Statistically Quality Design
Topics
–
–
–
–
–
–
Four perspectives in quality improvement
Review DOE topics and terminologies
Implementation plan and procedure for
experimental design
Full factorial design and Yates’ algorithm
Full factorial design example: improving wave
solder process at TDY company
Concepts and examples for conducting Block and
Latin square designs
Four Perspectives in Quality Improvement
1. Downstream Perspective: Customer’s Quality
– such as: fuel consumption, noise, failure rates, pollution, etc.
2. Midstream Perspective: Manufacturing Quality (spec.+ drawings)
– important for production or trading.
3. Upstream: Quality of Design (robustness of objective function)
– good for design & development after product planning
4. Origin: Quality of Technology (robustness of technology )
– good for technology development prior to product planning
– functionality of generic function
– example: Hook’s Law for spring Y    M
 Quality Engineering
提供在產品設計及生產過程中對市場可能會發生故障之預測技術/方法
Characteristics of Technology Development
1. Technology Readiness (先行性): 若產品計畫前作技術開發
則設計完後僅須作適當調整
2. Flexibility (汎用性):針對一系列或下一代產品作品質改善
3. Reproducibility (再現性): R & D  Manufacturing  Market
Statistically Quality Design
Topics
–
–
–
–
–
–
Four perspectives in quality improvement
Review DOE topics and terminologies
Implementation plan and procedure for
experimental design
Full factorial design and Yates’ algorithm
Full factorial design example: improving wave
solder process at TDY company
Concepts and examples for conducting Block and
Latin square designs
Design of Experiments (Topics)
CRD
- completely randomized design
Full Factorial
- all possible combinations of factors and levels
Fractional Factorial
- assume some interaction will not occur, a factor is assigned
Latin Square
- each level of each factor appears only once with each level of two
other factors
Yates’ Notation & Algorithm
On-line V.S. Off-line Quality Control
Design of Experiments (Topics cont.)
System Design
– the selection of materials, parts, equipment, and process parameters
Parameter Design
– study the effect of noise factor in DOE
Tolerance Design
– the specification of appropriate tolerance, product and process parameters
Signal to Noise Ratio (S/N Ratio)
– a comparison of the influence of control factors (signal) to that of
noise factors
Orthogonal Array
− a design where correlation between factors is zero
Outer Array
– in parameter design to identify the combination of noise factors
Experimental Design Terminology
ANOVA
– Analysis of Variance
Experimental Unit
– largest collection of experiment material
Treatment
– what is done to the experiment materials
Sampling Unit
– a part of experimental unit
Experimental Design Terminology (cont.)
Experimental/ Sampling Errors
– a measure of variation
Randomization
– a system of using random number
Replication
– number of times a specific combination of factor level is run during an
experiment
Factor
– an input to a process produces an effect. controllable factors vs. noise
factors
Level
– a setting or value of a factor
Run
– number of trials for each condition of an experiment
Experimental Design Terminology (cont.)
Quality Characteristic
– the response variable (output)
Interaction
– the combination of two factors generates a result that is different
from individual factor.
– main effect vs. interaction effect
DOF
– independent piece of information
Resolution
– number of letters in the shortest length in defining relation
– the lower the number, the more saturated the design is
ANOVA (變異數分析)
The method of analyzing data collected by CRD / RCBD
ANOVA equation

 
Yij  Y ..  Y i.  Y ..  Yij  Y i.





  Yij  Y ..   Y i.  Y ..   Yij  Y i.
i
=>
2
j
SST
i
=
2
j
i
SStrt
+


2
j
SSE
ANOVA Table
source of variation
d.f.
Between Trt.
a -1
Error
n–a
Total
N-1
SS
 Y
MS

 Y  Y 
 Y  Y 
i.
 Y ..
ij
i.
ij
..
2
2
2
MStrt  SStrt a  1
MSE  SSE N  a 
F
MStrt MSE
ANOVA
Example:
一個製造紙袋用紙的工廠想改善其紙張強度，若已知紙張強度與紙漿之
濃度有關，今欲調查5種不同之濃度(Hardwood Concentration)，5%，
10%，15%，20%及25%，每一濃度取5個觀察值(obs.，tensile strength)，
其結果如下表：
Obs.
濃度
1
2
3
4
5
Yi.
5%
7
7
15
11
9
49
9.8
10%
12
17
12
18
18
77
15.4
15%
14
18
18
19
19
88
17.6
20%
19
25
22
19
23
108
21.6
25%
7
10
11
15
11
54
10.8
Y..  376
Y i.
ANOVA (cont.)
Yij     i   ij  i   ij
 ij ~ N (0,  2 )
H 0 : 1  2  3  4  5 (i.e. 1   2   3   4   5  0)
H 0 : 至少有一 i  0




SStrt   Y i.  Y ..  475.56
2


SSE   Yij  Y i.  161.20
2
SSTO   Yij  Y ..  636.96
2
source of variation
d.f.
SS
MS
F
Between Trt.
5 -1
475.56
118.89
14.75062
Error
25 – 5
161.20
8.06
Total
25 - 1
636.96
Fixed Effect vs. Random Effect
Fixed Effect
– 一個工廠有三部機器，觀察三者間有無顯著差異(chosen in a
nonrandom manner/ a small hand-selected factor level)
Radom Effect
– 某工廠有30部同類型機器，由其中隨機取出三台，由此三台機器
觀察工廠內30台機器之管理狀態(study the source of variability/
the variation associated with a factor)
Fixed Effect
Radom Effect
Effect A
F0  MS A MSE
F0  MS A MS AB
Effect B
F0  MS B MSE
F0  MS B MS AB
Effect AB
F0  MS AB MSE
F0  MS AB MSE
Statistically Quality Design
Topics
–
–
–
–
–
–
Four perspectives in quality improvement
Review DOE topics and terminologies
Implementation plan and procedure for
experimental design
Full factorial design and Yates’ algorithm
Full factorial design example: improving wave
solder process at TDY company
Concepts and examples for conducting Block and
Latin square designs
Implementation Plan and Procedure for
Experimental Design
Basically, a twelve-steps (procedure) approach
for conducting any experimental design can be
divided into the following three stages:
Stage 1 (準備及設計選擇階段)
1.
Define the problems and state the objective of the experiment
2.
Select quality characteristic (response) and input variables (factors)
3.
Determine the desired number of runs and replications
4.
Consider the randomization of runs during the selection of the best
design type
Stage 2 (實驗及分析資料階段)
1.
Conduct the experiment and record the data
2.
Analyze the data using analyze of mean, analysis of variance
3.
Use Yates’ algorithm and normal probability plot to determine the
significant main and interaction effects
Stage 3 (建立預估模式及確認評估階段)
1.
Develop a fitted model using regression analysis
2.
Draw conclusion and make prediction
3.
Perform confirmatory tests
4.
Assess results and make decision
Steps for Experimental Design
1. Statement of the problem:_________________________________
(During this step you should estimate your current level of quality by way of
Cpk, dpm, or total loss. This estimate will be compared with improvements
found after Step 7.)
2. Objective of the experiment:_______________________________
3. Star Date:_____________ ; End Date:_____________
4. Select quality characteristics
(also known as responses, dependent variables, or output variables). These
characteristics should be related to customer needs and expectations.
Response
1
2
3
Type
Anticipated Range
How will you measure the
response?
Steps for Experimental Design (cont.)
5. Select factors
(also know as parameters or input variables) which are anticipated to have an
effect on response.
Factor
Type
Controllable or
Noise
Range of
Interest
Levels
Anticipated
Interactions With
How
Measured
1
2
3
6. Determine the number of resources to be used in the experiment
(Consider the desired number, the cost per resource, time per experimental
trial, and maximum allowable number of resources.)
7. Which design types and analysis strategies are appropriate?
(Discuss advantage and disadvantages of each.)
Steps for Experimental Design (cont.)
8. Select the best design type and analysis strategy to suit your needs
9. Can all the runs be randomized?__________________________
Which factors are most difficult to randomize?________________
10. Conduct the experiment and record the data
(Monitor both of these events for accuracy)
11. Analyze the data, draw conclusions, mark predictions, and do
confirmatory tests
12. Assess results and make decisions
(Evaluate your new state of quality and compare with the quality level prior to the
improvement effort. If necessary, conduct more experimentation.)
Statistically Quality Design
Topics
–
–
–
–
–
–
Four perspectives in quality improvement
Review DOE topics and terminologies
Implementation plan and procedure for
experimental design
Full factorial design and Yates’ algorithm
Full factorial design example: improving wave
solder process at TDY company
Concepts and examples for conducting Block and
Latin square designs
Full Factorial Experimental Design
Principles:
Random Number Table
There are 400 digits in this random number table,
3 appears 41 times.
3 Factors, 2 Levels
Four dimensional visibility with
2  8 test combinations in a full factorial matrix
3
Label The Cells
8 Test
Combinations 23
A-
A+
B-
(1)
a
B+
b
ab
B-
c
ac
bc
abc
C-
C+
B+
Yates’ Notation
2 8
3
Cell
A
(1)
a
+
b
ab +
c
ac
+
bc
abc +
B
+
+
+
+
Test Combinations
AB
+
+
+
+
C
+
+
+
+
AC
+
+
+
+
BC
+
+
+
+
ABC
+
+
+
+
Yates’ Notation
2  16 Test Combinations
4
Yates’ Work Session
Y = yield strength , PSI
A, B and C are concentrations of 3 separate elements
A-
A+
B-
58
56
36
39
B+
51
53
34
32
B-
53
48
54
59
B+
49
49
55
61
C-
C+
Determine the size of each contrast using Yates’ algorithm
What combination of elements will give the highest yield strength?
The Yates’ Algorithm
two variables; A, B
number of variables, n = 2
number of columns, n = 2
For top ½ of each column: 1st  2nd


Yates’ Work Session
Yates’ Worksheet, 3 Variables
Cell
y
y
(1)
a
b
ab
c
ac
bc
abc
TOTAL
Y
1
2
3
 4 RANK
Analysis of Variance for a A×B Factorial
Experiment
ANOVA of factorial experiment:
The total sum of squares can be partitioned into :
Total SS = SS(A) + SS(B) + SS(AB) + SSE
ANOVA Table For AXB Factorial Experiment
Source
d.f.
SS
MS
Factor A
Factor B
Interaction AB
Error
(a-1)
(b-1)
(a-1)(b-1)
(n-ab)
SS(A)
SS(B)
SS(AB)
SSE
SS(A)/(a-1)
SS(B)/(b-1)
SS(AB)/((a-1)(b-1)
SSE/(n-ab)
Total
(n-1)
Total SS
n = rab
r = number of times each factorial treatment combination appears in the experiment
A×B Factorial Experiment (Cont.)
 A2
SS ( A) 
 CF
rb
 B2
SS ( B) 
 CF
ra
( AB) 2
SS ( AB) 
 CF  SS ( A)  SS ( B)
r
SSE  TOTAL SS  SSA  SSB  SS ( AB)
Test each null hypothesis:
MS ( A)
MS ( B)
MS ( AB)
F
and F 
and F 
MSE
MSE
MSE
Example: A × B Factorial Experiment
 The evaluation of a flame retardant was conducted at two
different laboratories on three different materials with the
following results
Materials
Laboratory
1
2
3
1
4.1 , 3.9
4.3
2.7 , 3.1
2.6
3.1 , 2.8
3.3
1.9 , 2.2
2.3
3.5 , 3.2
3.6
2.7 , 2.3
2.5
2
Example: A×B Factorial Experiment (Cont.)
Total For Calculating Sums of Squares
Material (B)
Laboratory
1
2
3
Total (A)
1
2
12.3
8.4
9.2
6.4
10.3
7.5
31.8
22.3
Total (B)
20.7
15.6
17.8
54.1
There are n = rab = (3)(2)(3) =18 observation
(54.1) 2
CF 
 162.6006
18
Total SS
 ( 4.12  3.9 2  ...  2.5 2 )  CF
 170.53  162.6006  7.9294
(31.8 2  22.32 )
 CF
SS(A) 
9
 167.6144  162.6006  5.0139
( 20.7 2  15.6 2  17.8 2 )
 CF
SS(B) 
6
 164.7817  162.6006  2.1811
(12.32  9.2 2  ...  7.5 2 )
SS(AB) 
 CF  SS ( A)  SS ( B )
3
 169.93  162.6006  5.0139  2.1311  .1344
SSE  TOTAL SS  SS ( A)  SS ( B )  SS ( AB)
 7.9294  5.0139  2.1811  .1344  .6000
Example: A×B Factorial Experiment (Cont.)
ANOVA Table
Source
d.f.
SS
MS
F
Laboratory (A)
Material (B)
Interaction (AB)
Error
1
2
2
12
5.0139
2.1811
.1344
.6000
5.0139
1.0906
0.672
0.0500
100.28
21.81
1.34
Total
17
7.9294
 Testing hypothesis to confirm interaction exists or not
MS ( AB) 0.0672
F

 1.34
MSE
0.05
Since F0.05
2 ,12
 3.89, the interaction is not significant
The null hypothesis is not rejected
No differences among interaction
MS ( B) 1.0906
F

 21.81
MSE
0.0500
 Since F0.05
2 ,12
 3.89, the null hypothesis is rejected.
The laboratory and material are important.
Main Effect Larger Than Interaction
Interaction Larger Than Main Effect
Two-way ANOVA
Open the two_way.mtw worksheet
Stat  ANOVA  Two-way Analysis of Variance
response
Materials
Materials
Enter OK
ANOVA Table
P-value < 0.05
1. The materials and laboratory are significant (important).
2. The interaction is not significant.
Main Effects Plot
Stat  ANOVA  Main Effects Plot
Main Effects Plot for response
Data Means
Materials
Laboratory
3.50
Mean
3.25
3.00
2.75
2.50
1
2
3
1
2
Interactions Plots
Stat  ANOVA  Interactions Plot
Interaction Plot for response
Data Means
Materials
1
2
3
4.0
Mean
3.5
3.0
2.5
2.0
1
2
Laboratory
Statistically Quality Design
Topics
–
–
–
–
–
–
Four perspectives in quality improvement
Review DOE topics and terminologies
Implementation plan and procedure for
experimental design
Full factorial design and Yates’ algorithm
Full factorial design example: improving wave
solder process at TDY company
Concepts and examples for conducting Block and
Latin square designs
Full Factorial Experiment Example:
Improving Wave Solder Process at Teledyne
• Objective :
– To determine the effect of flux type and lead length on the
DFDAU wave soldering (WS) defects
• Planted steps for statistically designed experiment
(1) Select output variables, 2 factors, 2 levels and 8 Runs
(2) Randomize the sequence of runs and labels 8 DFDAU boards
(3) Select two touchup operators to check the WS defects
consistency
(4) Iso-plot the major WS defects for top/rear sides to compare one
operator against another
(5) Analyze the data using ANOVA table with interactions or using
Yates' algorithm
(6) Plot/interpret the results and draw the conclusions
Wave Soldering Process Flow Chart
Statistically Designed Experiment
Run No.
Y1
Y2
Y3
Y4
Y5
Y6
Y7
Y8
Flux Type
New(OA)
Old(RMA)
Old(RMA)
New(OA)
Old(RMA)
New(OA)
New(OA)
Old(RMA)
Lead Length
Label
Trimmed Leads
ab
Std. Lead Length
(1)
Std. Lead Length
(1)
Std. Lead Length
b
Trimmed Leads
a
Trimmed Leads
ab
Std. Lead Length
b
Trimmed Leads
a
New Flux = Alpha # 857
Old Flux = RMA
Std. Lead Length (IC connect. point not trimmed;
IC接點處之引線未被切平)
Trimmed Leads ( about .045”)
Yates’ Algorithm
2 4
2
Notations
ANOVA Table
22 = 4 Combinations
Std. Leads
A-
Trimmed Leads
A+
Cell
(1)
Contrasts
A B AB
+
Old
Flux B-
(1)
a
a
+
-
-
New
B+
Flux
b
ab
b
-
+
-
ab
+
+
+
Operator #1
Std. Leads Trimmed Leads
12
10.5
Old Flux
New Flux
9
7.5
9
6
7
8
8
9
Avg. Operator
Std. Leads
7
6
10.5
Old Flux
Operator #2
11
9
9.5
Old Flux
7
8
5
7
7
7
New Flux
7
4.5
2
Trm. Leads
7.5
10
7.25
9.5
7
8
7
7.5
New Flux
7
5.75
4.5
ANOVA Analysis For Two Factorial
Experiment
Two way ANOVA for flux type and lead length
Interaction Plot for Flux Type and Lead
Length
Interaction Plot for response_aver
Data Means
Flux
Type_aver
new
old
10
Mean
9
8
7
6
Std
Trimmed
Lead Length_aver
2 Factor Full Factorial Experiment
Summary and Conclusions
Summary of Findings:
• ISPLOT reveals that 2 operators
were fairly consistent in calling
out VIA defects
Conclusions:
• The ANOVA/Yates’s Analysis
indicates lead length to be the most
significant factor
• VIA defects consist of 77 %
• Interaction between flux and lead
length proven to be the least
significant factor
vs. 93 % of total defects, 1 vs. 2
• Only the rear VIA defects are
considered for output measures
• Defect level : 2941 ppm
(Trimmed leads)
• Defect level : 3959 ppm
(Std. lead length)
• 26 % improvement can be expected
if using the “ Trimmed Leads“
• 23 % improvement can be expected
if using the “ OA “ Flux
• Optimal range for the board
temperature needs to be further
studied
Statistically Quality Design
Topics
–
–
–
–
–
–
Four perspectives in quality improvement
Review DOE topics and terminologies
Implementation plan and procedure for
experimental design
Full factorial design and Yates’ algorithm
Full factorial design example: improving wave
solder process at TDY company
Concepts and examples for conducting Block and
Latin square designs
Latin Square (拉丁方格 )
Operators
Processes
1
2
3
Model
I
II
III
A
B
C
B
C
A
C
A
B
Yijk     i   j   k   ijk i, j, k  1..., r
Operators
Processes
Material Source
I, II, III
1, 2, 3
A, B, C
Greco-Latin Square
I
II
III
1
A
B
C
2
B
C
A
3
C
A
B
Operators
Processes
Material # 1 Source
Material # 2 Source
I, II, III
1, 2, 3
A, B, C
 ,  ,
Latin Square Design
 By using a Latin square design, three sources of variation, A, B and
C, can be investigated simultaneously, providing there is no
interaction between the three factors and also that each of them has
the same number of levels r.
For example, suppose each factor has four levels denoted by
A1 , A2 , A3 , A4 , B1 , B2 , B3 , B4 and C1 , C2 , C3 , C4 . If factor A is associated
with the rows of the table and B with the columns of the table then
each levels of factor C must appear once in each row and once in
each column. In order to achieve this a systematic cyclic pattern, it can
be set down for the C’s as shown in the table. To randomize the design,
the allocation of the A’s and B’s to the rows and columns is then
carried out at random.
Ai
Bj
A2
A4
A3
A1
B4
B2
B1
B3
C1
C4
C2
C1
C3
C2
C4
C3
C3
C4
C1
C2
C2
C3
C4
C1
Latin Square Models
Latin Square Model:
Yijk     i   j   k   ijk i, j, k  1..., r
The α’s, β’s, γ’s and ε’s are mutually independent.
Analysis of variance for a Latin Square design
The total sum of squares is divided into four component parts, one
for each source of variation and one for the residual. 2
Y ...     yijk ,
N  r2,
C.F . 
Y ...
N
SST     y 2 ijk  C.F .
1
SSA   Y 2 i ..  C.F .
r
1
SSB   Y 2 .J .  C.F .
r
1
SSC   Y 2 ..k  C.F .
r
SSE  SST  SSA  SSB  SSC
Here Yi… is the sum of over the r observations in which factor A is at level
i, with similar interpretation for Y.j. and Y..k and Y… is the sum of all the r2
observations.
The analysis and test statistics are summarized in the
following ANOVA table.
ANOVA Table for Latin Square
Source
d.f
S.S
M.S
Factor A
r-1
SSA
2
ˆ
2
ˆ 2 2 ˆ12
Factor B
r-1
SSB
2
ˆ
3
ˆ 32 ˆ12
Factor C
r-1
SSC
ˆ 4 2
ˆ 4 2 ˆ12
r2 - 3r + 2
SSE
ˆ12
r2 - 1
SST
Residual
Total
F
Example
The following 4 X 4 Latin Square in which the effects of three factors, farm,
type of fertilizer applied, and method of application (C1, C2, C3 or C4) on
the yield crop are being investigated.
Fertilizer
B1 B2
Farm
A1
A2
A3
A4
B3
B4
33C4 33C3 33C1 35C2
38C2 33C1 37C3 32C4
33C1 36C2 35C4 32C3
32C3 32C4 37C2 29C1
To ease the calculations, the data can be coded by subtracting 33 from
each observation. Then the row and column totals and the totals for each
method of application are calculated. (扣除33, 不致影響ANOVA分析)
Farm
Total
Method
Total
1
2
3
4
1
2
0C4
5C2
0C1
0C3
0C1
3C2
 1C3
4
Fertilizer
3
4
0C1
4C3
2C4
2C2
 1C4
 1C3
 1C4
4C2
 4C1
2
10
-4
C1
C2
C4
-4
14
C3
2
10
Total
2
8
4
-2
12
12
Y ...  12
C.F .  12
N  r 2  16
2
 9.0
16
SST  0 2  0 2  0 2  2 2  ...(4) 2  C.F .  94  9  85
2
2
2
2
[
2

8

4

(

2
)
]
SSA 
 C.F .  22  9  13
4
2
2
2
2
SSB  [4  2  10  (4) ]  C.F .  34  9  25
4
2
2
2
2
[(

4
)

14

2

0
]  C.F .  54  9  45
SSC 
4
SSE  SST  SSA  SSB  SSC  2
The calculations necessary for testing the significant of the three
factors are summarized in the following ANOVA table.
Source
Farm
Fertilizer
Method
Residual
Total
d.f
3
3
3
6
15
S.S
13
25
45
2
85
M.S
4.33
8.33
15.00
0.333
F
13.0
25.0
45.0
Since the critical value are F0.99(3, 6) = 9.78 and F0.999(3, 6) = 23.70,
the farm effect is significant at 1 % level. The type of fertilizer used
and the method of application are both significant at the 0.1 % level.
Latin Square Models
Open the Latin Square.mtw worksheet
Stat  ANOVA  General Linear Model
P-value < 0.05
The farm effect, the type of fertilizer used and the
method of application are significant at α= 0.05