Chapter 13, Problem 52SE

a.

To determine

Find the number of levels of factor in the experiment.

Answer to Problem 52SE

There are 5 levels of factor in the experiment.

Explanation of Solution

Given info:

The data represents the ANOVA table with missing values.

Calculation:

The levels of factors can be obtained as follows:

$\text{levels of factor=df}\left(\text{Factor}\right)+1$

And, $\text{df}\left(\text{Factor}\right)\text{=df}\left(\text{Total}\right)-\text{df}\left(\text{Error}\right)$

Thus,

$\begin{array}{c}\text{df}\left(\text{Factor}\right)=19-15\\ =4\end{array}$

Thus,

$\begin{array}{c}\text{levels of factor=4+1}\\ \text{=5}\end{array}$

Thus, there are 5 levels of factor in the experiment.

b.

To determine

Find the numbers of replicates were used in the experiment.

Answer to Problem 52SE

There are four replicates of each levels are used.

Explanation of Solution

Justification:

From the ANOVA table, it is clear that total degree of freedom is 19. That is the total number of trials is 20$\left(=19+1\right)$. From part (a) it is clear that there are 5 levels for each factor. Therefore, there are 4 $\left(=\frac{20}{5}\right)$ replicates of each levels are used.

c.

To determine

Fill the missing information in the ANOVA table. Use Bound for the P-value.

Answer to Problem 52SE

The complete ANOVA table is given below:

Source	DF	SS	MS	F	P-value
Factor	4	158.7	39.675	3.533	$0.025<P\text{-value <0}\text{.05}$
Error	15	167.5	11.167
Total	19	326.2
$S=3.342$	$R-\text{Sq}=0.4865$	$R-\text{Sq}\left(\text{adj}\right)=34.96\%$

Explanation of Solution

Calculation:

From the ANOVA table, $\text{df}\left(\text{Error}\right)=15\text{and }M{S}_E=167.5$

From part (a), $\text{df}\left(\text{Factor}\right)=4$.

$\begin{array}{c}M{S}_E=\frac{S{S}_E}{\left(a-1\right)}\\ =\frac{167.5}{15}\\ =11.1667\end{array}$

$\begin{array}{c}\text{SS}\left(\text{Factor}\right)\text{=SS}\left(\text{Total}\right)-\text{SS}\left(\text{Error}\right)\\ =326.2-167.5\\ =158.7\end{array}$

$\begin{array}{c}M{S}_{\text{Factor}}=\frac{S{S}_{\text{Factor}}}{\left(a-1\right)}\\ =\frac{158.7}{4}\\ =39.675\end{array}$

$\begin{array}{c}{F}_0=\frac{M{S}_{\text{Factor}}}{M{S}_E}\\ =\frac{39.675}{11.167}\\ =3.553\end{array}$

$\begin{array}{c}R-\text{sq}=1-\frac{S{S}_E}{S{S}_{\text{Total}}}\\ =1-\frac{167.5}{326.2}\\ =1-0.5135\\ =0.4865\end{array}$

From Appendix table VI, the value F statistic with $\left(4,15\right)$ df is 3.06 and corresponding P-value is 0.05. The value F statistic with $\left(4,15\right)$ df is 3.8 and corresponding P-value is 0.025. Here, $F_0=3.553$. That is, $3.06<F_0\text{ < 3}\text{.8}$. Therefore, $0.025<P\text{-value <0}\text{.05}$.

d.

To determine

Find the conclusion can you draw about difference in the factor-level means if $\alpha =0.05$. Also find the conclusion if $\alpha =0.01$.

Answer to Problem 52SE

There is significant difference in the factor-level means if $\alpha =0.05$.

There are no significant difference in the factor-level means if $\alpha =0.01$.

Explanation of Solution

Calculation:

State the hypotheses:

Null hypothesis:

$H_0:Thereisnodifferenceinfactorlevel$

Alternative hypothesis:

$H_1:Thereisdifferenceinfactorlevel$

The level of significance is 0.05. From part (b), it is clear that $0.025<P\text{-value <0}\text{.05}$.

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 $P\text{-value}$ is less than the level of significance.

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

Therefore, the null hypothesis is rejected.

Thus, there is sufficient evidence to conclude that there is significant difference in the factor-level means at $\alpha =0.05$ level of significance.

Here, $0.025<P\text{-value <0}\text{.05}$ and $\alpha =0.01$. That is $P\text{-value}>\alpha$. Therefore the null hypothesis is failed to reject. Hence, there is sufficient evidence to conclude that there is no significant difference in the factor-level means at $\alpha =0.01$ level of significance.