Design and Analysis of Experiments Lecture 6 · Lecture 6.1 1Postgraduate Certificate in Statistics © 2016 Michael Stuart Design and Analysis of Experiments Lecture 6.1 1. Review

Lecture 6.1 1

© 2016 Michael Stuart

Design and Analysis of Experiments

Lecture 6.1

1. Review of split unit experiments

− Why split?

− Why block?

2. Review of Laboratory 2

− Cambridge grassland experiment

− Soup mix packet filling

3. Extending plot and treatment structures

− Wood stain experiment

4. Robust Product Design

5. An interesting interaction?

Postgraduate Certificate in Statistics


Lecture 6.1 2


Split units experiments

arise when

– one set of treatment factors is applied to

experimental units,

– a second set of factors is applied to sub units of

these experimental units.

Originated in agriculture where they are referred to as

split plot experiments.

Whole units may be regarded as blocks

"Most industrial experiments are ... split plot in their

design.“ C. Daniel (1976) p. 175



Lecture 6.1 3


Why split?

• Adding another factor after the experiment

started

• Changing one factor is

– more difficult

– more expensive

– more time consuming

than changing others

• Some factors require better precision than others



Cambridge grassland

Component lifetimes

Water resistance

Corrosion resistance

Lecture 6.1 4


Why block?

• Blocking is useful when there are

known external factors (covariates)

that affect variation between plots.

• Blocking reduces bias arising due to

block effects disproportionately affecting factor effects

due to levels disproportionally allocated to blocks.

• Neighbouring plots are likely to be

more homogeneous than separated plots, so that

– blocking reduces variation affecting comparisons

when treatments are compared within blocks

– (precision is increased

when results are combined across blocks).



Lecture 6.1 5


Block or Not?

• Not blocking when there is a block effect implies

reduced power for treatment effects test;

because Error term includes block variation.

• Blocking when there is no block effect implies


because Error degrees of freedom reduced



Lecture 6.1 6



Lecture 6.1


− Why split?

− Why block?










Lecture 6.1 7


Laboratory 2, Exercise 1

Cambridge Grassland Experiment

3 grassland treatments

Rejuvenator R

Harrow H

no treatment C

randomly allocated to 3 neighbouring plots,

replicated in 6 neighbouring blocks

4 fertilisers

Farmyard manure F

Straw S

Artificial fertiliser A

no fertiliser C

randomly allocated to 4 sub plots within each plot.



Lecture 6.1 8


Cambridge Grassland Experiment

Blocks 1 2 3 4 5 6

Whole Plots 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

Treatments H C R H R C C H R H R C C H R C R H

Sub Plot 1 C A A C F F A A A A F F F C A F F C

Sub Plot 2 A S C A S A C C F F A S S A S A S S

Sub Plot 3 F C F F C C S F S C S A C S C C C F

Sub Plot 4 S F S S A S F S C S C C A F F S A A



Lecture 6.1 9


Experimental results, Yields in pound (lbs)



Block 1 Block 2 Block 3 Block 4 Block 5 Block 6

C H R C H R C H R C H R C H R C H R

A 266 213 208 210 222 266 220 184 184 216 178 207 202 175 184 169 142 151

C 165 127 155 150 167 163 155 118 153 159 125 135 147 118 98 132 104 69

F 198 180 200 247 203 228 190 168 174 225 149 162 184 175 144 164 145 116

S 184 127 150 188 167 157 140 128 141 174 107 113 154 112 113 116 89 101

Lecture 6.1 10


Treatment yields vs Layout yields

(Block 1)



Block 1

C H R

A 266 213 208

C 165 127 155

F 198 180 200

S 184 127 150

Block 1

Whole Plot 1 2 3

Treatment H C R

Sub Plot 1 C

127

A

266

A

208

Sub Plot 2 A

213

S

184

C

155

Sub Plot 3 F

180

C

165

F

200

Sub Plot 4 S

127

F

198

S

150

Lecture 6.1 11


3-Step Decomposition of Total Variation

Step 1: Two components of total variation

Step 2: Analysis of whole plot total variation

Step 3: Analysis of subplot total variation



Lecture 6.1 12


Units

Blocks

Whole Plots

Subplots

Plot Structure



Lecture 6.1 13


Step 1: Two components of total variation

Mintab model: Plot Subplot(Plot) Source DF SS MS F P

Plot 17 54577 3210.4 2.63 0.004

Subplot(Plot) 54 65896 1220.3 **

Error 0 * *

Total 71 120473

Note: DF for Plot Variation: 18 − 1 = 17

DF for Subplot Variation: (4 − 1) x 18 = 54

Minitab model: Plot Source DF SS MS F P

Plot 17 54577 3210 2.63 0.004

Error 54 65896 1220

Total 71 120473



Lecture 6.1 14


Units

Blocks

Whole Plots

Subplots


Treatment

Factors

Treatment



ANOVA

MS(Blocks)

MS(Treatments)

MS(Whole Plot Error)

Whole Plot and Treatment Structure

Lecture 6.1 15



Minitab model: Block Treatment Source DF SS MS F P

Block 5 37425 7485 6.79 0.000

Treatment 2 12471 6236 5.65 0.005

Error 64 70577 1103

Total 71 120473

Minitab model: Plot (see Slide 13) Source DF SS MS F P

Plot 17 54577 3210 2.63 0.004

Error 54 65896 1220

Total 71 120473

Plot Error DF = 17 – 5 – 2 = 10

Plot Error SS = 54577 – 37425 – 12471 = 4681



Lecture 6.1 16


Classwork 1



• From the values on Slide 15, construct an

analysis of variance table for whole plots

variation.

Lecture 6.1 17


Units

Blocks

Whole Plots

Subplots

Whole Plot and Treatment Structure

Treatment

Factors

Treatment

ANOVA

MS(Blocks)

MS(Treatments)


B x T



Lecture 6.1 18


Minitab model: Block Treatment Block*Treatment

Source DF SS MS F P

Block 5 37425 7485 6.13 0.000

Treatment 2 12471 6236 5.11 0.009

Block*Treatment 10 4681 468 0.38 0.949

Error 54 65896 1220

Total 71 120473




Lecture 6.1 19


Units

Blocks

Whole Plots

Subplots

Step 3: Split Plot Analysis

Plot and Treatment Structure

Treatment

Factors

Treatment

Fertiliser

ANOVA

MS(Blocks)

MS(Treatments)


B x T

MS(Fertiliser)

MS(Interactions)

MS(Subplot Error) Postgraduate Certificate in Statistics


Subplots

Lecture 6.1 20


Split Plots Analysis

Minitab model: B + T + B*T

+ F + T*F + B*F

Random effect(s) B Fixed effects T F

Source DF SS MS F P

B 5 37425 7485 21.37 0.002 x

T 2 12471 6236 13.32 0.002

B*T 10 4681 468 1.94 0.079

F 3 56023 18674 151.24 0.000

T*F 6 782 130 0.54 0.774

B*F 15 1852 123 0.51 0.914

Error 30 7240 241

Total 71 120473



Lecture 6.1 21


Expected Mean Squares

Source Expected Mean Square

Block + 3 + 4 + 12

Treatment + 4 + Treatment effect

Plot + 4

Fertiliser + 3 + Fertiliser effect

Treatment*Fertiliser + Treatment x Fertiliser effect

Block*Fertiliser + 3

Error / Subplot



2S

2S

2S

2S

2P

2P

1I

)(J

2i

2S

2S

2P

2B

2FB

2FB

2FB

2S

Lecture 6.1 22


Classwork 2

• Identify the mean squares and F-ratios for testing

– treatment effects,

– fertiliser effects and

– treatment by fertiliser interaction effects.

• Confirm the values of the F-ratios



Lecture 6.1 23



Minitab model: B + T + B*T

+ F + T*F + B*F

Random effect(s) B Fixed effects T F

Source DF SS MS F P

B 5 37425 7485 15.99 0.000

T 2 12471 6236 13.32 0.002

B*T 10 4681 468 2.32 0.027

F 3 56023 18674 92.43 0.000

T*F 6 782 130 0.64 0.594

Error 45 7240 202

Total 71 120473



Lecture 6.1 24



Source Expected Mean Square

Block + 4 + 12

Treatment + 4 + Treatment effect

Plot + 4

Fertiliser + Fertiliser effect

Treatment*Fertiliser + Treatment x Fertiliser effect

Error / Subplot



2S

2S

2S

2S

2P

2P

2S

2S

2P

2B

Lecture 6.1 25


Decomposition Summary

Step 1

Source DF SS

Plot Total 17 54577

Subplot Total 54 65896

Total 71 120473



Lecture 6.1 26



Step 2

Source DF SS MS F P

Block 5 37425 7485 15.99 0.000

Treatment 2 12471 6236 13.32 0.002

Plot Error 10 4681 468 2.32 0.027

Plot Total 17 54577


Total 71 120473



Lecture 6.1 27



Step 3

Source DF SS MS F P

Block 5 37425 7485 15.99 0.000

Treatment 2 12471 6236 13.32 0.002

Plot Error 10 4681 468 2.32 0.027

Plot Total 17 54577 3210 2.63 0.004

Fertiliser 3 56023 18674 92.43 0.000

T*F 6 782 130 0.64 0.694

Subplot Error 45 9092 202


Total 71 120473



Lecture 6.1 28



Whole Plots

Source DF SS MS F P

Block 5 37425 7485 15.99 0.000

Treatment 2 12471 6236 13.32 0.002

Plot Error 10 4681 468 2.32 0.027

Plot Total 17 54577 3210 2.63 0.004

Fertiliser 3 56023 18674 92.43 0.000

T*F 6 782 130 0.64 0.694



Total 71 120473



Lecture 6.1 29



Sub plots

Source DF SS MS F P

Block 5 37425 7485 15.99 0.000

Treatment 2 12471 6236 13.32 0.002

Plot Error 10 4681 468 2.32 0.027

Plot Total 17 54577 3210 2.63 0.004

Fertiliser 3 56023 18674 92.43 0.000

T*F 6 782 130 0.64 0.694



Total 71 120473



Lecture 6.1 30



Source DF SS MS F P

Block 5 37425 7485 15.99 0.000

Treatment 2 12471 6236 13.32 0.002

Plot Error 10 4681 468 2.32 0.027

Plot Total 17 54577 3210 2.63 0.004

Fertiliser 3 56023 18674 92.43 0.000

T*F 6 782 130 0.64 0.694



Total 71 120473



Lecture 6.1 31


Subplots Residuals vs Fitted Values



Lecture 6.1 32


Same diagnostic, Different interpretation?



Lecture 6.1 33


Subplots Residuals Normal Plot



Lecture 6.1 34


Whole Plots Residuals vs Fitted Values



Lecture 6.1 35


Whole Plots Residuals Normal Plot



2

1

0

-1

-2

-3210-1-2

Dele

ted

Resi

du

al

Score

Lecture 6.1 36


Check Interactions



Lecture 6.1 37


Check Interactions



Lecture 6.1 38


Check Interactions



Lecture 6.1 39


Check Interactions



Lecture 6.1 40


Interaction plots for Grassland experiment

Treatments



Lecture 6.1 41

© 2016 Michael Stuart Postgraduate Certificate in Statistics


Lecture 6.1 42



Lecture 6.1


− Why split?

− Why block?










Lecture 6.1 43


Laboratory 2, Exercise 2:

Soup mix packet filling machine

Questions:

What factors affect soup powder fill variation?

How can fill variation be minimised?

Potential factors

A: Number of ports for adding oil, 1 or 3,

B: Mixer vessel temperature, ambient or cooled,

C: Mixing time, 60 or 80 seconds,

D: Batch weight, 1500 or 2000 lbs,

E: Delay between mixing and packaging, 1 or 7 days.

Response: Spread of weights of 5 sample packets



Lecture 6.1 44


Minitab analysis



Lecture 6.1 45


Minitab analysis

Normal plot vs Pareto Principle vs Lenth?



Lecture 6.1 46


Alias analysis

Estimated Effects

Term Effect Alias

E -0.470 E + A*B*C*D

B*E 0.405 B*E + A*C*D

D*E -0.315 D*E + A*B*C

E is aliased with or confounded with A*B*C*D



Lecture 6.1 47


Graphical and numerical summaries

B D

– + – +

E – 1.71 1.22

E – 1.31 1.60

+ 0.83 1.15 + 1.17 0.82



Lecture 6.1 48


Best conditions

Best conditions:

Temp Low, Weight High, Delay High.

Best conditions with Delay Low:

Temp High, Weight Low. Postgraduate Certificate in Statistics


Lecture 6.1 49


Reduced model

Fit model using active terms:

B + D + E + BE + DE

DE confirmed as active. Postgraduate Certificate in Statistics


Lecture 6.1 50


Diagnostics

1.81.61.41.21.00.8

2

1

0

-1

-2

-3

Fitted Value

Dele

ted

Resid

ual

Diagnostic Plot



Lecture 6.1 51


Diagnostics

3

2

1

0

-1

-2

-3

210-1-2

Dele

ted

Resid

ual

Score

Normal Probability Plot



Lecture 6.1 52


Delete Design point 5, iterate analysis

• Effect estimates similar

• Interaction patterns similar

• s = 0.15, df = 9 ( = 14 – 5 )

Mean SE Mean

B*D*E

- - - 1.700 0.153

+ - - 1.205 0.108

- + - 1.975 0.108

+ + - 1.225 0.108

- - + 0.975 0.108

+ - + 1.360 0.108

- + + 0.690 0.108

+ + + 0.940 0.108 0.69 2.26×0.15/√2 = 0.45 to 0.93

1.205 2.26×0.15/√2 = 0.965 to 1.445



Lecture 6.1 53



Lecture 6.1


− Why split?

− Why block?










Lecture 6.1 54


Water Resistance of Wood Stains

• Testing water resistance of four wood stains

– Pretreatments applied to whole boards

– Pretreated boards cut into 4 panels

– Stains applied to panels

– Replicated 3 times



Lecture 6.1 55


Results

Pretreatment 1 Pretreatment 2

Board 1 2 3 4 5 6

Panels 1-4 5-8 9-12 13-16 17-20 21-24

Stain 1 43.0 57.4 52.8 46.6 52.2 32.1

Stain 2 51.8 60.9 59.2 53.5 48.3 34.4

Stain 3 40.8 51.1 51.7 35.4 45.9 32.2

Stain 4 45.5 55.3 55.3 32.5 44.6 30.1



Lecture 6.1 56


Extending the unit structure

Suppose the 6 boards were in 3 blocks of 2

e.g. 2 boards selected from each of 3 production runs,

or 2 boards treated on each of 3 successive days

Note: Boards nested in Blocks

Block Board Pretreatment

1 1 1

4 2

2 2 1

5 2

3 3 1

6 2



Lecture 6.1 57


Factor

Pretreatment

Stain

Units

Blocks

Boards

Panels

Unit / Treatment Structure Diagram



Lecture 6.1 58


Results of Water Resistance Experiment

Block Board Pretreatment Stain

1 2 3 4

1 1 1 43.0 51.8 40.8 45.5

4 2 46.6 53.5 35.4 32.5

2 2 1 57.4 60.9 51.1 55.3

5 2 52.2 48.3 45.9 44.6

3 3 1 52.8 59.2 51.7 55.3

6 2 32.1 34.4 32.2 30.1



Lecture 6.1 59


Factor

Pretreatment

Stain

Units

Blocks

Boards

Panels

Extended Unit / Treatment Structure

and Analysis of Variance

ANOVA

MS(Blocks)

MS(Pretreatment)

MS(Boards Residuals)

MS(Stain)

MS(P x S)

MS(Panels Residuals)



Lecture 6.1 60


Analysis of Variance for Water Resistance

Minitab model: Block

Pretreat Block * Pretreat

Stain Pretreat * Stain

Source DF SS MS F P

Block 2 376.99 188.49 0.95 0.514

Pretreat 1 782.04 782.04 3.93 0.186

Block*Pretreat 2 398.38 199.19 15.67 0.000

Stain 3 266.01 88.67 6.98 0.006

Pretreat*Stain 3 62.79 20.93 1.65 0.231

Error 12 152.52 12.71

Total 23 2038.72 Postgraduate Certificate in Statistics


Lecture 6.1 61





Lecture 6.1 62


Analysis of Variance for Water Resistance

Minitab model: Block

Pretreat Block * Pretreat


Source DF SS MS F P

Block 2 376.99 188.49 0.95 0.514

Pretreat 1 782.04 782.04 3.93 0.186

Boards 2 398.38 199.19 15.67 0.000

Stain 3 266.01 88.67 6.98 0.006

Pretreat*Stain 3 62.79 20.93 1.65 0.231

Error 12 152.52 12.71

Total 23 2038.72

Block*Pretreat



Lecture 6.1 63


Analysis ignoring blocks

Minitab model: Pretreat Board(Pretreat)


Source DF SS MS F P

Pretreat 1 782.04 782.04 4.03 0.115

Board(Pretreat) 4 775.36 193.84 15.25 0.000

Stain 3 266.00 88.67 6.98 0.006

Pretreat*Stain 3 62.79 20.93 1.65 0.231

Error 12 152.52 12.71

Total 23 2038.72



Lecture 6.1 64


Block or Not?

• Not blocking when there is a block effect implies


because Error term includes block variation.

• Blocking when there is no block effect implies


because Error degrees of freedom reduced



Lecture 6.1 65


Extending the treatment structure

Suppose the four Stain levels are combinations of

two 2-level factors:

– Stain type, 1 or 2,

– number of Coats applied, 1 or 2.

Factor Units

Blocks

↓

Pretreatment Boards

↓

Stain x Coats Panels



Lecture 6.1 66


Extending the Minitab model

Block

Pretreatment Block*Pretreatment

Stain Coat Stain*Coat

Pretreatment*Stain Pretreatment*Coat

Pretreatment*Stain*Coat



Lecture 6.1 67


Analysis of Variance

Source DF SS MS F P

Block 2 376.99 188.49 0.95 0.514

Pretreatment 1 782.04 782.04 3.93 0.186

Block*Pretreatment 2 398.38 199.19 15.67 0.000

Stain 1 38.00 38.00 2.99 0.109

Coat 1 214.80 214.80 16.90 0.001

Stain*Coat 1 13.20 13.20 1.04 0.328

Pretreatment*Stain 1 43.20 43.20 3.40 0.090

Pretreatment*Coat 1 18.38 18.38 1.45 0.252

Pretreatment*Stain*Coat 1 1.21 1.21 0.10 0.762

Error 12 152.52 12.71

Total 23 2038.72



Lecture 6.1 68



Lecture 6.1


− Why split?

− Why block?










Lecture 6.1 69


Robustness Studies

Seek optimal settings of experimental factors

that remain optimal,

irrespective of uncontrolled environmental factors.

Run the experimental design,

the inner array,

at fixed settings of the environmental variables,

the outer array.

Popularised by Taguchi.

Improved by Box et al



Lecture 6.1 70


Study of Detergent Robustness

• Detergent performance affected by

– Temperature of wash water, T ( + or − )

– Hardness of wash water, H ( + or − )

– concentration of detergent in water, R ( + or − )

• Key product design factors:

– amount of Ingredient 1 A ( + or − )

– amount of Ingredient 2 B ( + or − )

– process version 1 C ( + or − )

– process version 2 D ( + or − )

• Response: Whiteness,

measured by reflectometer



Lecture 6.1 71


Study of Detergent Robustness

• Design points in a 24−1 fractional factorial plan

used to produce batches of 8 variants of the

detergent;

• Design points in a 23−1 fractional factorial plan

used to set up 4 wash conditions;

• Samples of each detergent assessed under each

of the 4 wash conditions



Lecture 6.1 72


Results



Environmental factors

T – + – + H – – + +

Product Version

Design factors R + – – + A B C D i ii iii iv Mean Range

1 – – – – 88 85 88 85 86.50 3 2 + – – + 80 77 80 76 78.25 4 3 – + – + 90 84 91 86 87.75 7 4 + + – – 95 87 93 88 90.75 8 5 – – + + 84 82 83 84 83.25 2 6 + – + – 85 84 82 82 83.25 3 7 – + + – 91 93 92 92 92.00 2 8 + + + + 89 88 89 87 88.25 2

Lecture 6.1 73


Treatment

Factors

Design

factors

Environmental

factors

Experimental

Units

Detergent

Types

Detergent

Samples

Unit / Treatment Structure Diagram



Lecture 6.1 74


27−2 Estimated Effects



Term Effect Term Effect

T -2.5 R*A 0

H -0.25 R*B -0.125

R 0.25 R*C -0.125

A -2.25 R*D 0

B 6.875 A*B 1.875

C 0.875 A*C 0.375

D -3.75 A*D 0

T*A -0.5 T*A*B -0.375

T*B -0.625 T*A*C -0.125

T*C 2.125 T*A*D 0.75

T*D -0.25 H*A*B 0.125

H*A -0.75 H*A*C -0.125

H*B 0.375 H*A*D 0

H*C -0.375 R*A*B 0.375

H*D 0.5 R*A*C -0.125

R*A*D -0.75

Lecture 6.1 75


Split plots model analysis

B significant, positive,

set at high (+) level

T and TC interaction

significant



Lecture 6.1 76


Split plots model analysis

At low C, whiteness is highly sensitive to T.

At high C, whiteness is relatively insensitive to T.



Lecture 6.1 77


Conclusion

• Set B and C to high levels, A and D as convenient

Environmental factors

T – + – + H – – + + Design factors R + – – +

Product A B C D i ii iii iv Mean Range

1 – – – – 88 85 88 85 86.50 3 2 + – – + 80 77 80 76 78.25 4 3 – + – + 90 84 91 86 87.75 7 4 + + – – 95 87 93 88 90.75 8 5 – – + + 84 82 83 84 83.25 2 6 + – + – 85 84 82 82 83.25 3 7 – + + – 91 93 92 92 92.00 2 8 + + + + 89 88 89 87 88.25 2



Lecture 6.1 78



Lecture 6.1


− Why split?

− Why block?










Lecture 6.1 79


Interaction between Factors

Case study: Emotional Arousal

Male and female subjects presented with four

different visual stimuli,

pictures of

– an infant

– a landscape

– a male nude

– a female nude

Levels of subjects' emotional arousal were measured

Arousal.xls



Lecture 6.1 80




Infa

nt

Lan

dsd

cap

e

Nu

de

Fem

ale

Nu

de M

ale

10

15

20

25

Male

Pictures In

fan

t

Lan

dsd

cap

e

Nu

de

Fem

ale

Nu

de M

ale

10

15

20

25

Female

Pictures

Levels of Arousal of Males and Females to Different Visual Stimuli



Lecture 6.1 81




NMNFLI

25

20

15

10

Picture

Me

an

Aro

usa

l Le

ve

l

Picture Main Effects Plot

NMNFLI

25

20

15

10

Picture

Me

an

Aro

usa

l Le

ve

l

F

M

Gender

Interaction Plot



Lecture 6.1 82


Minute test

– How much did you get out of today's class?

– How did you find the pace of today's class?

– What single point caused you the most

difficulty?

– What single change by the lecturer would have

most improved this class?



Lecture 6.1 83


Reading

Lecture Notes: Split Units Design and Analysis

Lab 2 Feedback

(BHH §13.1 to p. 544)



Documents

Design and Analysis of Experiments Lecture 6 · Lecture 6.1 1Postgraduate Certificate in Statistics © 2016 Michael Stuart Design and Analysis of Experiments Lecture 6.1 1. Review