In my lecture notes it is used that if $x_1,...,x_n$ is an i.i.d. sample from a Poisson distribution with parameter $\lambda$ then a confidence interval for $\lambda$ is given as $$\left[ \bar{x}+\frac{q^2}{n}-\frac{q}{\sqrt{n}}\sqrt{\bar{x}-\frac{q^2}{4n}},\bar{x}+\frac{q^2}{n}+\frac{q}{\sqrt{n}}\sqrt{\bar{x}+\frac{q^2}{4n}}\right]$$ where $\bar{x}=\sum_{i=1}^nx_i$ and $q$ is the $1-\alpha/2$ quantile of a standard normal distribution.
So I tried to to show that myself but got stuck. Here is what I got: by CLT for large $n$ we have $$\frac{\bar{x} - \lambda}{\sqrt{\lambda /n}}$$is approximately standard normal. Then using that $P\left(q_{\alpha/2}\le \frac{\bar{x} - \lambda}{\sqrt{\lambda /n}}\le q\right)$ is a approximately $1-\alpha$ and the MLE for $\lambda$ is $\hat{\lambda}=1/\bar{x}$ I get a interval of the form $$\bar{x}\pm q\sqrt{\hat{\lambda}/n}.$$ However I can't seem to arrive at the interval given in the lecture notes and some help would be appreciated.
The point is that we want the interval to be independent of the estimated value.
We can rewrite the event $\left\{q_{\alpha/2}\leqslant \frac{\bar{x} - \lambda}{\sqrt{\lambda /n}}\leqslant q\right\}$ as $\left\{q_{\alpha/2}\sqrt \lambda\leqslant \sqrt n(\bar{x} - \lambda)\leqslant q\sqrt{\lambda}\right\}$.
Solving the inequations $q_{\alpha/2}x\leqslant \sqrt{n}(\overline{x}-x^2)$ and $\sqrt{n}(\overline{x}-x^2)\leqslant qx$ will give that $x$ belongs to some interval; then replace $x$ by $\sqrt \lambda$ to get an interval for $\lambda$.