Suppose I have a population, I know the variance $\sigma^2$, but I don't know the mean $\mu$. How to find the confidence interval for the square of the mean $\mu^2$ ?
It's well known that the CI of the mean $\mu$ is $P(a < \mu < b) = 1-\alpha$ where $a = \bar x- z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}$ and $b = \bar x+ z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}$.
Since $P(a < \mu < b)=P(2a < 2\mu < 2b) = 1-\alpha$, can I just use $P(a^2 < \mu^2 < b^2) = 1-\alpha$ to find the CI of $\mu^2$?
Similarly, if I want to find the CI of $\mu^2+2\mu$, is it correct to do this: $$P(a^2+2a < \mu^2+2\mu < b^2+2b) = 1-\alpha\,\,?$$
And also, what if I'm not sure that the population follows the standard normal distribution? Is there any generalized formula for the CI?
By CLT,
$$ \sqrt{n}\frac {\bar{X} - \mu} {\sigma} \to \mathcal{N}(0, 1) $$
Here the key is you need to recognize that this is a (asymtotic) pivotal quantity: the distribution is independent of the unknown parameter $\mu$, and thus we can compute the quantile $z_{\alpha/2}$, and then invert it to construct the CI.
To construct a similar pivotal quantity involving $\mu^2$, we may consider the sample $X_i^2$, with
$$ E[X^2] = Var[X] + E[X]^2 = \sigma^2 + \mu^2$$
Since $Var[X^2]$ involve the fourth moment, we just use the sample variance $S_{X^2}^2$ to estimate it, as it is the consistent estimator of $Var[X^2]$:
$$ S_{X^2}^2 = \frac {1} {n - 1} \sum_{i=1}^n \left(X_i^2 - \bar{X^2}\right)^2$$
where
$$ \bar{X^2} = \frac {1} {n} \sum_{i=1}^n X_i^2 $$
is the sample mean. Then by Slutsky and CLT,
$$ \sqrt{n} \frac {\bar{X^2} - \sigma^2 - \mu^2} {S_{X^2}} \to \mathcal{N}(0, 1)$$
Hence,
$$ \begin{align} \Pr\left\{-z_{\alpha/2} \leq \sqrt{n} \frac {\bar{X^2} - \sigma^2 - \mu^2} {S_{X^2}} \leq z_{\alpha/2} \right\} \to 1 - \alpha \\ \Rightarrow \Pr\left\{-z_{\alpha/2} \frac {S_{X^2}} {\sqrt{n}} \leq \bar{X^2} - \sigma^2 - \mu^2 \leq z_{\alpha/2} \frac {S_{X^2}} {\sqrt{n}} \right\} \to 1 - \alpha \\ \Rightarrow \Pr\left\{\bar{X^2} - \sigma^2 -z_{\alpha/2} \frac {S_{X^2}} {\sqrt{n}} \leq \mu^2 \leq \bar{X^2} - \sigma^2 + z_{\alpha/2} \frac {S_{X^2}} {\sqrt{n}} \right\} \to 1 - \alpha \\ \end{align} $$
As a result, $$ \left[\bar{X^2} - \sigma^2 - z_{\alpha/2} \frac {S_{X^2}} {\sqrt{n}}, \bar{X^2} - \sigma^2 + z_{\alpha/2} \frac {S_{X^2}} {\sqrt{n}} \right] $$
is a asymptotic $(1 - \alpha)$ confidence interval for $\mu^2$.
Note that it is just one way to construct the CI; there are other possibilities as well.