# Likelihood ratio test of Poisson distribution

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...

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 $$T, as $$\mathbb{P}(T<-1.6449\mid H_0)=0.05$$ (see e.g. quantiles of normal). Hence $$c=-1.6449\times 10+100$$.