For each of the following test, outline a research scenario and use any hypothetical data to show how the test is conducted. Paired two-sample t-test

Correlated samples t test or paired t test or repeated measures t test:
The paired t is used when the two sample are paired or related samples. When the sample is measured twice, or paired samples are taken, then the two samples would be dependent. This indicates that paired t test is used for comparing the dependent samples.
Conditions to perform the paired t-test:
Dependent variable is measured on a continuous scale.
Each score in one sample is paired with a particular score in the other sample.
The difference of the either group follows normal distribution.
Example:
Junior school research team would like to examine whether students performance in math is increased after recruiting two high qualified math teachers. Both pre-test and post-test scores of a random sample of 20 students were recorded. The mean difference between post-test and pre-test score was 15.4 with standard deviation 3.4. Does the data provide evidence to conclude that students performance in math is increased. Use 5% level of significance.
Hypotheses and level of significance:
The hypotheses to be tested are:
Null hypothesis:
$H_0:{\mu }_d=0$
Alternative hypothesis:
$H_1:{\mu }_d>0$
Assume the level of significance is given as $\alpha =0.05.$
Paired t-test statistic:
In order to test a hypothesis regarding whether the difference between a paired set of observations $(x_i,y_i)$ is significant or not, the paired t-test is used.
The test statistic for the paired t-test is given as follows:
$t=\frac{\bar{d}}{se(\bar{d})}$ where,
$\bar{d}=\frac{1}{n}\sum m_{i=1}^n(x_i-y_i)$
$se(\bar{d})=$
standard error of $\bar{d}$
$=\frac{standarderrorof\bar{d}}{n}$
$=\sqrt{\frac{\frac{1}{n-1}\sum m_{i=1}^n(d_i-\bar{d}{)}^2}{n}}$
The following table shows the necessary calculations for finding the test statistic:
$se\bar{d}=\sqrt{\frac{\frac{1}{n-1}\sum m_{i=1}^n(d_i-\bar{d}{)}^2}{n}}$
$=\frac{3.4}{\sqrt{20}}$
$\approx 0.7603$
$=20.2552$
$df=n-1$
$=20-1$
$=19$
Thus, $t=20.2552.$
Decision rule:
If the p-value is less than the level of significance then reject the null hypothesis. Otherwise fail to reject the null hypothesis
The p-value is given by:
The p-value $=0(Fromexcel=T.DIST.RT(20.2552,19))$
Since the p-value $(=0)$
less than level of significance $(=0.05)$, decision is to reject the null hypothesis.
There is sufficient evidence to conclude that student’s performance in math is increased.