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Spring 2010 Midterm 2 Form A Solutions

Spring 2010 Midterm 2 Form A Solutions - Midterm 2 Form A...

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Midterm 2 - Form A - Spring 2010 Economics 203 Instructors: Petry and Sahakyan Name: 1. B. A positive residual is more likely to follow a positive residual when positive autocorrelation is present. Similarly, a negative residual is more likely to follow a negative residual when positive autocorrelation is present. In contrast, negative autocorrelation is characterized by adjacent residuals being likely to be of opposite signs. 2. D. R 2 of 0 means your SSR is 0 (with SST =50,000). That means your F -statistic is 0, since it has SSR on top and something positive on bottom. 3. A. The pvalues for estimated coefficients are always two-tailed ( 6 =). The pvalue of 0.000402 is for the estimated coefficient b 1 yet hypotheses are always on the population parameters, thus β 1 . 4. A. The test implied by the Excel output is the two-sided test. To get the p-value for the one-sided test, we halve the p-value from the two-sided test. 5. C. The predicted sales is ˆ SALES = 8 . 7+5(4 . 2) = 29 . 7. The appropriate critical value is t 0 . 025 , 8 = 2 . 31. This gives everything that we need to put into the confidence interval formula: ˆ SALES ± t α/ 2 s ε s 1 /n + ( x g - ¯ x ) 2 ( n - 1) s 2 x = 29 . 7 ± (2 . 31)(3 . 234578798) s 1 / 10 + (5 - 5) 2 (10 - 1)2 . 22 = 29 . 7 ± (2 . 31)(3 . 234578798) p 1 / 10 = (27 . 337 , 32 . 063) 6. C. The SST for model 1 is SSE 1 + SSR 1 = 432 + 123 = 555. Since the models are applied to the same data, SST 2 = 555 also. Then SSE 2 = SST 2 - SSR 2 = 555 - 100 = 455. 7. C. ¯ R 2 1 = 1 - SSE 1 / ( n 1 - k 1 - 1) SST 1 / ( n 1 - 1) = 1 - 432 / (200 - 5 - 1) 555 / (200 - 1) = 0 . 2016 ¯ R 2 2 = 1 - SSE 2 / ( n 2 - k 2 - 1) SST 2 / ( n 2 - 1) = 1 - 455 / (200 - 3 - 1) 555 / (200 - 1) = 0 . 1676 8. C. The partial F statistic is: F = ( SSR f - SSR r ) /k d MSE f = ( SSR 1 - SSR 2 ) / 2 MSE 1 = (123 - 100) / 2 432 / (200 - 5 - 1) = 5 . 164 9. A. The divisor in the numerator is 2, so the numerator (first) degrees of freedom is 2. The divisor in the denominator is 194, so the denominator (second) degrees of freedom is 194. 10. D. The partial F test statistic of 5.164 is more extreme than the critical value of 3.042. Thus reject the null hypothesis. You have found at least one of the two variables deleted significanty explains the dependent variable. Thus the full model is preferred to the reduced model.

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