An exact and an approximate confidence interval for a Poisson distribution $X_{1}, ..., X_{10}$ ~ $Pois(\theta)$
Observations:
$x_{1} = x_{3} = x_{6} = x_{8} = x_{9} = 0$; $x_{2} = x_{5} = x_{10} = 1$; $x_{4} = 2$; $x_{7} = 3$.
I want to determine an exact (numerical) and an approximate (numerical) confidence interval for $\theta$ of confidence level 0.9:

*

*exact confidence interval: I don't know how to do this. On the internet I can only find things about an approximate confidence interval.


*approximate confidence interval:
$X = (X_{1}, ..., X_{10})$
MLE:  $\hat{\theta} = \bar{X} = 0.8$
And: $Var(X) = \theta$. So an estimate for the variance is also $\bar{X} = 0.8$.
Then according to the answer: An approximate confidence interval is $\bar{X}$ $\pm$ $\sqrt{\bar{X}/n}$ $\xi_{1 - \alpha/2}$.
But this means that: T = $\sqrt{n}$ $\frac{\theta - E[X]}{\sqrt{Var(X)}}$ ~ N(0,1).
But this does not hold for X ~ Pois($\theta$), right?
Because a Poisson distribution does not have to be symmetric.

Actually, according to the central limit theorem, it does (in terms of limit distribution). You can check that the poisson distribution checks every requirement :
$X_1,..X_n $ are independant, iid, $\bar{X}, \sigma$ are finite and $\sigma\neq0 $
