Stata Data Analysis Examples: Zero-inflated Poisson Regression



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Stata Data Analysis Examples
Zero-inflated Poisson Regression

Examples of Zero-inflated Poisson Regression

Example 1. School administrators study the attendance behavior of high school juniors at two schools. Predictors of the number of days of absence include gender of the student and standardized test scores in math and language arts.

Example 2. The state wildlife biologists want to model how many fish are being caught by fishermen at a state park. Visitors are asked how long they stayed, how many people were in the group, were there children in the group and how many fish were caught. Some visitors do not fish, but there is no data on whether a person fished or not. Some visitors who did fish did not catch any fish so there are excess zeros in the data because of the people that did not fish.

Description of the Data

Let's pursue Example 2 from above.

We have data on 250 groups that went to a park. Each group was questioned about how many fish they caught (count), how many children were in the group (child), how many people were in the group (persons), and whether or not they brought a camper to the park (camper).

In addition to predicting the number of fish caught, there is interest in predicting the existence of excess zeros, i.e. the zeroes that were not simply a result of bad luck fishing. We will use the variables child, persons, and camper in our model. Let's look at the data.

webuse fish

summarize count child persons camper


    Variable |       Obs        Mean    Std. Dev.       Min        Max
-------------+--------------------------------------------------------
       count |       250       3.296    11.63503          0        149
       child |       250        .684    .8503153          0          3
     persons |       250       2.528     1.11273          1          4
      camper |       250        .588    .4931824          0          1

histogram count, discrete freq
     


tab1 child persons camper

-> tabulation of child  

      child |      Freq.     Percent        Cum.
------------+-----------------------------------
          0 |        132       52.80       52.80
          1 |         75       30.00       82.80
          2 |         33       13.20       96.00
          3 |         10        4.00      100.00
------------+-----------------------------------
      Total |        250      100.00

-> tabulation of persons  

    persons |      Freq.     Percent        Cum.
------------+-----------------------------------
          1 |         57       22.80       22.80
          2 |         70       28.00       50.80
          3 |         57       22.80       73.60
          4 |         66       26.40      100.00
------------+-----------------------------------
      Total |        250      100.00

-> tabulation of camper  

     camper |      Freq.     Percent        Cum.
------------+-----------------------------------
          0 |        103       41.20       41.20
          1 |        147       58.80      100.00
------------+-----------------------------------
      Total |        250      100.00

Some Strategies You Might Be Tempted To Try

Before we show how you can analyze this with a zero-inflated Poisson analysis, let's consider some other methods that you might use.

OLS Regression - You could try to analyze these data using OLS regression. However, count data are highly non-normal and are not well estimated by OLS regression.
Zero-inflated Negative Binomial Regression - Negative binomial regression does better with over dispersed data, i.e. variance much larger than the mean.
Ordinary Count Models - Poisson or negative binomial models might be more appropriate if there are no excess zeros.

Stata Zero-inflated Poisson Analysis

The command below is for a zero-inflated Poisson model in which the number of fish caught is predicted with child and camper and the excessive zeroes in the outcome variable count are predicted with persons.

zip count child camper, inflate(persons) vuong

Fitting constant-only model:

Iteration 0:   log likelihood =  -1347.807  
Iteration 1:   log likelihood = -1315.5343  
Iteration 2:   log likelihood = -1126.3689  
Iteration 3:   log likelihood = -1125.5358  
Iteration 4:   log likelihood = -1125.5357  
Iteration 5:   log likelihood = -1125.5357  

Fitting full model:

Iteration 0:   log likelihood = -1125.5357  
Iteration 1:   log likelihood = -1044.8553  
Iteration 2:   log likelihood = -1031.8733  
Iteration 3:   log likelihood = -1031.6089  
Iteration 4:   log likelihood = -1031.6084  
Iteration 5:   log likelihood = -1031.6084  

Zero-inflated Poisson regression                  Number of obs   =        250
                                                  Nonzero obs     =        108
                                                  Zero obs        =        142

Inflation model = logit                           LR chi2(2)      =     187.85
Log likelihood  = -1031.608                       Prob > chi2     =     0.0000

------------------------------------------------------------------------------
             |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
count        |
       child |  -1.042838   .0999883   -10.43   0.000    -1.238812    -.846865
      camper |   .8340222   .0936268     8.91   0.000      .650517    1.017527
       _cons |   1.597889   .0855382    18.68   0.000     1.430237     1.76554
-------------+----------------------------------------------------------------
inflate      |
     persons |  -.5643472   .1629638    -3.46   0.001    -.8837503    -.244944
       _cons |   1.297439   .3738522     3.47   0.001     .5647022    2.030176
------------------------------------------------------------------------------
Vuong test of zip vs. standard Poisson:            z =     3.57  Pr>z = 0.0002

The output begins the iteration log giving the values of the log likelihoods starting with a model that has no predictors. The last value in the log is the final value of the log likelihood for the full model and is repeated below.

Next comes the header information. On the right-hand side the number of observations used (316) is given along with the likelihood ratio chi-squared. This compares the full model to a model without count predictors, giving a difference of two degrees of freedom. This is followed by the p-value for the chi-square. The model, as a whole, is statistically significant.

Below the header you will find the Poisson regression coefficients for each of the count predicting variables along with standard errors, z-scores, p-values and 95% confidence intervals for the coefficients. Following these are logit coefficients for the variable predicting excess zeros along with its standard errors, z-scores, p-values and confidence intervals.

Below the various coefficients you will find the results of the Vuong test. The Vuong test compares the zero-inflated model with an ordinary Poisson regression model. A significant z-test indicates that the zero-inflated model is better.

Now, just to be on the safe side, let's rerun the zip command with the robust option in order to obtain robust standard errors for the poisson regression coefficients. We cannot include the vuong option when using robust standard errors.

zip count child camper, inflate(persons) robust

Fitting constant-only model:

Iteration 0:   log pseudolikelihood =  -1347.807  
Iteration 1:   log pseudolikelihood = -1315.5343  
Iteration 2:   log pseudolikelihood = -1126.3689  
Iteration 3:   log pseudolikelihood = -1125.5358  
Iteration 4:   log pseudolikelihood = -1125.5357  
Iteration 5:   log pseudolikelihood = -1125.5357  

Fitting full model:

Iteration 0:   log pseudolikelihood = -1125.5357  
Iteration 1:   log pseudolikelihood = -1044.8553  
Iteration 2:   log pseudolikelihood = -1031.8733  
Iteration 3:   log pseudolikelihood = -1031.6089  
Iteration 4:   log pseudolikelihood = -1031.6084  
Iteration 5:   log pseudolikelihood = -1031.6084  

Zero-inflated Poisson regression                  Number of obs   =        250
                                                  Nonzero obs     =        108
                                                  Zero obs        =        142

Inflation model      = logit                      Wald chi2(2)    =       7.25
Log pseudolikelihood = -1031.608                  Prob > chi2     =     0.0266

------------------------------------------------------------------------------
             |               Robust
             |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
count        |
       child |  -1.042838   .3893772    -2.68   0.007    -1.806004   -.2796731
      camper |   .8340222   .4076029     2.05   0.041     .0351352    1.632909
       _cons |   1.597889   .2934631     5.44   0.000     1.022711    2.173066
-------------+----------------------------------------------------------------
inflate      |
     persons |  -.5643472   .2888849    -1.95   0.051    -1.130551    .0018567
       _cons |   1.297439    .493986     2.63   0.009     .3292445    2.265634
------------------------------------------------------------------------------

Using the robust option has resulted in a fairly large change in the model chi-square, which is now a Wald chi-square. This statistic is based on log pseudo-likelihoods instead of log-likelihoods.

The robust standard errors attempt to adjust for heterogeneity in the model.

Finally, we will use the prchange command (findit prchange) by J. Scott Long and Jeremy Freese to get the predicted change in days absent.

prchange

zip: Changes in Rate for count

        min->max      0->1     -+1/2    -+sd/2
 child   -4.1079   -2.7818   -2.2961   -1.9285
camper    1.6792    1.6792    1.8071    0.8720

exp(xb):   2.1051

base x values for count equation: 

          child   camper
    x=     .684     .588
sd(x)=  .850315  .493182

base x values for binary equation: 

        persons
    x=    2.528
sd(x)=  1.11273

Sample Write-Up of the Analysis

The zero-inflated Poisson regression model predicting fish caught (count) from child, camper, and persons was statistically significant (both with and without robust standard errors). The predictor of excess zeros, persons, was statistically significant. The count predictors child and camper were also each statically significant. For these data, the expected change in log(count) for a one-unit increase in child was -1.0428. Groups with campers (camper = 1) had an expected log(count) value 0.8340 higher than groups without campers (camper = 0). The Vuong test suggests that our zero-inflated model is a significant improvement over a standard Poisson model.

Cautions, Flies in the Ointment

It is not recommended that zero-inflated Poisson models be applied to small samples. What constitutes a small sample does not seem to be clearly defined in the literature.

Stata Data Analysis Examples Zero-inflated Poisson Regression