Answer
$1.54\leq \sigma^2\leq 5.31$
Work Step by Step
Given $n=22, s^2=2.6, c=0.95$
we use the $\chi^2$ table with $df=21, \alpha/2=0.025, 0.975$
to find $\chi^2_{right}=35.479$ and $\chi^2_{left}=10.283$
The confidence interval can be found as $\frac{21\times2.6}{35.479}$ to $\frac{21\times2.6}{10.283}$
The 95% confidence interval of the true variance is $1.54\leq \sigma^2\leq 5.31$
