How do you test if the average of a population is the same as the variance of the same population?

Question

What can be a statistical test to find out if a population has the mean equal to its own variance? I.e. Mean(X)=Var(X)?

I am interested in it because Poisson regression makes the assumption that the mean of the predicted variable is the same as the variance of the predicted variable, i.e. lambda = E(Y) = Var(Y) where lambda is the parameter for the Poisson distribution for Y.

How can I test if given samples are taken from a Poisson distribution? — user2974951, May 23 '22 at 10:59
Are you actually interested in testing $H_0: \mu =\sigma^2$, or are you interested in testing if a distribution is Poisson and see $\mu\ne\sigma^2$ as a way to rule out Poisson? After all, $N(1,1)$ has $\mu=\sigma^2$ but isn’t Poisson. — Dave, May 23 '22 at 11:08
You made me realize that testing if the distribution is Poisson is what I need; which also answers my question as others have already received answers to the same question. — Fanta, May 23 '22 at 11:21
Except with count data and data (such as proportions) that are unitless, it usually makes no sense to equate the mean of a variable with its variance, because the two quantities use different units of measurement. For instance, when $X$ is a length, you are asking questions of the form "is one meter equal to one hectare?" — whuber, May 23 '22 at 13:10

score 2 · Answer 1 · answered May 24 '22 at 01:11

Your real problem seems to be testing if you have overdispersion (less usually underdispersion, my last Poisson regression project actually saw underdispersion, counting territorial birds!) The usual procedure is to fit the Poisson regression model, and then test for overdispersion (or underdispersion) afterwards.

There are a lot of posts about this on site, look through this stored search.

Glen_b · Accepted Answer · 2022-05-24T09:07:22.830

Beware!

In Poisson regression, it's the conditional distribution of the response not its marginal distribution that is Poisson.

Consequently, the variance will be equal to the mean within each specific set of values of the predictors, but not across all of them. In general the marginal variance will exceed the marginal mean, even when the Poisson regression assumptions are exactly correct.
While there are tests for this, I would not advise using hypothesis tests as assumption checks. Among other things:
- it answers the wrong question; it gives a noisy answer to a question you already know the answer to (no, your data are not exactly drawn from a conditionally-Poisson-distributed population), instead of telling you about the thing you need to (whether the model inadequacy is consequential for your inference).
- using the same data to choose models (via assumption tests or by other means) and to perform inference (estimation, confidence intervals, tests etc) or prediction itself interferes with the very properties of those analyses you're seeking to guarantee.

With a Poisson regression, there's a simple check of whether the conditional variance and mean are consistent available (and indeed a test if you insist on doing one anyway), which is to compare the residual deviance to the degrees of freedom. [There are other checks and tests, naturally.]

How do you test if the average of a population is the same as the variance of the same population?

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