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# exam1solutions - ISYE 6414 - Spring 2009 Solution Exam 1...

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ISYE 6414 - Spring 2009 Solution Exam 1 Problem 1. We know that in this coeﬃcients output table t = estimate SE , then, the complete table is: Coeﬃcient Estimate SE t P( > | t | ) (Intercept) -64.34 23.05267 -2.791 0.005943 Solar.R 0.05934 0.023 2.58 0.013 Wind -3.33 0.65 -5.1231 9.1715 · 10 - 7 Temp 1.6275 0.25 6.51 0 And the estimates linear model is: ˆ Y i = - 64 . 34 + 0 . 05934 X i 1 - 3 . 33 X i 2 + 1 . 6275 X i 3 Where the index 1 , 2 and 3, represent the Solar.R , Wind and Temp variables respectively. In order to obtain the MSE value, we can use the ( X T X ) - 1 matrix, combined witht the previous table. That is: SE ( b 0 ) = 23 . 05267 = p V ar ( b 0 ) = q MSE · ( X T X ) - 1 11 = MSE · 0 . 896 MSE = 23 . 0527 2 0 . 896 = 593 . 1105 Then, if x h = (1 , 190 , 7 . 4 , 67), then: ˆ Y h = x T h b = 31 . 3351 V ar ± ˆ Y h ² = MSE · x T h ( X T X ) - 1 x h = MSE · 0 . 031086 = 18 . 437504 Then, the conﬁdence interval for ˆ Y h is: ˆ Y h ± t 0 . 975 , 149 s ± ˆ Y h ² = 31 .

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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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exam1solutions - ISYE 6414 - Spring 2009 Solution Exam 1...

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