Chapter 11, Problem 43CE

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 data on truck production (number of vehicles completed) during the second shift at five truck plants for each day in a randomly chosen week. Research question: Are the mean production rates the same by plant and by day?

To determine

Choose appropriate ANOVA model. State the hypothesis.

Answer to Problem 43CE

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

For factor A:

Null hypothesis:

$H_0:A_1=A_2=A_3=A_4=0$

Alternative hypothesis:

$H_1:$ Not all $A_j$’s are equal.

For factor B:

Null hypothesis:

$H_0:B_1=B_2=B_3=B_4=B_5=0$

Alternative hypothesis:

$H_1:$ Not all $B_k$’s are equal.

Explanation of Solution

The given information is a production rate for three plants.

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 plant and B denotes the day.

State the hypotheses:

For factor A:

Null hypothesis:

$H_0:A_1=A_2=A_3=A_4=0$

There is no effect for plant.

Alternative hypothesis:

$H_1:$ Not all $A_j$’s are equal.

At least one plant has effect.

For factor B:

Null hypothesis:

$H_0:B_1=B_2=B_3=0$

There is no effect for day.

Alternative hypothesis:

$H_1:$ Not all $B_k$’s are equal.

At least one day has effect.

To determine

Display the data for each plot and explain the plot.

Answer to Problem 43CE

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 production rate for Wednesday and Thursday. That is, the mean production rates are different. Therefore, mean production rate for different days 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 43CE

There is enough evidence to conclude that mean production rate is affected by factor day.

Explanation of Solution

Calculation:

From the MegaStat output of part (c),

The p-value for plant is 0.0912.

The p-value for factor days is 0.0012.

Rejection rule:

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

Conclusion:

For factor day:

From the MegaStat output, the p-value is 0.0012.

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, mean production rate is affected by factor day.

For factor plant:

From the MegaStat output, the p-value is 0.0912.

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, mean production rate is affected by factor plant.

Here, the p-values for day is very small. Therefore, samples were support the rejection of the null hypotheses. The p-values for plant 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 43CE

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 factor days.  That is factor days condition is significant. For manufacturer the production rate is important. 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 trucks. If the nature of the data is given then it is easy to collect samples from more trucks. 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 43CE

There is enough evidence to conclude that significant difference among all pair of days excluding ‘Thursday and Wednesday’ and ‘Friday and Tuesday’.

Explanation of Solution

Calculation:

Significance level is $\alpha =0.05$.

 State the hypotheses:

Null hypothesis:

$H_0:{\mu }_i={\mu }_{j}i=1,2,3,4,5$

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 Monday and Friday is 0.0392.

The value of the p-value for Monday and Tuesday is 0.0255.

The value of the p-value for Monday and Thursday is 0.0004.

The value of the p-value for Monday and Wednesday is 0.0002.

The value of the p-value for Friday and Tuesday is 0.8176.

The value of the p-value for Friday and Thursday is 0.0269.

The value of the p-value for Friday and Wednesday is 0.0112.

The value of the p-value for Tuesday and Thursday is 0.0414.

The value of the p-value for Tuesday and Wednesday is 0.0174.

The value of the p-value for Thursday and Wednesday is 0.6457.

Conclusion:

For Monday and Friday:

The p-value is 0.0392, 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 is significant difference among means of Monday and Friday.

For Monday and Tuesday:

The p-value is 0.0255, 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 is significant difference among means of Monday and Tuesday.

For Monday and Thursday:

The p-value is 0.0004, 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 is significant difference among means of Monday and Thursday.

For Monday and Wednesday:

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 is significant difference among means of Monday and Wednesday.

For Friday and Tuesday:

The p-value is 0.8176, 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 means of Friday and Tuesday.

For Friday and Thursday:

The p-value is 0.0269, 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 means of Friday and Thursday.

For Friday and Wednesday:

The p-value is 0.0112, 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 means of Friday and Wednesday.

For Tuesday and Thursday:

The p-value is 0.0414, 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 means of Tuesday and Thursday.

For Tuesday and Wednesday:

The p-value is 0.0174, 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 means of Tuesday and Wednesday.

For Thursday and Wednesday:

The p-value is 0.6457, 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 means of Thursday and Wednesday.

To determine

Perform test of homogeneity of variance and explain the results.

Answer to Problem 43CE

There is enough evidence to conclude that variances for days are not different.

Explanation of Solution

Calculation:

State the hypotheses:

For factor days:

Null hypothesis:

$H_0:{\sigma }_{B_1}^2={\sigma }_{B_2}^2={\sigma }_{B_3}^2={\sigma }_{B_4}^2={\sigma }_{B_5}^2$

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 5 days. That is $c=5$. Substitute this value in the above formula.

Therefore,

$\begin{array}{c}df\left(\text{Between group}\right)=df\left(\text{Numerator}\right)\\ =d{f}_1\\ =5\end{array}$

Thus, between group degrees of freedom are 5.

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.

$\begin{array}{c}n=n_1+n_2+n_3+n_4+n_5\\ =4+4+4+4+4\\ =20\end{array}$

Substitute $c=5\text{ and }n=20$ in the above formula.

Therefore,

$\begin{array}{c}df\left(\text{within group}\right)=df\left(\text{denominator}\right)\\ =d{f}_2\\ =\frac{20}{5}-1\\ =4-1\\ =3\end{array}$

Thus, within group degrees of freedom are 3.

Critical-value:

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

• • Go through the row corresponding to 5 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 5 and column corresponding to the number of groups 3.
• • Obtain the value corresponding to (5, 3) from the table.

Thus, the critical-value is 50.7.

For surface:

The sample variances for surface are: $s_1^2={33.091}^2,s_2^2={30.587}^2,s_3^2={33.601}^2,s_4^2={21.393}^2\text{and}{s}_5^2={15.777}^2$.

Here,

$\begin{array}{c}{s}_{\max }^2=\text{Max}\left({33.091}^2,{30.587}^2,{33.601}^2,{21.393}^2,{15.777}^2\right)\\ ={33.601}^2\\ =1,129.0272\end{array}$

$\begin{array}{c}{s}_{\min }^2=\text{Min}\left({33.091}^2,{30.587}^2,{33.601}^2,{21.393}^2,{15.777}^2\right)\\ ={15.777}^2\\ =248.9137\end{array}$

Substitute these values in the above equation.

Therefore,

$\begin{array}{c}H=\frac{1,129.0272}{248.9137}\\ =4.5358\end{array}$

Thus, the Hartley’s test statistics is 4.5358.

Conclusion:

Here, the H-test statistic is 4.5358.

Here, $H\text{-statistic}\left(=4.5358\right)<H_{\text{Critical}}\left(=50.7\right)$.

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 days are not different.