Design of Experiments

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Design of Engineering ExperimentsPart 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)

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Relationship between the tools and the practical problems

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.

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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.

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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.

A B-1 -11 -1-1 11 1

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Example Tea Experiment Data

F M Tea Type Sugar Cup Response Avg1 L 1.5 P 3 5 42 L 1.5 M 8 4 63 L 2 P 5 6 5.54 L 2 M 4 5 4.55 R 1.5 P 6 5 5.56 R 1.5 M 5 4 4.57 R 2 P 4 5 4.58 R 2 M 6 7 6.5

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Effect estimation






Determining the most influential factors—calculation of the main effect

F M Tea Type Sugar Cup Response Avg1 L (-1) 1.5 (-1) P (-1) 3 5 42 L (-1) 1.5 (-1) M (+1) 8 4 63 L (-1) 2 (+1) P (-1) 5 6 5.54 L (-1) 2 (+1) M (+1) 4 5 4.55 R(+1) 1.5 (-1) P (-1) 6 5 5.56 R(+1) 1.5 (-1) M (+1) 5 4 4.57 R(+1) 2 (+1) P (-1) 4 5 4.58 R(+1) 2 (+1) M (+1) 6 7 6.5

Avg (-1) 5 5 4.875Avg (+1) 5.25 5.25 5.375Effect 0.25 0.25 0.5

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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

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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

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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

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Estimate the main effects & interactions in the Tea experiments

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Plot the main effects in the Tea experiments

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Plot the Sugar-Tea type interaction

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Plot the Cup-Tea type interaction

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Effect estimation






Tea Experiment ANOVA

Source

Sum ofSquares

DF

MeanSquare

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

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Example Tea Experiment Data


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.

ANOVA• If two S2 are computed for two samples that

come from two normal populations with a common variance then their ratio S1

2/S22

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

F-distribution

F 0

Rejection Region

tableF

CalculatedF

Two-Factor Analysis of Variance

SST

SSA

SSB

SSAB

SSE

Factor A

Factor B

Interaction Between A and BInherent Variation (Error)

Two-Factor Analysis of Variance



Source

Sum ofSquares

DF

MeanSquare

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


Source

Sum ofSquares

DF

MeanSquare

F-Value

P-value

ABC 9 1 9 8.542373 0.0111

Residual 14.75 14 1.053571

Total 23.75 15

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

Residuals EstimationStandardOrder

ActualValue

PredictedValue

Residual

1 3 4.375 -1.3752 8 5.875 2.1253 5 5.875 -0.8754 4 4.375 -0.3755 6 5.875 0.1256 5 4.375 0.6257 4 4.375 -0.3758 6 5.875 0.1259 5 4.375 0.62510 4 5.875 -1.87511 6 5.875 0.12512 5 4.375 0.62513 5 5.875 -0.87514 4 4.375 -0.37515 5 4.375 0.62516 7 5.875 1.125

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

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)

Residuals Plot (Normality)Rank Reseduals (i-0.5)/161 -1.875 0.031252 -1.375 0.093753 -0.875 0.156254 -0.875 0.218755 -0.375 0.281256 -0.375 0.343757 -0.375 0.406258 0.125 0.468759 0.125 0.5312510 0.125 0.5937511 0.625 0.6562512 0.625 0.7187513 0.625 0.7812514 0.625 0.8437515 1.125 0.9062516 2.125 0.96875


Residuals Plot (Independence)

Residuals Plot: Variance Constance

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.

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.

ABC interaction plot

C = MC = P

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

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.

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

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)

Half-Normal Probability PlotRank (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



SourceSum of

Squares DFMean

SquareF

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


Source

Sum ofSquares

DF

MeanSquare

FValue

Prob > F

ABC 10.125 1 10.125 9 0.0240

Residual 6.75 6 1.125

Total 16.875 7

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.

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.