Size and Power of Test: Poisson Distribution In a homework problem, I am given a Poisson distribution with $\lambda = 1$ as null hypothesis, $\lambda$ greater than or equal to 2 as an alternate hypothesis, and 3 as a test statistic. I am instructed to reject the null hypothesis if I observe that test statistic. We are describing traffic accidents in the problem, so the question is "What is the probability that you are committing a Type I Error if you reject the null hypothesis ($\lambda=1$) when you see 3 accidents?" I'm then asked for the power of the test.
My first thought was to use $(e^{-\lambda})*((\lambda^3)/(3!))$ to get the probability of 3 given a $\lambda$ of 1. But it seems like it has to be more complicated than that. Would someone mind putting me on the right track?
(By the way, I know it's extremely annoying to have the formula written in words rather than symbols---I'm brand new to the site and under sort of a time crunch, but I'll get it right in future posts. I apologize for the irritation!).   
 A: It is unclear from your question whether you are just observing one Poisson random variable, or a sample of them.  I will assume the latter, but the former follows as a special case.  Anyway, under this view of the problem you have $X_1,...,X_n \sim \text{IID Pois}(\lambda)$ and your hypotheses are:
$$H_0: \lambda = 1 \quad \quad \quad H_A: \lambda \geqslant 2.$$
Your rejection region is $\bar{X} \geqslant 3$, so you have power function given by:
$$\begin{equation} \begin{aligned}
B(\lambda) 
&= \mathbb{P}(\text{Reject } H_0 | \lambda) \\[6pt]
&= \mathbb{P}(\bar{X} \geqslant 3 | \lambda) \\[6pt]
&= \mathbb{P}(n\bar{X} \geqslant 3n | \lambda) \\[6pt]
&= 1-\mathbb{P}(T < 3n | \lambda) \\[6pt]
&= 1- \exp(-\lambda) \sum_{k=0}^{3n-1} \frac{\lambda^k}{k!}. \\[6pt]
\end{aligned} \end{equation}$$
In the special case where $n=1$ (i.e., you have only one obesrvation) you have power:
$$\begin{equation} \begin{aligned}
B(\lambda) 
&= 1- \exp(-\lambda) \Big[ 1 + \lambda + \frac{\lambda^2}{2} \Big]. \\[6pt]
\end{aligned} \end{equation}$$
