Likelihood ratio test of Poisson distribution

Question

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...

Answer 1

From the properties of Poisson random variables, $S=X_1+\dots+X_{100}$ has Poisson$(100\lambda)$ distribution; underh $H_0$, $S\sim Po(100)$ with $\mathbb{E}(S)=100$ and $Var(S)=100$. Hence $T:=\frac{S-100}{\sqrt{100}}$ is asymptotically normal $\mathcal{N}(0,1)$; under the alternative hypothesis $T$ would be smaller than under $H_0$. Hence we reject when $T<t_0=-1.6449$, as $\mathbb{P}(T<-1.6449\mid H_0)=0.05$ (see e.g. quantiles of normal). Hence $c=-1.6449\times 10+100$.