Solved – Poisson Hypothesis Testing for Two Parameters

hypothesis testingpoisson distribution

So, for fun, I am taking some of the data of calls from the call center I work at and trying to do some hypothesis testing on them, specifically the number of calls received in a week, and using a Poisson distribution to fit it. Due to the subject matter of my job, there are two types of weeks, let's call one of them on-weeks where I hypothesize there are more calls, and off-weeks where I hypothesize there are fewer.

I have a theory that the $\lambda$ from on-weeks (let's call it $\lambda_1$) is larger than that of the one from off-weeks (let's call it $\lambda_2$)

So the hypothesis I want to test is $H_0: \lambda_1 > \lambda_2, H_1: \lambda_1 \leq \lambda_2 $

I know how to test for one parameter (say $H_0: \lambda_1 > 1, H_1: \lambda_1 \leq 1 $) but not so sure how to go about doing 2 given a data set. Let's say I take two week's worth of data from each one $X_1 = 2$ and $X_2 = 3$ for the off-week and $Y_1 = 2$ and $Y_2=6$ for the on-week. Can someone help walk me though this simpler version such that I can apply it to a larger data set? Any help is appreciated, thank you.

Best Answer

Note that normally the equality goes in the null (with good reason).

That issue aside, I'll mention a couple of approaches to a test of this kind of hypothesis

  1. A very simple test: condition on the total observed count $n$, which converts it to a binomial test of proportions. Imagine there are $w_\text{on}$ on-weeks and $w_\text{off}$ off-weeks and $w$ weeks combined.

Then under the null, the expected proportions are $\frac{w_\text{on}}{w}$ and $\frac{w_\text{off}}{w}$ respectively. You can do a one-tailed test of the proportion in the on-weeks quite easily.

  1. You could construct a one tailed test by adapting a statistic related to a likelihood-ratio test; the z-form of the Wald-test or a score test can be done one tailed for example and should work well for largish $\lambda$.

There are other takes on it.

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