Carlos,
Check out the annotated negative binomial output on the UCLA stats portal:
http://www.ats.ucla.edu/stat/stata/output/stata_nbreg_output.htm
The definition of alpha in the Parameter Estimates section may be what you
are after.
Carter
-----Original Message-----
From: [email protected]
[mailto:[email protected]] On Behalf Of Carlo Lazzaro
Sent: Friday, February 01, 2008 2:02 PM
To: [email protected]
Subject: st: R: RE: testing for overdispersione after Poisson Regression
Dear Nick,
Thanks a lot for your kindness and for your time.
For sake of completeness, I report below all the routine I have performed.
My purpose was to test for the propriety of Poisson regression in dealing
with the issue of which indvars should be considered relevant in explaining
the increased or reduced undergoing to hospital visit.
After the result of post estimation, I am not clear whether I should better
switch to a negative binomial regression.
Thanks a lot again and Kind Regards,
Carlo
----------------------------------------------------------------------------
poisson ___hospvisit ___drvisit ofnp opp opnp emr exchlth poorhlth numchron
adldiff noreast midwe
> st west age black male married school faminc employed privins medicaid
Iteration 0: log likelihood = -15443.546
Iteration 1: log likelihood = -9957.3521 (backed up)
Iteration 2: log likelihood = -7079.7823
Iteration 3: log likelihood = -3293.7693
Iteration 4: log likelihood = -2791.9417
Iteration 5: log likelihood = -2773.1699
Iteration 6: log likelihood = -2772.855
Iteration 7: log likelihood = -2772.8545
Iteration 8: log likelihood = -2772.8545
Poisson regression Number of obs =
4406
LR chi2(21) =
1063.31
Prob > chi2 =
0.0000
Log likelihood = -2772.8545 Pseudo R2 =
0.1609
----------------------------------------------------------------------------
--
___hospvisit | Coef. Std. Err. z P>|z| [95% Conf.
Interval]
-------------+--------------------------------------------------------------
--
___drvisit | .031564 .0025853 12.21 0.000 .0264969
.0366311
ofnp | -.0035584 .0036833 -0.97 0.334 -.0107775
.0036608
opp | .009755 .0050281 1.94 0.052 -.0000998
.0196098
opnp | .0085071 .0049302 1.73 0.084 -.0011558
.0181701
emr | .3080027 .0136605 22.55 0.000 .2812287
.3347768
exchlth | -.61551 .1764821 -3.49 0.000 -.9614086
-.2696114
poorhlth | .2268273 .0761728 2.98 0.003 .0775312
.3761233
numchron | .1681921 .0190159 8.84 0.000 .1309217
.2054626
adldiff | .1914079 .0698929 2.74 0.006 .0544204
.3283954
noreast | -.0116473 .082506 -0.14 0.888 -.1733561
.1500614
midwest | .1485223 .0726288 2.04 0.041 .0061724
.2908723
west | .0275293 .0823022 0.33 0.738 -.13378
.1888387
age | .1960624 .045575 4.30 0.000 .1067371
.2853877
black | .0360653 .0935412 0.39 0.700 -.1472722
.2194028
male | .1517583 .0644441 2.35 0.019 .0254502
.2780663
married | .000077 .0673596 0.00 0.999 -.1319454
.1320993
school | .0047826 .0083723 0.57 0.568 -.0116268
.0211919
faminc | .0051941 .0100271 0.52 0.604 -.0144586
.0248468
employed | -.076601 .10922 -0.70 0.483 -.2906683
.1374663
privins | .1028314 .0828429 1.24 0.215 -.0595376
.2652005
medicaid | .1541872 .1045082 1.48 0.140 -.050645
.3590195
_cons | -3.734123 .3769326 -9.91 0.000 -4.472897
-2.995348
----------------------------------------------------------------------------
--
. estat gof
Goodness-of-fit chi2 = 3599.159
Prob > chi2(4384) = 1.0000
. estat gof, pearson
Goodness-of-fit chi2 = 5438.241
Prob > chi2(4384) = 0.0000
----------------------------------------------------------------------------
-----Messaggio originale-----
Da: [email protected]
[mailto:[email protected]] Per conto di Nick Cox
Inviato: venerd� 1 febbraio 2008 19.41
A: [email protected]
Oggetto: st: RE: testing for overdispersione after Poisson Regression
Seems as if neither is! The key is surely what other output you have,
including any indications of major or minor problems.
-----Original Message-----
From: [email protected]
[mailto:[email protected]] On Behalf Of Carlo Lazzaro
testing for overdispersion after a Poisson regression,
I have obtained two opposite results:
estat gof
Goodness-of-fit chi2 = 3599.159
Prob > chi2(4384) = 1.0000
. estat gof, pearson
Goodness-of-fit chi2 = 5438.241
Prob > chi2(4384) = 0.0000
Which one is trustworthy??
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