Suppose we know the sampled variance s2s^2, we want to check if the population variance σ2\sigma^2 is as claimed.

The main test statistic is

χ2=(n−1)s2σ02\chi^2=\frac{(n-1)s^2}{\sigma_0^2}

where nn is the sample size, ss is the sampled standard deviation, σ0\sigma_0 is the proposed population stdev. This χ2\chi^2 should follow χn−12\chi_{n-1}^2 distribution under H0H_0, i.e., the degree of freedom is n−1n-1.

H0H_0 H1H_1 pp-value Decision Rule
σ2=σ02\sigma^2=\sigma_0^2 σ2>σ02\sigma^2\gt \sigma_0^2 P(χn−12>χ2)\mathbb{P}(\chi^2_{n-1}\gt \chi^2) χ2>χn−1,α2\chi^2\gt\chi^2_{n-1,\alpha}

Confidence Interval

This is not symmetric.

[(n−1)s2χn−1,α/22,(n−1)s2χn−1,1−α/22]\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>0.05Nn\gt 0.05N, we adjust the variance

σ^xˉ2=s2nN−nn\hat{\sigma}^2_{\bar{x}}=\frac{s^2}{n}\frac{N-n}{n}

and the 1−α1-\alpha CI is

xˉ±tn−1,α/2σ^xˉ2\bar{x} \plusmn t_{n-1,\alpha/2}\hat{\sigma}^2_{\bar{x}}