# Finding a p-value based on the Poisson distribution

I'm having trouble understanding what I need to do to solve this problem for homework:

Suppose a company which produces fire alarms has claimed that the fire alarms make only one false alarm per year, on average. Let $X$ denote the number of false alarms per year. Assume $X \sim Poisson(\lambda)$. Under the company's claim, the probability of observing $x$ fire alarms per year is

$P(X = x) = \frac{e^{-\lambda}\lambda^x}{x!} = \frac{e^{-1}}{x!}, x = 0,1,...$

A customer had a bad experience with the fire alarm he purchased before. He wants to conduct hypothesis testing $H_0: \lambda = 1$ versus $H_1: \lambda > 1$, and he purchased another fire alarm from the same company. He allows 1% chance for falsely rejecting $H_0$. After a year, he observed three false alarms.

a. Find the p-value based on the single observation of three false alarms for the year.

b. Draw a conclusion based on the p-value in part a.

For part a, I'm very unsure how to do this. In class, we've only done hypothesis testing and p-values on normal distributions so I'm not sure if I can follow the same method where I use a test statistic, and find a value based on that value? I'm just very confused on what to do.

Recall that the $p$-value is the probability of there being an outcome at least as extreme as observed. Here $X\sim\mathrm{Pois}(\lambda)$, and we observed a value of $3$. So the $p$-value is $$\mathbb P(X\geqslant 3) = 1 - e^{-1}\left(1 + 1 + \frac12\cdot1^2 + \frac1{3!}\cdot1^3\right)= 1 - \frac83 e^{-1}\approx 0.01899.$$ Since $0.01899 > 0.01$, we fail to reject the null hypothesis.
• How are you getting the last number $1^2$? isn't $e^{-1} \times (\frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!})$? Or am I not supposed to be using $\frac{e^{-1}}{x!}$