Chapter 11, Problem 35CE

Instructions: You may use Excel, MegaStat, Minitab, JMP, or another computer package of your choice. Attach appropriate copies of the output or capture the screens, tables, and relevant graphs and include them in a written report. Try to state your conclusions succinctly in language that would be clear to a decision maker who is a nonstatistician. Exercises marked * are based on optional material. Answer the following questions, or those your instructor assigns.

1. a. Choose an appropriate ANOVA model. State the hypotheses to be tested.
2. b. Display the data visually (e.g., dot plots or line plots by factor). What do the displays show?
3. c. Do the ANOVA calculations using the computer.
4. d. State the decision rule for α = .05 and make the decision. Interpret the p-value.
5. e. In your judgment, are the observed differences in treatment means (if any) large enough to be of practical importance?
6. f. Given the nature of the data, would more data collection be practical?
7. g. Perform Tukey multiple comparison tests and discuss the results.
8. h. Perform a test for homogeneity of variances. Explain fully.

Below are results of braking tests of the Ford Explorer on glare ice, packed snow, and split traction (one set of wheels on ice, the other on dry pavement), using three braking methods. Research question: Is the mean stopping distance affected by braking method and/or by surface type?

To determine

Choose appropriate ANOVA model. State the hypothesis.

Answer to Problem 35CE

Two-factor ANOVA without replication is used in this situation.

For factor A:

Null hypothesis:

$H_0:A_1=A_2=A_3=0$

Alternative hypothesis:

$H_1:$ Not all $A_j$’s are equal.

For factor B:

Null hypothesis:

$H_0:B_1=B_2=B_3=0$

Alternative hypothesis:

$H_1:$ Not all $B_k$’s are equal.

Explanation of Solution

The given information is a stopping time for ford explorer using three methods.

Two-factor ANOVA without replication is mainly used for comparing effect of two factors without testing the interaction. Therefore, the data comparison of effect of two factors can be model using two-factor ANOVA without replication.

Factor A denotes the methods and B denotes the Road conditions,

State the hypotheses:

For factor A:

Null hypothesis:

$H_0:A_1=A_2=A_3=0$

There is no effect for methods.

Alternative hypothesis:

$H_1:$ Not all $A_j$’s are equal.

At least one method has effect.

For factor B:

Null hypothesis:

$H_0:B_1=B_2=B_3=0$

There is no effect for road conditions or surface type.

Alternative hypothesis:

$H_1:$ Not all $B_k$’s are equal.

At least one road conditions has effect.

To determine

Display the data for each plot and explain the plot.

Answer to Problem 35CE

The dot plot is given by:

Explanation of Solution

Calculation:

Dot plot:

Software procedure:

Step-by-step procedure for Dot plot using the Mega Stat is given below:

• • Open an EXCEL file.
• • In Mega Stat, select Analysis of variance and then Randomized block ANOVA.
• • In Input range drop down box, select Values.
• • Click on Plot the data.
• • Click OK.

The output using the Mega Stat software is given below:

Observations:

From the dot plot it is clear that mean stopping time for surface ice is different from the other two surfaces. That is, the mean measurements are different. Therefore, mean stopping time for different are significant.

To determine

Perform ANOVA.

Explanation of Solution

Calculation:

Two-factor ANOVA without replication and Turkey’s comparison:

Software procedure:

Step-by-step procedure for Two-factor ANOVA without replication and Turkey’s comparison using the MegaStat is given below:

• • Open an EXCEL file.
• • In Mega Stat, select Analysis of variance and then Randomized block ANOVA.
• • In Input range drop down box, select Values.
• • In post-hoc analysis select p less than 0.05
• • Click OK.

The output using the Mega Stat software is given below:

To determine

Explain the conclusion.

Interpret the p-value.

Answer to Problem 35CE

There is enough evidence to conclude that mean stopping time affected by surface and but not by methods.

Explanation of Solution

Calculation:

From the MegaStat output,

The p-value for factor surface is 0.0002.

The p-value for factor methods is 0.5387.

Rejection rule:

If $p\text{-value}\le \alpha$, reject $H_0$. Otherwise do not reject $H_0$.

Conclusion:

For factor surface:

From the MegaStat output, the p-value is 0.0002.

That is less than the significance level $\alpha =0.05$.

That is, $p\text{-value}<\text{0}\text{.05}$.

Therefore, the null hypothesis is rejected.

Hence, means mean stopping time affected by surface.

For factor method:

From the MegaStat output, the p-value is 0.5387.

That is greater than the significance level $\alpha =0.05$.

That is, $p\text{-value}>\text{0}\text{.05}$.

Therefore, the null hypothesis is not rejected.

Hence, means mean stopping time is not affected by methods.

Here, the p-values for surface are very small. Therefore, samples were support the rejection of the null hypotheses. The p-values for methods are not very small. Therefore, samples were not support the rejection of the null hypotheses.

To determine

Check whether the observed difference in treatment means large enough to be of practical importance.

Answer to Problem 35CE

The observed difference in treatment means large enough to be of practical importance.

Explanation of Solution

From the dot plot or the individual value plot it clear that mean stopping time is affected by surface condition.  That is surface condition is significant. Therefore, observed difference in treatment means large enough to be of practical importance.

To determine

Check whether more data collection is practical if nature of the data is given.

Explanation of Solution

Answer will vary. One of the possible answers is given below:

The samples are taken from the Ford Explorer. If the nature of the data is given then it is easy to collect samples from more Ford Explorer. Thus, the data collection is practical if nature of the data is given.

To determine

Perform Turkeys test using explain the results.

Answer to Problem 35CE

There is enough evidence to conclude that significant difference among ice and split traction.

There is enough evidence to conclude that significant difference among ice and packed snow.

Explanation of Solution

Calculation:

Significance level is $\alpha =0.05$.

 State the hypotheses:

Null hypothesis:

$H_0:{\mu }_i={\mu }_{j}i=1,2,3$

That is, all the pair treatment means are equal.

Alternative hypothesis:

$H_0:{\mu }_i\ne {\mu }_{j}i=1,2,3$

That is, at least one pair of treatment means differ.

Here, the Turkey’s test statistic is $T_{\text{calc}}=\frac{\left|\overline{y_i}-\overline{y_j}\right|}{\sqrt{\text{MSE}\left[\frac{1}{n_i}+\frac{1}{n_i}\right]}}$  where $\overline{y_i}$ and $\overline{y_j}$ are the mean of i^th and j^th group, MSE is the pooled variance of the c samples, and $n_i$ and $n_j$ are sample size of the i^th and j^th group.

Rejection rule:

If $p\text{-value}\le \alpha$, reject $H_0$. Otherwise do not reject $H_0$.

From the MegaStat output in part (c),

The value of the p-value for ice and packed snow is 0.0001.

The value of the p-value for ice and split traction is 0.0002.

The value of the p-value for packed snow and split traction is 0.1999.

Conclusion:

For ice and packed snow:

The p-value is 0.0001, which less than the significance level is $\alpha =0.05$.

That is, $p\text{-value}<\text{0}\text{.05}$. Therefore, the null hypothesis is rejected.

Thus, there is enough evidence to conclude that there significant difference among ice and packed snow

For ice and split traction:

The p-value is 0.0002, which less than the significance level is $\alpha =0.05$.

That is, $p\text{-value}<\text{0}\text{.05}$. Therefore, the null hypothesis is rejected.

Thus, there is enough evidence to conclude that there significant difference among ice and split traction.

For packed snow and split traction:

The p-value is 0.1999, which greater than the significance level is $\alpha =0.05$.

That is, $p\text{-value}>\text{0}\text{.05}$. Therefore, the null hypothesis is not rejected.

Thus, there is enough evidence to conclude that there no significant difference among packed snow and split traction.

To determine

Perform test of homogeneity of variance and explain the results.

Answer to Problem 35CE

There is enough evidence to conclude that variances for surface and methods are not different.

Explanation of Solution

Calculation:

 State the hypotheses:

For factor method:

Null hypothesis:

$H_0:{\sigma }_{A_1}^2={\sigma }_{A_2}^2={\sigma }_{A_3}^2$

All the variances for methods are equal.

Alternative hypothesis:

$H_1:$ Not all ${\sigma }_j^2$ are equal.

At least one pair of variance differs.

For factor surface:

Null hypothesis:

$H_0:{\sigma }_{B_1}^2={\sigma }_{B_2}^2={\sigma }_{B_3}^2$

All the surface variances are equal.

Alternative hypothesis:

$H_1:$ Not all ${\sigma }_j^2$ are equal.

At least one pair of variance differs.

Rejection rule:

If $H\ge H_{\text{Crtical}}$, reject $H_0$. Otherwise do not reject $H_0$.

Test statistics:

The Hartley’s test statistics H is given by:

$H=\frac{s_{\max }^2}{s_{\min }^2}$

Where, $s_{\max }^2$ and $s_{\min }^2$ is the maximum and minimum of sample variance respectively.

Degrees of freedom:

For between group:

$df\left(\text{Between group}\right)=c$,  where c is the number of groups.

Here, there are 3 groups. That is $c=3$. Substitute this value in the above formula.

Therefore,

$\begin{array}{c}df\left(\text{Between group}\right)=df\left(\text{Numerator}\right)\\ =d{f}_1\\ =3\end{array}$

Thus, between group degrees of freedom are 3.

For within group:

$df\left(\text{within group}\right)=\frac{n}{c}-1$, where c is number of groups and n is the sample size.

$\begin{array}{c}n=n_1+n_2+n_3\\ =3+3+3\\ =9\end{array}$

Substitute $c=3\text{ and }n=9$ in the above formula.

Therefore,

$\begin{array}{c}df\left(\text{within group}\right)=df\left(\text{denominator}\right)\\ =d{f}_2\\ =\frac{9}{3}-1\\ =3-1\\ =2\end{array}$

Thus, within group degrees of freedom are 2.

Critical-value:

Procedure for the value of Hartley’s H using Table 11.5:

• • Go through the row corresponding to 3 in Table 11.5 of five percent critical values of Hartley’s $H=\frac{s_{\max }^2}{s_{\min }^2}$ .
• • Go through the row corresponding to 3 and column corresponding to the number of groups 2.
• • Obtain the value corresponding to (3, 2) from the table.

Thus, the critical-value is 87.5.

For surface:

The sample variances for surface are $s_1^2={9.849}^2,s_2^2={37.528}^2,\text{and}{s}_3^2={11.358}^2$.

Here,

$\begin{array}{c}{s}_{\max }^2=\text{Max}\left({9.849}^2,{37.528}^2,{11.358}^2\right)\\ ={37.528}^2\\ =1,408.350784\end{array}$

$\begin{array}{c}{s}_{\min }^2=\text{Min}\left({9.849}^2,{37.528}^2,{11.358}^2\right)\\ ={9.849}^2\\ =97.0028\end{array}$

Substitute these values in the above equation.

Therefore,

$\begin{array}{c}H=\frac{1,408.350784}{97.0028}\\ =14.5187\end{array}$

Thus, the Hartley’s test statistics is 14.5187.

For methods:

The sample variances for surface are $s_1^2={151.803}^2,s_2^2={177.827}^2,\text{and}{s}_3^2={164.739}^2$.

Here,

$\begin{array}{c}{s}_{\max }^2=\text{Max}\left({151.803}^2,{177.827}^2,{164.739}^2\right)\\ ={177.827}^2\\ =31,622.44193\end{array}$

$\begin{array}{c}{s}_{\min }^2=\text{Min}\left({151.803}^2,{177.827}^2,{164.739}^2\right)\\ ={151.803}^2\\ =23,044.15081\end{array}$

Substitute these values in the above equation.

Therefore,

$\begin{array}{c}H=\frac{31,622.44193}{23,044.15081}\\ =1.3723\end{array}$

Thus, the Hartley’s test statistics is 1.3723.

Conclusion:

Surface:

Here, the H-test statistic is 14.5.

Here, $H\text{-statistic}\left(=14.5\right)<H_{\text{Critical}}\left(=87.5\right)$.

That is test statistic is less than the critical value.

Therefore, the null hypothesis is not rejected.

Thus, there is enough evidence to conclude that variances for surface are not different.

Methods:

Here, the H-test statistic is 1.37.

Here, $H\text{-statistic}\left(=1.37\right)<H_{\text{Critical}}\left(=87.5\right)$.

That is test statistic is less than the critical value.

Therefore, the null hypothesis is not rejected.

Thus, there is enough evidence to conclude that variances for smethods are not different.