Answer
$t_0\gt t_α$: null hypothesis is rejected.
There is enough evidence to conclude that chocolate appears to improve teacher evaluations.
Work Step by Step
$x ̅_1,n_1~and~s_1$ refer to chocolate group and $x̅_2,n_2~and~s_2$ refer to nonchocolate group.
$H_0:~µ_1=µ_2$ versus $H_1:~µ_1\gt µ_2$
$t_0=\frac{(x ̅_1-x ̅_2)-(µ_1-µ_2)}{\sqrt {\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}=\frac{(4.2-3.9)-0}{\sqrt {\frac{0.8^2}{50}+\frac{0.8^2}{50}}}=1.875$
$n=50$, so:
$d.f.=n-1=49$
Right-tailed test:
$t_α=t_{0.1}=1.299$
(According to Table VI, for d.f. = 50, the closest value to 49, and area in right tail = 0.1)
Since $t_0\gt t_α$, we reject the null hypothesis.
