Chapter_6_Recommended_Homework_Solutions

Chapter_6_Recommended_Homework_Solutions - Chapter 6...

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Unformatted text preview: Chapter 6 Recommended Homework Solutions 6-1. a) The regression equation is Thermal = 0.0249 + 0.129 Density Predictor Coef StDev T P Constant 0.024934 0.001786 13.96 0.000 Density 0.128522 0.007738 16.61 0.000 S = 0.0005852 R-Sq = 98.6% R-Sq(adj) = 98.2% Analysis of Variance Source DF SS MS Regression 1 0.000094464 0.000094464 Residual Error 4 0.000001370 0.000000342 Total 5 0.000095833 F 275.84 P 0.000 ^ y = 0.0249 + 0.129 x b) 0.0005747 -0.0007088 0.0001486 -0.0004799 -0.0000644 0.0005298 c) SSE = 0.000001370 2 = 0.000000342 ^ ^ ^ d) se( 1 ) = 0.007738, se( 0 ) = 0.001786 e) SST = 0.000095833 SSR = 0.000094464, SSE = 0.000001370, and SSR + SSE = 0.000095834 SST = SSR + SSE f) R2 = 98.6%. This is interpreted as 98.6% of the total variability in thermal conductivity can be explained by the fitted regression model. g) See the Minitab output given in part a. Based on the t-tests, we conclude that the slope and intercept are nonzero. h) See the Minitab output in part a. Based on the analysis of variance, we can reject the null hypothesis and conclude that the regression is significant. i) 0: 0.024934 2.776(0.001786); 0.02, 0.03 1: 0.128522 2.776(0.007738); 0.107, 0.15 j) Residual Plots Residuals Versus Density (response is Conduct) 0.0005 Residual 0.0000 -0.0005 0.18 0.23 0.28 Density Residuals Versus the Fitted Values (response is Conduct) 0.0005 Residual 0.0000 -0.0005 0.050 0.055 0.060 Fitted Value Normal Probability Plot of the Residuals (response is Conduct) 1 Normal Score 0 -1 -0.0005 0.0000 0.0005 Residual k) r = 0.993, P-value = 0, Therefore, we conclude there is a significant correlation between density and conductivity. 6-5. a) The regression equation is permeability = 40.6 - 2.12 strength Predictor Constant strength S = 1.038 Coef 40.5536 -2.1232 SE Coef 0.7509 0.2313 T 54.00 -9.18 P 0.000 0.000 R-Sq = 86.6% SS 90.759 13.999 104.757 R-Sq(adj) = 85.6% MS 90.759 1.077 F 84.28 P 0.000 Analysis of Variance Source DF Regression 1 Residual Error 13 Total 14 ^ y = 40.6 - 2.12 x b) -0.97179 0.00066 1.56517 0.35430 0.21735 0.99417 0.79921 0.83762 0.24199 -1.22252 -0.49351 -0.38411 -0.74715 -2.15947 0.96808 c) SSE = 13.999 2 = 1.077 ^ ^ ^ d) se( 1 ) = 0.2313, se( 0 ) = 0.7509 e) SST = 104.757 SSR = 90.759, SSE = 13.999, and SSR + SSE = 104.757 SST = SSR + SSE f) R2 = 86.6%. This is interpreted as 86.6% of the total variability in permeability can be explained by the fitted regression model. g) See the Minitab output given in part a. Based on the t-tests, we conclude that the slope and intercept are nonzero. h) See the Minitab output in part a. Based on the analysis of variance, we can reject the null hypothesis and conclude that the regression is significant. i) 0: 40.5536 2.16(0.7509); 38.93, 41.18 1: -2.1232 2.16(0.2313); -2.62, -1.62 j) Residual Plots Residuals Versus strength (response is permeabi) 2 1 Residual 0 -1 -2 1 2 3 4 5 strength Residuals Versus the Fitted Values (response is permeabi) 2 1 Residual 0 -1 -2 29 30 31 32 33 34 35 36 37 38 Fitted Value Normal Probability Plot of the Residuals (response is permeabi) 2 1 Normal Score 0 -1 -2 -2 -1 0 1 2 Residual k) r = -0.931, P-value = 0; Therefore, we conclude there is a significant correlation between strength and temperature. 6-7. a) 0.055137 b) (0.054460, 0.055813) c) (0.053376, 0.056897) d) The prediction interval is wider than the confidence interval. 6-9. a) 36.095 b) (35.059, 37.131) c) (32.802, 39.388) d) The prediction interval is wider than the confidence interval. a) The regression equation is y = 351 - 1.27 x1 - 0.154 x2 Predictor Constant x1 x2 S = 25.50 Coef 350.99 -1.272 -0.15390 SE Coef 74.75 1.169 0.08953 T 4.70 -1.09 -1.72 P 0.018 0.356 0.184 VIF 2.6 2.6 6-11. R-Sq = 86.2% R-Sq(adj) = 77.0% Analysis of Variance Source Regression Residual Error Total Source X1 x2 DF 1 1 DF 2 3 5 SS 12161.6 1950.4 14112.0 Seq SS 10240.4 1921.2 MS 6080.8 650.1 F 9.35 P 0.051 b) -24.9866 24.3075 11.8203 -20.4595 12.8296 -3.5113 c) SSE = 1950.4 2 = 650.1 ^ d) R-Sq = 86.2%, R-Sq(adj) = 77.0%; R-Sq(adj) is less than R-Sq. because the model contains terms that are not contributing significantly to the model. The adjusted R2 value will penalize the user for adding terms to the model that are not significant. e) See part a. Based on the p-value from the ANOVA table, the regression model is significant at the 0.10 level of significance. ^ ^ ^ f) se( 0 ) = 74.75, se( 1 ) = 1.169, se( 2 ) = 0.08953 g) See part a. Based on the p-values for each coefficient, the regressors do not appear to be significant at the 0.05 level of significance. h) 0: 350.99 3.182(74.74); 113.17, 588.81 1: -1.272 3.182(1.169); -4.99, 2.45 2 : -0.1539 3.182(0.08953); -0.439, 0.131 i) Obs 1 2 3 4 5 6 SRES1 -1.65529 1.35770 0.51526 -1.05590 1.09375 -0.18436 COOK1 1.69265 0.63183 0.02083 0.27192 1.48548 0.00898 Residuals Versus x1 (response is y) 30 20 10 Residual 0 -10 -20 -30 0 10 20 30 40 x1 Residuals Versus the Fitted Values (response is y) 30 20 10 Residual 0 -10 -20 -30 100 150 200 Fitted Value Normal Probability Plot of the Residuals (response is y) 1 Normal Score 0 -1 -30 -20 -10 0 10 20 30 Residual j) The VIFs are 2.6. There is no indication of a problem with multicollinearity. 6-13. a) The regression equation is y = - 103 + 0.605 x1 + 8.92 x2 + 1.44 x3 + 0.014 x4 Predictor Constant x1 x2 x3 x4 S = 15.58 Coef -102.7 0.6054 8.924 1.437 0.0136 SE Coef 207.9 0.3689 5.301 2.392 0.7338 T -0.49 1.64 1.68 0.60 0.02 P 0.636 0.145 0.136 0.567 0.986 VIF 2.3 2.2 1.3 1.0 R-Sq = 74.5% SS 4957.2 1699.0 6656.3 Seq SS 3758.9 1109.4 88.9 0.1 R-Sq(adj) = 59.9% MS 1239.3 242.7 F 5.11 P 0.030 Analysis of Variance Source DF Regression 4 Residual Error 7 Total 11 Source x1 x2 x3 x4 b) -18.7580 1.8862 23.3109 -8.9565 9.1852 6.6436 4.8136 -0.1568 DF 1 1 1 1 -17.8502 -12.9376 6.6216 6.1980 c) SSE = 1699.0 , 2 = 242.7 ^ d) R-Sq = 74.5% R-Sq(adj) = 59.9%; not significant. e) see part a. Based on the p-value from the ANOVA table, the regression model is significant at the 0.05 level of significance. R-Sq(adj) is less than R-Sq. since there are terms in the model that are ^ ^ ^ ^ ^ f) se( 0 ) = 207.9, se( 1 ) = 0.3689, se( 2 ) = 5.301, se( 3 ) = 2.392, se( 4 ) = 0.7338 g) See part a. Based on the p-values for each coefficient, the regressors do not appear to be significant. h) 0: -102.7 2.365(207.9); -594.38, 388.98 1: 0.6054 2.365(0.3689); -0.267, 1.478 2 : 8.924 2.365(5.301); -3.613, 21.461 3: 1.437 2.365(2.392); -4.22, 7.094 4: 0.0136 2.365(0.7338); -1.722, 1.75 i) Residuals Versus x1 (response is y) 25 20 15 10 Residual 5 0 -5 -10 -15 -20 25 35 45 55 65 75 85 x1 Residuals Versus the Fitted Values (response is y) 25 20 15 10 Residual 5 0 -5 -10 -15 -20 230 240 250 260 270 280 290 300 310 Fitted Value Normal Probability Plot of the Residuals (response is y) 2 1 Normal Score 0 -1 -2 -20 -15 -10 -5 0 5 10 15 20 25 Residual k) The VIFs are all less than 10, there is no indication of a problem with multicollinearity. 6-15. a) 149.9 b) (85.1, 214.7) c) (-12.5, 312.3) d) The prediction interval is wider than the confidence interval. 6-17. a) 287.56 b) (263.77, 311.35) c) (243.69, 331.44) d) The prediction interval is wider than the confidence interval. a) 6-24. 2.7 2.6 kWh 2.5 2.4 2.3 3.9 4.0 4.1 4.2 4.3 4.4 4.5 4.6 4.7 Dollars No, a straight line relationship does not seem plausible. b) The regression equation is kWh = 2.43 + 0.012 Dollars c) Analysis of Variance Source Regression Residual Error Total DF 1 13 14 SS 0.00015 0.26369 0.26384 MS 0.00015 0.02028 F 0.01 P 0.933 2) H 0 :1 = 0 3) H 1:1 0 4) = 0.05 5) The test statistic is f0 = SS R / k SSE / ( n - p) 6) Reject H0 if f0 > f,1,8 where f0.05,1,13 = 4.67 7) Using the results from the ANOVA table f0 = 0.00015 / 1 = 0.01 0.2637 / 13 8) Since 0.01 < 4.67 do not reject H0 and conclude that the regression model is not significant at = 0.05. P-value > 0.10 (from the computer output the P-value is found to be 0.933 d) Predictor Constant Dollars Coef 2.4299 0.0119 StDev 0.6094 0.1389 T 3.99 0.09 P 0.002 0.933 0.0119 - t 0.025,13 (0.1389) 1 0.0119 + t 0.025,13 (0.1389) 0.0119 - 2.160(0.1389) 1 0.0119 + 2.160(0.1389) - 0.288 1 0.312 e) 2) H 0 : 1 = 0 3) H1 : 1 0 4) = 0.05 5) The test statistic is t0 = ^ 1 ^ se(1 ) 6) Reject H0 if t0 < - /2,n-2 where - 0.025,13 = - t t 2.160 or t0 > t0.025,13 = 2.160 7) Using the results from the table above t0 = 0.0119 = 0.0857 0.1389 8) Since 2.160 < 0.0857 < 2.160 do not reject H 0 and conclude the slope is practically 0. Dollars is not a significant predictor of electrical usage at = 0.05. f) 2) H 0 : 0 = 0 3) H1 : 0 0 4) = 0.05 5) The test statistic is t0 = ^ 0 ^ se(0 ) 6) Reject H0 if t0 < - /2,n-2 where - 0.025,13 = - t t 2.160 or t0 > t0.025,13 = 2.160 7) Using the results from the table above t0 = 2.4299 = 3.987 0.6094 8) Since 3.987 > 2.160 reject H 0 and conclude the intercept is not zero at = 0.05. 6-30. a) The regression equation is y = 3829 - 0.215 x3 + 21.2 x4 + 1.66 x5 Predictor Constant x3 x4 x5 S = 43.66 Coef 3829 -0.2149 21.2134 1.6566 SE Coef 2262 0.1088 0.9050 0.5502 T 1.69 -1.97 23.44 3.01 P 0.099 0.056 0.000 0.005 R-Sq = 99.3% R-Sq(adj) = 99.3% y = 3829.26 - 0.215x 3 +21213x 4 + 1.657 x 5 . b) Analysis of Variance Source Regression Residual Error Total DF 3 36 39 SS 9863398 68638 9932036 MS 3287799 1907 F 1724.42 P 0.000 2) H 0 : 3 = 4 = 5 = 0 3) H 1: j 0 for at least one j 4) = 0.01 5) The test statistic is f0 = SS R / k SSE / ( n - p) 6) Reject H0 if f0 > f,3,36 where f0.01,3,36 = 4.38 7) Using the results from the ANOVA table f0 = 9863398 / 3 = 1724.42 68638.2 / 36 8) Since 1724.42 > 4.38 reject H0 and conclude that the regression model is significant at = 0.01. P-value < 0.00001 c) All at = 0.01 H 0 : 3 = 0 H 1: 3 0 t0 = -1.97 | t 0 | > t /2 ,36 / Do not reject H 0 t0.005,36 = 2.72 H 0 : 4 = 0 H 1: 4 0 t0 =23.44 | t 0 | > t / 2 ,36 Reject H 0 H 0 : 5 = 0 H 1: 5 0 t0 = 3.01 | t 0 | > t / 2 ,36 Reject H 0 Do not need x3 term in the model at = 0.01. d) R2 = 0.993 R 2 = 0.9925 adj The slight decrease in R 2 may be reflective of the insignificant x3 term. adj e) Normal Probability Plot 99.9 99 95 cumulative percent 80 50 20 5 1 0.1 -80 -50 -20 10 40 70 100 Residuals Normality assumption appears reasonable. This is evident by the fact the residuals fall along a straight line. f) Residual Plot 100 70 40 Residuals 10 -20 -50 -80 3 3.3 3.6 3.9 4.2 4.5 4.8 (X 1000) Predicted Plot is satisfactory. There does not appear to be a nonrandom pattern in the residual vs. predicted plot. g) Residual Plot for y 100 70 40 10 -20 -50 -80 28 28.4 28.8 29.2 29.6 x3 30 30.4 (X 1000) Slight indication that variance increases as x3 increases. This is evident by the "fanning out" appearance of the residuals. h) Using the equation found in part a): y = 3829.26 - 0.215(1670) + 21213(170) + 1.657(1589) = 9709.39 . 6-31. a) The regression equation is y* = 19.7 - 1.27 x3* + 0.00541 x4 +0.000408 x5 Predictor Constant x3* x4 x5 S = 0.01314 Coef 19.690 -1.2673 0.0054140 0.0004079 SE Coef 9.587 0.9594 0.0002711 0.0001645 T 2.05 -1.32 19.97 2.48 P 0.047 0.195 0.000 0.018 R-Sq = 99.1% Residuals R-Sq(adj) = 99.0% Analysis of Variance Source Regression Residual Error Total 2) H 0: = 4 = 5 = 0 3 3) H 1: j 0 DF 3 36 39 SS 0.68611 0.00622 0.69233 MS 0.22870 0.00017 F 1323.62 P 0.000 for at least one j 4) = 0.01 5) The test statistic is f0 = SS R / k SSE / ( n - p) 6) Reject H0 if f0 > f,3,36 where f0.01,3,36 = 4.38 7) Using the results from the ANOVA table f0 = 0.686112 / 3 = 1323.62 0.00622033 / 36 8) Since 1323.62 > 4.38 reject H0 and conclude that the regression model is significant at = 0.01. P-value < 0.00001 b) = 0.01 H 0: 3 H1: 3 =0 0 t .005,36 = 2.72 H 0 : 4 = 0 H 1: 4 0 t 0 = 19.97 | t 0 | > t / 2 ,36 Reject H 0 H 0 : 5 = 0 H 1: 5 0 t 0 = 2.48 | t 0 | > t / 2 ,36 / Do not reject H 0 t 0 = -132 . | t 0 | > t /2 ,36 / Do not reject H 0 3 : Do not reject H0 and conclude that ln(x3) is not a significant regressor in the model at = 0.01. 4 : Reject H0 and conclude that x4 is a significant regressor in the model at = 0.01. 5 : Do not reject H0 and conclude that x5 is not a significant regressor in the model at = 0.01. c) Residual Plot for ln(y) (X 1E-3) 38 28 (X 1E-3) 38 28 18 8 -2 -12 -22 Residual Plot for ln(y) Residuals 8 -2 -12 -22 8 8.1 8.2 8.3 8.4 8.5 Residuals 18 1026 1027 1028 1029 1030 1031 1032 ln(x3) (X 0.01) Predicted Curvature is evident in the residuals plots from this model, whereas non-stable variance was evident in previous model. ...
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