How do you choose between Poisson and negative binomial models for discrete count outcomes?

One key criterion is the relative value of the variance to the mean after accounting for the effect of the predictors. A previous article discussed the concept of a variance that is larger than the model assumes: overdispersion.

(Underdispersion is also possible, but much less common).

There are two ways to check for overdispersion:

**The Pearson Chi**^{2}dispersion statistic

The Pearson Chi^{2} dispersion statistic for the model run in that article was 2.94. If the variance is equal to the mean, the dispersion statistic would equal one.

When the dispersion statistic is close to one, a Poisson model fits. If it is larger than one, a negative binomial model fits better.

**Residual Plots**

Plotting the standardized deviance residuals to the predicted counts is another method of determining which model, Poisson or negative binomial, is a better fit for the data.

Here is the plot using a Poisson model when regressing the number of visits to the doctor in a two week period on gender, income and health status.

The series of waves in the graph is not an unusual structure when graphing count model residuals and predicted outcomes.

Our primary focus is on the scale of the y axis. A good fitting model will have the majority of the points between negative 2 and positive 2. There should be few points below negative 3 and above positive 3.

Adding more predictors to the model can have an impact on improving the plot but the Poisson model is clearly a very poor fitting model for these data.

If we use the same predictors but use a negative binomial model, the graph improves significantly.

Notice now the maximum value for the standardized deviance residual is now 4 as compared to 8 for the Poisson model. The model still has room for improvement. That would require, if they are available, selecting better predictors of the outcome.

Now let’s compare the graphs when the Pearson Chi^{2} dispersion is closer to one. We will now regress the count of rabbits per 400 square yard plots on shrub coverage, density of shrubbery and variety of shrubbery. The Pearson Chi^{2} dispersion for this model is 1.15.

Using a Poisson model our graph looks like this:

Almost all of the residual points are now inside of negative 2 and positive 2.

Here is the graph of the negative binomial model using the same predictors:

The two graphs are nearly identical.

As you have seen, graphing the standardized deviance residuals by the predicted outcomes can help us verify which type of model is a better fit for your data.

### Related Posts

- Overdispersion in Count Models: Fit the Model to the Data, Don’t Fit the Data to the Model
- The Problem with Linear Regression for Count Data: Predicting Length of Stay in Hospital
- The Importance of Including an Exposure Variable in Count Models
- Analyzing Zero-Truncated Count Data: Length of Stay in the ICU for Flu Victims

{ 4 comments… read them below or add one }

Hello:

I have a question regarding this as well: how would you test for over dispersion in GAMMs? I have data that I believe is zero-inflated but I want to check whether to apply a negative binomial or a poisson would be better?

I’m very new with these types of statistics and have found myself very confused.

Thank you!

Hi Kelsey,

It’s my view that “zero inflated” is a misleading term. It actually does not mean that you have a lot of zeros in your outcome variable. Zero inflated means you have observations in your data set that can only be zeros. For example, let’s suppose your outcome is “number of fish caught” by people on the pier. There might be people on the pier that aren’t trying to catch fish, they might be there to enjoy the smell of the salt in the air. People not fishing would be considered to be zero inflated and should be accounted for differently.

Regarding how do you determine whether you should use a negative binomial model or a Poisson model, my first choice is to use a negative binomial model. If the data is not overdispersed the negative binomial model will most likely not converge. If it doesn’t converge I would then use a Poisson model.

Jeff

Hello:

I had a question. On this page you stated that a good fitting model would have the majority of its residual points between 2 and -2. I am wondering why this is the case? Any help you could provide would be fantastic.

Hi Benjamin,

Our goal for a good fitting model is a normal distribution of our residuals. In a standardized normal distribution (think of the bell curve) approximately 95% of all of the observations are between -2 and +2. We would like to see that pattern with our standardized residuals. In the case for count models with discrete outcomes, we use the standardized deviance residuals because they have been modified to reflect the distribution that we would have with a continuous outcome. Standardized Pearson residuals should not be used.

Jeff