Suppose $X_1, ..., X_n$ form a random sample from the Poisson Distribution with unknown mean $\lambda$. Let $\lambda_0$ and $\lambda_1$ be specific values such that $\lambda_1>\lambda_0>0$ and suppose that it is desired to test the hypotheses $H_0: \lambda=\lambda_0$ and $H_1:\lambda=\lambda_1$.
a. Construct a test procedure that has size $\alpha$ and that minimizes $\beta$
b. Suppose $\lambda_0=\frac{1}{4}$, $\lambda_1=\frac{1}{2}$, and $n=20$. For $\alpha=0.05$, find the minimum value of $\beta$.
I can solve part A using Neyman-Pearson Lemma and I come up with a test procedure of $P(\sum_{i=1}^{n}x_i \geq c|\lambda=\lambda_0)=\alpha$ where $c$ is a constant. I'm maybe 80% sure on that answer, so if someone else comes up with something totally different, I'd appreciate feedback.
However, part B has me stumped. I am not completely sure how to do it. I think I need to solve the probability equation in part A for the constant $c$ and then use that to solve a new probability equation, that is: $P(\sum_{i=1}^{n}x_i \geq c|\lambda=\lambda_1)=1-\beta$.
The part that is really throwing me off is how to solve for $c$.I know $\sum_{i=1}^{n}x_i$ follows Poisson Distribution with parameter $n\lambda$, but where do I go from there?
I don't necessarily need specific answers as much as I need a gentle push in the right direction :)
For (b), the value $n\lambda$ for both hypothesis is either $5$ (under $H_0$) or $10$ (under $H_!$). So the probabilities can be calculated directly.
First find $c$ as such integer value that $$1-\sum_{k=0}^{c-1}\frac{5^k}{k!}e^{-5}\geq 0.05$$ and $$1-\sum_{k=0}^{c}\frac{5^k}{k!}e^{-5}\leq 0.05$$
We find $c$ as a value where $P(\sum_{i=1}^nX_i\geq c|\lambda=\lambda_0)$ jumps over a value of $0.05$, since there is no hope that the probability can fall to $0.05$ exactly.
$$1-\sum_{k=0}^8\frac{5^k}{k!}e^{-5}\approx 0.0681>0.05, \quad 1-\sum_{k=0}^9\frac{5^k}{k!}e^{-5}\approx 0.031828<0.05,$$ and then $c=9$.
Next the minimal (over all possible tests) value of $\beta$ equals $$ \beta= \mathbb P\left(\sum_{i=1}^{n}X_i < 9\mid\lambda=\lambda_1\right)=\sum_{k=0}^8\frac{10^k}{k!}e^{-10} = 0,332819678750719. $$
For both cases it is possible to use CLT instead of direct calculations. The distribution of $$\frac{\sum_{i=1}^{20}X_i-20\lambda}{\sqrt{20\lambda}}$$ is approximately normal $N(0,1)$ and $$ \mathbb P\left(\sum_{i=1}^{20}X_i\geq c\biggm|\lambda=\frac14\right)=\mathbb P\left(\frac{\sum_{i=1}^{20}X_i-5}{\sqrt{5}}\geq \frac{c-5}{\sqrt{5}}\right)\approx 1-\Phi\left(\frac{c-5}{\sqrt{5}}\right)=0.05 $$ From the table of $\Phi(x)$ find $\frac{c-5}{\sqrt{5}}\approx 1.645$ and $c\approx 8.678$. And then $$ \beta= \mathbb P\left(\sum_{i=1}^{n}X_i < 9\mid\lambda=\lambda_1\right)=\mathbb P\left(\frac{\sum_{i=1}^{20}X_i-10}{\sqrt{10}} < \frac{8.678-10}{\sqrt{10}}\right)\approx 1-\Phi\left(-0.418\right)=0.338.$$
Here I do not take in mind that $c$ should be integer since at this case the values obtained by normal approximation will be very far from the exact values.