Design and Analysis of Experiments Lecture 6.1

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Design and Analysis of Experiments Lecture 6.1. Review of split unit experiments Review of Laboratory 2 Review of special topics (part). Minute Test: How Much. Minute Test: How Fast. Split units experiments. arise when one set of treatment factors is applied to experimental units, - PowerPoint PPT Presentation

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Lecture 6.1 1© 2012 Michael Stuart

Design and Analysis of ExperimentsLecture 6.1

1. Review of split unit experiments

2. Review of Laboratory 2

3. Review of special topics (part)

Diploma in StatisticsDesign and Analysis of Experiments

Lecture 6.1 2© 2012 Michael Stuart

Diploma in StatisticsDesign and Analysis of Experiments

Minute Test: How Much

Lecture 6.1 3© 2012 Michael Stuart

Diploma in StatisticsDesign and Analysis of Experiments

Minute Test: How Fast

Lecture 6.1 4© 2012 Michael Stuart

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 designs.

"Most industrial experiments are ... split plot in their design.“ C. Daniel (1976) p. 175

Diploma in StatisticsDesign and Analysis of Experiments

Lecture 6.1 5© 2012 Michael Stuart

Reasons for using split units

• 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

Diploma in StatisticsDesign and Analysis of Experiments

Lecture 6.1 6© 2012 Michael Stuart

Units

Blocks

Whole units

Subunits

Recognising Plot and Treatment Structure

Factor

Whole unitTreatment

SubunitTreatment

ANOVA

MS(Blocks)

MS(WTreatments)MS(Whole units)MS(B x WT)

MS(STreatments)MS(Interactions)MS(Subunits)

Diploma in StatisticsDesign and Analysis of Experiments

Lecture 6.1 7© 2012 Michael Stuart

Illustration: Water resistance of wood stains

Diploma in StatisticsDesign and Analysis of Experiments

Lecture 6.1 8© 2012 Michael Stuart

Units

Boards

Panels

Plot structure, Treatment structure

Factor

Pretreatment

Stain

Diploma in StatisticsDesign and Analysis of Experiments

24 subunits (panels) nested in 6 whole units (boards).2 pretreatments allocated to whole units.4 stains allocated to subunits within whole units.

Lecture 6.1 9© 2012 Michael Stuart

Assessing variation

• Variation between boards due to– chance– Pretreatments?

• Variation between panels due to– chance– stains?– pretreatment by stain interaction?

Diploma in StatisticsDesign and Analysis of Experiments

Lecture 6.1 10© 2012 Michael Stuart

Units

Boards

Panels

Analysis of Variance

Factor

Pretreatment

Stain

Diploma in StatisticsDesign and Analysis of Experiments

ANOVA

MS(Pretreatment)MS(Boards)

MS(Stain)MS(Interaction)MS(Panels)

Minitab modelPretreatment Board(Pretreatment)Stain Pretreatment*Stain

Lecture 6.1 11© 2012 Michael Stuart

Analysis of Variance

Source DF SS MS F P

Pretreatment 1 782.04 782.04 4.03 0.115Board(Pretreatment) 4 775.36 193.84 15.25 0.000

Stain 3 266.00 88.67 6.98 0.006Pretreatment*Stain 3 62.79 20.93 1.65 0.231Error 12 152.52 12.71

Total 23 2038.72

Diploma in StatisticsDesign and Analysis of Experiments

Lecture 6.1 12© 2012 Michael Stuart

Justifying the ANOVA Source Expected Mean Square1 Pretreat (5) + 4.0000 (2) + Q[1]2 Board(Pretreat) (5) + 4.0000 (2)3 Stain (5) + Q[3]4 Pretreat*Stain (5) + Q[4]5 Error (5)

Alternative notation:

Source Expected Mean Square

1 Pretreat + 4 + Pretreatment effect

2 Board(Pretreat) + 4

3 Stain + Stain effect

4 Pretreat*Stain + interaction effect

5 ErrorDiploma in StatisticsDesign and Analysis of Experiments

2P

2P

2P

2P

2P

2B

2B

Lecture 6.1 13© 2012 Michael Stuart

Extending the unit structureSuppose the 6 boards were in 3 blocks of 2

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

e.g. 2 boards treated on 3 successive days

Diploma in StatisticsDesign and Analysis of Experiments

Block 1 Block 2 Block 3B P S R B P S R B P S R1 1 1 43.0 2 1 1 57.4 3 1 1 52.81 1 2 51.8 2 1 2 60.9 3 1 2 59.21 1 3 40.8 2 1 3 51.1 3 1 3 51.71 1 4 45.5 2 1 4 55.3 3 1 4 55.3

4 2 1 46.6 5 2 1 52.2 6 2 1 32.14 2 2 53.5 5 2 2 48.3 6 2 2 34.44 2 3 35.4 5 2 3 45.9 6 2 3 32.24 2 4 32.5 5 2 4 44.6 6 2 4 30.1

Lecture 6.1 14© 2012 Michael Stuart

Units

Blocks

Boards

Panels

Expanded Unit and Treatment Structure

Factor

Pretreatment

Stain

ANOVA

MS(Blocks)

MS(Pretreatment)MS(Boards)MS(B x PT)

MS(Stain)MS(Interactions)MS(Panels)

Diploma in StatisticsDesign and Analysis of Experiments

Lecture 6.1 15© 2012 Michael Stuart

Analysis of VarianceMinitab model

BlockPretreatment Block*PretreatmentStain Block*Stain Pretreatment*StainSource DF SS MS F P

Block 2 376.99 188.49 0.95 0.514

Pretreatment 1 782.04 782.04 3.93 0.186Block*Pretreatment 2 398.38 199.19 15.67 0.000

Stain 3 266.01 88.67 6.98 0.006Pretreatment*Stain 3 62.79 20.93 1.65 0.231Error 12 152.52 12.71

Total 23 2038.72

Diploma in StatisticsDesign and Analysis of Experiments

Lecture 6.1 16© 2012 Michael Stuart

Extending the treatment structure

4 Stain levels ↔ two 2-level factors:

Stain type 1 or 2

number of Coats applied 1 or 2

Diploma in StatisticsDesign and Analysis of Experiments

Lecture 6.1 17© 2012 Michael Stuart

Units

Blocks

Boards

Panels

Expanded Unit and Treatment Structure

Factor

Pretreatment

Stain x Coat

ANOVA

MS(Blocks)

MS(Pretreatment)MS(B x PT)

MS(Stain)MS(Coat)MS(Interactions)MS(Panels)

Diploma in StatisticsDesign and Analysis of Experiments

Lecture 6.1 18© 2012 Michael Stuart

Minitab model

Block

Pretreatment Block*Pretreatment

Stain Coat Stain*Coat Pretreatment*Stain Pretreatment*CoatPretreatment*Stain*Coat

Diploma in StatisticsDesign and Analysis of Experiments

Lecture 6.1 19© 2012 Michael Stuart

Analysis of Variance

Source DF SS MS F P

Block 2 376.99 188.49 0.95 0.514Pretreatment 1 782.04 782.04 3.93 0.186Block*Pretreatment 2 398.38 199.19 15.67 0.000

Stain 1 38.00 38.00 2.99 0.109Coat 1 214.80 214.80 16.90 0.001Stain*Coat 1 13.20 13.20 1.04 0.328

Pretreatment*Stain 1 43.20 43.20 3.40 0.090Pretreatment*Coat 1 18.38 18.38 1.45 0.252Pretreatment*Stain*Coat 1 1.21 1.21 0.10 0.762Error 12 152.52 12.71

Total 23 2038.72

Diploma in StatisticsDesign and Analysis of Experiments

Lecture 6.1 20© 2012 Michael Stuart

Diploma in StatisticsDesign and Analysis of Experiments

Laboratory 2, Exercise 1: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 21© 2012 Michael Stuart

Diploma in StatisticsDesign and Analysis of Experiments

Minitab analysis

Lecture 6.1 22© 2012 Michael Stuart

Diploma in StatisticsDesign and Analysis of Experiments

Minitab analysis

Normal plot vs Pareto Principle vs Lenth?

Lecture 6.1 23© 2012 Michael Stuart

Diploma in StatisticsDesign and Analysis of Experiments

Minitab analysis

Estimated Effects for Y

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

Lecture 6.1 24© 2012 Michael Stuart

Diploma in StatisticsDesign and Analysis of Experiments

Graphical and numerical summaries

+-

1.71.61.51.41.31.21.11.00.90.8

E

Mea

n

-+

B

+-

1.71.61.51.41.31.21.11.00.90.8

E

Mea

n

-+

D

Interaction Plot for Y Interaction Plot for Y

E E – + – +

B – 1.71 0.83

D – 1.31 1.17

+ 1.22 1.15 + 1.60 0.82

Lecture 6.1 25© 2012 Michael Stuart

Diploma in StatisticsDesign and Analysis of Experiments

Best conditions

Best conditions:

B Low, D High, E High.

Best conditions with E Low:

B High, D Low.

+-

1.71.61.51.41.31.21.11.00.90.8

E

Mea

n

-+

B

+-

1.71.61.51.41.31.21.11.00.90.8

E

Mea

n

-+

D

Interaction Plot for Y Interaction Plot for Y

Lecture 6.1 26© 2012 Michael Stuart

Diploma in StatisticsDesign and Analysis of Experiments

Reduced modelFit model using active terms:

B + D + E + BE + DE

DE confirmed as active.

D

B

DE

BE

E

76543210

Term

Standardized Effect

2.262

Pareto Chart of the Standardized Effects

Lecture 6.1 27© 2012 Michael Stuart

Diagnostics

Diploma in StatisticsDesign and Analysis of Experiments

1.81.61.41.21.00.8

2

1

0

-1

-2

-3

Fitted Value

Dele

ted

Resi

dual

Diagnostic Plot

Lecture 6.1 28© 2012 Michael Stuart

Diploma in StatisticsDesign and Analysis of Experiments

Diagnostics

3

2

1

0

-1

-2

-3210-1-2

Dele

ted

Resi

dual

Score

Normal Probability Plot

Lecture 6.1 29© 2012 Michael Stuart

Diploma in StatisticsDesign and Analysis of Experiments

Delete Design point 5, iterate analysis

• Effect estimates similar• Interaction patterns similar• s = 0.15, df = 9 ( = 14 – 5 )Least Squares Means for Y Mean SE MeanB*D*E - - - 1.7000 0.1532 + - - 1.2050 0.1083 - + - 1.9750 0.1083 + + - 1.2250 0.1083 - - + 0.9750 0.1083 + - + 1.3600 0.1083 - + + 0.6900 0.1083 + + + 0.9400 0.1083

0.69 2.26×0.15/√2 = 0.37 to 1.01

1.205 2.26×0.15/√2 = 0.965 to 1.445

Lecture 6.1 30© 2012 Michael Stuart

Diploma in StatisticsDesign and Analysis of Experiments

Laboratory 2, Exercise 2Cambridge Grassland Experiment

3 grassland treatmentsRejuvenator RHarrow Hno treatment C

randomly allocated to 3 neighbouring plots,replicated in 6 neighbouring blocks

4 fertilisersFarmyard manure FStraw SArtificial fertiliser Ano fertiliser C

randomly allocated to 4 sub plots within each plot.

Lecture 6.1 31© 2012 Michael Stuart

Diploma in StatisticsDesign and Analysis of Experiments

Cambridge Grassland ExperimentBlocks 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 3Treatments 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 32© 2012 Michael Stuart

Diploma in StatisticsDesign and Analysis of Experiments

Randomised Blocks analysis forTreatments

Source DF SS MS F P

WB 5 149700 29940 15.99 0.000

WT 2 49884 24942 13.32 0.002

WB*WT 10 18725 1872 **

Error 0 * *

Total 17 218309

Model: WB + WT + WBxWT

Lecture 6.1 33© 2012 Michael Stuart

Diagnostics

Diploma in StatisticsDesign and Analysis of Experiments

Lecture 6.1 34© 2012 Michael Stuart

Diagnostics

Diploma in StatisticsDesign and Analysis of Experiments

Lecture 6.1 35© 2012 Michael Stuart

WB x WT Interaction Plot

Diploma in StatisticsDesign and Analysis of Experiments

Lecture 6.1 36© 2012 Michael Stuart

WT x WB Interaction Plot

Diploma in StatisticsDesign and Analysis of Experiments

Lecture 6.1 37© 2012 Michael Stuart

Diploma in StatisticsDesign and Analysis of Experiments

Split Plots AnalysisModel: B + T + B x T

+ F + B x F + T x F

Source DF SS MS F PB 5 37425.1 7485.0 21.37 0.002 xT 2 12471.0 6235.5 13.32 0.002B*T 10 4681.1 468.1 1.94 0.079

F 3 56022.7 18674.2 151.24 0.000B*F 15 1852.1 123.5 0.51 0.914T*F 6 781.5 130.3 0.54 0.774Error 30 7239.6 241.3

Total 71 120473.3

Lecture 6.1 38© 2012 Michael Stuart

Diploma in StatisticsDesign and Analysis of Experiments

250200150100

2

1

0

-1

-2

-3

Fitted Value

Dele

ted

Resi

dual

Residuals Versus Fitted Values

Diagnostics

Lecture 6.1 39© 2012 Michael Stuart

Diploma in StatisticsDesign and Analysis of Experiments

250200150100

2

1

0

-1

-2

-3

Fitted Value

Dele

ted

Resi

dual

Residuals Versus Fitted Values

Same diagnostic, Different interpretation?

Lecture 6.1 40© 2012 Michael Stuart

Treatment comparisons

• First step: summary statistics

Variable Treatment Count MeanY C 24 181.46 H 24 150.96 R 24 157.17

s2 = MS(B*T) = 468.1; df = 10

Diploma in StatisticsDesign and Analysis of Experiments

Lecture 6.1 41© 2012 Michael Stuart

Compare Harrow with Control

Diploma in StatisticsDesign and Analysis of Experiments

tsignificanllystatisticaisdifference

23.2t

88.425.6

5.30t

25.624

1.46824

1.468ns

nsSE

5.3046.18196.150YY

05,.10

C

2

H

2

CH

Lecture 6.1 42© 2012 Michael Stuart

Compare Harrow with Control

Diploma in StatisticsDesign and Analysis of Experiments

tsignificanllystatisticaisdifference

23.2t

88.425.6

5.30t

25.624

1.46824

1.468ns

nsSE

5.3046.18196.150YY

05,.10

C

2

H

2

CH

Lecture 6.1 43© 2012 Michael Stuart

Compare Harrow with Control

Diploma in StatisticsDesign and Analysis of Experiments

tsignificanllystatisticaisdifference

23.2t

88.425.6

5.30t

25.624

1.46824

1.468ns

nsSE

5.3046.18196.150YY

05,.10

C

2

H

2

CH

Lecture 6.1 44© 2012 Michael Stuart

Compare Harrow with Control

Diploma in StatisticsDesign and Analysis of Experiments

tsignificanllystatisticaisdifference

23.2t

88.425.6

5.30t

25.624

1.46824

1.468ns

nsSE

5.3046.18196.150YY

05,.10

C

2

H

2

CH

Lecture 6.1 45© 2012 Michael Stuart

Compare Harrow with Control

Diploma in StatisticsDesign and Analysis of Experiments

tsignificanllystatisticaisdifference

23.2t

88.425.6

5.30t

25.624

1.46824

1.468ns

nsSE

5.3046.18196.150YY

05,.10

C

2

H

2

CH

Lecture 6.1 46© 2012 Michael Stuart

Diploma in StatisticsDesign and Analysis of Experiments

Laboratory 2, Exercise 3Fertiliser experiment

Fertilisers potentially affecting bean yield

Low HighDung (D): none 10 tons per acreNitrochalk (N):none 0.4 cwt per acreSuperPhosphate (P):none 0.6 cwt per acreMuriate of Potash (K): none 1.0 cwt per acre

Questions:

What factors affect bean yield?

How can bean yield be maximised?

Lecture 6.1 47© 2012 Michael Stuart

Diploma in StatisticsDesign and Analysis of Experiments

Exercise 2, Minitab analysis

N is marginally significant.

Pareto Principle suggests adding NPK and DP.

ADC

CDA

ABBD

ABDABCACDBC

DAC

BCDB

86420

Term

Effect

7.880

A DB NC PD K

Factor Name5

0

-5

-10210-1-2

Effe

ctScore

Not SignificantSignificant

Effect Type

B

Pareto Chart of the Effects

Lenth's PSE = 3

Normal Plot of the Effects

Lecture 6.1 48© 2012 Michael Stuart

Term Effect CoefConstant 47.250Block -0.375D -0.750 -0.375N -8.000 -4.000P 0.250 0.125K -2.250 -1.125D*N 1.250 0.625D*P 4.500 2.250D*K 0.000 0.000N*P 2.250 1.125N*K -1.750 -0.875P*K 0.500 0.250D*N*P -2.000 -1.000D*N*K 2.000 1.000D*P*K 2.250 1.125N*P*K -5.500 -2.750

Diploma in StatisticsDesign and Analysis of Experiments

-0.75

Lecture 6.1 49© 2012 Michael Stuart

Diploma in StatisticsDesign and Analysis of Experiments

Reduced modelFit model using active terms:

D, N, P, K, DP, NP, NK, PK, NPK

Active effects confirmed.Diagnostics unremarkable

CCD

ABDBC

DAC

BCDB

543210

Term

Standardized Effect

2.447

A DB NC PD K

Factor Name

Pareto Chart of the Standardized Effects

Lecture 6.1 50© 2012 Michael Stuart

Diploma in StatisticsDesign and Analysis of Experiments

3-factor interaction

Adding N reduces yield overall, but has a small positive effect at high P and low K.

At low N, the P effect is negative at low K and positive at high K.

At high N, this interaction is reversed.

The best combination is no fertiliser.

+-

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5048

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4442

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K

Mea

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4442

40

K

Mea

n

-+

P

Interaction Plot for YN Low

Interaction Plot for YN High

Lecture 6.1 51© 2012 Michael Stuart

Least Squares Means for Y

Mean SE Mean N*P*K - - - 55.5 2.419 + - - 41.5 2.419 - + - 47.5 2.419 + + - 49.0 2.419 - - + 49.0 2.419 + - + 42.5 2.419 - + + 53.0 2.419 + + + 40.0 2.419

Diploma in StatisticsDesign and Analysis of Experiments

Lecture 6.1 52© 2012 Michael Stuart

Best combination

N*P*K Mean SEBest: - - - 55.5 2.419Next best: - + + 53.0 2.419

Difference: 2.5 3.42

t: 2.5/3.42 = 0.73,

not statistically significantly different.

Diploma in StatisticsDesign and Analysis of Experiments

Lecture 6.1 53© 2012 Michael Stuart

Review topics (part)

• Time related issues– repeated measures– cross-over designs

• Complex block structures• Analysis of Covariance• Robustness studies• Response surface designs

Diploma in StatisticsDesign and Analysis of Experiments

Lecture 6.1 54© 2012 Michael Stuart

Time related issues1. Repeated measures

Example:

Calves fed diet supplements to improve growth

Initial weight recorded, Y0

blocks formed based on initial weight,

weights recorded at

4 weeks, Y1

8 weeks Y2

12 weeks Y3

16 weeks Y4

Diploma in StatisticsDesign and Analysis of Experiments

Lecture 6.1 55© 2012 Michael Stuart

Split plots analysis?

• Calves are whole units• Time periods are sub units

Problems:

correlation structure

varying standard deviation

Solution:

Multivariate analysis

Diploma in StatisticsDesign and Analysis of Experiments

Lecture 6.1 56© 2012 Michael Stuart

Time related issues2. Crossover designs

• Repeated measures designs compares diets on different calves,

• reduce variation by comparing diets on same calves,

• e.g. diet A for weeks 1 to 4

diet B for weeks 5 to 8

diet C for weeks 9 to 12

diet D for weeks 13 to 16• requires attention to order of dietsDiploma in StatisticsDesign and Analysis of Experiments

Lecture 6.1 57© 2012 Michael Stuart

Crossover design

• Every diet occurs– once for each calf,– once in each time

period– Latin square

? correlation structure

? carry over

? experimental set up versus actual use

Diploma in StatisticsDesign and Analysis of Experiments

Calf Time Period 1 - 4 5 – 8 9 – 12 13 - 16

1 A B C D 2 B D A C 3 C A D B 4 D C B A

Lecture 6.1 58© 2012 Michael Stuart

Complex blocking

• 2 blocking factors– calves and time periods– Latin square– Latin rectangle

• Incomplete blocks– more treatments than plots in a block– balanced incomplete blocks

Diploma in StatisticsDesign and Analysis of Experiments

Lecture 6.1 59© 2012 Michael Stuart

Diploma in StatisticsDesign and Analysis of Experiments

Analysis of Covariance

Objective: take account of variation in uncontrolled environmental variables.

Solution: measure the environmental variables at each design point and incorporate in the analysis through regression methods (Analysis of Covariance)

Effects: reduces "error" variation, makes factor effects more significant

adjusts factor effect estimates to take account of extra variation source.

Lecture 6.1 60© 2012 Michael Stuart

Diploma in StatisticsDesign and Analysis of Experiments

Analysis of Covariance; Illustration

Breaking strength of monofilament fibre (Y)produced by three different machines (1, 2, 3)

allowing for variation in fibre thickness (X)

Machine 1 Machine 2 Machine 3

Y X Y X Y X 36 20 40 22 35 21 41 25 48 28 37 23 39 24 39 22 42 26 42 25 45 30 34 21 49 32 44 28 32 15

Lecture 6.1 61© 2012 Michael Stuart

Diploma in StatisticsDesign and Analysis of Experiments

Analysis of Covariance; Minitab

Lecture 6.1 62© 2012 Michael Stuart

Diploma in StatisticsDesign and Analysis of Experiments

Analysis of Covariance; MinitabGeneral Linear Model: Y versus Machine

Source DF Seq SS Adj SS Adj MS F PX 1 305.13 178.01 178.01 69.97 0.000Machine 2 13.28 13.28 6.64 2.61 0.118Error 11 27.99 27.99 2.54Total 14 346.40

S = 1.59505

One-way ANOVA: Y versus Machine

Source DF SS MS F PMachine 2 140.4 70.2 4.09 0.044Error 12 206.0 17.2Total 14 346.4

S = 4.143

Lecture 6.1 63© 2012 Michael Stuart

Diploma in StatisticsDesign and Analysis of Experiments

Analysis of Covariance; Minitab

32.530.027.525.022.520.017.515.0

50

45

40

35

30

X

Y

123

Machine

Scatterplot of Y vs X

Lecture 6.1 64© 2012 Michael Stuart

Covariance vs BlockingChance causes and assignable causes of variation

(W. Shewhart, 1931)

Chance causes of variation are themany individually negligible and unpredictable

butcollectively influential

factors that affect a process or system.

Assignable causes of variation are thefew individually influential and predictable effect

factors that affect a process or system.

Diploma i StatisticsDesign and Analysis of Experiments

Lecture 6.1 65© 2012 Michael Stuart

Covariance vs Blocking

Blocking Chance causes

Covariance Assignable causes

Diploma i StatisticsDesign and Analysis of Experiments

Lecture 6.1 66© 2012 Michael Stuart

Diploma in StatisticsDesign and Analysis of Experiments

Robustness StudiesSeek optimal settings of experimental factorsthat 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 67© 2012 Michael Stuart

Study of Detergent Robustness

Diploma in StatisticsDesign and Analysis of Experiments

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 68© 2012 Michael Stuart

Study of Detergent Robustness

Diploma in StatisticsDesign and Analysis of Experiments

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 69© 2012 Michael Stuart

Study of Detergent Robustness

Diploma in StatisticsDesign and Analysis of Experiments

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 70© 2012 Michael Stuart

Split plots model analysis

B significant, positive,

set at high (+) level

T and TC interaction

significant

Diploma in StatisticsDesign and Analysis of Experiments

Lecture 6.1 71© 2012 Michael Stuart

Split plots model analysis

At low C, whiteness is highly sensitive to T.

At high C, whiteness is relatively insensitive to T.

Diploma in StatisticsDesign and Analysis of Experiments

Lecture 6.1 72© 2012 Michael Stuart

Conclusion

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

Diploma in StatisticsDesign and Analysis of Experiments

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