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April 29, 2024
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We have data on the number of fish caught by visitors to a national park. Some of the visitors do not fish, but we do not have the data on whether a person fished or not; we merely have data on how many fish were caught together with several covariates. As our data have a preponderance of zeros (142 out of 250), we use the zip command to model the outcome.
. zip count persons livebait, inf(child camper) nolog vuong
Zero-inflated poisson regression Number of obs = 250
Nonzero obs = 108
Zero obs = 142
Inflation model = logit LR chi2(2) = 506.48
Log likelihood = -850.7014 Prob > chi2 = 0.0000
------------------------------------------------------------------------------
count | Coef. Std. Err. z P>|z| [95% Conf. Interval]
---------+--------------------------------------------------------------------
count |
persons | .8068853 .0453288 17.801 0.000 .7180424 .8957281
livebait | 1.757289 .2446082 7.184 0.000 1.277866 2.236713
_cons | -2.178472 .2860289 -7.616 0.000 -2.739078 -1.617865
---------+--------------------------------------------------------------------
inflate |
child | 1.602571 .2797719 5.728 0.000 1.054228 2.150913
camper | -1.015698 .365259 -2.781 0.005 -1.731593 -.2998038
_cons | -.4922872 .3114562 -1.581 0.114 -1.10273 .1181558
------------------------------------------------------------------------------
Vuong Test of Zip vs. Poisson: 3.946 Prob > Z 0.000
Since the vuong option was specified, the output includes a Vuong test comparing the zero-inflated Poisson model and the Poisson model. Vuong (1989) developed a general test for testing non-nested models. This test was applied to zero-inflated Poisson and negative binomial models by Greene (1994), although that paper was never published and a description of the work appears in Greene (1999). This topic is also covered by Long (1997). This test statistic has a standard normal distribution with large positive values favoring the zero inflated Poisson model and large negative values favoring the Poisson model. In this example, the zero inflated model is strongly favored.
We could also estimate this model using zinb:
. zinb count persons livebait, inf(child camper) nolog vuong zip
Zero-inflated negative binomial regression Number of obs = 250
Nonzero obs = 108
Zero obs = 142
Inflation model = logit LR chi2(2) = 82.23
Log likelihood = -401.5478 Prob > chi2 = 0.0000
------------------------------------------------------------------------------
count | Coef. Std. Err. z P>|z| [95% Conf. Interval]
---------+--------------------------------------------------------------------
count |
persons | .9742984 .1034938 9.414 0.000 .7714543 1.177142
livebait | 1.557523 .4124424 3.776 0.000 .7491503 2.365895
_cons | -2.730064 .476953 -5.724 0.000 -3.664874 -1.795253
---------+--------------------------------------------------------------------
inflate |
child | 3.185999 .7468551 4.266 0.000 1.72219 4.649808
camper | -2.020951 .872054 -2.317 0.020 -3.730146 -.3117567
_cons | -2.695385 .8929071 -3.019 0.003 -4.44545 -.9453189
---------+--------------------------------------------------------------------
/lnalpha | .5110429 .1816816 2.813 0.005 .1549535 .8671323
---------+--------------------------------------------------------------------
alpha | 1.667029 .3028685 1.167604 2.380076
------------------------------------------------------------------------------
Likelihood ratio test of alpha=0: chi2(1) = 898.307 Prob > chi2 = 0.0000
Vuong Test of Zinb vs. Neg. Bin: Std. Normal 5.591 Prob > Z 0.0000
Since the zip option was specified, a likelihood ratio test of the null that alpha=0 appears at the bottom of the output table. When alpha=0, the zero inflated negative binomial model reduces to the zero inflated Poisson model. In this example, that hypothesis is very strongly rejected.
As in the zip example, the Vuong test, obtained with the vuong option, strongly favors the zero inflated model.