Answer
$X_0^2\lt X_{1-α}^2$: null hypothesis is rejected.
There is enough evidence to conclude that the variability in wait time is less for a single line than for multiple lines.
Work Step by Step
$H_0:~σ=1.2$ versus $H_1:~σ\lt1.2$
$X_0^2=\frac{(n-1)s^2}{σ_0^2}=\frac{(50-1)0.84^2}{1.2^2}=24.01$
Left-tailed test:
$n=50$
$d.f.=n-1=49$
$X_{1-α}^2=X_{0.95}^2=34.764$
(According to Table VII, for d.f. = 50, the closest value to 49, and area to the right of critical value = 0.95)
Since $X_0^2\lt X_{1-α}^2$, we reject the null hypothesis.
