2014fall-exam1-solutions.pdf - ISyE4031-A Regression and...

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ISyE4031 - A Regression and Forecasting Early Exam 1 Solutions Fall 2014 1. a. 1 ˆ = SS xy / SS xx => SS xx = (25.438)( - 2.33) = - 59.27 b. Since 0 ˆ = y - 1 ˆ x , x = 7.15558/2.33 = 3.071. c. s = 2 s = MSE = ) 2 /( n SSE =1.357 t = ˆ 1 ˆ 1 s = xx SS s / ˆ 1 = 25.438 .357/ 1 2.33 - = 8.66. 2. a. s = 2 s = MSE = 18 / SSE = 0033 . 0 = 0.05749 b. R 2 = 1.1138/1.1733 = 0.9492 or 94.92%. 3. a. Yes. We test H 0 : 1 = 2 = 3 = 4 = 0 vs. H a : at least one is not 0. F (model) = 51.77 from the output and F .05,4,19 = 2.90 (or F .05,4,20 = 2.74). Since 51.77 > 2.90 (or 2.740), we reject H 0 , meaning that the model as a whole is useful. From ANOVA p = 0 (corresponding to F (model)) which is less than    . So we reject H 0 . b. y ˆ = 96 1.82(25) + 0.565(15) + 0.014(625) - 0.0118(225) = 65.07 Prediction error = 68 - 65.07=2.93. c. We test H 0 : 1 = 0 vs. H a : 1 0. Test statistic, | t| = 7.68 > t .025,19 = 2.093 (or t .025,20 = 2.086). So, we reject H 0 which implies “speed” is a significant predictor. Corresponding p -value = 0 < 0.05. So, we reject H 0 . It confirms the conclusion. d. Keep the predictor i in the model if Either significant: p -value < => Reject H 0 : i =0. Or because of the principle of hierarchy.
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