在用heckman二步法矫正工资性别差异研究中的就业选择偏差问题时:
1.数据处理:将显示有工作但工资信息缺失的数据,将工资补为平均值;将没有工作的数据的工资信息保留为" ."不变(还是补为0??)
2.对男性进行heckman矫正,结果如下:
heckman lnwm gender educ exp nature scale gh hk exp2,select( gender educ exp hk) twostep
note: gender omitted because of collinearity
note: gender omitted because of collinearity
note: two-step estimate of rho = 2.2865901 is being truncated to 1
Heckman selection model -- two-step estimates Number of obs = 1838
(regression model with sample selection) Censored obs = 98
Uncensored obs = 1740
Wald chi2(7) = 2.12
Prob > chi2 = 0.9530
------------------------------------------------------------------------------
lnwm | Coef. Std. Err. z P>|z| [95% Conf. Interval]
-------------+----------------------------------------------------------------
lnwm |
gender | 0 (omitted)
educ | .1523852 .2055005 0.74 0.458 -.2503884 .5551588
exp | -.001987 .0667335 -0.03 0.976 -.1327822 .1288083
nature | -.0768735 .6000659 -0.13 0.898 -1.252981 1.099234
scale | -.0170609 .5376126 -0.03 0.975 -1.070762 1.03664
gh | .1287478 .5940804 0.22 0.828 -1.035628 1.293124
hk | -.3742824 2.02939 -0.18 0.854 -4.351814 3.603249
exp2 | -.0005622 .0016192 -0.35 0.728 -.0037357 .0026114
_cons | -.2126611 3.135731 -0.07 0.946 -6.358581 5.933258
-------------+----------------------------------------------------------------
select |
gender | 0 (omitted)
educ | .0278653 .0164336 1.70 0.090 -.0043439 .0600746
exp | -.0126519 .0037872 -3.34 0.001 -.0200746 -.0052292
hk | -.3336197 .1292501 -2.58 0.010 -.5869453 -.080294
_cons | 1.902672 .2202098 8.64 0.000 1.471069 2.334275
-------------+----------------------------------------------------------------
mills |
lambda | 11.76152 32.77764 0.36 0.720 -52.48148 76.00453
-------------+----------------------------------------------------------------
rho | 1.00000
sigma | 11.761524
------------------------------------------------------------------------------
请问:
①所说的逆mills比率是看lambda数值么?数值大小有什么 含义呢?
②为什么这里上边的哪些变量都不显著了呢?用ols明明大部分都显著了呀?这说明什么?是处理出了问题么?