Chapter 10.3, Problem 93E

(a)

To determine

To explain why these are paired data.

Explanation of Solution

We need to use paired t procedures if the two samples contain the same subjects or if the subjects in one sample are related to the subjects in the other sample.

We need to use two sample t procedures if the subjects in the two sample are completely unrelated.

In this case, we have the distance thrown in baseball and the distance thrown is softball for $24$ students each. The first sample is thus the distance thrown in baseball and the second sample is distance thrown in softball.

Since the two sample contains the same subjects then we should use a paired t procedures.

(b)

To determine

To explain how the graph give some evidence that students like these can throw a baseball farther than the softball.

Explanation of Solution

It is claimed that: Students like these can throw a baseball further than a softball.

The data in the boxplot represents the difference between the distance in the baseball and the distance in softball for the $24$ students.

We note that most of the boxplot lies to the right of zero, which indicates that most of the distances are positive and thus the distance in the Baseball tends to be higher than the softball. This agrees with the claim and thus there is some evidence for the claim.

(c)

To determine

To state an appropriate hypothesis for performing a test about the true mean difference.

Answer to Problem 93E

The appropriate hypotheses for this is:

  $\begin{array}{l}{H}_0:{\mu }_D=0\\ H_a:{\mu }_D>0\end{array}$

Explanation of Solution

So, the given claim that: mean difference is positive.

Now, we have to find out the appropriate hypotheses for performing a significance test.

Thus, the claim is either the null hypothesis or the alternative hypothesis. The null hypothesis states that the population proportions are equal. If the null hypothesis is the claim then the alternative hypothesis states the opposite of the null hypothesis.

Therefore, the appropriate hypotheses for this is:

  $\begin{array}{l}{H}_0:{\mu }_D=0\\ H_a:{\mu }_D>0\end{array}$

Where we have,

  ${\mu }_D=$ themean difference between the baseball and the softball in the distance thrown.

(d)

To determine

To explain why the large/normal condition is not met in this case.

Explanation of Solution

The normal/large condition requires that either the sample is large or the distribution of the differences is approximately normal.

The sample is not large because the sample size is $24$ which is not at least $30$ .

The distribution of the differences is not approximately normal because the distribution is strongly skewed to the right due to the outlier to the right in the boxplot.

This then implies that the condition is then not met.

(e)

To determine

To use the result of simulation to estimate the P -value and what conclusion would you draw.

Answer to Problem 93E

There is convincing evidence at the $\alpha =0.05$ level that students like these can throw a baseball farther than the softball.

Explanation of Solution

The differences is given in the question then we note that:

  $n=24$

The mean is the calculated as:

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

As we know that the P -value is the probability of obtaining the sample results or results more extreme when the null hypothesis is true.

We note that $0\text{ of the 100}$ dots in the dot plot lie to the right of $6.5417$ , then,

  $\begin{array}{l}P\text{-value}=\frac{0}{100}\\ =0\end{array}$

And we know that if the P -value is less than or equal to the significance level then the null hypothesis is rejected, then,

  $P<0.05\Rightarrow \text{Reject }{H}_0$

Thus, we conclude that there is convincing evidence at the $\alpha =0.05$ level that students like these can throw a baseball farther than the softball.