Answer
$X_0^2\gt X_{1-α}^2$: null hypothesis is not rejected.
There is not enough evidence to conclude that the standard deviation wait-time is less than 18.0 seconds.
Work Step by Step
$x ̅=\frac{108.5+67.4+58.0+75.9+65.1+80.4+95.5+86.3+70.9+72.0}{10}=78$
$s^2=\frac{(108.5-78)^2+(67.4-78)^2+(50.8-78)^2+(75.9-78)^2+(65.1-78)^2+(80.4-78)^2+(95.5-78)^2+(86.3-78)^2+(70.9-78)^2+(72.0-78)^2}{10-1}=231.19$
$H_0:~σ=18.0$ versus $H_1:~σ\lt18.0$
$X_0^2=\frac{(n-1)s^2}{σ_0^2}=\frac{(10-1)231.19}{18.0^2}=6.422$
Left-tailed test:
$n=10$
$d.f.=n-1=9$
$X_{1-α}^2=X_{0.95}^2=3.325$
(According to Table VII, for d.f. = 9 and area to the right of critical value = 0.95)
Since $X_0^2\gt X_{1-α}^2$, we do not reject the null hypothesis.
