商务与经济统计习题答案（第8版，中文版）SBE8-SM15

Chapter 15 Multiple Regression Learning Objectives 1. Understand how multiple regression analysis can be used to develop relationships involving one dependent variable and several independent variables. 2. Be able to interpret the coefficients in a multiple regression analysis. 3. Know the assumptions necessary to conduct statistical tests involving the hypothesized regression model. 4. Understand the role of computer packages in performing multiple regression analysis. 5. Be able to interpret and use computer output to develop the estimated regression equation. 6. Be able to determine how good a fit is provided by the estimated regression equation. 7. Be able to test for the significance of the regression equation. 8. Understand how multicollinearity affects multiple regression analysis. 9. Know how residual analysis can be used to make a judgement as to the appropriateness of the model, identify outliers, and determine which observations are influential. Solutions: 1. a. b1 = .5906 is an estimate of the change in y corresponding to a 1 unit change in x1 when x2 is held constant. b2 = .4980 is an estimate of the change in y corresponding to a 1 unit change in x2 when x1 is held constant. 2. a. The estimated regression equation is = 45.06 + 1.94x1 An estimate of y when x1 = 45 is = 45.06 + 1.94(45) = 132.36 b. The estimated regression equation is = 85.22 + 4.32x2 An estimate of y when x2 = 15 is = 85.22 + 4.32(15) = 150.02 c. The estimated regression equation is = -18.37 + 2.01x1 + 4.74x2 An estimate of y when x1 = 45 and x2 = 15 is = -18.37 + 2.01(45) + 4.74(15) = 143.18 3. a. b1 = 3.8 is an estimate of the change in y corresponding to a 1 unit change in x1 when x2, x3, and x4 are held constant. b2 = -2.3 is an estimate of the change in y corresponding to a 1 unit change in x2 when x1, x3, and x4 are held constant. b3 = 7.6 is an estimate of the change in y corresponding to a 1 unit change in x3 when x1, x2, and x4 are held constant. b4 = 2.7 is an estimate of the change in y corresponding to a 1 unit change in x4 when x1, x2, and x3 are held constant. 4. a. = 235 + 10(15) + 8(10) = 255; sales estimate: $255,000 b. Sales can be expected to increase by $10 for every dollar increase in inventory investment when advertising expenditure is held constant. Sales can be expected to increase by $8 for every dollar increase in advertising expenditure when inventory investment is held constant. 5. a. The Minitab output is shown below: The regression equation is Revenue = 88.6 + 1.60 TVAdv Predictor Coef SE Coef T P Constant 88.638 1.582 56.02 0.000 TVAdv 1.6039 0.4778 3.36 0.015 S = 1.215 R-Sq = 65.3% R-Sq(adj) = 59.5% Analysis of Variance Source DF SS MS F P Regression 1 16.640 16.640 11.27 0.015 Residual Error 6 8.860 1.477 Total 7 25.500 b. The Minitab output is shown below: The regression equation is Revenue = 83.2 + 2.29 TVAdv + 1.30 NewsAdv Predictor Coef SE Coef T P Constant 83.230 1.574 52.88 0.000 TVAdv 2.2902 0.3041 7.53 0.001 NewsAdv 1.3010 0.3207 4.06 0.010 S = 0.6426 R-Sq = 91.9% R-Sq(adj) = 88.7% Analysis of Variance Source DF SS MS F P Regression 2 23.435 11.718 28.38 0.002 Residual Error 5 2.065 0.413 Total 7 25.500 Source DF Seq SS TVAdv 1 16.640 NewsAdv 1 6.795 c. No, it is 1.60 in part 2(a) and 2.99 above. In this exercise it represents the marginal change in revenue due to an increase in television advertising with newspaper advertising held constant. d. Revenue = 83.2 + 2.29(3.5) + 1.30(1.8) = $93.56 or $93,560 6. a. The Minitab output is shown below: The regression equation is Speed = 49.8 + 0.0151 Weight Predictor Coef SE Coef T P Constant 49.78 19.11 2.61 0.021 Weight 0.015104 0.006005 2.52 0.025 S = 7.000 R-Sq = 31.1% R-Sq(adj) = 26.2% Analysis of Variance Source DF SS MS F P Regression 1 309.95 309.95 6.33 0.025 Error 14 686.00 49.00 Total 15 995.95 b. The Minitab output is shown below: The regression equation is Speed = 80.5 - 0.00312 Weight + 0.105 Horsepwr Predictor Coef SE Coef T P Constant 80.487 9.139 8.81 0.000 Weight -0.003122 0.003481 -0.90 0.386 Horsepwr 0.10471 0.01331 7.86 0.000 S = 3.027 R-Sq = 88.0% R-Sq(adj) = 86.2% Analysis of Variance Source DF SS MS F P Regression 2 876.80 438.40 47.83 0.000 Residual Error 13 119.15 9.17 Total 15 995.95 7. a. The Minitab output is shown below: The regression equation is Sales = 66.5 + 0.414 Compet$ - 0.270 Heller$ Predictor Coef SE Coef T P Constant 66.52 41.88 1.59 0.156 Compet$ 0.4139 0.2604 1.59 0.156 Heller$ -0.26978 0.08091 -3.33 0.013 S = 18.74 R-Sq = 65.3% R-Sq(adj) = 55.4% Analysis of Variance Source DF SS MS F P Regression 2 4618.8 2309.4 6.58 0.025 Residual Error 7 2457.3 351.0 Total 9 7076.1 b. b1 = .414 is an estimate of the change in the quantity sold (1000s) of the Heller mower with respect to a $1 change in price in competitor’s mower with the price of the Heller mower held constant. b2 = -.270 is an estimate of the change in the quantity sold (1000s) of the Heller mower with respect to a $1 change in its price with the price of the competitor’s mower held constant. c. = 66.5 + 0.414(170) - 0.270(160) = 93.68 or 93,680 units 8. a. The Minitab output is shown below: The regression equation is Return = 247 - 32.8 Safety + 34.6 ExpRatio Predictor Coef SE Coef T P Constant 247.4 110.4 2.24 0.039 Safety -32.84 13.95 -2.35 0.031 ExpRatio 34.59 14.13 2.45 0.026 S = 16.98 R-Sq = 58.2% R-Sq(adj) = 53.3% Analysis of Variance Source DF SS MS F P Regression 2 6823.2 3411.6 11.84 0.001 Residual Error 17 4899.7 288.2 Total 19 11723.0 b. 9. a. The Minitab output is shown below: The regression equation is %College = 26.7 - 1.43 Size + 0.0757 SatScore Predictor Coef SE Coef T P Constant 26.71 51.67 0.52 0.613 Size -1.4298 0.9931 -1.44 0.170 SatScore 0.07574 0.03906 1.94 0.072 S = 12.42 R-Sq = 38.2% R-Sq(adj) = 30.0% Analysis of Variance Source DF SS MS F P Regression 2 1430.4 715.2 4.64 0.027 Residual Error 15 2312.7 154.2 Total 17 3743.1 b. = 26.7 - 1.43(20) + 0.0757(1000) = 73.8 Estimate is 73.8% 10. a. The Minitab output is shown below: The regression equation is Revenue = 33.3 + 7.98 Cars Predictor Coef SE Coef T P Constant 33.34 83.08 0.40 0.695 Cars 7.9840 0.6323 12.63 0.000 S = 226.7 R-Sq = 92.5% R-Sq(adj) = 91.9% Analysis of Variance Source DF SS MS F P Regression 1 8192067 8192067 159.44 0.000 Error 13 667936 51380 Total 14 8860003 b. An increase of 1000 cars in service will result in an increase in revenue of $7.98 million. c. The Minitab output is shown below: The regression equation is Revenue = 106 + 8.94 Cars - 0.191 Location Predictor Coef SE Coef T P Constant 105.97 85.52 1.24 0.239 Cars 8.9427 0.7746 11.55 0.000 Location -0.1914 0.1026 -1.87 0.087 S = 207.7 R-Sq = 94.2% R-Sq(adj) = 93.2% Analysis of Variance Source DF SS MS F P Regression 2 8342186 4171093 96.66 0.000 Error 12 517817 43151 Total 14 8860003 11. a. SSE = SST - SSR = 6,724.125 - 6,216.375 = 507.75 b. c. d. The estimated regression equation provided an excellent fit. 12. a. b. c. Yes; after adjusting for the number of independent variables in the model, we see that 90.5% of the variability in y has been accounted for. 13. a. b. c. The estimated regression equation provided an excellent fit. 14. a. b. c. The adjusted coefficient of determination shows that 68% of the variability has been explained by the two independent variables; thus, we conclude that the model does not explain a large amount of variability. 15. a. b. Multiple regression analysis is preferred since both R2 andshow an increased percentage of the variability of y explained when both independent variables are used. 16. Note: the Minitab output is shown with the solution to Exercise 6. a. No; R-Sq = 31.1% b. Multiple regression analysis is preferred since both R-Sq and R-Sq(adj) show an increased percentage of the variability of y explained when both independent variables are used. 17. a. b. The fit is not very good 18. Note: The Minitab output is shown with the solution to Exercise 10. a. R-Sq = 94.2% R-Sq(adj) = 93.2% b. The fit is very good. 19. a. MSR = SSR/p = 6,216.375/2 = 3,108.188 b. F = MSR/MSE = 3,108.188/72.536 = 42.85 F.05 = 4.74 (2 degrees of freedom numerator and 7 denominator) Since F = 42.85 > F.05 = 4.74 the overall model is significant. c. t = .5906/.0813 = 7.26 t.025 = 2.365 (7 degrees of freedom) Since t = 2.365 > t.025 = 2.365, b1 is significant. d. t = .4980/.0567 = 8.78 Since t = 8.78 > t.025 = 2.365, b2 is significant. 20. A portion of the Minitab output is shown below. The regression equation is Y = - 18.4 + 2.01 X1 + 4.74 X2 Predictor Coef SE Coef T P Constant -18.37 17.97 -1.02 0.341 X1 2.0102 0.2471 8.13 0.000 X2 4.7378 0.9484 5.00 0.002 S = 12.71 R-Sq = 92.6% R-Sq(adj) = 90.4% Analysis of Variance Source DF SS MS F P Regression 2 14052.2 7026.1 43.50 0.000 Residual Error 7 1130.7 161.5 Total 9 15182.9 a. Since the p-value corresponding to F = 43.50 is .000 < a = .05, we reject H0: b1 = b2 = 0; there is a significant relationship. b. Since the p-value corresponding to t = 8.13 is .000 < a = .05, we reject H0: b1 = 0; b1 is significant. c. Since the p-value corresponding to t = 5.00 is .002 < a = .05, we reject H0: b2 = 0; b2 is significant. 21. a. In the two independent variable case the coefficient of x1 represents the expected change in y corresponding to a one unit increase in x1 when x2 is held constant. In the single independent variable case the coefficient of x1 represents the expected change in y corresponding to a one unit increase in x1. b. Yes. If x1 and x2 are correlated one would expect a change in x1 to be accompanied by a change in x2. 22. a. SSE = SST - SSR = 16000 - 12000 = 4000 b. F = MSR/MSE = 6000/571.43 = 10.50 F.05 = 4.74 (2 degrees of freedom numerator and 7 denominator) Since F = 10.50 > F.05 = 4.74, we reject H0. There is a significant relationship among the variables. 23. a. F = 28.38 F.01 = 13.27 (2 degrees of freedom, numerator and 1 denominator) Since F > F.01 = 13.27, reject H0. Alternatively, the p-value of .002 leads to the same conclusion. b. t = 7.53 t.025 = 2.571 Since t > t.025 = 2.571, b1 is significant and x1 should not be dropped from the model. c. t = 4.06 t.025 = 2.571 Since t > t.025 = 2.571, b2 is significant and x2 should not be dropped from the model. 24. Note: The Minitab output is shown in part (b) of Exercise 6 a. F = 47.83 F.05 = 3.81 (2 degrees of freedom numerator and 13 denominator) Since F = 47.83 > F.05 = 3.81, we reject H0: b1 = b2 = 0. Alternatively, since the p-value = .000 < a = .05 we can reject H0. b. For Weight: H0: b1 = 0 Ha: b1 ¹ 0 Since the p-value = 0.386 > a = 0.05, we cannot reject H0 For Horsepower: H0: b2 = 0 Ha: b2 ¹ 0 Since the p-value = 0.000 < a = 0.05, we can reject H0 25. a. The Minitab output is shown below: The regression equation is P/E = 6.04 + 0.692 Profit% + 0.265 Sales% Predictor Coef SE Coef T P Constant 6.038 4.589 1.32 0.211 Profit% 0.6916 0.2133 3.24 0.006 Sales% 0.2648 0.1871 1.42 0.180 S = 5.456 R-Sq = 47.2% R-Sq(adj) = 39.0% Analysis of Variance Source DF SS MS F P Regression 2 345.28 172.64 5.80 0.016 Residual Error 13 387.00 29.77 Total 15 732.28 b. Since the p-value = 0.016 < a = 0.05, there is a significant relationship among the variables. c. For Profit%: Since the p-value = 0.006 < a = 0.05, Profit% is significant. For Sales%: Since the p-value = 0.180 > a = 0.05, Sales% is not significant. 26. Note: The Minitab output is shown with the solution to Exercise 10. a. Since the p-value corresponding to F = 96.66 is 0.000 < a = .05, there is a significant relationship among the variables. b. For Cars: Since the p-value = 0.000 < a = 0.05, Cars is significant c. For Location: Since the p-value = 0.087 > a = 0.05, Location is not significant 27. a. = 29.1270 + .5906(180) + .4980(310) = 289.8150 b. The point estimate for an individual value is = 289.8150, the same as the point estimate of the mean value. 28. a. Using Minitab, the 95% confidence interval is 132.16 to 154.16. b. Using Minitab, the 95% prediction interval is 111.13 to 175.18. 29. a. = 83.2 + 2.29(3.5) + 1.30(1.8) = 93.555 or $93,555 Note: In Exercise 5b, the Minitab output also shows that b0 = 83.230, b1 = 2.2902, and b2 = 1.3010; hence, = 83.230 + 2.2902x1 + 1.3010x2. Using this estimated regression equation, we obtain = 83.230 + 2.2902(3.5) + 1.3010(1.8) = 93.588 or $93,588 The difference ($93,588 - $93,555 = $33) is simply due to the fact that additional significant digits are used in the computations. From a practical point of view, however, the difference is not enough to be concerned about. In practice, a computer software package is always used to perform the computations and this will not be an issue. The Minitab output is shown below: Fit Stdev.Fit 95% C.I. 95% P.I. 93.588 0.291 ( 92.840, 94.335) ( 91.774, 95.401) Note that the value of FIT () is 93.588. b. Confidence interval estimate: 92.840 to 94.335 or $92,840 to $94,335 c. Prediction interval estimate: 91.774 to 95.401 or $91,774 to $95,401 30. a. Since weight is not statistically significant (see Exercise 24), we will use an estimated regression equation which uses only Horsepower to predict the speed at 1/4 mile. The Minitab output is shown below: The regression equation is Speed = 72.6 + 0.0968 Horsepwr Predictor Coef SE Coef T P Constant 72.650 2.655 27.36 0.000 Horsepwr 0.096756 0.009865 9.81 0.000 S = 3.006 R-Sq = 87.3% R-Sq(adj) = 86.4% Analysis of Variance Source DF SS MS F P Regression 1 869.43 869.43 96.21 0.000 Residual Error 14 126.52 9.04 Total 15 995.95 Unusual Observations Obs Horsepwr Speed Fit SE Fit Residual St Resid 2 290 108.000 100.709 0.814 7.291 2.52R 6 450 116.200 116.190 2.036 0.010 0.00 X R denotes an observation with a large standardized residual X denotes an observation whose X value gives it large influence. The output shows that the point estimate is a speed of 101.290 miles per hour. b. The 95% confidence interval is 99.490 to 103.089 miles per hour. c. The 95% prediction interval is 94.596 to 107.984 miles per hour. 31. a. Using Minitab the 95% confidence interval is 58.37% to 75.03%. b. Using Minitab the 95% prediction interval is 35.24% to 90.59%. 32. a. E(y) = b0 + b1 x1 + b2 x2 where x2 = 0 if level 1 and 1 if level 2 b. E(y) = b0 + b1 x1 + b2(0) = b0 + b1 x1 c. E(y) = b0 + b1 x1 + b2(1) = b0 + b1 x1 + b2 d. b2 = E(y | level 2) - E(y | level 1) b1 is the change in E(y) for a 1 unit change in x1 holding x2 constant. 33. a. two b. E(y) = b0 + b1 x1 + b2 x2 + b3 x3 where x2 x3 Level 0 0 1 1 0 2 0 1 3 c. E(y | level 1) = b0 + b1 x1 + b2(0) + b3(0) = b0+ b1 x1 E(y | level 2) = b0 + b1 x1 + b2(1) + b3(0) = b0 + b1 x1 + b2 E(y | level 3) = b0 + b1 x1 + b2(0) + b3(0) = b0 + b1 x1 + b3 b2 = E(y | level 2) - E(y | level 1) b3 = E(y | level 3) - E(y | level 1) b1 is the change in E(y) for a 1 unit change in x1 holding x2 and x3 constant. 34. a. $15,300 b. Estimate of sales = 10.1 - 4.2(2) + 6.8(8) + 15.3(0) = 56.1 or $56,100 c. Estimate of sales = 10.1 - 4.2(1) + 6.8(3) + 15.3(1) = 41.6 or $41,600 35. a. Let Type = 0 if a mechanical repair Type = 1 if an electrical repair The Minitab output is shown below: The regression equation is Time = 3.45 + 0.617 Type Predictor Coef SE Coef T P Constant 3.4500 0.5467 6.31 0.000 Type 0.6167 0.7058 0.87 0.408 S = 1.093 R-Sq = 8.7% R-Sq(adj) = 0.0% Analysis of Variance Source DF SS MS F P Regression 1 0.913 0.913 0.76 0.408 Residual Error 8 9.563 1.195 Total 9 10.476 b. The estimated regression equation did not provide a good fit. In fact, the p-value of .408 shows that the relationship is not significant for any reasonable value of a. c. Person = 0 if Bob Jones performed the service and Person = 1 if Dave Newton performed the service. The Minitab output is shown below: The regression equation is Time = 4.62 - 1.60 Person Predictor Coef SE Coef T P Constant 4.6200 0.3192 14.47 0.000 Person -1.6000 0.4514 -3.54 0.008 S = 0.7138 R-Sq = 61.1% R-Sq(adj) = 56.2% Analysis of Variance Source DF SS MS F P Regression 1 6.4000 6.4000 12.56 0.008 Residual Error 8 4.0760 0.5095 Total 9 10.4760 d. We see that 61.1% of the variability in repair time has been explained by the repair person that performed the service; an acceptable, but not good, fit. 36. a. The Minitab output is shown below: The regression equation is Time = 1.86 + 0.291 Months + 1.10 Type - 0.609 Person Predictor Coef SE Coef T P Constant 1.8602 0.7286 2.55 0.043 Months 0.29144 0.08360 3.49 0.013 Type 1.1024 0.3033 3.63 0.011 Person -0.6091 0.3879 -1.57 0.167 S = 0.4174 R-Sq = 90.0% R-Sq(adj) = 85.0% Analysis of Variance Source DF SS MS F P Regression 3 9.4305 3.1435 18.04 0.002 Residual Error 6 1.0455 0.1743 Total 9 10.4760 b. Since the p-value corresponding to F = 18.04 is .002 < a = .05, the overall model is statistically significant. c. The p-value corresponding to t = -1.57 is .167 > a = .05; thus, the addition of Person is not statistically significant. Person is highly correlated with Months (the sample correlation coefficient is -.691); thus, once the effect of Months has been accounted for, Person will not add much to the model. 37. a. Let Position = 0 if a guard Position = 1 if an offensive tackle. b. The Minitab output is shown below: The regression equation is Rating = 11.2 + 0.732 Position + 0.0222 Weight - 2.28 Speed Predictor Coef SE Coef T P Constant 11.223 4.523 2.48 0.022 Position 0.7324 0.2893 2.53 0.019 Weight 0.02219 0.01039 2.14 0.045 Speed -2.2775 0.9290 -2.45 0.023 S = 0.6936 R-Sq = 47.5% R-Sq(adj) = 40.1% Analysis of Variance Source DF SS MS F P Regression 3 9.1562 3.0521 6.35 0.003 Residual Error 21 10.1014 0.4810 Total 24 19.2576 c. Since the p-value corresponding to F = 6.35 is .003 < a = .05, there is a significant relationship between rating and the independent variables. d. The value of R-Sq (adj) is 40.1%; the estimated regression equation did not provide a very good fit. e. Since the p-value for Position is t = 2.53 < a = .05, position is a significant factor in the player’s rating. f. 38. a. The Minitab output is shown below: The regression equation is Risk = - 91.8 + 1.08 Age + 0.252 Pressure + 8.74 Smoker Predictor Coef SE Coef T P Constant -91.76 15.22 -6.03 0.000 Age 1.0767 0.1660 6.49 0.000 Pressure 0.25181 0.04523 5.57 0.000 Smoker 8.740 3.001 2.91 0.010 S = 5.757 R-Sq = 87.3% R-Sq(adj) = 85.0% Analysis of Variance Source DF SS MS F P Regression 3 3660.7 1220.2 36.82 0.000 Residual Error 16 530.2 33.1 Total 19 4190.9 b. Since the p-value corresponding to t = 2.91 is .010 < a = .05, smoking is a significant factor. c. Using Minitab, the point estimate is 34.27; the 95% prediction interval is 21.35 to 47.18. Thus, the probability of a stroke (.2135 to .4718 at the 95% confidence level) appears to be quite high. The physician would probably recommend that Art quit smoking and begin some type of treatment designed to reduce his blood pressure. 39. a. The Minitab output is shown below: The regression equation is Y = 0.20 + 2.60 X Predictor Coef SE Coef T P Constant 0.200 2.132 0.09 0.931 X 2.6000 0.6429 4.04 0.027 S = 2.033 R-Sq = 84.5% R-Sq(adj) = 79.3% Analysis of Variance Source DF SS MS F P Regression 1 67.600 67.600 16.35 0.027 Residual Error 3 12.400 4.133 Total 4 80.000 b. Using Minitab we obtained the following values: xi yi Standardized Residual 1 3 2.8 .16 2 7 5.4 .94 3 5 8.0 -1.65 4 11 10.6 .24 5 14 13.2 .62 The point (3,5) does not appear to follow the trend of remaining data; however, the value of the standardized residual for this point, -1.65, is not large enough for us to conclude that (3, 5) is an outlier. c. Using Minitab, we obtained the following values: xi yi Studentized Deleted Residual 1 3 .13 2 7 .91 3 5 - 4.42 4 11 .19 5 14 .54 t.025 = 4.303 (n - p - 2 = 5 - 1 - 2 = 2 degrees of freedom) Since the studentized deleted residual for (3, 5) is -4.42 < -4.303, we conclude that the 3rd observation is an outlier. 40. a. The Minitab output is shown below: The regression equation is Y = -53.3 + 3.11 X Predicator Coef Stdev t-ratio p Constant -53.280 5.786 -9.21 0.003 X 3.1100 0.2016 15.43 0.001 s = 2.851 R-sq = 98.8% R-sq (adj) = 98.3% Analysis of Variance SOURCE DF SS MS F p Regression 1 1934.4 1934.4 238.03 0.001 Error 3 24.4 8.1 Total 4 1598.8 b. Using the Minitab we obtained the following values: xi yi Studentized Deleted Residual 22 12 -1.94 24 21 -.12 26 31 1.79 28 35 .40 40 70 -1.90 t.025 = 4.303 (n - p - 2 = 5 - 1 - 2 = 2 degrees of freedom) Since none of the studentized deleted residuals are less than -4.303 or greater than 4.303, none of the observations can be classified as an outlier. c. Using Minitab we obtained the following values: xi yi hi 22 12 .38 24 21 .28 26 31 .22 28 35 .20 40 70 .92 The critical value is Since none of the values exceed 1.2, we conclude that there are no influential observations in the data. d. Using Minitab we obtained the following values: xi yi Di 22 12 .60 24 21 .00 26 31 .26 28 35 .03 40 70 11.09 Since D5 = 11.09 > 1 (rule of thumb critical value), we conclude that the fifth observation is influential. 41. a. The Minitab output appears in the solution to part (b) of Exercise 5; the estimated regression equation is: Revenue = 83.2 + 2.29 TVAdv + 1.30 NewsAdv b. Using Minitab we obtained the following values: Standardized Residual 96.63 -1.62 90.41 -1.08 94.34 1.22 92.21 - .37 94.39 1.10 94.24 - .40 94.42 -1.12 93.35 1.08 With the relatively few observations, it is difficult to determine if any of the assumptions regarding the error term have been violated. For instance, an argument could be made that there does not appear to be any pattern in the plot; alternatively an argument could be made that there is a curvilinear pattern in the plot. c. The values of the standardized residuals are greater than -2 and less than +2; thus, using test, there are no outliers. As a further check for outliers, we used Minitab to compute the following studentized deleted residuals: Observation Studentized Deleted Residual 1 -2.11 2 -1.10 3 1.31 4 - .33 5 1.13 6 - .36 7 -1.16 8 1.10 t.025 = 2.776 (n - p - 2 = 8 - 2 - 2 = 4 degrees of freedom) Since none of the studentized deleted residuals is less tan -2.776 or greater than 2.776, we conclude that there are no outliers in the data. d. Using Minitab we obtained the following values: Observation hi Di 1 .63 1.52 2 .65 .70 3 .30 .22 4 .23 .01 5 .26 .14 6 .14 .01 7 .66 .81 8 .13 .06 The critical average value is Since none of the values exceed 1.125, we conclude that there are no influential observations. However, using Cook’s distance measure, we see that D1 > 1 (rule of thumb critical value); thus, we conclude the first observation is influential. Final Conclusion: observations 1 is an influential observation. 42. a. The Minitab output is shown below: The regression equation is Speed = 71.3 + 0.107 Price + 0.0845 Horsepwr Predictor Coef SE Coef T P Constant 71.328 2.248 31.73 0.000 Price 0.10719 0.03918 2.74 0.017 Horsepwr 0.084496 0.009306 9.08 0.000 S = 2.485 R-Sq = 91.9% R-Sq(adj) = 90.7% Analysis of Variance Source DF SS MS F P Regression 2 915.66 457.83 74.12 0.000 Residual Error 13 80.30 6.18 Total 15 995.95 Source DF Seq SS Price 1 406.39 Horsepwr 1 509.27 Unusual Observations Obs Price Speed Fit SE Fit Residual St Resid 2 93.8 108.000 105.882 2.007 2.118 1.45 X X denotes an observation whose X value gives it large influence. b. The standardized residual plot is shown below. There appears to be a very unusual trend in the standardized residuals. - - xx x 1.2+ - x SRES1 - x - x - x 0.0+ x x - x - x x x - - -1.2+ x - - x - x - --------+---------+---------+---------+---------+--------FITS1 90.0 96.0 102.0 108.0 114.0 c. The Minitab output shown in part (a) did not identify any observations with a large standardized residual; thus, there does not appear to be any outliers in the data. d. The Minitab output shown in part (a) identifies observation 2 as an influential observation. 43. a. The Minitab output is shown below: The regression equation is %College = - 26.6 + 0.0970 SatScore Predictor Coef SE Coef T P Constant -26.61 37.22 -0.72 0.485 SatScore 0.09703 0.03734 2.60 0.019 S = 12.83 R-Sq = 29.7% R-Sq(adj) = 25.3% Analysis of Variance Source DF SS MS F P Regression 1 1110.8 1110.8 6.75 0.019 Residual Error 16 2632.3 164.5 Total 17 3743.1 Unusual Observations Obs SatScore %College Fit SE Fit Residual St Resid 3 716 40.00 42.86 10.79 -2.86 -0.41 X X denotes an observation whose X value gives it large influence. b. The Minitab output shown in part a identifies observation 3 as an influential observation. c. The Minitab output appears in the solution to Exercise 9; the estimates regression equation is %College = 26.7 - 1.43 Size + 0.0757 SATScore d. The following Minitab output was also provided as part of the regression output for part c. Unusual Observations Obs. Size %College Fit Stdev.Fit Residual St.Resid 3 30.0 40.0 38.04 10.97 1.96 0.34 X X denotes an obs. whose X value gives it large influence. Observation 3 is still identified as an influential observation. 44. 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. = -1.41 + .0235(84) + .00486(540) = 3.19 45. 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. = 14.4 - 8.69(4) + 13.5(6.5) = 67.39 46. 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 Predicator Coef Stdev t-ratio 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 Error 12 134,67 11.2225 Total 14 1746.67 b. t.025 = 2.179 (12 DF) for b1: 3.61 > 2.179; reject H0 : b1 = 0 for b2: 5.08 > 2.179; reject H0 : b2 = 0 c. See computer output. d. 47. a. The regression equation is Y = -1.41 + .0235 X1 + .00486 X2 Predictor Coef Stdev t-ratio 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 Error 7 .1179 .0168 Total 9 1.88000 b. F.05 = 4.74 (2 DF numerator, 7 DF denominator) F = 52.44 > F.05; significant relationship. c. good fit d. t.025 = 2.365 (7 DF) for B1: t = 2.71 > 2.365; reject H0 : B1 = 0 for B2: t = 4.51 > 2.365; reject H0 : B2 = 0 48. a. The regression equation is Y = 14.4 - 8.69 X1 + 13.52 X2 Predictor Coef Stdev t-ratio 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 Error 5 71.17 14.234 Total 7 720.00 b. F.05 = 5.79 (5 DF) F = 22.79 > F.05; significant relationship. c. good fit d. t.025 = 2.571 (5 DF) for b1: t = -5.59 < -2.571; reject H0 : b1 = 0 for b2: t = 6.48 > 2.571; reject H0 : b2 = 0 49. a. The Minitab output is shown below: The regression equation is Price = 12.8 + 2.26 BookVal Predictor Coef SE Coef T P Constant 12.793 6.624 1.93 0.064 BookVal 2.2649 0.6631 3.42 0.002 S = 19.50 R-Sq = 29.4% R-Sq(adj) = 26.9% Analysis of Variance Source DF SS MS F P Regression 1 4433.9 4433.9 11.67 0.002 Error 28 10642.3 380.1 Total 29 15076.1 b. The value of R-sq is 29.4%; the estimated regression equation does not provide a good fit. c. The Minitab output is shown below: The regression equation is Price = 5.88 + 2.54 BookVal + 0.484 ReturnEq Predictor Coef SE Coef T P Constant 5.877 5.545 1.06 0.299 BookVal 2.5356 0.5331 4.76 0.000 ReturnEq 0.4841 0.1174 4.12 0.000 S = 15.55 R-Sq = 56.7% R-Sq(adj) = 53.5% Analysis of Variance Source DF SS MS F P Regression 2 8544.2 4272.1 17.66 0.000 Error 27 6531.9 241.9 Total 29 15076.1 Since the p-value corresponding to the F test is 0.000, the relationship is significant. 50. a. The Minitab output is shown below: The regression equation is Speed = 97.6 + 0.0693 Price - 0.00082 Weight + 0.0590 Horsepwr - 2.48 Zero60 Predictor Coef SE Coef T P Constant 97.57 11.79 8.27 0.000 Price 0.06928 0.03805 1.82 0.096 Weight -0.000816 0.002593 -0.31 0.759 Horsepwr 0.05901 0.01543 3.82 0.003 Zero60 -2.4836 0.9601 -2.59 0.025 S = 2.127 R-Sq = 95.0% R-Sq(adj) = 93.2% Analysis of Variance Source DF SS MS F P Regression 4 946.18 236.55 52.28 0.000 Residual Error 11 49.77 4.52 Total 15 995.95 b. Since the p-value corresponding to the F test is 0.000, the relationship is significant. c. Since the p-values corresponding to the t test for both Horsepwr (p-value = .003) and Zero60 (p-value = .025) are less than .05, both of these independent variables are significant. d. The Minitab output is shown below: The regression equation is Speed = 103 + 0.0558 Horsepwr - 3.19 Zero60 Predictor Coef SE Coef T P Constant 103.103 9.448 10.91 0.000 Horsepwr 0.05582 0.01452 3.84 0.002 Zero60 -3.1876 0.9658 -3.30 0.006 S = 2.301 R-Sq = 93.1% R-Sq(adj) = 92.0% Analysis of Variance Source DF SS MS F P Regression 2 927.12 463.56 87.54 0.000 Residual Error 13 68.84 5.30 Total 15 995.95 Source DF Seq SS Horsepwr 1 869.43 Zero60 1 57.68 Unusual Observations Obs Horsepwr Speed Fit SE Fit Residual St Resid 2 290 108.000 103.352 1.015 4.648 2.25R 12 155 84.600 82.747 1.773 1.853 1.26 X R denotes an observation with a large standardized residual X denotes an observation whose X value gives it large influence. e. The standardized residual plot is shown below: - SRES - x - - 1.5+ - x x - - - 2 x x 0.0+ x x 2 - x - - xx - -1.5+ - x x - ----+---------+---------+---------+---------+---------+--FIT 84.0 90.0 96.0 102.0 108.0 114.0 There is an unusual trend in the plot and one observation appears to be an outlier. f. The Minitab output indicates that observation 2 is an outlier g. The Minitab output indicates that observation 12 is an influential observation. 51. a. The Minitab output is shown below: 640+ - x Exposure- - - 480+ - x - - - x 320+ - - - - x 160+ x 3 x - x ------+---------+---------+---------+---------+---------+TimesAir 15 30 45 60 75 90 b. The Minitab output is shown below: The regression equation is Exposure = 53.2 + 6.74 TimesAir Predictor Coef SE Coef T P Constant 53.24 16.53 3.22 0.012 TimesAir 6.7427 0.4472 15.08 0.000 S = 31.70 R-Sq = 96.6% R-Sq(adj) = 96.2% Analysis of Variance Source DF SS MS F P Regression 1 228520 228520 227.36 0.000 Error 8 8041 1005 Total 9 236561 Since the p-value is 0.000, the relationship is significant. c. The Minitab output is shown below: The regression equation is Exposure = 73.1 + 5.04 TimesAir + 101 BigAds Predictor Coef SE Coef T P Constant 73.063 7.507 9.73 0.000 TimesAir 5.0368 0.3268 15.41 0.000 BigAds 101.11 15.99 6.32 0.000 S = 13.08 R-Sq = 99.5% R-Sq(adj) = 99.3% Analysis of Variance Source DF SS MS F P Regression 2 235363 117682 687.84 0.000 Error 7 1198 171 Total 9 236561 d. The p-value corresponding to the t test for BigAds is 0.000; thus, the dummy variable is significant. e. The dummy variable enables us to fit two different lines to the data; this approach is referred to as piecewise linear approximation. 52. a. The Minitab output is shown below: Resale% = 38.8 +0.000766 Price Predictor Coef SE Coef T P Constant 38.772 4.348 8.92 0.000 Price 0.0007656 0.0001900 4.03 0.000 S = 5.421 R-Sq = 36.7% R-Sq(adj) = 34.4% Analysis of Variance Source DF SS MS F P Regression 1 477.25 477.25 16.24 0.000 Residual Error 28 822.92 29.39 Total 29 1300.17 Since the p-value corresponding to F = 16.24 is .000 < a = .05, there is a significant relationship between Resale% and Price. b. R-Sq = 36.7%; not a very good fit. c. Let Type1 = 0 and Type2 = 0 if a small pickup; Type1 = 1 and Type2 = 0 if a full-size pickup; and Type1 = 0 and Type2 = 1 if a sport utility. The Minitab output using Type1, Type2, and Price is shown below: The regression equation is Resale% = 42.6 + 9.09 Type1 + 7.92 Type2 +0.000341 Price Predictor Coef SE Coef T P Constant 42.554 3.562 11.95 0.000 Type1 9.090 2.248 4.04 0.000 Type2 7.917 2.163 3.66 0.001 Price 0.0003415 0.0001800 1.90 0.069 S = 4.298 R-Sq = 63.1% R-Sq(adj) = 58.8% Analysis of Variance Source DF SS MS F P Regression 3 819.77 273.26 14.79 0.000 Residual Error 26 480.40 18.48 Total 29 1300.17 d. Since the p-value corresponding to F = 14.79 is .000 < a = .05, there is a significant relationship between Resale% and the independent variables. Note that individually, Price is not significant at the .05 level of significance. If we rerun the regression using just Type1 and Type2 the value of R-Sq (adj) decreases to 54.4%, a drop of only 4%. Thus, it appears that for these data, the type of vehicle is the strongest predictor of the resale value.