商务与经济统计习题答案（第8版中文版）SBE8-SM16.doc

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1、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 mode

2、l.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

3、 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 = -

4、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

5、 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

6、= - 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

7、.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 0.4

8、242 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 Mini

9、tab 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

10、 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 sh

11、own 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.Th

12、e 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 t

13、he 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

14、 - 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+ * - * - -

15、- * 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

16、% 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 0.05

17、5X 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 significa

18、nt. Note also that since the p-value of .005 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 = 17

19、60/4 = 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