用非线性sur回归之后怎么对某个参数的估计值进行检验-经管之家官网！

用非线性sur回归之后怎么对某个参数的估计值进行检验

求教，我用非线性sur方法估计了一个三方程的系统，之后我想检验系数a2=1这个假设，请问应该如何做？
回归结果如下：
. nlsur(x1=ln({a1})+1/{a2}*(x2-{a3}*x3)+({a2}-1)/{a2}*ln({a4}))(x4=ln(1-{a1})+1/{a
> 2}*(x5-{a5}*x3)+({a2}-1)/{a2}*ln({a4}))(x6=ln({a4})+{a2}/({a2}-1)*ln({a1}*exp({a
> 3}*({a2}-1)*x3/{a2})*x7^(({a2}-1)/{a2})+(1-{a1})*exp({a5}*({a2}-1)*x3/{a2}) *x8^
> (({a2}-1)/{a2}))), initial(a1 0.2 a2 0.2 a3 0.2 a4 0.2 a5 0.2) vce(bootstrap)
(running nlsur on estimation sample)
Bootstrap replications (50)
----+--- 1 ---+--- 2 ---+--- 3 ---+--- 4 ---+--- 5
.................................................. 50
FGNLS regression
---------------------------------------------------------------------
Equation | ObsParms RMSE R-sq Constant
----------------+----------------------------------------------------
1 x1 | 37 4 .1398538 -6.9577 a1
2 x4 | 37 4 .5979394 -18.9117 a1
3 x6 | 37 5 .4347721 -9.9394 a4
---------------------------------------------------------------------
Bootstrap results
------------------------------------------------------------------------------
| Observed Bootstrap Normal-based
| Coef. Std. Err. z P>|z| [95% Conf. Interval]
-------------+----------------------------------------------------------------
/a1 | .4768049 .0190288 25.06 0.000 .4395091 .5141007
/a2 | .4189816 .0153873 27.23 0.000 .388823 .4491403
/a3 | .0016004 .0002748 5.82 0.000 .0010618 .002139
/a4 | .6618533 .0178606 37.06 0.000 .6268473 .6968594
/a5 |-.0241542 .0022701 -10.64 0.000 -.0286034 -.0197049
------------------------------------------------------------------------------