Let there be a random sample $X_1,...,X_n \sim Poison(\theta)$, where $\theta>0$ is unknown. Show that $P(\mathbf{X},\theta)=\frac{\bar{X}-\theta}{\sqrt{\bar{X}/n}}$ is asymptotically pivotal, then construct as asymptotic $1-\alpha$ confidence interval for $\theta$. Also, construct an asymptotic $1-\alpha$ confidence interval for $\theta$ by inversion of the acceptance region provided by the score test.
my work:
We know that $\bar{X} \sim AN(\theta,\frac{\theta}{n})$.
$P(\mathbf{X},\theta)=\frac{\sqrt{n}(\bar{X}-\theta)/\sqrt{\theta}}{\sqrt{\bar{X}/\theta}}$, where $\frac{\sqrt{n}(\bar{X}-\theta)}{\sqrt{\theta}} \sim AN(0,1)$ and $\bar{X}/\theta \sim AN(0,\frac{1}{n})$.
However, how can I find the distribution of $P(\mathbf{X},\theta)$ given that I know this. I'm not sure what the asymptotic distribution is of the denominator.
Regarding the score test method, I have the following:
We reject $H_0:\theta=\theta_0$ in favor of $H_1:\theta \ne \theta_0$ when $\frac{S^2(\theta_0)}{ni(\theta_0)}>\chi^2_{1;\alpha}$.
We find $i(\theta)=-E[\frac{\partial}{\partial \theta}(-1+\frac{x}{\theta})]=-E[-\frac{x}{\theta^2}]=\frac{1}{\theta}$, since $E(X)=\theta$.
$S(\theta)=\frac{\partial}{\partial \theta} (-n\theta +ln(\theta)\sum x_i -\sum ln(x_i!))=-n + \frac{\sum x_i}{\theta}$.
Thus, we have $\frac{(-n + \frac{\sum x_i}{\theta_0})^2}{\frac{n}{\theta_0}}=\frac{\theta_0n^2-2n\sum x_i +(\sum x_i)^2/\theta_0}{n}>\chi^2_{1;\alpha}$ as our rejection region.
Our acceptance region is $\theta_0^2n^2-2\theta_0n\sum x_i +(\sum x_i)^2 \le \chi^2_{1;\alpha}$. Solving for $\theta_0$, I get $\theta_0=\frac{\sum x_i}{n}$. Where do I go from here to determine the asymptotic $1-\alpha$ confidence interval for $\theta$?