商务与经济统计习题答案（第8版，中文版）SBE8-SM14

Chapter 14 Simple Linear Regression Learning Objectives 1. Understand how regression analysis can be used to develop an equation that estimates mathematically how two variables are related. 2. Understand the differences between the regression model, the regression equation, and the estimated regression equation. 3. Know how to fit an estimated regression equation to a set of sample data based upon the least-squares method. 4. Be able to determine how good a fit is provided by the estimated regression equation and compute the sample correlation coefficient from the regression analysis output. 5. Understand the assumptions necessary for statistical inference and be able to test for a significant relationship. 6. Learn how to use a residual plot to make a judgement as to the validity of the regression assumptions, recognize outliers, and identify influential observations. 7. Know how to develop confidence interval estimates of y given a specific value of x in both the case of a mean value of y and an individual value of y. 8. Be able to compute the sample correlation coefficient from the regression analysis output. 9. Know the definition of the following terms: independent and dependent variable simple linear regression regression model regression equation and estimated regression equation scatter diagram coefficient of determination standard error of the estimate confidence interval prediction interval residual plot standardized residual plot outlier influential observation leverage Solutions: 1 a. b. There appears to be a linear relationship between x and y. c. Many different straight lines can be drawn to provide a linear approximation of the relationship between x and y; in part d we will determine the equation of a straight line that “best” represents the relationship according to the least squares criterion. d. Summations needed to compute the slope and y-intercept are: e. 2. a. b. There appears to be a linear relationship between x and y. c. Many different straight lines can be drawn to provide a linear approximation of the relationship between x and y; in part d we will determine the equation of a straight line that “best” represents the relationship according to the least squares criterion. d. Summations needed to compute the slope and y-intercept are: e. 3. a. b. Summations needed to compute the slope and y-intercept are: c. 4. a. b. There appears to be a linear relationship between x and y. c. Many different straight lines can be drawn to provide a linear approximation of the relationship between x and y; in part d we will determine the equation of a straight line that “best” represents the relationship according to the least squares criterion. d. Summations needed to compute the slope and y-intercept are: e. pounds 5. a. b. There appears to be a linear relationship between x and y. c. Many different straight lines can be drawn to provide a linear approximation of the relationship between x and y; in part d we will determine the equation of a straight line that “best” represents the relationship according to the least squares criterion. Summations needed to compute the slope and y-intercept are: d. A one million dollar increase in media expenditures will increase case sales by approximately 14.42 million. e. 6. a. b. There appears to be a linear relationship between x and y. c. Summations needed to compute the slope and y-intercept are: d. A one percent increase in the percentage of flights arriving on time will decrease the number of complaints per 100,000 passengers by 0.07. e 7. a. b. Let x = DJIA and y = S&P. Summations needed to compute the slope and y-intercept are: c. or approximately 1500 8. a. Summations needed to compute the slope and y-intercept are: b. Increasing the number of times an ad is aired by one will increase the number of household exposures by approximately 3.07 million. c. 9. a. b. Summations needed to compute the slope and y-intercept are: c. 10. a. b. Let x = performance score and y = overall rating. Summations needed to compute the slope and y-intercept are: c. or approximately 84 11. a. b. There appears to be a linear relationship between the variables. c. The summations needed to compute the slope and the y-intercept are: d. 12. a. b. There appears to be a positive linear relationship between the number of employees and the revenue. c. Let x = number of employees and y = revenue. Summations needed to compute the slope and y-intercept are: d. 13. a. b. The summations needed to compute the slope and the y-intercept are: c. or approximately $13,080. The agent＇s request for an audit appears to be justified. 14. a. b. The summations needed to compute the slope and the y-intercept are: c. 15. a. The estimated regression equation and the mean for the dependent variable are: The sum of squares due to error and the total sum of squares are Thus, SSR = SST - SSE = 80 - 12.4 = 67.6 b. r2 = SSR/SST = 67.6/80 = .845 The least squares line provided a very good fit; 84.5% of the variability in y has been explained by the least squares line. c. 16. a. The estimated regression equation and the mean for the dependent variable are: The sum of squares due to error and the total sum of squares are Thus, SSR = SST - SSE = 114.80 - 6.33 = 108.47 b. r2 = SSR/SST = 108.47/114.80 = .945 The least squares line provided an excellent fit; 94.5% of the variability in y has been explained by the estimated regression equation. c. Note: the sign for r is negative because the slope of the estimated regression equation is negative. (b1 = -1.88) 17. The estimated regression equation and the mean for the dependent variable are: The sum of squares due to error and the total sum of squares are Thus, SSR = SST - SSE = 11.2 - 5.3 = 5.9 r2 = SSR/SST = 5.9/11.2 = .527 We see that 52.7% of the variability in y has been explained by the least squares line. 18. a. The estimated regression equation and the mean for the dependent variable are: The sum of squares due to error and the total sum of squares are Thus, SSR = SST - SSE = 335,000 - 85,135.14 = 249,864.86 b. r2 = SSR/SST = 249,864.86/335,000 = .746 We see that 74.6% of the variability in y has been explained by the least squares line. c. 19. a. The estimated regression equation and the mean for the dependent variable are: The sum of squares due to error and the total sum of squares are Thus, SSR = SST - SSE = 47,582.10 - 7547.14 = 40,034.96 b. r2 = SSR/SST = 40,034.96/47,582.10 = .84 We see that 84% of the variability in y has been explained by the least squares line. c. 20. a. Let x = income and y = home price. Summations needed to compute the slope and y-intercept are: b. The sum of squares due to error and the total sum of squares are Thus, SSR = SST - SSE = 11,373.09 – 2017.37 = 9355.72 r2 = SSR/SST = 9355.72/11,373.09 = .82 We see that 82% of the variability in y has been explained by the least squares line. c. or approximately $173,500 21. a. The summations needed in this problem are: b. $7.60 c. The sum of squares due to error and the total sum of squares are: Thus, SSR = SST - SSE = 5,648,333.33 - 233,333.33 = 5,415,000 r2 = SSR/SST = 5,415,000/5,648,333.33 = .9587 We see that 95.87% of the variability in y has been explained by the estimated regression equation. d. 22. a. The summations needed in this problem are: b. The sum of squares due to error and the total sum of squares are: Thus, SSR = SST - SSE = 1998 - 1272.4495 = 725.5505 r2 = SSR/SST = 725.5505/1998 = 0.3631 Approximately 37% of the variability in change in executive compensation is explained by the two-year change in the return on equity. c. It reflects a linear relationship that is between weak and strong. 23. a. s2 = MSE = SSE / (n - 2) = 12.4 / 3 = 4.133 b. c. d. t.025 = 3.182 (3 degrees of freedom) Since t = 4.04 > t.05 = 3.182 we reject H0: b1 = 0 e. MSR = SSR / 1 = 67.6 F = MSR / MSE = 67.6 / 4.133 = 16.36 F.05 = 10.13 (1 degree of freedom numerator and 3 denominator) Since F = 16.36 > F.05 = 10.13 we reject H0: b1 = 0 Source of Variation Sum of Squares Degrees of Freedom Mean Square F Regression 67.6 1 67.6 16.36 Error 12.4 3 4.133 Total 80.0 4 24. a. s2 = MSE = SSE / (n - 2) = 6.33 / 3 = 2.11 b. c. d. t.025 = 3.182 (3 degrees of freedom) Since t = -7.18 < -t.025 = -3.182 we reject H0: b1 = 0 e. MSR = SSR / 1 = 8.47 F = MSR / MSE = 108.47 / 2.11 = 51.41 F.05 = 10.13 (1 degree of freedom numerator and 3 denominator) Since F = 51.41 > F.05 = 10.13 we reject H0: b1 = 0 Source of Variation Sum of Squares Degrees of Freedom Mean Square F Regression 108.47 1 108.47 51.41 Error 6.33 3 2.11 Total 114.80 4 25. a. s2 = MSE = SSE / (n - 2) = 5.30 / 3 = 1.77 b. t.025 = 3.182 (3 degrees of freedom) Since t = 1.82 < t.025 = 3.182 we cannot reject H0: b1 = 0; x and y do not appear to be related. c. MSR = SSR/1 = 5.90 /1 = 5.90 F = MSR/MSE = 5.90/1.77 = 3.33 F.05 = 10.13 (1 degree of freedom numerator and 3 denominator) Since F = 3.33 < F.05 = 10.13 we cannot reject H0: b1 = 0; x and y do not appear to be related. 26. a. s2 = MSE = SSE / (n - 2) = 85,135.14 / 4 = 21,283.79 t.025 = 2.776 (4 degrees of freedom) Since t = 3.43 > t.025 = 2.776 we reject H0: b1 = 0 b. MSR = SSR / 1 = 249,864.86 / 1 = 249.864.86 F = MSR / MSE = 249,864.86 / 21,283.79 = 11.74 F.05 = 7.71 (1 degree of freedom numerator and 4 denominator) Since F = 11.74 > F.05 = 7.71 we reject H0: b1 = 0 c. Source of Variation Sum of Squares Degrees of Freedom Mean Square F Regression 249864.86 1 249864.86 11.74 Error 85135.14 4 21283.79 Total 335000 5 27. The sum of squares due to error and the total sum of squares are: SSE = SST = 2442 Thus, SSR = SST - SSE = 2442 - 170 = 2272 MSR = SSR / 1 = 2272 SSE = SST - SSR = 2442 - 2272 = 170 MSE = SSE / (n - 2) = 170 / 8 = 21.25 F = MSR / MSE = 2272 / 21.25 = 106.92 F.05 = 5.32 (1 degree of freedom numerator and 8 denominator) Since F = 106.92 > F.05 = 5.32 we reject H0: b1 = 0. Years of experience and sales are related. 28. SST = 411.73 SSE = 161.37 SSR = 250.36 MSR = SSR / 1 = 250.36 MSE = SSE / (n - 2) = 161.37 / 13 = 12.413 F = MSR / MSE = 250.36 / 12.413= 20.17 F.05 = 4.67 (1 degree of freedom numerator and 13 denominator) Since F = 20.17 > F.05 = 4.67 we reject H0: b1 = 0. 29. SSE = 233,333.33 SST = 5,648,333.33 SSR = 5,415,000 MSE = SSE/(n - 2) = 233,333.33/(6 - 2) = 58,333.33 MSR = SSR/1 = 5,415,000 F = MSR / MSE = 5,415,000 / 58,333.25 = 92.83 Source of Variation Sum of Squares Degrees of Freedom Mean Square F Regression 5,415,000.00 1 5,415,000 92.83 Error 233,333.33 4 58,333.33 Total 5,648,333.33 5 F.05 = 7.71 (1 degree of freedom numerator and 4 denominator) Since F = 92.83 > 7.71 we reject H0: b1 = 0. Production volume and total cost are related. 30. Using the computations from Exercise 22, SSE = 1272.4495 SST = 1998 SSR = 725.5505 = 45,833.9286 t.025 = 2.571 Since t = 1.69 < 2.571, we cannot reject H0: b1 = 0 There is no evidence of a significant relationship between x and y. 31. SST = 11,373.09 SSE = 2017.37 SSR = 9355.72 MSR = SSR / 1 = 9355.72 MSE = SSE / (n - 2) = 2017.37/ 16 = 126.0856 F = MSR / MSE = 9355.72/ 126.0856 = 74.20 F.01 = 8.53 (1 degree of freedom numerator and 16 denominator) Since F = 74.20 > F.01 = 8.53 we reject H0: b1 = 0. 32. a. s = 2.033 b. 10.6 ± 3.182 (1.11) = 10.6 ± 3.53 or 7.07 to 14.13 c. d. 10.6 ± 3.182 (2.32) = 10.6 ± 7.38 or 3.22 to 17.98 33. a. s = 1.453 b. 24.69 ± 3.182 (.68) = 24.69 ± 2.16 or 22.53 to 26.85 c. d. 24.69 ± 3.182 (1.61) = 24.69 ± 5.12 or 19.57 to 29.81 34. s = 1.33 2.28 ± 3.182 (.85) = 2.28 ± 2.70 or -.40 to 4.98 2.28 ± 3.182 (1.58) = 2.28 ± 5.03 or -2.27 to 7.31 35. a. s = 145.89 2,033.78 ± 2.776 (68.54) = 2,033.78 ± 190.27 or $1,843.51 to $2,224.05 b. 2,033.78 ± 2.776 (161.19) = 2,033.78 ± 447.46 or $1,586.32 to $2,481.24 36. a. b. s = 3.5232 80.859 ± 2.160 (1.055) = 80.859 ± 2.279 or 78.58 to 83.14 c. 80.859 ± 2.160 (3.678) = 80.859 ± 7.944 or 72.92 to 88.80 37. a. s2 = 1.88 s = 1.37 13.08 ± 2.571 (.52) = 13.08 ± 1.34 or 11.74 to 14.42 or $11,740 to $14,420 b. sind = 1.47 13.08 ± 2.571 (1.47) = 13.08 ± 3.78 or 9.30 to 16.86 or $9,300 to $16,860 c. Yes, $20,400 is much larger than anticipated. d. Any deductions exceeding the $16,860 upper limit could suggest an audit. 38. a. b. s2 = MSE = 58,333.33 s = 241.52 5046.67 ± 4.604 (267.50) = 5046.67 ± 1231.57 or $3815.10 to $6278.24 c. Based on one month, $6000 is not out of line since $3815.10 to $6278.24 is the prediction interval. However, a sequence of five to seven months with consistently high costs should cause concern. 39. a. Summations needed to compute the slope and y-intercept are: b. SST = 39,065.14 SSE = 4145.141 SSR = 34,920.000 r2 = SSR/SST = 34,920.000/39,065.141 = 0.894 The estimated regression equation explained 89.4% of the variability in y; a very good fit. c. s2 = MSE = 4145.141/8 = 518.143 270.63 ± 2.262 (8.86) = 270.63 ± 20.04 or 250.59 to 290.67 d. 270.63 ± 2.262 (24.42) = 270.63 ± 55.24 or 215.39 to 325.87 40. a. 9 b. = 20.0 + 7.21x c. 1.3626 d. SSE = SST - SSR = 51,984.1 - 41,587.3 = 10,396.8 MSE = 10,396.8/7 = 1,485.3 F = MSR / MSE = 41,587.3 /1,485.3 = 28.00 F.05 = 5.59 (1 degree of freedom numerator and 7 denominator) Since F = 28 > F.05 = 5.59 we reject H0: B1 = 0. e. = 20.0 + 7.21(50) = 380.5 or $380,500 41. a. = 6.1092 + .8951x b. t.025 = 2.306 (1 degree of freedom numerator and 8 denominator) Since t = 6.01 > t.025 = 2.306 we reject H0: B1 = 0 c. = 6.1092 + .8951(25) = 28.49 or $28.49 per month 42 a. = 80.0 + 50.0x b. 30 c. F = MSR / MSE = 6828.6/82.1 = 83.17 F.05 = 4.20 (1 degree of freedom numerator and 28 denominator) Since F = 83.17 > F.05 = 4.20 we reject H0: B1 = 0. Branch office sales are related to the salespersons. d. = 80 + 50 (12) = 680 or $680,000 43. a. The Minitab output is shown below: The regression equation is Price = - 11.8 + 2.18 Income Predictor Coef SE Coef T P Constant -11.80 12.84 -0.92 0.380 Income 2.1843 0.2780 7.86 0.000 S = 6.634 R-Sq = 86.1% R-Sq(adj) = 84.7% Analysis of Variance Source DF SS MS F P Regression 1 2717.9 2717.9 61.75 0.000 Residual Error 10 440.1 44.0 Total 11 3158.0 Predicted Values for New Observations New Obs Fit SE Fit 95.0% CI 95.0% PI 1 75.79 2.47 ( 70.29, 81.28) ( 60.02, 91.56) b. r2 = .861. The least squares line provided a very good fit. c. The 95% confidence interval is 70.29 to 81.28 or $70,290 to $81,280. d. The 95% prediction interval is 60.02 to 91.56 or $60,020 to $91,560. 44. a/b. The scatter diagram shows a linear relationship between the two variables. c. The Minitab output is shown below: The regression equation is Rental$ = 37.1 - 0.779 Vacancy% Predictor Coef SE Coef T P Constant 37.066 3.530 10.50 0.000 Vacancy% -0.7791 0.2226 -3.50 0.003 S = 4.889 R-Sq = 43.4% R-Sq(adj) = 39.8% Analysis of Variance Source DF SS MS F P Regression 1 292.89 292.89 12.26 0.003 Residual Error 16 382.37 23.90 Total 17 675.26 Predicted Values for New Observations New Obs Fit SE Fit 95.0% CI 95.0% PI 1 17.59 2.51 ( 12.27, 22.90) ( 5.94, 29.23) 2 28.26 1.42 ( 25.26, 31.26) ( 17.47, 39.05) Values of Predictors for New Observations New Obs Vacancy% 1 25.0 2 11.3 d. Since the p-value = 0.003 is less than a = .05, the relationship is significant. e. r2 = .434. The least squares line does not provide a very good fit. f. The 95% confidence interval is 12.27 to 22.90 or $12.27 to $22.90. g. The 95% prediction interval is 17.47 to 39.05 or $17.47 to $39.05. 45. a. b. The residuals are 3.48, -2.47, -4.83, -1.6, and 5.22 c. With only 5 observations it is difficult to determine if the assumptions are satisfied. However, the plot does suggest curvature in the residuals that would indicate that the error term assumptions are not satisfied. The scatter diagram for these data also indicates that the underlying relationship between x and y may be curvilinear. d. The standardized residuals are 1.32, -.59, -1.11, -.40, 1.49. e. The standardized residual plot has the same shape as the original residual plot. The curvature observed indicates that the assumptions regarding the error term may not be satisfied. 46. a. b. The assumption that the variance is the same for all values of x is questionable. The variance appears to increase for larger values of x. 47. a. Let x = advertising expenditures and y = revenue b. SST = 1002 SSE = 310.28 SSR = 691.72 MSR = SSR / 1 = 691.72 MSE = SSE / (n - 2) = 310.28/ 5 = 62.0554 F = MSR / MSE = 691.72/ 62.0554= 11.15 F.05 = 6.61 (1 degree of freedom numerator and 5 denominator) Since F = 11.15 > F.05 = 6.61 we conclude that the two variables are related. c. d. The residual plot leads us to question the assumption of a linear relationship between x and y. Even though the relationship is significant at the .05 level of significance, it would be extremely dangerous to extrapolate beyond the range of the data. 48. a. b. The assumptions concerning the error term appear reasonable. 49. a. Let x = return on investment (ROE) and y = price/earnings (P/E) ratio. b. c. There is an unusual trend in the residuals. The assumptions concerning the error term appear questionable. 50. a. The MINITAB output is shown below: The regression equation is Y = 66.1 + 0.402 X Predictor Coef Stdev t-ratio p Constant 66.10 32.06 2.06 0.094 X 0.4023 0.2276 1.77 0.137 s = 12.62 R-sq = 38.5% R-sq(adj) = 26.1% Analysis of Variance SOURCE DF SS MS F p Regression 1 497.2 497.2 3.12 0.137 Error 5 795.7 159.1 Total 6 1292.9 Unusual Observations Obs. X Y Fit Stdev.Fit Residual St.Resid 1 135 145.00 120.42 4.87 24.58 2.11R R denotes an obs. with a large st. resid. The standardized residuals are: 2.11, -1.08, .14, -.38, -.78, -.04, -.41 The first observation appears to be an outlier since it has a large standardized residual. b. 2.4+ - * STDRESID- - - 1.2+ - - - - * 0.0+ * - - * * - * - -1.2+ * - --+---------+---------+---------+---------+---------+----YHAT 110.0 115.0 120.0 125.0 130.0 135.0 The standardized residual plot indicates that the observation x = 135,y = 145 may be an outlier; note that this observation has a standardized residual of 2.11. c. The scatter diagram is shown below - Y - * - - 135+ - - * * - - 120+ * * - - - * - 105+ - - * ----+---------+---------+---------+---------+---------+--X 105 120 135 150 165 180 The scatter diagram also indicates that the observation x = 135,y = 145 may be an outlier; the implication is that for simple linear regression an outlier can be identified by looking at the scatter diagram. 51. a. The Minitab output is shown below: The regression equation is Y = 13.0 + 0.425 X Predictor Coef Stdev t-ratio p Constant 13.002 2.396 5.43 0.002 X 0.4248 0.2116 2.01 0.091 s = 3.181 R-sq = 40.2% R-sq(adj) = 30.2% Analysis of Variance SOURCE DF SS MS F p Regression 1 40.78 40.78 4.03 0.091 Error 6 60.72 10.12 Total 7 101.50 Unusual Observations Obs. X Y Fit Stdev.Fit Residual St.Resid 7 12.0 24.00 18.10 1.20 5.90 2.00R 8 22.0 19.00 22.35 2.78 -3.35 -2.16RX R denotes an obs. with a large st. resid. X denotes an obs. whose X value gives it large influence. The standardized residuals are: -1.00, -.41, .01, -.48, .25, .65, -2.00, -2.16 The last two observations in the data set appear to be outliers since the standardized residuals for these observations are 2.00 and -2.16, respectively. b. Using MINITAB, we obtained the following leverage values: .28, .24, .16, .14, .13, .14, .14, .76 MINITAB identifies an observation as having high leverage if hi > 6/n; for these data, 6/n = 6/8 = .75. Since the leverage for the observation x = 22, y = 19 is .76, MINITAB would identify observation 8 as a high leverage point. Thus, we conclude that observation 8 is an influential observation. c. 24.0+ * - Y - - - 20.0+ * - * - * - - 16.0+ * - * - - * - 12.0+ * - +---------+---------+---------+---------+---------+------X 0.0 5.0 10.0 15.0 20.0 25.0 The scatter diagram indicates that the observation x = 22, y = 19 is an influential observation. 52. a. The Minitab output is shown below: The regression equation is Amount = 4.09 + 0.196 MediaExp Predictor Coef SE Coef T P Constant 4.089 2.168 1.89 0.096 MediaExp 0.19552 0.03635 5.38 0.001 S = 5.044 R-Sq = 78.3% R-Sq(adj) = 75.6% Analysis of Variance Source DF SS MS F P Regression 1 735.84 735.84 28.93 0.001 Residual Error 8 203.51 25.44 Total 9 939.35 Unusual Observations Obs MediaExp Amount Fit SE Fit Residual St Resid 1 120 36.30 27.55 3.30 8.75 2.30R R denotes an observation with a large standardized residual b. Minitab identifies observation 1 as having a large standardized residual; thus, we would consider observation 1 to be an outlier. 53. a. The Minitab output is shown below: The regression equation is Exposure = - 8.6 + 7.71 Aired Predictor Coef SE Coef T P Constant -8.55 21.65 -0.39 0.703 Aired 7.7149 0.5119 15.07 0.000 S = 34.88 R-Sq = 96.6% R-Sq(adj) = 96.2% Analysis of Variance Source DF SS MS F P Regression 1 276434 276434 227.17 0.000 Residual Error 8 9735 1217 Total 9 286169 Unusual Observations Obs Aired Exposure Fit SE Fit Residual St Resid 1 95.0 758.8 724.4 32.0 34.4 2.46RX R denotes an observation with a large standardized residual X denotes an observation whose X value gives it large influence. b. Minitab identifies observation 1 as having a large standardized residual; thus, we would consider observation 1 to be an outlier. Minitab also identifies observation 1 as an influential observation. 54. a. The Minitab output is shown below: The regression equation is Salary = 707 + 0.00482 MktCap Predictor Coef SE Coef T P Constant 707.0 118.0 5.99 0.000 MktCap 0.0048154 0.0008076 5.96 0.000 S = 379.8 R-Sq = 66.4% R-Sq(adj) = 64.5% Analysis of Variance Source DF SS MS F P Regression 1 5129071 5129071 35.55 0.000 Residual Error 18 2596647 144258 Total 19 7725718 Unusual Observations Obs MktCap Salary Fit SE Fit Residual St Resid 6 507217 3325.0 3149.5 338.6 175.5 1.02 X 17 120967 116.2 1289.5 86.4 -1173.3 -3.17R R denotes an observation with a large standardized residual X denotes an observation whose X value gives it large influence. b. Minitab identifies observation 6 as having a large standardized residual and observation 17 as an observation whose x value gives it large influence. A standardized residual plot against the predicted values is shown below: 55. No. Regression or correlation analysis can never prove that two variables are casually related. 56. The estimate of a mean value is an estimate of the average of all y values associated with the same x. The estimate of an individual y value is an estimate of only one of the y values associated with a particular x. 57. To determine whether or not there is a significant relationship between x and y. However, if we reject B1 = 0, it does not imply a good fit. 58. a. The Minitab output is shown below: The regression equation is Price = 9.26 + 0.711 Shares Predictor Coef SE Coef T P Constant 9.265 1.099 8.43 0.000 Shares 0.7105 0.1474 4.82 0.001 S = 1.419 R-Sq = 74.4% R-Sq(adj) = 71.2% Analysis of Variance Source DF SS MS F P Regression 1 46.784 46.784 23.22 0.001 Residual Error 8 16.116 2.015 Total 9 62.900 b. Since the p-value corresponding to F = 23.22 = .001 < a = .05, the relationship is significant. c. = .744; a good fit. The least squares line explained 74.4% of the variability in Price. d. 59. a. The Minitab output is shown below: The regression equation is Options = - 3.83 + 0.296 Common Predictor Coef SE Coef T P Constant -3.834 5.903 -0.65 0.529 Common 0.29567 0.02648 11.17 0.000 S = 11.04 R-Sq = 91.9% R-Sq(adj) = 91.2% Analysis of Variance Source DF SS MS F P Regression 1 15208 15208 124.72 0.000 Residual Error 11 1341 122 Total 12 16550 b. ; approximately 40.6 million shares of options grants outstanding. c. = .919; a very good fit. The least squares line explained 91.9% of the variability in Options. 60. a. The Minitab output is shown below: The regression equation is IBM = 0.275 + 0.950 S&P 500 Predictor Coef StDev T P Constant 0.2747 0.9004 0.31 0.768 S&P 500 0.9498 0.3569 2.66 0.029 S = 2.664 R-Sq = 47.0% R-Sq(adj) = 40.3% Analysis of Variance Source DF SS MS F P Regression 1 50.255 50.255 7.08 0.029 Error 8 56.781 7.098 Total 9 107.036 b. Since the p-value = 0.029 is less than a = .05, the relationship is significant. c. r2 = .470. The least squares line does not provide a very good fit. d. Woolworth has higher risk with a market beta of 1.25. 61. a. b. It appears that there is a positive linear relationship between the two variables. c. The Minitab output is shown below: The regression equation is High = 23.9 + 0.898 Low Predictor Coef SE Coef T P Constant 23.899 6.481 3.69 0.002 Low 0.8980 0.1121 8.01 0.000 S = 5.285 R-Sq = 78.1% R-Sq(adj) = 76.9% Analysis of Variance Source DF SS MS F P Regression 1 1792.3 1792.3 64.18 0.000 Residual Error 18 502.7 27.9 Total 19 2294.9 d. Since the p-value corresponding to F = 64.18 = .000 < a = .05, the relationship is significant. e. = .781; a good fit. The least squares line explained 78.1% of the variability in high temperature. f. 62. The MINITAB output is shown below: The regression equation is Y = 10.5 + 0.953 X Predictor Coef Stdev t-ratio p Constant 10.528 3.745 2.81 0.023 X 0.9534 0.1382 6.90 0.000 s = 4.250 R-sq = 85.6% R-sq(adj) = 83.8% Analysis of Variance SOURCE DF SS MS F p Regression 1 860.05 860.05 47.62 0.000 Error 8 144.47 18.06 Total 9 1004.53 Fit Stdev.Fit 95% C.I. 95% P.I. 39.13 1.49 ( 35.69, 42.57) ( 28.74, 49.52) a. = 10.5 + .953 x b. Since the p-value corresponding to F = 47.62 = .000 < a = .05, we reject H0: b1 = 0. c. The 95% prediction interval is 28.74 to 49.52 or $2874 to $4952 d. Yes, since the expected expense is $3913. 63. a. The Minitab output is shown below: The regression equation is Defects = 22.2 - 0.148 Speed Predictor Coef SE Coef T P Constant 22.174 1.653 13.42 0.000 Speed -0.14783 0.04391 -3.37 0.028 S = 1.489 R-Sq = 73.9% R-Sq(adj) = 67.4% Analysis of Variance Source DF SS MS F P Regression 1 25.130 25.130 11.33 0.028 Residual Error 4 8.870 2.217 Total 5 34.000 Predicted Values for New Observations New Obs Fit SE Fit 95.0% CI 95.0% PI 1 14.783 0.896 ( 12.294, 17.271) ( 9.957, 19.608) b. Since the p-value corresponding to F = 11.33 = .028 < a = .05, the relationship is significant. c. = .739; a good fit. The least squares line explained 73.9% of the variability in the number of defects. d. Using the Minitab output in part (a), the 95% confidence interval is 12.294 to 17.271. 64. a. There appears to be a negative linear relationship between distance to work and number of days absent. b. The MINITAB output is shown below: The regression equation is Y = 8.10 - 0.344 X Predictor Coef Stdev t-ratio p Constant 8.0978 0.8088 10.01 0.000 X -0.34420 0.07761 -4.43 0.002 s = 1.289 R-sq = 71.1% R-sq(adj) = 67.5% Analysis of Variance SOURCE DF SS MS F p Regression 1 32.699 32.699 19.67 0.002 Error 8 13.301 1.663 Total 9 46.000 Fit Stdev.Fit 95% C.I. 95% P.I. 6.377 0.512 ( 5.195, 7.559) ( 3.176, 9.577) c. Since the p-value corresponding to F = 419.67 is .002 < a = .05. We reject H0 : b1 = 0. d. r2 = .711. The estimated regression equation explained 71.1% of the variability in y; this is a reasonably good fit. e. The 95% confidence interval is 5.195 to 7.559 or approximately 5.2 to 7.6 days. 65. a. Let X = the age of a bus and Y = the annual maintenance cost. The MINITAB output is shown below: The regression equation is Y = 220 + 132 X Predictor Coef Stdev t-ratio p Constant 220.00 58.48 3.76 0.006 X 131.67 17.80 7.40 0.000 s = 75.50 R-sq = 87.3% R-sq(adj) = 85.7% Analysis of Variance SOURCE DF SS MS F p Regression 1 312050 312050 54.75 0.000 Error 8 45600 5700 Total 9 357650 Fit Stdev.Fit 95% C.I. 95% P.I. 746.7 29.8 ( 678.0, 815.4) ( 559.5, 933.9) b. Since the p-value corresponding to F = 54.75 is .000 < a = .05, we reject H0: b1 = 0. c. r2 = .873. The least squares line provided a very good fit. d. The 95% prediction interval is 559.5 to 933.9 or $559.50 to $933.90 66. a. Let X = hours spent studying and Y = total points earned The MINITAB output is shown below: The regression equation is Y = 5.85 + 0.830 X Predictor Coef Stdev t-ratio p Constant 5.847 7.972 0.73 0.484 X 0.8295 0.1095 7.58 0.000 s = 7.523 R-sq = 87.8% R-sq(adj) = 86.2% Analysis of Variance SOURCE DF SS MS F p Regression 1 3249.7 3249.7 57.42 0.000 Error 8 452.8 56.6 Total 9 3702.5 Fit Stdev.Fit 95% C.I. 95% P.I. 84.65 3.67 ( 76.19, 93.11) ( 65.35, 103.96) b. Since the p-value corresponding to F = 57.42 is .000 < a = .05, we reject H0: b1 = 0. c. 84.65 points d. The 95% prediction interval is 65.35 to 103.96 67. a. The Minitab output is shown below: The regression equation is Audit% = - 0.471 +0.000039 Income Predictor Coef SE Coef T P Constant -0.4710 0.5842 -0.81 0.431 Income 0.00003868 0.00001731 2.23 0.038 S = 0.2088 R-Sq = 21.7% R-Sq(adj) = 17.4% Analysis of Variance Source DF SS MS F P Regression 1 0.21749 0.21749 4.99 0.038 Residual Error 18 0.78451 0.04358 Total 19 1.00200 Predicted Values for New Observations New Obs Fit SE Fit 95.0% CI 95.0% PI 1 0.8828 0.0523 ( 0.7729, 0.9927) ( 0.4306, 1.3349) b. Since the p-value = 0.038 is less than a = .05, the relationship is significant. c. r2 = .217. The least squares line does not provide a very good fit. d. The 95% confidence interval is .7729 to .9927.