This page shows an example of zero-inflated Poisson regression analysis with footnotes
explaining the output in Stata. The data collected were academic information on
316 students at two different schools. The response variable is days absent during the school year (daysabs).
We explore its relationship with math standardized test scores (mathnce),
language standardized test scores (langnce), and gender (female).

As assumed for a Poisson model,
our response variable is a count variable, each subject has the same length of
observation time, and the variance of the response variable is relatively close
to the mean of the response variable. In a dataset in which the response
variable is a count, the number of zeroes may seem excessive. With the
example dataset in mind, consider the
processes that could lead to a response variable value of zero. A student might
be absent zero days during the school year if he never gets sick and never skips
school. Another student might be absent zero days during the school year because
her parents insist she go to school every day, regardless of illness or desire
to skip school. These two students will look identical in the response variable,
but they have arrived at the same outcome through two different processes.
The first student could have been absent during the school year (had he
become ill or opted to skip school), but was not.  The second student was
certain to be absent zero days.  The second student will be referred to
from this point forward as a “certain zero”.
Thus, the number of zeroes may be inflated and the number of students absent for
zero days cannot be explained in the same manner as the number of students that
were absent for more than zero days.  Some students were absent zero days
for the same reasons other students were absent one, two, or three days (health
and truancy) and while some students were absent zero days for a different set
of reasons.

A standard Poisson model would
not distinguish between the two processes causing an excessive number of zeroes, but a zero-inflated model allows
for and
accommodates this complication.  When analyzing a dataset with an excessive
number of outcome zeros and two possible processes that arrive at a zero
outcome, a zero-inflated model should be considered.  We can look at a histogram
of the response variable to try to gauge if the number of zeros is excessive.
(If two processes generated the zeroes in the response variable but there is
not an excessive number of zeroes, a zero-inflated model may or may not be
used.)

use https://stats.idre.ucla.edu/stat/stata/notes/lahigh, clear
generate female = (gender == 1)
histogram daysabs, discrete freq

While zero is the most common number of days absent, it is difficult to see from
this histogram if the number of zeroes is excessive.  Thus, we can run a zero-inflated Poisson model and test
whether it better predicts our response variable than a standard Poisson model.

The zero-inflated Poisson regression generates two separate models and then
combines them. First, a logit model is generated for the “certain zero” cases
described above, predicting whether or not a student would be in this group.
Then, a Poisson model is generated to predict the counts for those students who
are not certain zeros.  Finally, the two models are combined.  When
running a zero-inflated Poisson model in Stata, you must specify both models: first the
count model, then the model predicting the certain zeros.  In this example,
we are predicting count with mathnce, langnce and female, and
predicting the certain zeros with mathnce and langnce.

zip daysabs mathnce langnce female, inflate(mathnce langnce)

Fitting constant-only model:

Iteration 0:   log likelihood = -1494.2292  
Iteration 1:   log likelihood =  -1388.076  
Iteration 2:   log likelihood =   -1385.31  
Iteration 3:   log likelihood = -1385.2992  
Iteration 4:   log likelihood = -1385.2992  

Fitting full model:

Iteration 0:   log likelihood = -1385.2992  
Iteration 1:   log likelihood = -1351.1453  
Iteration 2:   log likelihood = -1350.7936  
Iteration 3:   log likelihood = -1350.7935  

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

Inflation model = logit                           LR chi2(3)      =      69.01
Log likelihood  = -1350.794                       Prob > chi2     =     0.0000

------------------------------------------------------------------------------
             |      Coef.   Std. Err.      z    P>|z|     [95% Conf. Interval]
-------------+----------------------------------------------------------------
daysabs      |
     mathnce |  -.0002822   .0018619    -0.15   0.880    -.0039314    .0033671
     langnce |  -.0095004   .0019094    -4.98   0.000    -.0132428    -.005758
      female |   .2473714   .0488047     5.07   0.000     .1517161    .3430268
       _cons |   2.297473   .0693793    33.11   0.000     2.161492    2.433453
-------------+----------------------------------------------------------------
inflate      |
     mathnce |   .0188534    .011087     1.70   0.089    -.0028766    .0405834
     langnce |   .0110383   .0111014     0.99   0.320    -.0107201    .0327966
       _cons |  -2.963912   .5315156    -5.58   0.000    -4.005664   -1.922161
------------------------------------------------------------------------------

Iteration History

Fitting constant-only model:a

Iteration 0:   log likelihood = -1494.2292  
Iteration 1:   log likelihood =  -1388.076  
Iteration 2:   log likelihood =   -1385.31  
Iteration 3:   log likelihood = -1385.2992  
Iteration 4:   log likelihood = -1385.2992  

Fitting full model:b

Iteration 0:   log likelihood = -1385.2992  
Iteration 1:   log likelihood = -1351.1453  
Iteration 2:   log likelihood = -1350.7936  
Iteration 3:   log likelihood = -1350.7935

a. Fitting constant-only model – This is a listing of the log likelihoods at each iteration
for the logistic model predicting whether
or not a student is a certain zero. Remember that
logistic regression
uses maximum likelihood estimation, which is an iterative procedure. The first
iteration (called Iteration 0) is the log likelihood of the “null” or “empty”
model; that is, a model with intercept only model for the count model and intercept set to zero for inflated model or logistic model.  At the next iteration (called
Iteration 1), the variables specified for predicting certain zeroes are included in the model. In this example, the predictors
for the constant-only model are mathnce and langnce.  At each
iteration, the log likelihood increases because the goal is to maximize the log
likelihood. When the
difference between successive iterations is very small, the model is said to
have “converged” and the iterating stops. For more
information on this process for binary outcomes, see

Regression Models for Categorical and Limited Dependent Variables by J.
Scott Long (page 52-61).

b. Fitting full model – This is a listing of the log likelihoods at each iteration
for the full model, combining the constant-only model with the count model.
Again, the fitting of this model is an iterative procedure. Note that the log
likelihood of Iteration 0 for the full model is equal to the log likelihood at
which the constant-only model had converged.  This illustrates that the
full model begins with the fitted constant-only model stopped and improves on it
with the count model.

Model Summary

Zero-inflated Poisson regression                  Number of obse   =        316
                                                  Nonzero obsf     =        254
                                                  Zero obsg        =         62

Inflation modelc = logit                          LR chi2(3)h      =      69.01
Log likelihoodd  = -1350.794                      Prob > chi2i     =     0.0000

c. Inflation model – This indicates that the inflated model is a logit
model, predicting a latent binary outcome: whether or not a student is a certain zero. This also informs the interpretation of the parameter estimates.

d. Log Likelihood – This is the log likelihood of the fitted full model. It
is used in the Likelihood Ratio Chi-Square test of whether all predictors’
regression coefficients in the count model are simultaneously zero.

e. Number of obs – This is the number of observations in the dataset
for which all of the response and predictor variables are non-missing.

f. Nonzero obs – This is the number of observations in the dataset for
which the response variable is not equal to zero.

g. Zero obs – This is the number of observations in the dataset for
which the response variable is equal to zero.

h. LR chi2(3) – This is the Likelihood Ratio (LR) Chi-Square test that at least one of the predictors’ regression
coefficient in the count model is not equal to zero. The number in the parentheses indicates the
degrees of freedom of the Chi-Square distribution used to test the LR Chi-Square
statistic and is defined by the number
of predictors in the model (3). The LR Chi-Square statistic can be calculated by
-2*( L(null model of full model) – L(fitted model of full model)) = -2*((-1385.2992) – (-1350.7935)) =
69.01.

i. Prob > chi2 – This is the probability of getting a LR test
statistic as extreme as, or more so, than the observed statistic under the null
hypothesis; the null hypothesis is that all of the regression coefficients
for count model are simultaneously equal to zero. In other words, this is the
probability of obtaining this chi-square statistic (69.01) or one more extreme if there is in fact
no effect of the predictor variables in the count model. This p-value is compared to a specified
alpha level, our willingness to accept a type I error, which is typically set at
0.05 or 0.01. The small p-value from the LR test,  <0.00001, would lead us to
conclude that at least one of the regression coefficients in the count model is not
equal to zero. The parameter of the chi-square distribution used to test the
null hypothesis is defined by the degrees of freedom in the prior line,
chi2(3).

Parameter Estimates

------------------------------------------------------------------------------
             |      Coef.l  Std. Err.m     zn   P>|z|o    [95% Conf. Interval]p
-------------+----------------------------------------------------------------
daysabsj     |
     mathnce |  -.0002822   .0018619    -0.15   0.880    -.0039314    .0033671
     langnce |  -.0095004   .0019094    -4.98   0.000    -.0132428    -.005758
      female |   .2473714   .0488047     5.07   0.000     .1517161    .3430268
       _cons |   2.297473   .0693793    33.11   0.000     2.161492    2.433453
-------------+----------------------------------------------------------------
inflatek     |
     mathnce |   .0188534    .011087     1.70   0.089    -.0028766    .0405834
     langnce |   .0110383   .0111014     0.99   0.320    -.0107201    .0327966
       _cons |  -2.963912   .5315156    -5.58   0.000    -4.005664   -1.922161
------------------------------------------------------------------------------

j. daysabs – This is the response variable predicted by the full
model.

k. inflate – This portion of the output refers to the logistic model predicting whether or not a student is a certain zero.

l. Coef. – These are the regression coefficients.
The coefficients in the daysabs section of the output are interpreted as
you would interpret coefficients from a standard Poisson model: the expected
number of days absent changes by exp(Coef.) for each unit increase in the
corresponding predictor.

    Predicting Days Absent for For Students Not in the
“Certain Zero” Group

    mathnce – If a subject
were to increase his mathnce score by one point, the expected number of
days absent in a year would decrease by a factor of exp(-.0002822) = .99983821 while holding all other variables in the model constant.
Thus, the higher a student’s mathnce score, the fewer predicted days
absent.

    langnce – If a subject were to increase his
langnce score by one point, the expected number of days absent in a year
would decrease by a factor of exp(-.0095004) = .99046 while holding all other
variables in the model constant. Thus, the higher a student’s langnce
score, the fewer predicted days absent.

    female – The expected number of days absent in a
year for a female student is exp(.2473714) = 1.2764187 times the expected number
of days in a year for a male student while holding all other variables in the
model constant.  If female student and male student are not certain zeros
and have identical mathnce and langnce scores, the expected number
of days absent for the female student would be 1.2764187 times the expected
number of days absent for the male student.

    _cons – If all of the predictor variables in the model are
evaluated at zero, the predicted number of days absent would be calculated as
exp(_cons) = exp(2.297473). For males (the variable female evaluated at zero) with
zero mathnce and langnce scores, the predicted number of days
absent would be 9.9677412. This may seem very high, considering the mean number
of days absent is less than 6, but note that evaluating mathnce and
langnce at zero is out of the range of plausible scores.

    Predicting Membership in the “Certain Zero” Group

    mathnce – If a subject were to increase her mathnce score by one point, the odds that
she would
be in the “Certain Zero” group would increase by a factor of exp(0.0188534) =
1.0190322. In other words, the higher a student’s mathnce score, the more
likely the student is a certain zero.

    langnce – If a subject were to increase her
langnce score by one point, the odds that she would be in the “Certain Zero”
group would increase by a factor of exp(0.0110383) = 1.0110994. In other words,
the higher a student’s langnce score, the more likely the student is a
certain zero.

    _cons – If all of the predictor variables in the model are
evaluated at zero, the logit for being in the “Certain Zero” group is
exp(-2.963912) = 0.0516166.  This means that the predicted odds of a
student with mathnce and langnce scores of zero being a certain
zero are 0.0516166 (though remember that evaluating mathnce and
langnce at zero is out of the range of plausible scores).  Odds of
0.05 are very close to 0, meaning that it is very unlikely students (if there
were any) with mathnce and
langnce scores of 0 would be in “certain zero” group.

m. Std. Err. – These are the standard errors of the individual
regression coefficients for the two models. They are used
in both the calculation of the z test statistic, superscript n, and the
confidence interval of the regression coefficient, superscript p.

n. z – The test statistic z is the ratio of the Coef. to the Std. Err. of the respective predictor. The z value follows a standard normal distribution which is used to test against a two-sided alternative hypothesis that the
Coef. is not equal to zero.

o. P>|z| – This is the probability the z test statistic (or a more extreme test statistic) would be observed under the null hypothesis
that a particular predictor’s regression coefficient is zero, given that the
rest of the predictors are in the model. For a given alpha level, P>|z| determines whether of not the null hypothesis
can be rejected. If P>|z|
is less than alpha, then the null hypothesis can be rejected and the parameter
estimate is considered statistically significant at that alpha level.

    Predicting Days Absent for For Students Not in the
“Certain Zero” Group

mathnce – The z test
statistic for the predictor mathnce is (-0.0002822/0.0018619) = -0.15 with an
associated p-value of 0.880. If we set our alpha level to 0.05, we would fail to
reject the null hypothesis and conclude that the regression coefficient for
mathnce has not been
found to be statistically different from zero given langnce and female
are in the model.

    langnce –The z test
statistic for the predictor langnce is (-0.0095004/0.0019094) = -4.98 with an
associated p-value of < 0.001. If we set our alpha level to 0.05, we would
reject the null hypothesis and conclude that the regression coefficient for
langnce has been
found to be statistically different from zero given mathnce and female
are in the model.

    female –
The z test
statistic for the predictor female is (0.2473714/0.0488047) = 5.07 with an
associated p-value of < 0.001. If we again set our alpha level to 0.05, we would
reject the null hypothesis and conclude that the difference between males and
females has been found to be statistically different given that mathnce and
langnce are in the
model.

    _cons – The z test
statistic for the intercept, _cons, is (2.297473/0.0693793) = 33.11 with
an associated p-value of < 0.001. If we set our alpha level at 0.05, we would
reject the null hypothesis and conclude that _cons has been found to be
statistically different from zero given mathnce, langnce and
female are in the model and evaluated at zero.

    Predicting Membership in the “Certain Zero” Group

    mathnce –
The z test
statistic for the predictor mathnce is (0.0188534/0.011087) = 1.70 with an
associated p-value of 0.089. If we set our alpha level to 0.05, we would
fail to reject the null hypothesis and conclude that the regression coefficient for
mathnce has not been
found to be statistically different from zero given langnce is in the
model.

    langnce – The z test
statistic for the predictor langnce is (0.0110383/0.0111014) = 0.99 with an
associated p-value of 0.320. If we set our alpha level to 0.05, we would fail to reject
the null hypothesis and conclude that the regression coefficient for langnce has
not been
found to be statistically different from zero given mathnce is in the model.

_cons -The z test
statistic for the intercept, _cons, is (-2.963912/0.5315156) = -5.58 with an
associated p-value of < 0.001. With an alpha level of 0.05, we would reject the
null hypothesis and conclude that _cons has been found to be
statistically different from zero given mathnce and langnce are in
the model and evaluated at zero.

p. [95% Conf. Interval] – This is the Confidence Interval (CI) for an
individual coefficient given that the other predictors are in the model. For a
given predictor with a level of 95% confidence, we’d say that we are 95%
confident that the “true” coefficient lies between the lower and upper limit of
the interval.  It is calculated as the Coef. (z_α/2)*(Std.Err.),
where z_α/2 is a critical value on the standard normal distribution.
The CI is equivalent to the z test statistic: if the CI includes zero,
we’d fail to reject the null hypothesis that a particular regression coefficient
is zero given the other predictors are in the model. An advantage of a CI is
that it is illustrative; it provides a range where  the “true” parameter may
lie.

For more information

In cases where there is a question as to
which count model to use, the countfit
command is helpful for comparing the range of
count models.  You can download countfit from within Stata by typing search countfit
(see
How can I used the search command to search for programs and get additional
help? for more information about using search).

In times past, the Vuong test had been used to test whether a zero-inflated Poisson model or a Poisson model (without the zero-inflation) was a better fit for the data. However, this test is no longer considered valid. Please see The Misuse of The Vuong Test For Non-Nested Models to Test for Zero-Inflation by Paul Wilson for further information.