Question

Have a ball! Can students throw a baseball farther than a softball? To find out, researchers conducted a study involving $24$randomly selected students from a large high school. After warming up, each student threw a baseball as far as he or she could and threw a softball as far as he she could, in a random order. The distance in yards for each throw was recorded. Here are the data, along with the difference (Baseball – Softball) in distance thrown, for each student:

a. Explain why these are paired data.

b. A boxplot of the differences is shown. Explain how the graph gives some evidence that students like these can throw a baseball farther than a softball.

c. State appropriate hypotheses for performing a test about the true mean difference. Be sure to define any parameter(s) you use.

d. Explain why the Normal/Large Sample condition is not met in this case. The mean difference (Baseball−Softball) in distance thrown for these $24$students is $\overline{x}$diff = $6.54$yards. Is this a surprisingly large result if the null hypothesis is true? To find out, we can perform a simulation assuming that students have the same ability to throw a baseball and a softball. For each student, write the two distances thrown on different note cards. Shuffle the two cards and designate one distance to baseball and one distance to softball. Then subtract the two distances (Baseball−Softball) . Do this for all the students and find the simulated mean difference. Repeat many times. Here are the results of $100$trials of this simulation

e. Use the results of the simulation to estimate the P-value. What conclusion would you draw ?

Short answer

Part(a) These are to utilise paired t methods because the two samples contain the identical participants.

Part(b) 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.

Part(c) The appropriate hypotheses is :

$$\begin{gathered} {\mathrm{H}}_0:{\mathrm{\mu }}_{\mathrm{D}}=0 \\ {\mathrm{H}}_{\mathrm{a}}:{\mathrm{\mu }}_{\mathrm{D}}>0 \end{gathered}$$

Part(d) 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.

Part(e) There is convincing evidence that student like these can throw baseball farther than soft ball.

Step-by-step solution

Part(a) Step 1 : Given information

We need to explain the paired data.

Part(a) Step 2 : Simplify

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.

Part(b) Step 1 : Given information

We need to explain given boxplot.

Part(b) Step 2 : Simplify

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.

Part(c) Step 1 : Given information

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

Part(c) Step 2 : Simplify

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:

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

Part(d) Step 1 : Given information

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

Part(d) Step 2 : Simplify

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.

Part(e) Step 1 : Given information

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

Part(e) Step 2 : Simplify

As given in the question , $n=24$

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

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.

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

Therefore, At the $\alpha =0.05$ level , students like these can throw baseball faster than softball.