# I am given a sample size of 9 from a normal population, as well as the 95% confidence interval for population mean. I want to calculate the 95% population variance.

Calculating the confidence interval for population variance given the confidence interval for population mean
I am given a sample size of 9 from a normal population, as well as the 95% confidence interval for population mean. I want to calculate the 95% population variance.
I see that the confidence interval for an unknown population mean is $\overline{Y}±{t}_{\frac{\alpha }{2}}\frac{S}{\sqrt{n}}$. Then since the midpoint of the confidence interval for population mean is the sample mean, I can plug that into $\overline{Y}$, 9 into n, and 0.025 into $\alpha$. From there, I can solve for S. How can I find the confidence interval for population variance from there?
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Step 1
A two-sided interval for the variance is
$\left[\frac{\left(n-1\right){S}^{2}}{{\chi }_{1-\alpha /2,n-1}^{2}},\frac{\left(n-1\right){S}^{2}}{{\chi }_{\alpha /2,n-1}^{2}}\right]$
where ${S}^{2}$ is the sample variance, n is the sample size, and ${\chi }_{\alpha /2,n-1}^{2}$ is the $\alpha /2$ quantile of the chi-square distribution with $n-1$ degrees of freedom; i.e., it is the value at which the CDF of a ${\chi }_{n-1}^{2}$ random variable equals $\alpha /2$; similarly, ${\chi }_{1-\alpha /2,n-1}^{2}$ is the $1-\alpha /2$ quantile.
Step 2
For example, with $n=9$ and $\alpha =0.05$, we have
${\chi }_{0.025,8}^{2}\approx 2.17973,\phantom{\rule{0ex}{0ex}}{\chi }_{0.975,8}^{2}\approx 17.5345.$
Notice that because these are in the denominators of the confidence limits, the larger quantile is used for the lower confidence limit, and the smaller quantile is used for the upper confidence limit.