x 1 + β 2 x 2 1 + ɛ,We fit the reduced model with lm and store the results:> treesreduced.lm anova(treesreduced.lm, treesfull.lm)Analysis of Variance TableModel 1: Volume ~ -1 + Girth + I(Girth^2)Model 2: Volume ~ Girth + I(Girth^2) + Height + I(Height^2)Res.Df RSS Df Sum of Sq F Pr(>F)1 29 321.652 26 185.86 3 135.79 6.3319 0.002279 **26
---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1We see from the output that the complete model is highly significant compared to the modelthat does not incorporate Height or the Intercept. We wonder (with our tongue in our cheek) ifthe Height^2 term in the full model is causing all of the trouble. We will fit an alternative reducedmodel that only deletes Height^2.> treesreduced2.lm anova(treesreduced2.lm, treesfull.lm)Analysis of Variance TableModel 1: Volume ~ Girth + I(Girth^2) + HeightModel 2: Volume ~ Girth + I(Girth^2) + Height + I(Height^2)Res.Df RSS Df Sum of Sq F Pr(>F)1 27 186.012 26 185.86 1 0.14865 0.0208 0.8865In this case, the improvement to the reduced model that is attributable to Height^2 is notsignificant, so we can delete Height^2 from the model with a clear conscience. We notice that thep-value for this latest partial F test is 0.8865, which seems to be remarkably close to the p-valuewe saw for the univariate t test of Height^2 at the beginning of this example. In fact, the p-valuesare exactly the same. Perhaps now we gain some insight into the true meaning of the univariatetests.8 Residual Analysis and Diagnostic ToolsAll of the tools from SLR are valid in multiple linear regression, too, but there are some slightchanges for the multivariate case. We list these below, and apply them to the trees example.Shapiro-Wilk, Breusch-Pagan, Durbin-Watson: unchanged from SLR, but we are now equippedto talk about the Shapiro-Wilk test statistic for the residuals. It is defined by the formulawhere E ∗ is the sorted residuals and a 1×n is defined byW = aT E ∗E T E , (58)a =m T V −1√mT V −1 V −1 m , (59)27
Page 5: We see from the output that for the
Page 10 and 11: predict(trees.lm, newdata = new, in
Page 12 and 13: 3.3 Overall F-TestAnother way to as
Page 14 and 15: ●Volume10 30 50 70● ●●●
Page 16 and 17: Coefficients:Estimate Std. Error t
Page 18 and 19: F-statistic: 350.5 on 2 and 28 DF,p
Page 21 and 22: Height by a new variable Tall which
Page 23 and 24: eforehand and wrote the discussion
Page 25: for this hypothesis test there are
Page 29 and 30: In the trees example we may take th
Page 31 and 32: Here is an example of how it works,

Multiple Linear Regression

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