Two level factorial designs (chap6)

1

Chapter 6 The 2k Factorial Design

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6.1 Introduction

• The special cases of the general factorial design • k factors and each factor has only two levels• Levels:

– quantitative (temperature, pressure,…), or qualitative (machine, operator,…)

– High and low– Each replicate has 2 2 = 2k observations

3

• Assumptions: (1) the factor is fixed, (2) the design is completely randomized and (3) the usual normality assumptions are satisfied

• Wildly used in factor screening experiments

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6.2 The 22 Factorial Design • Two factors, A and B, and each factor has two

levels, low and high.• Example: the concentration of reactant v.s. the

amount of the catalyst (Page 219)

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• “-” And “+” denote the low and high levels of a factor, respectively

• Low and high are arbitrary terms

• Geometrically, the four runs form the corners of a square

• Factors can be quantitative or qualitative, although their treatment in the final model will be different

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• Average effect of a factor = the change in response produced by a change in the level of that factor averaged over the levels if the other factors.

• (1), a, b and ab: the total of n replicates taken at the treatment combination.

• The main effects:

AAyy

n

b

n

aab

baabn

ababn

A

2

)1(

2

)]1([2

1)]}1([]{[

2

1

BByy

n

a

n

bab

ababn

baabn

B

2

)1(

2

)]1([2

1)]}1([]{[

2

1

7

• The interaction effect:

• In that example, A = 8.33, B = -5.00 and AB = 1.67

• Analysis of Variance• The total effects:

n

ab

n

ab

baabn

ababn

AB

22

)1(

])1([2

1)]}1([]{[

2

1

baabContrast

ababContrast

baabContrast

AB

B

A

)1(

)1(

)1(

8

• Sum of squares:

ABBATE

i j

n

kijkT

AB

B

A

SSSSSSSSSS

n

yySS

n

ababSS

n

ababSS

n

baabSS

4

4

])1([

4

)]1([

4

)]1([

22

1

2

1 1

2

2

2

2

9

Response:Conversion ANOVA for Selected Factorial ModelAnalysis of variance table [Partial sum of squares]

Sum of Mean FSource Squares DF Square Value Prob > FModel 291.67 3 97.22 24.82 0.0002A 208.33 1 208.33 53.19 < 0.0001B 75.00 1 75.00 19.15 0.0024AB 8.33 1 8.33 2.13 0.1828Pure Error 31.33 8 3.92Cor Total 323.00 11

Std. Dev. 1.98 R-Squared 0.9030Mean 27.50 Adj R-Squared 0.8666C.V. 7.20 Pred R-Squared 0.7817

PRESS 70.50 Adeq Precision 11.669

The F-test for the “model” source is testing the significance of the overall model; that is, is either A, B, or AB or some combination of these effects important?

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• Table of plus and minus signs:

I A B AB

(1) + – – +

a + + – –

b + – + –

ab + + + +

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• The regression model:

– x1 and x2 are coded variables that represent the

two factors, i.e. x1 (or x2) only take values on –

1 and 1.– Use least square method to get the estimations

of the coefficients– For that example,

– Model adequacy: residuals (Pages 224~225) and normal probability plot (Figure 6.2)

22110 xxy

21 2

00.5

2

33.85.27ˆ xxy

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6.3 The 23 Design

• Three factors, A, B and C, and each factor has two levels. (Figure 6.4 (a))

• Design matrix (Figure 6.4 (b))• (1), a, b, ab, c, ac, bc, abc• 7 degree of freedom: main effect = 1, and

interaction = 1

13

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• Estimate main effect:

• Estimate two-factor interaction: the difference between the average A effects at the two levels of B

])1([4n

1

4

)1(

4

abcacaba

])1([4

1

bccbabcacaba

n

bccb

n

yy

bcabccacbaban

A

AA

n

aacbbc

n

cababc

acacbabbcabcn

AB

44

)1(

)]1([4

1

15

• Three-factor interaction:

• Contrast: Table 6.3– Equal number of plus and minus– The inner product of any two columns = 0– I is an identity element– The product of any two columns yields another

column– Orthogonal design

• Sum of squares: SS = (Contrast)2/8n

)]1([4n

1

)]}1([][][]{[4

1

ababcacbcabc

ababcacbcabcn

ABC

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Factorial Effect

TreatmentCombination

I A B AB C AC BC ABC

(1) + – – + – + + –

a + + – – – – + +

b + – + – – + – +

ab + + + + – – – –

c + – – + + – – +

ac + + – – + + – –

bc + – + – + – + –

abc + + + + + + + +Contrast   24 18 6 14 2 4 4

Effect   3.00 2.25 0.75 1.75 0.25 0.50 0.50

Table of – and + Signs for the 23 Factorial Design (pg. 231)

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• Example 6.1

A = carbonation, B = pressure, C = speed, y = fill deviation

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Term Effect SumSqr % ContributionModel InterceptError A 3 36 46.1538Error B 2.25 20.25 25.9615Error C 1.75 12.25 15.7051Error AB 0.75 2.25 2.88462Error AC 0.25 0.25 0.320513Error BC 0.5 1 1.28205Error ABC 0.5 1 1.28205Error LOF 0Error P Error 5 6.41026

Lenth's ME 1.25382 Lenth's SME 1.88156

• Estimation of Factor Effects

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• ANOVA Summary – Full Model

Response:Fill-deviation ANOVA for Selected Factorial ModelAnalysis of variance table [Partial sum of squares]

Sum of Mean FSource Squares DF Square Value Prob > FModel 73.00 7 10.43 16.69 0.0003A 36.00 1 36.00 57.60 < 0.0001B 20.25 1 20.25 32.40 0.0005C 12.25 1 12.25 19.60 0.0022AB 2.25 1 2.25 3.60 0.0943AC 0.25 1 0.25 0.40 0.5447BC 1.00 1 1.00 1.60 0.2415ABC 1.00 1 1.00 1.60 0.2415Pure Error 5.00 8 0.63Cor Total 78.00 15

Std. Dev. 0.79 R-Squared 0.9359Mean 1.00 Adj R-Squared 0.8798C.V. 79.06 Pred R-Squared 0.7436

PRESS 20.00 Adeq Precision 13.416

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• The regression model and response surface:– The regression model:

– Response surface and contour plot (Figure 6.7)

21321 2

75.0

2

75.1

2

25.2

2

00.300.1ˆ xxxxxy

Coefficient Standard 95% CI 95% CI

Factor Estimate DF Error Low High Intercept 1.00 1 0.20 0.55 1.45 A-Carbonation 1.50 1 0.20 1.05 1.95 B-Pressure 1.13 1 0.20 0.68 1.57 C-Speed 0.88 1 0.20 0.43 1.32 AB 0.38 1 0.20 -0.072 0.82

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• Contour & Response Surface Plots – Speed at the High Level

DESIGN-EXPERT Plot

Fill-deviationX = A: CarbonationY = B: Pressure

Design Points

Actual FactorC: Speed = 250.00

Fill-deviation

A: Carbonation

B: P

res

su

re

10.00 10.50 11.00 11.50 12.00

25.00

26.25

27.50

28.75

30.00

0.5

1.375

2.25

3.125

2 2

2 2

DESIGN-EXPERT Plot

Fill-deviationX = A: CarbonationY = B: Pressure

Actual FactorC: Speed = 250.00

-0.375

0.9375

2.25

3.5625

4.875

F

ill-

de

via

tio

n

10.00

10.50

11.00

11.50

12.00

25.00

26.25

27.50

28.75

30.00

A: Carbonation B: Pressure

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• Refine Model – Remove Nonsignificant Factors

Response: Fill-deviation ANOVA for Selected Factorial ModelAnalysis of variance table [Partial sum of squares]

Sum of Mean FSource Squares DF Square Value Prob > FModel 70.75 4 17.69 26.84 < 0.0001A 36.00 1 36.00 54.62 < 0.0001B 20.25 1 20.25 30.72 0.0002C 12.25 1 12.25 18.59 0.0012AB 2.25 1 2.25 3.41 0.0917Residual 7.25 11 0.66LOF 2.25 3 0.75 1.20 0.3700Pure E 5.00 8 0.63C Total 78.00 15

Std. Dev. 0.81 R-Squared 0.9071Mean 1.00 Adj R-Squared 0.8733C.V. 81.18 Pred R-Squared 0.8033

PRESS 15.34 Adeq Precision 15.424

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6.4 The General 2k Design • k factors and each factor has two levels• Interactions• The standard order for a 24 design: (1), a, b, ab, c,

ac, bc, abc, d, ad, bd, abd, cd, acd, bcd, abcd

two-factor interactions2

three-factor interactions3

1 factor interaction

k

k

k

24

• The general approach for the statistical analysis:– Estimate factor effects– Form initial model (full model)– Perform analysis of variance (Table 6.9)– Refine the model– Analyze residual– Interpret results

•

2

...

)(2

12

2

)1()1)(1(

KABCkKABC

KABCk

KABC

Contrastn

SS

Contrastn

KABC

kbaContrast

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6.5 A Single Replicate of the 2k Design• These are 2k factorial designs

with one observation at each corner of the “cube”

• An unreplicated 2k factorial design is also sometimes called a “single replicate” of the 2k

• If the factors are spaced too closely, it increases the chances that the noise will overwhelm the signal in the data

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• Lack of replication causes potential problems in statistical testing– Replication admits an estimate of “pure error”

(a better phrase is an internal estimate of error)

– With no replication, fitting the full model results in zero degrees of freedom for error

• Potential solutions to this problem– Pooling high-order interactions to estimate

error (sparsity of effects principle)– Normal probability plotting of effects

(Daniels, 1959)

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• Example 6.2 (A single replicate of the 24 design)– A 24 factorial was used to investigate the effects

of four factors on the filtration rate of a resin– The factors are A = temperature, B = pressure,

C = concentration of formaldehyde, D= stirring rate

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• Estimates of the effects

Term Effect SumSqr % ContributionModel InterceptError A 21.625 1870.56 32.6397Error B 3.125 39.0625 0.681608Error C 9.875 390.062 6.80626Error D 14.625 855.563 14.9288Error AB 0.125 0.0625 0.00109057Error AC -18.125 1314.06 22.9293Error AD 16.625 1105.56 19.2911Error BC 2.375 22.5625 0.393696Error BD -0.375 0.5625 0.00981515Error CD -1.125 5.0625 0.0883363Error ABC 1.875 14.0625 0.245379Error ABD 4.125 68.0625 1.18763Error ACD -1.625 10.5625 0.184307Error BCD -2.625 27.5625 0.480942Error ABCD 1.375 7.5625 0.131959

Lenth's ME 6.74778 Lenth's SME 13.699

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• The normal probability plot of the effectsDESIGN-EXPERT PlotFiltration Rate

A: TemperatureB: PressureC: ConcentrationD: Stirring Rate

Normal plot

No

rma

l % p

rob

ab

ility

Effect

-18.12 -8.19 1.75 11.69 21.62

1

5

10

20

30

50

70

80

90

95

99

A

CD

AC

AD

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DESIGN-EXPERT Plot

Filtration Rate

X = A: TemperatureY = C: Concentration

C- -1.000C+ 1.000

Actual FactorsB: Pressure = 0.00D: Stirring Rate = 0.00

C: ConcentrationInteraction Graph

Filt

ratio

n R

ate

A: Temperature

-1.00 -0.50 0.00 0.50 1.00

41.7702

57.3277

72.8851

88.4426

104

DESIGN-EXPERT Plot

Filtration Rate

X = A: TemperatureY = D: Stirring Rate

D- -1.000D+ 1.000

Actual FactorsB: Pressure = 0.00C: Concentration = 0.00

D: Stirring RateInteraction Graph

Filt

ratio

n R

ate

A: Temperature

-1.00 -0.50 0.00 0.50 1.00

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58.25

73.5

88.75

104

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• B is not significant and all interactions involving B are negligible

• Design projection: 24 design => 23 design in A,C and D

• ANOVA table (Table 6.13)

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Response:Filtration Rate ANOVA for Selected Factorial ModelAnalysis of variance table [Partial sum of squares]

Sum of Mean FSource Squares DF Square Value Prob >FModel 5535.81 5 1107.16 56.74 < 0.0001A 1870.56 1 1870.56 95.86 < 0.0001C 390.06 1 390.06 19.99 0.0012D 855.56 1 855.56 43.85 < 0.0001AC 1314.06 1 1314.06 67.34 < 0.0001AD 1105.56 1 1105.56 56.66 < 0.0001Residual 195.12 10 19.51Cor Total 5730.94 15

Std. Dev. 4.42 R-Squared 0.9660Mean 70.06 Adj R-Squared 0.9489C.V. 6.30 Pred R-Squared 0.9128

PRESS 499.52 Adeq Precision 20.841

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• The regression model:

• Residual Analysis (P. 251)• Response surface (P. 252)

Final Equation in Terms of Coded Factors:

Filtration Rate =+70.06250+10.81250 * Temperature+4.93750 * Concentration+7.31250 * Stirring Rate-9.06250 * Temperature * Concentration+8.31250 * Temperature * Stirring Rate

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DESIGN-EXPERT PlotFiltration Rate

Studentized Residuals

No

rma

l % p

rob

ab

ility

Normal plot of residuals

-1.83 -0.96 -0.09 0.78 1.65

1

5

10

20

30

50

70

80

90

95

99

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• Half-normal plot: the absolute value of the effect estimates against the cumulative normal probabilities.

DESIGN-EXPERT PlotFiltration Rate

A: TemperatureB: PressureC: ConcentrationD: Stirring Rate

Half Normal plot

Ha

lf N

orm

al %

pro

ba

bility

|Effect|

0.00 5.41 10.81 16.22 21.63

0

20

40

60

70

80

85

90

95

97

99

A

CD

AC

AD

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• Example 6.3 (Data transformation in a Factorial Design)

A = drill load, B = flow, C = speed, D = type of mud, y = advance rate of the drill

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• The normal probability plot of the effect estimates

DESIGN-EXPERT Plotadv._rate

A: loadB: flowC: speedD: mud

Half Normal plot

Ha

lf N

orm

al %

pro

ba

bili

ty

|Effect|

0.00 1.61 3.22 4.83 6.44

0

20

40

60

70

80

85

90

95

97

99

B

C

D

BCBD

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• Residual analysisDESIGN-EXPERT Plotadv._rate

Residual

No

rma

l % p

rob

ab

ility

Normal plot of residuals

-1.96375 -0.82625 0.31125 1.44875 2.58625

1

5

10

20

30

50

70

80

90

95

99

DESIGN-EXPERT Plotadv._rate

Predicted

Re

sid

ua

ls

Residuals vs. Predicted

-1.96375

-0.82625

0.31125

1.44875

2.58625

1.69 4.70 7.70 10.71 13.71

40

• The residual plots indicate that there are problems with the equality of variance assumption

• The usual approach to this problem is to employ a transformation on the response

• In this example, yy ln*

41

DESIGN-EXPERT PlotLn(adv._rate)

A: loadB: flowC: speedD: mud

Half Normal plotH

alf

No

rma

l % p

rob

ab

ility

|Effect|

0.00 0.29 0.58 0.87 1.16

0

20

40

60

70

80

85

90

95

97

99

B

C

D

Three main effects are large

No indication of large interaction effects

What happened to the interactions?

42

Response: adv._rate Transform: Natural log Constant: 0.000

ANOVA for Selected Factorial Model

Analysis of variance table [Partial sum of squares]Sum of Mean F

Source Squares DF Square Value Prob > FModel 7.11 3 2.37 164.82 < 0.0001B 5.35 1 5.35 371.49 < 0.0001C 1.34 1 1.34 93.05 < 0.0001D 0.43 1 0.43 29.92 0.0001Residual 0.17 12 0.014Cor Total 7.29 15

Std. Dev. 0.12 R-Squared 0.9763Mean 1.60 Adj R-Squared 0.9704C.V. 7.51 Pred R-Squared 0.9579

PRESS 0.31 Adeq Precision 34.391

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• Following Log transformation

Final Equation in Terms of Coded Factors:

Ln(adv._rate) =+1.60+0.58 * B+0.29 * C+0.16 * D

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DESIGN-EXPERT PlotLn(adv._rate)

Residual

No

rma

l % p

rob

ab

ility

Normal plot of residuals

-0.166184 -0.0760939 0.0139965 0.104087 0.194177

1

5

10

20

30

50

70

80

90

95

99

DESIGN-EXPERT PlotLn(adv._rate)

PredictedR

es

idu

als

Residuals vs. Predicted

-0.166184

-0.0760939

0.0139965

0.104087

0.194177

0.57 1.08 1.60 2.11 2.63

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• Example 6.4:– Two factors (A and D) affect the mean number

of defects– A third factor (B) affects variability– Residual plots were useful in identifying the

dispersion effect– The magnitude of the dispersion effects:

– When variance of positive and negative are equal, this statistic has an approximate normal distribution

)(

)(ln

2

2*

iS

iSFi

46

6.6 The Addition of Center Points to the 2k Design • Based on the idea of replicating some of the runs

in a factorial design• Runs at the center provide an estimate of error and

allow the experimenter to distinguish between two possible models:

01 1

20

1 1 1

First-order model (interaction)

Second-order model

k k k

i i ij i ji i j i

k k k k

i i ij i j ii ii i j i i

y x x x

y x x x x

47

no "curvature"F Cy y

The hypotheses are:

01

11

: 0

: 0

k

iii

k

iii

H

H

2

Pure Quad

( )F C F C

F C

n n y ySS

n n

This sum of squares has a single degree of freedom

48

• Example 6.6

5Cn

Usually between 3 and 6 center points will work well

Design-Expert provides the analysis, including the F-test for pure quadratic curvature

49

Response: yield ANOVA for Selected Factorial ModelAnalysis of variance table [Partial sum of squares]

Sum of Mean FSource Squares DF Square Value Prob > FModel 2.83 3 0.94 21.92 0.0060A 2.40 1 2.40 55.87 0.0017B 0.42 1 0.42 9.83 0.0350AB 2.500E-003 1 2.500E-003 0.058 0.8213Curvature 2.722E-003 1 2.722E-003 0.063 0.8137Pure Error 0.17 4 0.043Cor Total 3.00 8

Std. Dev. 0.21 R-Squared 0.9427Mean 40.44 Adj R-Squared 0.8996

C.V. 0.51 Pred R-Squared N/A

PRESS N/A Adeq Precision 14.234

50

• If curvature is significant, augment the design with axial runs to create a central composite design. The CCD is a very effective design for fitting a second-order response surface model