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)


The test statistic under the null hypothesis $H_0$ is

$$ \chi^2_{obs} = \frac{(n-1)s^2}{\sigma^2_0} $$


  • $s^2 =\dfrac{1}{n-1} \sum (x_i -\overline{x})^2$ is the sample variance.

The test statistic follows $\chi^2$ distribution with $n-1$ degrees of freedom.

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