Is Poisson appropriate and how to interpret?

[Sarah Brown]

I have a dataset that, among other variables, has acceptance rate (as a percentage, 0-1 scale) and number of payments (0, 1, 2, 3). I want to assess whether or not the leniency in acceptances (that is, a higher acceptance rate) is associated with the likelihood that payments are made.
At first, I ran an ordered logit regression and calculated the marginal effects at the 25th, 50th, and 75 percentile of acceptance rate. However, because my dependent variable is a count, I'm thinking I may need to use a Poisson or negative binomial model. The mean of my DV is not equal to the variance, but I think to solve this, I can just use "poisson y x, vce(robust)."
This is the exact question: Present one table and one graph assessing whether the leniency of a firm is associated with the likelihood that renewal fees are paid.
When I run the poisson regression, I get this:

. poisson number_payments acceptance_rate, irr vce(robust)

Iteration 0: log pseudolikelihood = -6399.7702
Iteration 1: log pseudolikelihood = -6399.7702

Poisson regression Number of obs = 4,017
Wald chi2(1) = 13.32
Prob > chi2 = 0.0003
Log pseudolikelihood = -6399.7702 Pseudo R2 = 0.0006

---------------------------------------------------------------------------------
| Robust
number_payments | IRR Std. Err. z P>|z| [95% Conf. Interval]
----------------+----------------------------------------------------------------
acceptance_rate | .8221783 .0441114 -3.65 0.000 .740112 .9133444
_cons | 2.44222 .0869474 25.08 0.000 2.277616 2.61872
---------------------------------------------------------------------------------
I'm not sure how to interpret this in the context of the question.

Thanks for your help!

[Nick Cox]

This appears to be an assignment; if so, our policy on homework is pertinent.

But regardless of that, you can draw a graph to see what your regression is doing with a scatter plot and fitted exponential. Despite the significance levels, it is not obvious that the fit is good substantively, so that there may be less to interpret than you think.

If 3 is an upper bound as a matter of principle, Poisson is a dubious choice.

I can't comment on the underlying economics here.

[Sarah Brown]

Hi Nick,

Thanks for your reply. The question is from a task sent by my research advisor for an RA task, not a homework assignment.

Three is an upper bound because there can only be three payments before expiration. So it's a "true maximum" in the population sense. I'm mostly confused here on the difference between poisson with robust SEs and nbreg. I get the same marginal effects and coefficients, but the robust standard errors in the Poisson model are slightly lower than the standard errors of nbreg. I was always taught that you can correct for violation of the mean=variance assumption by using either, but I'm not sure what the practical difference is.

[Nick Cox]

As the response is a counted fraction 0/3, 1/3, 2/3, 3/3 I recommend fracreg or a binomial model.

I can't see that any of the other models you mention make sense. The fact that Poisson and nbreg results are similar is not surprising but not cogent either.

For a bounded variable quite different variance-mean relations take over. Consider that if the mean approached 0 or 3 it could only do so if all values approached 0 or 3 so that variance must approach 0 at either bound.

I am still eager to see a graph.

[Maarten Buis]

Your variable is a count of the number of "successes" in a given number of trials (in your case 3). In that case the binomial distribution is a more natural model than the Poisson. In Stata that is implemented via binreg

binreg number_payments acceptance_rate, or vce(robust) n(3)

[Nick Cox]

I'd back Maarten here. The command name binreg had slipped my mind, although clearly not the idea of a binomial model.

[Sarah Brown]

Ah! Thank you for the help!

[Nick Cox]

Never allowed to go above 3 is a bound, regardless of whether the researcher or the economy or whatever else is the bounding power. Also, if you fit Poisson regression you're fitting an exponential: you really think that is the right functional form here?

Why is ordered logit sensible if the data aren't bounded?

You're counting payments out of 3; that's just as binary and binomial as heads in 3 coin tosses. Otherwise, please define "truly binary".

I don't find quartiles useful for small counts, the answer is either one of the values (here 0 1 2 3) or exceptionally half-way between them. It's easy to show that you can get the same summaries for very different samples and very different summaries for almost identical samples.