lecture6solutions - Problem 6.8 Pag.249 1. x h = (1 , 5 ,...

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ISYE 6414 - Spring 2009 Solution lecture 5 The Linear Model: Multiple Regression Problem 6.1 Pag.248 In solution lecture 5. .. Problem 6.2 Pag.248 In solution lecture 5. .. Problem 6.3 Pag.248 In solution lecture 5. .. Problem 6.4 Pag.248 In solution lecture 5. .. Problem 6.6 Pag.248 1. H 0 : β 1 = β 2 = 0 vs. H a : { β 1 6 = 0 } ∪ { β 2 6 = 0 } . MSR = 936 . 35, MSE = 7 . 254. F * = MSR MSE = 129 . 083. F (0 . 99 , 2 , 13) = 6 . 70. And given that F * > 6 . 7, we reject the null hypothesis and accept that there is a regression relation. 2. P - value = P ( F (2 , 13) > 129 . 083 ) 0. 3. s ( b 1 ) = 0 . 301 , s ( b 0 ) = 0 . 673. B = t (0 . 9975 , 13) = 3 . 372. 3 . 41 β 1 5 . 44 2 . 106 β 2 6 . 644 4. Ellipsoidal joint confidence area: Using in R: g=lm(y~x1+x2,data=data) v=vcov(g) v=v[2:3,2:3] plot(ellipse(v,centre=c,level=0.99),type="l") The joint confidence area is plotted in figure 1.
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Unformatted text preview: Problem 6.8 Pag.249 1. x h = (1 , 5 , 4). Y h = x T h b = 77 . 275. V ar Y h = MSE x T h ( X T X )-1 x h = 1 . 127 2 Then the 99% condence interval is: 77 . 275 3 . 012 1 . 127 73 . 88 E ( Y h ) 80 . 670 2. V ar Y new = MSE h 1 + x T h ( X T X )-1 x h i = 2 . 919 2 Then the 99% prediction interval is: 77 . 275 3 . 012 2 . 919 68 . 483 ( Y new ) 86 . 067 1 Figure 1: Joint condence area (99%). Problem 6.22 Pag.253 In solution lecture 5. .. Problem 6.23 Pag.253 In solution lecture 5. .. 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 Institute of Technology.

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lecture6solutions - Problem 6.8 Pag.249 1. x h = (1 , 5 ,...

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