Chapter 9.3, Problem 14E

The article “Use of Taguchi Methods and Multiple Regression Analysis for Optimal Process Development of High Energy Electron Beam Case Hardening of Cast Iron” (M. Jean and Y. Treng, Surface Engineering. 2003:150–156) describes a factorial experiment designed to determine factors in a high-energy electron beam process that affect hardness in metals. Results for two factors, each with three levels, are presented in the following table. Factor A is the travel speed in mm/s, and factor B is accelerating voltage in volts. The outcome is Vickers hardness. There were six replications for each treatment. In the article, a total of seven factors were studied: the two presented here are those that were found to be the most significant.

A	B	Hardness
10	10	875	896	921	686	642	613
10	25	712	719	698	621	632	645
10	50	568	546	559	757	723	734
20	10	876	835	868	812	796	772
20	25	889	876	849	768	706	615
20	50	756	732	723	681	723	712
30	10	901	926	893	856	832	841
30	25	789	801	776	845	827	831
30	50	792	786	775	706	675	568

1. a. Estimate all main effects and interactions.
2. b. Construct an ANOVA table. You may give ranges for the P-values.
3. c. Is the additive model plausible? Provide the value of the test statistic and the P-value.
4. d. Can the effect of travel speed on the hardness be described by interpreting the main effects of travel speed? If so, interpret the main effects, using multiple comparisons at the 5% level if necessary. If not, explain why not.
5. e. Can the effect of accelerating voltage on the hardness be described by interpreting the main effects of accelerating voltage? If so, interpret the main effects, using multiple comparisons at the 5% level if necessary. If not, explain why not.

To determine

Find all the main and interaction effects.

Answer to Problem 14E

The interaction effects are:

${\gamma }_{11}=9.0556,{\gamma }_{12}=-22.944,{\gamma }_{13}=13.889,{\gamma }_{21}=-16.722,{\gamma }_{22}=9.611,{\gamma }_{23}=7.1111$,${\gamma }_{31}=7.6667,{\gamma }_{32}=13.333\text{and}{\gamma }_{33}=-21$

The main effects are:

${\widehat{\alpha }}_1=-61.389,{\widehat{\alpha }}_2=18.722\text{and }{\widehat{\alpha }}_2=42.667,{\widehat{\beta }}_1=66.056,{\widehat{\beta }}_2=-2.94444\text{and }{\widehat{\beta }}_3=-63.111$.

Explanation of Solution

Calculation:

The given information is that the experiment involves the response of two factors (A (travel speed) and B (accelerating voltage)).

The first cell refers to travel speed 10 and accelerating voltage 10.

The first cell mean can be obtained as follows:

$\begin{array}{c}{\overline{X}}_{11.}=\frac{1}{K}{\displaystyle \sum _{i=1}^K{X}_{11k}}\\ =\frac{1}{6}\left(875+896+921+686+642+613\right)\\ =\frac{4,633}{6}\\ =772.1667\end{array}$

Similarly the means of remaining cells are given in the below table:

Here, the row means refers to the factor travel speed.

The first row mean can be obtained as follows:

$\begin{array}{c}{\overline{X}}_{i\mathrm{..}}=\frac{1}{J}{\displaystyle \sum _{j=1}^J{\overline{X}}_{ij.}}\\ =\frac{1}{3}\left(772.1667+671.1667+647.8333\right)\\ =697.0556\end{array}$

Similarly the means of remaining rows are given in the below table:

Here, the column means refers to the factor accelerating voltage.

The first column mean can be obtained as follows:

$\begin{array}{c}{\overline{X}}_{.j.}=\frac{1}{I}{\displaystyle \sum _{i=1}^I{\overline{X}}_{ij.}}\\ =\frac{1}{3}\left(772.1667+826.5+874.83333\right)\\ =824.5\end{array}$

Similarly the means of remaining columns are given in the below table:

The remaining row and column mean can be obtained as shown in the table:

	10	25	50	Row Mean ${\overline{X}}_{i\mathrm{..}}$
10	772.1667	671.1667	647.8333	697.0556
20	826.5	783.8333	721.1667	777.1667
30	874.8333	811.5	717	801.1111
Column Mean ${\overline{X}}_{.j.}$	824.5	755.5	695.3333	758.4444

The row effects can be obtained as follows:

${\widehat{\alpha }}_i={\overline{X}}_{i\mathrm{..}}-{\overline{X}}_{\mathrm{...}}$

Here, ${\widehat{\alpha }}_i{}^{\prime }s$ are the main effects of the factor travel speed.

Substitute ${\overline{X}}_{\mathrm{1..}}=697.0556\text{and }{\overline{X}}_{\mathrm{...}}=758.4444$ in the above formula.

$\begin{array}{c}{\widehat{\alpha }}_1={\overline{X}}_{\mathrm{1..}}-{\overline{X}}_{\mathrm{...}}\\ =697.0556-758.4444\\ =-61.389\end{array}$

Substitute ${\overline{X}}_{\mathrm{2..}}=777.16667\text{and }{\overline{X}}_{\mathrm{...}}=758.4444$ in the above formula.

$\begin{array}{c}{\widehat{\alpha }}_2={\overline{X}}_{\mathrm{2..}}-{\overline{X}}_{\mathrm{...}}\\ =777.1667-758.4444\\ =18.722\end{array}$

Substitute ${\overline{X}}_{\mathrm{3..}}=801.1111\text{and }{\overline{X}}_{\mathrm{...}}=758.4444$ in the above formula.

$\begin{array}{c}{\widehat{\alpha }}_3={\overline{X}}_{\mathrm{3..}}-{\overline{X}}_{\mathrm{...}}\\ =801.1111-758.4444\\ =42.667\end{array}$

Thus, the row effects are ${\widehat{\alpha }}_1=-61.389,{\widehat{\alpha }}_2=18.722\text{and }{\widehat{\alpha }}_2=42.667$.

The column effects can be obtained as follows:

${\widehat{\beta }}_j={\overline{X}}_{.j.}-{\overline{X}}_{\mathrm{...}}$

Here, ${\widehat{\beta }}_j{}^{\prime }s$ are the main effects of the factor accelerating voltage.

Substitute ${\overline{X}}_{\mathrm{.1.}}=824.5\text{and }{\overline{X}}_{\mathrm{...}}=758.4444$ in the above formula.

$\begin{array}{c}{\widehat{\beta }}_1={\overline{X}}_{\mathrm{.1.}}-{\overline{X}}_{\mathrm{...}}\\ =824.5-758.4444\\ =66.056\end{array}$

Substitute ${\overline{X}}_{\mathrm{.2.}}=755.5\text{and }{\overline{X}}_{\mathrm{...}}=758.4444$ in the above formula.

$\begin{array}{c}{\widehat{\beta }}_2={\overline{X}}_{\mathrm{.2.}}-{\overline{X}}_{\mathrm{...}}\\ =755.5-758.4444\\ =-2.9444\end{array}$

Substitute ${\overline{X}}_{\mathrm{.3.}}=695.3333\text{and }{\overline{X}}_{\mathrm{...}}=758.4444$ in the above formula.

$\begin{array}{c}{\widehat{\beta }}_3={\overline{X}}_{\mathrm{.3.}}-{\overline{X}}_{\mathrm{...}}\\ =695.3333-758.4444\\ =-63.111\end{array}$

Thus, the column effects are ${\widehat{\beta }}_1=66.056,{\widehat{\beta }}_2=-2.94444\text{and }{\widehat{\beta }}_3=-63.111$.

The interaction effects can be obtained as follows:

${\gamma }_{ij}={\overline{X}}_{ij.}-{\overline{X}}_{i\mathrm{..}}-{\overline{X}}_{.j.}+{\overline{X}}_{\mathrm{...}}$

Substitute ${\overline{X}}_{11.}=772.1667,{\overline{X}}_{\mathrm{1..}}=697.0556,{\overline{X}}_{\mathrm{.1.}}=824.5,{\overline{X}}_{\mathrm{...}}=758.4444$ in the above formula.

$\begin{array}{c}{\gamma }_{11}={\overline{X}}_{11.}-{\overline{X}}_{\mathrm{1..}}-{\overline{X}}_{\mathrm{.1.}}+{\overline{X}}_{\mathrm{...}}\\ =772.1667-697.0556-824.5+758.4444\\ =9.0556\end{array}$

Substitute ${\overline{X}}_{12.}=671.1667,{\overline{X}}_{\mathrm{1..}}=697.0556,{\overline{X}}_{\mathrm{.2.}}=755.5,{\overline{X}}_{\mathrm{...}}=758.4444$ in the above formula.

$\begin{array}{c}{\gamma }_{12}={\overline{X}}_{12.}-{\overline{X}}_{\mathrm{1..}}-{\overline{X}}_{\mathrm{.2.}}+{\overline{X}}_{\mathrm{...}}\\ =671.1667-697.0556-755.5+758.4444\\ =-22.944\end{array}$

Substitute ${\overline{X}}_{13.}=647.8333,{\overline{X}}_{\mathrm{1..}}=697.0556,{\overline{X}}_{\mathrm{.3.}}=695.3333,{\overline{X}}_{\mathrm{...}}=758.4444$ in the above formula.

$\begin{array}{c}{\gamma }_{13}={\overline{X}}_{13.}-{\overline{X}}_{\mathrm{1..}}-{\overline{X}}_{\mathrm{.3.}}+{\overline{X}}_{\mathrm{...}}\\ =647.8333-697.0556-695.3333+758.4444\\ =13.889\end{array}$

Substitute ${\overline{X}}_{21.}=826.5,{\overline{X}}_{\mathrm{2..}}=777.1667,{\overline{X}}_{\mathrm{.1.}}=824.5,{\overline{X}}_{\mathrm{...}}=758.4444$ in the above formula.

$\begin{array}{c}{\gamma }_{21}={\overline{X}}_{21.}-{\overline{X}}_{\mathrm{2..}}-{\overline{X}}_{\mathrm{.1.}}+{\overline{X}}_{\mathrm{...}}\\ =826.5-777.1667-824.5+758.4444\\ =-16.722\end{array}$

Substitute ${\overline{X}}_{22.}=783.8333,{\overline{X}}_{\mathrm{2..}}=777.1667,{\overline{X}}_{\mathrm{.2.}}=755.5,{\overline{X}}_{\mathrm{...}}=758.4444$ in the above formula.

$\begin{array}{c}{\gamma }_{22}={\overline{X}}_{22.}-{\overline{X}}_{\mathrm{2..}}-{\overline{X}}_{\mathrm{.2.}}+{\overline{X}}_{\mathrm{...}}\\ =783.8333-777.1667-755.5+758.4444\\ =9.611\end{array}$

Substitute ${\overline{X}}_{23.}=721.1667,{\overline{X}}_{\mathrm{2..}}=777.1667,{\overline{X}}_{\mathrm{.3.}}=695.3333,{\overline{X}}_{\mathrm{...}}=758.4444$ in the above formula.

$\begin{array}{c}{\gamma }_{23}={\overline{X}}_{23.}-{\overline{X}}_{\mathrm{2..}}-{\overline{X}}_{\mathrm{.3.}}+{\overline{X}}_{\mathrm{...}}\\ =721.1667-777.1667-695.3333+758.4444\\ =7.1111\end{array}$

Substitute ${\overline{X}}_{31.}=874.8333,{\overline{X}}_{\mathrm{3..}}=801.1111,{\overline{X}}_{\mathrm{.1.}}=824.5,{\overline{X}}_{\mathrm{...}}=758.4444$ in the above formula.

$\begin{array}{c}{\gamma }_{31}={\overline{X}}_{31.}-{\overline{X}}_{\mathrm{3..}}-{\overline{X}}_{\mathrm{.1.}}+{\overline{X}}_{\mathrm{...}}\\ =874.8333-801.1111-824.5+758.4444\\ =7.6667\end{array}$

Substitute ${\overline{X}}_{32.}=811.5,{\overline{X}}_{\mathrm{3..}}=801.1111,{\overline{X}}_{\mathrm{.2.}}=755.5,{\overline{X}}_{\mathrm{...}}=758.4444$ in the above formula.

$\begin{array}{c}{\gamma }_{32}={\overline{X}}_{32.}-{\overline{X}}_{\mathrm{3..}}-{\overline{X}}_{\mathrm{.2.}}+{\overline{X}}_{\mathrm{...}}\\ =811.5-801.1111-755.5+758.4444\\ =13.333\end{array}$

Substitute ${\overline{X}}_{33.}=717,{\overline{X}}_{\mathrm{3..}}=801.1111,{\overline{X}}_{\mathrm{.3.}}=695.3333,{\overline{X}}_{\mathrm{...}}=758.4444$ in the above formula.

$\begin{array}{c}{\gamma }_{33}={\overline{X}}_{33.}-{\overline{X}}_{\mathrm{3..}}-{\overline{X}}_{\mathrm{.3.}}+{\overline{X}}_{\mathrm{...}}\\ =717-801.1111-695.3333+758.4444\\ =-21\end{array}$

Thus, the interaction effects are ${\gamma }_{11}=9.0556,{\gamma }_{12}=-22.944,{\gamma }_{13}=13.889,{\gamma }_{21}=-16.722,{\gamma }_{22}=9.611,{\gamma }_{23}=7.1111$${\gamma }_{31}=7.6667,{\gamma }_{32}=13.333\text{and}{\gamma }_{33}=-21$ and the main effects are ${\widehat{\alpha }}_1=-61.389,{\widehat{\alpha }}_2=18.722,{\widehat{\alpha }}_3=42.667,{\widehat{\beta }}_1=66.056,{\widehat{\beta }}_2=-2.9444\text{and }{\widehat{\beta }}_3=-63.111$.

To determine

Construct an ANOVA table and the find the ranges for the P-values.

Answer to Problem 14E

The ANOVA table is,

Source	DF	SS	MS	F	P
A	2	106,912	53,456	8.74	0.001
B	2	150,390	75,195.2	12.29	0.000
Interaction	4	11,409	2,852.2	0.47	0.760
Error	45	275,228	6,116.2
Total	53	543,939

For Factor A, the P-value is 0.001.

For Factor B, the P-value is 0.000.

For interaction, the P-value is 0.760.

Explanation of Solution

Calculation:

The factor A is travel speed and factor B is accelerating voltage.

Step-by-step procedure for finding the Two-Way ANOVA table is as follows:

Software procedure:

• Choose Stat > ANOVA > Two-Way.
• In Response, enter the column of Hardness.
• In Row Factor, enter the column of A.
• In Column Factor, enter the column of B.
• Click OK.

Output obtained by MINITAB procedure is as follows:

For Factor A, the F-test statistic is 8.74 and the P-value is 0.001.

For Factor B, the F-test statistic is 12.29and the P-value is 0.000.

For interaction, the F-test statistic is 0.47 and the P-value is 0.760.

To determine

Explain whether the additive model is plausible.

Answer to Problem 14E

The additive model is plausible.

Explanation of Solution

Calculation:

Interaction:

Null hypothesis:

$H_0:{\gamma }_{11}={\gamma }_{12}=\mathrm{...}={\gamma }_{33}=0$

Alternative hypothesis:

$H_1:\text{At least one}{\gamma }_{ij}\ne 0$

For interaction, the F-test statistic is 0.47 and the P-value is 0.760.

Decision:

If $P\text{-value}\le \alpha ,$ reject the null hypothesis $H_0$.

If $P\text{-value}>\alpha ,$ fail to reject the null hypothesis $H_0$.

Conclusion:

Interaction:

Here, the P-value is greater than the level of significance.

That is, $P\text{-value}\left(0.760\right)>\alpha \left(=0.05\right)$.

Therefore, the null hypothesis is not rejected.

Thus, the interaction is not significant at $\alpha =0.05$ level of significance.

Therefore all the interactions are equal to zero.

Thus, the additive model is plausible.

To determine

Check whether the effects of travel speed on the hardness can be described by the main effects of travel speed. If so, interpret the main effects by multiple comparisons at the 5% level. If not explain the reason.

Answer to Problem 14E

Yes, the effects of travel speed on the hardness can be described by the main effects of travel speed.

There is sufficient evidence to conclude that the effect of a travel speed of 10 differs from those of both 20 and 30 at $\alpha =0.05$ level of significance.

Explanation of Solution

Calculation:

Factor A is travel speed.

Main effect of factor A:

Null hypothesis:

$H_0:{\alpha }_1={\alpha }_2={\alpha }_3=0$

Alternative hypothesis:

$H_1:\text{ At least one }{\alpha }_i\ne 0$

For Factor A, the F-test statistic is 8.74 and P- value is 0.001.

Decision:

If $P\text{-value}\le \alpha ,$ reject the null hypothesis $H_0$.

If $P\text{-value}>\alpha ,$ fail to reject the null hypothesis $H_0$.

Conclusion:

Factor A:

Here, the P-value is less than the level of significance.

That is, $P\text{-value}\left(<0.001\right)<\alpha \left(=0.05\right)$.

Therefore, the null hypothesis is rejected.

Thus, some of the main effects of factor A are non-zero.

Hence, it is not plausible that the main effects of travel speed on the hardness are equal to zero at $\alpha =0.05$ level of significance.

Since, the main effects of travel speed on the hardness are not all equal to zero, the effects of travel speed on the hardness can be described by the main effects of travel speed.

Thus, the effects of travel speed on the hardness can be described by the main effects of travel speed.

The main effects can be interpret using Tukey’s method.

State the hypotheses:

Null hypothesis:

$H_0\text{:}$ There is no travel speed affect the hardness.

Alternative hypothesis:

$H_1:$ There is travel speed affect the hardness.

Decision:

By Tukey’s method for multiple comparisons,

If $\left|{\widehat{\alpha }}_i-{\widehat{\alpha }}_j\right|>q_{I,IJ\left(K-1\right),\alpha }\sqrt{\frac{MSE}{JK}}$ reject the null hypothesis $H_0$

If $\left|{\widehat{\alpha }}_i-{\widehat{\alpha }}_j\right|\le q_{I,IJ\left(K-1\right),\alpha }\sqrt{\frac{MSE}{JK}}$ fail to reject the null hypothesis $H_0$

Here $I=3,J=3,\alpha =0.05,MSE=6116.2$

From Appendix A table A.9, the upper 5% point of the $q_{3,45,0.05}$ distribution is not found in the table. Then approximate it to $q_{3,40,0.05}$. The upper 5% point of the $q_{3,40,0.05}$ distribution is 3.44.

For comparing travel speed in 10 mm/s and 20 mm/s:

The 5% critical value is,

$\begin{array}{c}{q}_{I,IJ\left(K-1\right),\alpha }\sqrt{\frac{MSE}{JK}}=q_{3,45,0.05}\sqrt{\frac{MSE}{JK}}\\ \approx q_{3,40,0.05}\sqrt{\frac{MSE}{JK}}\end{array}$

Substitute $q_{3,40,0.05}=3.44,J=3,K=6\text{and }MSE=6,116.2$

$\begin{array}{c}{q}_{3,45,0.05}\sqrt{\frac{MSE}{JK}}=\left(3.44\right)\sqrt{\frac{6,116.2}{\left(3\right)\left(6\right)}}\\ =\left(3.44\right)\sqrt{\frac{6,116.2}{18}}\\ =\left(3.44\right)\sqrt{339.789}\\ =\left(3.44\right)\left(18.4334\right)\\ =63.41\end{array}$

From part (a), the row effects are ${\widehat{\alpha }}_1=-61.389,{\widehat{\alpha }}_2=18.722,{\widehat{\alpha }}_3=42.667$.

$\begin{array}{c}\left|{\widehat{\alpha }}_1-{\widehat{\alpha }}_2\right|=\left|-61.389-18.722\right|\\ =80.111\end{array}$

Which is greater than 63.41.

Thus, reject the null hypothesis $H_0$.

Hence, for travel speed in 10 mm/s and 20 mm/s there is travel speed affect the hardness.

For comparing travel speed in 10 mm/s and 30 mm/s:

$\begin{array}{c}\left|{\widehat{\alpha }}_1-{\widehat{\alpha }}_3\right|=\left|-61.389-42.667\right|\\ =104.056\end{array}$

Which is greater than 63.41.

Thus, reject the null hypothesis $H_0$.

Hence, for travel speed in 10 mm/s and 30 mm/s there is travel speed affect the hardness.

For comparing travel speed in 20 mm/s and 30 mm/s:

$\begin{array}{c}\left|{\widehat{\alpha }}_2-{\widehat{\alpha }}_3\right|=\left|18.722-42.667\right|\\ =23.945\end{array}$

Which is less than 63.41.

Thus, fail to reject the null hypothesis $H_0$.

Hence, for travel speed in 20 mm/s and 30 mm/s there is no travel speed affect the hardness.

Conclusion:

There is sufficient evidence to conclude that the effect of a travel speed of 10 differs from those of both 20 and 30 at $\alpha =0.05$ level of significance.

To determine

Check whether the effects of accelerating voltage on the hardness can be described by the main effects of accelerating voltage. If so, interpret the main effects by multiple comparisons at the 5% level. If not explain the reason.

Answer to Problem 14E

Yes, the effects of accelerating voltage on the hardness can be described by the main effects of accelerating voltage.

There is sufficient evidence to conclude that the effect of an accelerating voltage in 10 volts differs from those of both 25 volts and 50 volts at $\alpha =0.05$ level of significance.

Explanation of Solution

Calculation:

Factor B is accelerating voltage.

Main effect of factor B:

Null hypothesis:

$H_0:{\beta }_1={\beta }_2={\beta }_3=0$

Alternative hypothesis:

$H_1:\text{ At least one }{\beta }_i\ne 0$

For Factor B, the F-test statistic is 12.29 and the P-value is 0.000.

Decision:

If $P\text{-value}\le \alpha ,$ reject the null hypothesis $H_0$.

If $P\text{-value}>\alpha ,$ fail to reject the null hypothesis $H_0$.

Conclusion:

Factor B:

Here, the P-value is less than the level of significance.

That is, $P\text{-value}\left(0.000\right)<\alpha \left(=0.05\right)$.

Therefore, the null hypothesis is rejected.

Thus, some of the main effects of factor B are zero.

Hence, it is not plausible that the main effect of accelerating voltage on the hardness are equal to zero at $\alpha =0.05$ level of significance.

Since, the main effects of accelerating voltage on the hardness are not equal to zero, the effects of accelerating voltage on the hardness can be described by the main effects of accelerating voltage.

Thus, the effects of accelerating voltage on the hardness can be described by the main effects of accelerating voltage.

The main effects can be interpret using Tukey’s method.

State the hypotheses:

Null hypothesis:

$H_0\text{:}$ There is no accelerating voltage affect the hardness.

Alternative hypothesis:

$H_1:$ There is accelerating voltage affect the hardness.

Decision:

By Tukey’s method for multiple comparisons,

If $\left|{\widehat{\beta }}_i-{\widehat{\beta }}_j\right|>q_{J,IJ\left(K-1\right),\alpha }\sqrt{\frac{MSE}{IK}}$ reject the null hypothesis $H_0$

If $\left|{\widehat{\beta }}_i-{\widehat{\beta }}_j\right|\le q_{J,IJ\left(K-1\right),\alpha }\sqrt{\frac{MSE}{IK}}$ fail to reject the null hypothesis $H_0$

Here $I=3,J=3,\alpha =0.05,MSE=6116.2$

From Appendix A table A.9, the upper 5% point of the $q_{3,45,0.05}$ distribution is not found in the table. Then approximate it to $q_{3,40,0.05}$. The upper 5% point of the $q_{3,40,0.05}$ distribution is 3.44.

For comparing accelerating in 10 volts and 25 volts:

The 5% critical value is,

$\begin{array}{c}{q}_{J,IJ\left(K-1\right),\alpha }\sqrt{\frac{MSE}{IK}}=q_{3,45,0.05}\sqrt{\frac{MSE}{IK}}\\ \approx q_{3,40,0.05}\sqrt{\frac{MSE}{IK}}\end{array}$

Substitute $q_{3,40,0.05}=3.44,I=3,K=6\text{and }MSE=6,116.2$

$\begin{array}{c}{q}_{3,45,0.05}\sqrt{\frac{MSE}{IK}}=\left(3.44\right)\sqrt{\frac{6,116.2}{\left(3\right)\left(6\right)}}\\ =\left(3.44\right)\sqrt{\frac{6,116.2}{18}}\\ =\left(3.44\right)\sqrt{339.789}\\ =\left(3.44\right)\left(18.4334\right)\\ =63.41\end{array}$

From part (a), the row effects are ${\widehat{\beta }}_1=66.056,{\widehat{\beta }}_2=-2.9444\text{and }{\widehat{\beta }}_3=-63.111$.

$\begin{array}{c}\left|{\widehat{\beta }}_1-{\widehat{\beta }}_2\right|=\left|66.056+2.9444\right|\\ =69.00\end{array}$

Which is greater than 63.41.

Thus, reject the null hypothesis $H_0$.

Hence, for accelerating in 10 volts and 25 volts there is accelerating voltage affect the hardness.

For comparing accelerating in 10 volts and 50 volts:

$\begin{array}{c}\left|{\widehat{\beta }}_1-{\widehat{\beta }}_3\right|=\left|66.056+63.111\right|\\ =129.167\end{array}$

Which is greater than 63.41.

Thus, reject the null hypothesis $H_0$.

Hence, for accelerating in 10 volts and 50 volts there is accelerating voltage affect the hardness.

For comparing accelerating in 25 volts and 50 volts:

$\begin{array}{c}\left|{\widehat{\beta }}_2-{\widehat{\beta }}_3\right|=\left|-2.9444+63.111\right|\\ =60.1666\end{array}$

Which is less than 63.41.

Thus, fail to reject the null hypothesis $H_0$.

Hence, for accelerating in 25 volts and 50 volts there is no accelerating voltage affect the hardness.

Conclusion:

There is sufficient evidence to conclude that the effect of an accelerating voltage in 10 volts differs from those of both 25 volts and 50 volts at $\alpha =0.05$ level of significance.