# 1 1 1 x x e e y e β β β β b a portion of the

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1 0 1 ( ) 1 x x e E y e β β β β + + = + b. A portion of the Minitab binary logistic regression output follows: Logistic Regression Table Odds 95% CI Predictor Coef SE Coef Z P Ratio Lower Upper Constant -2.805 1.432 -1.96 0.050 Price 1.1492 0.5143 2.23 0.025 3.16 1.15 8.65 Log-Likelihood = -8.200 Test that all slopes are zero: G = 9.465, DF = 1, P-Value = 0.002 Thus, the estimated logit is ˆ( ) g x = -2.805 + 1.1492 x c. For chocolates that have a price per serving of \$4.00 ˆ(4) g = -2.805 + 1.1492(4) = 1.7918 and ˆ (4) 1.7918 ˆ 1.7918 (4) 6.0002 ˆ 0.86 1 6.0002 1 1 g g e e y e e = = = = + + + d. Using the Minitab output shown in part (b), the estimated odds ratio is 3.16. We can conclude that the estimated odds of having a quality rating of very good or excellent for a chocolate that has a price of \$4.00 per serving is 3.16 times greater than the estimated odds for a chocolate with a price of \$3.00 per serving. Moreover, this interpretation is true for any one dollar difference in the price per serving. 49. a. The expected increase in final college grade point average corresponding to a one point increase in high school grade point average is .0235 when SAT mathematics score does not change. Similarly, the expected increase in final college grade point average corresponding to a one point increase in the SAT mathematics score is .00486 when the high school grade point average does not change. b. ˆ y = -1.41 + .0235(84) + .00486(540) = 3.19 50. a. Job satisfaction can be expected to decrease by 8.69 units with a one unit increase in length of service if the wage rate does not change. A dollar increase in the wage rate is associated with a 13.5 point increase in the job satisfaction score when the length of service does not change. b. ˆ y = 14.4 - 8.69(4) + 13.5(6.5) = 67.39

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Multiple Regression 15 - 25 51. a. The computer output with the missing values filled in is as follows: The regression equation is Y = 8.103 + 7.602 X1 + 3.111 X2 Predictor Coef SE Coef T Constant 8.103 2.667 3.04 X1 7.602 2.105 3.61 X2 3.111 0.613 5.08 S = 3.35 R-sq = 92.3% R-sq (adj) = 91.0% Analysis of Variance SOURCE DF SS MS F Regression 2 1612 806 71.82 Residual Error 12 134,67 11.2225 Total 14 1746.67 b. Using t table (12 degrees of freedom), area in tail corresponding to t = 3.61 is less than .005; p -value is less than .01 Because p -value , α reject H 0 : β 1 = 0 Using t table (12 degrees of freedom), area in tail corresponding to t = 5.08 is less than .005; p -value is less than .01 Because p -value , α reject H 0 : β 2 = 0 c. See computer output. d. 2 14 1 (1 .923) .91 12 a R = = 52. a. The regression equation is Y = -1.41 + .0235 X1 + .00486 X2 Predictor Coef SE Coef T Constant -1.4053 0.4848 -2.90 X1 0.023467 0.008666 2.71 X2 .00486 0.001077 4.51 S = 0.1298 R-sq = 93.7% R-sq (adj) = 91.9% Analysis of Variance SOURCE DF SS MS F Regression 2 1.76209 .881 52.44 Residual Error 7 .1179 .0168 Total 9 1.88000
Chapter 15 15 - 26 b. Using F table (2 degrees of freedom numerator and 7 degrees of freedom denominator), p -value is less than .01 Because p -value , α there is a significant relationship. c. 2 SSR .937 SST R = = 2 9 1 (1 .937) .919 7 a R = = good fit d. t .025 = 2.365 (7 DF) for 1 β : p -value is between .02 and .05; reject H 0 : 1 β = 0 for 2 β : p -value is less than .01; reject H 0 : 2 β = 0 53. a. The regression equation is Y = 14.4 - 8.69 X1 + 13.52 X2 Predictor Coef SE Coef T Constant 14.448 8.191 1.76 X1 -8.69 1.555 -5.59 X2 13.517 2.085 6.48 S = 3.773 R-sq = 90.1% R-sq (adj) = 86.1% Analysis of Variance SOURCE DF SS MS F Regression 2 648.83 324.415 22.79 Residual Error 5 71.17 14.234 Total 7 720.00 b. F .05 = 5.79 (5 DF) F = 22.79 > F .05 ; significant relationship.

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