Answer
$t_0\gt -t_α$: null hypothesis is not rejected.
There is not enough eveidence to conclude that students are studying less than 480 minutes each week.
Work Step by Step
$x ̅=\frac{504+267+220+322+538+542+428+481+413+302+602}{11}=419.9$
$s=\sqrt {\frac{(504-419.9)^2+(267-419.9)^2+(220-419.9)^2+(322-419.9)^2+(538-419.9)^2+(542-419.9)^2+(428-419.9)^2+(481-419.9)^2+(413-419.9)^2+(302-419.9)^2+(602-419.9)^2}{11-1}}=126.4$
$H_0:~µ=480$ versus $H_1:~µ\lt480$
Requirement:
The population from which the sample is extracted is normally distributed with no outliers.
$n=11$, so:
$d.f.=n-1=10$
$t_0=\frac{x ̅-µ_0}{\frac{s}{\sqrt n}}=\frac{419.9-480}{\frac{126.4}{\sqrt {11}}}=-1.577$
Left-tailed test:
$t_α=t_{0.05}=1.812$
(According to Table VI, for d.f. = 10 and area in right tail = 0.05)
So, $-t_α=-1.812$
Since $t_0\gt -t_α$, we do not reject the null hypothesis.
