Let $X$ be a Poi($\nu$) distributed rv with unknown $\nu \in (0,\infty)$ and we set $ \theta:=\nu $. We want to test the hypotheses $ H_0:\nu_0=1 $ against $ H_1:\nu_1=\nu_0=2 $
a) We look at the (randomized) Neyman - Pearson - Test which has the form
$T(k) = \begin{cases} 1 & R(k) > c \\ \gamma & R(k)=c \\ 0 & R(k)<c \end{cases}$ , with $c \in \mathbb{R} $ and $\gamma \in [0,1]$
$ R(k) := \frac{P_{\nu_1}(X=k)}{P_{\nu_0}(X=k)} $ is the Maximum Liklihood quotient.
Compute $c^*$ and $ \gamma^* $ such that $T$ is a test with niveau $\alpha = 0.05$ i.e. $E_{\nu_0}[T]=\alpha$.
b) You see that $X=3$. How do you decide in terms of the test wich you construct in a)?
Values of a Poi($\nu$) distributed rv for task a')
$ P(X\leq 0) = 0.368$
$P(X\leq 1) = 0.736$
$P(X\leq 2) = 0.92$
$P(X\leq 3) = 0.981$
$P(X\leq 4) = 0.996$
$P(X\leq 5) = 0.999$
I have computed $ R(k) = \frac{e^{-2n} \prod_{i=1}^n \frac{2^{k_i}}{k_i !} }{e^{-n}\prod_{i=1}^n \frac{1}{k_i !}} = e^2 2^{\sum_{i=1}^n k_i} $.
But how can i now compute $c^*$? If i have $c^*$ then i can compute $\gamma^*$ with the fact that
$ E_{\nu_0}[T] = P(R(k)>c^*) + \gamma^* P(R(k)=c^*) = 0.05 = \alpha $