I would like to test in what regression fits my data best. My dependent variable is a count, and has a lot of zeros.

And I would need some help to determine what model and family to use (poisson or quasipoisson, or zero-inflated poisson regression), and how to test the assumptions.

  1. Poisson Regression: as far as I understand, the strong assumption is that dependent variable mean = variance. How do you test this? How close together do they have to be? Are unconditional or conditional mean and variance used for this? What do I do if this assumption does not hold?
  2. I read that if variance is greater than mean we have overdispersion, and a potential way to deal with this is including more independent variables, or family=quasipoisson. Does this distribution have any other requirements or assumptions? What test do I use to see whether (1) or (2) fits better - simply anova(m1,m2)?
  3. I also read that negative-binomial distribution can be used when overdispersion appears. How do I do this in R? What is the difference to quasipoisson?
  4. Zero-inflated Poisson Regression: I read that using the vuong test checks what models fits better.

    > vuong (model.poisson, model.zero.poisson)

    Is that correct? What assumptions does a zero-inflated regression have?

  5. UCLA's Academic Technology Services, Statistical Consulting Group has a section about zero-inflated Poisson Regressions, and test the zeroinflated model (a) against the standard poisson model (b):

    > m.a <- zeroinfl(count ~ child + camper | persons, data = zinb)
    > m.b <- glm(count ~ child + camper, family = poisson, data = zinb)
    > vuong(m.a, m.b)

I don't understand what the | persons part of the first model does, and why you can compare these models. I had expected the regression to be the same and just use a different family.


1) Calculate the mean and the sample variance. $\frac{\bar{X}}{S^2}$ should be $\mathrm{F}(1,n-1)$ distributed, where $n$ is the size of the sample and the process is truly Poisson - since they are independent estimates of the same variance.

Note that this test ignores the covariates - so probably not the best way to check over-dispersion in that situation.

Note also that this test is probably weak against the zero-inflated hypothesis.

3) negative binomial in R: use glm.nb from the MASS package, or use the zeroinfl function from the pscl package using the negative binomial link.

4) zip (zero-inflated Poisson) is a mixture model. You have a binary outcome, according to which a subject belongs to group A (where a 0 is certain) or to group B (where counts are Poisson or neg binomial distributed). An observed 0 is due to subjects from group A + subjects from group B who just happened to be lucky. Both aspects of the model can depend on covariates: group membership is modeled like a logistic (log odds is linear in the covariates) and the Poisson part is modeled in the usual way: log mean is linear in the covariates. So you need the usual assumptions for a logistic (for the certain 0 part) and the usual assumptions for a Poisson. In other words, a zip model will not cure your overdispersion problems - it only cures a big gomp of zeroes.

5) not sure what the data set is and couldn't find the reference. zeroinfl needs a model for both the poisson part and the binary (certain 0 or not) part. The certain 0 part goes second. So m.a is saying that whether the person is a certain 0 or not depends on "persons" - and assuming the subject is not a certain 0, count is a function of camper and child. In other words log(mean) is a linear function of camper and child for those subjects not requiring a 0 count.

m.b is just a general linear model of count in terms of camper and child - both assumed to be fixed effects. The link function is Poisson.

| cite | improve this answer | |
  • $\begingroup$ Thank you! A quick question: is there a way to produce r^2 or pseudo-r^2 like Nagelkerke in glm using family=poisson in R? Thank you! $\endgroup$ – Torvon Oct 21 '12 at 1:23
  1. library(pastecs)

stat.desc(dep_var) - and then take a look if the mean and the variance are equal. From here you can also calculate the % of zeroes in your vector.

| cite | improve this answer | |
  • 3
    $\begingroup$ Welcome to the site. This is more like a comment than an answer; also, it is better to use proper spelling and so on - this isn't texting and many people who read this site have English as a 2nd or 3rd or .... language. $\endgroup$ – Peter Flom Nov 9 '12 at 12:21
  • 3
    $\begingroup$ Please, work on improving this quick reply. $\endgroup$ – chl Nov 9 '12 at 14:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.