17.3.5.2 Algorithms (One Sample Test for Variance)

Let $\sigma\,\!$ be the variance of sample $x\,\!$ and $\sigma_0\,\!$ is the hypothetical variance, this function tests the hypotheses:

$H_0:\sigma = \sigma_0\,\!$ vs $H_1:\sigma \ne \sigma_0\,\!$

$H_0:\sigma\le\sigma_0$ vs $H_1:\sigma > \sigma_0\,\!$

$H_0:\sigma\ge\sigma_0$ vs $H_1:\sigma < \sigma_0\,\!$

To compare the variance, we firstly compute the Chi-square value by:

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

where $s^2\,\!$ is the sample variance. For a given significance level, $\alpha\,\!$ , we will reject the null hypothesis $H_0\,\!$ when:

$|x^2| \ne \chi_{\alpha/2}^2\,\!$ , for two tailed test

$x^2>\chi_\alpha^2$ , for upper tailed test

$x^2<\chi_{1-\alpha}^2$ , for lower tailed test

And the confidence interval for the sample variance can be generated by:

Null Hypothesis	Confidence Interval
$H_0:\sigma = \sigma_0\,\!$	$\sqrt{\frac{(n-1)s^2}{\chi_{(\alpha/2,n-1)}^2}} \le \sigma \le \sqrt{\frac{(n-1)s^2}{\chi_{(1-\alpha/2,n-1)}^2}}$
$H_0:\sigma\le\sigma_0$	$\sqrt{\frac{(n-1)s^2}{\chi_{(\alpha,n-1)}^2}} \le \sigma \le \infty$
$H_0:\sigma\ge\sigma_0$	$0 \le \sigma \le \sqrt{\frac{(n-1)s^2}{\chi_{(1-\alpha,n-1)}^2}}$

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