One way, would be to compare the predicted probabilities which can
conveniently done with -prvalue- (which is part of J. Scott Long's
-spostado- package).
For example:
. sysuse auto
(1978 Automobile Data)
. gen foreXmpg = foreign*mpg
. poisson rep mpg foreign foreXmpg
Iteration 0: log likelihood = -112.63814
Iteration 1: log likelihood = -112.63814
Poisson regression Number of obs =
69
LR chi2(3) =
7.08
Prob > chi2 =
0.0694
Log likelihood = -112.63814 Pseudo R2 =
0.0305
----------------------------------------------------------------------------
rep78 | Coef. Std. Err. z P>|z| [95% Conf.
Interval]
-------------+--------------------------------------------------------------
mpg | .0113925 .0172281 0.66 0.508 -.0223738
.0451589
foreign | .4614341 .5658458 0.82 0.415 -.6476034
1.570472
foreXmpg | -.0069618 .024157 -0.29 0.773 -.0543087
.0403851
_cons | .8814441 .3510788 2.51 0.012 .1933423
1.569546
----------------------------------------------------------------------------
. prvalue , x(foreign = 0 foreX = 0) rest(mean)
poisson: Predictions for rep78
Predicted rate: 3.08 95% CI [2.6 , 3.65]
Predicted probabilities:
Pr(y=0|x): 0.0461 Pr(y=1|x): 0.1418
Pr(y=2|x): 0.2182 Pr(y=3|x): 0.2238
Pr(y=4|x): 0.1722 Pr(y=5|x): 0.1060
Pr(y=6|x): 0.0543 Pr(y=7|x): 0.0239
Pr(y=8|x): 0.0092 Pr(y=9|x): 0.0031
mpg foreign foreXmpg
x= 21.289855 0 0
. qui sum mpg
. local mean = r(mean)
. prvalue , x(foreign = 1 foreX = `mean') rest(mean)
poisson: Predictions for rep78
Predicted rate: 4.21 95% CI [3.28 , 5.4]
Predicted probabilities:
Pr(y=0|x): 0.0149 Pr(y=1|x): 0.0626
Pr(y=2|x): 0.1317 Pr(y=3|x): 0.1847
Pr(y=4|x): 0.1943 Pr(y=5|x): 0.1636
Pr(y=6|x): 0.1147 Pr(y=7|x): 0.0690
Pr(y=8|x): 0.0363 Pr(y=9|x): 0.0170
mpg foreign foreXmpg
x= 21.289855 1 21.297297
Hope this helps,
Scott
> -----Original Message-----
> From: [email protected] [mailto:owner-
> [email protected]] On Behalf Of Sheng Wang
> Sent: Monday, April 11, 2005 2:42 PM
> To: [email protected]
> Subject: st: interaction term in negative binomial regression
>
> Dear all:
>
> I have a question about how to interpret the interaction term in negative
> binomial regression results.
>
> Below is the section from the stata output, gender and usage are two
> control
> variables. Dummy is a dummy variable created for 2 conditions (0 or 1).
> Extraversion is considered a continuous variable (1-5), and interaction is
> a
> product of the dummy variable and the mean-centered extraversion. I'd like
> to understand the different relationships between extraversion and
> quantity
> under condition =0 or condition =1? How can I calculate if there is a
> stronger relationship between extraversion and quantity under the two
> different conditions?
>
>
> Quantity | Coef. Std. Err. z P>|z| [95% Conf.
> Interval]
> -------------+------------------------------------------------------------
> --
> --
> gender | 1.215867 .3982474 3.05 0.002 .4353161
> 1.996417
> usage | .2103553 .1310798 1.60 0.109 -.0465563
> .467267
> dummy | 4.035392 .6155144 6.56 0.000 2.829006
> 5.241778
> extraversion | 1.946443 1.131335 1.72 0.085 -.2709335
> 4.163819
> interaction | -2.616264 1.203618 -2.17 0.030 -4.975313
> -.2572159
> _cons | -10.07717 4.202655 -2.40 0.016 -18.31423
> -1.840122
> -------------+------------------------------------------------------------
> --
> --
>
>
> I would really appreciate any assistance with this issue.
>
> Sincerely,
>
> Sheng Wang
> The Ohio State University
>
>
*
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