Note: For $S^2 = \frac{1}{n-1}\sum^n_{i=1}(x_i - \bar x)^2$
If $s^2$ is the variance of a random sample of size $n$ from a normal population, a $100(1-\alpha)\%$ C.I. for $\sigma^2$ is
$$ \frac{(n-1)S^2}{\chi^2_{\alpha / 2}} \lt \sigma^2 \lt \frac{(n-1)S^2}{\chi^2_{1-\alpha / 2}} $$
where $\chi^2_{\alpha / 2}$ and $\chi^2_{1 - \alpha / 2}$ are $\chi^2$-values with $n-1$ df, leaving areas of $\alpha / 2$$1-\alpha / 2$, respectively, to the right.
Given $n = 10$ and all the data
Task: find the $\sigma$ for 95% C.I.