This question asks:

$N_A$ and $N_B$ are variables of the counts of the number of events 'A' and events 'B' respectively. Those variables follow Poisson distributions with parameters $\lambda_A$ and $\lambda_B$.

In nature I made one observation of each variable; I observed $n_A$ events 'A' and $n_B$ events 'B'. From those, I am asking: Are $\lambda_A$ and $\lambda_B$ equal or different?

How can I make a hypothesis testing on the null that the rates $\lambda_A$ and $\lambda_B$ are the same?

The answer suggest to use a Wald test with this formula:


Based on the comments here, I suppose $n_1, n_2$ are what the questioner calls $n_A, n_B$. But I'm not sure about $\widehat{\lambda}_1$ – I would assume that it is the estimate for $\lambda_1$, but isn't that just $n_1$ (as the highest probability that one sees $n$ occurrences is when $\lambda = n$)?

  • 1
    $\begingroup$ Your post is highly confusing. You start by defining the $\lambda_i$ as random variables, but then you talk about the $\hat{\lambda}_i$ (i.e. as if the $\lambda$s were parameters -- which is consistent with the conventional usage). You then talk about the $n_i$ (which is defined in the question at the link you gave as observed values -- realizations of $N_1$ and $N_2$) as if they were random variables, which they aren't. Please distinguish your random variables from your parameters & your observations, and check the question at the page you link to for a definition of the notation it uses. $\endgroup$
    – Glen_b
    Mar 11, 2019 at 22:48
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    $\begingroup$ I agree with @Glen_b that your question is confusing. Maybe a look at this page (or its links) will help you get oriented, and maybe provide a useful test. If you get oriented without finding a solution, perhaps edit your question so someone can help. $\endgroup$
    – BruceET
    Mar 13, 2019 at 6:33
  • $\begingroup$ Thanks Glenn and Bruce – I have tried to update my question, does this help? Basically: it seems like this equation requires four variables but I only know two. $\endgroup$
    – Xodarap
    Mar 13, 2019 at 20:53
  • 1
    $\begingroup$ To apply the answer you referenced, you need to take $n_1=n_2=1$ because you have exactly one observation from each distribution. $\endgroup$
    – whuber
    Mar 13, 2019 at 21:40


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