From Wackerly, Mathematical Statistics with Applications, 7th Ed. (p548):

Let $\displaystyle Y_1,Y_2,\cdots, Y_n$ denote a random sample from a population having a Poisson distribution with mean $\displaystyle \lambda_1$. Let $\displaystyle X_1,X_2,\cdots,X_m$ denote an independent random sample from a population having a Poisson distribution with mean $\displaystyle \lambda_2$. Derive the most powerful test for testing $\displaystyle H_0:\lambda_1=\lambda_2=2$ versus $\displaystyle H_a:\lambda_1=1/2,\lambda_2=3$.
According to the likelihood function,

$\displaystyle L(\lambda_1,\lambda_2)=\prod_{i=1}^n \frac{\lambda_1^{y_i} e^{-\lambda_1}}{y_i!}\prod_{j=1}^m \frac{\lambda_2^{x_j} e^{-\lambda_2}}{x_j!}=\frac{\lambda_1^{n\sum_{i=1}^n y_i}\lambda_2^{m\sum_{j=1}^m x_j}e^{-n\lambda_1-m\lambda_2}}{\prod_{i=1}^n y_i!\prod_{j=1}^m x_j!}$.

So by Neyman-Pearson, our rejection region for the most powerful test is given by

$\displaystyle \frac{L(\lambda_1=2,\lambda_2=2)}{L(\lambda_1=1/2,\lambda_2=3)}=\frac{2^{n\sum_{i=1}^n y_i}2^{m\sum_{j=1}^m x_j}e^{-2n-2m}}{(1/2)^{n\sum_{i=1}^n y_i}3^{m\sum_{j=1}^m x_j}e^{-(n/2)-3m}}$

$\displaystyle =\frac{2^{2n\sum_{i=1}^n y_i}(2/3)^{m\sum_{j=1}^m x_j}}{e^{(3/2)n-m}}<k$.

Substituting $\displaystyle k'=ke^{(3/2)n-m}$, we have the rejection region

$\displaystyle 2^{2n\sum_{i=1}^n y_i}(2/3)^{m\sum_{j=1}^m x_j}<k'$,

where $\displaystyle k'$ we define by

$\displaystyle \alpha=P\left[2^{2n\sum_{i=1}^n y_i}(2/3)^{m\sum_{j=1}^m x_j}<k'\Big|\lambda_1=\lambda_2=2\right]$,

and $\displaystyle \alpha$ is the desired probability of a type I error, that is, the "level" of the test.

Assuming I have not made any errors in the above analysis, then I am curious, can I do anything more towards defining $\displaystyle k'$? In particular, is there any way to define $\displaystyle k'$ as a function of $\displaystyle \alpha$?

Thanks!