2021年商务与经济统计习题答案（第8版中文版）SBE8-SM15

1、chapter 15 multiple regression learning objectives understand how multiple regression analysis can be used to develop relationships involving one dependent variable and several independent variables. be able to interpret the coefficients in a multiple regression analysis. know the assumptions necess

2、ary to conduct statistical tests involving the hypothesized regression model. understand the role of computer packages in performing multiple regression analysis. be able to interpret and use computer output to develop the estimated regression equation. be able to determine how good a fit is provide

3、d by the estimated regression equation. be able to test for the significance of the regression equation. understand how multicollinearity affects multiple regression analysis. know how residual analysis can be used to make a judgement as to the appropriateness of the model, identify outliers, and de

4、termine which observations are influential. solutions: a. b1=.596 is an estimate of the change in y corresponding to a 1 unit change in x1 when x2 is held constant. b2=.498 is an estimate of the change in y corresponding to a 1 unit change in x2 when x1 is held constant. a. the estimated regression

5、equation is=46 + 94x1 an estimate of y when x1=45 is=46 + 94(45)=1336 b. the estimated regression equation is=822 + 32x2 an estimate of y when x2=15 is=822 + 32(15)=15.2 c. the estimated regression equation is=-137 + 1x1 + 74x2 an estimate of y when x1=45 and x2=15 is=-137 + 1(45) + 74(15)=1418 a. b

6、1=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=-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=6 is an estimate of the change in y corresponding to a 1 unit c

7、hange in x3 when x1, x2, and x4 are held constant. b4=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. a.=235 + 1(15) + 8(1)=255; sales estimate: $255, b. sales can be expected to increase by $1 for every dollar increase in inventory i

8、nvestment 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. a. the minitab output is shown below: the regression equation is revenue=86 + 6 tvadv predictor coef se co

9、ef t p constant 8638 582 52 . tvadv 639 .4778 36 .15 s=215 r-sq=63% r-sq(adj)=55% analysis of variance source df ss ms f p regression 1 164 164 127 .15 residual error 6 86 477 total 7 25 b. the minitab output is shown below: the regression equation is revenue=82 + 29 tvadv + 3 newsadv predictor coef

10、 se coef t p constant 823 574 588 . tvadv 292 .341 53 .1 newsadv 31 .327 6 .1 s=.6426 r-sq=99% r-sq(adj)=87% analysis of variance source df ss ms f p regression 2 2435 1718 238 .2 residual error 5 65 .413 total 7 25 source df seq ss tvadv 1 164 newsadv 1 795 c. no, it is 6 in part 2(a) and 99 above.

11、 in this exercise it represents the marginal change in revenue due to an increase in television advertising with newspaper advertising held constant. d. revenue=82 + 29(5) + 3(8)=$956 or $93,56 a. the minitab output is shown below: the regression equation is speed=48 + .151 weight predictor coef se

12、coef t p constant 478 111 61 .21 weight .1514 .65 52 .25 s= r-sq=31% r-sq(adj)=22% analysis of variance source df ss ms f p regression 1 395 395 33 .25 error 14 68 4 total 15 9995 b. the minitab output is shown below: the regression equation is speed=8.5 - .312 weight + .15 horsepwr predictor coef s

13、e coef t p constant 8.487 139 81 . weight -.3122 .3481 -.9 .386 horsepwr .1471 .1331 86 . s=27 r-sq=8% r-sq(adj)=82% analysis of variance source df ss ms f p regression 2 878 434 483 . residual error 13 1115 17 total 15 9995 a. the minitab output is shown below: the regression equation is sales=65 +

14、 .414 compet$ - .27 heller$ predictor coef se coef t p constant 652 488 59 .156 compet$ .4139 .264 59 .156 heller$ -.26978 .891 -33 .13 s=174 r-sq=63% r-sq(adj)=54% analysis of variance source df ss ms f p regression 2 4618 234 58 .25 residual error 7 2453 35 total 9 771 b. b1=.414 is an estimate of

15、 the change in the quantity sold (1s) 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=-.27 is an estimate of the change in the quantity sold (1s) of the heller mower with respect to a $1 change in its price with the p

16、rice of the competitors mower held constant. c.=65 + .414(17) - .27(16)=968 or 93,68 units a. the minitab output is shown below: the regression equation is return=247 - 38 safety + 36 expratio predictor coef se coef t p constant 244 11.4 24 .39 safety -384 195 -35 .31 expratio 359 113 45 .26 s=198 r

17、-sq=52% r-sq(adj)=53% analysis of variance source df ss ms f p regression 2 6822 3416 184 .1 residual error 17 4897 282 total 19 1172 b. a. the minitab output is shown below: the regression equation is %college=27 - 43 size + .757 satscore predictor coef se coef t p constant 271 567 .52 .613 size -4

18、298 .9931 -44 .17 satscore .7574 .396 94 .72 s=142 r-sq=32% r-sq(adj)=3.% analysis of variance source df ss ms f p regression 2 143.4 712 64 .27 residual error 15 2317 152 total 17 3741 b.=27 - 43(2) + .757(1)=78 estimate is 78% 1. a. the minitab output is shown below: the regression equation is rev

19、enue=33 + 98 cars predictor coef se coef t p constant 334 88 .4 .695 cars 984 .6323 163 . s=227 r-sq=95% r-sq(adj)=99% analysis of variance source df ss ms f p regression 1 819267 819267 1544 . error 13 667936 5138 total 14 8863 b. an increase of 1 cars in service will result in an increase in reven

20、ue of $98 million. c. the minitab output is shown below: the regression equation is revenue=16 + 94 cars - .191 location predictor coef se coef t p constant 197 852 24 .239 cars 9427 .7746 155 . location -.1914 .126 -87 .87 s=27 r-sq=92% r-sq(adj)=92% analysis of variance source df ss ms f p regress

21、ion 2 8342186 417193 966 . error 12 517817 43151 total 14 8863 1 a. sse=sst - ssr=6,72125 - 6,21375=575 b. c. d. the estimated regression equation provided an excellent fit. 1 a. b. c. yes; after adjusting for the number of independent variables in the model, we see that 9.5% of the variability in y

22、 has been accounted for. 1 a. b. c. the estimated regression equation provided an excellent fit. 1 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 amoun

23、t of variability. 1 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. 1 note: the minitab output is shown with the solution to exercise a. no; r-sq=31% b. multiple regression analy

24、sis 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. 1 a. b. the fit is not very good 1 note: the minitab output is shown with the solution to exercise 1. a. r-sq=92% r-sq(adj)=92% b. the fit is very g

25、ood. 1 a. msr=ssr/p=6,21375/2=3,1188 b. f=msr/mse=3,1188/7536=485 f.5=74 (2 degrees of freedom numerator and 7 denominator) since f=485 f.5=74 the overall model is significant. c. t=.596/.813=26 t.25=365 (7 degrees of freedom) since t=365 t.25=365, b1 is significant. d. t=.498/.567=78 since t=78 t.2

26、5=365, b2 is significant. 2. a portion of the minitab output is shown below. the regression equation is y=- 14 + 1 x1 + 74 x2 predictor coef se coef t p constant -137 197 -2 .341 x1 12 .2471 13 . x2 7378 .9484 .2 s=171 r-sq=96% r-sq(adj)=9.4% analysis of variance source df ss ms f p regression 2 145

27、2 721 45 . residual error 7 113.7 165 total 9 15189 a. since the p-value corresponding to f=45 is . a=.5, we reject h: b1=b2=; there is a significant relationship. b. since the p-value corresponding to t=13 is . a=.5, we reject h: b1=; b1 is significant. c. since the p-value corresponding to t= is .

28、2 f.5=74, we reject h. there is a significant relationship among the variables. 2 a. f=238 f.1=127 (2 degrees of freedom, numerator and 1 denominator) since f f.1=127, reject h. alternatively, the p-value of .2 leads to the same conclusion. b. t=53 t.25=571 since t t.25=571, b1 is significant and x1

29、 should not be dropped from the model. c. t=6 t.25=571 since t t.25=571, b2 is significant and x2 should not be dropped from the model. 2 note: the minitab output is shown in part (b) of exercise 6 a. f=483 f.5=81 (2 degrees of freedom numerator and 13 denominator) since f=483 f.5=81, we reject h: b

30、1=b2=. alternatively, since the p-value=. a=.5, we cannot reject h for horsepower: h: b2= ha: b2 1 since the p-value=. a=.5, we can reject h 2 a. the minitab output is shown below: the regression equation is p/e=4 + .692 profit% + .265 sales% predictor coef se coef t p constant 38 589 32 .211 profit

31、% .6916 .2133 24 .6 sales% .2648 .1871 42 .18 s=456 r-sq=42% r-sq(adj)=3% analysis of variance source df ss ms f p regression 2 3428 1764 8 .16 residual error 13 38 277 total 15 7328 b. since the p-value=.16 a=.5, there is a significant relationship among the variables. c. for profit%: since the p-v

32、alue=.6 a=.5, sales% is not significant. 2 note: the minitab output is shown with the solution to exercise 1. a. since the p-value corresponding to f=966 is . a=.5, there is a significant relationship among the variables. b. for cars: since the p-value=. a=.5, location is not significant 2 a.=2127 +

33、 .596(18) + .498(31)=28815 b. the point estimate for an individual value is=28815, the same as the point estimate of the mean value. 2 a. using minitab, the 95% confidence interval is 1316 to 151 b. using minitab, the 95% prediction interval is 1113 to 171 2 a.=82 + 29(5) + 3(8)=9555 or $93,555 note

34、: in exercise 5b, the minitab output also shows that b=823, b1=292, and b2=31; hence,=823 + 292x1 + 31x using this estimated regression equation, we obtain=823 + 292(5) + 31(8)=9588 or $93,588 the difference ($93,588 - $93,555=$33) is simply due to the fact that additional significant digits are use

35、d 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. 95

36、88 .291 ( 984, 9335) ( 9774, 941) note that the value of fit () is 958 b. confidence interval estimate: 984 to 9335 or $92,84 to $94,335 c. prediction interval estimate: 9774 to 941 or $91,774 to $95,41 3. a. since weight is not statistically significant (see exercise 24), we will use an estimated r

37、egression equation which uses only horsepower to predict the speed at 1/4 mile. the minitab output is shown below: the regression equation is speed=76 + .968 horsepwr predictor coef se coef t p constant 765 655 236 . horsepwr .96756 .9865 81 . s=6 r-sq=83% r-sq(adj)=84% analysis of variance source d

38、f ss ms f p regression 1 8643 8643 921 . residual error 14 1252 4 total 15 9995 unusual observations obs horsepwr speed fit se fit residual st resid 2 29 1 1.79 .814 291 52r 6 45 112 1119 36 .1 . x r denotes an observation with a large standardized residual x denotes an observation whose x value giv

39、es it large influence. the output shows that the point estimate is a speed of 129 miles per hour. b. the 95% confidence interval is 949 to 189 miles per hour. c. the 95% prediction interval is 9596 to 1984 miles per hour. 3 a. using minitab the 95% confidence interval is 537% to 73%. b. using minita

40、b the 95% prediction interval is 324% to 9.59%. 3 a. e(y)=b + b1 x1 + b2 x2 where x2= if level 1 and 1 if level 2 b. e(y)=b + b1 x1 + b2()=b + b1 x1 c. e(y)=b + b1 x1 + b2(1)=b + 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.

41、3 a. two b. e(y)=b + b1 x1 + b2 x2 + b3 x3 where x2 x3 level 1 1 2 1 3 c. e(y | level 1)=b + b1 x1 + b2() + b3()=b+ b1 x1 e(y | level 2)=b + b1 x1 + b2(1) + b3()=b + b1 x1 + b2 e(y | level 3)=b + b1 x1 + b2() + b3()=b + b1 x1 + b3 b2=e(y | level 2) - e(y | level 1) b3=e(y | level 3) - e(y | level 1)

42、 b1 is the change in e(y) for a 1 unit change in x1 holding x2 and x3 constant. 3 a. $15,3 b. estimate of sales=1.1 - 2(2) + 8(8) + 13()=51 or $56,1 c. estimate of sales=1.1 - 2(1) + 8(3) + 13(1)=46 or $41,6 3 a. let type= if a mechanical repair type=1 if an electrical repair the minitab output is s

43、hown below: the regression equation is time=45 + .617 type predictor coef se coef t p constant 45 .5467 31 . type .6167 .758 .87 .48 s=93 r-sq=7% r-sq(adj)=.% analysis of variance source df ss ms f p regression 1 .913 .913 .76 .48 residual error 8 563 195 total 9 1.476 b. the estimated regression eq

44、uation did not provide a good fit. in fact, the p-value of .48 shows that the relationship is not significant for any reasonable value of a. c. person= 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

45、time=62 - 6 person predictor coef se coef t p constant 62 .3192 147 . person -6 .4514 -54 .8 s=.7138 r-sq=61% r-sq(adj)=52% analysis of variance source df ss ms f p regression 1 4 4 156 .8 residual error 8 76 .595 total 9 1.476 d. we see that 61% of the variability in repair time has been explained by the repair person that performed the service; an acceptable, but not good, fit. 3 a. the minitab output is shown below: the regression equation is time=86 + .291 months +