Answer
$F_0\gt F_{1-α,n_1-1,n_2-1}$: null hypothesis is rejected.
There is enough evidence to conclude that the variability in wait time in the single line is less than that for the multiple lines.
Work Step by Step
$s_1,n_1~and~d.f._1$ refer to single line and $s_2,n_2~and~d.f._2$ refer to multiple lines.
$H_0:~σ_1=σ_2$ versus $H_1:σ_1\ltσ_2$
$s_1=\sqrt {\frac{∑(x_{1_i}-x ̅_1)^2}{n_1-1}}=0.574$
$s_2=\sqrt {\frac{∑(x_{2_i}-x ̅_2)^2}{n_2-1}}=1.012$
$F_0=\frac{s_1^2}{s_2^2}=\frac{0.574^2}{1.012^2}=0.32$
$d.f_1=n_1-1=20-1=19$
$d.f_2=n_2-1=20-1=19$
Left-tailed test:
$F_{α,n_1-1,n_2-1}=F_{0.05,19,19}=2.12$
(According to table VIII, for $d.f._1=20$, the closest value to 19, $d.f._2=20$, the closest value to 19, and area in the right tail = 0.05)
$F_{1-α,n_1-1,n_2-1}=F_{0.95,19,19}=\frac{1}{F_{0.05,19,19}}=\frac{1}{2.12}=0.47$
Since $F_0\lt F_{1-α,n_1-1,n_2-1}$, we reject the null hypothesis.
