Residuals sum of squares 153758603 standard error of

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Residuals Sum of squares = 1537.58603 Standard error of e = 6.82594 Fit R-squared = .98284 --------+------------------------------------------------------------- Variable| Coefficient Standard Error b/St.Er. P[|Z|>z] Mean of X --------+------------------------------------------------------------- |Covariance matrix based on 50 replications. Constant| -84.0258*** 16.08614 -5.223 .0000 Y| .03784*** .00271 13.952 .0000 9232.86 PG| -17.0990*** 4.37160 -3.911 .0001 2.31661 --------+------------------------------------------------------------- Ordinary least squares regression ............ Residuals Sum of squares = 1472.79834 Standard error of e = 6.68059 Standard errors are based on Fit R-squared = .98356 50 bootstrap replications --------+------------------------------------------------------------- Variable| Coefficient Standard Error t-ratio P[|T|>t] Mean of X --------+------------------------------------------------------------- Constant| -79.7535*** 8.67255 -9.196 .0000 Y| .03692*** .00132 28.022 .0000 9232.86 PG| -15.1224*** 1.88034 -8.042 .0000 2.31661 --------+------------------------------------------------------------- ™    19/35
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Part 7: Estimating the Variance of  b Multicollinearity Not “short rank,” which is a deficiency in the model. A characteristic of the data set which affects the covariance matrix. Regardless, is unbiased. Consider one of the unbiased coefficient estimators of k. E[bk] = k Var[ b ] = 2( X’X )-1 . The variance of bk is the k th diagonal element of 2( X’X )-1 . We can isolate this with the result in your text. Let [X,z ] be [Other x s , x k] = [ X 1, x 2] (a convenient notation for the results in the text). We need the residual maker, MX . The general result is that the diagonal element we seek is [ z M 1 z ]-1 , which we know is the reciprocal of the sum of squared residuals in the regression of z on X . ™    20/35
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Part 7: Estimating the Variance of  b Variance of Least Squares 1 2 2 2 1 Model: = + + Variance of c Variance of c is the lower right element of this matrix. Var[c] = [ ] * * where * = the vector of residuals from the - - γ = σ ÷ σ σ = X y z b X X X z z X z z z M z z z z ε ( 29 ( 29 2 2 n 2 i i 1 n 2 2 i i 1 2 2 1 n 2 2 i i 1 regression of on . * * The R in that regression is R = 1 - , so (z z) * * 1 R (z z) . Therefore, Var[c] = [ ] 1 R (z z) = = - = - = - - σ σ = - - z|X X z|X z X z z z z z M z z|X ™    21/35
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Part 7: Estimating the Variance of  b Multicollinearity ( 29 2 2 1 n 2 2 i i 1 Var[c] = [ ] 1 R (z z) All else constant, the variance of the coefficient on rises as the fit in the regression of on the other variables goes up. If the fit is perfect, the - = σ σ = - - X z|X z M z z z ( 29 2 variance becomes infinite. 1 Variance inflation factor: VIF(z) = . 1 R - z|X ™    22/35
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Part 7: Estimating the Variance of  b Gasoline Market Regression Analysis: logG versus logIncome, logPG The regression equation is logG = - 0.468 + 0.966 logIncome - 0.169 logPG Predictor Coef SE Coef T P Constant -0.46772 0.08649 -5.41 0.000 logIncome 0.96595 0.07529 12.83 0.000 logPG -0.16949 0.03865 -4.38 0.000 S = 0.0614287 R-Sq = 93.6% R-Sq(adj) = 93.4% Analysis of Variance Source DF SS MS F P Regression 2 2.7237 1.3618 360.90 0.000 Residual Error 49 0.1849 0.0038 Total 51 2.9086 ™    23/35
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Part 7: Estimating the Variance of  b Gasoline Market Regression Analysis: logG versus logIncome, logPG, ... The regression equation is logG = - 0.558 + 1.29 logIncome - 0.0280 logPG - 0.156 logPNC + 0.029 logPUC - 0.183 logPPT Predictor Coef SE Coef T P Constant -0.5579 0.5808 -0.96 0.342 logIncome 1.2861 0.1457 8.83 0.000 logPG -0.02797 0.04338 -0.64 0.522 logPNC -0.1558 0.2100 -0.74 0.462 logPUC 0.0285 0.1020 0.28 0.781 logPPT -0.1828 0.1191 -1.54 0.132 S = 0.0499953 R-Sq = 96.0% R-Sq(adj) = 95.6% Analysis of Variance Source DF SS MS F P Regression 5 2.79360 0.55872 223.53 0.000 Residual Error 46 0.11498 0.00250 Total 51 2.90858 The standard error on logIncome doubles when the three variables are added to the equation.
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