Answer
$X_{1-\frac{α}{2}}^2\lt X_0^2\lt X_{\frac{α}{2}}^2$: null hypothesis is not rejected.
Work Step by Step
$H_0:~σ=0.5$ versus $H_1:~σ\ne0.5$
$X_0^2=\frac{(n-1)s^2}{σ_0^2}=\frac{(18-1)0.62^2}{0.5^2}=26.1392$
Two-tailed test:
$n=18$
$d.f.=n-1=17$
$X_{1-\frac{α}{2}}^2=X_{0.975}^2=7.564$
(According to Table VII, for d.f. = 17 and area to the right of critical value = 0.975)
$X_{\frac{α}{2}}^2=X_{0.025}^2=30.191$
(According to Table VII, for d.f. = 17 and area to the right of critical value = 0.025)
Since $X_{1-\frac{α}{2}}^2\lt X_0^2\lt X_{\frac{α}{2}}^2$, we do not reject the null hypothesis.
