91Ó°ÊÓ

We need to explain the paired data.

If the two samples contain the same individuals or if the subjects in one sample are connected to the subjects in the other sample, we must employ paired t methods.
If the subjects in the two samples are fully unrelated, we must employ two sample t methods.
In this scenario, we have the baseball distance thrown and the softball distance thrown for $24$students each.
As a result, the first sample is the baseball distance thrown, whereas the second sample is the softball distance thrown.
We should utilise paired t methods because the two samples contain the identical participants.

We need to explain given boxplot.

Students like them, it is believed, can throw a baseball farther than a softball.
The data in the boxplot reflects the difference in distance between the $24$students' baseball and softball distances.
We can see that the majority of the boxplot is to the right of zero, indicating that the majority of the distances are positive, and so the distance in baseball is higher than in softball.
This supports the allegation, thus there is some evidence to back it up.

We need to state hypotheses for performing a test about the true mean difference.

As a result, the claim that the mean difference is positive is correct.
Now we must determine the most relevant hypotheses for a significance test.
As a result, either the null hypothesis or the alternative hypothesis is the claim.
According to the null hypothesis, the population proportions are equal.
If the claim is the null hypothesis, the alternative hypothesis is the polar opposite of the null hypothesis.
As a result, the following assumptions are appropriate:

$$\begin{gathered} {\mathrm{H}}_0:{\mathrm{μ}}_{\mathrm{D}}=0 \\ {\mathrm{H}}_{\mathrm{a}}:{\mathrm{μ}}_{\mathrm{D}}>0 \end{gathered}$$
Here, ${\mathrm{μ}}_D$is the mean difference in thrown distance between the baseball and the softball.

We need to explain why Normal/Large Sample condition is not met in this case.

The normal or big condition necessitates either a large sample or a distribution of differences that is roughly normal.
The sample isn't big enough because the sample size is only $24$, which isn't even close to $30$.
Because the distribution of the differences is considerably skewed to the right due to the outlier to the right in the boxplot, the distribution is not approximately normal.
This indicates that the condition has not been met.

We need to estimate the P-value to draw conclusion.

As given in the question , $n=24$

Mean : $$\begin{gathered} \overline{x}=\frac{{\displaystyle \underset{i=1}{\overset{n}{∑}}}{x}_i}{n} \\ =\frac{8+32+9+12+14+12+5+0+12+5+7+2+18-3+9+6+2+1+3+1+3+1+3-5}{24} \\ =\frac{157}{24}=6.5417 \end{gathered}$$

P-value is probability of obtaining sample results.

P-value $=\frac{0}{10}=0$

As P-value is less than or equal to significance level then null hypothesis is rejected.

$\mathrm{P}<0.05→\mathrm{Reject}{\mathrm{H}}_0$

Therefore, At the $\mathrm{Î\pm }=0.05$ level , students like these can throw baseball faster than softball.