Answer
$t_0\lt -t_α$: null hypothesis is rejected.
There is statistical evidence that the scores decreased.
Practically, the decrease is not very significant.
Work Step by Step
$H_0:~µ=73.2$ versus $H_1:~µ\lt73.2$
Requirement:
$n\geq30$
$n=3851$, so:
$d.f.=n-1=3850$
$t_0=\frac{x ̅-µ_0}{\frac{s}{\sqrt n}}=\frac{72.8-73.2}{\frac{12.3}{\sqrt {3851}}}=-2.018$
Left-tailed test:
When the sample size is very large, we can use the approximation $t_α\approx z_α$:
$t_α=t_{0.05}\approx z_{0.05}$
If the area of the standard normal curve to the right of $z_{0.05}$ is 0.05, then the area of the standard normal curve to the left of $z_{0.05}$ is $1−0.05=0.95$
According to Table V, there are 2 z-scores which give the closest value to 0.95: 1.64 and 1.65. So, let's find the mean of these z-scores: $\frac{1.64+1.65}{2}=1.645$
So, $-t_α=-1.645$
Since $t_0\lt -t_α$, we reject the null hypothesis.
