Chapter 11, Problem 48DE

a.

To determine

Check whether there is evidence that there is a difference in the mean salary of teams in the American League versus teams in the National League.

Answer to Problem 48DE

The conclusion is that there is no evidence of a difference in the mean salary of teams in the American League versus teams in the National League.

Explanation of Solution

In this context, let ${\mu }_1$ denote the mean salary of teams in the American League and ${\mu }_2$ denote the mean salary of teams in the National League.

The hypotheses are given below:

Null hypothesis:

$H_0:{\mu }_1={\mu }_2$

Alternative hypothesis:

$H_1:{\mu }_1\ne {\mu }_2$

Significance level, $\alpha$:

It is given that the significance level, $\alpha =0.05$.

Degrees of freedom:

The degrees of freedom is as follows:

$\begin{array}{c}df=n_1+n_2-2\\ =14+16-2\\ =28\end{array}$

Step-by-step procedure to obtain the critical values using MINITAB software:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘t’ distribution.
• In Degrees of freedom, enter 28.
• Click the Shaded Area tab.
• Choose Probability and Both Tails for the region of the curve to shade.
• Enter the Probability as 0.05.
• Click OK.

Output obtained using MINITAB software is given below:

From the MINITAB output, the critical values are $\pm 1.983$.

The decision rule is as follows:

If $t<-1.983$, then reject the null hypothesis $H_0$.

If $t\text{>}1.983$, then reject the null hypothesis $H_0$.

If $-1.983<t<1.983$, then fail to reject the null hypothesis $H_0$.

Test statistic:

Step-by-step procedure to obtain the P-value and test statistic using MINITAB software:

• Choose Stat > Basic Statistics > 2 sample t.
• Choose Each sample in its own column.
• In Sample 1, enter the column of American.
• In Sample 2, enter the column of National.
• Choose Assume equal variance.
• Choose Options.
• In Confidence level, enter 95.
• In Alternative, select not equal.
• Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

From the given MINITAB output, the value of the test statistic is –0.94.

Decision:

The value of test statistic lies between the critical values.

From the decision rule, fail to reject the null hypothesis.

Therefore, there is no evidence of a difference in the mean salary of teams in the American League versus teams in the National League.

b.

To determine

Check whether there is evidence that there is a difference in the mean home attendance of team in the American League versus teams in the National League.

Answer to Problem 48DE

The conclusion is that there is no evidence of a difference in the mean home attendance of teams in the American League versus teams in the National League.

Explanation of Solution

In this context, let ${\mu }_1$ denote the mean home attendance of teams in the American League and ${\mu }_2$ denote the mean home attendance of teams in the National League.

The hypotheses are given below:

Null hypothesis:

$H_0:{\mu }_1={\mu }_2$

Alternative hypothesis:

$H_1:{\mu }_1\ne {\mu }_2$

Significance level, $\alpha$:

It is given that the significance level, $\alpha =0.05$.

From the MINITAB output in Part (a), the critical values are $\pm 2.048$.

Test statistic:

Step-by-step procedure to obtain the P-value and test statistic using MINITAB software:

• Choose Stat > Basic Statistics > 2 sample t.
• Choose Each sample in its own column.
• In Sample 1, enter the column of American.
• In Sample 2, enter the column of National.
• Choose Assume equal variance.
• Choose Options.
• In Confidence level, enter 95.
• In Alternative, select not equal.
• Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

From the given MINITAB output, the value of the test statistic is 0.88.

Decision:

The value of test statistic lies between the critical values.

From the decision rule, fail to reject the null hypothesis.

Therefore, there is no evidence of a difference in the mean home attendance of teams in the American League versus teams in the National League.

c.

To determine

Check whether there is evidence of a difference in the mean number of wins for teams in the American League versus teams in the National League.

Answer to Problem 48DE

The conclusion is that there is no evidence of difference in the mean number of wins for teams in the American League versus teams in the National League.

Explanation of Solution

In this context, let ${\mu }_1$ denotes the mean number of wins for teams in the American League and ${\mu }_2$ denotes the mean number of wins for teams in the National League.

The hypotheses are given below:

Null hypothesis:

$H_0:{\mu }_1={\mu }_2$

Alternative hypothesis:

$H_1:{\mu }_1\ne {\mu }_2$

Significance level, $\alpha$:

It is given that the significance level, $\alpha =0.05$.

Degrees of freedom:

The degrees of freedom is as follows:

$\begin{array}{c}df=n_1+n_2-2\\ =10+10-2\\ =18\end{array}$

Step-by-step procedure to obtain the critical values using MINITAB software:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘t’ distribution.
• In Degrees of freedom, enter 18.
• Click the Shaded Area tab.
• Choose Probability and Both Tails for the region of the curve to shade.
• Enter the Probability as 0.05.
• Click OK.

Output obtained using MINITAB software is given below:

From the MINITAB output, the critical values are $\pm 2.101$.

The decision rule is as follows:

If $t<-2.101$, then reject the null hypothesis $H_0$.

If $t\text{>}2.101$, then reject the null hypothesis $H_0$.

If $-2.101<t<2.101$, then fail to reject the null hypothesis $H_0$.

Test statistic:

Step-by-step procedure to obtain the P-value and test statistic using MINITAB software:

• Choose Stat > Basic Statistics > 2 sample t.
• Choose Each sample in its own column.
• In Sample 1, enter the column of American.
• In Sample 2, enter the column of National.
• Choose Assume equal variance.
• Choose Options.
• In Confidence level, enter 95.
• In Alternative, select not equal.
• Click OK in all the dialogue boxes.

Output obtained using MINITAB software is given below:

From the given MINITAB output, the value of the test statistic is –13.

Decision:

The value of test statistic does not lie between the critical values.

That is, $\text{Test statistic}\left(-13\right)<\text{Critical}\text{value}\left(\pm 2.101\right)$

From the decision rule, reject the null hypothesis.

Therefore, there is evidence of a difference in the mean number of wins for teams in the American League versus teams in the National League.