这是一份warwick华威大学ST230-10的成功案例

If the value of $\mu$ is lnown, then
$$
\left[\frac{\sum_{i=1}^{n}\left(X_{i}-\mu\right)^{2}}{\chi_{n}^{2}(\alpha / 2)}, \frac{\sum_{i=1}^{n}\left(X_{i}-\mu\right)^{2}}{\chi_{n}^{2}(1-\alpha / 2)}\right]
$$
is a $100(1-\alpha) \%$ confidence interval for $\sigma^{2}$, where $\chi_{n}^{2}(p)$ is the $100(1-p) \%$ point of the chi-square distribution with $n$ degrees of freedom.

If the value of $\mu$ is estimated from the data, then Theorem $6.1$ can be used to demonstrate that
$$
\left[\frac{\sum_{i=1}^{n}\left(X_{i}-\bar{X}\right)^{2}}{\chi_{n-1}^{2}(\alpha / 2)}, \frac{\sum_{i=1}^{n}\left(X_{i}-\bar{X}\right)^{2}}{\chi_{n-1}^{2}(1-\alpha / 2)}\right]
$$
is a $100(1-\alpha) \%$ confidence interval for $\sigma^{2}$, where $\chi_{n-1}^{2}(p)$ is the $100(1-p) \%$ point of the chi-square distribution with $(n-1)$ degrees of freedom.

ST230-10 COURSE NOTES ：

If $X_{1}, X_{2}, \ldots, X_{n}$ is a random sample from a distribution with parameter $\theta$, and $\mu_{k}=E\left(X^{k}\right)$ is a function of $\theta$ for some $k$, then a method of moments estimator (or MOM estimator) of $\theta$ is obtained using the following procedure:
Solve $\mu_{k}=\widehat{\mu_{k}}$ for the parameter $\theta$.
For example, let $X$ be a uniform random variable on the interval $[0, b]$, and assume that $b>0$ is unknown. Since $E(X)=b / 2$, a MOM estimator is obtained as follows:
$$
\mu_{1}=\widehat{\mu_{1}} \Longrightarrow E(X)=\frac{1}{n} \sum_{i=1}^{n} X_{i} \Longrightarrow \frac{b}{2}=\bar{X} \Longrightarrow \widehat{b}=2 \bar{X}
$$