Variables $X_1 \dots X_n$ ~ Pois($\theta$), i.i.d. Find the best test of size $\alpha$ of $H_0 : \theta = 1$ against $H_1 : \theta = 1.21$. Show that the smallest value of n to make $\alpha = 0.05$ and $\beta \le 0.1$ is about 213. ($\beta$ is the probability of type II error)
The way I approached it:
$f(x|\theta)=\prod_{i=1}^{n}e^{-\theta}\theta^{x_i}/x_i!=e^{-n\theta}\theta^{\sum_ix_i}/\prod_ix_i!$
Then I used the Neyman-Pearson lemma to construct a test $\phi(x)=1$ if $f_1/f_2\ge k$, $\phi(x)=0$ otherwise. Since $f_1/f_2\ge k$ is monotonically increasing with $\sum_i x_i$, this is equivalent to $S_n = \sum_i x_i\ge c$. Using the Central limit theorem, for large n $\frac{S_n-n\theta}{\sqrt{n\theta}}$~N(0,1) (using the fact that for X~Poisson, $Var(X)=EX=n\theta$). Hence the size of the test is $P_{\theta=\theta_0}(\frac{S_n-n\theta}{\sqrt{n\theta}}\ge z_\alpha)=0.05$. The probability of type II error is: $P_{\theta=\theta_1}(Z\lt z_\alpha)=P_{\theta=\theta_1}(\frac{S_n-n\theta}{\sqrt{n\theta}})\le0.1.$
Could anyone tell me how to continue from this point? Do I have to go back to the Poisson distribution to work out the smallest n or do I use the normal distribution?
