I have the following problem for which I do not know if my solution is correct or not.
Let $X_1 , ... , X_n \sim Bernoulli(p)$.
Let $logit(p) = \frac{p}{1-p}$, let $S_n = \sum_{j=1}^n X_j$.
I want to find of the limit of logit($\frac{S_n}{n}$) and its confidence interval of level 95% I call $\frac{S_n}{n} = \overline{S_n}$ This is what I have done:
Using the TLC, I can write: $$ \sqrt{n} ( \overline{S_n} - p) \to \mathcal{N}(0,p(1-p)) $$
So using delta method I find that:
$$ \sqrt{n} ( logit(\overline{S_n}) - log(\frac{p}{1-p})) \to \mathcal{N}(0,p(1-p)[\frac{d}{dp}logit(p)]^2) $$
$$ \sqrt{n} ( logit(\overline{S_n}) - \log{\frac{1}{1-p}}) \to \mathcal{N}(0,\frac{1}{p(1-p)}) $$
I now assume that I have n = 500 and $S_{500} = 380$
I know that confidence interval are of the form:
$$ \overline{X} \pm Z_{\alpha/2} \sqrt{\frac{\sigma^2}{n}} $$
In my case, I have that $\overline{X} = logit(\frac{380}{500})$, $n=500$ and $\sigma^2 = \frac{1}{p(1-p)}$ with $p = \frac{380}{500}$.
Can someone check if this is right?