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lecture8solutions - ISYE 6414 - Summer 2009 Solution...

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ISYE 6414 - Summer 2009 Solution Lecture 8. Problem 10.23 Pag. 419 We know that ˆ Y = Xb and ˆ Y ( i ) = Xb ( i ) . Then, from 10.33a: D i = ( Xb - Xb ( i ) ) 0 ( Xb - Xb ( i ) ) pMSE = ( b - b ( i ) ) 0 X 0 X ( b - b ( i ) ) pMSE Problem 10.24 Pag. 419 If n = p and X is invertible, then: H = X ( X’X ) - 1 X 0 = X ( X ) - 1 ( X 0 ) - 1 X 0 = I n I n = I p I p = I h ii = 1 and ˆ Y i = Y i . Problem 10.26 Pag. 419 σ 2 ± ˆ Y ² = σ 2 H . Then, σ 2 ± ˆ Y i ² = σ 2 h ii . n X i =1 σ 2 ± ˆ Y i ² = σ 2 n X i =1 h ii = σ 2 p Problem 10.6 Pag. 414 1. Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 3.995e+03 3.378e+02 11.829 5.72e-16 *** x1 9.192e-04 6.312e-04 1.456 0.152 x2 1.212e+01 3.977e+01 0.305 0.762 --- Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 Residual standard error: 248.3 on 49 degrees of freedom Multiple R-squared: 0.04494, Adjusted R-squared: 0.005953 F-statistic: 1.153 on 2 and 49 DF, p-value: 0.3242 2. Added X 1 plot: Figure 1. Added X 2 plot: Figure 2.
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This note was uploaded on 09/01/2011 for the course ISYE 6414 taught by Professor Staff during the Fall '08 term at Georgia Tech.

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lecture8solutions - ISYE 6414 - Summer 2009 Solution...

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