Chapter 9.1, Problem 17E

a.

To determine

Construct an ANOVA table and give a range for the P-value.

Answer to Problem 17E

The ANOVA table is,

Source	DF	SS	MS	F	P
Machine	4	6,862	1,715.5	7.8823	P-value<0.001
Error	30	6,529.1	217.64
Total	34	13,391

The range of P-value is $P\text{-value <0}\text{.001}$

Explanation of Solution

Given info:

The design variable is the machine and the response is the processing time. The table provides the summary statistics of processing time corresponding to machine.

Calculation:

The ANOVA table can be obtained as follows:

There are four machines, therefore $I=5$.

The total number of observations is,

$\begin{array}{c}N=7+7+7+7+7\\ =35\end{array}$.

The degrees of freedom corresponding to the machine can be obtained as follows:

$\begin{array}{c}\text{Degrees of freedom}=\text{I}-1\\ =5-1\\ =4\end{array}$

The degrees of freedom corresponding to the total can be obtained as follows:

$\begin{array}{c}\text{Degrees of freedom}=N-1\\ =35-1\\ =34\end{array}$

The degrees of freedom corresponding to the error can be obtained as follows:

$\begin{array}{c}\text{Degrees of freedom}=\left(N-1\right)-\left(I-2\right)\\ =34-4\\ =30\end{array}$

Total mean can be obtained as follows:

$\overline{X}=\frac{{\overline{X}}_1+{\overline{X}}_2+{\overline{X}}_3+{\overline{X}}_4}{I}$

Substitute ${\overline{X}}_1=25.43,{\overline{X}}_2=23.71,{\overline{X}}_3=44.57,{\overline{X}}_4=23.14,{\overline{X}}_5=58\text{and }I=5$

$\begin{array}{c}\overline{X}=\frac{25.43+23.71+44.57+23.14+58}{5}\\ =\frac{174.85}{5}\\ =34.97\end{array}$

The treatment sum of squares (SSTr) can be obtained as follows:

$\text{SSTr}={\displaystyle \sum _{i=1}^4{J}_i}{\left({\overline{X}}_i-\overline{X}\right)}^2$

Substitut e

${\overline{X}}_1=25.43,{\overline{X}}_2=23.71,{\overline{X}}_3=44.57,{\overline{X}}_4=23.14,{\overline{X}}_5=58\text{and }{J}_1=J_2=J_3=J_4=7$

$\begin{array}{c}\text{SSTr}=\left(\begin{array}{l}7{\left(25.43-34.97\right)}^2+7{\left(25.43-34.97\right)}^2+7{\left(25.43-34.97\right)}^2\\ +7{\left(25.43-34.97\right)}^2+7{\left(25.43-34.97\right)}^2\end{array}\right)\\ =7{\left(-9.54\right)}^2+7{\left(-11.26\right)}^2+7{\left(9.6\right)}^2+7{\left(-11.83\right)}^2+7{\left(23.03\right)}^2\\ =7\left(91.0116\right)+7\left(126.7876\right)+7\left(92.16\right)+7\left(139.9489\right)+7\left(530.3809\right)\\ =637.0812+887.5132+645.12+979.6423+3,712.666\\ \approx 6,862\end{array}$

The error sum of squares (SSE) can be obtained as follows:

$SSE={\displaystyle \sum _{i=1}^I\left(J_i-1\right)s_i^2}$

Substitute $s_1=10.67,s_2=13.92,s_3=15.90,s_4=12.75,s_4=19.11\text{and }{J}_1=J_2=J_3=J_4=7$

$\begin{array}{c}SSE={\displaystyle \sum _{i=1}^I\left(J_i-1\right)s_i^2}\\ =6{\left(10.67\right)}^2+6{\left(13.92\right)}^2+6{\left(15.90\right)}^2+6{\left(12.75\right)}^2+6{\left(19.11\right)}^2\\ =6\left(113.8489\right)+6\left(193.7664\right)+6\left(252.81\right)+6\left(162.5625\right)+6\left(365.1921\right)\\ =683.0934+1,162.5984+1,516.86+975.375+2,191.1526\\ \approx 6,529.1\end{array}$

The total sum of squares (SST) can be obtained as follows:

$SST=SSTr+SSE$

Substitute $SSTr=6862\text{and }SSE=6529.1$.

$\begin{array}{c}SST=6862+6529.1\\ =13,391.1\\ \approx 13,391\end{array}$

The treatment mean square can be obtained as follows:

$MSTr=\frac{SSTr}{I-1}$

Substitute $SSTr=6862\text{and I=5}$.

$\begin{array}{c}MSTr=\frac{6862}{5-1}\\ =\frac{6862}{4}\\ =1,715.5\end{array}$

The error mean square can be obtained as follows:

$MSE=\frac{SSE}{N-1}$

Substitute $SSE=6529.1\text{and N=35}$.

$\begin{array}{c}MSE=\frac{6529.1}{35-5}\\ =\frac{6529.1}{30}\\ =217.64\end{array}$

The F-value can be obtained as follows:

$F=\frac{MSTr}{MSE}$

Substitute $MSTr=1,715.5\text{and }MSE=217.64$.

$\begin{array}{c}F=\frac{1,715.5}{217.64}\\ =7.8823\end{array}$

Thus, the F-value is 7.8823.

From Appendix A table A.8, the upper 0.1% point of the $F_{4,30}$ distribution is 6.12. Thus, the P-value is less than 0.001.

Therefore, the range of P-value is $P\text{-value <0}\text{.001}$.

Thus, the ANOVA table is,

Source	DF	SS	MS	F	P
Machine	4	6,862	1,715.5	7.8823	P-value<0.001
Error	30	6,529.1	217.64
Total	34	13,391

b.

To determine

Check whether the mean processing time differs for different machines.

Answer to Problem 17E

There is sufficient evidence to conclude that the mean processing time differs for different machines.

Explanation of Solution

Calculation:

State the hypotheses:

Null hypothesis:

$H_0:{\mu }_1={\mu }_2={\mu }_3={\mu }_4={\mu }_5$

Alternative hypothesis:

$H_a:$ At least two of the ${\mu }_i{}^{\prime }s$ differ from each other.

From part (a), the F-ratio is 7.8823.

P-value:

From part (a), the P-value is less than 0.001.

Since, the level of significance is not specified; the prior level of significance $\alpha =0.05$ can be used.

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:

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

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

By rejection rule, reject the null hypothesis.

There is sufficient evidence to conclude that the mean processing time differs for different machines at $\alpha =0.05$ level of significance.