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Unformatted text preview: Homework 6 – solution 1. a) The fitted line equation is y = 27 . 2 . 298 x . The estimated expected value of the strength at carbonation depth of 25.0 mm is ˆ y = 27 . 2 . 298 · 25 . 0 = 19 . 7. b) We test H : β 1 = 0 versus H 1 : β 1 6 = 0. We obtain t = ˆ β 1 se ( ˆ β 1 ) = . 298 . 041 = 7 . 23 < t . 005 , 16 . So the regression is significant. The Pvalue is zero up to 3 decimal points. c) We obtain f = MS R MS E = 52 . 25 > f . 01 , 1 , 16 . So the regression is significant according to this test as well. The Pvalue is zero up to 3 decimal points. d) We test H : β 1 = 320 versus H 1 : β 1 6 = . 320. We obtain t = ˆ β 1 +0 . 320 se ( ˆ β 1 ) = . 022 . 041 = 0 . 54 < t . 025 , 16 . So we can’t reject H and the model has to be deemed valid. e) This time, we test H : β = 28 . 0 versus H 1 : β > 28 . 0. We obtain t = ˆ β 28 . se ( ˆ β ) = . 8 1 . 65 < t . 05 , 16 . So we can’t reject H and, therefore, conclude that the claim is false at α = 0 . 05....
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This note was uploaded on 03/17/2008 for the course IE 121 taught by Professor Perevalov during the Spring '08 term at Lehigh University .
 Spring '08
 Perevalov

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