Stata置信区间为_variable

Question

是否有可能获得回归上下限作为变量，以便我可以预测回归的上下限而无需硬编码。

即

reg y x 
gen ypred = _b[_cons] + _b[x]*48 \\ prediction for x = 48 
gen ypred_beta95u= 0.8401741 + 0.0202769*48 \\ prediction upper bound 
gen ypred_beta95l= 0.550594 + 0.0097727*48 \\ prediction lower bound 

我可以得到回归系数没有硬编码，我想知道是否有可能做同样的事情有上限和下限在Stata。

Answer 1

这样做的两种方法依赖于Stata的margins，它计算在某些协变量的固定值下的先前拟合模型的预测。第一个原则是从第一原则开始。第二个依赖于提取margins留下的存储估计结果。

. sysuse auto, clear 
(1978 Automobile Data) 

. reg price mpg 

     Source |  SS   df  MS  Number of obs =  74 
-------------+---------------------------------- F(1, 72)  =  20.26 
     Model | 139449474   1 139449474 Prob > F  = 0.0000 
    Residual | 495615923  72 6883554.48 R-squared  = 0.2196 
-------------+---------------------------------- Adj R-squared = 0.2087 
     Total | 635065396  73 8699525.97 Root MSE  = 2623.7 

------------------------------------------------------------------------------ 
     price |  Coef. Std. Err.  t P>|t|  [95% Conf. Interval] 
-------------+---------------------------------------------------------------- 
     mpg | -238.8943 53.07669 -4.50 0.000 -344.7008 -133.0879 
     _cons | 11253.06 1170.813  9.61 0.000  8919.088 13587.03 
------------------------------------------------------------------------------ 

. margins, at(mpg=48) level(95) 

Adjusted predictions       Number of obs  =   74 
Model VCE : OLS 

Expression : Linear prediction, predict() 
at   : mpg    =   48 

------------------------------------------------------------------------------ 
      |   Delta-method 
      |  Margin Std. Err.  t P>|t|  [95% Conf. Interval] 
-------------+---------------------------------------------------------------- 
     _cons | -213.8679 1449.736 -0.15 0.883 -3103.864 2676.128 
------------------------------------------------------------------------------ 

. /* Mostly By Hand */ 
. gen ll1 = _b[_cons] - invttail(`e(df_r)',0.025)*_se[_cons] 

. gen ub1 = _b[_cons] + invttail(`e(df_r)',0.025)*_se[_cons] 

. /* From Stata's own stored output */ 
. matrix E = r(table) 

. matrix list E 

E[9,1] 
      _cons 
    b -213.86793 
    se 1449.7361 
    t -.14752197 
pvalue .88313237 
    ll -3103.864 
    ul 2676.1282 
    df   72 
    crit 1.9934636 
eform   0 

. gen ll2 = E[5,1] 

. gen ub2 = E[6,1] 

. list ll2 ub1 ll2 ub2 in 1/2, clean noobs 

      ll2  ub1   ll2  ub2 
    -3103.864 2676.128 -3103.864 2676.128 
    -3103.864 2676.128 -3103.864 2676.128