ECON 4706 B Winter 2022 Solution Set 3

ECON 4706 B Winter 2022 Simon Power Solution Set 3: Total Marks = 120 1. [10 marks] Answer Question 6.15 on p. 288. Originally, the reparameterization involved substituting ? 3 = 𝜃 − ? 2 into the regular 3-variable model, yielding ? = ? 1 + ? 2 ? 2 + (𝜃 − ? 2 )? 3 + ? = ? 1 + ? 2 (? 2 − ? 3 ) + 𝜃? 3 + ? (6.34) This last can be estimated by generating a new variable ?1 = ? 2 − ? 3 and then running a regression of ? on ?1 and ? 3 (and a constant). The estimated parameter associated with ? 3 can then be interpreted as an estimate of 𝜃 . If you instead substitute using ? 2 = 𝜃 − ? 3 , this yields ? = ? 1 + (𝜃 − ? 3 )? 2 + ? 3 ? 3 + ? = ? 1 + 𝜃? 2 + ? 3 (? 3 − ? 2 ) + ? This last can be estimated by generating a new variable ?2 = ? 3 − ? 2 and then running a regression of ? on ? 2 and ?2 (and a constant). The estimated parameter associated with ? 2 can then be interpreted as an estimate of 𝜃 . Therefore, the only difference would be that in the first case you need to look at the estimated parameter associated with ? 3 , whereas in the second case you need to look at the estimated parameter associated with ? 2 . (10 marks) 2. [60 marks] A researcher suggests that the following model might be useful for analyzing earnings ???????? 𝑖 = ? 1 + ? 2 ? 𝑖 + ? 3 ??? 𝑖 + ? 4 ?????? 𝑖 + ? 5 ???? 𝑖 + ? 6 ???????? 𝑖 + ? 7 ??????? 𝑖 + ? 𝑖 𝑖 = 1, 2, … , 500 where the variables are as defined in Appendix B (pp. 565-569). Using the EAWE17.dta dataset: a) Estimate the model, and then copy and paste your output into your assignment. . reg EARNINGS S EXP ASVABC MALE ETHBLACK ETHHISP

2 Source | SS df MS Number of obs = 500 -------------+---------------------------------- F(6, 493) = 9.62 Model | 6176.2891 6 1029.38152 Prob > F = 0.0000 Residual | 52757.3313 493 107.012842 R-squared = 0.1048 -------------+---------------------------------- Adj R-squared = 0.0939 Total | 58933.6204 499 118.103448 Root MSE = 10.345 ------------------------------------------------------------------------------ EARNINGS | Coefficient Std. err. t P>|t| [95% conf. interval] -------------+---------------------------------------------------------------- S | .8889189 .2429088 3.66 0.000 .4116548 1.366183 EXP | .375916 .1947544 1.93 0.054 -.006735 .758567 ASVABC | 1.711967 .6423381 2.67 0.008 .4499087 2.974024 MALE | 3.034885 .948896 3.20 0.001 1.170506 4.899264 ETHBLACK | -1.214694 1.435579 -0.85 0.398 -4.035301 1.605913 ETHHISP | 1.598289 1.510863 1.06 0.291 -1.370237 4.566814 _cons | .995592 4.411777 0.23 0.822 -7.672613 9.663797 ------------------------------------------------------------------------------ (5 marks) b) Another researcher now comes along and wonders whether work experience with previous employer(s) is of equal value to a worker as work experience with current employer. In order to investigate this issue, estimate a new version of the model in which work experience, EXP, is divided into two parts, namely, work experience with previous employer(s), PREVEXP, and work experience with current employer, TENURE, where PREVEXP = EXP – TENURE. Copy and paste your output into your assignment. . gen PREVEXP = EXP - TENURE . reg EARNINGS S PREVEXP TENURE ASVABC MALE ETHBLACK ETHHISP Source | SS df MS Number of obs = 500 -------------+---------------------------------- F(7, 492) = 8.32 Model | 6240.01803 7 891.431147 Prob > F = 0.0000 Residual | 52693.6023 492 107.100818 R-squared = 0.1059 -------------+---------------------------------- Adj R-squared = 0.0932 Total | 58933.6204 499 118.103448 Root MSE = 10.349 ------------------------------------------------------------------------------ EARNINGS | Coefficient Std. err. t P>|t| [95% conf. interval] -------------+---------------------------------------------------------------- S | .8500501 .2481777 3.43 0.001 .3624313 1.337669 PREVEXP | .319232 .2082312 1.53 0.126 -.0899001 .7283642 TENURE | .4634436 .2254671 2.06 0.040 .0204464 .9064408 ASVABC | 1.727731 .642927 2.69 0.007 .4645101 2.990952 MALE | 3.020357 .9494728 3.18 0.002 1.154836 4.885879 ETHBLACK | -1.202964 1.436249 -0.84 0.403 -4.024903 1.618974 ETHHISP | 1.596846 1.511485 1.06 0.291 -1.372917 4.566609 _cons | 1.428917 4.449195 0.32 0.748 -7.31285 10.17068 ------------------------------------------------------------------------------ (5 marks) NOTE: For parts c), d), and e), you may NOT use the STATA “test” or “lincom” commands. c) Test to see whether the impact of work experience with previous employer(s) is equal to that of work experience with current employer, that is, test the null hypothesis ? 0 : ? 1 = ? 2 against the

4 which we can rewrite as ? 1 = 𝜃 + ? 2 This can now be substituted into the model from part b), namely, the model ???????? 𝑖 = ? 1 + ? 2 ? 𝑖 + ? 1 ??????? 𝑖 + ? 2 ?????? 𝑖 + ? 4 ?????? 𝑖 + ? 5 ???? 𝑖 + ? 6 ???????? 𝑖 + ? 7 ??????? 𝑖 + ? 𝑖 yielding ???????? 𝑖 = ? 1 + ? 2 ? 𝑖 + (𝜃 + ? 2 )??????? 𝑖 + ? 2 ?????? 𝑖 + ? 4 ?????? 𝑖 + ? 5 ???? 𝑖 + ? 6 ???????? 𝑖 + ? 7 ??????? 𝑖 + ? 𝑖 which we can rewrite as ???????? 𝑖 = ? 1 + ? 2 ? 𝑖 + 𝜃??????? 𝑖 + ? 2 ??? 𝑖 + ? 4 ?????? 𝑖 + ? 5 ???? 𝑖 + ? 6 ???????? 𝑖 + ? 7 ??????? 𝑖 + ? 𝑖 where we use the fact that ??? = ??????? + ?????? . We can now run this last model, in which case the t-statistic associated with the estimated slope parameter for PREVEXP will be the relevant t-statistic for testing the null hypothesis of interest. . reg EARNINGS S PREVEXP EXP ASVABC MALE ETHBLACK ETHHISP Source | SS df MS Number of obs = 500 -------------+---------------------------------- F(7, 492) = 8.32 Model | 6240.01802 7 891.431146 Prob > F = 0.0000 Residual | 52693.6023 492 107.100818 R-squared = 0.1059 -------------+---------------------------------- Adj R-squared = 0.0932 Total | 58933.6204 499 118.103448 Root MSE = 10.349 ------------------------------------------------------------------------------ EARNINGS | Coefficient Std. err. t P>|t| [95% conf. interval] -------------+---------------------------------------------------------------- S | .8500501 .2481777 3.43 0.001 .3624313 1.337669 PREVEXP | -.1442116 .1869512 -0.77 0.441 -.5115329 .2231097 EXP | .4634436 .2254671 2.06 0.040 .0204464 .9064408 ASVABC | 1.727731 .642927 2.69 0.007 .4645101 2.990952 MALE | 3.020357 .9494728 3.18 0.002 1.154836 4.885879 ETHBLACK | -1.202964 1.436249 -0.84 0.403 -4.024903 1.618974 ETHHISP | 1.596846 1.511485 1.06 0.291 -1.372917 4.566609

5 _cons | 1.428917 4.449195 0.32 0.748 -7.31285 10.17068 ------------------------------------------------------------------------------ Looking at the “reg” output, we can see that the relevant t -statistic is equal to -0.77 (which matches up with the t-statistic from part c) above). Comparing this t-statistic of -0.77 with the critical values of ±1.965 (500 degrees of freedom, closest to the correct 492 degrees of freedom, in Table A.2), we conclude that we should not reject the null hypothesis at the 5% significance level. This suggests that the impact of work experience with previous employer(s) is equal to that of work experience with current employer. (10 marks) e) Repeat part c) using the general F-test approach, described on pp. 281-282. Here we are interested in testing the single linear restriction ? 1 − ? 2 = 0 . As there is only a single linear restriction, the general F-statistic formula simplifies to the formula given in equation (6.27) ? = (???? − ????) ???? (? − ?) ⁄ where, as usual, ???? denotes the ??? from the restricted model and ???? denotes the ??? from the unrestricted model. To make progress, we need to work out how to estimate the restricted model. But, recall that the original model, the model from part a), imposes this very restriction. So, we can just use the RSS from the model in part a) as our ???? . Of course, the unrestricted model in this case is just the model that we estimated in part b). Therefore, the F-statistic can now be calculated as ? = (???? − ????) ???? (? − ?) ⁄ = (52757.3313 − 52693.6023) 52693.6023 (500 − 8) ⁄ ≈ 0.5950 Comparing this F-statistic of 0.5950 with the critical value of 3.86 (1 and 500 degrees of freedom, closest to the correct 1 and 492 degrees of freedom, in Table A.3), we conclude that we should not reject the null hypothesis at the 5% significance level. This suggests that the impact of work experience with previous employer(s) is equal to that of work experience with current employer. (10 marks) f) Check your ans wer to part e) by using the STATA “test” command. Copy and paste your output into your assignment. . quietly reg EARNINGS S PREVEXP TENURE ASVABC MALE ETHBLACK ETHHISP . test PREVEXP - TENURE = 0 ( 1) PREVEXP - TENURE = 0

7 We need to test the null hypothesis ? 0 : ℎ??????????𝑖?𝑖?𝑦 against the alternative hypothesis ? 1 : ℎ????????????𝑖?𝑖?𝑦 To implement the (regular) White Test, we need to save the residuals from the original regression, and then run an auxiliary regression of the squared residuals on a constant, the original explanatory variables, their squares, and cross-products, eliminating any duplicates or redundancies, and finally calculate the White statistic and assess its significance or otherwise. . quietly reg EARNINGS S EXP MALE . predict uhat, residuals . gen uhat2 = uhat*uhat . gen S2 = S*S . gen EXP2 = EXP*EXP . gen SEXP = S*EXP . gen SMALE = S*MALE . gen EXPMALE = EXP*MALE . reg uhat2 S EXP MALE S2 EXP2 SEXP SMALE EXPMALE Source | SS df MS Number of obs = 500 -------------+---------------------------------- F(8, 491) = 0.59 Model | 845687.293 8 105710.912 Prob > F = 0.7886 Residual | 88379048.5 491 179998.062 R-squared = 0.0095 -------------+---------------------------------- Adj R-squared = -0.0067 Total | 89224735.8 499 178807.086 Root MSE = 424.26 ------------------------------------------------------------------------------ uhat2 | Coefficient Std. err. t P>|t| [95% conf. interval] -------------+---------------------------------------------------------------- S | -16.04391 89.54707 -0.18 0.858 -191.9866 159.8988 EXP | 19.77196 73.21361 0.27 0.787 -124.0787 163.6226 MALE | 442.0115 319.9239 1.38 0.168 -186.5773 1070.6 S2 | 1.194691 2.704523 0.44 0.659 -4.119176 6.508557 EXP2 | .4142353 2.337004 0.18 0.859 -4.177527 5.005997 SEXP | -.9127528 3.561481 -0.26 0.798 -7.910377 6.084872 SMALE | -25.55789 17.2505 -1.48 0.139 -59.4518 8.336016 EXPMALE | -10.22291 15.75875 -0.65 0.517 -41.18582 20.74 _cons | 5.622095 783.5953 0.01 0.994 -1533.992 1545.236 ------------------------------------------------------------------------------ Calculating the White statistic, we obtain ? = ?? 2 = (500)(0.0095) ≈ 4.75

8 Comparing this White statistic of approximately 4.75 with the critical value of 15.507 from the chi-squared distribution (8 degrees of freedom in Table A.4), we conclude that we should not reject the null hypothesis at the 5% significance level. This suggests that we do not have a problem with heteroskedasticity in this model. (10 marks) b) Repeat part a), but this time use the variation of the White test in which the auxiliary regression consists in a regression of the squared residuals (from the original regression) on just a constant, the predictions (from the original regression), and the squared predictions (from the original regression). (Be sure to include a clear statement of the appropriate null and alternative hypotheses, the formula for the test statistic, and the necessary calculations for your test in your answer.) Explain. We need to test the null hypothesis ? 0 : ℎ??????????𝑖?𝑖?𝑦 against the alternative hypothesis ? 1 : ℎ????????????𝑖?𝑖?𝑦 . quietly reg EARNINGS S EXP MALE . predict EARNINGShat (option xb assumed; fitted values) . gen EARNINGShat2 = EARNINGShat*EARNINGShat . reg uhat2 EARNINGShat EARNINGShat2 Source | SS df MS Number of obs = 500 -------------+---------------------------------- F(2, 497) = 0.36 Model | 129699.51 2 64849.7551 Prob > F = 0.6966 Residual | 89095036.3 497 179265.667 R-squared = 0.0015 -------------+---------------------------------- Adj R-squared = -0.0026 Total | 89224735.8 499 178807.086 Root MSE = 423.4 ------------------------------------------------------------------------------ uhat2 | Coefficient Std. err. t P>|t| [95% conf. interval] -------------+---------------------------------------------------------------- EARNINGShat | 38.29927 47.69783 0.80 0.422 -55.41497 132.0135 EARNINGShat2 | -1.113954 1.339153 -0.83 0.406 -3.745054 1.517146 _cons | -209.0442 419.1176 -0.50 0.618 -1032.505 614.4164 ------------------------------------------------------------------------------ Calculating the White statistic, we obtain ? = ?? 2 = (500)(0.0015) ≈ 0.75 Comparing this White statistic of approximately 0.75 with the critical value of 5.991 from the chi- squared distribution (2 degrees of freedom in Table A.4), we conclude that we should not reject

10 -------------+---------------------------------------------------------------- S | 1.015831 .6350025 1.60 0.111 -.2371755 2.268837 EXP | .8307698 .3449942 2.41 0.017 .1500166 1.511523 MALE | 1.302787 1.344268 0.97 0.334 -1.349764 3.955337 _cons | -1.695978 11.87901 -0.14 0.887 -25.136 21.74405 ------------------------------------------------------------------------------ Calculating the Goldfeld-Quandt test statistic, we obtain ?? = (??? 2 𝑑𝑓 2 ⁄ ) (??? 1 𝑑𝑓 1 ⁄ ) = (14400.7005 180 ⁄ ) (17057.2484 180 ⁄ ) ≈ 0.844 Comparing this GQ-statistic of approximately 0.844 with the critical value of 1.26 from the F- distribution (200 and 200 degrees of freedom, the closest to the correct 180 and 180 degrees of freedom, in Table A.3), we conclude that we should not reject the null hypothesis at the 5% significance level. This suggests that we do not have a problem with heteroskedasticity in this model. IMPORTANT NOTE: The numerical answer to part c) of this question may vary somewhat due to the random way in which STATA can break ties when executing the gsort command. (10 marks) NOTE: For parts d) and e), do NOT include the MALE variable in your wage equation model. d) Use the Chow test, as described on pp. 255- 258, to test the “parameter stability” of this cross - sectional model. Divide the sample into two halves, so that the first half consists of the first 250 observations and the second half consists of the remaining 250 observations. (Be sure to include a clear statement of the appropriate null and alternative hypotheses, the formula for the test statistic, and the necessary calculations for your test in your answer.) Explain. NOTE: Be very careful that you have the observations in the correct order, that is, the same order in which they appeared in the original data set. HINT: One way to estimate a model in STATA using only the first 250 observations would be to use the following STATA code: gen Z = _n reg Y X2 X3 X4 if Z <= 250 This code creates a new variable, Z, which is equal to the observation number, and then estimates a (general) model using only the first 250 observations in the sample. You can adapt this code, as necessary, for the present problem. We need to test the null hypothesis ? 0 : ????????? ????𝑖?𝑖?𝑦 against the alternative hypothesis

11 ? 1 : ??ℎ???𝑖?? For convenience, let the first sub-sample regression be called Regression A, the second sub-sample regression be called Regression B, and the overall regression be called Regression P. We can estimate Regression A as . gen R = _n . reg EARNINGS S EXP if R <= 250 Source | SS df MS Number of obs = 250 -------------+---------------------------------- F(2, 247) = 5.34 Model | 1209.9506 2 604.975298 Prob > F = 0.0053 Residual | 27960.1397 247 113.198946 R-squared = 0.0415 -------------+---------------------------------- Adj R-squared = 0.0337 Total | 29170.0903 249 117.148957 Root MSE = 10.639 ------------------------------------------------------------------------------ EARNINGS | Coefficient Std. err. t P>|t| [95% conf. interval] -------------+---------------------------------------------------------------- S | .970185 .2981759 3.25 0.001 .3828933 1.557477 EXP | .3782088 .2705952 1.40 0.163 -.1547595 .9111771 _cons | 3.304496 5.475831 0.60 0.547 -7.480782 14.08977 ------------------------------------------------------------------------------ and Regression B as . reg EARNINGS S EXP if R >= 251 Source | SS df MS Number of obs = 250 -------------+---------------------------------- F(2, 247) = 14.41 Model | 3006.62224 2 1503.31112 Prob > F = 0.0000 Residual | 25759.8099 247 104.290728 R-squared = 0.1045 -------------+---------------------------------- Adj R-squared = 0.0973 Total | 28766.4321 249 115.52784 Root MSE = 10.212 ------------------------------------------------------------------------------ EARNINGS | Coefficient Std. err. t P>|t| [95% conf. interval] -------------+---------------------------------------------------------------- S | 1.534816 .2924356 5.25 0.000 .9588303 2.110801 EXP | .5379028 .2715091 1.98 0.049 .0031345 1.072671 _cons | -9.382973 5.540385 -1.69 0.092 -20.2954 1.529451 ------------------------------------------------------------------------------ and Regression P as . reg EARNINGS S EXP Source | SS df MS Number of obs = 500 -------------+---------------------------------- F(2, 497) = 15.83 Model | 3528.51731 2 1764.25865 Prob > F = 0.0000 Residual | 55405.1031 497 111.479081 R-squared = 0.0599 -------------+---------------------------------- Adj R-squared = 0.0561 Total | 58933.6204 499 118.103448 Root MSE = 10.558 ------------------------------------------------------------------------------ EARNINGS | Coefficient Std. err. t P>|t| [95% conf. interval] -------------+----------------------------------------------------------------

13 EARNINGS | Coefficient Std. err. t P>|t| [95% conf. interval] -------------+---------------------------------------------------------------- S | 1.534816 .2986151 5.14 0.000 .9481035 2.121528 EXP | .5379028 .2772464 1.94 0.053 -.0068247 1.08263 D | 12.68747 7.798188 1.63 0.104 -2.634238 28.00917 DS | -.5646308 .4178295 -1.35 0.177 -1.385573 .2563113 DEXP | -.159694 .3836746 -0.42 0.677 -.9135294 .5941414 _cons | -9.382973 5.657459 -1.66 0.098 -20.49862 1.732677 ------------------------------------------------------------------------------ Using these results, together with the results from the original (restricted) model from above, we can calculate the appropriate F-statistic [analogous to equation (5.40)] ? = (???? − ????) ? ⁄ ???? (? − 2?) ⁄ = (55405.1031 − 53719.9496) 3 ⁄ 53719.9496 (500 − 6) ⁄ ≈ 5.17 Comparing this F-statistic of approximately 5.17 with the critical value of 2.62 (3 and 500 degrees of freedom, the closest to the correct 3 and 494 degrees of freedom, in Table A.3), we conclude that we should eject the null hypothesis at the 5% significance level. This suggests that the model does not exhibit parameter stability. And, yes, we do get the same result as above in part d). This is not surprising, because the two methods of implementing the Chow test should be equivalent. (10 marks)