Stata Data Analysis Examples: Zero-inflated Poisson Regression



Stat Computing > Stata > Data Analysis Examples

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 1 from above. This example uses the same data as the poisson, negative binomial and zero-inflated negative binomial regressions.

We have attendance data on 316 high school juniors from two urban high schools in the file poissonreg.dta The response variable of interest is days absent, daysabs. The variables math and langarts give the standardized test scores for math and language arts respectively. The variable male is a binary indicator of student gender.

In addition to predicting the number of days absent there is interest in predicting the existence of excess zeros, i.e., the probability that a student will have zero absences. We will use both male and school to investigate this.

Let's look at the data.

use http://www.ats.ucla.edu/stat/stata/dae/poissonreg, clear

summarize daysabs math langarts male

    Variable |       Obs        Mean    Std. Dev.       Min        Max
-------------+--------------------------------------------------------
     daysabs |       316    5.810127    7.449003          0         45
        math |       316    48.75101    17.88076   1.007114   98.99289
    langarts |       316    50.06379    17.93921   1.007114   98.99289
        male |       316    .4873418    .5006325          0          1
      school |       316    1.496835     .500783          1          2

tabstat daysabs, stat(n mean var)

    variable |         N      mean  variance
-------------+------------------------------
     daysabs |       316  5.810127  55.48764
--------------------------------------------

histogram daysabs, discrete freq



tab1 male school

-> tabulation of male  

       male |      Freq.     Percent        Cum.
------------+-----------------------------------
          0 |        162       51.27       51.27
          1 |        154       48.73      100.00
------------+-----------------------------------
      Total |        316      100.00

-> tabulation of school  

     school |      Freq.     Percent        Cum.
------------+-----------------------------------
          1 |        159       50.32       50.32
          2 |        157       49.68      100.00
------------+-----------------------------------
      Total |        316      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

zip daysabs math langarts male, inflate(school male) vuong

Fitting constant-only model:

Iteration 0:   log likelihood = -1494.2292  
Iteration 1:   log likelihood = -1384.2385  
Iteration 2:   log likelihood = -1380.4502  
Iteration 3:   log likelihood = -1380.4264  
Iteration 4:   log likelihood = -1380.4264  

Fitting full model:

Iteration 0:   log likelihood = -1380.4264  
Iteration 1:   log likelihood = -1346.2718  
Iteration 2:   log likelihood = -1345.9222  
Iteration 3:   log likelihood = -1345.9221  

Zero-inflated Poisson regression                  Number of obs   =        316
                                                  Nonzero obs     =        254
                                                  Zero obs        =         62

Inflation model = logit                           LR chi2(3)      =      69.01
Log likelihood  = -1345.922                       Prob > chi2     =     0.0000

------------------------------------------------------------------------------
             |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
daysabs      |
        math |  -.0003056   .0018612    -0.16   0.870    -.0039535    .0033423
    langarts |  -.0094936   .0019086    -4.97   0.000    -.0132344   -.0057528
        male |   -.246477   .0487593    -5.05   0.000    -.3420436   -.1509105
       _cons |    2.54522   .0732565    34.74   0.000      2.40164      2.6888
-------------+----------------------------------------------------------------
inflate      |
      school |   1.151285   .3143818     3.66   0.000     .5351077    1.767462
        male |   .8692552   .3046072     2.85   0.004      .272236    1.466274
       _cons |  -3.704412   .5943179    -6.23   0.000    -4.869254   -2.539571
------------------------------------------------------------------------------
Vuong test of zip vs. standard Poisson:            z =     5.48  Pr>z = 0.0000

The output looks very much like the output from an OLS regression. 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 with three degrees of freedom for the full model, followed by the p-value for the chi-square. The model, as a whole, is statistically significant. The header also includes a pseudo-R² which is 0.0536 in this example.

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

Since math is clearly not significant, let's rerun the model without it.

zip daysabs langarts male, inflate(school male) vuong

Fitting constant-only model:

Iteration 0:   log likelihood = -1494.2292  
Iteration 1:   log likelihood = -1384.2385  
Iteration 2:   log likelihood = -1380.4502  
Iteration 3:   log likelihood = -1380.4264  
Iteration 4:   log likelihood = -1380.4264  

Fitting full model:

Iteration 0:   log likelihood = -1380.4264  
Iteration 1:   log likelihood = -1346.2822  
Iteration 2:   log likelihood = -1345.9356  
Iteration 3:   log likelihood = -1345.9356  

Zero-inflated Poisson regression                  Number of obs   =        316
                                                  Nonzero obs     =        254
                                                  Zero obs        =         62

Inflation model = logit                           LR chi2(2)      =      68.98
Log likelihood  = -1345.936                       Prob > chi2     =     0.0000

------------------------------------------------------------------------------
             |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
daysabs      |
    langarts |   -.009721   .0013138    -7.40   0.000     -.012296   -.0071461
        male |  -.2470039   .0486519    -5.08   0.000    -.3423598    -.151648
       _cons |   2.542103   .0707735    35.92   0.000     2.403389    2.680816
-------------+----------------------------------------------------------------
inflate      |
      school |   1.151284   .3143835     3.66   0.000     .5351039    1.767465
        male |   .8693162    .304612     2.85   0.004     .2722876    1.466345
       _cons |  -3.704446   .5943313    -6.23   0.000    -4.869313   -2.539578
------------------------------------------------------------------------------
Vuong test of zip vs. standard Poisson:            z =     5.66  Pr>z = 0.0000

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 daysabs langarts male, inflate(school male) robust

Fitting constant-only model:

Iteration 0:   log pseudolikelihood = -1494.2292  
Iteration 1:   log pseudolikelihood = -1384.2385  
Iteration 2:   log pseudolikelihood = -1380.4502  
Iteration 3:   log pseudolikelihood = -1380.4264  
Iteration 4:   log pseudolikelihood = -1380.4264  

Fitting full model:

Iteration 0:   log pseudolikelihood = -1380.4264  
Iteration 1:   log pseudolikelihood = -1346.2822  
Iteration 2:   log pseudolikelihood = -1345.9356  
Iteration 3:   log pseudolikelihood = -1345.9356  

Zero-inflated Poisson regression                  Number of obs   =        316
                                                  Nonzero obs     =        254
                                                  Zero obs        =         62

Inflation model      = logit                      Wald chi2(2)    =      12.58
Log pseudolikelihood = -1345.936                  Prob > chi2     =     0.0019

------------------------------------------------------------------------------
             |               Robust
             |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
daysabs      |
    langarts |   -.009721   .0032475    -2.99   0.003     -.016086   -.0033561
        male |  -.2470039   .1262323    -1.96   0.050    -.4944147    .0004069
       _cons |   2.542103   .1749726    14.53   0.000     2.199163    2.885042
-------------+----------------------------------------------------------------
inflate      |
      school |   1.151284   .3096867     3.72   0.000     .5443094    1.758259
        male |   .8693162   .3009801     2.89   0.004     .2794061    1.459226
       _cons |  -3.704446   .5694312    -6.51   0.000     -4.82051   -2.588381
------------------------------------------------------------------------------

Using the robust option has resulted in a fairly large change in the model chi-square, which is now a Wald chi-square, based on log pseudolikelihoods, instead of a likelihood ratio chi-square.

In the main body of the output contains the poisson and probit coefficients, robust standard errors, z-scores, p-values and 95% confidence intervals for the coefficients. 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 daysabs

          min->max      0->1     -+1/2    -+sd/2
langarts   -5.6598   -0.0900   -0.0556   -0.9987
    male   -2.2729   -2.2729   -2.2767   -1.1402

exp(xb):   5.7195

base x values for count equation: 

        langarts      male
    x=   50.0638   .487342
sd(x)=   17.9392   .500633

base x values for binary equation: 

         school     male
    x=  1.49684  .487342
sd(x)=  .500783  .500633

Sample Write-Up of the Analysis

Before we begin the sample write-up we need to get the output into a form more acceptable for publication. The estout command (findit estout by Ben Jann of ETH Zurich), will get us close to what we want.

estout, cells(b(star fmt(%8.2f)) se(par fmt(%8.2f))) stats(ll chi2, fmt(%8.2f))

        .
        b/se
daysabs 
langarts        -0.01**
        (0.00)
male    -0.25*
        (0.13)
_cons   2.54***
        (0.17)
inflate 
school  1.15***
        (0.31)
male    0.87**
        (0.30)
_cons   -3.70***
        (0.57)
ll      -1345.94
chi2    12.58

With a little bit of manual editing we can produce an acceptable table of the output.

model 
count coefficients
language arts     -0.01**
                  (0.00)
male              -0.25*
                  (0.13)
constant           2.54***
                  (0.17)
excess zero probit coefficients
school             1.15***
                  (0.31)
male               0.87**
                  (0.30)
constant          -3.70***
                  (0.57)
log psuedo-
   likelihood  -1345.94
chi-squared       12.58
legend: coefficient/(standard error)  * p<=.05 ** p<0.01 *** p<0.001

The zero-inflated poisson regression model predicting days absent from language arts and gender was statistically significant (chi-squared = 12.58, df = 2, p<.01). The predictors of excess zeros, school (1.15) and male (0.87) were both statistically significant. The count predictors langarts and male were also each statically significant. For these data, the expected log count for a one-unit increase in language arts was -0.01. This translates to a decrease of almost one day (0.999) absence for a one standard deviation increase in language arts when gender is held constant. Male students had an expected log count -0.25 less than female students which amounts to about 2.27 fewer days absent than females while holding language arts constant.

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