Lecture 9 slides(Categories)

# Lecture 9 slides(Categories) - Last time Discussed the...

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Last time Discussed the basics of multiple linear regression, from both algebriac and geometric perspectives Showed that the least squares estimators are unbiased and found their variance-covariance matrix Lecture 9 – p. 1/33 Today’s material Finish 3.4 (have now covered 3.1 - 3.4) Start 3.10 (will come back to other sections later) Lecture 9 – p. 2 / Partitioning Regression Sums of Squares If we order the covariates so that the the intercept vector ( X 0 ) and the other ˜ p variables are numbered X 0 , X 1 , ... X ˜p Can then write SS Reg = y t b X ( X t X ) - 1 X t ˜ X ( ˜ X t ˜ X ) - 1 ˜ X t B y + ˜ b t ˜ X t ˜ X ˜ b n ¯ y 2 = R ( β p , β p - 1 , ...β ˜ p +1 | β ˜ p , ..., β 0 ) + R ( β ˜ p , ..., β 0 ) Lecture 9 – p. 3/33 Partitioning Regression Sums of Squares R ( β p , β p - 1 , ...β ˜ p +1 | β ˜ p , ..., β 0 ) is the additional regression sums of squares explained by covariates X ˜p + 1 , ..., X p beyond the smaller model using only ˜ X Partitioning allows us to test complex hypotheses R ( β ˜ p , ..., β 0 ) = ˜ b t ˜ X t ˜ X ˜ b n ¯ y 2 is equivalent to the regression sums of squares under a null hypothesis that ( β ˜p + 1 , ..., β p ) = β * = 0 , i.e. for y = ˜ X ˜ β + X * β * + ǫ where X * contain the covariate vectors from X not contained in ˜ X , β * is fixed to be 0 Lecture 9 – p. 4 /

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Testing complex hypotheses To test a null hypothesis that a subset of the parameters, β * = 0 , intuitively want to look for significant explanation beyond the best explanation by maximizing only over ˜ X Another way of looking at it is to examine the amount of residual sums of squares reduced by allowing β * n = 0 Can write R ( β * | ˜ β ) = SS Reg,H 1 SS Reg,H 0 That means R ( β * | ˜ β ) = SS Tot SS Res,H 1 ( SS Tot SS Res,H 0 ) = SS Res,H 0 SS Res,H 1 Lecture 9 – p. 5/33 Testing subsets of parameters Under H 0 , R ( β * | ˜ β ) 2 has a χ 2 p 1 distribution, where p 1 = p ˜ p Also under H 0 , s 2 ( n k ) , where k = p + 1 , has a χ 2 n - k distribution Therefore, under H 0 , F = R ( β * | ˜ β ) /p 1 s 2 ∼ F p 1 ,n - k Lecture 9 – p. 6 / Back to Stat course example Assume that we’re testing for a significant association of either the homework grade OR the midterm grade Therefore, our hypotheses would be: H 0 : β hw = β midterm = 0 H 1 : β hw n = 0 OR β midterm n = 0 Lecture 9 – p. 7/33 Example: Stat500 > none.mod<-lm(final˜1) > all.mod<-lm(final˜hw+midterm) > anova(none.mod,all.mod) Analysis of Variance Table Model 1: final ˜ 1 Model 2: final ˜ hw + midterm Res.Df RSS Df Sum of Sq F Pr(>F) 1 54 1325.25 2 52 925.96 2 399.28 11.211 8.948e-05 *** Lecture 9 – p. 8 /
Example: Stat500 > ### Same test appears at bottom > summary(all.mod)

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## Lecture 9 slides(Categories) - Last time Discussed the...

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