You have a paired design. It is the same $n = 15$ students
taking the test both times. Let's call the first score for the $i$th
subject $X_i$ and the second score $Y_i.$ You want to do a
one-sample z-test of the differences $D_i = X_i - Y_i.$
You don't give the individual scores, but the averages are
$\bar D = \bar X - \bar Y.$
The null hypothesis is $H_O: \mu_D = 0$ (no different after
playing the game) and $H_a: \mu_d > 0$ (better scores after playing
the game).
The test statistic is $$Z = \frac{\bar D - 0}{\sigma/\sqrt{n}} = 1.67.$$ The critical value at the 5% level is the value $c = 1.645$ that cuts 5% from the upper tail of the standard normal curve.
Because $T = 1.67 > c = 1.645,$ you reject the null hypothesis
and conclude that the game might have enabled the students to
get better scores on the second test. (Or maybe learned something
from taking the first test!)
However, $T$ exceeds $c$ by only
a little, and evidence is not 'strong'. If you subject the
findings to a more stringent standard and test at the 1% level,
then the critical new value $c^\prime = 3.326$ that cuts 1% from
the upper tail of the standard normal distribution.
According to this more stringent standard, you do not reject
the null hypothesis.
The P-value is the probability to the right of $Z = 1.67$
under the standard normal curve. That probability is 0.47.
With the p-value, we can test at any desired level of significance.
In particular, at the 5% level, we reject because $.047 < .05 = 5\%$.
However, at the 1% level, we do not reject because $.047 > .01 = 1\%.$
In case it is useful, I pasted output below (somewhat abridged) from doing this
test in Minitab statistical software:
One-Sample Z
Test of mu = 0 vs > 0
The assumed standard deviation = 3.7
N Mean SE Mean Z P
15 1.600 0.955 1.67 0.047
Best Answer
Once you select the type-I error level ($\alpha$) the two approaches are strictly equivalent. This is explained in most introductory stats books.
For a one-sided test like yours: $$ t>t_{n-1,\alpha} \text{ iff } p<\alpha. $$