Suppose we have a sample $X_1, X_2, \ldots, X_n \stackrel{\text{iid}}{\sim} Binomial(\theta)$, where $n$ is known to be large. I would like to calculate the 95% confidence interval for $\theta$, and I am given that the sample mean is some constant $k$.
Notably, I am not given the sample variance. How do I calculate the confidence interval?
My current approach is to note that the confidence interval is of the form $\left(\bar{x} - t_{\frac{1 + \gamma}{2}}^{(n-1)} \cdot \frac{s}{\sqrt{n}}, \bar{x} + t_{\frac{1 + \gamma}{2}}^{(n-1)} \cdot \frac{s}{\sqrt{n}}\right)$. Now, since $n$ is large, we replace $s$ with an unbiased estimator; in this case, $\sqrt{\bar{x} ( 1 - \bar{x})}$. So finally, we get that the confidence interval is
$$\left(k - t_{\frac{1 + \gamma}{2}}^{(n-1)} \cdot \frac{\sqrt{k ( 1 - k)}}{\sqrt{n}}, k + t_{\frac{1 + \gamma}{2}}^{(n-1)} \cdot \frac{\sqrt{k ( 1 - k)}}{\sqrt{n}}\right),$$
which seems to have gotten us a confidence interval purely in terms of $n$ and $k$.
Is this correct?
