Chapter 12, Problem 3SR

a.

To determine

Obtain the null and the alternative hypotheses.

Explanation of Solution

Denote ${\mu }_1,{\mu }_2,\text{ and }{\mu }_3$ the means of tuition rates for northeast, southeast, and west, respectively.

The null and alternative hypotheses are given below:

Null Hypothesis

$H_0:{\mu }_1={\mu }_2={\mu }_3$

All the means of tuition rates for the various regions are equal.

Alternative Hypothesis

$H_1:$ At least one mean tuition rates in the various regions is different from others.

b.

To determine

Give the decision rule.

Explanation of Solution

The degrees of freedom for the numerator is obtained as given below:

$\begin{array}{c}\text{Degrees}\text{of}\text{freedom = }k-1\\ =3-1\\ =2\end{array}$

The degrees of freedom for the denominator is obtained as given below:

$\begin{array}{c}\text{Degrees}\text{of}\text{freedom = }n-k\\ =14-3\\ =11\end{array}$

The critical F value is as follows:

Here, the test is one-tailed and the level of significance $\left(\alpha \right)$ is 0.05.

Use Table - B.6A Critical Values of the F Distribution (α = .05) to obtained critical F value.

• Locate 2 in the column of Degrees of Freedom for the Numerator
• Locate 11 in the row of Degrees of Freedom for the Denominator
• The intersection of row and column is 3.98

Thus, the critical F value is 3.98.

Decision rule:

If $F>3.98$, then reject the null hypothesis.

Therefore, the decision rule is reject $H_0$ if the computed value of F exceeds 3.98.

c.

To determine

Obtain an ANOVA table.

Find the value of the test statistic.

Answer to Problem 3SR

The ANOVA table is,

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F
Treatments	44.16	2	22.082	25.43
Error	9.55	11	0.868
Total	53.71	13

The value of the test statistics is 25.43.

Explanation of Solution

Here, the level of significance $\left(\alpha \right)$ is 0.05.

Step-by-step procedure to obtain ANOVA table using Excel-MegaStat:

• In EXCEL, Select Add-Ins > Mega Stat > Analysis of Variance.
• Choose One-Factor ANOVA.
• In Input Range, enter the column of Northeast, Southeast and West.
• Choose When p < 0.05.
• Click OK.

Output using the Excel-MegaStat software is given below:

From the output, the ANOVA table is obtained as given below:

Source of Variation	Sum of Squares	Degrees of Freedom	Mean Square	F
Treatments	44.16	2	22.082	25.43
Error	9.55	11	0.868
Total	53.71	13

From the output, the value of the test statistics is 25.43.

d.

To determine

Find the decision regarding the null hypothesis.

Answer to Problem 3SR

The null hypothesis is rejected.

Explanation of Solution

Decision:

The F value is 25.43 and the F critical value is 3.98.

Here, F value is greater than F critical value. That is, 25.43 > 3.98.

Using decision rule, reject the null hypothesis.

The decision is, reject the null hypothesis because the computed F value (25.43) is greater than the F critical value (3.81).

Thus, the null hypothesis $\left(H_0\right)$ is rejected.

Therefore, there is sufficient evidence that there is a difference in the mean tuition rates for the various regions.

e.

To determine

Check if there be a significant difference between the mean tuition in the Northeast and that of the West.

Construct a 95% confidence interval for the difference.

Answer to Problem 3SR

There be a significant difference between the mean tuition in the Northeast and that of the West.

A 95% confidence interval for the difference is 2.9 and 5.5.

Explanation of Solution

A 95% confidence interval is as follows:

$\left(\begin{array}{l}\text{Confidence interval}\text{for}\\ \text{the}\text{difference}\text{in}\text{treatment}\\ \text{means}\end{array}\right)=\left({\overline{x}}_1-{\overline{x}}_3\right)\pm t\sqrt{\text{MSE}\left(\frac{1}{n_1}+\frac{1}{n_3}\right)}$

Where,

${\overline{x}}_1$ is the mean of northeast, ${\overline{x}}_3$ is the mean of west, $n_1$ is the number of observations in the northeast, $n_3$ is the number of observations in the west, t is the critical value with $\left(n-k=11\right)$ degrees of freedom and MSE is mean square term obtained from the ANOVA table.

Step-by-step procedure to obtain ANOVA table using Excel-MegaStat:

• In EXCEL, Select Add-Ins > Mega Stat > Analysis of Variance.
• Choose One-Factor ANOVA.
• In Input Range, enter the column of Northeast, Southeast and West.
• In Post-Hoc Analysis, Choose When p < 0.05.
• Click OK.

Output using the Excel-MegaStat software is given below:

From the output, mean of northwest is 41, mean of west is 36.8, and MSE is 0.868.

The value of t is as follows:

Step-by-step procedure to obtain ANOVA table using Excel-MegaStat:

• In EXCEL, Select Add-Ins > MegaStat > Probability.
• Choose Continuous probability distributions.
• Select t-distribution and select calculate t given probability and enter probability as 0.05.
• Enter df as 11.
• Click Ok.

Output using the Excel-MegaStat software is given below:

From the output, the t is $\pm 2.20$1. Now the 95% confidence interval is,

$\begin{array}{c}\left(\begin{array}{l}\text{Confidence interval}\text{for}\\ \text{the}\text{difference}\text{in}\text{treatment}\\ \text{means}\end{array}\right)=\left({\overline{x}}_1-{\overline{x}}_3\right)\pm t\sqrt{\text{MSE}\left(\frac{1}{n_1}+\frac{1}{n_3}\right)}\\ =\left(41-36.8\right)\pm 2.201\sqrt{0.868\left(\frac{1}{5}+\frac{1}{5}\right)}\\ =4.2\pm 1.30\\ =\left(4.2-1.30,4.2+1.30\right)\\ =\left(2.9,5.5\right)\end{array}$

Therefore, a 95% confidence interval for the difference is between 2.9 and 5.5. Both the end points are positive and does not include zero. Thus, there is a significant difference between the mean tuition rates in the Northeast and that of the West.

Thus, the 95% confidence interval for the difference is 2.9 and 5.5.