Answer
Confidence interval: $13.25\lt σ^2\lt56.98$
Work Step by Step
We want to estimate the population variance using a sample obtained from a population that is normally distributed.
$n=12$. So:
$d.f.=n-1=11$
$level~of~confidence=(1-α).100$%
$90$% $=(1-α).100$%
$0.90=1-α$
$α=0.1$
$X_{1-\frac{α}{2}}^2=X_{0.95}^2=4.575$
(According to Table VII, for d.f. = 11 and area to the right of critical value = 0.95)
$X_{\frac{α}{2}}^2=X_{0.05}^2=19.675$
(According to Table VII, for d.f. = 11 and area to the right of critical value = 0.05)
$Lower~bound=\frac{(n-1)s^2}{X_{\frac{α}{2}}^2}=\frac{11\times23.7}{19.675}=13.25$
$Upper~bound=\frac{(n-1)s^2}{X_{1-\frac{α}{2}}^2}=\frac{11\times23.7}{4.575}=56.98$
