How would one find the probability of a type one error in a poisson distribution?
I know the PDF is f(x/lambda) = ((lambda^x)/(x!))*(e^negativelambda)
Sample Size = 1
My test hypotheses are H0: lambda = 4 and H1: lambda < 4, Rejecting H0 if X <= 1
I'm particular struggling with the integral that I have set up, which is
integral(-infinity, 1) ((4^x)/(x!))*(e^-4) dx
The Type I error is the probability that, given the null hypothesis is true, the rejection criterion is met. In your case, this means that $X$ is Poisson distributed with $\lambda = 4$, and you find that $X \le 1$; i.e. $$\Pr[X \le 1 \mid \lambda = 4] = \Pr[X = 0 \mid \lambda = 4] + \Pr[X = 1 \mid \lambda = 4] = e^{-4} \frac{4^0}{0!} + e^{-4} \frac{4^1}{1!}.$$ That's it. There is no integration involved because $X$ is a discrete-valued random variable, being Poisson distributed. It is not possible for $X$ to be anything other than a nonnegative integer.