Advanced Econometrics Instructor: Takashi Yamano 11/14/2003 Due: 11/21/2003 Homework 5 (30 points) Sample Answers 1. (16 points) Read Example 13.4 and an AER paper by Meyer, Viscusi, and Durbin (1995). We will reproduce their results by using data called INJURY.dta. (a) Meyer, Viscusi, and Durbin (1995) want to measure impacts of a policy change on compensation laws in Kentucky. Who are in control groups and treatment groups? Why? Control groups: low-income workers Treatment groups: high-income workers In 1980, Kentucky raised the maximum benefit from $131 to $217. In 1982, Michigan raised it from $181 to $307. Before the changes in the benefits in Kentucky and Michigan, high-income workers, whose weekly earnings were higher than the maximum benefits, had an incentive to go back to work because their benefits were significantly lower than their weekly earnings. The increases in the maximum benefits, however, made it less costly for high-income workers to stay away from work. In contrast, the increases in the maximum benefits did not affect low-income workers incentive to stay away from work. Thus, we can think high-income workers as treatment groups and low-income workers as control groups for the changes in compensation laws in Kentucky and Michigan. Kentucky Mean duration (weeks) Mean ln(duration) (b) By using the data, fill in the shaded areas in the following table for Kentucky and Michigan. (use Kentucky dummy: ky=1 for Kentucky) Treatment Group Control Group Differences DID Before (1) After (2) Before (3) After (4) (2)-(1) (5) (4)-(3) (6) (5)-(6) (7) 11.18 (29.0) 1.382 (1.30) 12.89 (28.3) 1.580 (1.30) 6.271 (12.4) 1.126 (1.22) 7.037 (16.1) 1.133 (1.27) 1.717 [1.47] 0.198** [3.73] 0.766 [1.52] 0.008 [0.17] 0.951 0.190 Sample size 1,233 1,161 1,705 1,527 Michigan Mean duration 14.78 19.43 10.96 13.65 4.655 2.692 1.963 (weeks) (34.7) (39.5) (26.4) (34.1) [1.34] [1.45] Mean 1.582 1.871 1.413 1.510 0.289* 0.097 0.192 1
ln(duration) (1.42) (1.45) (1.35) (1.35) [2.16] [1.17] Sample size 239 219 589 477 (c) Estimate the following equation for Kentucky and Michigan separately and report the results: log(duration) = $ 0 + $ 1 afchnge + $ 2 highearn + $ 3 (afchnge x highearn) + $ 4 male + $ 5 married + $ 6 age + u Kentucky log(duration) = 0.866 + 0.014 afchnge + 0.202 highearn + 0.218 (afchnge x highearn) (12.8)** (0.31) (3.91)** (3.13)** - 0.039 male + 0.087 married + 0.007 age + u (0.87) (2.14)* (4.86)** Michigan log(duration) = 1.118 + 0.082 afchnge + 0.111 highearn + 0.186 (afchnge x highearn) (7.81)** (0.96) (0.99) (1.20) - 0.207 male - 0.030 married + 0.014 age + u (2.14)* (0.37) (4.64)** (d) Estimate the same model in (c) for samples before (afchnge=0) and after (afchnge=1) the policy change separately. Do the Chow test. Has there been a structural change? Kentucky All log(duration) = 0.867 + 0.309 highearn - 0.037 male + 0.088 married + 0.007 age + u (13.5)** (7.81)** (0.82) (2.17)* (4.92)** SSR=8488.8, R 2 =0.0254, n=5360, k+1=5 Kentucky Before log(duration) = 0.896 + 0.215 highearn - 0.101 male + 0.109 married + 0.007 age + u (10.4)** (4.04)** (1.66) (1.96)* (3.48)** SSR=4229.9, R 2 =0.0188, n=2787, k+1=5 Kentucky After log(duration) = 0.850 + 0.404 highearn + 0.030 male + 0.061 married + 0.007 age + u (8.91)** (6.88)** (0.45) (1.03) (3.35)** SSR=4224.5, R 2 =0.0357, n=2573, k+1=5 F = [8488.8 (4229.9 + 4224.5)]/ 5 = 4.35 (4229.9 + 4224.5) /[5360 10)] This is larger than the critical F-value for degrees of freedom of five and a large number, 2.21. Thus, the result indicates a structural change before and after the change in the compensation law in Kentucky. 2
Michigan All log(duration) = 1.150 + 0.200 highearn - 0.210 male - 0.023 married + 0.014 age + u (8.24)** (2.36)* (2.16)* (0.29) (4.65)** SSR=2740.3, R 2 =0.0273, n=1484, k+1=5 Michigan Before log(duration) = 1.132 + 0.117 highearn - 0.169 male - 0.105 married + 0.014 age + u (6.05)** (1.01) (1.29) (0.97) (3.44)** SSR=1466.8, R 2 =0.0217, n=804, k+1=5 Michigan After log(duration) = 1.173 + 0.280 highearn - 0.254 male + 0.065 married + 0.014 age + u (5.58)** (2.26)* (1.76) (0.54) (3.14)** SSR=1261.2, R 2 =0.0365, n=680, k+1=5 F = [2740.3 (1466.8+ 1261.2)]/ 5 = 1.33 (1466.8+ 1261.2) /[1484 10)] This is smaller than the critical F-value for degrees of freedom of five and a large number, 2.21. Thus, the result indicates no structural changes before and after the change in the compensation law in Michigan. 2. (14 points) For this exercise use the data called TRAFFIC1_vertical.dta (I will send you this data set via email). Read Example 13.7 in Wooldridge. (a) Theory says that the FE estimators and FD estimators should be identical. Estimate a Fixed Effect model by using TRAFIC1_vertical.dta and report the results. (Use stateid for identification; use xtreg command.) Your answers should be identical to the ones in Equation 13.27 in page 447 in Wooldridge. The key to this question is that you need specify a year dummy for observations in 1990 in FE. In FD models, a constant term is eliminated. The constant term in Equation 13.27 in Wooldridge is not a constant term but a year dummy for 1990.. xtreg dthrte open admn year90, fe Fixed-effects (within) regression Number of obs = 102 Group variable (i): stateid Number of groups = 51 R-sq: within = 0.7375 Obs per group: min = 2 between = 0.0005 avg = 2.0 overall = 0.1186 max = 2 F(3,48) = 44.96 corr(u_i, Xb) = -0.2455 Prob > F = 0.0000 dthrte Coef. Std. Err. t P> t [95% Conf. Interval] 3
open -.4196787.2055948-2.04 0.047 -.8330547 -.0063028 admn -.1506024.1168223-1.29 0.204 -.3854894.0842846 year90 -.4967872.0524256-9.48 0.000 -.6021959 -.3913784 _cons 2.918364.0924787 31.56 0.000 2.732423 3.104304 sigma_u.57970511 sigma_e.24289261 rho.85066169 (fraction of variance due to u_i) F test that all u_i=0: F(50, 48) = 9.46 Prob > F = 0.0000 (b) Estimate a Random Effect model and conduct the Hausman test between the FE and RE models. Report the Hausman test statistic. (Hint: use the help command to learn how to conduct the Hausman test in STATA.). xtreg dthrte open admn year90, fe Fixed-effects (within) regression Number of obs = 102 Group variable (i): stateid Number of groups = 51 R-sq: within = 0.7375 Obs per group: min = 2 between = 0.0005 avg = 2.0 overall = 0.1186 max = 2 F(3,48) = 44.96 corr(u_i, Xb) = -0.2455 Prob > F = 0.0000 dthrte Coef. Std. Err. t P> t [95% Conf. Interval] open -.4196787.2055948-2.04 0.047 -.8330547 -.0063028 admn -.1506024.1168223-1.29 0.204 -.3854894.0842846 year90 -.4967872.0524256-9.48 0.000 -.6021959 -.3913784 _cons 2.918364.0924787 31.56 0.000 2.732423 3.104304 sigma_u.57970511 sigma_e.24289261 rho.85066169 (fraction of variance due to u_i) F test that all u_i=0: F(50, 48) = 9.46 Prob > F = 0.0000. est store all. xtreg dthrte open admn year90, re Random-effects GLS regression Number of obs = 102 Group variable (i): stateid Number of groups = 51 R-sq: within = 0.7236 Obs per group: min = 2 between = 0.0001 avg = 2.0 overall = 0.1816 max = 2 Random effects u_i ~ Gaussian Wald chi2(3) = 125.97 corr(u_i, X) = 0 (assumed) Prob > chi2 = 0.0000 dthrte Coef. Std. Err. z P> z [95% Conf. Interval] open -.1883466.1275398-1.48 0.140 -.43832.0616269 admn -.0316602.097594-0.32 0.746 -.222941.1596206 year90 -.5290525.0516877-10.24 0.000 -.6303585 -.4277465 _cons 2.783205.0992372 28.05 0.000 2.588704 2.977706 sigma_u.50356947 sigma_e.24289261 rho.81125791 (fraction of variance due to u_i) 4
. hausman all ---- Coefficients ---- (b) (B) (b-b) sqrt(diag(v_b-v_b)) all. Difference S.E. open -.4196787 -.1883466 -.2313322.1612539 admn -.1506024 -.0316602 -.1189422.0642095 year90 -.4967872 -.5290525.0322654.0087652 b = consistent under Ho and Ha; obtained from xtreg B = inconsistent under Ha, efficient under Ho; obtained from xtreg Test: Ho: difference in coefficients not systematic chi2(3) = (b-b)'[(v_b-v_b)^(-1)](b-b) = 6.24 Prob>chi2 = 0.1007 Thus, the Hausman test is not significant even at the 10 percent level. Thus, it can not reject the null hypothesis that the Random Effect model are biased because of time-invariant unobserved variables. (c) By using Example 9-1 in the lecture notes as an example, transfer TRAFIC1_vertical.dta to a linearized data (where each state has one row). And estimate the First Difference model. Report the results.. reg ddthrte dadmn dopen Source SS df MS Number of obs = 51 -------------+------------------------------ F( 2, 48) = 3.23 Model.762579785 2.381289893 Prob > F = 0.0482 Residual 5.66369475 48.117993641 R-squared = 0.1187 -------------+------------------------------ Adj R-squared = 0.0819 Total 6.42627453 50.128525491 Root MSE =.3435 ddthrte Coef. Std. Err. t P> t [95% Conf. Interval] dopen -.4196787.2055948-2.04 0.047 -.8330547 -.0063028 dadmn -.1506024.1168223-1.29 0.204 -.3854894.0842846 _cons -.4967872.0524256-9.48 0.000 -.6021959 -.3913784 The results are exactly same as in the FE model in (b). Here is a STATA do file that was used to manage the data. set more off set matsize 800 clear use c:\docs\fasid\classes\econometrics\wooldridge_data\traffic1_vertical.dt a keep if year90==0 keep state admn open dthrte stateid rename admn admn85 rename open open85 rename dthrte dthrte85 sort stateid 5
save c:\docs\tmp\traffic85.dta, replace clear use c:\docs\fasid\classes\econometrics\wooldridge_data\traffic1_vertical.dt a keep if year90==1 keep state admn open dthrte stateid rename admn admn90 rename open open90 rename dthrte dthrte90 sort stateid save c:\docs\tmp\traffic90.dta, replace clear use c:\docs\tmp\traffic85.dta sort stateid merge stateid using c:\docs\tmp\traffic90.dta gen dadmn=admn90-admn85 gen dopen=open90-open85 gen ddthrte=dthrte90-dthrte85 reg ddthrte dopen dadmn 6