Chapter 9.2, Problem 5E

Refer to Exercise 13 in Section 9.1.

1. a. Use the Bonferroni method to determine which pairs of means, if any, are different at the 5% level.
2. b. Use the Tukey-Kramer method to determine which pairs of means, if any, are different at the 5% level.
3. c. Which is the more powerful method to find all the pairs of treatments whose means are different, the Bonferroni method or the Tukey–Kramer method?

To determine

Compare the pairs of treatments and conclude that any of the mean temperatures are different at 5% level of significance level of significance using Bonferroni method.

Answer to Problem 5E

The conclusion is that the mean temperature of $750°C$ is different from the mean temperature of $850°C\text{and }900°C$.

The mean temperature of $800°C$ is different from the mean temperature of $900°C$.

Explanation of Solution

Calculation:

The given information is based on conducting experiment on tensile strength of iron. A sample of 5 specimens with four temperatures has been taken to test its tensile strength.

The aim is to compare the means of temperatures.

Bonferroni method for the comparison of multiple means:

Comparison of two means:

The mean effects of the i^th and j^th groups are considered to be significantly different, if the following condition is satisfied:

$\left|{\overline{X}}_{i.}-{\overline{X}}_{j.}\right|>t_{N-I,\frac{\alpha }{2C}}\sqrt{MSE\left(\frac{1}{J_i}+\frac{1}{J_j}\right)}$

Where, $C=\frac{I\left(I-1\right)}{2}$

Where, N is the sample size, C be the number of pairs of difference to compare such that $C=I\left(I-2\right)$, I denotes the number of treatments, $J_i$ and $J_j$ are sizes of the i^th and j^th groups respectively, ${\overline{X}}_{i.}$ and ${\overline{X}}_{j.}$ are means of the i^th and j^th groups respectively, MSE is the mean square error obtained from the analysis and $\alpha$ is the level of significance at which the test is performed, $t_{N-I,\frac{\alpha }{2C}}$ is the upper $\alpha$ point for the given degrees of freedom.

The value of C is:

Substitute $I=4$ for 4 temperatures in the study, to find the value of C:

$\begin{array}{c}C=\frac{I\left(I-1\right)}{2}\\ =\frac{4\left(4-1\right)}{2}\\ =2\left(3\right)\\ =6\end{array}$

The number of pairs to compare is 6.

The critical value at 5% level of significance is:

Substitute $N=20$,$\alpha =0.05$,$I=4$ and $C=6$,

$\begin{array}{c}{t}_{N-I,\frac{\alpha }{2C}}=t_{20-4,\frac{0.05}{2\left(6\right)}}\\ =t_{16,0.004167}\end{array}$

Software procedure:

Step by step procedure to obtain the critical-value using the MINITAB software is given below:

• Choose Graph > Probability Distribution Plot choose View Probability> OK.
• From Distribution, choose ‘t’ distribution.
• Enter Degrees of freedom as 16.
• Click the Shaded Area tab.
• Choose P Value and Right Tail for the region of the curve to shade.
• Enter the data value as 0.004167.
• Click OK.

Output using the MINITAB software is given below:

From the MINITAB output, the value of $t_{16,0.004167}$ is 3.008.

From the Exercise 13 in section 9.1, the value of MSE is 2.30.

The value of $t_{N-1,\frac{\alpha }{2C}}\sqrt{MSE\left(\frac{1}{J_i}+\frac{1}{J_j}\right)}$ is:

Substitute the $t_{16,0.004167}=3.008$, $MSE=2.30$ and $J_i=J_j=5$,

$\begin{array}{c}{t}_{N-1,\frac{\alpha }{2C}}\sqrt{MSE\left(\frac{1}{J_i}+\frac{1}{J_j}\right)}=t_{16,0.004167}\sqrt{2.30\left(\frac{1}{5}+\frac{1}{5}\right)}\\ =3.008\sqrt{2.30\left(\frac{1}{5}+\frac{1}{5}\right)}\\ =3.008\left(0.9592\right)\\ =2.8852\end{array}$

Descriptive measures of temperatures:

Software procedure:

Step by step procedure to obtain the descriptive measures using the MINITAB software is given below:

• Choose Stat > Basic Statistics > Display Descriptive Statistics.
• In Variables enter the columns $750°C,800°C,850°C,900°C$.
• Choose option statistics, and select N total and Mean.
• Click OK.

Output obtained using MINITAB is given below:

From the MINITAB, the mean value of $750°C$ is 19.36, mean $800°C$ is 18.992, mean $850°C$ is 16.464 and mean $900°C$ is 15.27.

Compare the means of $750°C$ and $800°C$:

State the hypotheses:

Null hypothesis:

$H_0:{\mu }_i={\mu }_j$

That is, there is no significant difference between the mean effects of i^th and j^th treatments.

Alternative hypothesis:

$H_a:{\mu }_i\ne {\mu }_j$

That is, there is significant difference between the mean effects of i^th and j^th treatments.

Substitute ${\overline{X}}_{i.}=19.36$ and ${\overline{X}}_{j.}=18.992$,

$\begin{array}{c}\left|{\overline{X}}_{i.}-{\overline{X}}_{j.}\right|=\left|19.36-18.992\right|\\ =\left|0.368\right|\\ =0.368\\ \cancel{>}2.885\left(=t_{N-1,\frac{\alpha }{2C}}\sqrt{MSE\left(\frac{1}{J_i}+\frac{1}{J_j}\right)}\right)\end{array}$

It has been found that the difference of mean $750°C$ and mean $800°C$ is less than the critical value.

Thus, the means of the $750°C$ is not different from the mean of $800°C$.

Compare the means of $750°C$ and $850°C$:

Substitute ${\overline{X}}_{i.}=19.36$ and ${\overline{X}}_{j.}=16.464$,

$\begin{array}{c}\left|{\overline{X}}_{i.}-{\overline{X}}_{j.}\right|=\left|19.36-16.464\right|\\ =\left|2.894\right|\\ =2.894\\ >2.885\left(=t_{N-1,\frac{\alpha }{2C}}\sqrt{MSE\left(\frac{1}{J_i}+\frac{1}{J_j}\right)}\right)\end{array}$

It has been found that the difference of mean $750°C$ and mean $850°C$ is greater than the critical value.

Thus, the means of the $750°C$ is different from the mean of $850°C$.

Compare the means of $750°C$ and $900°C$:

Substitute ${\overline{X}}_{i.}=19.36$ and ${\overline{X}}_{j.}=15.27$,

$\begin{array}{c}\left|{\overline{X}}_{i.}-{\overline{X}}_{j.}\right|=\left|19.36-15.27\right|\\ =\left|4.09\right|\\ =4.09\\ >2.885\left(=t_{N-1,\frac{\alpha }{2C}}\sqrt{MSE\left(\frac{1}{J_i}+\frac{1}{J_j}\right)}\right)\end{array}$

It has been found that the difference of mean $750°C$ and mean $900°C$ is greater than the critical value.

Thus, the means of the $750°C$ is different from the mean of $900°C$.

Compare the means of $800°C$ and $850°C$:

Substitute ${\overline{X}}_{i.}=18.992$ and ${\overline{X}}_{j.}=16.464$,

$\begin{array}{c}\left|{\overline{X}}_{i.}-{\overline{X}}_{j.}\right|=\left|18.992-16.464\right|\\ =\left|2.528\right|\\ =2.528\\ \cancel{>}2.885\left(=t_{N-1,\frac{\alpha }{2C}}\sqrt{MSE\left(\frac{1}{J_i}+\frac{1}{J_j}\right)}\right)\end{array}$

It has been found that the difference of mean $800°C$ and mean $850°C$ is less than the critical value.

Thus, the mean of the $800°C$ is not different from the mean of $850°C$.

Compare the means of $800°C$ and $900°C$:

Substitute ${\overline{X}}_{i.}=18.992$ and ${\overline{X}}_{j.}=15.27$,

$\begin{array}{c}\left|{\overline{X}}_{i.}-{\overline{X}}_{j.}\right|=\left|18.992-15.27\right|\\ =\left|3.722\right|\\ =3.722\\ >2.885\left(=t_{N-1,\frac{\alpha }{2C}}\sqrt{MSE\left(\frac{1}{J_i}+\frac{1}{J_j}\right)}\right)\end{array}$

It has been found that the difference of mean $800°C$ and mean $900°C$ is greater than the critical value.

Thus, the means of the $800°C$ is different from the mean of $900°C$.

Compare the means of $850°C$ and $900°C$:

Substitute ${\overline{X}}_{i.}=16.464$ and ${\overline{X}}_{j.}=15.27$,

$\begin{array}{c}\left|{\overline{X}}_{i.}-{\overline{X}}_{j.}\right|=\left|16.464-15.27\right|\\ =\left|1.194\right|\\ =1.194\\ \cancel{>}2.885\left(=t_{N-1,\frac{\alpha }{2C}}\sqrt{MSE\left(\frac{1}{J_i}+\frac{1}{J_j}\right)}\right)\end{array}$

It has been found that the difference of mean $850°C$ and mean $900°C$ is less than the critical value.

Thus, the means of the $850°C$ is not different from the mean of $900°C$.

Thus, by comparing the mean temperature of $750°C,800°C,850°C,900°C$, the difference of the mean $750°C$ is greater than the mean $850°C$ and mean $900°C$, the difference of the mean $800°C$ is greater than the mean $900°C$

Hence it is clear that the mean temperature of $750°C$ is different from the mean temperature of $850°C\text{and }900°C$ and the mean temperature of $800°C$ is different from the mean temperature of $900°C$.

To determine

Compare the pairs of treatments and conclude that any of the mean temperatures are different at 5% level of significance level of significance using Tukey-Kramer method.

Answer to Problem 5E

The conclusion is that the mean temperature of $750°C$ is different from the mean temperature of $850°C\text{and }900°C$.

The mean temperature of $800°C$ is different from the mean temperature of $900°C$.

Explanation of Solution

Calculation:

The aim is to compare the means of temperature.

Tukey-Kramer method  for the comparison of multiple means:

Comparison of two means:

The mean effects of the i^th and j^th groups are considered to be significantly different, if the following condition is satisfied:

$\left|{\overline{X}}_{i.}-{\overline{X}}_{j.}\right|>q_{I,N-I,\alpha }\sqrt{\frac{MSE}{2}\left(\frac{1}{J_i}+\frac{1}{J_j}\right)}$

Where, N is the sample size, I denotes the number of treatments, $J_i$ and $J_j$ are sizes of the i^th and j^th groups respectively, ${\overline{X}}_{i.}$ and ${\overline{X}}_{j.}$ are means of the i^th and j^th groups respectively, MSE is the mean square error obtained from the analysis and $\alpha$ is the level of significance at which the test is performed and $q_{I,N-I,\alpha }$ denotes the studentized range distribution with I and $N-I$ degrees of freedom.

The studentized range value at 5% level of significance is:

Substitute $N=20$,$\alpha =0.05$ and $I=4$,

$\begin{array}{c}{q}_{I,N-I,\alpha }=q_{4,20-4,0.05}\\ =q_{4,16,0.05}\end{array}$

Use Table A.9: upper percentage points for the studentized range value.

Procedure:

For $q_{4,16,0.05}$,

• Locate $v_1=4$ in the top column of the table.
• Locate $v_2=16$ in the left column of the table.
• Obtain the value in the corresponding row below 0.05.

That is, $q_{4,16,0.05}=4.05$

From the Exercise 13 in section 9.1, the value of MSE is 2.30.

The value of $q_{I,N-1,\alpha }\sqrt{MSE\left(\frac{1}{J_i}+\frac{1}{J_j}\right)}$ is:

Substitute the $q_{4,16,0.05}=4.05$, $MSE=2.30$ and $J_i=J_j=5$,

$\begin{array}{c}{q}_{I,N-1,\alpha }\sqrt{MSE\left(\frac{1}{J_i}+\frac{1}{J_j}\right)}=q_{4,16,0.05}\sqrt{\frac{2.30}{2}\left(\frac{1}{5}+\frac{1}{5}\right)}\\ =4.05\sqrt{\frac{2.30}{2}\left(\frac{1}{5}+\frac{1}{5}\right)}\\ =4.05\left(0.6782\right)\\ =2.747\end{array}$

From the previous part (a), the mean value of $750°C$ is 19.36, mean $800°C$ is 18.992, mean $850°C$ is 16.464 and mean $900°C$ is 15.27.

Compare the means of $750°C$ and $800°C$:

Substitute ${\overline{X}}_{i.}=19.36$ and ${\overline{X}}_{j.}=18.992$,

$\begin{array}{c}\left|{\overline{X}}_{i.}-{\overline{X}}_{j.}\right|=\left|19.36-18.992\right|\\ =\left|0.368\right|\\ =0.368\\ \cancel{>}2.747\left(=q_{I,N-I,\alpha }\sqrt{\frac{MSE}{2}\left(\frac{1}{J_i}+\frac{1}{J_j}\right)}\right)\end{array}$

It has been found that the difference of mean $750°C$ and mean $800°C$ is less than the critical value.

Thus, the means of the $750°C$ is not different from the mean of $800°C$.

Compare the means of $750°C$ and $850°C$:

Substitute ${\overline{X}}_{i.}=19.36$ and ${\overline{X}}_{j.}=16.464$,

$\begin{array}{c}\left|{\overline{X}}_{i.}-{\overline{X}}_{j.}\right|=\left|19.36-16.464\right|\\ =\left|2.894\right|\\ =2.894\\ >2.747\left(=q_{I,N-I,\alpha }\sqrt{\frac{MSE}{2}\left(\frac{1}{J_i}+\frac{1}{J_j}\right)}\right)\end{array}$

It has been found that the difference of mean $750°C$ and mean $850°C$ is greater than the critical value.

Thus, the means of the $750°C$ is different from the mean of $850°C$.

Compare the means of $750°C$ and $900°C$:

Substitute ${\overline{X}}_{i.}=19.36$ and ${\overline{X}}_{j.}=15.27$,

$\begin{array}{c}\left|{\overline{X}}_{i.}-{\overline{X}}_{j.}\right|=\left|19.36-15.27\right|\\ =\left|4.09\right|\\ =4.09\\ >2.747\left(=q_{I,N-I,\alpha }\sqrt{\frac{MSE}{2}\left(\frac{1}{J_i}+\frac{1}{J_j}\right)}\right)\end{array}$

It has been found that the difference of mean $750°C$ and mean $900°C$ is greater than the critical value.

Thus, the means of the $750°C$ is different from the mean of $900°C$.

Compare the means of $800°C$ and $850°C$:

Substitute ${\overline{X}}_{i.}=18.992$ and ${\overline{X}}_{j.}=16.464$,

$\begin{array}{c}\left|{\overline{X}}_{i.}-{\overline{X}}_{j.}\right|=\left|18.992-16.464\right|\\ =\left|2.528\right|\\ =2.528\\ \cancel{>}2.747\end{array}$

It has been found that the difference of mean $800°C$ and mean $850°C$ is less than the critical value.

Thus, the means of the $800°C$ is not different from the mean of $850°C$.

Compare the means of $800°C$ and $900°C$:

Substitute ${\overline{X}}_{i.}=18.992$ and ${\overline{X}}_{j.}=15.27$,

$\begin{array}{c}\left|{\overline{X}}_{i.}-{\overline{X}}_{j.}\right|=\left|18.992-15.27\right|\\ =\left|3.722\right|\\ =3.722\\ >2.747\left(=q_{I,N-I,\alpha }\sqrt{\frac{MSE}{2}\left(\frac{1}{J_i}+\frac{1}{J_j}\right)}\right)\end{array}$

It has been found that the difference of mean $800°C$ and mean $900°C$ is greater than the critical value.

Thus, the means of the $800°C$ is different from the mean of $900°C$.

Compare the means of $850°C$ and $900°C$:

Substitute ${\overline{X}}_{i.}=16.464$ and ${\overline{X}}_{j.}=15.27$,

$\begin{array}{c}\left|{\overline{X}}_{i.}-{\overline{X}}_{j.}\right|=\left|16.464-15.27\right|\\ =\left|1.194\right|\\ =1.194\\ \cancel{>}2.747\left(=q_{I,N-I,\alpha }\sqrt{\frac{MSE}{2}\left(\frac{1}{J_i}+\frac{1}{J_j}\right)}\right)\end{array}$

It has been found that the difference of mean $850°C$ and mean $900°C$ is less than the critical value.

Thus, the means of the $850°C$ is not different from the mean of $900°C$.

Thus, by comparing the mean temperature of $750°C,800°C,850°C,900°C$, the differences of the mean $750°C$ is greater than the mean $850°C$ and mean $900°C$, the difference of the mean $800°C$ is greater than the mean $900°C$

Hence it is clear that the mean temperature of $750°C$ is different from the mean temperatures of $850°C\text{and }900°C$ and the mean temperature of $800°C$ is different from the mean temperature of $900°C$.

To determine

Identify the more powerful method among the Bonferroni method and Tukey-Kramer method.

Answer to Problem 5E

The Tukey-Kramer method is more powerful than the Bonferroni method.

Explanation of Solution

Interpretation:

Condition for more powerful method of multiple comparisons:

The method of multiple comparisons, which has a smaller critical value for a given dataset, is the more powerful method of multiple comparisons for that particular dataset.

From the previous parts (a) and (b), the critical value of the Bonferroni method at 5% level of significance is 2.885 and the critical value of the Tukey-Kramer method at 5% level of significance is 2.75.

Therefore, the critical value of Tukey-Kramer method holds the minimum critical value and it is said to be more powerful than the Bonferroni method.