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# Solucionario Walpole 8 ED

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forward selection, variable x1 is entered first, and no other variables are entered at 0.05 level. Hence the final model is y\u2c6 = \u22126.33592 + 0.33738x1. (b) For the backward elimination, variable x3 is eliminated first, then variable x4 and then variable x2, all at 0.05 level of significance. Hence only x1 remains in the model and the final model is the same one as in (a). Solutions for Exercises in Chapter 12 177 (c) For the stepwise regression, after x1 is entered, no other variables are entered. Hence the final model is still the same one as in (a) and (b). 12.49 Using computer output, with \u3b1 = 0.05, x4 was removed first, and then x1. Neither x2 nor x3 were removed and the final model is y\u2c6 = 2.18332 + 0.95758x2 + 3.32533x3. 12.50 (a) y\u2c6 = \u221229.59244 + 0.27872x1 + 0.06967x2 + 1.24195x3 \u2212 0.39554x4 + 0.22365x5. (b) The variables x3 and x5 were entered consecutively and the final model is y\u2c6 = \u221256.94371 + 1.63447x3 + 0.24859x5. (c) We have a summary table displayed next. Model s2 PRESS R2 \u2211 i |\u3b4i| x2x5 x1x5 x1x3x5 x3x5 x3x4x5 x2x4x5 x2x3x5 x3x4 x1x2x5 x5 x3 x2x3x4x5 x1 x2x3 x1x3 x1x3x4x5 x2 x4x5 x1x2x3x5 x2x3x4 x1x2 x1x4x5 x1x3x4 x2x4 x1x2x4x5 x1x2x3x4x5 x1x4 x1x2x3 x4 x1x2x3x4 x1x2x4 176.735 174.398 174.600 194.369 192.006 196.211 186.096 249.165 184.446 269.355 257.352 197.782 274.853 264.670 226.777 188.333 328.434 289.633 195.344 269.800 297.294 192.822 240.828 352.781 207.477 214.602 287.794 249.038 613.411 266.542 317.783 2949.13 3022.18 3207.34 3563.40 3637.70 3694.97 3702.90 3803.00 3956.41 3998.77 4086.47 4131.88 4558.34 4721.55 4736.02 4873.16 4998.07 5136.91 5394.56 5563.87 5784.20 5824.58 6564.79 6902.14 7675.70 7691.30 7714.86 7752.69 8445.98 10089.94 10591.58 0.7816 0.7845 0.8058 0.7598 0.7865 0.7818 0.7931 0.6921 0.7949 0.6339 0.6502 0.8045 0.6264 0.6730 0.7198 0.8138 0.5536 0.6421 0.8069 0.7000 0.6327 0.7856 0.7322 0.5641 0.7949 0.8144 0.6444 0.7231 0.1663 0.7365 0.6466 151.681 166.223 174.819 189.881 190.564 170.847 184.285 192.172 189.107 189.373 199.520 192.000 202.533 210.853 219.630 207.542 217.814 209.232 216.934 234.565 231.374 216.583 248.123 248.621 249.604 257.732 249.221 264.324 259.968 297.640 294.044 178 Chapter 12 Multiple Linear Regression and Certain Nonlinear Regression Models (d) It appears that the model with x2 = LLS and x5 = Power is the best in terms of PRESS, s2, and \u2211 i |\u3b4i|. 12.51 (a) y\u2c6 = \u2212587.21085 + 428.43313x. (b) y\u2c6 = 1180.00032\u2212 192.69121x+ 35.20945x2. (c) The summary of the two models are given as: Model s2 R2 PRESS µY = \u3b20 + \u3b21x 1,105,054 0.8378 18,811,057.08 µY = \u3b20 + \u3b21x+ \u3b211x 2 430,712 0.9421 8,706,973.57 It appears that the model with a quadratic term is preferable. 12.52 The parameter estimate for \u3b24 is 0.22365 with a standard error of 0.13052. Hence, t = 1.71 with P -value = 0.6117. Fail to reject H0. 12.53 \u3c3\u2c62b1 = 20, 588.038, \u3c3\u2c6 2 b11 = 62.650, and \u3c3\u2c6b1b11 = \u22121, 103.423. 12.54 (a) The following is the summary of the models. Model s2 R2 PRESS Cp x2x3 x2 x1x2 x1x2x3 x3 x1 x1x3 8094.15 8240.05 8392.51 8363.55 8584.27 8727.47 8632.45 0.51235 0.48702 0.49438 0.51292 0.46559 0.45667 0.47992 282194.34 282275.98 289650.65 294620.94 297242.74 304663.57 306820.37 2.0337 1.5422 3.1039 4.0000 2.8181 3.3489 3.9645 (b) The model with ln(x2) appears to have the smallest Cp with a small PRESS. Also, the model ln(x2) and ln(x3) has the smallest PRESS. Both models appear to better than the full model. 12.55 (a) There are many models here so the model summary is not displayed. By using MSE criterion, the best model, contains variables x1 and x3 with s 2 = 313.491. If PRESS criterion is used, the best model contains only the constant term with s2 = 317.51. When the Cp method is used, the best model is model with the constant term. (b) The normal probability plot, for the model using intercept only, is shown next. We do not appear to have the normality. \u22122 \u22121 0 1 2 \u2212 2. 0 \u2212 1. 5 \u2212 1. 0 \u2212 0. 5 0. 0 0. 5 1. 0 Normal Q\u2212Q Plot Theoretical Quantiles Sa m pl e Qu an tile s Solutions for Exercises in Chapter 12 179 12.56 (a) V\u302olt = \u22121.64129 + 0.000556 Speed \u2212 67.39589 Extension. (b) P -values for the t-tests of the coefficients are all < 0.0001. (c) The R2 = 0.9607 and the model appears to have a good fit. The residual plot and a normal probability plot are given here. 5500 6000 6500 7000 7500 8000 8500 \u2212 40 0 \u2212 20 0 0 20 0 40 0 y^ R es id ua l \u22122 \u22121 0 1 2 \u2212 40 0 \u2212 20 0 0 20 0 40 0 Normal Q\u2212Q Plot Theoretical Quantiles Sa m pl e Qu an tile s 12.57 (a) y\u2c6 = 3.13682 + 0.64443x1 \u2212 0.01042x2 + 0.50465x3 \u2212 0.11967x4 \u2212 2.46177x5 + 1.50441x6. (b) The final model using the stepwise regression is y\u2c6 = 4.65631 + 0.51133x3 \u2212 0.12418x4. (c) Using Cp criterion (smaller the better), the best model is still the model stated in (b) with s2 = 0.73173 and R2 = 0.64758. Using the s2 criterion, the model with x1, x3 and x4 has the smallest value of 0.72507 and R 2 = 0.67262. These two models are quite competitive. However, the model with two variables has one less variable, and thus may be more appealing. (d) Using the model in part (b), displayed next is the Studentized residual plot. Note that observations 2 and 14 are beyond the 2 standard deviation lines. Both of those observations may need to be checked. 8 9 10 11 12 \u2212 2 \u2212 1 0 1 2 y^ St ud en tiz ed R es id ua l 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 12.58 The partial F -test shows a value of 0.82, with 2 and 12 degrees of freedom. Conse- quently, the P -value = 0.4622, which indicates that variables x1 and x6 can be excluded from the model. 180 Chapter 12 Multiple Linear Regression and Certain Nonlinear Regression Models 12.59 (a) y\u2c6 = 125.86555 + 7.75864x1 + 0.09430x2 \u2212 0.00919x1x2. (b) The following is the summary of the models. Model s2 R2 PRESS Cp x2 x1 x1x2 x1x2x3 680.00 967.91 650.14 561.28 0.80726 0.72565 0.86179 0.92045 7624.66 12310.33 12696.66 15556.11 2.8460 4.8978 3.4749 4.0000 It appears that the model with x2 alone is the best. 12.60 (a) The fitted model is y\u2c6 = 85.75037\u221215.93334x1+2.42280x2+1.82754x3+3.07379x4. (b) The summary of the models are given next. Model s2 PRESS R2 Cp x1x2x4 x4 x3x4 x1x2x3x4 x1x4 x2x4 x2x3x4 x1x3x4 x2 x3 x2x3 x1 x1x3 x1x2 x1x2x3 9148.76 19170.97 21745.08 10341.20 10578.94 21630.42 25160.18 12341.87 160756.81 171264.68 183701.86 95574.16 107287.63 109137.20 125126.59 447, 884.34 453, 304.54 474, 992.22 482, 210.53 488, 928.91 512, 749.78 532, 065.42 614, 553.42 1, 658, 507.38 1, 888, 447.43 1, 896, 221.30 2, 213, 985.42 2, 261, 725.49 2, 456, 103.03 2, 744, 659.14 0.9603 0.8890 0.8899 0.9626 0.9464 0.8905 0.8908 0.9464 0.0695 0.0087 0.0696 0.4468 0.4566 0.4473 0.4568 3.308 8.831 10.719 5.000 3.161 10.642 12.598 5.161 118.362 126.491 120.349 67.937 68.623 69.875 70.599 When using PRESS as well as the s2 criterion, a model with x1, x2 and x4 appears to be the best, while when using the Cp criterion, the model with x1 and x4 is the best. When using the model with x1, x2 and x4, we find out that the P -value for testing \u3b22 = 0 is 0.1980 which implies that perhaps x2 can be excluded from the model. (c) The model in part (b) has smaller Cp as well as competitive PRESS in comparison to the full model. 12.61 Since H = X(X\u2018X)\u22121X\u2018, and n\u2211 i=1 hii = tr(H), we have n\u2211 i=1 hii = tr(X(X\u2018X) \u22121X\u2018) = tr(X\u2018X(X\u2018X)\u22121) = tr(Ip) = p, where