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Stata Data Analysis Examples
Zero-inflated Negative Binomial Regression

Examples of Zero-inflated Negative Binomial 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 negative binomial analysis, let's consider some other methods that you might use.

Stata Zero-inflated Negative Binomial Analysis

The output looks very much like the output from an OLS regression. It begins with 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 negative binomial 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 logit coefficients for predicting excess zeros along with their standard errors, z-scores, p-values and confidence intervals.  Additionally, there will be an estimate of the natural log of the over dispersion coefficient, alpha, along with the untransformed value. If the alpha coefficient is zero then the model is better estimated using an Poisson regression model.

Below the various coefficients you will find the results of the Vuong test. The Vuong test compares the zero-inflated model with an ordinary negative binomial 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 zinb 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.

zinb count child camper, inflate(persons) robust
Fitting constant-only model:

Iteration 0:   log pseudolikelihood = -519.33992  
Iteration 1:   log pseudolikelihood = -471.96077  
Iteration 2:   log pseudolikelihood = -465.38193  
Iteration 3:   log pseudolikelihood = -464.39882  
Iteration 4:   log pseudolikelihood = -463.92704  
Iteration 5:   log pseudolikelihood = -463.79248  
Iteration 6:   log pseudolikelihood = -463.75773  
Iteration 7:   log pseudolikelihood =  -463.7518  
Iteration 8:   log pseudolikelihood = -463.75119  
Iteration 9:   log pseudolikelihood = -463.75118  

Fitting full model:

Iteration 0:   log pseudolikelihood = -463.75118  (not concave)
Iteration 1:   log pseudolikelihood = -440.43162  
Iteration 2:   log pseudolikelihood = -434.96651  
Iteration 3:   log pseudolikelihood = -433.49903  
Iteration 4:   log pseudolikelihood = -432.89949  
Iteration 5:   log pseudolikelihood = -432.89091  
Iteration 6:   log pseudolikelihood = -432.89091  

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

Inflation model      = logit                      Wald chi2(2)    =      40.15
Log pseudolikelihood = -432.8909                  Prob > chi2     =     0.0000

------------------------------------------------------------------------------
             |               Robust
             |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
count        |
       child |  -1.515255   .2419059    -6.26   0.000    -1.989381   -1.041128
      camper |   .8790514   .4740421     1.85   0.064     -.050054    1.808157
       _cons |   1.371048   .4240159     3.23   0.001     .5399923    2.202104
-------------+----------------------------------------------------------------
inflate      |
     persons |  -1.666563   .5200032    -3.20   0.001     -2.68575   -.6473755
       _cons |   1.603104   .6748018     2.38   0.018     .2805164    2.925691
-------------+----------------------------------------------------------------
    /lnalpha |   .9853533   .0968016    10.18   0.000     .7956256    1.175081
-------------+----------------------------------------------------------------
       alpha |   2.678758   .2593081                      2.215827    3.238405
------------------------------------------------------------------------------

Using the robust option has resulted in a some 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 model is still statistically significant. 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 number of fish caught.

prchange

zinb: Changes in Rate for count

        min->max      0->1     -+1/2    -+sd/2
 child   -6.0879   -4.8010   -3.6329   -3.0108
camper    1.8336    1.8336    1.9810    0.9537

exp(xb):   2.1826

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 negative binomial regression model predicting number of 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 -1.515255. This amounts to a decrease of about 3.6 in the expected count for fish for each child in the party. Groups with campers (camper = 1) had an expected log(count) 0.879051 higher than groups without campers (camper = 0), i.e., beign a camper increases the expected coount of fish caught by 1.83.  We can see in our model that the dispersion parameter alpha is significantly different from zero.  This suggests that our data is overdispersed and that a negative binomial model is more appropriate than a Poisson model.  The Vuong suggests that our zero-inflated model is a significant improvement over a standard negative binomial model.

Cautions, Flies in the Ointment

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

    • See Also

     

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