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.
+-
5654
52
5048
46
4442
40
K
Mea
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-+
P
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52
5048
46
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