Chapter 12.2, Problem 8E

For Exercises 3 through 8, the null hypothesis was rejected. Use the Scheffe test when sample sizes are unequal or the Tukey test when sample sizes are equal, to test the differences between the pairs of means. Assume all variables are normally distributed, samples are independent, and the population variances are equal.

8. Exercise 20 in Section 12-1.

20. Average Debt of College Graduates Kiplinger’s listed the top 100 public colleges based on many factors. From that list, here is the average debt at graduation for various schools in four selected states. At a = 0.05, can it be concluded that the average debt at graduation differs for these four states?

Source: www.Kiplinger.com

To determine

To test: The difference between the means.

Answer to Problem 8E

There is significant difference between the means ${\overline{X}}_1$ and ${\overline{X}}_4$, ${\overline{X}}_2$ and ${\overline{X}}_4$, and ${\overline{X}}_3$ and ${\overline{X}}_4$.

Explanation of Solution

Given info:

The table shows the average debt at graduation for various schools in four selected states. The level of significance is 0.05.

Calculation:

Consider, ${\overline{X}}_1$, ${\overline{X}}_2$, ${\overline{X}}_3$ and ${\overline{X}}_4$ represents the means of New York, Virginia, California and Pennsylvania, $s_1^2,s_2^2\text{, }{s}_3^2\text{ and }{s}_4^2$ represents the variances of samples of New York, Virginia, California and Pennsylvania.

Step-by-step procedure to obtain the test mean and standard deviation using the MINITAB software:

• Choose Stat > Basic Statistics > Display Descriptive Statistics.
• In Variables enter the columns New York, Virginia, California and Pennsylvania.
• Choose option statistics, and select Mean, Variance and N total.
• Click OK.

Output using the MINITAB software is given below:

The sample sizes $n_1,n_2\text{,}{n}_3\text{ and }{n}_4$ are 5, 5, 5 and 5.

The means are ${\overline{X}}_1,{\overline{X}}_2\text{ and }{\overline{X}}_3$ are 14395, 14668, 14275 and 18327.

The sample variances are $s_1^2,s_2^2\text{, }{s}_3^2\text{ and }{s}_4^2$ are 1571071, 5605207, 674050 and 5834042.

Here, the samples of sizes of four states are equal. So, the test used here is Tukey test.

Tukey test:

Critical value:

Here, k is 4 and degrees of freedom $v=N-k$,

Where,

$\begin{array}{c}N=n_1+n_2+n_3+n_4\\ =5+5+5+5\\ =20\end{array}$

Substitute 20 for N and 4 for k in v

$\begin{array}{c}v=N-k\\ =20-4\\ =16\end{array}$

The critical F-value is obtained using the Table N: Critical Values for the Tukey test with the level of significance $\alpha =0.05$.

Procedure:

• Locate 16 in the column of v of the Table H.
• Obtain the value in the corresponding row below 4.

That is, the critical value is 4.05.

Comparison of the means:

The formula for finding $s_W^2$ is,

$s_W^2=\frac{{\displaystyle \sum \left(n_i-1\right)s_i^2}}{{\displaystyle \sum \left(n_i-1\right)}}$

That is,

$\begin{array}{c}{s}_W^2=\frac{(5-1)1,571,071+(5-1)5,605,207+(5-1)674,050+\left(5-1\right)5,834,042}{\left(5-1\right)+\left(5-1\right)+\left(5-1\right)+\left(5-1\right)}\\ =\frac{6,284,284+22,420,828+2,696,200+23,336,168}{16}\\ =\frac{54,737,480}{16}\\ =3,421,093\end{array}$

Comparison between the means ${\overline{X}}_1$ and ${\overline{X}}_2$:

The hypotheses are given below:

Null hypothesis:

$H_0:$ There is no significant difference between ${\overline{X}}_1$ and ${\overline{X}}_2$.

Alternative hypothesis:

$H_1:$ There is significant difference between ${\overline{X}}_1$ and ${\overline{X}}_2$.

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

Absolute value:

The formula for comparing the means ${\overline{X}}_1$ and ${\overline{X}}_2$ is,

$q_1=\frac{{\overline{X}}_1-{\overline{X}}_2}{\sqrt{\frac{s_W^2}{n}}}$

Substitute 14,395, 14,668 for ${\overline{X}}_1$ and ${\overline{X}}_2$ and 3,421,093 for $s_W^2$

$\begin{array}{c}{q}_1=\frac{14,395-14,668}{\sqrt{\frac{3,421,093}{5}}}\\ =-\frac{273}{\sqrt{684,218.6}}\\ =-\frac{273}{827.1751}\\ =-0.33\end{array}$

Thus, the value of $q_1$ is –0.33.

Hence, the absolute value of $q_1$ is 0.33.

Conclusion:

The absolute value is 0.33.

Here, the absolute value is lesser than the critical value.

That is, $0.33<4.05$.

Thus, the null hypothesis is not rejected.

Hence, there is no significant difference between the means ${\overline{X}}_1$ and ${\overline{X}}_2$.

Comparison between the means ${\overline{X}}_1$ and ${\overline{X}}_3$:

The hypotheses are given below:

Null hypothesis:

$H_0:$ There is no significant difference between ${\overline{X}}_1$ and ${\overline{X}}_3$.

Alternative hypothesis:

$H_1:$ There is significant difference between ${\overline{X}}_1$ and ${\overline{X}}_3$.

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

Absolute value:

The formula for comparing the means ${\overline{X}}_1$ and ${\overline{X}}_3$ is,

$q_2=\frac{{\overline{X}}_1-{\overline{X}}_3}{\sqrt{\frac{s_W^2}{n}}}$

Substitute 14,395, 14,275 for ${\overline{X}}_1$ and ${\overline{X}}_3$ and 3,421,093 for $s_W^2$

$\begin{array}{c}{q}_2=\frac{14,395-14,275}{\sqrt{\frac{3,421,093}{5}}}\\ =\frac{120}{\sqrt{684,218.6}}\\ =\frac{120}{827.1751}\\ =0.15\end{array}$

Thus, the value of $q_2$ is 0.15.

Hence, the absolute value of $q_2$ is 0.15.

Conclusion:

The absolute value is 0.15.

Here, the absolute value is lesser than the critical value.

That is, $0.15<4.05$.

Thus, the null hypothesis is not rejected.

Hence, there is no significant difference between the means ${\overline{X}}_1$ and ${\overline{X}}_3$.

Comparison between the means ${\overline{X}}_1$ and ${\overline{X}}_4$:

The hypotheses are given below:

Null hypothesis:

$H_0:$ There is no significant difference between ${\overline{X}}_1$ and ${\overline{X}}_4$.

Alternative hypothesis:

$H_1:$ There is significant difference between ${\overline{X}}_1$ and ${\overline{X}}_4$.

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

Absolute value:

The formula for comparing the means ${\overline{X}}_1$ and ${\overline{X}}_4$ is,

$q_3=\frac{{\overline{X}}_1-{\overline{X}}_4}{\sqrt{\frac{s_W^2}{n}}}$

Substitute 14,395, 18,327 for ${\overline{X}}_1$ and ${\overline{X}}_4$ and 3,421,093 for $s_W^2$

$\begin{array}{c}{q}_3=\frac{14,395-18,327}{\sqrt{\frac{3,421,093}{5}}}\\ =-\frac{3932}{\sqrt{684,218.6}}\\ =-\frac{3932}{827.1751}\\ =-4.75\end{array}$

Thus, the value of $q_3$ is –4.75.

Hence, the absolute value of $q_3$ is 4.75.

Conclusion:

The absolute value is 4.75.

Here, the absolute value is greater than the critical value.

That is, $4.75>4.05$.

Thus, the null hypothesis is rejected.

Hence, there is significant difference between the means ${\overline{X}}_1$ and ${\overline{X}}_4$.

Comparison between the means ${\overline{X}}_2$ and ${\overline{X}}_3$:

The hypotheses are given below:

Null hypothesis:

$H_0:$ There is no significant difference between ${\overline{X}}_2$ and ${\overline{X}}_3$.

Alternative hypothesis:

$H_1:$ There is significant difference between ${\overline{X}}_2$ and ${\overline{X}}_3$.

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

Absolute value:

The formula for comparing the means ${\overline{X}}_2$ and ${\overline{X}}_3$ is,

$q_4=\frac{{\overline{X}}_2-{\overline{X}}_3}{\sqrt{\frac{s_W^2}{n}}}$

Substitute 14,668 and 14,275 for ${\overline{X}}_2$ and ${\overline{X}}_3$ and 3,421,093 for $s_W^2$

$\begin{array}{c}{q}_4=\frac{14,668-14,275}{\sqrt{\frac{3,421,093}{5}}}\\ =\frac{393}{\sqrt{684,218.6}}\\ =\frac{393}{827.1751}\\ =0.48\end{array}$

Thus, the value of $q_4$ is 0.48.

Hence, the absolute value of $q_4$ is 0.48.

Conclusion:

The absolute value is 0.48.

Here, the absolute value is lesser than the critical value.

That is, $0.48<4.05$.

Thus, the null hypothesis is rejected.

Hence, there is no significant difference between the means ${\overline{X}}_2$ and ${\overline{X}}_3$.

Comparison between the means ${\overline{X}}_2$ and ${\overline{X}}_4$:

The hypotheses are given below:

Null hypothesis:

$H_0:$ There is no significant difference between ${\overline{X}}_2$ and ${\overline{X}}_4$.

Alternative hypothesis:

$H_1:$ There is significant difference between ${\overline{X}}_2$ and ${\overline{X}}_4$.

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

Absolute value:

The formula for comparing the means ${\overline{X}}_2$ and ${\overline{X}}_4$ is,

$q_5=\frac{{\overline{X}}_2-{\overline{X}}_4}{\sqrt{\frac{s_W^2}{n}}}$

Substitute 14,668 and 18,327 for ${\overline{X}}_2$ and ${\overline{X}}_4$ and 3,421,093 for $s_W^2$

$\begin{array}{c}{q}_5=\frac{14,668-18,327}{\sqrt{\frac{3,421,093}{5}}}\\ =-\frac{3659}{\sqrt{684,218.6}}\\ =-\frac{3659}{827.1751}\\ =-4.42\end{array}$

Thus, the value of $q_5$ is –4.42.

Hence, the absolute value of $q_5$ is 4.42.

Conclusion:

The absolute value is 4.42.

Here, the absolute value is greater than the critical value.

That is, $4.42>4.05$.

Thus, the null hypothesis is rejected.

Hence, there is significant difference between the means ${\overline{X}}_2$ and ${\overline{X}}_4$.

Comparison between the means ${\overline{X}}_3$ and ${\overline{X}}_4$:

The hypotheses are given below:

Null hypothesis:

$H_0:$ There is no significant difference between ${\overline{X}}_3$ and ${\overline{X}}_4$.

Alternative hypothesis:

$H_1:$ There is significant difference between ${\overline{X}}_3$ and ${\overline{X}}_4$.

Rejection region:

The null hypothesis would be rejected if absolute value greater than the critical value.

Absolute value:

The formula for comparing the means ${\overline{X}}_3$ and ${\overline{X}}_4$ is,

$q_5=\frac{{\overline{X}}_3-{\overline{X}}_4}{\sqrt{\frac{s_W^2}{n}}}$

Substitute 14,275 and 18,327 for ${\overline{X}}_3$ and ${\overline{X}}_4$ and 3,421,093 for $s_W^2$

$\begin{array}{c}{q}_6=\frac{14,275-18,327}{\sqrt{\frac{3,421,093}{5}}}\\ =-\frac{4052}{\sqrt{684,218.6}}\\ =-\frac{4052}{827.1751}\\ =-4.90\end{array}$

Thus, the value of $q_6$ is –4.90.

Hence, the absolute value of $q_6$ is 4.90.

Conclusion:

The absolute value is 4.90.

Here, the absolute value is greater than the critical value.

That is, $4.90>4.05$.

Thus, the null hypothesis is rejected.

Hence, there is significant difference between the means ${\overline{X}}_3$ and ${\overline{X}}_4$.

Justification:

Here, there is significant difference between the means ${\overline{X}}_1$ and ${\overline{X}}_4$, ${\overline{X}}_2$ and ${\overline{X}}_4$, and ${\overline{X}}_3$ and ${\overline{X}}_4$.