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Let $X_i \sim Pois(\lambda)$, $i=1, ...,10$. We have the two hypothesis:

$H_0 : \lambda = 0.1$ and $H_{\alpha} : \lambda = 0.5$. We reject $H_0$ if $\sum_{i=1}^{10} X_i \geq 3$.

I am trying to calculate the significance level $\alpha$ and the power of test $1 - \beta$. I was thinking about calculating the type 1 and 2 error, but I think my results aren't correct:

$\alpha = \mathbb{P} [\sum_{i=1}^{10} X_i \geq 3 | H_0] = \sum_{k=3}^{10} \frac{0.1^k}{k!} \approx 0.000155$

and

$1 - \beta = 1 - \mathbb{P} [\sum_{i=1}^{10} X_i \leq 2 | H_{\alpha}] = 1 - 0.9856123$, but this result seems not to be really plausible...

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Let $T=\sum_{i=1}^{10} X_i$. Supposing that the $X_i$ are independent, under the null, it is the case that $T\sim \text{Poisson}(1)$ (where we are using the fact that in general if $X\sim \text{Poi}(\lambda_1)$ and $Y\sim\text{Poi}(\lambda_2)$ and $X, Y$ are independent then $X+Y\sim \text{Poi}(\lambda_1+\lambda_2)$.

In particular, the probability of type I error is given by $$ \alpha=P(T\geq 3)=1-P(T=0)-P(T=1)-P(T=2) $$ where $T\sim \text{Poi}(1)$. Similarly, the power of the test, i.e. the probability that we reject the null given that it is false is given by $$ 1-\beta=P(T\geq 3)=1-P(T=0)-P(T=1)-P(T=2) $$ where $T\sim \text{Poi}(5)$

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