# PS9s - < 1 309 of Model B and< 1 341(= √ 1 79743 of...

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SOLUTIONS TO ISMT111 PROBLEM SET #9 (1) (C) Not reject the null hypothesis because b α 1 has a large observed signiﬁcance level of 60.6%. We are testing H o : α 1 = 0 vs.H a : α 1 6 = 0 at the signiﬁcance level of 0.05. (2) Answer = 48,563; b Y = ± - 306 . 75 + (0 . 3452)(900) + (5 . 7565)(55) - (0 . 0055)(900)(55) ² × 1 , 000 (3) (D) Reject the null hypothesis at α = 0 . 10 but not reject the null hypothesis at α = 0 . 05. The null being tested is H o : ‘MILES × SPEED’ is not important given ‘MILES’ and ‘SPEED’. Equivalently, we can test H o : β 3 = 0 (Two-tailed). From Model 2 output, the two-tail p -value is 0.062, which is between 0.05 and 0.10. (4) (D) Between 0.050 to 0.100 We are testing H o : β ≤ - 0 . 010 vs . H a : β > - 0 . 010 i.e. a right-tailed test; The observed test statistic is t obs = b β - β o S b β = - 0 . 0055 - ( - 0 . 010) 0 . 0024 = 1 . 875 with df = 6 The p -value is between 0.05 to 0.10. (5) Model C is a better model because (1) Model C has the highest R-Sq(adj); (2) Model C has the smallest standard error S e = 1 . 267
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Unformatted text preview: , < 1 . 309 of Model B, and < 1 . 341(= √ 1 . 79743) of Model A ); (3) Model C has all its coeﬃcient estimates (the beta hats) signiﬁcant at, say, α = 5%, (i.e., all p-values smaller than 0.05). (6) (a) True. S 2 Y = SST/ ( n-1) = 166 . 364 / 76 = 2 . 189 (b) False. The coeﬃcient of SENIOR in Model B has a p-value of 6.5%, greater than 5%. (c) False. Note: The reciprocal of SENIOR is used in Model C. (d) False. The correct value is √ MSE = √ 1 . 79743 = 1 . 341 . (e) True. = b β 5 ± t . 025 ,df =69 S b β 5 =-. 06391 ± (1 . 994945)(0 . 04800) ≈ (-. 1597 , . 0319) (Note: t . 025 ,df =69 = 1 . 994945 is obtained from Excel. Using z = 1 . 96 would produce a less accurate answer of (-. 15799 , . 03017). I should have set this question (e) with this answer instead.) * * * 1 ATChan 2008...
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