Chapter 12, Problem 12.1.1RE

If the null hypothesis is rejected in Exercises 1 through 8, use the Scheffé test when the sample sizes are unequal to test the differences between the means, and use the Tukey test when the sample sizes are equal. For these exercises, perform these steps. Assume the assumptions have been met.

a. State the hypotheses and identify the claim.

b. Find the critical value(s).

c. Compute the test value.

d. Make the decision.

e. Summarize the results.

Use the traditional method of hypothesis testing unless otherwise specified.

1. Lengths of Various Types of Bridges The data represent the lengths in feet of three types of bridges in the United States. At a = 0.01, test the claim that there is no significant difference in the means of the lengths of the types of bridges.

Simple truss	Segmented concrete	Continuous plate
745	820	630
716	750	573
700	790	525
650	674	510
647	660	480
625	640	460
608	636	451
598	620	450
550	520	450
545	450	425
534	392	420
528	370	360

To determine

To state: The hypothesis $H_0\text{ and }{H}_1$.

To identify: The claim.

Answer to Problem 12.1.1RE

The hypothesis $H_0$ and $H_1$ are,

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

$H_1:$ At least one mean is different from the others.

The claim is that, all the means are same.

Explanation of Solution

Given info:

The data shows the lengths in feet of three types of bridges in the United States. The level of significance is 0.01.

Calculation:

The hypotheses are given below:

Null hypothesis:

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

Alternative hypothesis:

$H_1:$ At least one mean is different from the others

Here, all the means are same is tested. Hence, the claim is that, all the means are same.

To determine

To find: The critical value.

Answer to Problem 12.1.1RE

The critical value is 5.285.

Explanation of Solution

Given info:

The level of significance is 0.01. The number of samples k is 3, the sample sizes $n_1$, $n_2$ and $n_3$ are 12, 12 and 12.

Calculation:

The degrees of freedom are $\text{d}\text{.f}\text{.N}=k-1\text{ and d}\text{.f}\text{.D}=N-k$.

Where

$\begin{array}{c}N=n_1+n_2+n_3\\ =12+12+12\\ =36\end{array}$

Substitute 3 for k in $\text{d}\text{.f}\text{.N}$

$\begin{array}{c}\text{d}\text{.f}\text{.N}=k-1\\ \text{=3-1}\\ \text{=2}\end{array}$

Substitute 36 for N and 3 for k in $\text{d}\text{.f}\text{.D}$

$\begin{array}{c}\text{d}\text{.f}\text{.D}=N-k\\ =36-3\\ =33\end{array}$

Critical value:

The critical F-value is obtained using the Table H: The F-Distribution with the level of significance $\alpha =0.01$.

Procedure:

• Locate 30 and 40 in the degrees of freedom, denominator row of the Table H.
• Obtain the value in the corresponding degrees of freedom, numerator column below 2.

That is,

$\begin{array}{c}\text{Critical value=}\frac{5.39+5.18}{2}\\ =\frac{10.57}{2}\\ =5.285\end{array}$

That is, the critical value is 5.285.

Rejection region:

The null hypothesis would be rejected if $F>5.285$.

To determine

To compute: The test value.

Answer to Problem 12.1.1RE

The test value is 6.94.

Explanation of Solution

Calculation:

Software procedure:

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

• Choose Stat > ANOVA > One-Way.
• In Response, enter the Temperatures.
• In Factor, enter the Factor.
• Click OK.

Output using the MINITAB software is given below:

From the MINITAB output, the test value F is 6.94.

To determine

To make: The decision.

Answer to Problem 12.1.1RE

The null hypothesis is rejected.

Explanation of Solution

Conclusion:

From the result of part (c), the test value is 6.94.

Here, the F-statistic value is greater than the critical value.

That is, $6.94>5.285$.

Thus, it can be concluding that, the null hypothesis is rejected.

To determine

To explain: The results.

Answer to Problem 12.1.1RE

The result concludes that, there is a significant difference between the means ${\overline{X}}_1$ and ${\overline{X}}_3$

Explanation of Solution

Calculation:

From the results, it can be observed that the null hypothesis is rejected. Thus, it can be concluding that there is evidence to reject the claim that all means are same.

Consider, ${\overline{X}}_1$, ${\overline{X}}_2$ and ${\overline{X}}_3$ represents the means of Florida, Pennsylvania and Maine, $s_1^2,s_2^2\text{ and }{s}_3^2$ represents the variances of samples of Florida, Pennsylvania and Maine.

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 Florida, Pennsylvania and Maine.
• Choose option statistics, and select Mean, Variance and N total.
• Click OK.

Output using the MINITAB software is given below:

The sample sizes $n_1,n_2{\text{ and n}}_3$ are 12, 12 and 12.

The means are ${\overline{X}}_1,{\overline{X}}_2\text{ and }{\overline{X}}_3$ are 620.5, 610.2 and 477.8.

The sample variances are $s_1^2,s_2^2\text{ and }{s}_3^2$ are 5445.9, 22108.7 and 5280.3.

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

Tukey test:

Critical value:

Here, k is 3 and degrees of freedom $v=N-k$

Substitute 36 for N and 3 for k in v

$\begin{array}{c}v=N-k\\ =36-3\\ =33\end{array}$

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 nearest value of 33 in the column of v of the Table H.
• Obtain the value in the corresponding row below 3.

That is, the critical value is 4.45.

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,

$\begin{array}{c}{s}_W^2=\frac{(12-1)5445.9+(12-1)22,108.7+(12-1)5280.3}{\left(12-1\right)+\left(12-1\right)+\left(12-1\right)}\\ =\frac{59904.9+243,195.7+58,083.3}{33}\\ =\frac{361,183.9}{33}\\ =10,944.97\end{array}$

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 620.5 and 610.2 for ${\overline{X}}_1$ and ${\overline{X}}_2$ and 10,944.97 for $s_W^2$

$\begin{array}{c}{q}_1=\frac{620.5-610.2}{\sqrt{\frac{10,944.97}{12}}}\\ =\frac{10.3}{\sqrt{912.081}}\\ =\frac{10.3}{30.2007}\\ =0.34\end{array}$

Thus, the value of $q_1$ is 0.34.

Hence, the absolute value of $q_1$ is 0.34.

Conclusion:

The absolute value is 0.34.

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

That is, $0.34<4.45$.

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 620.5 and 477.8 for ${\overline{X}}_1$ and ${\overline{X}}_3$ and 10,944.97 for $s_W^2$

$\begin{array}{c}{q}_2=\frac{620.5-477.8}{\sqrt{\frac{10,944.97}{12}}}\\ =\frac{142.7}{\sqrt{912.081}}\\ =\frac{142.7}{30.2007}\\ =4.72\end{array}$

Thus, the value of $q_2$ is 4.72.

Hence, the absolute value of $q_2$ is 4.72.

Conclusion:

The absolute value is 4.72.

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

That is, $4.72>4.45$.

Thus, the null hypothesis is rejected.

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

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,

$q_3=\frac{{\overline{X}}_2-{\overline{X}}_3}{\sqrt{\frac{s_W^2}{n}}}$

Substitute 610.2 and 477.8 for ${\overline{X}}_2$ and ${\overline{X}}_3$ and 10,944.97 for $s_W^2$

$\begin{array}{c}{q}_3=\frac{610.2-477.8}{\sqrt{\frac{10,944.97}{12}}}\\ =\frac{132.4}{\sqrt{912.081}}\\ =\frac{132.4}{30.2007}\\ =4.38\end{array}$

Thus, the value of $q_3$ is 4.38.

Hence, the absolute value of $q_3$ is 4.38.

Conclusion:

The absolute value is 4.38.

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

That is, $4.38<4.45$.

Thus, the null hypothesis is not rejected.

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