The document discusses the 2k factorial design, which is a special case of the general factorial design where k factors are each studied at two levels (usually labeled as low and high). The 2k factorial design is widely used in industrial experimentation and forms a basic building block for other experimental designs. Key aspects covered include orthogonality, estimating main effects and interactions between factors, ANOVA analysis to determine significant effects, and evaluating residuals to validate model assumptions.

1. DOE 6E Montgomery
Design of Engineering Experiments
Part 5 – The 2k Factorial Design
• Text reference, Chapter 6
• Special case of the general factorial design; k factors,
all at two levels
• The two levels are usually called low and high (they
could be either quantitative or qualitative)
• Very widely used in industrial experimentation
• Form a basic “building block” for other very useful
experimental designs (DNA)

2. Relationship between the tools
and the practical problems
DOE 6E Montgomery
Practical Questions Statistical Answer
We have varied all the controllable factors at the
same time how can we estimate the effect of each
factor separately?
Orthogonality Concept
Do all the controllable factors have the same effect
on the examined performance measure?
How can we quantify the effect controllable factors
of rank them based on their importance?
Effect estimation
Do some of the factors depend on others in
affecting the performance measure?
Interaction Estimation
If we repeat the experiment, another time do we
get the same conclusions?
To what extent can we generalize the conclusions of
our experiment?
ANOVA but its assumptions
must be met.

3. Orthogonallity & Orthogonal Arrays
(OA)
• Orthogonality comes from the Greek orthos, meaning "straight", and
gonia, meaning "angle". It has somewhat different meanings depending
on the context, but most involve the idea of perpendicular, non-
overlapping, varying independently, or uncorrelated.
• In mathematics, two lines or curves are orthogonal if they are perpendicular at
their point of intersection. Two vectors are orthogonal if and only if their dot
product is zero.
• Typically in Cartesian coordinates, one considers primarily bound vectors. A
bound vector is determined by the coordinates of the terminal point, its initial
point always having the coordinates of the origin O = (0,0). Thus the bound
vector represented by (1,0) is a vector of unit length pointing from the origin
up the positive x-axis.
DOE 6E Montgomery

4. Orthogonallity & Orthogonal Arrays
• Using the 22=4 Orthogonal array,
vector A (-1,1,-1,1) and B (-1,-1,1,1)
are orthogonal because
– Their inner product is zero
– All the possible pair of levels appear
the same number of time
• It is essential to estimate the effect of
each of the studied factors
independently.
DOE 6E Montgomery
A B
-1 -1
1 -1
-1 1
1 1

5. Example Tea Experiment Data
DOE 6E Montgomery
F M
Tea Type Sugar Cup Response Avg
1 L 1.5 P 3 5 4
2 L 1.5 M 8 4 6
3 L 2 P 5 6 5.5
4 L 2 M 4 5 4.5
5 R 1.5 P 6 5 5.5
6 R 1.5 M 5 4 4.5
7 R 2 P 4 5 4.5
8 R 2 M 6 7 6.5

6. Relationship between the tools
and the practical problems
DOE 6E Montgomery
Practical Questions Statistical Answer
We have varied all the controllable factors at the
same time how can we estimate the effect of each
factor separately?
Orthogonality Concept
Do all the controllable factors have the same effect
on the examined performance measure?
How can we quantify the effect controllable factors
of rank them based on their importance?
Effect estimation
Do some of the factors depend on others in
affecting the performance measure?
Interaction Estimation
If we repeat the experiment, another time do we
get the same conclusions?
To what extent can we generalize the conclusions of
our experiment?
ANOVA but its assumptions
must be met.

7. Determining the most influential factors—
calculation of the main effect
F M
Tea Type Sugar Cup Response Avg
1 L (-1) 1.5 (-1) P (-1) 3 5 4
2 L (-1) 1.5 (-1) M (+1) 8 4 6
3 L (-1) 2 (+1) P (-1) 5 6 5.5
4 L (-1) 2 (+1) M (+1) 4 5 4.5
5 R(+1) 1.5 (-1) P (-1) 6 5 5.5
6 R(+1) 1.5 (-1) M (+1) 5 4 4.5
7 R(+1) 2 (+1) P (-1) 4 5 4.5
8 R(+1) 2 (+1) M (+1) 6 7 6.5
Avg (-1) 5 5 4.875
Avg (+1) 5.25 5.25 5.375
Effect 0.25 0.25 0.5

8. Evaluating the Interdependence
• What is interaction? Does tea depend on sugar
in impacting the taste?
• Fix tea at its (-1) level (Lepton) and vary the
sugar-compute the effect of sugar
• Fix tea at its (+1) level Rabee and vary the
sugar-compute the effect of sugar
• Compare the effects calculated in the above
two steps-if the are equal, then there is no
interdependence
DOE 6E Montgomery

9. Evaluating the Interdependence
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10. Evaluating the Interdependence
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11. Evaluating the Interdependence
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12. Generating Interactions Columns in OAs
• No. of two-factor interaction in 2k design =
– k!/(2!*(k-2)!)
• No. of three-factor interaction in 2k design =
– k!/(3!*(k-3)!)
• No. of X-factor interaction in 2k design =
– k!/(x!*(k-x)!)
• Each interaction column is obtained by multiplying
the elements of the factors that comprise (form) it
DOE 6E Montgomery

13. Interactions in Tea experiments
• Three factors (k) were studied.
• No. of two-factor interaction in 2k design =
– 3!/(2!*(3-2)!) = 3
• No. of three-factor interaction in 2k design =
– 3!/(3!*(3-3)!) = 1
• No. of X-factor interaction in 2k design =
– k!/(x!*(k-x)!)
• Each interaction column is obtained by multiplying
the elements of the factors that comprise (form) it
DOE 6E Montgomery

14. Estimate the main effects & interactions in
the Tea experiments
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15. Plot the main effects in the Tea
experiments
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16. Plot the Sugar-Tea type interaction
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17. Plot the Cup-Tea type interaction
DOE 6E Montgomery

18. Relationship between the tools
and the practical problems
DOE 6E Montgomery
Practical Questions Statistical Answer
We have varied all the controllable factors at the
same time how can we estimate the effect of each
factor separately?
Orthogonality Concept
Do all the controllable factors have the same effect
on the examined performance measure?
How can we quantify the effect controllable factors
of rank them based on their importance?
Effect estimation
Do some of the factors depend on others in
affecting the performance measure?
Interaction Estimation
If we repeat the experiment, another time do we
get the same conclusions?
To what extent can we generalize the conclusions of
our experiment?
ANOVA but its assumptions
must be met.

19. Tea Experiment ANOVA
Source
Sum of
Squares DF
Mean
Square
F-
Value P-value
A 0.25 1 0.25 0.153846154 0.7051
B 0.25 1 0.25 0.153846154 0.7051
C 1 1 1 0.615384615 0.4554
AB 0.25 1 0.25 0.153846154 0.7051
AC 0 1 0 0 1.0000
BC 0 1 0 0 1.0000
ABC 9 1 9 5.538461538 0.0464
Pure Error 13 8 1.625
Total 23.75 15

20. Example Tea Experiment Data
DOE 6E Montgomery
F M
Tea Type Sugar Cup Response Avg
1 L 1.5 P 3 5 4
2 L 1.5 M 8 4 6
3 L 2 P 5 6 5.5
4 L 2 M 4 5 4.5
5 R 1.5 P 6 5 5.5
6 R 1.5 M 5 4 4.5
7 R 2 P 4 5 4.5
8 R 2 M 6 7 6.5

21. ANOVA
• Tool for identifying the causes of variances and
quantifying their contributions.
• Vertical vs. horizontal variance
• Variance equation:
• Every value in a random sample provide a piece
of information about the population & to do so it
must be unpredictable.
• Number of data points that provide information -
the number of points that can freely vary.

22. ANOVA
• If two S2 are computed for two samples that
come from two normal populations with a
common variance then their ratio S1
2/S2
2
follows Fdf1,df2.
• If an effect is not significant, its variance
should be close to the error variance.
• In hypothesis testing, it is assumed that the
effect is zero and the calculated F-value is
evaluated on this basis

23. F-distribution
F
0
Rejection Region
tableF
CalculatedF

24. Two-Factor Analysis of Variance
SST
SSA
SSB
SSAB
SSE
Factor A
Factor B
Interaction
Between A
and B
Inherent
Variation
(Error)

25. Two-Factor Analysis of Variance

26. Tea Experiment ANOVA

27. Tea Experiment ANOVA
Source
Sum of
Squares DF
Mean
Square
F-
Value P-value
A 0.25 1 0.25 0.153846154 0.7051
B 0.25 1 0.25 0.153846154 0.7051
C 1 1 1 0.615384615 0.4554
AB 0.25 1 0.25 0.153846154 0.7051
AC 0 1 0 0 1.0000
BC 0 1 0 0 1.0000
ABC 9 1 9 5.538461538 0.0464
Pure Error 13 8 1.625
Total 23.75 15

28. Tea Experiment ANOVA
Source
Sum of
Squares DF
Mean
Square
F-
Value P-value
ABC 9 1 9 8.542373 0.0111
Residual 14.75 14 1.053571
Total 23.75 15

29. Tea Experiment Regression
• Y = average + (Effect/2)X
• Y =5.13+0.75ABC
• This should be used to predict the values of the
responses and estimate the error

30. Residuals Estimation
Standard
Order
Actual
Value
Predicted
Value Residual
1 3 4.375 -1.375
2 8 5.875 2.125
3 5 5.875 -0.875
4 4 4.375 -0.375
5 6 5.875 0.125
6 5 4.375 0.625
7 4 4.375 -0.375
8 6 5.875 0.125
9 5 4.375 0.625
10 4 5.875 -1.875
11 6 5.875 0.125
12 5 4.375 0.625
13 5 5.875 -0.875
14 4 4.375 -0.375
15 5 4.375 0.625
16 7 5.875 1.125

31. Residuals (error) assessment
• Subtracting the actual values from the predicted ones (YA-
Yp) e
• Errors average should be zero (the positive impacts cancel the
negative ones).
• It should be normally distributed  any linear function of
normally distributed variable is normal (Central limit theory)
• It should be independent of the run order
• Considerably large values should be examined
• Its variance with the predicted values should be the same
(homogeneous variance)
• So errors should be independent, Normally distributed with
constant variance

32. Residuals Plot (Normality)

33. Residuals Plot (Normality)
• Sort the calculated errors from the smallest
to the largest
• Assign a rank value (i) to each of them
• Calculate (i-0.5)/(No. of residuals)

34. Residuals Plot (Normality)
Rank Reseduals (i-0.5)/16
1 -1.875 0.03125
2 -1.375 0.09375
3 -0.875 0.15625
4 -0.875 0.21875
5 -0.375 0.28125
6 -0.375 0.34375
7 -0.375 0.40625
8 0.125 0.46875
9 0.125 0.53125
10 0.125 0.59375
11 0.625 0.65625
12 0.625 0.71875
13 0.625 0.78125
14 0.625 0.84375
15 1.125 0.90625
16 2.125 0.96875

35. Residuals Plot (Normality)

36. Residuals Plot (Independence)

37. Residuals Plot: Variance Constance

38. Use the table of the data and locate
the maximum (or the minimum)
F M
Tea Type Sugar Cup Response Avg
1 L 1.5 P 3 5 4
2 L 1.5 M 8 4 6
3 L 2 P 5 6 5.5
4 L 2 M 4 5 4.5
5 R 1.5 P 6 5 5.5
6 R 1.5 M 5 4 4.5
7 R 2 P 4 5 4.5
8 R 2 M 6 7 6.5

39. If the main effects are the only significant effects
use their plot to identify the best settings.

40. Conclusions From the
Experiments
• None of the studied factors has an independent
effect on the tea taste as all the main effects were
statistically not significant.
• Because of the significance of the interaction ABC,
the impact of tea type on the taste depends on both
the sugar quantity and the cup type.
• The residuals analysis revealed no serious violation
of the ANOVA assumption. In fact only the
variance constancy assumption is suspicious.
• The best performance was attained with Rabee
tea when used with two sugar cubes and a Mug.

41. ABC interaction plot
C = M
C = P

42. Single Replicate Experiments
Tea Type Sugar Cup Res
1 L 1.5 P 3
2 L 1.5 M 8
3 L 2 P 5
4 L 2 M 4
5 R 1.5 P 6
6 R 1.5 M 5
7 R 2 P 4
8 R 2 M 6

43. Single Replicate Experiments
• Estimate the effects and divide them into
two groups-the small verses the large ones.
• Use half-normal probability plot
• Use ANOVA –values of the sums of
squares should be divided into two groups.

44. Estimate the effects and divide them into
two groups-the small verses the large ones
Term Effect
A 0.25
B -0.75
C 1.25
AB 0.25
AC -0.75
BC -0.75
ABC 2.25

45. Half-Normal Probability Plot
• Obtain the absolute values of the effects
• Sort the calculated absolute Effects from the
smallest to the largest.
• Assign a rank value (i) to each Effect.
• Calculate (i-0.5)/(No. of Effects)

46. Half-Normal Probability Plot
Rank (i) Term Effect (i-0.5)/7
1 A 0.25 0.071429
2 AB 0.25 0.214286
3 B 0.75 0.357143
4 AC 0.75 0.5
5 BC 0.75 0.642857
6 C 1.25 0.785714
7 ABC 2.25 0.928571

47. Half-Normal Probability Plot

48. Half-Normal Probability Plot
Source
Sum of
Squares DF
Mean
Square
F
Value Prob > F
C 3.125 1 3.125 4.310345 0.0925
ABC 10.125 1 10.125 13.96552 0.0135
Error 3.625 5 0.725
Total 16.875 7

49. Half-Normal Probability Plot
Source
Sum of
Squares DF
Mean
Square
F
Value Prob > F
ABC 10.125 1 10.125 9 0.0240
Residual 6.75 6 1.125
Total 16.875 7

50. What to do if the Residuals plots
reveal unusual pattern
• Consider one of the conventional
transformations of the response such as: log,
ln, square root, inverse square root
• Use Power transformation –Box-Cox plot
• The above procedures are normally
implemented using a software package.
• If these transformations are not effective,
the experiments must be repeated.

51. Selecting Best Settings
• Use the table of the data and locate the
maximum (or the minimum)
• If the main effects are the only significant
effects use their plot to identify the best
settings.
• If there is one or more significant two or
higher factor interactions, use their plots along
with the main effects that are not involved in
them to identify the best settings.