Advanced Econometrics

Transcription

1 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) (29.0) (1.30) (28.3) (1.30) (12.4) (1.22) (16.1) (1.27) [1.47] 0.198** [3.73] [1.52] [0.17] Sample size 1,233 1,161 1,705 1,527 Michigan Mean duration (weeks) (34.7) (39.5) (26.4) (34.1) [1.34] [1.45] Mean *

2 ln(duration) (1.42) (1.45) (1.35) (1.35) [2.16] [1.17] Sample size (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) = afchnge highearn (afchnge x highearn) (12.8)** (0.31) (3.91)** (3.13)** male married age + u (0.87) (2.14)* (4.86)** Michigan log(duration) = afchnge highearn (afchnge x highearn) (7.81)** (0.96) (0.99) (1.20) male married 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) = highearn male married 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) = highearn male married 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) = highearn male married 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 = [ ( )]/ 5 = 4.35 ( ) /[ )] This is larger than the critical F-value for degrees of freedom of five and a large number, Thus, the result indicates a structural change before and after the change in the compensation law in Kentucky. 2

3 Michigan All log(duration) = highearn male married 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) = highearn male married 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) = highearn male married 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 = [ ( )]/ 5 = 1.33 ( ) /[ )] This is smaller than the critical F-value for degrees of freedom of five and a large number, 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 ). 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 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 in Wooldridge is not a constant term but a year dummy for 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 = Obs per group: min = 2 between = avg = 2.0 overall = max = 2 F(3,48) = corr(u_i, Xb) = Prob > F = dthrte Coef. Std. Err. t P> t [95% Conf. Interval] 3

4 open admn year _cons sigma_u sigma_e rho (fraction of variance due to u_i) F test that all u_i=0: F(50, 48) = 9.46 Prob > F = (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 = Obs per group: min = 2 between = avg = 2.0 overall = max = 2 F(3,48) = corr(u_i, Xb) = Prob > F = dthrte Coef. Std. Err. t P> t [95% Conf. Interval] open admn year _cons sigma_u sigma_e rho (fraction of variance due to u_i) F test that all u_i=0: F(50, 48) = 9.46 Prob > F = 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 = Obs per group: min = 2 between = avg = 2.0 overall = max = 2 Random effects u_i ~ Gaussian Wald chi2(3) = corr(u_i, X) = 0 (assumed) Prob > chi2 = dthrte Coef. Std. Err. z P> z [95% Conf. Interval] open admn year _cons sigma_u sigma_e rho (fraction of variance due to u_i) 4

5 . hausman all ---- Coefficients ---- (b) (B) (b-b) sqrt(diag(v_b-v_b)) all. Difference S.E. open admn year 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 = 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 = F( 2, 48) = 3.23 Model Prob > F = Residual R-squared = Adj R-squared = Total Root MSE =.3435 ddthrte Coef. Std. Err. t P> t [95% Conf. Interval] dopen dadmn _cons 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

6 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