# Probability of Type I Error with 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)*((lamda^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 I need to get the sum of the probabilities of 1 and 2, with lambda = 1, and subtract that from one?

Would someone mind putting me on the right track? (I hope this question was intelligible).

(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!).

Thanks very much!

Your second suggestion is right, with a little detail incorrect. A reasonable test is to reject the null hypothesis if the observation is $\ge 3$. So the probability of Type I error is $1$ minus the probability that, if $\lambda=1$, our observation is $0$ or $1$ or $2$.
Thus the probability of Type I error is $1-e^{-1}\left(1+1+\frac{1}{2}\right)$.
• You are welcome. It is all too easy to forget about $0$, it is so tiny. I have seen this little slip a few hundred times. – André Nicolas Oct 19 '14 at 17:26