Finding the error of type I 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 $\overline{x}=1$
 A: 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.
