Answer
$X^2\gt X_α^2$: the null hypothesis is rejected.
There is sufficient evidence to conclude that the likelihood of a fatality involving a pedestrian is not equal to every day of the week.
Work Step by Step
$H_0:$ the likelihood of a fatality involving a pedestrian is equal to every day of the week.
That is, $P(Sunday)=P(Monday)=P(Tuesday)=...=\frac{1}{7}$
$H_1:$ the likelihood of a fatality involving a pedestrian is not equal to every day of the week.
Total:300 pedestrians
Expected count of Sunday: $300\times\frac{1}{7}=42.857$
Expected count of Monday: $300\times\frac{1}{7}=42.857$
...
Expected count of Saturday: $300\times\frac{1}{7}=42.857$
$X^2=Σ\frac{(O_i-E_i)^2}{E_1}=\frac{(39-42.857)^2}{42.857}+\frac{(40-42.857)^2}{42.857}+\frac{(30-42.857)^2}{42.857}+\frac{(40-42.857)^2}{42.857}+\frac{(41-42.857)^2}{42.857}+\frac{(49-42.857)^2}{42.857}+\frac{(61-42.857)^2}{42.857}=13.227$
$k=7$. So, $d.f.=7-1=6$
$X_α^2=X_{0.05}^2=12.592$
(According to Table VII, for d.f. = 6 and area to the right of critical value = 0.05)
Since $X^2\gt X_α^2$, we reject the null hypothesis.
