Chapter 9.1, Problem 8E

The article “Impact of Free Calcium Oxide Content of Fly Ash on Dust and Sulfur Dioxide Emissions in a Lignite-Fired Power Plant” (D. Sotiropoulos, A. Georgakopoulos, and N. Kolovos, Journal of Air and Waste Management, 2005:1042–1049) presents measurements of dust emissions, in mg/m³, for four power plants. Thirty measurements were taken for each plant. The sample means and standard deviations are presented in the following table:

1. a. Construct an ANOVA table. You may give a range for the P-value.
2. b. Can you conclude that there are differences among the mean emission levels?

To determine

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

Answer to Problem 8E

The ANOVA table is,

Source	DF	SS	MS	F	P
Plant	3	12,712.7	4,237.6	4.8179	0.01<P-value<0.001
Error	116	102,027.8	879.55
Total	119	114,740.5

The range of P-value is $0.010<P\text{-value <0}\text{.001}$

Explanation of Solution

Given info:

The data represents the means and standard deviations of 30 measurements of dust emissions taken on four types of plants.

Calculation:

The ANOVA table can be obtained as follows:

There are four samples, therefore $I=4$.

The total number of observations is,

$\begin{array}{c}N=30+30+30+30\\ =120\end{array}$.

The degrees of freedom corresponding to the plant is obtained as follows:

$\begin{array}{c}\text{Degrees of freedom}=\text{I}-1\\ =4-1\\ =3\end{array}$

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

$\begin{array}{c}\text{Degrees of freedom}=N-1\\ =120-1\\ =119\end{array}$

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

$\begin{array}{c}\text{Degrees of freedom}=\left(N-1\right)-\left(I-2\right)\\ =119-3\\ =116\end{array}$

Total mean can be obtained as follows:

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

Substitute ${\overline{X}}_1=211.50,{\overline{X}}_2=214,{\overline{X}}_3=211.75,{\overline{X}}_4=236.08\text{and }I=4$

$\begin{array}{c}\overline{X}=\frac{211.50+214+211.75+236.08}{4}\\ =\frac{873.33}{4}\\ =218.3325\end{array}$

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

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

Substitute ${\overline{X}}_1=211.50,{\overline{X}}_2=214,{\overline{X}}_3=211.75,{\overline{X}}_4=236.08\text{and }{J}_1=J_2=J_3=J_4=30$

$\begin{array}{c}\text{SSTr}=\left(\begin{array}{l}30{\left(211.50-218.3325\right)}^2+30{\left(214-218.3325\right)}^2\\ +30{\left(211.75-218.3325\right)}^2+30{\left(236.09-218.3325\right)}^2\end{array}\right)\\ =\left(30{\left(-6.8325\right)}^2+30{\left(-4.3325\right)}^2+30{\left(-6.5825\right)}^2+30{\left(17.7475\right)}^2\right)\\ =30\left(46.68306\right)+30\left(18.77056\right)+30\left(43.32931\right)+30\left(314.9738\right)\\ =1,400.492+563.1167+1,299.879+9,449.213\\ =12,712.7\end{array}$

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

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

Substitute $s_1=24.85,s_2=35.26,s_3=33.53,s_4=23.09\text{and }{J}_1=J_2=J_3=J_4=30$

$\begin{array}{c}SSE={\displaystyle \sum _{i=1}^I\left(J_i-1\right)s_i^2}\\ =29{\left(24.85\right)}^2+29{\left(35.26\right)}^2+29{\left(33.53\right)}^2+29{\left(23.09\right)}^2\\ =29\left(617.5225\right)+29\left(1,243.2676\right)+29\left(1,124.2609\right)+29\left(533.1481\right)\\ =17,908.1525++36,054.7604+32,603.5661+15,461.2949\\ =102,027.8\end{array}$

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

$SST=SSTr+SSE$

Substitute $SSTr=12,712.7\text{and }SSE=102,027.8$.

$\begin{array}{c}SST=12,712.7+102,027.8\\ =114,740.5\end{array}$

The treatment mean square is obtained as follows:

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

Substitute $SSTr=12,712.7\text{and I=4}$.

$\begin{array}{c}MSTr=\frac{12,712.7}{4-1}\\ =\frac{12,712.7}{3}\\ =4,237.6\end{array}$

The error mean square is obtained as follows:

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

Substitute $SSE=102,027.8,\text{ I=4and N=120}$.

$\begin{array}{c}MSE=\frac{102,027.8}{120-4}\\ =\frac{102,027.8}{116}\\ =879.55\end{array}$

The F-value is obtained as follows:

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

Substitute $MSTr=4,237.6\text{and }MSE=879.55$.

$\begin{array}{c}F=\frac{4,237.6}{879.55}\\ =4.8179\end{array}$

Thus, the F-value is 4.8179.

From Appendix A table A.8, the upper 1% point of the $F_{3,116}$ distribution is 3.95. The upper 0.1% point of the $F_{3,116}$ distribution is 5.78. Thus, the P-value is between 0.01 and 0.001.

Therefore, the range of P-value is $0.010<P\text{-value <0}\text{.001}$.

Thus, the ANOVA table is,

Source	DF	SS	MS	F	P
Plant	3	12,712.7	4,237.6	4.8179	0.01<P-value<0.001
Error	116	102,027.8	879.55
Total	119	114,740.5

To determine

Check whether the mean emission levels differ for four types of plants.

Answer to Problem 8E

There is sufficient evidence to conclude that the mean emission levels differ for four types of plants.

Explanation of Solution

Calculation:

State the hypotheses:

Null hypothesis:

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

Alternative hypothesis:

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

From part (a), the F-ratio is 4.8179.

P-value:

From part (a), the P-value is between 0.01 and 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 emission levels differ for four types of plants at $\alpha =0.05$ level of significance.