Answer
$X_0^2\lt X_{1-α}^2$: reject the null hypothesis.
There is enough evidence to conclude that the variance is less than 95.
Work Step by Step
$H_0:~σ^2=95$ versus $H_1:~σ^2\lt95$
$X_0^2=\frac{(n-1)s^2}{σ_0^2}=\frac{(20-1)49.3}{95}=9.86$
Left-tailed test:
$n=20$
$d.f.=n-1=19$
$X_{1-α}^2=X_{0.90}^2=11.651$
(According to Table VII, for d.f. = 19 and area to the right of critical value = 0.90)
Since $X_0^2\lt X_{1-α}^2$, we reject the null hypothesis.
