Chapter 11.2, Problem 18E

The accompanying data resulted from an experiment to investigate whether yield from a certain chemical process depended either on the formulation of a particular input or on mixer speed.

A statistical computer package gave SS(Form) = 2253.44, SS(Speed) = 230.81, SS(Form*Speed) = 18.58, and SSE = 71.87.

a. Does there appear to be interaction between the factors?

b. Does yield appear to depend on either formulation or speed?

c. Calculate estimates of the main effects.

d. The fitted values are ${\hat{x}}_{ijk}=\hat{\mu }+{\hat{\alpha }}_i+{\hat{\beta }}_j+{\hat{\gamma }}_{ij}$, and the residuals are $x_{ijk}-{\hat{x}}_{ijk}$. Verify that the residuals are .23, −.87, .63, 4.50, −1.20, −3.30, −2.03, 1.97,.07, −1.10, −.30, 1.40, .67, −1.23, .57, −3.43, −.13,and 3.57.

e. Construct a normal probability plot from the residuals given in part (d). Do the ϵ_ijk’s appear to be normally distributed?

To determine

Identify whether there is a significant effect from interaction between speed and formulation.

Answer to Problem 18E

There is no sufficient evidence to conclude that there is an effect of interaction between mix speed and formulation on the chemical process.

Explanation of Solution

Given info:

An experiment was carried out to test the effect of formulation and speed on the chemical press. The formulation has two levels and speed has three levels.

The sum of squares due to formulation is 2,253.44, sum of squares due to speed is 230.81, sum of squares due to error is 71.87, sum of squares due to interaction between formulation and speed is 18.58.

Calculation:

Testing the Hypothesis:

Null hypothesis:

$H_{0AB}:{\gamma }_{ij}=0$

That is, the interaction effect between mix speed and formulation has no significant effect on chemical process.

Alternative hypothesis:

$H_{aAB}:\text{At least one of }{\gamma }_{ij}\text{'}s\ne 0$

That is, the interaction effect between mix speed and formulation has no significant effect on chemical process.

Test statistic:

Software procedure:

Step by step procedure to find the test statistic using Minitab is given below:

• Choose Stat > ANOVA > General Linear Model.
• In Responses, enter the column of Chemical process.
• In Model, enter the column of Speed, Formulation, Speed*Formulation.
• In Results, choose “Analysis of variance table”.
• Click OK in all dialog boxes.

Output obtained from MINITAB is given below:

Conclusion:

Interaction effect of AB:

The P- value for the interaction effect AB is 0.252 and the level of significance is 0.01.

The P- value is lesser than the level of significance.

That is, $0.252>\alpha =0.01$.

Thus, the null hypothesis is not rejected,

Hence, there is no sufficient evidence to conclude that there is an effect of interaction between mix speed and formulation on the chemical process.

To determine

Identify whether the yield of the chemical process depends on speed or formulation.

Answer to Problem 18E

The yield of a chemical process depends on both speed and formulation.

Explanation of Solution

From the MINITAB output obtained in part (a), the following can be observed.

For Main effect of factor A speed:

The P- value for the factor A (speed) is 0.000 and the level of significance is 0.01.

The P- value is lesser than the level of significance.

That is,.$0.000\left(=P-\text{value}<\alpha =0.01\right)$

Thus, the null hypothesis is rejected,

Hence, there is sufficient evidence to conclude that there is an effect of speed on the yield of chemical process.

For Main effect of factor B formulation:

The P- value for the factor B (formulation) is 0.000 and the level of significance is 0.01.

The P- value is lesser than the level of significance.

That is, $0.000\left(=P-\text{value}<\alpha =0.01\right)$.

Thus, the null hypothesis is rejected,

Hence, there is sufficient evidence to conclude that there is an effect of formulation on the yield of chemical process.

To determine

Find the estimates for the main effects.

Explanation of Solution

Calculation:

$X_{ijk}=\mu +{\alpha }_i+{\beta }_j+{\gamma }_{ij}+{\epsilon }_{ijk}$

Where,

$\mu$ is the overall mean.

${\alpha }_i$ is the main effect from factor A.

${\beta }_j$ is the main effect from factor A.

${\gamma }_{ij}$ is the interaction effect AB.

${\epsilon }_{ijk}$ is the error component.

Overall mean effect:

$\begin{array}{c}\mu =\frac{{\displaystyle \sum _i^{}{\displaystyle \sum _j^{}{\displaystyle \sum _k^{}{X}_{ijk}}}}}{n}\\ =\frac{189.7+185.1+189.0+\mathrm{...}+167.6+161.6+170.3}{18}\\ =\frac{3,156.20}{18}\\ =175.84\end{array}$

Mean due to the first level of factor A:

$\begin{array}{c}{\widehat{\mu }}_{1\cdot }=\frac{1}{3}{\displaystyle \sum _j^3{\overline{x}}_{1j}}\\ =\frac{1}{3}\left({\overline{x}}_{11}+{\overline{x}}_{12}+{\overline{x}}_{13}\right)\\ =\frac{1}{3}\left(187.33+187+186.17\right)\\ =187\end{array}$

Mean due to the second level of factor A:

$\begin{array}{c}{\widehat{\mu }}_{2\cdot }=\frac{1}{3}{\displaystyle \sum _j^3{\overline{x}}_{2j}}\\ =\frac{1}{3}\left({\overline{x}}_{21}+{\overline{x}}_{22}+{\overline{x}}_{23}\right)\\ =\frac{1}{3}\left(163.37+164.1+166.5\right)\\ =164.66\end{array}$

Main effect of factor A:

At first level:

$\begin{array}{c}{\widehat{\alpha }}_1=187-175.84\\ =11.16\end{array}$

At second level:

$\begin{array}{c}{\widehat{\alpha }}_2=164.66-175.84\\ =-11.18\end{array}$

Mean due to the first level of factor B:

$\begin{array}{c}{\widehat{\mu }}_{\cdot 1}=\frac{1}{2}{\displaystyle \sum _i^2{\overline{x}}_{i1}}\\ =\frac{1}{2}\left({\overline{x}}_{11}+{\overline{x}}_{12}\right)\\ =\frac{1}{2}\left(189.47+166.2\right)\\ =177.84\end{array}$

Mean due to the second level of factor B:

$\begin{array}{c}{\widehat{\mu }}_{\cdot 2}=\frac{1}{2}{\displaystyle \sum _i^2{\overline{x}}_{i2}}\\ =\frac{1}{2}\left({\overline{x}}_{12}+{\overline{x}}_{22}\right)\\ =\frac{1}{2}\left(180.6+161.03\right)\\ =170.82\end{array}$

Mean due to the third level of factor B

$\begin{array}{c}{\widehat{\mu }}_{\cdot 3}=\frac{1}{2}{\displaystyle \sum _i^2{\overline{x}}_{i3}}\\ =\frac{1}{2}\left({\overline{x}}_{13}+{\overline{x}}_{23}\right)\\ =\frac{1}{2}\left(191.03+166.73\right)\\ =119.25\end{array}$

Main effect of factor B:

$\begin{array}{c}{\widehat{\beta }}_1=177.84-175.84\\ =2\end{array}$

$\begin{array}{c}{\widehat{\beta }}_2=170.6-175.84\\ =-5.02\end{array}$

$\begin{array}{c}{\widehat{\beta }}_3=178.9-175.84\\ =3.06\end{array}$

To determine

Verify whether the calculated residuals are equal to the given residual values.

Explanation of Solution

Calculation:

${\gamma }_{ij}={\mu }_{ij}-\left(\mu +{\alpha }_i+{\beta }_j\right)$

The interaction effect for the first level of factor A and the first level of factor B

$\begin{array}{c}{\widehat{\gamma }}_{11}={\mu }_{11}-\left(\mu +{\alpha }_1+{\beta }_1\right)\\ =\left[\frac{189.7+188.6+190.1}{3}\right]-\left(175.84+11.16+2\right)\\ =189.47-189\\ =0.47\end{array}$

The interaction effect for the first level of factor A and the second level of factor B

$\begin{array}{c}{\widehat{\gamma }}_{12}={\mu }_{12}-\left(\mu +{\alpha }_1+{\beta }_2\right)\\ =\left[\frac{185.1+179.4+177.3}{3}\right]-\left(175.84+11.16-5.02\right)\\ =180.6-189\\ =-1.40\end{array}$

The interaction effect for the first level of factor A and the third level of factor B

$\begin{array}{c}{\widehat{\gamma }}_{13}={\mu }_{13}-\left(\mu +{\alpha }_1+{\beta }_3\right)\\ =\left[\frac{189.0+193.0+191.1}{3}\right]-\left(175.84+11.16+3.06\right)\\ =191.03-190.06\\ =0.97\end{array}$

Similarly, the remaining values are given below:

S. No	Values
1	${\widehat{\gamma }}_{11}=0.47$
2	${\widehat{\gamma }}_{12}=-1.40$
3	${\widehat{\gamma }}_{13}=0.97$
4	${\widehat{\gamma }}_{21}=-0.45$
5	${\widehat{\gamma }}_{22}=1.39$
6	${\widehat{\gamma }}_{23}=-0.97$

The fitted values are calculated by using the formula:

${\widehat{x}}_{ijk}=\widehat{\mu }+{\widehat{\alpha }}_i+{\widehat{\beta }}_j+{\gamma }_{ij}$

Where,

i represents the levels of factor A.

j represents the levels of factor B.

k represents the observation.

The predicted value when $i=1,j=1,k=1$ is calculated as follows:

$\begin{array}{c}{\widehat{x}}_{111}=175.84+11.16+2+0.47\\ =189.47\end{array}$

$\begin{array}{c}{\widehat{x}}_{122}=175.84+11.16-5.02-1.40\\ =180.6\end{array}$

Similarly, the other fitted values are calculated, the table shows the fitted values:

S. No	1	2	3	4	5	6	7	8	9
Fitted values ${\widehat{x}}_{ijk}$	189.47	189.47	189.47	166.20	166.20	166.20	180.60	180.60	180.60
S. No	10	11	12	13	14	15	16	17	18
Fitted values ${\widehat{x}}_{ijk}$	161.03	161.03	161.03	191.03	191.03	191.03	166.73	166.73	166.73

The residuals values are calculated by using the formula:

${\epsilon }_{ijk}=x_{ijk}-{\widehat{x}}_{ijk}$

The table shows below gives the residuals for each observation:

S. No	Observed $x_{ijk}$	Fitted values ${\widehat{x}}_{ijk}$	${\epsilon }_{ijk}=x_{ijk}-{\widehat{x}}_{ijk}$
1	189.7	189.47	0.23
2	188.6	189.47	–0.87
3	190.1	189.47	0.63
4	165.1	166.20	–1.1
5	165.9	166.20	–0.3
6	167.6	166.20	1.4
7	185.1	180.60	4.5
8	179.4	180.60	–1.2
9	177.3	180.60	–3.3
10	161.7	161.03	0.67
11	159.8	161.03	–1.23
12	161.6	161.03	0.57
13	189	191.03	–2.03
14	193	191.03	1.97
15	191.1	191.03	0.07
16	163.3	166.73	–3.43
17	166.6	166.73	–0.13
18	170.3	166.73	3.57

Hence, the calculated residuals values are equal to given residual values.

To determine

Construct a normal probability plot for the residuals obtained in part (d).

Identify whether the residuals are normally distributed.

Answer to Problem 18E

The normal probability of residuals is given below:

The residuals are normally distributed.

Explanation of Solution

Calculation:

Normal probability plot of residuals:

Software procedure:

Step-by-step procedure to construct a normal probability plot of residuals is given below:

• Click on Graph>Probability plot>Single.
• Click OK.
• Under Graph variables, select the column containing the residual values.
• Click on Distribution and selection Normal.
• Click OK.

Interpretation:

The normal probability plot of residuals suggests that the residuals are normally distributed because the residuals fall approximately on a straight line.