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lecture3solutions

# lecture3solutions - ISYE 6414 Spring 2009 Solution lecture...

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ISYE 6414 - Spring 2009 Solution lecture 3 - The Linear Model: OLS Regression Problem 2.1 Pag.89 1. The student conclusion is supported by the data with a 0.05 level of signiﬁcance. 2. The intercept helps to deﬁne the linear fucntion inside the range of the predictor X , but X = 0 is outside this range, so it’s interpretation is not valid. Problem 2.2 Pag.89 No, if β 1 < 0 there exits a negative linear relation between Y and X . Problem 2.3 Pag.90 The slope is not signiﬁcant diﬀerent than zero. The statement is not valid. Problem 2.4 Pag.90 1. We know that t { 0 . 995 , 118 } = 2 . 61814. Then, the cvonﬁdence interval (99%) is: 0 . 03883 ± 2 . 61814(0 . 01277) = [0 . 0054 ; 0 . 07226] And the zero is not in the interval. 2. H 0 : β 1 = 0 vs. H a : β 1 6 = 0. The test statistic is t * = 0 . 03883 0 . 01270 = 3 . 04072. Due to | t * | > t { 0 . 995 , 118 } = 2 . 61814, we reject H 0 and conclude H a . 3. P - value = 2 · P ( t { 118 } > 3 . 04072 ) = 0 . 00291 Problem 2.5 Pag.90 1. t { 0 . 95 , 43 } = 1 . 6811. The 90% conﬁdence interval for β 1 is: 15 . 0352 ± 1 . 6811(0 . 4831) = [14 . 2231 ; 15 . 8473] 2. H 0 : β 1 = 0 vs. H a : β 1 6 = 0. The test statistic is t * = 15 . 0352 0 . 4831 = 31 . 122. Due to | t * | > t { 0 . 95 , 43 } = 1 . 6811, we reject H 0 and conclude H a . P - value = 2 · P ( t { 43 } > 31 . 122 ) 0 3. Yes, they are consistent because the interval in part (1) does not contain the zero. 4. H 0 : β 1 14 vs. H a : β 1 > 14. The test statistic is t * = 15 . 0352 - 14 0 . 4831 = 2 . 1428. Due to | t * | > t { 0 . 95 , 43 } = 1 . 6811, we reject H 0 and conclude H a . P - value = 2 · P ( t { 43 } > 2 . 1428 ) = 0 . 0189 5. No Problem 2.18 Pag.92 Because with the t test we can contrast any kind of hypothesis about the value of the parameter β 1 ; but with the F test, we just can test H 0 : β 1 = 0 vs. H a : β 1 6 = 0. 1

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Problem 2.19 Pag.92 H 0 : β 1 = 0 vs. H a : β 1 6 = 0. The testing statistic is F = MSR MSE . But E ( MSE ) = σ 2 and E ( MSR ) = σ 2 + β 2 1 S XX . This implies that if H 0 is true E ( F ) = 1, but if H a is true E ( F ) > 1. Problem 2.20 Pag.92 No, there is no parameter in the model being estimated by R 2 Problem 2.21 Pag.92 No Problem 2.22 Pag.92 Could R 2 be zero for the ﬁrst 10 cases and non-zero for the 30 cases? Yes.
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lecture3solutions - ISYE 6414 Spring 2009 Solution lecture...

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