Stata Data Analysis Examples: Zero-inflated Negative Binomial Regression



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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 1 from above. This example uses the same data as the poisson, negative binomial and zero-inflated poisson 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 the variable 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 negative binomial 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 Poisson Regression - Zero-inflated Poisson regression does better when there is no over dispersion 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 Negative Binomial Analysis

zinb daysabs math langarts male, inflate(school) vuong

Fitting constant-only model:

Iteration 0:   log likelihood = -989.38831  
Iteration 1:   log likelihood = -898.18259  
Iteration 2:   log likelihood = -889.68818  
Iteration 3:   log likelihood = -888.03208  
Iteration 4:   log likelihood =  -887.7173  
Iteration 5:   log likelihood = -887.66641  
Iteration 6:   log likelihood = -887.65515  
Iteration 7:   log likelihood = -887.65269  
Iteration 8:   log likelihood = -887.65226  
Iteration 9:   log likelihood = -887.65216  
Iteration 10:  log likelihood = -887.65214  

Fitting full model:

Iteration 0:   log likelihood = -887.65214  
Iteration 1:   log likelihood = -879.10154  
Iteration 2:   log likelihood =  -878.2199  
Iteration 3:   log likelihood = -878.21612  
Iteration 4:   log likelihood = -878.21612  

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

Inflation model = logit                           LR chi2(3)      =      18.87
Log likelihood  = -878.2161                       Prob > chi2     =     0.0003

------------------------------------------------------------------------------
             |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
daysabs      |
        math |  -.0009049   .0047223    -0.19   0.848    -.0101605    .0083506
    langarts |  -.0135826   .0054659    -2.48   0.013    -.0242956   -.0028696
        male |  -.4124475    .135929    -3.03   0.002    -.6788635   -.1460316
       _cons |   2.694427   .2236049    12.05   0.000     2.256169    3.132684
-------------+----------------------------------------------------------------
inflate      |
      school |   19.31956   5997.093     0.00   0.997    -11734.77    11773.41
       _cons |   -40.6701   11994.19    -0.00   0.997    -23548.84     23467.5
-------------+----------------------------------------------------------------
    /lnalpha |   .0967353   .1146672     0.84   0.399    -.1280084    .3214789
-------------+----------------------------------------------------------------
       alpha |   1.101569   .1263138                       .879846    1.379166
------------------------------------------------------------------------------
Vuong test of zinb vs. standard negative binomial: z =     1.12  Pr>z = 0.1311

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.

zinb daysabs langarts male, inflate(school) vuong zip

Fitting constant-only model:

Iteration 0:   log likelihood = -989.38831  
Iteration 1:   log likelihood = -898.18259  
Iteration 2:   log likelihood = -889.68818  
Iteration 3:   log likelihood = -888.03208  
Iteration 4:   log likelihood =  -887.7173  
Iteration 5:   log likelihood = -887.66641  
Iteration 6:   log likelihood = -887.65515  
Iteration 7:   log likelihood = -887.65269  
Iteration 8:   log likelihood = -887.65226  
Iteration 9:   log likelihood = -887.65216  
Iteration 10:  log likelihood = -887.65214  

Fitting full model:

Iteration 0:   log likelihood = -887.65214  
Iteration 1:   log likelihood = -879.11709  
Iteration 2:   log likelihood = -878.23821  
Iteration 3:   log likelihood = -878.23446  
Iteration 4:   log likelihood = -878.23446  

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

Inflation model = logit                           LR chi2(2)      =      18.84
Log likelihood  = -878.2345                       Prob > chi2     =     0.0001

------------------------------------------------------------------------------
             |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
daysabs      |
    langarts |  -.0143223   .0038697    -3.70   0.000    -.0219068   -.0067378
        male |  -.4123582   .1359164    -3.03   0.002    -.6787495   -.1459669
       _cons |   2.687697   .2207211    12.18   0.000     2.255092    3.120302
-------------+----------------------------------------------------------------
inflate      |
      school |   19.32026   5999.448     0.00   0.997    -11739.38    11778.02
       _cons |  -40.66774    11998.9    -0.00   0.997    -23558.07    23476.74
-------------+----------------------------------------------------------------
    /lnalpha |   .0963472   .1146165     0.84   0.401    -.1282969    .3209913
-------------+----------------------------------------------------------------
       alpha |   1.101141   .1262089                      .8795922    1.378494
------------------------------------------------------------------------------
Likelihood-ratio test of alpha=0: chibar2(01) =   943.87 Pr>=chibar2 =  0.0000
Vuong test of zinb vs. standard negative binomial: z =     1.13  Pr>z = 0.1298

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

zinb: Changes in Rate for daysabs

          min->max      0->1     -+1/2    -+sd/2
langarts   -8.9375   -0.1710   -0.0841   -1.5121
    male   -2.4247   -2.4247   -2.4374   -1.2138

exp(xb):   5.8691

base x values for count equation: 

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

base x values for binary equation: 

         school
    x=  1.49684
sd(x)=  .500783

Sample Write-Up of the Analysis

Instead of the usual interpretation of the coefficients in the model, we will explain why this is not a good model to use in an analysis. To see an example of a sample write-up of a zero-inflated model please see the page for zip.

The first indication that something is amiss is that the standard errors for the constant and coefficient for the inflation part of the model are so large. The associated z-scores are zero and the p-values are close to 1.0. The other indicator is that the Vuong test is not significant, i.e., the zero-inflated negative binomial model is not significantly better than the standard negative binomial model.

Although the count part of the model is still valid and you can obtain predicted counts you are probably better off running this model as an ordinary negative binimial regression 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 Negative Binomial Regression