Finding the error of type I

Question

let $x_1.... x_n$ be a random sample from a Poisson distribution with mean $\theta$, that is

$$f(x ;\theta) = \theta^x e^{-\theta}/x!.$$

We use a test that accepts the null hypothesis if $(1/3)≤ \overline{x} ≤ (2/3)$ and reject otherwise.

For $n=9$ what is the error type I?

My attempt: The idea of this question is to find the region that rejects the null hypothesis when it is true.

$$\begin{align}p(\text{type I error${}\mid H_0$ is true})&=p(\text{reject $H_0$ when $\theta=0.4$})\\ &=p(\overline{x}<(1/3)\mid\theta=0.4)+p(\overline{x}>(2/3)\mid\theta=0.4)\end{align}$$

then I use another idea if $x_i$ with parameter $\theta$ belong to poison then $\sum_{i=1}^n(x_i)$ is also belong to Poisson with parameter $n\theta$

then I choose $\overline{x}= 0$ and $\overline{x}=1$

Answer 1

Thanks for showing what you have done so far.

With your second method, letting $$T = \sum_{i=1}^9 X_i = 9\bar X \sim \mathsf{Pois}(9\theta),$$ you can get an exact Poisson probability:

When $\theta = 0.4,$ you fail to reject $H_0$ if $3 \le T \le 6,$ so the probability of rejection when $H_0$ is true is $$1 - P(3 \le T \le 6 \,|\, \lambda = 9\theta = 9(.4) = 3.6)\\ = 1 - P(3 \le T \le 6\,|\,\lambda=3.6) = 0.3760.$$

Using R:

1 - sum(dpois(3:6, 3.6))
[1] 0.3760203

In the figure below, the rejection regiou consists of integers outside the vertical (red) dotted lines.

R code for figure:

t = 0:10;  PDF = dpois(t, 3.6)
plot(t, PDF, type="h", col="blue", lwd=3, main="POIS(3.6)")
 abline(v=0, col="green2");  abline(h=0, col="green2")
 abline(v = c(2.5, 6.5), col="red", lwd=2, lty="dotted")

Note: Using $\bar X,$ you might try to get the Type I error probability, from a normal approximation, but as you can see from the figure the mean of the underlying Poisson distribution is not large enough for a normal approximation to be accurate.