design and analysis of experiments lecture 3...design and analysis of experiments lecture 3.1 1....
TRANSCRIPT
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Lecture 3.1 1
© 2016 Michael Stuart
Design and Analysis of Experiments
Lecture 3.1
1. Review of Lecture 2.2
– 2-level factors
– Homework 2.2.1
2. A 23 experiment
3. 24 in 16 runs with no replicates
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Lecture 3.1 2
© 2016 Michael Stuart
2k Factorial Designs
• Designs with k factors each at 2 levels
• 2k factor level combinations
– 22 = 4
– 23 = 8
L H
A
H
B
L
L H
A
H
B
L
H
L C
Ref: Lecture 1.1
Multi-factor Designs
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Lecture 3.1 3
© 2016 Michael Stuart
Why use 2- level factorial designs?
3 factors each with 2 levels: 23 = 8
3 factors each with 3 levels: 33 = 27
3 factors each with 4 levels: 43 = 64
3 factors with levels 3, 4 and 5, respectively:
3x4x5 = 60
More reasons later!
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Lecture 3.1 4
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A 22 experiment
Project:
optimisation of a chemical process yield
Factors (with levels):
operating temperature (Low, High)
catalyst (C1, C2)
Design:
Process run at all four possible combinations of
factor levels, in duplicate, in random order.
2k Factorial Designs
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Lecture 3.1 5
© 2016 Michael Stuart
Design matrix
Replicate 1
Replicate 2
Columns designate experimental factors
Rows designate experimental conditions
Design
Point Temperature Catalyst
1 Low 1
2 High 1
3 Low 2
4 High 2
5 Low 1
6 High 1
7 Low 2
8 High 2
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Lecture 3.1 6
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Results (standard order)
Standard Order
Run Order
Temperature Catalyst Yield
1 6 Low 1 60
2 8 High 1 72
3 1 Low 2 52
4 4 High 2 83
5 3 Low 1 54
6 7 High 1 68
7 2 Low 2 45
8 5 High 2 80
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Lecture 3.1 7
© 2016 Michael Stuart
Effects as calculated by Minitab
Estimated Effects and Coefficients for Yield
Term Effect Coef SE Coef T P
Constant 64.25 1.31 49.01 0.000
Temperature 23.0 11.50 1.31 8.77 0.001
Catalyst 1.5 0.75 1.31 0.57 0.598
Temperature*Catalyst 10.0 5.00 1.31 3.81 0.019
S = 3.70810
Effect = Coef x 2
SE(Effect) = SE(Coef) x 2
T effect: 23 = Mean Yield at HighT – Mean Yield at Low T
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Lecture 3.1 8
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Direct Calculation
Temp Cat Rep
1 Rep
2 Rep1 – Rep2
Variance =
½(Rep1–Rep2)² Degrees of Freedom
Low 1 60 54 6 18 1
High 1 72 68 4 8 1
Low 2 52 45 7 24.5 1
High 2 83 80 3 4.5 1
Sum 55 4
Mean = s² 13.75
s 3.7
LowY HighY LowHigh YY52.75 75.75 23
2/s4/s2)YY(SE 2LowHigh 2.6
6.2
23
)YY(SE
YYt
LowHigh
LowHigh8.8
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Lecture 3.1 9
© 2016 Michael Stuart
Direct Calculation
Temp Cat Rep
1 Rep
2 Rep1 – Rep2
Variance =
½(Rep1–Rep2)² Degrees of Freedom
Low 1 60 54 6 18 1
High 1 72 68 4 8 1
Low 2 52 45 7 24.5 1
High 2 83 80 3 4.5 1
Sum 55 4
Mean = s² 13.75
s 3.7
LowY HighY LowHigh YY52.75 75.75 23
2/s4/s2)YY(SE 2LowHigh 2.6
6.2
23
)YY(SE
YYt
LowHigh
LowHigh8.8
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Lecture 3.1 10
© 2016 Michael Stuart
Classwork 2.2.2
Calculate a confidence interval for the Temperature
effect.
t4, .05 = 2.78
CI: 23 2.78 x 2.6
23 7.2
15.8 to 30.2
23YY LowHigh
6.2)YY(SE LowHigh
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Lecture 3.1 11
© 2016 Michael Stuart
Design matrix
Design
Point Temperature Catalyst
1 Low 1
2 High 1
3 Low 2
4 High 2
5 Low 1
6 High 1
7 Low 2
8 High 2
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Lecture 3.1 12
© 2016 Michael Stuart
Design matrix:
generic notation
+ = "High"; – = "Low"
Design
Point
Temperature
A
Catalyst
B
1 – –
2 + –
3 – +
4 + +
5 – –
6 + –
7 – +
8 + +
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Lecture 3.1 13
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Design Matrix with Y’s
Main effect estimates:
¼ (Y2+Y4+Y6+Y8) – ¼ (Y1+Y3+Y5+Y7)
¼ (Y3+Y4+Y7+Y8) – ¼ (Y1+Y2+Y5+Y6)
Design
Point
Temperature
A
Catalyst
B Yield
1 – – Y1
2 + – Y2
3 – + Y3
4 + + Y4
5 – – Y5
6 + – Y6
7 – + Y7
8 + + Y8
Â
B̂Postgraduate Certificate in Statistics
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Lecture 3.1 14
© 2016 Michael Stuart
Design Matrix with Y’s
Main effect estimates:
¼ (Y2+Y4+Y6+Y8) – ¼ (Y1+Y3+Y5+Y7)
¼ (Y3+Y4+Y7+Y8) – ¼ (Y1+Y2+Y5+Y6)
Design
Point
Temperature
A
Catalyst
B Yield
1 – – Y1
2 + – Y2
3 – + Y3
4 + + Y4
5 – – Y5
6 + – Y6
7 – + Y7
8 + + Y8
Â
B̂Postgraduate Certificate in Statistics
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Lecture 3.1 15
© 2016 Michael Stuart
Design Matrix with Y’s
Interaction effect estimates:
½ (Y4+Y8) – ½ (Y3+Y7)
½ (Y2+Y6) – ½ (Y1+Y5)
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Design
Point
Temperature
A
Catalyst
B Yield
1 – – Y1
2 + – Y2
3 – + Y3
4 + + Y4
5 – – Y5
6 + – Y6
7 – + Y7
8 + + Y8
BHighatÂ
BLowatÂ
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Lecture 3.1 16
© 2016 Michael Stuart
The interaction effect
A effect at High B: ½ (Y4 + Y8) – ½ (Y3 + Y7)
A effect at Low B: ½ (Y2 + Y6) – ½ (Y1 + Y5)
½ difference: ¼ (Y4 + Y8) – ¼ (Y3 + Y7)
– ¼ (Y2 + Y6) + ¼ (Y1 + Y5)
= ¼ (Y4 + Y8 + Y1 + Y5) – ¼ (Y3 + Y7 + Y2 + Y6)
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Lecture 3.1 17
© 2016 Michael Stuart
Design Matrix: applying the signs
Effect estimates: Sum/4 Sum/4
Classwork 3.1.1: Check correspondence with Slide 14.
 B̂
Temperature
A
Catalyst
B
– Y1 – Y1
+ Y2 – Y2
– Y3 + Y3
+ Y4 + Y4
– Y5 – Y5
+ Y6 – Y6
– Y7 + Y7
+ Y8 + Y8
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Lecture 3.1 18
© 2016 Michael Stuart
Augmented Design Matrix with Y’s
Interaction effect:
(Y1 – Y2 – Y3 + Y4 + Y5 – Y6 – Y7 + Y8)/4
Classwork 3.1.2: Check correspondence with Slide 16
Design Point
A B AB Yield
1 – – + Y1 2 + – – Y2 3 – + – Y3
4 + + + Y4
5 – – + Y5
6 + – – Y6
7 – + – Y7
8 + + + Y8
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Lecture 3.1 19
© 2016 Michael Stuart
Augmented Design Matrix with Data
Classwork 3.1.3: Calculate the estimate of the AB
interaction.
Design Point
A B AB Yield
1 – – + 60
2 + – – 72
3 – + – 52
4 + + + 83
5 – – + 54
6 + – – 68
7 – + – 45
8 + + + 80
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Lecture 3.1 20
© 2016 Michael Stuart
Dual role of the design matrix
• Prior to the experiment, the rows designate the
design points, the sets of conditions under which
the process is to be run.
• After the experiment, the columns designate the
contrasts, the combinations of design point
means which measure the main effects of the
factors.
• The extended design matrix facilitates the
calculation of interaction effects
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Lecture 3.1 21
© 2016 Michael Stuart
Lecture 3.1
1. Review of Lecture 2.2
– 2-level factors
– Homework 2.2.1
2. A 23 experiment
3. 24 in 16 runs with no replicates
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Lecture 3.1 22
© 2016 Michael Stuart
Homework 2.2.1
Test the statistical significance of and
calculate confidence intervals for
the Catalyst effect and
the Temperature by Catalyst interaction effect.
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Lecture 3.1 23
© 2016 Michael Stuart
B AB Yield
– + 60
– – 72
+ – 52
+ + 83
– + 54
– – 68
+ – 45
+ + 80
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)YY(SE
YYt
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Lecture 3.1 24
© 2016 Michael Stuart
Lecture 3.1
1. Review of Lecture 2.2
– 2-level factors
– Homework 2.2.1
2. A 23 experiment
3. 24 in 16 runs with no replicates
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Lecture 3.1 25
© 2016 Michael Stuart
Part 2 A 23 experiment
3 factors each at 2 levels
Project:
optimisation of a chemical process yield
Factors (with levels):
operating Temperature T (°C) (160, 180)
raw material Concentration C (%) (20, 40)
catalyst K (A, B)
Design:
Process run at all eight possible combinations of
factor levels, in duplicate, in random order.
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Lecture 3.1 26
© 2016 Michael Stuart
Design matrix
(standard order)
Run T C K
1 – + –
2 + – –
3 – + +
4 + – –
5 + + –
6 – – –
7 + + +
8 – – +
9 + – +
10 + + –
11 – + +
12 – – +
13 – – –
14 + – +
15 + + +
16 – + –
A 23 experiment
Run order for design points
(in duplicate)
Factor
Design Point
A = T B = C C = K
1 – – –
2 + – –
3 – + –
4 + + –
5 – – +
6 + – +
7 – + +
8 + + +
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Lecture 3.1 27
© 2016 Michael Stuart
A 23 experiment
Results, in standard order T C K Yield Mean SD
– – – 59 61 60 1.41
+ – – 74 70 72 2.83
– + – 50 58 54 5.66
+ + – 69 67 68 1.41
– – + 50 54 52 2.83
+ – + 81 85 83 2.83
– + + 46 44 45 1.41
+ + + 79 81 80 1.41
Ref: PilotPlant.xls
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Lecture 3.1 28
© 2016 Michael Stuart
A 23 experiment
– Initial analysis
– Calculating effects
– Calculating s
– Minitab analysis
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Lecture 3.1 29
© 2016 Michael Stuart
Initial analysis
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Lecture 3.1 30
© 2016 Michael Stuart
Initial analysis
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Lecture 3.1 31
© 2016 Michael Stuart
Calculating main effects
Classwork 3.1.4: Calculate K main effect
T C K Mean – – – 60 + – – 72 – + – 54 + + – 68 – – + 52 + – + 83 – + + 45 + + + 80
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Lecture 3.1 32
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Calculating interaction effects,
the extended design matrix
Classwork 3.1.5: Complete the missing columns.
Calculate the TK and TCK interactions
Design Point
T C K TC TK CK TCK Mean
1 – – – + + 60 2 + – – – – 72 3 – + – – + 54 4 + + – + – 68 5 – – + + – 52 6 + – + – + 83 7 – + + – – 45 8 + + + + + 80
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Lecture 3.1 33
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Calculating s
T C K Yield Variance =½(diff)2
– – – 59 61
+ – – 74 70
– + – 50 58
+ + – 69 67
– – + 50 54
+ – + 81 85
– + + 46 44
+ + + 79 81
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Lecture 3.1 34
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Calculating s
T C K Yield Variance =½(diff)2
– – – 59 61
+ – – 74 70
– + – 50 58
+ + – 69 67
– – + 50 54
+ – + 81 85
– + + 46 44
+ + + 79 81
2
8
32
2
8
8
2
2
Total 64
s2 8
s 2.828
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Lecture 3.1 35
© 2016 Michael Stuart
Classwork 3.1.6
Calculate the t-ratio for the Temperature effect and
the 3-factor interaction. What conclusions do you
draw?
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Lecture 3.1 36
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Minitab analysis
Estimated Effects for Yield
Term Effect SE T P
T 23.0 1.414 16.26 0.000
C -5.0 1.414 -3.54 0.008
K 1.5 1.414 1.06 0.320
T*C 1.5 1.414 1.06 0.320
T*K 10.0 1.414 7.07 0.000
C*K 0.0 1.414 0.00 1.000
T*C*K 0.5 1.414 0.35 0.733
S = 2.82843
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Lecture 3.1 37
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Diagnostic analysis
De
lete
d R
es
idu
al
Score
3
2
1
0
-1
-2
-3
210-1-2
Normal Probability Plot of the Residuals
(response is Y)
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Lecture 3.1 38
© 2016 Michael Stuart
Diagnostic analysis
Fitted Value
De
lete
d R
es
idu
al
8070605040
3
2
1
0
-1
-2
-3
Residuals Versus the Fitted Values
(response is Y)
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Lecture 3.1 39
© 2016 Michael Stuart
Exercise 3.1.1
An experiment was run to assess the effects of
three factors on the life of a cutting tool
A: Cutting speed
B: Tool geometry
C: Cutting angle.
The full 23 design was replicated three times. The
results are shown in the next slide and are available
in Excel file Tool Life.xls.
Carry out a full analysis and report.
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Lecture 3.1 40
© 2016 Michael Stuart
Exercise 3.1.1
Cutting Speed
Tool Geometry
Cutting Angle
Tool Life
- - - 22 31 25 + - - 32 43 29 - + - 35 34 50 + + - 55 47 46 - - + 44 45 38 + - + 40 37 36 - + + 60 50 54 + + + 39 41 47
Exercise 3.1.2:
Web Exercises
Ref: Tool Life.xls
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Lecture 3.1 41
© 2016 Michael Stuart
Lecture 3.1
1. Review of Lecture 2.2
2. A 23 experiment
3. 24 in 16 runs with no replicates
– Normal plot, Pareto chart, Lenth's method
– Reduced model method
– Design projection method
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Lecture 3.1 42
© 2016 Michael Stuart
Project: Improving the filtration rate of a
chemical manufacturing process
Key factor: Formaldehyde concentration.
Problem: Reducing the level of formaldehyde
concentration reduces the filtration rate
to an unacceptably low level.
Proposal: Raise levels of
– Temperature,
– Pressure and
– Stirring rate.
Design: 24 unreplicated, random run order
Part 3 24 in 16 runs, no replicates
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Lecture 3.1 43
© 2016 Michael Stuart
Data
Temperature Pressure Formaldehyde Concentration
Stirring Rate
Filtration Rate
Low Low Low Low 45 High Low Low Low 71 Low High Low Low 48 High High Low Low 65 Low Low High Low 68 High Low High Low 60 Low High High Low 80 High High High Low 65 Low Low Low High 43 High Low Low High 100 Low High Low High 45 High High Low High 104 Low Low High High 75 High Low High High 86 Low High High High 70 High High High High 96
Ref: Formaldehyde.xls
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Lecture 3.1 44
© 2016 Michael Stuart
Initial analysis
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Lecture 3.1 45
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Initial analysis
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Lecture 3.1 46
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Minitab / DOE / Factorial /
Analyse Factorial Design
Estimated Effects Term Effect
T 21.625
P 3.125
F 9.875
S 14.625
T*P 0.125
T*F -18.125
T*S 16.625
P*F 2.375
P*S -0.375
F*S -1.125
T*P*F 1.875
T*P*S 4.125
T*F*S -1.625
P*F*S -2.625
T*P*F*S 1.375 Postgraduate Certificate in Statistics
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Lecture 3.1 47
© 2016 Michael Stuart
No replication: alternative analyses
• Normal plots of effects
– if no effects present, estimated effects reflect
chance variation, follow Normal model
– a few real effects will appear as exceptions in a
Normal plot
• Lenth method
– alternative estimate of s, given few real effects
• Best approach: combine both!
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Lecture 3.1 48
© 2016 Michael Stuart
Normal Effects Plot
20
10
0
-10
-20
210-1-2
Eff
ect
Score
A T
B P
C F
D S
Factor Name
Not Significant
Significant
Effect TypeAD
AC
D
C
A
Normal Plot of the Effects
Lenth's PSE = 2.625
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Lecture 3.1 49
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Alternative view: Pareto Chart
• vital few versus trivial many (Juran) Postgraduate Certificate in Statistics
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Lecture 3.1 50
© 2016 Michael Stuart
Minitab Pareto Chart
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Lecture 3.1 51
© 2016 Michael Stuart
Lenth's method
Given several Normal values with mean 0
and given their absolute values (magnitudes, or values
without signs), then it may be shown that
SD(Normal values) ≈ 1.5 × median(Absolute values).
Denote this by s0
Given a small number of effects with mean ≠ 0, then
SD(Normal values) is inflated.
Refinement: PSE ≈ 1.5 × median(Abs values < 2.5 × s0 )
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Lecture 3.1 52
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Calculating PSE
Term Effect
T 21.625
P 3.125
F 9.875
S 14.625
T*P 0.125
T*F -18.125
T*S 16.625
P*F 2.375
P*S -0.375
F*S -1.125
T*P*F 1.875
T*P*S 4.125
T*F*S -1.625
P*F*S -2.625
T*P*F*S 1.375
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Lecture 3.1 53
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Calculating PSE, ignore signs
Term Effect
T 21.625
P 3.125
F 9.875
S 14.625
T*P 0.125
T*F 18.125
T*S 16.625
P*F 2.375
P*S 0.375
F*S 1.125
T*P*F 1.875
T*P*S 4.125
T*F*S 1.625
P*F*S 2.625
T*P*F*S 1.375
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Lecture 3.1 54
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No. Term Effect
1 T*P 0.125
2 P*S 0.375
3 F*S 1.125
4 T*P*F*S 1.375
5 T*F*S 1.625
6 T*P*F 1.875
7 P*F 2.375
8 P*F*S 2.625
9 P 3.125
10 T*P*S 4.125
11 F 9.875
12 S 14.625
13 T*S 16.625
14 T*F 18.125
15 T 21.625
Calculating PSE, order the effects
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Lecture 3.1 55
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No. Term Effect
1 T*P 0.125
2 P*S 0.375
3 F*S 1.125
4 T*P*F*S 1.375
5 T*F*S 1.625
6 T*P*F 1.875
7 P*F 2.375
8 P*F*S 2.625
9 P 3.125
10 T*P*S 4.125
11 F 9.875
12 S 14.625
13 T*S 16.625
14 T*F 18.125
15 T 21.625
x 2.5 = 9.84 x 1.5 = 3.94
x 1.5 = 2.625 avge = 1.75
Calculating PSE, order the effects
s0
PSE
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Lecture 3.1 56
© 2016 Michael Stuart
From Excel, find median(Absolute Values) = 2.625,
so initial SE is s0 = 1.5 × 2.625 = 3.9375.
2.5 × s0 = 9.84375, 5 values exceed.
The median of the remaining 10 is
mean of 1.625 and 1.875 = 1.75
Hence, PSE = 1.5 × 1.75 = 2.625.
Check Slide 50
Calculating PSE
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Lecture 3.1 57
© 2016 Michael Stuart
Assessing statistical significance
Critical value for effect is t.05,df × PSE
df ≈ (number of effects)/3
t.05,5 = 2.57
PSE = 2.625
Critical value = 6.75
Check Slide 50
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Lecture 3.1 58
© 2016 Michael Stuart
Estimating s
PSE = 2.625 is the (pseudo) standard error of an
estimated effect.
SE(effect) = (s2/8 + s2/8) = s/2.
s ≈ 2 × 2.625 = 5.25
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Lecture 3.1 59
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Lecture 3.1
1. Review of Lecture 2.2
2. A 23 experiment
3. 24 in 16 runs with no replicates
– Normal plot, Pareto chart, Lenth's method
– Reduced model method
– Design projection method
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Lecture 3.1 60
© 2016 Michael Stuart
Reduced Model method
• Select identified terms for a fitted model
– omitted terms provide basis for estimating s
→check degrees of freedom
• Estimate effects
– ANOVA used to calculate s
• Check diagnostics
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Lecture 3.1 61
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Reduced Model method
Y equals overall mean
plus
T effect
plus
F effect
plus
S effect
plus
TF interaction effect
plus
TS interaction effect
plus
chance variation Postgraduate Certificate in Statistics
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Lecture 3.1 62
© 2016 Michael Stuart
Estimated effects
(Minitab)
Estimated Effects and Coefficients for R (coded units)
Term Effect Coef SE Coef T P
Constant 70.063 1.104 63.44 0.000
T 21.625 10.812 1.104 9.79 0.000
F 9.875 4.938 1.104 4.47 0.001
S 14.625 7.312 1.104 6.62 0.000
T*F -18.125 -9.062 1.104 -8.21 0.000
T*S 16.625 8.313 1.104 7.53 0.000
S = 4.41730
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Lecture 3.1 63
© 2016 Michael Stuart
Analysis of Variance (basis for s)
Source DF SS MS F-Value P-Value
T 1 1870.6 1870.56 95.86 0.000
F 1 390.1 390.06 19.99 0.001
S 1 855.6 855.56 43.85 0.000
T*F 1 1314.1 1314.06 67.34 0.000
T*S 1 1105.6 1105.56 56.66 0.000
Error 10 195.1 19.51
Total 15 5730.9
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Lecture 3.1 64
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Diagnostics
100908070605040
2
1
0
-1
-2
Fitted Value
De
lete
d R
esid
ua
l
Versus Fits(response is R)
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Lecture 3.1 65
© 2016 Michael Stuart
Diagnostics
3
2
1
0
-1
-2
-3
210-1-2
De
lete
d R
esid
ua
l
Score
Normal Probability Plot(response is R)
Postgraduate Certificate in Statistics
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Lecture 3.1 66
© 2016 Michael Stuart
Lecture 3.1
1. Review of Lecture 2.2
2. A 23 experiment
3. 24 in 16 runs with no replicates
– Normal plot, Pareto chart
– Lenth's method
– Reduced model method
– Design projection method
Postgraduate Certificate in Statistics
Design and Analysis of Experiments
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Lecture 3.1 67
© 2016 Michael Stuart
Design Projection Method
Pressure not
statistically significant,
exclude
leave 23, duplicated
T F S FR
– – – 45
– – – 48
+ – – 71
+ – – 65
– + – 68
– + – 80
+ + – 60
+ + – 65
– – + 43
– – + 45
+ – + 100
+ – + 104
– + + 75
– + + 70
+ + + 86
+ + + 96 Postgraduate Certificate in Statistics
Design and Analysis of Experiments
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Lecture 3.1 68
© 2016 Michael Stuart
Design Projection Method
Initial analysis
Postgraduate Certificate in Statistics
Design and Analysis of Experiments
-
Lecture 3.1 69
© 2016 Michael Stuart
Design Projection Method
Initial analysis
Postgraduate Certificate in Statistics
Design and Analysis of Experiments
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Lecture 3.1 70
© 2016 Michael Stuart
Design Projection Method
Minitab analyis
Estimated Effects and Coefficients for R (coded units)
Term Effect Coef SE Coef T P
Constant 70.063 1.184 59.16 0.000
T 21.625 10.812 1.184 9.13 0.000
F 9.875 4.938 1.184 4.17 0.003
S 14.625 7.312 1.184 6.18 0.000
T*F -18.125 -9.062 1.184 -7.65 0.000
T*S 16.625 8.313 1.184 7.02 0.000
F*S -1.125 -0.562 1.184 -0.48 0.647
T*F*S -1.625 -0.813 1.184 -0.69 0.512
S = 4.73682
Postgraduate Certificate in Statistics
Design and Analysis of Experiments
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Lecture 3.1 71
© 2016 Michael Stuart
Design Projection Method
Minitab analyis
Postgraduate Certificate in Statistics
Design and Analysis of Experiments
-
Lecture 3.1 72
© 2016 Michael Stuart
Design Projection Method
Minitab analyis
Postgraduate Certificate in Statistics
Design and Analysis of Experiments
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Lecture 3.1 73
© 2016 Michael Stuart
Identify optimal operating conditions
List means and SE's for 8 design points:
T*F*S Mean SE Mean
– – – 46.50 3.349
+ – – 68.00 3.349
– + – 74.00 3.349
+ + – 62.50 3.349
– – + 44.00 3.349
+ – + 102.00 3.349
– + + 72.50 3.349
+ + + 91.00 3.349
Postgraduate Certificate in Statistics Design and Analysis of Experiments
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Lecture 3.1 74
© 2016 Michael Stuart
Identify optimal operating conditions
Calculate confidence interval for optimum yield.
CI = 102.0 2.2 × 3.35 = ( 94.63 , 109.37 )
Classwork 3.1.7
Test the statistical significance of the difference
between the optimal yield and the 'next best' yield
Calculate a confidence interval for the 'next best'
yield.
Postgraduate Certificate in Statistics
Design and Analysis of Experiments
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Lecture 3.1 75
© 2016 Michael Stuart
Identify optimal operating conditions
T*F*S Mean SE Mean
– – – 46.50 3.349
+ – – 68.00 3.349
– + – 74.00 3.349
+ + – 62.50 3.349
– – + 44.00 3.349
+ – + 102.00 3.349
– + + 72.50 3.349
+ + + 91.00 3.349
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Lecture 3.1 76
© 2016 Michael Stuart
Lecture 3.1
1. Review of Lecture 2.2
2. A 23 experiment
3. 24 in 16 runs with no replicates
Postgraduate Certificate in Statistics
Design and Analysis of Experiments
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Lecture 3.1 77
© 2016 Michael Stuart
Laboratory 1, Thursday February 4, 6-8
Postgraduate Certificate in Statistics
Design and Analysis of Experiments
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Lecture 3.1 78
© 2016 Michael Stuart
Laboratory 1, Thursday February 4, 6-8
• Students will work in pairs (or threes where
necessary) but not singly, with a view to
– exploiting synergy in solving Laboratory
problems,
– promoting collaborative learning.
• Laboratory tasks involve analysing, discussing
and reporting on the results of designed
experiments.
• Laboratory handouts give detailed guidance on
the use of Mintab.
Postgraduate Certificate in Statistics
Design and Analysis of Experiments
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Lecture 3.1 79
© 2016 Michael Stuart
Laboratory 1, Thursday February 4, 6-8
• The laboratory handouts include the following:
Invitations to consider the results of Minitab
analysis and their statistical and substantive
interpretations are printed in italics.
Take some time for this; consult your neighbour
or tutor.
Enter your responses in a Word document, as a
log of your work and as draft contributions to a
report on the experiments and their analyses.
Postgraduate Certificate in Statistics
Design and Analysis of Experiments
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Lecture 3.1 80
© 2016 Michael Stuart
Laboratory 1, Thursday February 4, 6-8
• The laboratory handouts include Learning
Objectives.
• Students should check these objectives as they
proceed through the Laboratory.
• At the end of each Laboratory, students are
invited to
– review the Learning Objectives
and
– check whether they have been achieved.
Postgraduate Certificate in Statistics Design and Analysis of Experiments
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Lecture 3.1 81
© 2016 Michael Stuart
Laboratory 1, Thursday February 4, 6-8
• A Feedback document will be circulated at the
end of the Laboratory, containing
– solutions to the Laboratory tasks
– asides, constituting Extra Notes on relevant
topics.
• Laboratory handout will appear on the module
web page tomorrow,
• Feedback document will appear a week later.
• There will be a Minute Test at the end.
Postgraduate Certificate in Statistics
Design and Analysis of Experiments
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Lecture 3.1 82
© 2016 Michael Stuart
Reading
EM §5.3, §5.4, §5.6
DCM §6-2, §6-3 to p.221, §6.5 to p. 237
(BHH § 5.14 (Lenth plots) and all of Ch. 5!)
Extra Notes:
Lenth’s PSE
Postgraduate Certificate in Statistics
Design and Analysis of Experiments