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Chapter 11: Simple Linear Regression Analysis CHAPTER 11 – Simple Linear Regression Analysis 11.1 When there appears to be a linear relationship between y and x 11.2 y y x the observed value of the dependent variable y x 0 1 x the mean value of the y when the value of the independent variable is x = error term 11.3 1 : the change in the mean value of the dependent variable that is associated with a one-unit increase in the value of the independent variable 0 : the mean value of the dependent variable when the value of the independent variable is zero 11.4 When data is observed in time sequence, the data is called time series data. Cross-sectional data is observed at a single point in time. 11.5 The straight line appearance of this data plot suggests that the simple linear regression model with a positive slope might be appropriate. 11.6 a. y x 4.00 0 1 (4.00) is the mean of the starting salaries of all marketing graduates having a 4.00 GPA. b. y x 2.50 0 1 (2. 50) is the mean of the starting salaries of all marketing graduates having a 2.50 GPA. c. 1 the change in mean starting salary associated with a one point increase in the grade point average. d. 0 the mean starting salary for marketing graduates with a grade point average of 0.00. The interpretation of 0 fails to make practical sense because it requires that a marketing graduate have a grade point average of 0.00 in which case, of course, the marketing student would not have graduated. e. All factors other than the grade point average. For example, extra-curricular activities and the type of minor (if any) the graduate has. 11.7 The straight line appearance on this data plot suggest that the simple linear regression model with a positive slope might be appropriate. 11.8 a. It is the mean of the service times required when the number of copiers is 4. b. It is the mean of the service times required when the number of copiers is 6. c. The slope parameter equals the change in the mean service time that is associated with each additional copier serviced. d. The intercept is the mean service time when there are no copiers. It fails to make practical sense because it requires service time when no copiers exist. e. All factors other than the number of copiers serviced. 152 Chapter 11: Simple Linear Regression Analysis 11.9 The plot looks reasonably linear. 11.10 a. Mean demand when price difference is .10. b. Mean demand when price difference is –.05. c. Change in mean demand per dollar increase in price difference d. Mean demand when price difference = 0; yes e. Factors other than price difference; answers will vary. 11.11 a. 1000 y 500 0 10 20 30 40 50 60 70 80 90 100 0 x b. Yes, the plot looks linear, positive slope 11.12 a. Mean labor cost when batch size = 60 b. Mean labor cost when batch size = 30 c. Change in mean labor cost per unit increase in batch size d. Mean labor cost when batch size = 0; questionable e. Factors other than batch size; answers will vary. 11.13 a. 200 Price 150 100 10 15 20 25 size 153 Chapter 11: Simple Linear Regression Analysis b. Yes, the relationship looks to be linear with a positive slope. 11.14 a. Mean sales price when home size = 20 (2000 square feet) b. Mean sales price when home size = 18 (1800 square feet) c. Change in mean sales price per one unit (100 square foot) increase in home size d. Mean sales price when home size = 0; no e. Factors other than home size; answers will vary. 11.15 The quality or “goodness” of the fit of the least squares line to the observed data. 11.16 The “best” line that can be fitted to the observed data. The slope and the intercept of the least squares line. 11.17 Evaluate y b0 b1x for the given value of x. ˆ 11.18 Because we do not know how y and x are related outside the experimental region. 11.19 a. b0 = 14.8156 b1 = 5.70657 No. The interpretation of b0 does not make practical sense since it indicates that someone with a GPA = 0 would have a starting salary of $14,816, when in fact they would not have graduated with a GPA = 0. b. ˆ y = 14.8156 + 5.70657(3.25) = 33.362 That is, $33,362 11.20 a. b0 = 11.4641 b1 = 24.6022 No. The interpretation of b0 does not make practical sense since it indicates that 11.46 minutes of service would be required for a customer with no copiers. b. ˆ y = 11.4641 + 24.6022(4) = 109.873, or 109.9 minutes 11.21 a. b0 = 7.81409 b1 = 2.66522 Yes. The interpretation of b0 does make practical sense since it indicates that 781,409 bottles of detergent would be demanded when the price difference with other products is zero. b. ˆ y = 7.814088 + 2.665214(.10) = 8.081 c. y b0 b1 x ˆ 8.5 7.81409 2.6652 x .68591 x .257 , or about 26 cents 2.6652 11.22 a. 154 Chapter 11: Simple Linear Regression Analysis xi yi xi2 xi yi 5 71 25 355 62 663 3844 41106 35 381 1225 13335 12 138 144 1656 83 861 6889 71463 14 145 196 2030 46 493 2116 22678 52 548 2704 28496 23 251 529 5773 100 1024 10000 102400 41 435 1681 17835 75 772 5625 57900 xi 548 yi 5782 xi2 34978 xi yi 365027 ( xi )( yi ) SS xy xi yi n (548)(5,782 ) 365,027 100,982 .32 12 ( xi ) 2 SS xx xi2 n (548) 2 34,978 9,952 .667 12 SS xy 100,982 .32 b1 10.1463 SS xx 9,952 .667 5782 548 b0 y b1 x 10.1463 18.4880 12 12 b. b1 is the estimated increase in mean labor cost (10.1463) for every 1 unit increase in the batch size. b0 is the estimated mean labor cost (18.4880) when batch size = 0; no. c. y 18.488010.1463x ˆ d. ˆ y = 18.4880 + 10.1463(60) = 627.266 155 Chapter 11: Simple Linear Regression Analysis 11.23 a. MINITAB output Regression Analysis: Sale Price versus Size The regression equation is Sale Price = 48.0 + 5.70 Size Predictor Coef SE Coef T P Constant 48.02 14.41 3.33 0.010 Size 5.7003 0.7457 7.64 0.000 S = 10.59 R-Sq = 88.0% R-Sq(adj) = 86.5% b. b1 is the estimated increase in mean sales price (5.7003) for every hundred square foot increase in home size. b0 is the estimated mean sales price when square footage = 0. No, the interpretation of b0 makes no practical sense. c. y 48.02 5.7003 x. ˆ d. ˆ y = 48.02 + 5.7003 (20) = 162.026 That is, $162,026. 11.24 (1) Mean of error terms = 0 (2) Constant variance (3) Normality (4) Independence See page 466 in the text. 11.25 2 ; That is, the constant variance and standard deviation of the error term populations. SSE 1.438 11.26 s2 .2876 n2 72 s s 2 .2876 .5363 SSE 191.70166 11.27 s2 21.30018 n2 11 2 s s 2 21.30018 4.61521 11.28 s 2 2. 8059 .1002, s 28 s2 .3166 11.29 s 2 747 74.7, s 8.643 10 11.30 SSE 896 .8 s2 112 .1 n 2 10 2 s s 2 112 .1 10.588 156 Chapter 11: Simple Linear Regression Analysis 11.31 SSE 222.8242 s2 27.8530 n2 10 2 s s 2 27.8530 5.2776 11.32 Strong ( = .05) or very strong ( = .01) evidence that the regression relationship is significant. 11.33 Explanations will vary. 11.34 a. b0 = 14.816 b1 = 5.7066 b. SSE = 1.4381 s2 = .2876 s = .5363 c. sb1 = .3953 t = 14.44 t = b1 / sb1 = 5.7066 /.3953 = 14.44 d. df = 5 t.025 = 2.571 Reject Ho, Strong evidence of a significant relationship between x and y. e. t.005 = 4.032 Reject Ho, Very strong evidence of a significant relationship between x and y. f. p-value=.000 Reject at all , Extremely strong evidence of a significant relationship between x and y. g. 95% Cl: [ b1 ± t.025 s b1 ] = 5.7066 ± (2.571)(.3953) = [4.690, 6.723] We are 95% confident that the mean starting salary increases by between $4690 and $6723 for each 1.0 increase in GPA. h. 99% Cl: [ b1 ± t.005 sb1 ] = 5.7066 ± (4.032)(.3953) = [4.113, 7.300] We are 99% confident that the mean starting salary increases by between $4113 and $7300 for each 1.0 increase in GPA. i. sb0 = 1.235 t = 12.00 t = b0 / sb0 = 14.816 / 1.235 = 12.00 j. p-value=.000 Reject at all , Extremely strong evidence that the y-intercept is significant. s .5363 s b1 .3953 SS xx 1.8407 k. 1 x2 1 (3.0814) 2 s b0 s .5363 1.235 n SS xx 7 1.8407 157 Chapter 11: Simple Linear Regression Analysis 11.35 a. b0 = 11.4641 b1 = 24.6022 b. SSE = 191.7017 s2 = 21.3002 s = 4.615 c. sb1 = .8045 t = 30.580 t = b1 / sb1 = 24.602 /.8045 = 30.580 d. df = 9 t.025 = 2.262 Reject Ho, strong evidence of a significant relationship between x and y. e. t.005 = 3.250 Reject Ho, very strong evidence of a significant relationship between x and y. f. p-value=.000 Reject at all , extremely strong evidence of a significant relationship between x and y. g. [24.6022 2.262(.8045)] = [22.782, 26.422] h. [24.6022 3.250(.8045)] = [21.987,27.217] i. sb0 = 3.4390 t = 3.334 t = b0 / sb0 = 11.464 / 3.439 = 3.334 j. p-value=.0087 Reject at all except .001 s 4.61521 k. sb1 .8045 SS xx 32.909 1 x2 1 3.909 2 sb0 s 4.61521 3.439 n SS xx 11 32.909 11.36 See the solutions to 11.34 for guidance. a. b0 7.814, b1 2.665 b. SSE 2.806, s 2 .100, s .3166 c. s b1 .2585, t 10.31 d. Reject H 0 . e. Reject H 0 . f. p-value = less than .001; reject H 0 at each value of g. [2.665 ± 2.048(.2585)] = [2.136, 3.194] h. [2.665 ± 2.763(.2585)] = [1.951, 3.379] 158 Chapter 11: Simple Linear Regression Analysis i. s b0 .0799, t 97.82 j. p-value = less than .001; reject H 0 . s .31656 k. sb1 .2585 SS xx 1.49967 1 x2 1 .2133 2 sb0 s .31656 .079883 n SS xx 30 1.49967 11.37 See the solutions to 11.34 for guidance. a. b0 18.488, b1 10.1463 b. SSE = 747, s 2 75, s = 8.642 c. sb1 .0866, t 117.13 d. Reject H 0 . e. Reject H 0 . f. p-value = .000; reject H 0 at each value of g. [10.1463 ± 2.228(.0866)] = [9.953, 10.339] h. [10.1463 ± 3.169(.0866)] = [9.872, 10.421] i. sb0 4.677, t 3.95 j. p-value = .003; fail to reject H 0 at = .001. Reject H 0 at all other values of s 8.64154 k. sb1 .086621 SS xx 9952 .667 1 x2 1 45.667 2 sb0 s 8.64154 4.67658 n SS xx 12 9952 .667 11.38 See the solutions to 11.34 for guidance. a. b0 = 48.02 b1 = 5.7003 b. SSE = 896.8 s2 = 112.1 s = 10.588 c. sb1 = .7457 t = 7.64 t = b1 / sb1 = 5.7003 /.7457 = 7.64 d. df = 8 t.025 = 2.306 Reject Ho 159 Chapter 11: Simple Linear Regression Analysis e. t.005 = 3.355 Reject Ho f. p-value=.000 Reject at all g. [3.9807,7.4199] h. [3.198,8.202] i. sb0 = 14.41 t = 3.33 t = b0 / sb0 = 48.02 / 14.41 = 3.33 j. p-value=.010 Reject at all except .01 and .001 s 10.588 k. sb1 .7457 SS xx 201 .6 1 x2 1 18.8 2 sb0 s 10.588 14.41 10 SS xx 10 201 .6 11.39 Find sb1 from Minitab The regression equation is sales = 66.2 + 4.43 ad exp Predictor Coef SE Coef T P Constant 66.212 5.767 11.48 0.000 Ad exp 4.4303 0.5810 7.62 0.000 95% C.I. for β1 [ 4.4303 2.306(.5810) ] = [3.091,5.770] 11.40 a. b0 = -.1602 b1 = 1.2731 b. SSE = .1343 c. s2= .0336 s = .1833 d. sb1 = .1025 e. t = 12.4209 f. p-value=.00024 Reject at all g. [.9885, 1.5577], 95% confident the mean preference would increase. 11.41 The distance between xo and x , the average of the previously observed values of x. 11.42 A confidence interval is for the mean value of y. A prediction interval is for an individual value of y. 11.43 The smaller the distance value, the shorter the lengths of the intervals. 11.44 a. 33.362, [32.813, 33.911] 160 Chapter 11: Simple Linear Regression Analysis b. 33.362, [31.878, 34.846] 1 (3.25 3.0814 ) 2 c. Distance Value .1583 7 1.8407 [33.362 ± 2.571(.5363) .1583 ] = [32.813, 33.911] [33.362 ± 2.571(.5363) 1 .1583 ] = [31.878, 34.846] 11.45 a. 109.873, [106.721, 113.025] b. 109.873, [98.967, 120.779] c. 113 minutes 11.46 a. 8.0806; [7.948, 8.213] b. 8.0806; [7.419, 8.743] c. See graph in text, page 483 2 .065 d. s dist .065, s .3166 , dist .04215 .3166 99% C.I.: [8.0806 ± 2.763(.065)] = [7.9010, 8.2602] 99% P.I.: 8.0806 2.763(.3166 ) 1.04215 7.1877 ,8.9735 e. (1) 8.4804; [8.360, 8.600] (2) 8.4804; [7.821, 9.140] 2 .059 (3) s dist .059, s .3166 , dist .03473 .3166 99% C.I.: [8.4804 ± 2.763(.059)] = [8.3174, 8.6434] 99% P.I.: 8.4804 2.763(.3166 ) 1.03473 7.5907 ,9.3701 11.47 a. 627.26, [621.05, 633.47] b. 627.26, [607.03, 647.49] 2 2.79 c. s dist 2.79, s 8.642, dist .104227 8.642 99% C.I.: [627.26 ± 3.169(2.79)] = [(618.42, 636.10)] 99% P.I.: 627.26 3.169(8.642) 1.104227 598.48,656.04 161 Chapter 11: Simple Linear Regression Analysis 11.48 a. 162.03, [154.04, 170.02] b. 162.03, [136.33, 187.73] 11.49 2.3429, [1.7367, 2.9491] 11.50 Use a computer program to find the prediction equation y 109 1.075x . ˆ Point predictions and prediction intervals are: a. 87.5, [57.28, 117.72] b. 76.75, [48.04, 105.46] c. 66.00, [37.82, 94.18] d. 55.25, [26.55, 83.96] e. 44.5, [14.28, 74.72] 11.51 From Minitab The regression equation is Market Rate = 0.85 + 0.610 Accounting Rate Predictor Coef SE Coef T P Constant 0.847 1.975 0.43 0.670 Accounti 0.6105 0.1431 4.27 0.000 Predicted Values for New Observations New Obs Fit SE Fit 95.0% CI 95.0% PI 1 10.004 0.753 ( 8.494, 11.514) ( -0.310, 20.318) Values of Predictors for New Observations New Obs Accounti 1 15.0 a. y b0 b1 (15 .00 ) .847 .6105 (15 .00 ) 10 .0045 ˆ 95% C.I.: [8.494, 11.514] b. 10.0045 95% P.I.: [–.310, 20.318] 11.52 Total variation: measures the total amount of variation exhibited by the observed values of y. Unexplained variation: measures the amount of variation in the values of y that is not explained by the model (predictor variable). Explained variation: measures the amount of variation in the values of y that is explained by the predictor variable. 11.53 Proportion of the total variation in the n observed values of y that is explained by the simple linear regression model. 11.54 a. 61.380, 1.438, 59.942, .977, .988; 97.7% of the variation in the observations has been explained by the regression model. b. t = 14.44, p-value =.000; Reject Ho at both 11.55 a. 20110.54, 191.702, 19918.84, .990, .995; 99.0% of the variation in the observations has been explained by the regression model. 162 Chapter 11: Simple Linear Regression Analysis b. t = 30.58, p-value = 000; Reject Ho at both 11.56 a. 13.459; 2.806; 10.653; r 2 .792 ; r = .890; 79.2% of the variation in the observations has been explained by the regression model. b. t = 10.31, p-value = .000; Reject H 0 . 11.57 a. 9952.667; 7.2486; 9945.418; r2 .999; r = .999; 99.9% of the variation in the observations has been explained by the regression model. b. t = 117.134, p-value = .000; Reject H 0 . 11.58 a. 7447.5, 896.8, 6550.7, .880, .9381; 88.0% of the variation in the observations has been explained by the regression model. b. t=7.64,p-value=.000; Reject Ho at both 11.59 a. 5.316, .1343, 5.1817, .975, .987; 97.5% of the variation in the observations has been explained by the regression model. b. t = 12.421, p-value .000; Reject Ho at both 11.60 From Minitab The regression equation is sales = 66.2 + 4.43 ad exp Predictor Coef SE Coef T P Constant 66.212 5.767 11.48 0.000 d exp 4.4303 0.5810 7.62 0.000 S = 5.278 R-Sq = 87.9% R-Sq(adj) = 86.4% Analysis of Variance Source DF SS MS F P Regression 1 1619.3 1619.3 58.14 0.000 Residual Error 8 222.8 27.9 Total 9 1842.1 a. 1842.1, 222.8, 1619.3, r 2 .879 , r .9375 ; 87.9% of the variation in the observations has been explained by the regression model. b. t = 7.62, p-value = .000, Reject Ho. 11.61 From Minitab The regression equation is Market Rate = 0.85 + 0.610 Accounting Rate Predictor Coef SE Coef T P Constant 0.847 1.975 0.43 0.670 Accounti 0.6105 0.1431 4.27 0.000 S = 5.085 R-Sq = 25.9% R-Sq(adj) = 24.5% Analysis of Variance Source DF SS MS F P 163 Chapter 11: Simple Linear Regression Analysis Regression 1 470.74 470.74 18.21 0.000 Residual Error 52 1344.33 25.85 Total 53 1815.07 a. 1,815.07; 1,344.933; 470.74 r2 = .259 r = .509 b. t = 4.27, p-value = .000, Reject H 0 . 11.62 From Minitab The regression equation is HeatLoss = 109 - 1.08 Temperature Predictor Coef SE Coef T P Constant 109.000 9.969 10.93 0.000 Temperat -1.0750 0.2307 -4.66 0.002 S = 11.30 R-Sq = 75.6% R-Sq(adj) = 72.1% Analysis of Variance Source DF SS MS F P Regression 1 2773.5 2773.5 21.70 0.002 Residual Error 7 894.5 127.8 Total 8 3668.0 a. 3,688; 894.5; 2773.5 r2 = .756 r = –.870 b. t = -4.66, p-value = .002, Reject H 0 . 11.63 Mildly similar views. 11.64 H0: 1 0 versus Ha: 1 0 . 11.65 t-test on 1 11.66 a. F = 59.942 / (1.438 / 5) = 208.39 b. F.05 = 6.61 df1 = 1, df2 = 5 Since 208.39 > 6.61, reject H0 with strong evidence of a significant relationship between x and y. c. F.01 = 16.26 df1 = 1, df2 = 5 Since 208.39 > 16.26, reject H0 with very strong evidence of a significant relationship between x and y. d. p-value =.000; Reject H0 at all levels of , extremely strong evidence of a significant relationship between x and y. e. t2 = (14.44)2 = 208.51 (approximately equals F = 208.39) 164 Chapter 11: Simple Linear Regression Analysis (t.025)2 = (2.571)2 = 6.61 = F.05 11.67 a. F = 19918.844 / (21.30018 / 9) = 935.149 b. F.05 = 5.12 df1 = 1, df2 = 9 Since 935.149 > 5.12, reject H0 with strong evidence of a significant relationship between x and y. c. F.01 = 10.56 df1 = 1, df2 = 9 Since 935.149 > 10.56, reject H0 with very strong evidence of a significant relationship between x and y. d. p-value =less than .001; Reject H0 at all levels of , extremely strong evidence of a significant relationship between x and y. e. t2 = (30.58)2 = 935.14 (approximately equals F = 935.149) (t.025)2 = (2.262)2 = 5.12 = F.05 11.68 a. F = 106.303 b. F.05 =4.20, reject H 0 (df1 = 1, df2 = 28). Strong evidence of a significant relationship between x and y. c. F.01 =7.64, reject H 0 (df1 = 1, df2 = 28). Very strong evidence of a significant relationship between x and y. d. p-value = less than .001, reject H 0 . Extremely strong evidence of a significant relationship between x and y. e. (10.310)2 106.303 (within rounding error) (t.025)2 = 4.19 = F.05 11.69 a. F = 13,720.47 b. Reject H 0 . c. Reject H 0 . d. p-value = .000; reject H 0 . e. (117.13)2 13,720.47 (within rounding error) 165 Chapter 11: Simple Linear Regression Analysis 11.70 a. F = 6550.7 / (896.8 / 8) = 58.43 b. F.05 = 5.32 df1 = 1, df2 = 8 Since 58.43 > 5.32, reject H0. c. F.01 = 11.26 df1 = 1, df2 = 8 Since 58.43 > 11.3, reject H0. d. p-value =.000; Reject H0 at all levels of e. t2 = (7.64)2 = 58.37 (approximately equals F = 58.43) (t.025)2 = (2.306)2 = 5.32 = F.05 11.71 a. F = 5.1817 / (.13435 / 4) = 154.279 b. F.05 = 7.71 df1 = 1, df2 = 4 Since 154.279 > 7.71, reject H0. c. F.01 = 21.2 df1 = 1, df2 = 4 Since 154.279 > 21.2, reject H0. d. p-value =.0002; Reject H0 at all levels of e. t2 = (12.4209)2 = 154.279 (approximately equals F = 154.279) (t.025)2 = (2.776)2 = 7.71 = F.05 11.72 ˆ They should be plotted against each independent variable and against y . Funneling or curved patterns indicate violations of the regression assumptions. 11.73 Create a histogram, stem-and-leaf, and normal plot. 11.74 Transforming the dependent variable. 11.75 Approximate horizontal band appearance. No violations indicated 11.76 Possible violations of the normality and constant variance assumptions. 11.77 No. 11.78 a. 3(i ) 1 3(4) 1 .3235 3n 1 33 1 .5000 .3235 .1765, z .46 3(i ) 1 3(10) 1 .8529 3n 1 33 1 .8529 .5000 .3529, z 1.05 166 Chapter 11: Simple Linear Regression Analysis b. No 11.79 The residual plot has somewhat of a cyclical appearance. Since d=.473 is less than dL, 05=1.27, we conclude there is positive autocorrelation and since 4 - .473 = 3.527 and this is greater than dU,.05 = 1.45 we conclude that there is not negative autocorrelation. 11.80 The data plot in Figure 11.40b indicates that as x increases, y increases and becomes more variable. The residual plot in Figure 11.40c fans out as x increases, again indicating that y becomes more variables as x increases. 11.81 a. (i) y* 5.0206 ˆ 95% P.I. for y* = [4.3402, 5.7010] (ii) y e 5.0206 151.5022 ˆ 95% P.I . for y e 4.3402 , e 5.7010 76.7229,299.1664 c. The residual plot is curved indicating the straight line model does not fit the data appropriately. 11.82 a. Yes y 7(25.0069) 175.048 ˆ b. 95%C.I .7(21.4335),7(28.5803) 150.0345,200.0621 95% P.I .7(13.3044),7(36.7094) 93.1308,256.9658 Allow 200 minutes. 11.83 a. Yes; see the plot in part c. b. b0 306.619, b1 27.714 c. y 306.619 27.714x ˆ 260 240 y 220 2.0 2.4 2.8 3.2 x 167 Chapter 11: Simple Linear Regression Analysis d. p-value = .000, reject H 0 , significant e. x0 $2.10; y 248.420; [244.511, 252.327] ˆ x0 $2.75; y 230.405; [226.697, 234.112] ˆ x0 $3.10; y 220.705; [216.415, 224.994] ˆ 11.84 a. b1 = -6.4424 For every unit increase in width difference, the mean number of accidents are reduced by 6.4 per 100 million vehicles. b. p-value = .000 Reject H0 at all levels of c. r2 = .984 98.4% of the variation in accidents is explained by the width difference. 11.85 a. No b. Possibly not; Don’t take up smoking 11.86 a. Using Figure 11.48, there does seem to be a negative relationship between temperature and o-ring failure. b. The temperature of 31 was outside the experimental region. 11.87 Explanations will vary 11.88 a. Argentina, Turkey, Brazil, and Taiwan b. Pakistan and Jordan 11.89 a. There is a relationship since F = 21.13 with a p-value of .0002. b. b1 = 35.2877, [19.2202,51.3553] 11.90 For aggressive stocks a 95% confidence interval for 1 is * t .0163 .025(.003724) [.0163 2.365(.003724)] [.00749, .02512] where t.025 is based on 7 degrees of freedom. We are 95% confident that the effect of a one-month increase in the return length time for an aggressive stock is to increase the mean value of the average estimate of 1 by between .00749 and .02512. For defensive stocks a 95% confidence interval for 1 is * [–0.00462 ± 2.365(.00084164)] = [–.00661, –.00263]. For neutral stocks a 95% confidence interval for 1 is * [.0087255 ± 2.365(.001538)] = [.005088, .01236]. 11.91 ˆ y = 2.0572 + 6.3545(1/5) = 3.3281 168 Chapter 11: Simple Linear Regression Analysis Internet Exercise 11.92 Scatter Plot GMAT vs GPA 700 GMAT 650 600 3.1 3.2 3.3 3.4 3.5 3.6 GPA The regression equation is GMAT = 184 + 141 GPA Predictor Coef SE Coef T P Constant 184.27 84.63 2.18 0.034 GPA 141.08 25.36 5.56 0.000 S = 21.50 R-Sq = 39.2% R-Sq(adj) = 37.9% Analysis of Variance Source DF SS MS F P Regression 1 14316 14316 30.96 0.000 Residual Error 48 22197 462 Total 49 36513 Predicted Values for New Observations New Obs Fit SE Fit 95.0% CI 95.0% PI 1 678.06 5.16 ( 667.68, 688.45) ( 633.60, 722.53) Values of Predictors for New Observations New Obs GPA 1 3.50 169

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