Design and analysis of experiments (2014)

Class date(s):	27 February to 26 March 2014
(PDF)	Course slides
Download video: Link (plays in Google Chrome) [227M]
Download video: Link (plays in Google Chrome) [299M]
Download video: Link (plays in Google Chrome) [296M]
Download video: Link (plays in Google Chrome) [286M]
Download video: Link (plays in Google Chrome) [311M]
Download video: Link (plays in Google Chrome) [296M]
Download video: Link (plays in Google Chrome) [280M]
Download video: Link (plays in Google Chrome) [305 M]
Download video: Link (plays in Google Chrome) [301 M]

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)
17 March	10A	Video (280 M)	Audio (42 M)	See the source code below.
19 March	10B	Video (305 M)	Audio (42 M)	None
20 March	10C	Video (301 M)	Audio (42 M)	None

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)
)

Class 10A, 17 March

To find all the 2-factor and 3-factor coefficients in a linear model

terms.formula( y ~ (A+B+C+D)^2 )
terms.formula( y ~ (A+B+C+D)^3 )