商务与经济统计习题答案（第8版中文版）SBE8-SM15

1、Chapter 15Multiple RegressionLearning Objectives1.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 neces

2、sary 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 pro

3、vided 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,

4、 and determine which observations are influential.15 - 25Multiple RegressionSolutions: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 c

5、onstant.2.a.The estimated regression equation is= 45.06 + 1.94x1 An estimate of y when x1 = 45 is = 45.06 + 1.94(45) = 132.36b.The estimated regression equation is= 85.22 + 4.32x2An estimate of y when x2 = 15 is = 85.22 + 4.32(15) = 150.02c.The estimated regression equation is= -18.37 + 2.01x1 + 4.7

6、4x2An estimate of y when x1 = 45 and x2 = 15 is = -18.37 + 2.01(45) + 4.74(15) = 143.183.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 x

7、1, 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) =

8、 255; sales estimate: $255,000b.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 con

9、stant.5.a.The Minitab output is shown below:The regression equation isRevenue = 88.6 + 1.60 TVAdvPredictor Coef SE Coef T PConstant 88.638 1.582 56.02 0.000TVAdv 1.6039 0.4778 3.36 0.015S = 1.215 R-Sq = 65.3% R-Sq(adj) = 59.5%Analysis of VarianceSource DF SS MS F PRegression 1 16.640 16.640 11.27 0.

10、015Residual Error 6 8.860 1.477Total 7 25.500b.The Minitab output is shown below:The regression equation isRevenue = 83.2 + 2.29 TVAdv + 1.30 NewsAdvPredictor Coef SE Coef T PConstant 83.230 1.574 52.88 0.000TVAdv 2.2902 0.3041 7.53 0.001NewsAdv 1.3010 0.3207 4.06 0.010S = 0.6426 R-Sq = 91.9% R-Sq(a

11、dj) = 88.7%Analysis of VarianceSource DF SS MS F PRegression 2 23.435 11.718 28.38 0.002Residual Error 5 2.065 0.413Total 7 25.500Source DF Seq SSTVAdv 1 16.640NewsAdv 1 6.795c.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 increas

12、e in television advertising with newspaper advertising held constant.d.Revenue = 83.2 + 2.29(3.5) + 1.30(1.8) = $93.56 or $93,5606.a.The Minitab output is shown below:The regression equation isSpeed = 49.8 + 0.0151 WeightPredictor Coef SE Coef T PConstant 49.78 19.11 2.61 0.021Weight 0.015104 0.0060

13、05 2.52 0.025S = 7.000 R-Sq = 31.1% R-Sq(adj) = 26.2%Analysis of VarianceSource DF SS MS F PRegression 1 309.95 309.95 6.33 0.025Error 14 686.00 49.00Total 15 995.95b.The Minitab output is shown below:The regression equation isSpeed = 80.5 - 0.00312 Weight + 0.105 HorsepwrPredictor Coef SE Coef T PC

14、onstant 80.487 9.139 8.81 0.000Weight -0.003122 0.003481 -0.90 0.386Horsepwr 0.10471 0.01331 7.86 0.000S = 3.027 R-Sq = 88.0% R-Sq(adj) = 86.2%Analysis of VarianceSource DF SS MS F PRegression 2 876.80 438.40 47.83 0.000Residual Error 13 119.15 9.17Total 15 995.957.a.The Minitab output is shown belo

15、w:The regression equation isSales = 66.5 + 0.414 Compet$ - 0.270 Heller$Predictor Coef SE Coef T PConstant 66.52 41.88 1.59 0.156Compet$ 0.4139 0.2604 1.59 0.156Heller$ -0.26978 0.08091 -3.33 0.013S = 18.74 R-Sq = 65.3% R-Sq(adj) = 55.4%Analysis of VarianceSource DF SS MS F PRegression 2 4618.8 2309

16、.4 6.58 0.025Residual Error 7 2457.3 351.0Total 9 7076.1b.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 competitors mower with the price of the Heller mower held constant. b2 = -.270 is an estimate of the change in the

17、 quantity sold (1000s) of the Heller mower with respect to a $1 change in its price with the price of the competitors mower held constant.c. = 66.5 + 0.414(170) - 0.270(160) = 93.68 or 93,680 units8.a.The Minitab output is shown below:The regression equation isReturn = 247 - 32.8 Safety + 34.6 ExpRa

18、tioPredictor Coef SE Coef T PConstant 247.4 110.4 2.24 0.039Safety -32.84 13.95 -2.35 0.031ExpRatio 34.59 14.13 2.45 0.026S = 16.98 R-Sq = 58.2% R-Sq(adj) = 53.3%Analysis of VarianceSource DF SS MS F PRegression 2 6823.2 3411.6 11.84 0.001Residual Error 17 4899.7 288.2Total 19 11723.0b.9.a.The Minit

19、ab output is shown below:The regression equation is%College = 26.7 - 1.43 Size + 0.0757 SatScorePredictor Coef SE Coef T PConstant 26.71 51.67 0.52 0.613Size -1.4298 0.9931 -1.44 0.170SatScore 0.07574 0.03906 1.94 0.072S = 12.42 R-Sq = 38.2% R-Sq(adj) = 30.0%Analysis of VarianceSource DF SS MS F PRe

20、gression 2 1430.4 715.2 4.64 0.027Residual Error 15 2312.7 154.2Total 17 3743.1b.= 26.7 - 1.43(20) + 0.0757(1000) = 73.8Estimate is 73.8%10.a.The Minitab output is shown below:The regression equation isRevenue = 33.3 + 7.98 CarsPredictor Coef SE Coef T PConstant 33.34 83.08 0.40 0.695Cars 7.9840 0.6

21、323 12.63 0.000S = 226.7 R-Sq = 92.5% R-Sq(adj) = 91.9%Analysis of VarianceSource DF SS MS F PRegression 1 8192067 8192067 159.44 0.000Error 13 667936 51380Total 14 8860003b. An increase of 1000 cars in service will result in an increase in revenue of $7.98 million.c.The Minitab output is shown belo

22、w:The regression equation isRevenue = 106 + 8.94 Cars - 0.191 LocationPredictor Coef SE Coef T PConstant 105.97 85.52 1.24 0.239Cars 8.9427 0.7746 11.55 0.000Location -0.1914 0.1026 -1.87 0.087S = 207.7 R-Sq = 94.2% R-Sq(adj) = 93.2%Analysis of VarianceSource DF SS MS F PRegression 2 8342186 4171093

23、 96.66 0.000Error 12 517817 43151Total 14 886000311.a.SSE = SST - SSR = 6,724.125 - 6,216.375 = 507.75b.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 be

24、en 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 variabil

25、ity.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 pref

26、erred 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 good18.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

27、.MSR = SSR/p = 6,216.375/2 = 3,108.188b.F = MSR/MSE = 3,108.188/72.536 = 42.85F.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 = 7.26t.025 = 2.365 (7 degrees of freedom)Since t = 2.365 t.025 = 2.365, b1 is si

28、gnificant.d.t = .4980/.0567 = 8.78Since t = 8.78 t.025 = 2.365, b2 is significant.20.A portion of the Minitab output is shown below.The regression equation isY = - 18.4 + 2.01 X1 + 4.74 X2Predictor Coef SE Coef T PConstant -18.37 17.97 -1.02 0.341X1 2.0102 0.2471 8.13 0.000X2 4.7378 0.9484 5.00 0.00

29、2S = 12.71 R-Sq = 92.6% R-Sq(adj) = 90.4%Analysis of VarianceSource DF SS MS F PRegression 2 14052.2 7026.1 43.50 0.000Residual Error 7 1130.7 161.5Total 9 15182.9a.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 th

30、e 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 F.05 = 4.74, we reject H0. There is a significant relationship among the variables.23.a.F = 28.38F.01 = 13.27 (2 degrees of freedom, numerator and 1 deno

31、minator)Since F F.01 = 13.27, reject H0.Alternatively, the p-value of .002 leads to the same conclusion.b.t = 7.53t.025 = 2.571Since t t.025 = 2.571, b1 is significant and x1 should not be dropped from the model.c.t = 4.06t.025 = 2.571Since t t.025 = 2.571, b2 is significant and x2 should not be dro

32、pped from the model.24.Note: The Minitab output is shown in part (b) of Exercise 6a.F = 47.83F.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 = 0.05, we cannot reject H0For Horsepower:H0: b

33、2 = 0 Ha: b2 0Since the p-value = 0.000 a = 0.05, we can reject H025.a.The Minitab output is shown below:The regression equation isP/E = 6.04 + 0.692 Profit% + 0.265 Sales%Predictor Coef SE Coef T PConstant 6.038 4.589 1.32 0.211Profit% 0.6916 0.2133 3.24 0.006Sales% 0.2648 0.1871 1.42 0.180S = 5.45

34、6 R-Sq = 47.2% R-Sq(adj) = 39.0%Analysis of VarianceSource DF SS MS F PRegression 2 345.28 172.64 5.80 0.016Residual Error 13 387.00 29.77Total 15 732.28b.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, Sa

35、les% 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, Location is not significant27.a.= 29.

36、1270 + .5906(180) + .4980(310) = 289.8150b.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 + 2.2

37、9(3.5) + 1.30(1.8) = 93.555 or $93,555Note: 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 $93,588The difference (

38、$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 thi

39、s 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 $94,335c.Prediction interval estimate: 91.774 to 95.401 or

40、 $91,774 to $95,40130.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 isSpeed = 72.6 + 0.0968 HorsepwrPredictor Coef

41、SE Coef T PConstant 72.650 2.655 27.36 0.000Horsepwr 0.096756 0.009865 9.81 0.000S = 3.006 R-Sq = 87.3% R-Sq(adj) = 86.4%Analysis of VarianceSource DF SS MS F PRegression 1 869.43 869.43 96.21 0.000Residual Error 14 126.52 9.04Total 15 995.95Unusual ObservationsObs Horsepwr Speed Fit SE Fit Residual

42、 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 XR denotes an observation with a large standardized residualX 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%

43、 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 wherex2 = 0 if

44、level 1 and 1 if level 2b.E(y) = b0 + b1 x1 + b2(0) = b0 + b1 x1 c.E(y) = b0 + b1 x1 + b2(1) = b0 + b1 x1 + b2d.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.twob.E(y) = b0 + b1 x1 + b2 x2 + b3 x3 wherex2x3Level001102013c.E(y | level 1)