Let $B(n,p)$ be a Binomial Distribution. Consider the following estimate of $p$ :
\begin{equation} \hat{p} = \frac{\sum_{i=1}^N X_i}{nN} \end{equation} where $X_i \sim B(n,p)$ are independent samples. We have \begin{equation} \mathrm{Var}[\hat{p}] = \frac{p(1-p)}{n N} \end{equation} It is not hard to see that this estimate for $p$ has optimal rate of convergence.
Question
Consider now the situation where we need to approximate $q = \sqrt{p}$ given samples from $B(n,p)$. Is it possible to get an estimate $\hat{q}$ that achieves $\mathrm{Var}[\hat{q}] = q (1-q)/(n N) $? Is there a better solution than to simply take $\hat{q} = \sqrt{\hat{p}}$ ?
The Wald 95% confidence interval for $p$ is of the form $$\hat p \pm 1.96\sqrt{\hat p(1-\hat p)/n},$$ where $\hat p = X/n,$ the proportion of Successes in $n$ trials.
To get a 95% CI for $q = \sqrt{p}$ take the square root of the endpoints of the CI for $p.$
Especially for small $n$ or for $p$ near to $0$ or $1$ a the more accurate Argesti 95% CI for $p$ is of the form
$$\check p \pm 1.96\sqrt{\check p(1-\check p)/\check n},$$ where $\check n = n+2$ and $\check p = (X + 2)/\check n.$ By 'more accurate' I mean that the true probability of covering $p$ is closer to the claimed 95% than for the Wald interval. So you could get a more accurate 95% CI for $q$ by taking square roots of the endpoints of the Argreti interval.
Note: The same idea is used to get a confidence interval for the population standard deviation $\sigma$ from a confidence interval for the population variance $\sigma^2.$