Chi-square Test for the Variance
Let $X_1, X_2, \cdots, X_n$ be a random sample from a normal population with mean $\mu$ and variance $\sigma^2$. Let $\overline{x}=\frac{1}{n} \sum x_i$ be the sample mean and $s^2=\frac{1}{n-1} \sum (x_i-\overline{x})^2$ be the sample variance.
The hypothesis testing problem can be setup as :
| Situation | Hypothesis Testing Problem |
|---|---|
| Situation A : | $H_0: \sigma^2=\sigma^2_0$ against $H_a : \sigma^2 < \sigma^2_0$ (Left-tailed) |
| Situation B : | $H_0: \sigma^2=\sigma^2_0$ against $H_a : \sigma^2 > \sigma^2_0$ (Right-tailed) |
| Situation C : | $H_0: \sigma^2=\sigma^2_0$ against $H_a : \sigma^2 \neq \sigma^2_0$ (Two-tailed) |
Formula
The test statistic under the null hypothesis $H_0$ is
$$ \chi^2_{obs} = \frac{(n-1)s^2}{\sigma^2_0} $$
where,
$s^2 =\dfrac{1}{n-1} \sum (x_i -\overline{x})^2$is the sample variance.
The test statistic follows $\chi^2$ distribution with $n-1$ degrees of freedom.
