小弟我根据这个model 要分析foreign与emp short term和long term的关系 以下是我的分析 我怎么看他们之间的关系呢? 谢谢~
empit=β0+αemp(it-1)+β1wageit+β2ageit+β3foreignit+fi+εit
不好意思 不会打下标 model里的it β后的数字都是下标
. xtdpd emp L.emp wage age foreign, dgmm(emp)
Dynamic panel-data estimation Number of obs = 8659
Group variable: firm Number of groups = 1444
Time variable: year
Obs per group: min = 4
avg = 5.996537
max = 6
Number of instruments = 16 Wald chi2(4) = 38.36
Prob > chi2 = 0.0000
One-step results
------------------------------------------------------------------------------
emp | Coef. Std. Err. z P>|z| [95% Conf. Interval]
-------------+----------------------------------------------------------------
emp |
L1. | .0319678 .0631332 0.51 0.613 -.0917709 .1557065
wage | -1.787048 .500081 -3.57 0.000 -2.767189 -.806907
age | .5291213 .4608174 1.15 0.251 -.3740643 1.432307
foreign | .6019111 .272765 2.21 0.027 .0673015 1.136521
_cons | -1.67914 1.169679 -1.44 0.151 -3.971669 .6133897
------------------------------------------------------------------------------
Instruments for differenced equation
GMM-type: L(2/.).emp
Instruments for level equation
Standard: _cons
. xtdpd emp L.emp wage age foreign, dgmm(emp) two
Dynamic panel-data estimation Number of obs = 8659
Group variable: firm Number of groups = 1444
Time variable: year
Obs per group: min = 4
avg = 5.996537
max = 6
Number of instruments = 16 Wald chi2(4) = 13.04
Prob > chi2 = 0.0111
Two-step results
------------------------------------------------------------------------------
emp | Coef. Std. Err. z P>|z| [95% Conf. Interval]
-------------+----------------------------------------------------------------
emp |
L1. | .0764218 .0442999 1.73 0.085 -.0104044 .1632481
wage | -.175313 .6632977 -0.26 0.792 -1.475353 1.124727
age | .2582434 .4843928 0.53 0.594 -.6911491 1.207636
foreign | .477978 .3105539 1.54 0.124 -.1306964 1.086652
_cons | -1.065822 1.374842 -0.78 0.438 -3.760463 1.628818
------------------------------------------------------------------------------
Warning: gmm two-step standard errors are biased; robust standard
errors are recommended.
Instruments for differenced equation
GMM-type: L(2/.).emp
Instruments for level equation
Standard: _cons
. estat sargan
Sargan test of overidentifying restrictions
H0: overidentifying restrictions are valid
chi2(11) = 5.907061
Prob > chi2 = 0.8795
. estat abond
Arellano-Bond test for zero autocorrelation in first-differenced errors
+-----------------------+
|Order | z Prob > z|
|------+----------------|
| 1 |-2.2347 0.0254 |
| 2 | -.9199 0.3576 |
+-----------------------+
H0: no autocorrelation