Design and Analysis of Experiments Lecture 4.1
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Transcript of Design and Analysis of Experiments Lecture 4.1
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 1© 2010 Michael Stuart
Design and Analysis of ExperimentsLecture 4.1
1. Review of Lecture 3.2
2. More on Variance Components
3. Measurement System Analysis
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 2© 2010 Michael Stuart
Minute Test: How Much
54321
16
14
12
10
8
6
4
2
0
How Much
Count
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 3© 2010 Michael Stuart
Minute Test: How Fast
5432
18
16
14
12
10
8
6
4
2
0
How Fast
Count
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 4© 2010 Michael Stuart
Homework 3.2.1
Process improvement study, reduced model:
Y = + B + C + D + BC +
Set up a "design matrix" with columns for the significant effects, headed by the effect coefficients. Calculate a fitted value for each design point by applying the rows of signs to the effect coefficients and adding the overall mean. Crosscheck with the fitted values calculated by Minitab
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 5© 2010 Michael Stuart
Homework 3.2.1
Minitab provides estimated effects:
Term Effect
NaOHCon -17.250Speed 9.750Temp 21.750NaOHCon*Speed -7.500
Model?
Y = + B + C + D + BC +
Coef 49.750 -8.625 4.875 10.875 -3.750
Excel
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 6© 2010 Michael Stuart
Homework 3.2.1
Minitab provides fitted values,
residuals,
estimate via ANOVA
Analysis of Variance for Impurity (coded units)
Source DF SS MS F PMain Effects 3 3462.75 1154.25 381.86 0.0002-Way Interactions 1 225.00 225.00 74.44 0.000Residual Error 11 33.25 3.02Total 15 3721.00
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 7© 2010 Michael Stuart
Comparison of fits
All effect estimates are the same; SE's vary.
Lenth: s = 2.25, PSE = 1.125
Reduced: s = 1.74, SE(effect) = 0.87
Projected: s = 1.87, SE(effect) = 0.94
"Projected" model has 3 interactions missing from the "Reduced" model.
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 8© 2010 Michael Stuart
Degrees of freedom
"Error" degrees of freedom relevant for t
– check ANOVA table
– count estimated effects
– use replication structure
t5, .05 = 2.57
t8, .05 = 2.31
t11,.05 = 2.20
s = 2.25
s = 1.87
s = 1.74
Ref: EM Notes Ch. 4 pp. 3, 6
Ref: Extra Notes, Models for Experiments and Lab 2 Feedback
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 9© 2010 Michael Stuart
Design Point
A B C Y
1 – – – Y1
2 + – – Y2 3 – + – Y3 4 + + – Y4 5 – – + Y5 6 + – + Y6 7 – + + Y7 8 + + + Y8
Review of Lecture 3.2 Introduction to Fractional Factorial Designs
Each row gives design points for a 4-factor experiment
Fourth column estimates D main effect.
Fourth column also estimates ABC interaction effect.
In fact, fourth column estimates D + ABC in 24-1.
D=
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 10© 2010 Michael Stuart
Fractional factorial designs
Full 24 requires 16 runs
Half the full 24 requires 8 runs
Saves resources, including time
Sacrifices high order interactions via aliasing
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 11© 2010 Michael Stuart
Fractional factorial designs
Check D = ABC, design generator
Derive ABC from first principles.
D aliased with ABC
4th column estimates D + ABC, = D if ABC = 0
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 12© 2010 Michael Stuart
Fractional factorial designs
Alias List
A = BCD
B = ACD
C = ABD
D = ABC
AB = CD
AC = BD
AD = BC
ABCD = Y
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 13© 2010 Michael Stuart
Part 2: More on Variance Components
• Identifying sources of variation
• Hierarchical design for variance component estimation
• Hierarchical ANOVA
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 14© 2010 Michael Stuart
Values
Values
•
S
T
eB
eS
B
eT
e = eB + eS + eT
Sources of variation in moisture content
Batchvariation
Samplingvariation
Testingvariation
y
•
•
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 15© 2010 Michael Stuart
Components of Variance
Recall basic model:
Y = + eB + eS + eT
Components of variance:2T
2S
2B
2Y
2T
2S
2BY
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 16© 2010 Michael Stuart
Hierarchical Design forVariance Component Estimation
Batch 1 2 3 4 5 Sample 1 2 3 4 5 6 7 8 9 10 Test 40 39 30 30 26 28 25 26 29 28 14 15 30 31 24 24 19 20 17 17 Batch 6 7 8 9 10 Sample 11 12 13 14 15 16 17 18 19 20 Test 33 32 26 24 23 24 32 33 34 34 29 29 27 27 31 31 13 16 27 24 Batch 11 12 13 14 15 Sample 21 22 23 24 25 26 27 28 29 30 Test 25 23 25 27 29 29 31 32 19 20 29 30 23 23 25 25 39 37 26 28
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 17© 2010 Michael Stuart
Minitab Nested ANOVA
Analysis of Variance for Test
Source DF SS MS F PBatch 14 1216.2333 86.8738 1.495 0.224Sample 15 871.5000 58.1000 64.556 0.000Error 30 27.0000 0.9000Total 59 2114.7333
Variance Components % ofSource Var Comp. Total StDevBatch 7.193 19.60 2.682Sample 28.600 77.94 5.348Error 0.900 2.45 0.949Total 36.693 6.058
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 18© 2010 Michael Stuart
Model for Nested ANOVA
Yijk = m + Bi + Sj(i) + Tk(ij)
SS(TO) = SS(B) + SS(S) + SS(T)
59 = 14 + 15 + 30
)YY()YY()YY(YY .ijijk..i.ij..iijk
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 19© 2010 Michael Stuart
Minitab Nested ANOVA
Expected Mean Squares
1 Batch 1.00(3) + 2.00(2) + 4.00(1)2 Sample 1.00(3) + 2.00(2)3 Error 1.00(3)
Translation:
EMS(Batch) =
EMS(Sample) =
EMS(Test) =
2B
2S
2T 42
2S
2T 2
2T
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 20© 2010 Michael Stuart
Calculation
= EMS(Test)
= ½[EMS(Sample) – EMS(Test)]
= ¼[EMS(Batch) – EMS(Sample)]
Estimation
= MS(Test)
= ½[MS(Sample) – MS(Test)]
= ¼[MS(Batch) – MS(Sample)]
2B
2S
2T
2Bs
2Ss
2Ts
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 21© 2010 Michael Stuart
Conclusions fromVariance Components Analysis
Variance Components % ofSource Var Comp. Total StDevBatch 7.193 19.60 2.682Sample 28.600 77.94 5.348Error 0.900 2.45 0.949Total 36.693 6.058
Sampling variation dominates, testing variation is relatively small.
Investigate sampling procedure.
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 22© 2010 Michael Stuart
Part 3: Measurement System Analysis
• Accuracy and Precision• Repeatability and Reproducibility• Components of measurement variation• Analysis of Variance• Case study: the MicroMeter
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 23© 2010 Michael Stuart
The MicroMeter optical comparator
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 24© 2010 Michael Stuart
The MicroMeter optical comparator
• Place object on stage of travel table
• Align cross-hair with one edge
• Move and re-align cross-hair with other edge
• Read the change in alignment
• Sources of variation:
– instrument error
– operator error
– parts (manufacturing process) variation
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 25© 2010 Michael Stuart
Precise
Biased
Accurate
Characterising measurement variation;Accuracy and Precision
Imprecise
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 26© 2010 Michael Stuart
Characterising measurement variation;Accuracy and Precision
Centre and Spread
• Accurate means centre of spread is on target;
• Precise means extent of spread is small;
• Averaging repeated measurements improves precision, SE = /√n
– but not accuracy; seek assignable cause.
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 27© 2010 Michael Stuart
Accuracy and Precision: Example
Each of four technicians made six measurements of a standard (the 'true' measurement was 20.1), resulting in the following data:
Technician Data
1 20.2 19.9 20.1 20.4 20.2 20.4
2 19.9 20.2 19.5 20.4 20.6 19.4
3 20.6 20.5 20.7 20.6 20.8 21.0
4 20.1 19.9 20.2 19.9 21.1 20.0
Exercise: Make dotplots of the data. Assess the technicians for accuracy and precision
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 28© 2010 Michael Stuart
Accuracy and Precision: Example
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 29© 2010 Michael Stuart
Repeatability and Reproducability
Factors affecting measurement accuracy and precision may include:
– instrument
– material
– operator
– environment
– laboratory
– parts (manufacturing)
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 30© 2010 Michael Stuart
Repeatability and Reproducibility
Repeatability:
precision achievable under constant conditions:
– same instrument
– same material
– same operator
– same environment
– same laboratory
• How variable is measurement under these conditions
Reproducibility:
precision achievable under varying conditions:
– different instruments
– different material
– different operators
– changing environment
– different laboratories
• How much more variable is measurement under these conditions
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 31© 2010 Michael Stuart
Measurement Capability of the MicroMeter
4 operators measured each of 8 parts twice, with random ordering of parts, separately for each operator.
Three sources of variation:
– instrument error
– operator variation
– parts(manufacturing process) variation.
Data follow
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 32© 2010 Michael Stuart
Measurement Capability of the MicroMeter
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 33© 2010 Michael Stuart
Quantifying the variation
Each measurement incorporates components of variation from
– Operator error
– Parts variation
– Instrument error
and also
– Operator by Parts Interaction
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 34© 2010 Michael Stuart
Measurement Differences
Part Operator Repeats Diffs Part Operator Repeats Diffs
1 1 96.3 95.4 0.9 5 1 99.4 99.9 -0.5 2 97.0 96.9 0.1 2 100.1 99.8 0.3 3 98.2 97.4 0.8 3 100.9 99.4 1.5 4 97.4 99.6 -2.2 4 100.0 99.4 0.6
2 1 95.5 95.8 -0.3 6 1 93.8 94.9 -1.1 2 96.1 96.8 -0.7 2 95.9 95.8 0.1 3 97.9 99.4 -1.5 3 96.3 98.5 -2.2 4 97.3 100.0 -2.7 4 94.5 94.5 0
3 1 102.8 100.3 2.5 7 1 86.4 85.4 1 2 101.5 101.4 0.1 2 86.8 86.7 0.1 3 102.6 104.3 -1.7 3 88.2 89.6 -1.4 4 101.9 101.9 0 4 88.6 89.0 -0.4
4 1 94.6 96.2 -1.6 8 1 90.5 90.5 0 2 97.8 95.5 2.3 2 89.1 90.2 -1.1 3 96.0 94.3 1.7 3 92.9 92.1 0.8 4 95.3 94.4 0.9 4 92.1 92.4 -0.3
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 35© 2010 Michael Stuart
Graphical Analysis of Measurement Differences
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 36© 2010 Michael Stuart
Average measurementsby Operators and Parts
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 37© 2010 Michael Stuart
Graphical Analysis of Operators & Parts
Part
Measu
rem
ent
876543210
105
100
95
90
85
Operator
34
12
Interaction Plot
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 38© 2010 Michael Stuart
Graphical Analysis of Operators & Ordered Parts
PartOrder
Meas
876543210
105
100
95
90
85
Operator
34
12
Interaction Plot, ordered by Parts
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 39© 2010 Michael Stuart
Quantifying the variation
Notation:
E: SD of instrument error variation
P: SD of parts (manufacturing process) variation
O: SD of operator variation
OP: SD of operator by parts interaction variation
T: SD of total measurement variation
N.B.:
so
2E
2OP
2P
2O
2T
2E
2OP
2P
2OT
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 40© 2010 Michael Stuart
Calculating sE
Part Operator Repeats ½(diff)2 Part Operator Repeats ½(diff)2
1 1 96.3 95.4 0.40 5 1 99.4 99.9 0.13 2 97.0 96.9 0.00 2 100.1 99.8 0.04 3 98.2 97.4 0.32 3 100.9 99.4 1.13 4 97.4 99.6 2.42 4 100.0 99.4 0.18
2 1 95.5 95.8 0.04 6 1 93.8 94.9 0.61 2 96.1 96.8 0.25 2 95.9 95.8 0.01 3 97.9 99.4 1.13 3 96.3 98.5 2.42 4 97.3 100.0 3.65 4 94.5 94.5 0.00
3 1 102.8 100.3 3.13 7 1 86.4 85.4 0.50 2 101.5 101.4 0.00 2 86.8 86.7 0.00 3 102.6 104.3 1.45 3 88.2 89.6 0.98 4 101.9 101.9 0.00 4 88.6 89.0 0.08
4 1 94.6 96.2 1.28 8 1 90.5 90.5 0.00 2 97.8 95.5 2.64 2 89.1 90.2 0.61 3 96.0 94.3 1.45 3 92.9 92.1 0.32 4 95.3 94.4 0.40 4 92.1 92.4 0.05
sum = 18.6 sum = 7.0
s2 = (18.6 + 7.0)/32 = 0.8
sE = 0.89
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 41© 2010 Michael Stuart
Analysis of Variance
Analysis of Variance for Diameter
Source DF SS MS F P
Operator 3 32.403 10.801 6.34 0.003Part 7 1193.189 170.456 100.02 0.000Operator*Part 21 35.787 1.704 2.13 0.026
Error 32 25.600 0.800
Total 63 1286.979
S = 0.894427
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 42© 2010 Michael Stuart
Basis for Random Effects ANOVA
F-ratios in ANOVA are ratios of Mean Squares
Check: F(O) = MS(O) / MS(O*P)
F(P) = MS(P) / MS(O*P)
F(OP) = MS(OP) / MS(E)
Why?
MS(O) estimates E2 + 2OP
2 + 16O2
MS(P) estimates E2 + 2OP
2 + 8P2
MS(OP) estimates E2 + 2OP
2
MS(E) estimates E2
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 43© 2010 Michael Stuart
Variance Components
Estimated StandardSource Value Deviation
Operator 0.5686 0.75
Part 21.0939 4.59
Operator*Part 0.4521 0.67
Error 0.8000 0.89
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 44© 2010 Michael Stuart
Diagnostic Analysis
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 45© 2010 Michael Stuart
Diagnostic Analysis
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 46© 2010 Michael Stuart
Measurement system capability
E P means measurement system cannot distinguish between different parts.
Need E << P .
Define TP = sqrt(E2 + P
2).
Capability ratio = TP / E should exceed 5
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 47© 2010 Michael Stuart
Repeatability and Reproducibility
Repeatabilty SD = E
Reproducibility SD = sqrt(O2 + OP
2)
Total R&R = sqrt(O2 + OP
2 + E2)
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Diploma in StatisticsDesign and Analysis of Experiments
Lecture 4.1 48© 2010 Michael Stuart
Reading
EM §5.7, §7.5, §8.2
BHH, Ch. 5, §§6.1 - 6.3, §9.3