Chapter 9.5, Problem 4E

The article “Efficient Pyruvate Production by a Multi-Vitamin Auxotroph of Torulopsis glabrata: Key Role and Optimization of Vitamin Levels” (Y. Li. J. Chen, ct al. Applied Microbiology and Biotechnology, 200l:680–685) investigates the effects of the levels of several vitamins in a cell culture on the yield (in g/L) of pyruvate, a useful organic acid. The data in the following table are presented as two replicates of a 2³ design. The factors are A: nicotinic acid. B: thiamine, and C: biotin. (Two statistically insignificant factors have been dropped. In the article, each factor was tested at four levels: we have collapsed these to two.)

1. a. Compute estimates of the main effects and interactions, and the sum of squares and P-value for each.
2. b. Is the additive model appropriate?
3. c. What conclusions about the factors can be drawn from these results?

To determine

Obtain the estimates of the main effects and interaction, and sum of squares and P- value for the each treatment combination.

Answer to Problem 4E

The estimates of the main effects and interaction, and sum of squares and P- value for the each treatment combination are given below:

ANOVA table:

Variable	Effect	DF	Sum of squares	Mean Square	F	P
A	–0.04625	1	0.008556	0.008556	1.57	0.245
B	–0.19875	1	0.158006	0.158006	29.03	0.001
C	0.01625	1	0.001056	0.001056	0.19	0.671
AB	–0.02375	1	0.002256	0.002256	0.41	0.538
AC	–0.02375	1	0.002256	0.002256	0.41	0.538
BC	0.00875	1	0.000306	0.000306	0.06	0.818
ABC	–0.01125	1	0.000506	0.000506	0.09	0.768
Error		8	0.043550	0.005444
Total		15

Explanation of Solution

Given info:

The information is based on conducting the experiment on cell culture on the yield of pyruvate which has two levels for the factors of nicotinic acid (A), thiamine (B), and biotin (C). The experiment of chemical reaction has been conducted twice.

Calculation:

Denote the average yield of the factor A as ${\overline{X}}_A,{\overline{X}}_{AB},{\overline{X}}_{AC},{\overline{X}}_{ABC}$, average yield of the factor B as ${\overline{X}}_B,{\overline{X}}_{AB},{\overline{X}}_{BC},{\overline{X}}_{ABC}$ and average yield of the factor C as ${\overline{X}}_C,{\overline{X}}_{BC},{\overline{X}}_{AC},{\overline{X}}_{ABC}$

Factor A denotes the nicotinic, Factor B denotes the thiamine and Factor C denotes the biotin.

The effect estimate of the treatments is:

$\begin{array}{c}A\text{effect}=\frac{1}{4}\left(-{\overline{X}}_1+{\overline{X}}_A-{\overline{X}}_B+{\overline{X}}_{AB}-{\overline{X}}_C+{\overline{X}}_{AC}-{\overline{X}}_{BC}+{\overline{X}}_{ABC}\right)\\ B\text{effect}=\frac{1}{4}\left(-{\overline{X}}_1-{\overline{X}}_A+{\overline{X}}_B+{\overline{X}}_{AB}-{\overline{X}}_C-{\overline{X}}_{AC}+{\overline{X}}_{BC}+{\overline{X}}_{ABC}\right)\\ C\text{effect}=\frac{1}{4}\left(-{\overline{X}}_1-{\overline{X}}_A-{\overline{X}}_B-{\overline{X}}_{AB}+{\overline{X}}_C+{\overline{X}}_{AC}+{\overline{X}}_{BC}+{\overline{X}}_{ABC}\right)\\ AB\text{effect}=\frac{1}{4}\left({\overline{X}}_1-{\overline{X}}_A-{\overline{X}}_B+{\overline{X}}_{AB}+{\overline{X}}_C-{\overline{X}}_{AC}-{\overline{X}}_{BC}+{\overline{X}}_{ABC}\right)\\ BC\text{effect}=\frac{1}{4}\left({\overline{X}}_1+{\overline{X}}_A-{\overline{X}}_B-{\overline{X}}_{AB}-{\overline{X}}_C-{\overline{X}}_{AC}+{\overline{X}}_{BC}+{\overline{X}}_{ABC}\right)\\ AC\text{effect}=\frac{1}{4}\left({\overline{X}}_1-{\overline{X}}_A+{\overline{X}}_B-{\overline{X}}_{AB}-{\overline{X}}_C+{\overline{X}}_{AC}-{\overline{X}}_{BC}+{\overline{X}}_{ABC}\right)\\ ABC\text{effect}=\frac{1}{4}\left(-{\overline{X}}_1+{\overline{X}}_A+{\overline{X}}_B-{\overline{X}}_{AB}+{\overline{X}}_C-{\overline{X}}_{AC}-{\overline{X}}_{BC}+{\overline{X}}_{ABC}\right)\end{array}$

The effect estimate of treatment A is:

Substitute the corresponding values of mean yield,

$\begin{array}{c}A\text{effect}=\frac{1}{4}\left(-{\overline{X}}_1+{\overline{X}}_A-{\overline{X}}_B+{\overline{X}}_{AB}-{\overline{X}}_C+{\overline{X}}_{AC}-{\overline{X}}_{BC}+{\overline{X}}_{ABC}\right)\\ =\frac{1}{4}\left(-0.520+0.510-0.325+0.290-0.540+0.505-0.385+0.280\right)\\ =\frac{1}{4}\left(-0.185\right)\\ =-0.04625\end{array}$

The effect estimate of treatment B is:

$\begin{array}{c}B\text{effect}=\frac{1}{4}\left(-{\overline{X}}_1-{\overline{X}}_A+{\overline{X}}_B+{\overline{X}}_{AB}-{\overline{X}}_C-{\overline{X}}_{AC}+{\overline{X}}_{BC}+{\overline{X}}_{ABC}\right)\\ =\frac{1}{4}\left(-0.520-0.510+0.325+0.290-0.540-0.505+0.385+0.280\right)\\ =\frac{1}{4}\left(-0.795\right)\\ =-0.19875\end{array}$

The effect estimate of treatment C is:

$\begin{array}{c}C\text{effect}=\frac{1}{4}\left(-{\overline{X}}_1-{\overline{X}}_A-{\overline{X}}_B-{\overline{X}}_{AB}+{\overline{X}}_C+{\overline{X}}_{AC}+{\overline{X}}_{BC}+{\overline{X}}_{ABC}\right)\\ =\frac{1}{4}\left(-0.520-0.510-0.325-0.290+0.540+0.505+0.385+0.280\right)\\ =\frac{1}{4}\left(0.065\right)\\ =0.01625\end{array}$

The effect estimate of treatment AB is:

$\begin{array}{c}AB\text{effect}=\frac{1}{4}\left({\overline{X}}_1-{\overline{X}}_A-{\overline{X}}_B+{\overline{X}}_{AB}+{\overline{X}}_C-{\overline{X}}_{AC}-{\overline{X}}_{BC}+{\overline{X}}_{ABC}\right)\\ =\frac{1}{4}\left(0.520-0.510-0.325+0.290+0.540-0.505-0.385+0.280\right)\\ =\frac{1}{4}\left(-0.095\right)\\ =-0.02375\end{array}$

The effect estimate of treatment AC is:

$\begin{array}{c}AC\text{effect}=\frac{1}{4}\left({\overline{X}}_1-{\overline{X}}_A+{\overline{X}}_B-{\overline{X}}_{AB}-{\overline{X}}_C+{\overline{X}}_{AC}-{\overline{X}}_{BC}+{\overline{X}}_{ABC}\right)\\ =\frac{1}{4}\left(0.520-0.510+0.325-0.290-0.540+0.505-0.385+0.280\right)\\ =\frac{1}{4}\left(-0.095\right)\\ =-0.02375\end{array}$

The effect estimate of treatment BC is:

$\begin{array}{c}BC\text{effect}=\frac{1}{4}\left({\overline{X}}_1+{\overline{X}}_A-{\overline{X}}_B-{\overline{X}}_{AB}-{\overline{X}}_C-{\overline{X}}_{AC}+{\overline{X}}_{BC}+{\overline{X}}_{ABC}\right)\\ =\frac{1}{4}\left(0.520+0.510-0.325-0.290-0.540-0.505+0.385+0.280\right)\\ =\frac{1}{4}\left(0.035\right)\\ =0.00875\end{array}$

The effect estimate of treatment ABC is:

$\begin{array}{c}ABC\text{effect}=\frac{1}{4}\left(-{\overline{X}}_1+{\overline{X}}_A+{\overline{X}}_B-{\overline{X}}_{AB}+{\overline{X}}_C-{\overline{X}}_{AC}-{\overline{X}}_{BC}+{\overline{X}}_{ABC}\right)\\ =\frac{1}{4}\left(0.520+0.510+0.325-0.290+0.540-0.505-0.385+0.280\right)\\ =\frac{1}{4}\left(-0.045\right)\\ =-0.01125\end{array}$

The estimates of the main effects and interaction, and sum of squares and P- value for the each treatment combination are given below:

Step-by-step procedure for finding the factorial design table is as follows:

Software procedure:

• Choose Stat > DOE > Factorial > Create Factorial Design.
• Under Type of Design, choose General full factorial design.
• From Number of factors, choose 3.
• Click Designs.
• In Factor A, type A under Name and type 2 Under Number of Levels.
• In Factor B, type B under Name and type 2 Under Number of Levels.
• . In Factor C, type C under Name and type 2 Under Number of Levels.
•  From Number of replicates, choose 2.
•  Click OK.
• Select Summary table under Results.
• Click OK.
• Enter the corresponding Yield in the newly created factorial design worksheet based on the levels of each factor.

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

• Choose Stat > DOE > Factorial > Analyze Factorial Design.
• In Response, enter Yield.
• In Terms, select all the terms.
• In Results, choose “Model summary and ANOVA table”.
• Click OK in all the dialog boxes.

Output obtained by MINITAB procedure is as follows:

The sum of squares and P- values has been obtained for each treatment combination by using MINITAB.

Thus, the estimates of the main effects and interaction, and sum of squares and P- value for the each treatment combination are given below:

ANOVA table:

Variable	Effect	DF	Sum of squares	Mean Square	F	P
A	–0.04625	1	0.008556	0.008556	1.57	0.245
B	–0.19875	1	0.158006	0.158006	29.03	0.001
C	0.01625	1	0.001056	0.001056	0.19	0.671
AB	–0.02375	1	0.002256	0.002256	0.41	0.538
AC	–0.02375	1	0.002256	0.002256	0.41	0.538
BC	0.00875	1	0.000306	0.000306	0.06	0.818
ABC	–0.01125	1	0.000506	0.000506	0.09	0.768
Error		8	0.043550	0.005444
Total		15

To determine

Check whether the additive model is appropriate.

Answer to Problem 4E

Yes, the additive model is appropriate .

Explanation of Solution

Justification:

Principle rule to hold an additive model:

The additive model is acceptable when the interactions are small.

Hence the additive model is appropriate but interaction effect obtained in previous part (a) does not provide significant results.

To determine

State whether any main effects and interaction are important.

Answer to Problem 4E

There is sufficient evidence to conclude that there is no significant difference between the means of two levels in main effect A at $\alpha =0.05$ level of significance.

There is sufficient evidence to conclude that there is significant difference between the means of two levels in main effect B at $\alpha =0.05$ level of significance.

There is sufficient evidence to conclude that there is no significant difference between the means of two levels in main effect C at $\alpha =0.05$ level of significance.

There is sufficient evidence to conclude that the interaction is not significant at $\alpha =0.05$ level of significance.

Explanation of Solution

Calculation:

The testing of hypotheses is as follows:

State the hypotheses:

Main factor A:

Null hypothesis:

$H_0:$ There is no significant difference between the means of two levels in factor A.

Alternative hypothesis:

$H_a:$ There is significant difference between the means of two levels in factor A.

Main factor B:

Null hypothesis:

$H_0:$ There is no significant difference between the means of two levels in factor B.

Alternative hypothesis:

$H_a:$ There is significant difference between the means of two levels in factor B.

Interaction AB:

Null hypothesis:

$H_0:$ There is no interaction between the factor A and factor B.

Alternative hypothesis:

$H_a:$ There is interaction between the factor A and factor B.

Interaction AC:

Null hypothesis:

$H_0:$ There is no interaction between the factor A and factor C.

Alternative hypothesis:

$H_a:$ There is interaction between the factor A and factor C.

Interaction BC:

Null hypothesis:

$H_0:$ There is no interaction between the factor B and factor C.

Alternative hypothesis:

$H_a:$ There is interaction between the factor B and factor C.

Interaction ABC:

Null hypothesis:

$H_0:$ There is no interaction between the factor A, factor B and factor C.

Alternative hypothesis:

$H_a:$ There is interaction between the factor A, factor B and factor C.

Assume that the level of significance as 0.05.

From the MINITAB output obtained in previous part (a), the P- value for main effects and interaction are given below:

Treatment	P
A	0.245
B	0.001
C	0.671
AB	0.538
AC	0.538
BC	0.818
ABC	0.768

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 greater than the level of significance.

That is, $P\text{-value}\left(=0.245\right)>\alpha \left(=0.05\right)$.

By rejection rule, fail to reject the null hypothesis.

Thus, there is sufficient evidence to conclude that there is no significant difference between the means of two levels in main effect A at $\alpha =0.05$ level of significance.

Factor B:

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

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

By rejection rule, reject the null hypothesis.

Thus, there is sufficient evidence to conclude that there is significant difference between the means of two levels in main effect B at $\alpha =0.05$ level of significance.

Factor C:

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

That is, $P\text{-value}\left(=0.671\right)>\alpha \left(=0.05\right)$.

By rejection rule, fail to reject the null hypothesis.

Thus, there is sufficient evidence to conclude that there is no significant difference between the means of two levels in main effect C at $\alpha =0.05$ level of significance.

Interaction AB:

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

That is, $P\text{-value}\left(=0.538\right)>\alpha \left(=0.05\right)$.

By rejection rule, fail to reject the null hypothesis.

Thus, there is sufficient evidence to conclude that the interaction AB is not significant at $\alpha =0.05$ level of significance.

Interaction AC:

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

That is, $P\text{-value}\left(=0.538\right)>\alpha \left(=0.05\right)$.

By rejection rule, fail to reject the null hypothesis.

Thus, there is sufficient evidence to conclude that the interaction AC is not significant at $\alpha =0.05$ level of significance.

Interaction BC:

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

That is, $P\text{-value}\left(=0.818\right)>\alpha \left(=0.05\right)$.

By rejection rule, fail to reject the null hypothesis.

Thus, there is sufficient evidence to conclude that the interaction BC is not significant at $\alpha =0.05$ level of significance.

Interaction ABC:

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

That is, $P\text{-value}\left(=0.768\right)>\alpha \left(=0.05\right)$.

By rejection rule, fail to reject the null hypothesis.

Thus, there is sufficient evidence to conclude that the interaction ABC is not significant at $\alpha =0.05$ level of significance.

Hence the P-value of Factor B states that it has more effect on the yield than Factor A and Factor C.