Chapter 1.8, Problem 17E

a.

To determine

Plot the interaction plot with pH being the trace factor.

Explanation of Solution

The interaction plot with pH being the trace factor is given as follows:

b.

To determine

Check whether there is interaction between the factors pH and temperature.

Answer to Problem 17E

There is interaction between the factors pH and temperature.

Explanation of Solution

From the interaction plot obtained in Part (a), it is clear that the lines of pH I and pH II are not parallel to each other. Therefore, there is interaction between the factors pH and temperature.

c.

To determine

Calculate the main pH effects and the main temperature effects.

Answer to Problem 17E

The main pH effects are ${\alpha }_I=-0.125\text{ and }{\alpha }_{II}=0.125$.

The temperature effects are ${\beta }_A=6.375,{\beta }_B\text{=0}\text{.375,}{\beta }_C=-\text{2}\text{.625, and }{\beta }_D=-4.125$.

Explanation of Solution

The overall mean is computed as follows:

$\begin{array}{c}\overline{\mu }\mathrm{..}=\frac{108+103+101+100+111+104+100+98}{8}\\ =103.125\end{array}$

The main pH effects are calculated as follows:

$\begin{array}{c}{\alpha }_I={\overline{\mu }}_{I.}-\overline{\mu }\mathrm{..}\\ =\frac{108+103+101+100}{4}-103.125\\ =-0.125\\ {\alpha }_{II}={\overline{\mu }}_{II.}-\mu \mathrm{..}\\ =\frac{111+104+100+98}{4}-103.125\\ =0.125\end{array}$

Therefore, the main pH effects are ${\alpha }_I=-0.125\text{ and }{\alpha }_{II}=0.125$.

The main temperature effects are calculated as follows:

$\begin{array}{c}{\beta }_A={\overline{\mu }}_{A.}-\overline{\mu }\mathrm{..}\\ =\frac{108+111}{2}-103.125\\ =6.375\\ {\beta }_B={\overline{\mu }}_{B.}-\mu \mathrm{..}\\ =\frac{103+104}{2}-103.125\\ =0.375\\ {\beta }_C={\overline{\mu }}_{C.}-\mu \mathrm{..}\\ =\frac{101+100}{2}-103.125\\ =-2.625\\ {\beta }_D={\overline{\mu }}_{D.}-\mu \mathrm{..}\\ =\frac{100+98}{2}-103.125\\ =-4.125\end{array}$

Therefore, the main Pygmalion effects are ${\beta }_A=6.375,{\beta }_B\text{=0}\text{.375,}{\beta }_C=-\text{2}\text{.625, and }{\beta }_D=-4.125$.

d.

To determine

Calculate the interaction effects.

Answer to Problem 17E

The interaction effects are ${\gamma }_{IA}=-1.375,{\gamma }_{IB}=-0.375,{\gamma }_{IC}=0.625,{\gamma }_{ID}=1.125,{\gamma }_{IIA}=1.375,{\gamma }_{IIB}=0.375,{\gamma }_{IIC}=-0.625,$$\text{and }{\gamma }_{IID}=-1.125$.

Explanation of Solution

The interaction effects are calculated as follows:

$\begin{array}{c}{\gamma }_{IA}={\mu }_{IA}-\left(\overline{\mu }\mathrm{..}+{\alpha }_I+{\beta }_A\right)\\ =108-\left(103.125-0.125+6.375\right)\\ =-1.375\\ {\gamma }_{IB}={\mu }_{IB}-\left(\overline{\mu }\mathrm{..}+{\alpha }_I+{\beta }_B\right)\\ =103-\left(103.125-0.125+0.375\right)\\ =-0.375\\ {\gamma }_{IC}={\mu }_{IC}-\left(\overline{\mu }\mathrm{..}+{\alpha }_I+{\beta }_C\right)\\ =101-\left(103.125-0.125-2.625\right)\\ =0.625\\ {\gamma }_{ID}={\mu }_{ID}-\left(\overline{\mu }\mathrm{..}+{\alpha }_I+{\beta }_D\right)\\ =100-\left(103.125-0.125-4.125\right)\\ =1.125\\ {\gamma }_{IIA}={\mu }_{IIA}-\left(\overline{\mu }\mathrm{..}+{\alpha }_{II}+{\beta }_A\right)\\ =111-\left(103.125+0.125+6.375\right)\\ =1.375\\ {\gamma }_{IIB}={\mu }_{IIB}-\left(\overline{\mu }\mathrm{..}+{\alpha }_{II}+{\beta }_B\right)\\ =104-\left(103.125+0.125+0.375\right)\\ =0.375\\ {\gamma }_{IIC}={\mu }_{IIC}-\left(\overline{\mu }\mathrm{..}+{\alpha }_{II}+{\beta }_C\right)\\ =100-\left(103.125+0.125-2.625\right)\\ =-0.625\\ {\gamma }_{IID}={\mu }_{IID}-\left(\overline{\mu }\mathrm{..}+{\alpha }_{II}+{\beta }_D\right)\\ =98-\left(103.125+0.125-4.125\right)\\ =-1.125\end{array}$

Therefore, the interaction effects are ${\gamma }_{IA}=-1.375,{\gamma }_{IB}=-0.375,{\gamma }_{IC}=0.625,{\gamma }_{ID}=1.125,$${\gamma }_{IIA}=1.375,{\gamma }_{IIB}=0.375,{\gamma }_{IIC}=-0.625,\text{and }{\gamma }_{IID}=-1.125$.