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Econometrics-I-8

2550 part 8 hypothesis testing application time

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™    25/50
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Part 8: Hypothesis Testing Application Time series regression, LogG = 1 + 2logY + 3logPG + 4logPNC + 5logPUC + 6logPPT + 7logPN + 8logPD + 9logPS + Period = 1960 - 1995. Note that all coefficients in the model are elasticities. ™    26/50
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Part 8: Hypothesis Testing Full Model ---------------------------------------------------------------------- Ordinary least squares regression ............ LHS=LG Mean = 5.39299 Standard deviation = .24878 Number of observs. = 36 Model size Parameters = 9 Degrees of freedom = 27 Residuals Sum of squares = .00855 <******* Standard error of e = .01780 <******* Fit R-squared = .99605 <******* Adjusted R-squared = .99488 <******* --------+------------------------------------------------------------- Variable| Coefficient Standard Error t-ratio P[|T|>t] Mean of X --------+------------------------------------------------------------- Constant| -6.95326*** 1.29811 -5.356 .0000 LY| 1.35721*** .14562 9.320 .0000 9.11093 LPG| -.50579*** .06200 -8.158 .0000 .67409 LPNC| -.01654 .19957 -.083 .9346 .44320 LPUC| -.12354* .06568 -1.881 .0708 .66361 LPPT| .11571 .07859 1.472 .1525 .77208 LPN| 1.10125*** .26840 4.103 .0003 .60539 LPD| .92018*** .27018 3.406 .0021 .43343 LPS| -1.09213*** .30812 -3.544 .0015 .68105 --------+------------------------------------------------------------- ™    27/50
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Part 8: Hypothesis Testing Test About One Parameter Is the price of public transportation really relevant? H0 : 6 = 0. Confidence interval: b6 t(.95,27)  Standard error = .11571  2.052(.07859) = .11571  .16127 = (-.045557 ,.27698) Contains 0.0. Do not reject hypothesis Regression fit if drop? Without LPPT, R-squared= .99573 Compare R2, was .99605, F(1,27) = [(.99605 - .99573)/1]/[(1-.99605)/(36-9)] = 2.187 = 1.4722 (with some rounding difference) ™    28/50
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Part 8: Hypothesis Testing Hypothesis Test: Sum of Coefficients Do the three aggregate price elasticities sum to zero? H0 :β7 + β8 + β9 = 0 R = [0, 0, 0, 0, 0, 0, 1, 1, 1], q = [0] Variable| Coefficient Standard Error t-ratio P[|T|>t] Mean of X --------+------------------------------------------------------------- LPN| 1.10125*** .26840 4.103 .0003 .60539 LPD| .92018*** .27018 3.406 .0021 .43343 LPS| -1.09213*** .30812 -3.544 .0015 .68105 ™    29/50
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Part 8: Hypothesis Testing Imposing the Restriction ---------------------------------------------------------------------- Linearly restricted regression LHS=LG Mean = 5.392989 Standard deviation = .2487794 Number of observs. = 36 Model size Parameters = 8 <*** 9 – 1 restriction Degrees of freedom = 28 Residuals Sum of squares = .0112599 <*** With the restriction Residuals Sum of squares = .0085531 <*** Without the restriction Fit R-squared = .9948020 Restrictns. F[ 1, 27] (prob) = 8.5(.01) Not using OLS or no constant.R2 & F may be < 0 --------+------------------------------------------------------------- Variable| Coefficient Standard Error t-ratio P[|T|>t] Mean of X --------+------------------------------------------------------------- Constant| -10.1507*** .78756 -12.889 .0000 LY| 1.71582*** .08839 19.412 .0000 9.11093 LPG| -.45826*** .06741 -6.798 .0000 .67409 LPNC| .46945*** .12439 3.774 .0008 .44320 LPUC| -.01566 .06122 -.256 .8000 .66361 LPPT| .24223*** .07391 3.277 .0029 .77208 LPN| 1.39620*** .28022 4.983 .0000 .60539 LPD| .23885 .15395 1.551 .1324 .43343 LPS| -1.63505*** .27700 -5.903 .0000 .68105 --------+------------------------------------------------------------- F = [(.0112599 - .0085531)/1] / [.0085531/(36 – 9)] = 8.544691 ™    30/50
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Part 8: Hypothesis Testing Joint Hypotheses Joint hypothesis: Income elasticity = +1, Own price elasticity = -1. The hypothesis implies that logG = β1 + logY – logPg + β4 logPNC + ... Strategy: Regress logG – logY + logPg on the other variables and Compare the sums of squares With two restrictions imposed Residuals Sum of squares = .0286877 Fit R-squared = .9979006 Unrestricted Residuals Sum of squares = .0085531 Fit R-squared = .9960515 F = ((.0286877 - .0085531)/2) / (.0085531/(36-9)) = 31.779951 The critical F for 95% with 2,27 degrees of freedom is 3.354. The hypothesis is rejected. ™    31/50
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Part 8: Hypothesis Testing Basing the Test on R2 After building the restrictions into the model and computing restricted and unrestricted regressions: Based on R2s, F = ((.9960515 - .997096)/2)/((1-.9960515)/(36-9)) = -3.571166 (!) What's wrong? The unrestricted model used LHS = logG.
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