Factorial - Four Way ANOVA - How to find a statistically effective combination - r

Question

I have to analyze an experiment data set to find a most effective combination of a molecular biology reaction.
The experiment has four factors: Temperature, RPM, Time, Catalytic activity. And I am measuring the Efficiency of a reaction (EE). How can I find an effective combination of four factors for the highest efficiency(EE)?
No repeated measurements. All data are independent experimental data
As I understood - EE is parametric data, factors are categorical data (Fixed combinations).
Do I have to go for a Fourway ANOVA?
if so is this model correct for the analysis
library(lsmeans)
lm(EE ~ Temperature + RPM + Time+ Catalytic +
Temperature:RPM +
Temperature:Time +
Temperature:Catalytic +
RPM:Time+
RPM+Catalytic+
Time+Catalytic+
Temperature:RPM:Time +
Temperature:RPM:Catalytic+
Temperature:Time:Catalytic+
RPM:Time:Catalytic+
Temperature:RPM:Time:Catalytic, "data")
And, then how can I get the significant values for each pairwise comparison?
Here is the sample data set for an example.
> dput(df)
structure(list(TEMPERATURE = c(40, 40, 40, 40, 40, 40, 40, 40,
40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40,
40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40,
42.5, 42.5, 42.5, 42.5, 42.5, 42.5, 42.5, 42.5, 42.5, 42.5, 42.5,
42.5, 42.5, 42.5, 42.5, 42.5, 42.5, 42.5, 42.5, 42.5, 42.5, 42.5,
42.5, 42.5, 42.5, 42.5, 42.5, 42.5, 42.5, 42.5, 42.5, 42.5, 42.5,
42.5, 42.5, 42.5, 42.5, 42.5, 42.5, 42.5, 45, 45, 45, 45, 45,
45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45,
45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45,
45, 45, 45), RPM = c(150, 150, 150, 150, 150, 150, 150, 150,
150, 150, 150, 150, 150, 150, 150, 150, 150, 150, 150, 150, 200,
200, 200, 200, 200, 200, 200, 200, 200, 200, 200, 200, 200, 200,
200, 200, 200, 200, 200, 200, 150, 150, 150, 150, 150, 150, 150,
150, 150, 150, 150, 150, 150, 150, 150, 150, 150, 150, 150, 150,
200, 200, 200, 200, 200, 200, 200, 200, 200, 200, 200, 200, 200,
200, 200, 200, 200, 200, 200, 200, 150, 150, 150, 150, 150, 150,
150, 150, 150, 150, 150, 150, 150, 150, 150, 150, 150, 150, 150,
150, 200, 200, 200, 200, 200, 200, 200, 200, 200, 200, 200, 200,
200, 200, 200, 200, 200, 200, 200, 200), TIME = c(24, 24, 24,
24, 24, 48, 48, 48, 48, 48, 72, 72, 72, 72, 72, 96, 96, 96, 96,
96, 24, 24, 24, 24, 24, 48, 48, 48, 48, 48, 72, 72, 72, 72, 72,
96, 96, 96, 96, 96, 24, 24, 24, 24, 24, 48, 48, 48, 48, 48, 72,
72, 72, 72, 72, 96, 96, 96, 96, 96, 24, 24, 24, 24, 24, 48, 48,
48, 48, 48, 72, 72, 72, 72, 72, 96, 96, 96, 96, 96, 24, 24, 24,
24, 24, 48, 48, 48, 48, 48, 72, 72, 72, 72, 72, 96, 96, 96, 96,
96, 24, 24, 24, 24, 24, 48, 48, 48, 48, 48, 72, 72, 72, 72, 72,
96, 96, 96, 96, 96), CAT = c(4, 6, 8, 10, 12, 4, 6, 8, 10, 12,
4, 6, 8, 10, 12, 4, 6, 8, 10, 12, 4, 6, 8, 10, 12, 4, 6, 8, 10,
12, 4, 6, 8, 10, 12, 4, 6, 8, 10, 12, 4, 6, 8, 10, 12, 4, 6,
8, 10, 12, 4, 6, 8, 10, 12, 4, 6, 8, 10, 12, 4, 6, 8, 10, 12,
4, 6, 8, 10, 12, 4, 6, 8, 10, 12, 4, 6, 8, 10, 12, 4, 6, 8, 10,
12, 4, 6, 8, 10, 12, 4, 6, 8, 10, 12, 4, 6, 8, 10, 12, 4, 6,
8, 10, 12, 4, 6, 8, 10, 12, 4, 6, 8, 10, 12, 4, 6, 8, 10, 12),
EE = c(50, 53, 54, 57, 59, 53, 56, 59, 61, 64, 57, 58, 60,
62, 63, 56, 54, 52, 55, 55, 44, 48, 50, 50, 54, 49, 52, 56,
57, 56, 52, 56, 57, 58, 66, 46, 48, 48, 52, 49, 53, 57, 59,
62, 64, 54, 58, 60, 64, 66, 55, 59, 61, 63, 65, 54, 59, 64,
65, 67, 49, 51, 53, 54, 59, 50, 54, 63, 64, 64, 52, 56, 56,
59, 57, 52, 55, 58, 60, 63, 52, 56, 58, 61, 63, 54, 55, 58,
63, 63, 56, 58, 62, 62, 65, 57, 59, 62, 63, 66, 42, 42, 51,
54, 56, 46, 50, 52, 56, 58, 48, 51, 54, 55, 57, 48, 53, 56,
57, 61)), class = c("spec_tbl_df", "tbl_df", "tbl", "data.frame"
), row.names = c(NA, -120L), spec = structure(list(cols = list(
TEMPERATURE = structure(list(), class = c("collector_double",
"collector")), RPM = structure(list(), class = c("collector_double",
"collector")), TIME = structure(list(), class = c("collector_double",
"collector")), CAT = structure(list(), class = c("collector_double",
"collector")), EE = structure(list(), class = c("collector_double",
"collector"))), default = structure(list(), class = c("collector_guess",
"collector")), skip = 1), class = "col_spec"))

Answer 1

Try something like this:
library(rsm)
mod = rsm(EE ~ SO(Temperature, RPM, Time, Catalytic), data = data)
summary(mod)
This will fit a second-order surface (model equation includes all predictors, two-way interactions, and squares). The summary shows a stationary point and related statistics. If all the eigenvalues are negative, then it is a peak. Otherwise you have some kind of saddle point.
This model differs from the one in the OP in that it has no three-way or four-way interactions, but includes the squares of the predictors, which are really essential for fitting a second-order response surface.
Lots more details
I had to revise this a bit to account for the fact that R is case-sensitive!
> mod = rsm(EE ~ SO(TEMPERATURE, RPM, TIME, CAT), data = df)
There is an issue here in that RPM has only two values, so we can't estimate a pure quadratic effect. So there is one coefficient of NA and that messes up the computation of the stationary point. However, we can still plot the fitted surface (despite a few warning messages)
> par(mfrow = c(2,3))
> contour(mod, ~TEMPERATURE+RPM+TIME+CAT)
It looks like we are best off with large CAT and lower RPM (see that plot), so look again:
> par(mfrow=c(1,1))
> contour(mod, ~ TEMPERATURE + TIME, at = list(CAT = 12, RPM = 150))
So visually, we seem to get the best response at around temperature 43.5, time 65, catalyst 12, and rpm 150.
If you insist on modeling these as factors, it can be done, but you need to convert all the predictors to factors. This is a common error; you can have a designed experiment with only a few distinct values of a quantitative variable, but R does not read your mind and assume it's a factor; you have to convert it to one. In the following I have opted to fit a model with up to 2-way interactions.
> facmod = lm(EE ~ (factor(TEMPERATURE) + factor(TIME) + factor(RPM) + factor(CAT))^2, data = df)
> library(emmeans)
> emmip(facmod, TIME ~ TEMPERATURE | CAT*RPM)
The highest fitted response is at catalyst 12, RPM 150, temperature 42.5 or larger, and time 96. It is clear that 150 RPM is better (left vs. right comoparisons) and the high CAT is better (comparing panels vertically). These are different models and somewhat different results. I like the rsm approach better as it is more systematic.

Answer 2

For a screening DOE you collected more data than what was needed.
Here is a starting point, I welcome additional comments.
I would model the linear combination of all of your factors:
model <-lm(EE ~ TEMPERATURE + RPM + TIME +CAT , data=df)
summary(model)
Call:
lm(formula = EE ~ TEMPERATURE + RPM + TIME + CAT, data = df)
Residuals:
Min 1Q Median 3Q Max
-10.1850 -1.5742 0.3383 1.7767 9.7033
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 51.50833 6.47932 7.950 1.42e-12 ***
TEMPERATURE 0.27000 0.14245 1.895 0.06056 .
RPM -0.10533 0.01163 -9.056 4.10e-15 ***
TIME 0.03639 0.01084 3.358 0.00107 **
CAT 1.20417 0.10281 11.713 < 2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 3.185 on 115 degrees of freedom
Multiple R-squared: 0.6706, Adjusted R-squared: 0.6591
F-statistic: 58.52 on 4 and 115 DF, p-value: < 2.2e-16
Looking at the signs for the slope estimates. While this simple model is assuming a linear relationship, the physical significance of the slope is comparing the average of low values with average of the high values.
For example the the slope of Temperature term is positive. This means when the temperature increases form the low value (40C) to the high value (45C)the Efficiency will increase.
Temperature, TIME and CAT as positive slopes I would take the largest value available.
RPM had a negative slope so I would choose the lowest value available.
Thus my prediction from the experiments my prediction would yield EE=66,
While the highest result in from the experiment was:
df[which.max(df$EE),]
# A tibble: 1 x 5
TEMPERATURE RPM TIME CAT EE
<dbl> <dbl> <dbl> <dbl> <dbl>
1 42.5 150 96 12 67
Now you could investigate the non-linear relationships by looking at the results from this model:
Call:
lm(formula = EE ~ TEMPERATURE * RPM * TIME * CAT, data = df)
Here the slopes of the interaction terms are orders of magnitudes smaller than the linear terms. This could be misleading is the variables are not normalized.
Good luck.