Design and analysis of experiments (2013)

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Class date(s): 08 March 2013 to 03 April 2013
Nuvola mimetypes pdf.png (PDF) Course slides
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Course notes and slides

Class materials

Software source code

Use R 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

You could also use MATLAB if you prefer:

% Set up the X matrix:

n = 4;
temperature = [-1, +1, -1, +1];
substrate = [-1, -1, +1, +1];

X = [ ones(n, 1), temperature', substrate', (temperature .* substrate)'];

% or you can type it in directly (but this is error prone)
X = [+1 -1 -1 +1; ...
     +1 +1 -1 -1; ...
     +1 -1 +1 -1; ...
     +1 +1 +1 +1];

% Conversion is your y-variable
y = [69, 60, 64, 53]';

% Regression coefficients = b = [Intercept, b_T, b_S, b_{TS}]
b = inv(X'*X)*X'*y

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>

# 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 (error prone)
# mod.full <- lm(y ~ C + T + S + C*T + C*S + S*T + C*T*S)

# or the much more 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)
barchart(sort(abs(coeff.full)), xlab="Magnitude of effect", groups=coeff.full>0, col=c("lightblue", "orange"))