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How can I obtain the standard error of the regression with streg?

 Title Obtaining the standard error of the regression with streg Author William Gould, StataCorp

(Note: In previous versions, the generalized gamma distribution was specified as gamma, and renamed to ggamma in Stata 14.)

Question:

I am using streg, dist(ggamma) to estimate an AFT model. How can I obtain the standard error of the regression? I thought this could be done by using _b[_se]. [...]

Rather than _b[_se], type

    [ln_sig]_b[_cons]


to obtain the ln().

 .  sysuse auto, clear
(1978 Automobile Data)

.  stset mpg, f(foreign)

failure event:  foreign != 0 & foreign < .
obs. time interval:  (0, mpg]
exit on or before:  failure

------------------------------------------------------------------------------
74  total observations
0  exclusions
------------------------------------------------------------------------------
74  observations remaining, representing
22  failures in single-record/single-failure data
1576  total analysis time at risk and under observation
at risk from t =         0
earliest observed entry t =         0
last observed exit t =        41

.  streg weight, dist(gamma) nolog
failure _d:  foreign
analysis time _t:  mpg

Generalized gamma regression -- accelerated failure-time form

No. of subjects =           74                  Number of obs    =          74
No. of failures =           22
Time at risk    =         1576
LR chi2(1)       =        0.30
Log likelihood  =    -14.77069                  Prob > chi2      =      0.5842

------------------------------------------------------------------------------
_t |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
weight |  -.0000453   .0000776    -0.58   0.559    -.0001974    .0001068
_cons |   3.456707   .1853193    18.65   0.000     3.093488    3.819927
-------------+----------------------------------------------------------------
/ln_sig |  -1.425659    .201243    -7.08   0.000    -1.820088    -1.03123
/kappa |   .1663058   .5811509     0.29   0.775    -.9727291    1.305341
-------------+----------------------------------------------------------------
sigma |     .24035   .0483688                      .1620115    .3565681
------------------------------------------------------------------------------


When you see /something, the coefficient is [something]_b[_cons] and the standard error is [something]_se[_cons]:

 . display [ln_sig]_b[_cons]
-1.4256592


From the output above, you might also guess that the _b[sigma] would work, but it does not.

 . display _b[sigma]
r(111);


sigma is derived from ln_sig. I admit this can be confusing, and the way to resolve that confusion is to display the coefficient vector:

 . matrix list e(b)

e(b)[1,4]
_t:         _t:     ln_sig:      kappa:
weight       _cons       _cons       _cons
y1  -.00004532   3.4567075  -1.4256592   .16630579


From the above, I can see that the coefficients are

    You can type this                 or this

[_t]_b[weight]                    _b[_t:weight]
[_t]_b[_cons]                     _b[_t:_cons]
[ln_sig]_b[_cons]                 _b[ln_sig:_cons]
[kappa]_b[_cons]                  _b[kappa:_cons]


Whether you type the form on the left or the form the right makes no difference to Stata. I rather like the form on the left, but that is an aesthetic judgment, as one is a synonym for the other.

You can also type _b[weight] rather than [_t]_b[weight] (or _b[_t:weight]), because Stata assumes that you are referring to the first equation (in this case, _t) when you do not specify the name of the equation.

See [U] 13.5 Accessing coefficients and standard errors for more information and type help _variables to see the help file.