0
$\begingroup$

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:

$$Z=\frac{\widehat{\lambda}_1-\widehat{\lambda}_2}{\sqrt{\frac{{\widehat{\lambda}_1}}{n_1}+\frac{{\widehat{\lambda}_2}}{n_2}}}$$

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 $\lambda = n$)?

Improve this question
$\endgroup$
4
  • 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
  • 1
    $\begingroup$ 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

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.