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IV/2SLS models. Z i =0. Z i =1. 0 =0.80. 1 =0.57. 0 =3186. 1 =3278. Right hand term is (1/F) for the null hypothesis That the coefficients in the 1 st stage are all zero. 1. o. Β iv = ( 1 - 0 )/( 1 - 0 ) = -487.8/0.159 = $3067.9
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Zi=0 Zi=1 0=0.80 1=0.57
0=3186 1=3278
Right hand term is (1/F) for the null hypothesis That the coefficients in the 1st stage are all zero
1 o
Βiv= (1 - 0)/(1 - 0) = -487.8/0.159 = $3067.9 CPI78 = 65.2 CPI81=90.9 65.2/90.9 = .7173 .717*3067.92 = $2199
OLS of bivariate model IV of bivariate Model (Wald Est) 0.0020246/0.0291243 = 0.0695 Ratio of std errors should equal corr coef From previous page
Notice t-stat on Reduced form Is almost the same As t-stat in 2SLS 0.12/.028 = 4.285 IV estimate -0.0083481/0.0694 = -0.12031
LIST OF EXOGENOUS VARIABLES ALL VARIABLES NOT IN LIST ARE CONSIDERED ENDOGENOUS STRUCTURAL MODEL
Can reject at 5.1 percent the null the coefficients are The same
R2 is useless because of Rounding – must calculate yourself Output residuals from 2LSL model Regress on all exo factors Get test of overid by brute force
SSM = 1.467 • SST = 60444.5 • R2 = SSM/SST = 2.43E-5 • N = 254654 • NR2 = 6.18 • Dist as χ2(1) • P-value of 6.18 is 0.0129
Example • Suppose a school district requires that a child turn 6 by October 31 in the 1st grade • Has compulsory education until age 18 • Consider two kids • One born Oct 1, 1960 • Another born Nov 1,1960
Oct 1, 1960 • Starts school in 1966 (age 5) • Turns 6 a few months into school • Starts senior year in 1977 (age 16) • Does not turn 18 until after HS school is over • Nov 1, 1960 • Start school in 1967 (age 6) • Turns 7 a few months into school • Starts senior year in 1978 (age 17) • Turns 18 midway through senior year
Ratio of Std errors is 0.0003386/0.0239 = 0.0125 Abs[Rho(qob1,educ)] =0.0142
First-stage Reduced-forms -0.0110989/-0.1088179=.101995
