We are struggling with this question in Larson Introduction to Probability Theory and Statistical Inference. It's question #7 in section 8.3 (Some two-sample tests).
The question boils down to the following:
You have two Poisson random variables $X$ and $Y$. You have one observation of each, $x=47$ and $y=68$. We're asked to test the hypothesis that $E[X]=E[Y]$ and we're told to do it by developing a test based on the approximation of a Poisson by a normal.
I see other ways to attack this problem, but not by using normal approximations. The problem is we don't have a random sample, and we have just one observation of each normal distribution. So how can I construct a test with just that information? All the material in the section relies on having a full random sample from each of two populations.
The answer in the back of the book says "Reject $H_0$ for any $\alpha\geq0.006$".
Any suggestions, tips, leads or answers will be greatly appreciated. This is not homework, this came up in a group study section of bioinformaticians. Thank you.
To begin, please see this page for a variety of suggestions. Googling 'test difference in Poisson rates' fetches yet additional tests--including some authoritative ones, if you focus on notes from major universities.
An elementary method, related to testing equality of two binomial random variables and meeting the requirement to use normal approximation, is to observe that $$Y - X \stackrel{aprx}{\sim} \mathsf{Norm}(\mu = \lambda_y - \lambda_x,\, \sigma = \sqrt{\lambda_y + \lambda_x}),$$ for $\lambda_x$ and $\lambda_y$ sufficiently large. So one can show $$Z = \frac{Y - X}{\sqrt{Y+X}} \stackrel{aprx}{\sim} \mathsf{Norm}(0,1).$$ under $H_0: \lambda_y = \lambda_x.$
Then to test $H_0: \lambda_y = \lambda_x$ against $H_a: \lambda_y \ne \lambda_x$ at level $\alpha = .05$ one would reject for $|Z| > 1.96.$ For your data $Z \approx 1.96,$ leading to rejection with P-value about 5%.
To verify one instance of the null distribution, a simulation of 100,000 $Z$'s with $\lambda_X = \lambda_y = 50$ gives the following histogram shown along with the standard normal density curve.
One of the tests mentioned in the link above is implemented in R statistical software:
poisson.test(47,68)returns P-value about 0.01.Note: I can't make sense of the 'Answer' $\alpha \ge 0.006.$ My guess is that the text may intend you to do a likelihood ratio test and that some symbol other than $\alpha$ may have been intended.