One Normal Variance

Suppose we know the sampled variance $s^2$ , we want to check if the population variance $\sigma^2$ is as claimed.

The main test statistic is

\chi^2=\frac{(n-1)s^2}{\sigma_0^2}

where $n$ is the sample size, $s$ is the sampled standard deviation, $\sigma_0$ is the proposed population stdev. This $\chi^2$ should follow $\chi_{n-1}^2$ distribution under $H_0$ , i.e., the degree of freedom is $n-1$ .

$H_0$	$H_1$	$p$ -value	Decision Rule
$\sigma^2=\sigma_0^2$	$\sigma^2\gt \sigma_0^2$	$\mathbb{P}(\chi^2_{n-1}\gt \chi^2)$	$\chi^2\gt\chi^2_{n-1,\alpha}$

Hypotheses	p-value	Decision Rule (Reject $H_0$ if…)
$H_0: \sigma^2 \leq \sigma_0^2$ vs $H_1: \sigma^2 > \sigma_0^2$	$P(\chi^2_{n-1} > \chi^2)$	$\chi^2 > \chi^2_{n-1,\alpha}$
$H_0: \sigma^2 \geq \sigma_0^2$ vs $H_1: \sigma^2 < \sigma_0^2$	$P(\chi^2_{n-1} < \chi^2)$	$\chi^2 < \chi^2_{n-1,1-\alpha}$
$H_0: \sigma^2 = \sigma_0^2$ vs $H_1: \sigma^2 \ne \sigma_0^2$	$2 \times \min\{P(\chi^2_{n-1} > \chi^2),\ P(\chi^2_{n-1} < \chi^2)\}$	$\chi^2 > \chi^2_{n-1,\alpha/2}$ or $\chi^2 < \chi^2_{n-1,1-\alpha/2}$

Confidence Interval

This is not symmetric.

\Big[ \frac{(n-1)s^2}{\chi^2_{n-1,\alpha/2}}, \frac{(n-1)s^2}{\chi^2_{n-1,1-\alpha/2}} \Big]

Finite Population Correction

If $n\gt 0.05N$ , we adjust the variance

\hat{\sigma}^2_{\bar{x}}=\frac{s^2}{n}\frac{N-n}{n}

and the $1-\alpha$ CI is

\bar{x} \plusmn t_{n-1,\alpha/2}\hat{\sigma}^2_{\bar{x}}