Design and analysis of experiments (2014)
| Class date(s): | 27 February to 26 March 2014 | ||||
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Course notes and slides
- Course textbook (print chapter 5)
- Slides for class
| Date | Class number | Video and audio files | Other materials | Reading (PID) | Slides | |
|---|---|---|---|---|---|---|
| 27 February | 07C | Video (227 M) | Audio (40 M) | None | Chapter 5 | Slides for class
|
| 03 March | 08A | Video (299 M) | Audio (40 M) | None | ||
| 05 March | 08B | Video (296 M) | Audio (42 M) | None | ||
| 06 March | 08C | Video (286 M) | Audio (42 M) | See the code and derivations below | ||
| 10 March | 09A | Video (311 M) | Audio (42 M) | |||
| 13 March | 09B | Video (296 M) | Audio (42 M) | |||
Software source code
Class 08C, 06 March
Source code to estimate the DOE model
T <- c(-1, +1, -1, +1) # centered and scaled temperature
S <- c(-1, -1, +1, +1) # centered and scaled substrate concentration
y <- c(69, 60, 64, 53) # conversion is the response, y
mod <- lm(y ~ T + S + T * S)
summary(mod)
Call:
lm(formula = y ~ T + S + T * S)
Residuals:
ALL 4 residuals are 0: no residual degrees of freedom!
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 61.5 NA NA NA
T -5.0 NA NA NA
S -3.0 NA NA NA
T:S -0.5 NA NA NA
Residual standard error: NaN on 0 degrees of freedom
Multiple R-squared: 1, Adjusted R-squared: NaN
F-statistic: NaN on 3 and 0 DF, p-value: NA
Class 08C, 06 March
3-factor example <rst> <rst-options: 'toc' = False/> The data are from a plastics molding factory which must treat its waste before discharge. The :math:`y` variable represents the average amount of pollutant discharged [lb per day], while the 3 factors that were varied were:
- :math:`C`: the chemical compound added [A or B]
- :math:`T`: the treatment temperature [72°F or 100°F] - :math:`S`: the stirring speed [200 rpm or 400 rpm] - :math:`y`: the amount of pollutant discharged [lb per day]
.. tabularcolumns:: |l|l||c|c|c||c|
+-----------+-------+---------------+-----------------+-----------------+-----------------+ | Experiment| Order | :math:`C` | :math:`T` [°F] | :math:`S` [rpm] | :math:`y` [lb] | +===========+=======+===============+=================+=================+=================+ | 1 | 5 | A | 72 | 200 | 5 | +-----------+-------+---------------+-----------------+-----------------+-----------------+ | 2 | 6 | B | 72 | 200 | 30 | +-----------+-------+---------------+-----------------+-----------------+-----------------+ | 3 | 1 | A | 100 | 200 | 6 | +-----------+-------+---------------+-----------------+-----------------+-----------------+ | 4 | 4 | B | 100 | 200 | 33 | +-----------+-------+---------------+-----------------+-----------------+-----------------+ | 5 | 2 | A | 72 | 400 | 4 | +-----------+-------+---------------+-----------------+-----------------+-----------------+ | 6 | 7 | B | 72 | 400 | 3 | +-----------+-------+---------------+-----------------+-----------------+-----------------+ | 7 | 3 | A | 100 | 400 | 5 | +-----------+-------+---------------+-----------------+-----------------+-----------------+ | 8 | 8 | B | 100 | 400 | 4 | +-----------+-------+---------------+-----------------+-----------------+-----------------+
We showed the cube plot for this system in the class on 10 March. From the cube plot we could already see the main factors, and even the **CS** interaction was noticeable.
- **C effect**: There are 4 estimates of :math:`C = \displaystyle \frac{(+25) + (+27) + (-1) + (-1)}{4} = \frac{50}{4} = \bf{12.5}`
- **T effect**: There are 4 estimates of :math:`T = \displaystyle \frac{(+1) + (+3) + (+1) + (+1)}{4} = \frac{6}{4} = \bf{1.5}`
- **S effect**: There are 4 estimates of :math:`S = \displaystyle \frac{(-27) + (-1) + (-29) + (-1)}{4} = \frac{-58}{4} = \bf{-14.5}`
- **CT interaction**: There are 2 estimates of :math:`CT`. Recall that interactions are calculated as the half difference going from high to low. Consider the change in :math:`C` when
- :math:`T_\text{high}` (at :math:`S` high) = 4 - 5 = -1 - :math:`T_\text{low}` (at :math:`S` high) = 3 - 4 = -1 - First estimate = [(-1) - (-1)]/2 = 0 - :math:`T_\text{high}` (at :math:`S` low) = 33 - 6 = +27 - :math:`T_\text{low}` (at :math:`S` low) = 30 - 5 = +25 - Second estimate = [(+27) - (+25)]/2 = +1
- Average **CT** interaction = (0 + 1)/2 = **0.5** - You can interchange :math:`C` and :math:`T` and still get the same result.
- **CS interaction**: There are 2 estimates of :math:`CS`. Consider the change in :math:`C` when
- :math:`S_\text{high}` (at :math:`T` high) = 4 - 5 = -1 - :math:`S_\text{low}` (at :math:`T` high) = 33 - 6 = +27 - First estimate = [(-1) - (+27)]/2 = -14 - :math:`S_\text{high}` (at :math:`T` low) = 3 - 4 = -1 - :math:`S_\text{low}` (at :math:`T` low) = 30 - 5 = +25 - Second estimate = [(-1) - (+25)]/2 = -13
- Average **CS** interaction = (-13 - 14)/2 = **-13.5** - You can interchange :math:`C` and :math:`S` and still get the same result.
- **ST interaction**: There are 2 estimates of :math:`ST`: (-1 + 0)/2 = **-0.5**, calculate in the same way as above.
- **CTS interaction**: There is only a single estimate of :math:`CTS`:
- :math:`CT` effect at high :math:`S` = 0 - :math:`CT` effect at low :math:`S` = +1 - :math:`CTS` interaction = [(0) - (+1)] / 2 = **-0.5**
- You can calculate this also by considering the :math:`CS` effect at the two levels of :math:`T` - Or, you can calculate this by considering the :math:`ST` effect at the two levels of :math:`C`. - All 3 approaches give the same result.
Next, use computer software to verify that
.. math::
y = 11.25 + 6.25x_C + 0.75x_T -7.25x_S + 0.25 x_C x_T -6.75 x_C x_S -0.25 x_T x_S - 0.25 x_C x_T x_S
The :math:`\mathbf{X}` matrix and :math:`\mathbf{y}` vector used to calculate the least squares model:
.. math::
\begin{bmatrix} 5\\30\\6\\33\\4\\3\\5\\4 \end{bmatrix} &= \begin{bmatrix} +1 & -1 & -1 & -1 & +1 & +1 & +1 & -1\\ +1 & +1 & -1 & -1 & -1 & -1 & +1 & +1\\ +1 & -1 & +1 & -1 & -1 & +1 & -1 & +1\\ +1 & +1 & +1 & -1 & +1 & -1 & -1 & -1\\ +1 & -1 & -1 & +1 & +1 & -1 & -1 & +1\\ +1 & +1 & -1 & +1 & -1 & +1 & -1 & -1\\ +1 & -1 & +1 & +1 & -1 & -1 & +1 & -1\\ +1 & +1 & +1 & +1 & +1 & +1 & +1 & +1\\ \end{bmatrix} \begin{bmatrix} b_0 \\ b_C \\ b_T \\ b_{S} \\ b_{CT} \\ b_{CS} \\ b_{TS} \\ b_{CTS} \end{bmatrix} \\ \mathbf{y} &= \mathbf{X} \mathbf{b} </rst>
Class 08C, 06 March
# Create the design matrix in a quick way in R
C <- T <- S <- c(-1, +1)
design <- expand.grid(C=C, T=T, S=S)
design
C T S
1 -1 -1 -1
2 1 -1 -1
3 -1 1 -1
4 1 1 -1
5 -1 -1 1
6 1 -1 1
7 -1 1 1
8 1 1 1
C <- design$C
T <- design$T
S <- design$S
y <- c(5, 30, 6, 33, 4, 3, 5, 4)
# Full factorial model (you could make errors typing this in)
mod.full <- lm(y ~ C + T + S + C*T + C*S + S*T + C*T*S)
# This powerful notation will expand all terms up to the 3rd order interactions
mod.full <- lm( y ~ (C + T + S)^3 )
summary(mod.full)
Call:
lm(formula = y ~ C + T + S + C * T + C * S + S * T + C * T * S)
Residuals:
ALL 8 residuals are 0: no residual degrees of freedom!
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 11.25 NA NA NA
C 6.25 NA NA NA
T 0.75 NA NA NA
S -7.25 NA NA NA
C:T 0.25 NA NA NA
C:S -6.75 NA NA NA
T:S -0.25 NA NA NA
C:T:S -0.25 NA NA NA
# Guide to estimating significant coefficients, ignoring the intercept (1st coefficient)
coeff.full <- coef(mod.full)[2:length(coef(mod.full))]
# Pareto plot of the absolute coefficients
library(lattice)
coeff <- sort(abs(coeff.full), index.return=TRUE)
barchart(coeff$x,
xlim=c(0, max(abs(coeff.full))+0.1),
xlab=list("Magnitude of effect", cex=1.5),
ylab = list("Effect", cex=1.5),
groups=(coeff.full>0)[coeff$ix], col=c("lightblue", "orange"),
scales=list(cex=1.5)
)
# Eliminate some factors: use the regular linear model notation
# Eliminate C*T*S, C*T and S*T
mod.sub <- lm(y ~ C + T + S + C*S)
summary(mod.sub)
confint(mod.sub)
coeff.sub <- coef(mod.sub)[2:length(coef(mod.sub))]
coeff <- sort(abs(coeff.sub), index.return=TRUE)
barchart(coeff$x,
xlim=c(0, max(abs(coeff.sub))+0.1),
xlab=list("Magnitude of effect", cex=1.5),
ylab = list("Effect", cex=1.5),
groups=(coeff.sub>0)[coeff$ix], col=c("lightblue", "orange"),
scales=list(cex=1.5)
)