I'm given this problem:
Let $X_1,...X_{100}$ be a random sample from a Poisson distribution with mean $\lambda$. Consider testing the hypothesis $H_0$: $\lambda=1$ vs $H_1$: $\lambda<1$. Consider the rejection region $\sum_{i=1}^{n}x_{i}\leq c$. Using Central Limit Theorem to find $c$ s.t. $\alpha=0.05$.
My attempt:
The likelihood function is $L(\lambda)= \Pi_{i=1}^{100}\dfrac{{\lambda}^{X_{i}}e^{\lambda}}{\Pi_{i}X_{i}!}$.
Hence the likelihood ratio is $\dfrac{L(1)}{L(\bar{X})}$.
And we kind of have to 'solve' for $\sum_{i}(X_i)$ inside $P(\dfrac{L(1)}{L(\bar{X})}\leq k\mid \lambda=1)=0.05$.
Yet I got something like $\mathbb{P}(-100-\sum_{i}X_{i}(\ln\sum_{i}X_{i}-100-\ln 100)\leq \ln{k} | \lambda=1)$ and I'm not sure how to continue...