商务与经济统计习题答案（第8版中文版）SBE8-SM16.doc

Regression Analysis: Model BuildingChapter 16Regression Analysis: Model BuildingLearning Objectives1.Learn how the general linear model can be used to model problems involving curvilinear relationships.2.Understand the concept of interaction and how it can be accounted for in the general linear model.3.Understand how an F test can be used to determine when to add or delete one or more variables.4.Develop an appreciation for the complexities involved in solving larger regression analysis problems.5.Understand how variable selection procedures can be used to choose a set of independent variables for an estimated regression equation.6.Know how the Durban-Watson test can be used to test for autocorrelation.7.Learn how analysis of variance and experimental design problems can be analyzed using a regression model.Solutions:1.a.The Minitab output is shown below: The regression equation isY = - 6.8 + 1.23 X Predictor Coef Stdev t-ratio pConstant -6.77 14.17 -0.48 0.658X 1.2296 0.4697 2.62 0.059 s = 7.269 R-sq = 63.1% R-sq(adj) = 53.9% Analysis of Variance SOURCE DF SS MS F pRegression 1 362.13 362.13 6.85 0.059Error 4 211.37 52.84Total 5 573.50 b.Since the p-value corresponding to F = 6.85 is 0.59 > a = .05, the relationship is not significant.c. - 40+ * - Y - * * - * - 30+ - - - - * 20+ - - - - * 10+ -+-+-+-+-+-+X 20.0 25.0 30.0 35.0 40.0 45.0The scatter diagram suggests that a curvilinear relationship may be appropriate.d.The Minitab output is shown below:The regression equation isY = - 169 + 12.2 X - 0.177 XSQ Predictor Coef Stdev t-ratio pConstant -168.88 39.79 -4.24 0.024X 12.187 2.663 4.58 0.020XSQ -0.17704 0.04290 -4.13 0.026 s = 3.248 R-sq = 94.5% R-sq(adj) = 90.8% Analysis of Variance SOURCE DF SS MS F pRegression 2 541.85 270.92 25.68 0.013Error 3 31.65 10.55Total 5 573.50 e. Since the p-value corresponding to F = 25.68 is .013 < a = .05, the relationship is significant.f. = -168.88 + 12.187(25) - 0.17704(25)2 = 25.145 2.a.The Minitab output is shown below:The regression equation isY = 9.32 + 0.424 X Predictor Coef Stdev t-ratio pConstant 9.315 4.196 2.22 0.113X 0.4242 0.1944 2.18 0.117 s = 3.531 R-sq = 61.4% R-sq(adj) = 48.5% Analysis of Variance SOURCE DF SS MS F pRegression 1 59.39 59.39 4.76 0.117Error 3 37.41 12.47Total 4 96.80 The high p-value (.117) indicates a weak relationship; note that 61.4% of the variability in y has been explained by x.b.The Minitab output is shown below:The regression equation isY = - 8.10 + 2.41 X - 0.0480 XSQ Predictor Coef Stdev t-ratio pConstant -8.101 4.104 -1.97 0.187X 2.4127 0.4409 5.47 0.032XSQ -0.04797 0.01050 -4.57 0.045 s = 1.279 R-sq = 96.6% R-sq(adj) = 93.2% Analysis of Variance SOURCE DF SS MS F pRegression 2 93.529 46.765 28.60 0.034Error 2 3.271 1.635Total 4 96.800 At the .05 level of significance, the relationship is significant; the fit is excellent.c.= -8.101 + 2.4127(20) - 0.04797(20)2 = 20.965 3.a.The scatter diagram shows some evidence of a possible linear relationship.b.The Minitab output is shown below:The regression equation isY = 2.32 + 0.637 X Predictor Coef Stdev t-ratio pConstant 2.322 1.887 1.23 0.258X 0.6366 0.3044 2.09 0.075 s = 2.054 R-sq = 38.5% R-sq(adj) = 29.7% Analysis of Variance SOURCE DF SS MS F pRegression 1 18.461 18.461 4.37 0.075Error 7 29.539 4.220Total 8 48.000 c.The following standardized residual plot indicates that the constant variance assumption is not satisfied. - - * - 1.2+ * - - - * - * * 0.0+ - - * * - - -1.2+ - * * - - +-+-+-+-+-+-YHAT 3.0 4.0 5.0 6.0 7.0 8.0 d.The logarithmic transformation does not appear to eliminate the wedged-shaped pattern in the above residual plot. The reciprocal transformation does, however, remove the wedge-shaped pattern. Neither transformation provides a good fit. The Minitab output for the reciprocal transformation and the corresponding standardized residual pot are shown below.The regression equation is1/Y = 0.275 - 0.0152 X Predictor Coef Stdev t-ratio pConstant 0.27498 0.04601 5.98 0.000X -0.015182 0.007421 -2.05 0.080 s = 0.05009 R-sq = 37.4% R-sq(adj) = 28.5%Analysis of Variance SOURCE DF SS MS F pRegression 1 0.010501 0.010501 4.19 0.080Error 7 0.017563 0.002509Total 8 0.028064 - * - - - 1.0+ * - * - - - * 0.0+ * - - - * * - -1.0+ - * * - -+-+-+-+-+-+-YHAT 0.140 0.160 0.180 0.200 0.220 0.240 4.a.The Minitab output is shown below: The regression equation isY = 943 + 8.71 X Predictor Coef Stdev t-ratio pConstant 943.05 59.38 15.88 0.000X 8.714 1.544 5.64 0.005 s = 32.29 R-sq = 88.8% R-sq(adj) = 86.1% Analysis of Variance SOURCE DF SS MS F pRegression 1 33223 33223 31.86 0.005Error 4 4172 1043Total 5 37395 b.The p-value of .005 < a = .01; reject H05.The Minitab output is shown below:The regression equation isY = 433 + 37.4 X - 0.383 1/Y Predictor Coef Stdev t-ratio pConstant 432.6 141.2 3.06 0.055X 37.429 7.807 4.79 0.0171/Y -0.3829 0.1036 -3.70 0.034 s = 15.83 R-sq = 98.0% R-sq(adj) = 96.7% Analysis of Variance SOURCE DF SS MS F pRegression 2 36643 18322 73.15 0.003Error 3 751 250Total 5 37395 b.Since the linear relationship was significant (Exercise 4), this relationship must be significant. Note also that since the p-value of .005 < a = .05, we can reject H0.c.The fitted value is 1302.01, with a standard deviation of 9.93. The 95% confidence interval is 1270.41 to 1333.61; the 95% prediction interval is 1242.55 to 1361.47.6.a.The scatter diagram is shown below: - * 1.60+ - DISTANCE- - - 1.20+ - * - - - 0.80+ * - - * - - * * 0.40+ +-+-+-+-+-+-NUMBER 8.0 12.0 16.0 20.0 24.0 28.0 b.No; the relationship appears to be curvilinear.c.Several possible models can be fitted to these data, as shown below:= 2.90 - 0.185x + .00351x2 7.a.The Minitab output is shown below: - 36+ x - Shipment- - - 24+ - x - - x - x 12+ - - x x - 2x x - 0+ +-+-+-+-+-+-Media$ 0 25 50 75 100 125b.The Minitab output is shown below:The regression equation isShipment = 4.09 + 0.196 Media$Predictor Coef SE Coef T PConstant 4.089 2.168 1.89 0.096Media$ 0.19552 0.03635 5.38 0.000S = 5.044 R-Sq = 78.3% R-Sq(adj) = 75.6%Analysis of VarianceSource DF SS MS F PRegression 1 735.84 735.84 28.93 0.000Error 8 203.51 25.44Total 9 939.35Unusual ObservationsObs Media$ Shipment Fit StDev Fit Residual St Resid 1 120 36.30 27.55 3.30 8.75 2.30R R denotes an observation with a large standardized residualSimple linear regression appears to do good job in explaining the variability in shipments. However, the scatter diagram in part (a) indicates that a curvilinear relationship may be more appropriate.c.The Minitab output is shown below: The regression equation is Shipment = 5.51 + 0.00182 Media$Sq Predictor Coef SE Coef T P Constant 5.506 1.686 3.27 0.011 Media$Sq 0.0018225 0.0002792 6.53 0.000 S = 4.308 R-Sq = 84.2% R-Sq(adj) = 82.2% Analysis of Variance Source DF SS MS F P Regression 1 790.88 790.88 42.62 0.000 Error 8 148.47 18.56 Total 9 939.35 Unusual Observations Obs Media$Sq Shipment Fit StDev Fit Residual St Resid 3 10020 15.90 23.77 2.26 -7.87 -2.15R R denotes an observation with a large standardized residual8.a.The scatter diagram is shown below:It appears that a simple linear regression model is not appropriate because there is curvature in the plot.b.The Minitab output is shown below:The regression equation is2005% = 17.1 + 3.15 1999% - 0.0445 1999%SqPredictor Coef SE Coef T PConstant 17.099 4.639 3.69 0.0031999% 3.1462 0.4971 6.33 0.0001999%Sq -0.04454 0.01018 -4.37 0.001S = 5.667 R-Sq = 89.7% R-Sq(adj) = 88.1%Analysis of VarianceSource DF SS MS F PRegression 2 3646.3 1823.2 56.78 0.000Residual Error 13 417.4 32.1Total 15 4063.8c.The Minitab output is shown below:The regression equation isLog2000% = 1.17 + 0.449 Log1999%Predictor Coef SE Coef T PConstant 1.17420 0.07468 15.72 0.000Log1999% 0.44895 0.05978 7.51 0.000S = 0.08011 R-Sq = 80.1% R-Sq(adj) = 78.7%Analysis of VarianceSource DF SS MS F PRegression 1 0.36199 0.36199 56.40 0.000Residual Error 14 0.08985 0.00642Total 15 0.45184d.The estimated regression in part (b) is preferred because it explains a higher percentage of the variability in the dependent variable.9.a.The Minitab output is shown below:The regression equation isDistance = 268 + 1.52 Wind - 0.0177 WindSqPredictor Coef SE Coef T PConstant 267.841 2.358 113.60 0.000Wind 1.52143 0.07718 19.71 0.000WindSq -0.017698 0.004456 -3.97 0.001S = 7.073 R-Sq = 95.7% R-Sq(adj) = 95.3%Analysis of VarianceSource DF SS MS F PRegression 2 20233 10117 202.20 0.000Error 18 901 50Total 20 21134b.The estimated distance is 267.841 + 1.52143(-15) - 0.017698(15)2 = 241.04.c.The estimated distance is 267.841 + 1.52143(25) - 0.017698(25)2 = 269.34.10.a.SSR = SST - SSE = 1030MSR = 1030MSE = 520/25 = 20.8F = 1030/20.8 = 49.52F.05 = 4.24 (25 DF)Since 49.52 > 4.24 we reject H0: b1 = 0 and conclude that x1 is significant. b.F.05 = 3.42 (2 degrees of freedom numerator and 23 denominator)Since 48.3 > 3.42 the addition of variables x2 and x3 is statistically significant11.a.SSE = SST - SSR = 1805 - 1760 = 45MSR = 1760/4 = 440MSE =45/25 = 1.8F = 440/1.8 = 244.44F.05 = 2.76 (4 degrees of freedom numerator and 25 denominator)Since 244.44 > 2.76, variables x1 and x4 contribute significantly to the modelb.SSE(x1, x2, x3, x4) = 45c. SSE(x2, x3) = 1805 - 1705 = 100d.F.05 = 3.39 (2 numerator and 25 denominator DF)Since 15.28 > 3.39 we conclude that x1 and x3 contribute significantly to the model.12.a.The Minitab output is shown below.The regression equation isPoints = 170 + 6.61 TeamIntP