The university conducted a study to observe the efficiency of students to see whether students do better when they study all at once or in intervals.

The objective is to check whether the mean test scores of the two groups significantly differ at the 0.05 level.
Let ${\mu }_1$ denote the mean of the participants participated in the survey after studying for one hour continuously.
Let ${\mu }_2$ denote the mean of the participants participated in the survey after studying for three twenty-minute sessions.
The null hypothesis is that there is no difference between the mean test of scores between the two groups. That is,
$H_0:{\mu }_1=m{u}_2$
The alternative hypothesis is that there is a difference between the mean test of the score between the two groups. That is,
$H_1:{\mu }_1\ne {\mu }_2$
The level of significance is,
$\alpha =0.05$
Group 1:
Sample mean, ${\overline{x}}_1=75$
Sample variance, $S^{\frac{2}{1}}=120$
Sample Size, $n1=12$
Group 2:
Sample mean, ${\overline{x}}_2=86$
Sample variance, $S^{\frac{2}{2}}=100$
Sample Size, $n2=12$
$t=\frac{({\overline{x}}_1-{\overline{x}}_2)-({\mu }_1-{\mu }_2)}{\sqrt{S^{\frac{2}{p}}(\frac{1}{n}1+\frac{1}{n}2)}}$
The formula for pooled variance is,
$S^{\frac{2}{p}}=\frac{(n1-1)S^{\frac{2}{1}}+(n2-1)S^{\frac{2}{2}}}{n1+n2-2}$
$=\frac{(12-1)120+(12-1)100}{12+12-2}$
$=\frac{11(120)+11(100)}{24-2}$
$=\frac{1320+1100}{22}$
$=\frac{2420}{22}$
$=110$
The test statistic is computed below.
$t=\frac{({\overline{x}}_1-{\overline{x}}_2)-({\mu }_1-{\mu }_2)}{\sqrt{S^{\frac{2}{p}}(\frac{1}{n}1+\frac{1}{n}2)}}$
$=\frac{75-86-0}{\sqrt{110(\frac{1}{12}+\frac{1}{12})}}$
$=\frac{-11}{\sqrt{110(0.08333+0.08333)}}$
$=\frac{-11}{\sqrt{110\left(166667\right)}}$
$=\frac{-11}{\sqrt{18.33333}}$
$=\frac{-11}{4.281744}$
$=-2.569047$
The degrees of freedom is,
$df=n1+n2-2=12+12-2=24-2=22$
Computation of p value:
from the alternative hypothesis,it is clear that the test is a two- tailed test.
t represents the test statistic and df is the degrees of freedom p value
$=tdist(t,dftails)$
$=tdist(2.569047,22,2)$
$-0.017503$
Decision rule:
Reject the null hypothesis if p value is less than the level of significance.
That is, p value $<0.05$
Conclusion:
Here the p value is 0.017503 is less than 0.05. So, reject the null hypothesis and conclude that there is a significant difference between the means of the two groups.