Analysis of Variance for Attribute Data
Analysis of Variance for Attribute Data in Taguchi ‘s Approach
Zurnila Marli Kesuma, M.Si
Department of mathematics, Syiah Kuala University, Banda aceh, Indonesia
kesumaku@yahoo.com
Abstract. Taguchi has extended the audio concept of the signal-to-noise ratio (SN ratio) to
experiments involving many factors. The formula for signal-to-noise ratio are designed so
that an experimenter can always select the largest factor level setting to optimize the quality
characteristic of an experiment. When the quality characteristic is a proportion, such as the
fraction defective, denoted p, which can take values between 0 and 1, we use the omega (Ω)
transformation as an objective characteristics. The selection of important factor is not only
based on the response table, but also uses attribute accumulation analysis - analysis of
variance and the contribution ratio to establish significant factors. The method of conducting
the analysis of variance is create a table of data, Calculate the overall total sum of squares
due to class I and class II. Calculate the total degree of freedom. Draw the analysis of
variance. Calculate the predicted optimum process and calculate the confidence interval of a
predicted mean.
2000 Mathematics Subject Classification: 30C45, Secondary 30C80
Keywords : omega(Ω) transformation, analysis of variance,contribution ratio.
1. Introduction
Analysis variance (ANOVA) was first introduced by Sir Ronald Fisher. It is a method of
partitioning variability into identifiable sources of variation and the associated degrees of
freedom in an experiment. Taguchi recommends that statistical experimental design method
be employed to assist in quality improvement, particularly during parameter design and
tolerance design, specifically to reduce the variability. He proposed some steps in analysis
varians, that is calculate the percent contribution and pool insignificant factors.
Attribute data analysis such as the fraction defective, denoted p, can take values between 0
and 1. Frequently, p is expressed as a percentage where it can take values between 0% and
100%. With fraction defective, the best value for p is zero. Attribute accumulation is used
when the experimental data can be ranked or categorized.
Attribute accumulation analysis uses analysis of variance and the contribution ratio to
establish significant factors.
2. Fraction Defective Analysis and Weighting
Fraction defective type of data can be analyzed through fraction defective analysis. A better
method of analysis is to use the omega ((Ω) transformation for improving the additivity of
such characteristics. The respon table for fraction defective data is constructed by
determining the totals for each factor level in each category. For each factor, the total of the
data in level 1 and level 2 should be the same.
Fraction defective of each class can be formulated as follows :
pI =
fI
f I + f II + f III
pI =
p III =
=
fI
f ( III )
f II
f
= II
f I + f II + f III f ( III )
f III
f
= III
f I + f II + f III f ( III )
(2.1)
Accumulation analysis needs some understanding of the binomial distribution. If the fraction
defective is p, then the corresponding variance is:
σ 2 = p x (1-p)
(2.2)
This implies that the variance depends on p. However, if we are to compare two distributions
(corresponding to classes or categories in an experiment), we can only make a fair
comparison if the varians are approximately the same. Since, the sum of squares of different
Classes in accumulation analysis analysis would have different bases, it is important to
normalize these basis, by dividing the sum of squares of each class by its variance. This
procedur is frequently called weighting. The weight of each class is:
ωI =
1
σ
=
2
1
1
PI x(1 − P1 )
(2.3)
For ease of calculation, it may be better to use:
ωI =
1
σ
2
1
=
1
=
fI
f ( III )
=
fI
f ( III )
=
1
PI x(1 − P1 )
⎛
f
x ⎜1 − I
⎜
f ( III )
⎝
⎞
⎟
⎟
⎠
1
⎛ f ( III ) − f I ⎞
⎟
x⎜
⎜ f
⎟
( III )
⎝
⎠
f ( 2III )
f I x( f ( III ) − f I )
(2.4)
ω II =
f (2III )
(2.5)
f II x( f ( III ) − f II )
3. The Analysis of Variance
From the attribute accumulation, The fraction defective analysis needs some quantities to
conduct the analysis of Variance. And the following quantity should be calculated.
3.1 The Total Sum of Squares Due To Class I
S1= total sum of squares of class I
⎛
f2
= ⎜ f1 − 1
⎜
f ( III )
⎝
=
=
⎞
⎟ω I
⎟
⎠
f I f ( III ) − f 12
f ( III )
x
f I x( f ( III ) − f I )
f ( III )
f ( 2III )
f1 x( f ( III ) − f I )
x
f ( 2III )
f I x( f ( III ) − f I )
= f ( III )
(3.1)
Similarly, the total sum of squares of class II is also f ( III )
a.
The Overall Total Sum of Squares Due To Class I and Class II
ST = total sum of squares of class I and class II
= f ( III ) + f ( III )
(3.2)
In general :
ST = total sum of squares of class I and class II
=( total number of measurements) x (number of classes - 1)
3.3 The Degrees of Freedom Due To Class I
The degrees of freedom of class I is vI.
v I = f ( III ) − 1
Similarly, the degrees of freedom of class II is also f ( III ) − 1
3.4 The Total Degrees of Freedom
The total degrees of freedom due to class I and class II.
(3.3)
vT = total degrees of freedom of class I dan class II
= ( f ( III ) − 1) + ( f ( III ) − 1)
In general:
(3.4)
vT = total degrees of freedom of class I dan class II
= (total number of measurements -1) x (number of classes -1)
3.5 The Sum of Squares Due to The Mean of Each Class
Sm1 = sum of squares due to mean of class I
=
f12
f ( III )
ωI
(3.5)
3.6 The Total Sum of Squares Due to The Mean
Sm = sum of squares due to mean of class I and class II
=
f12
f ( III )
ωI +
f II2
f ( III )
ω II
= Sm1 + Sm II
(3.6)
3.7 The Overall Total Sum of Squares Due to Class I and Class II
ST = total sum of squares of class I and class II
(3.7)
= f ( III ) + f ( III )
In general :
ST = total sum of squares of class I and class II
=( total number of measurements) x (number of classes - 1)
3.8 The Degrees of Freedom Due to Class I
The degrees of freedom of class I is vI.
v I = f ( III ) − 1
(3.8)
Similarly, the degrees of freedom of class II is also f ( III ) − 1
3.9 The Total Degrees of Freedom
The total degrees of freedom due to class I and class II.
vT = total degrees of freedom of class I dan class II
= ( f ( III ) − 1) + ( f ( III ) − 1)
In general:
(3.9)
vT = total degrees of freedom of class I dan class II
= (total number of measurements -1) x (number of classes -1)
3.10 The Sum of Squares Due to The Mean of Each Class
Sm1 = sum of squares due to mean of class I
=
f12
f ( III )
ωI
(3.10)
3.11 The Total Sum of Squares Due to The Mean
Sm = sum of squares due to mean of class I and class II
=
f12
f ( III )
ωI +
f II2
f ( III )
ω II
= Sm1 + Sm II
(3.11)
3.12 The Sum of Squares Due to A Factor
SA = sum of squares due to factor A
⎛ f I2:AI
⎛ f II2: AI
f I2: A 2
f II2: A2
f I2 ⎞
f II2 ⎞
⎟
⎟ω II
⎜
⎜
=⎜
+
−
+
−
⎟ω I + ⎜ n
⎟
n
n
n
n
n
I :A2
I:A ⎠
II : A 2
II : A ⎠
⎝ I : AI
⎝ II : AI
(3.12)
In this case,
SA =
( f I1:AI + f I2: A 2 ) + ω1 + ( f II2: AI + f II2: A 2 )ω II
− Sm
4n
(3.13)
Similarly, the sums of squares due to the remaining factors (or interactions) are calculated.
3.13 The Degrees of Freedom for A Factor
The degrees of freedom for a factor, say factor A, is:
vA = (number of classes – 1) x (number of levels – 1)
(3.14)
The degrees of freedom of the remaining factors (or interactions) are similarly calculated.
3.14 The Error Sum of Squares
Since more than one measurement is made in each experiment, it is necessary to calculate
the error sum of squares, Se.
S2 = ST – (SA + SB + SC + SD + SE + SF + SG)
(3.15)
The degrees of freedom for ve is:
ve = vT – (vA + vB + vC + vD + vE + vF + vG)
4. Illustration.
(3.16)
Defects known as streaks occur during the polishing operation on optical lenses. An
experiment was designed to minimize these streaks. Seven factors were studied at two levels
each. The lenses were graded as Good, Fair and Bad, where only the Bad lenses were
unacceptable to the customer. Hence, the objective is to maximize Good and Fair lenses.
From the data below:
Figure 4.1 Orthogonal Array and Result
Exp A
B
1
1
1
2
1
1
3
1
2
4
1
2
5
2
1
6
2
1
7
2
2
8
2
2
Column Total
C
1
1
2
2
2
2
1
1
D
1
2
1
2
1
2
1
2
E
1
2
1
2
2
1
2
1
F
1
2
2
1
1
2
2
1
G
1
2
2
1
2
1
1
2
I
1
1
4
8
11
1
3
14
43
II
9
10
10
7
3
4
10
1
54
III
5
4
1
0
1
10
2
0
23
(I)
1
1
4
8
11
1
3
14
43
(II)
10
11
14
15
14
5
13
15
97
(III)
15
15
15
15
15
15
15
15
120
This is best done by first calculating the response table for the factors (or interactions) using
the frequencies.
Figure 4.2 Response Table of Factor Effect
(I)
(II)
Level 1
Level 2
Level 1
Level 2
A
14
29
36
18
B
14
29
26
28
C
19
24
30
24
D
19
24
32
22
E
20
23
24
30
F
34
9
20
34
G
13
30
30
24
Figure 4.3 Analysis of Variance
Source
A
B
C
D
E
F
G
e
Pool
Y
Y
Y
Y
Y
Sq
0.29
15.70
0.03
41.03
0.28
13.50
0.77
249.5
v
2
2
2
2
2
2
2
304
Mq
0.15
7.88
0.01
20.56
0.15
6.75
0.33
0.82
F-ratio
-
Sq’
10.96
31.224
11.88
-
rho %
4.41
12.36
3.72
-
Pooled e
ST
249.7
240
249.7
238
0.80
-
254.6
320.0
79.5
100
5. References
[1]
N. Belavendram, Quality by Design, Prentice Hall, 1995.
[2]
D.C.Montgomery, Design and Analysis of Experiments, John Wiley & Sons,
Singapore,1991
[3]
J.S. Simonoff,. Analyzing Categorical , Springer-Verlag, New York, 2003
Zurnila Marli Kesuma, M.Si
Department of mathematics, Syiah Kuala University, Banda aceh, Indonesia
kesumaku@yahoo.com
Abstract. Taguchi has extended the audio concept of the signal-to-noise ratio (SN ratio) to
experiments involving many factors. The formula for signal-to-noise ratio are designed so
that an experimenter can always select the largest factor level setting to optimize the quality
characteristic of an experiment. When the quality characteristic is a proportion, such as the
fraction defective, denoted p, which can take values between 0 and 1, we use the omega (Ω)
transformation as an objective characteristics. The selection of important factor is not only
based on the response table, but also uses attribute accumulation analysis - analysis of
variance and the contribution ratio to establish significant factors. The method of conducting
the analysis of variance is create a table of data, Calculate the overall total sum of squares
due to class I and class II. Calculate the total degree of freedom. Draw the analysis of
variance. Calculate the predicted optimum process and calculate the confidence interval of a
predicted mean.
2000 Mathematics Subject Classification: 30C45, Secondary 30C80
Keywords : omega(Ω) transformation, analysis of variance,contribution ratio.
1. Introduction
Analysis variance (ANOVA) was first introduced by Sir Ronald Fisher. It is a method of
partitioning variability into identifiable sources of variation and the associated degrees of
freedom in an experiment. Taguchi recommends that statistical experimental design method
be employed to assist in quality improvement, particularly during parameter design and
tolerance design, specifically to reduce the variability. He proposed some steps in analysis
varians, that is calculate the percent contribution and pool insignificant factors.
Attribute data analysis such as the fraction defective, denoted p, can take values between 0
and 1. Frequently, p is expressed as a percentage where it can take values between 0% and
100%. With fraction defective, the best value for p is zero. Attribute accumulation is used
when the experimental data can be ranked or categorized.
Attribute accumulation analysis uses analysis of variance and the contribution ratio to
establish significant factors.
2. Fraction Defective Analysis and Weighting
Fraction defective type of data can be analyzed through fraction defective analysis. A better
method of analysis is to use the omega ((Ω) transformation for improving the additivity of
such characteristics. The respon table for fraction defective data is constructed by
determining the totals for each factor level in each category. For each factor, the total of the
data in level 1 and level 2 should be the same.
Fraction defective of each class can be formulated as follows :
pI =
fI
f I + f II + f III
pI =
p III =
=
fI
f ( III )
f II
f
= II
f I + f II + f III f ( III )
f III
f
= III
f I + f II + f III f ( III )
(2.1)
Accumulation analysis needs some understanding of the binomial distribution. If the fraction
defective is p, then the corresponding variance is:
σ 2 = p x (1-p)
(2.2)
This implies that the variance depends on p. However, if we are to compare two distributions
(corresponding to classes or categories in an experiment), we can only make a fair
comparison if the varians are approximately the same. Since, the sum of squares of different
Classes in accumulation analysis analysis would have different bases, it is important to
normalize these basis, by dividing the sum of squares of each class by its variance. This
procedur is frequently called weighting. The weight of each class is:
ωI =
1
σ
=
2
1
1
PI x(1 − P1 )
(2.3)
For ease of calculation, it may be better to use:
ωI =
1
σ
2
1
=
1
=
fI
f ( III )
=
fI
f ( III )
=
1
PI x(1 − P1 )
⎛
f
x ⎜1 − I
⎜
f ( III )
⎝
⎞
⎟
⎟
⎠
1
⎛ f ( III ) − f I ⎞
⎟
x⎜
⎜ f
⎟
( III )
⎝
⎠
f ( 2III )
f I x( f ( III ) − f I )
(2.4)
ω II =
f (2III )
(2.5)
f II x( f ( III ) − f II )
3. The Analysis of Variance
From the attribute accumulation, The fraction defective analysis needs some quantities to
conduct the analysis of Variance. And the following quantity should be calculated.
3.1 The Total Sum of Squares Due To Class I
S1= total sum of squares of class I
⎛
f2
= ⎜ f1 − 1
⎜
f ( III )
⎝
=
=
⎞
⎟ω I
⎟
⎠
f I f ( III ) − f 12
f ( III )
x
f I x( f ( III ) − f I )
f ( III )
f ( 2III )
f1 x( f ( III ) − f I )
x
f ( 2III )
f I x( f ( III ) − f I )
= f ( III )
(3.1)
Similarly, the total sum of squares of class II is also f ( III )
a.
The Overall Total Sum of Squares Due To Class I and Class II
ST = total sum of squares of class I and class II
= f ( III ) + f ( III )
(3.2)
In general :
ST = total sum of squares of class I and class II
=( total number of measurements) x (number of classes - 1)
3.3 The Degrees of Freedom Due To Class I
The degrees of freedom of class I is vI.
v I = f ( III ) − 1
Similarly, the degrees of freedom of class II is also f ( III ) − 1
3.4 The Total Degrees of Freedom
The total degrees of freedom due to class I and class II.
(3.3)
vT = total degrees of freedom of class I dan class II
= ( f ( III ) − 1) + ( f ( III ) − 1)
In general:
(3.4)
vT = total degrees of freedom of class I dan class II
= (total number of measurements -1) x (number of classes -1)
3.5 The Sum of Squares Due to The Mean of Each Class
Sm1 = sum of squares due to mean of class I
=
f12
f ( III )
ωI
(3.5)
3.6 The Total Sum of Squares Due to The Mean
Sm = sum of squares due to mean of class I and class II
=
f12
f ( III )
ωI +
f II2
f ( III )
ω II
= Sm1 + Sm II
(3.6)
3.7 The Overall Total Sum of Squares Due to Class I and Class II
ST = total sum of squares of class I and class II
(3.7)
= f ( III ) + f ( III )
In general :
ST = total sum of squares of class I and class II
=( total number of measurements) x (number of classes - 1)
3.8 The Degrees of Freedom Due to Class I
The degrees of freedom of class I is vI.
v I = f ( III ) − 1
(3.8)
Similarly, the degrees of freedom of class II is also f ( III ) − 1
3.9 The Total Degrees of Freedom
The total degrees of freedom due to class I and class II.
vT = total degrees of freedom of class I dan class II
= ( f ( III ) − 1) + ( f ( III ) − 1)
In general:
(3.9)
vT = total degrees of freedom of class I dan class II
= (total number of measurements -1) x (number of classes -1)
3.10 The Sum of Squares Due to The Mean of Each Class
Sm1 = sum of squares due to mean of class I
=
f12
f ( III )
ωI
(3.10)
3.11 The Total Sum of Squares Due to The Mean
Sm = sum of squares due to mean of class I and class II
=
f12
f ( III )
ωI +
f II2
f ( III )
ω II
= Sm1 + Sm II
(3.11)
3.12 The Sum of Squares Due to A Factor
SA = sum of squares due to factor A
⎛ f I2:AI
⎛ f II2: AI
f I2: A 2
f II2: A2
f I2 ⎞
f II2 ⎞
⎟
⎟ω II
⎜
⎜
=⎜
+
−
+
−
⎟ω I + ⎜ n
⎟
n
n
n
n
n
I :A2
I:A ⎠
II : A 2
II : A ⎠
⎝ I : AI
⎝ II : AI
(3.12)
In this case,
SA =
( f I1:AI + f I2: A 2 ) + ω1 + ( f II2: AI + f II2: A 2 )ω II
− Sm
4n
(3.13)
Similarly, the sums of squares due to the remaining factors (or interactions) are calculated.
3.13 The Degrees of Freedom for A Factor
The degrees of freedom for a factor, say factor A, is:
vA = (number of classes – 1) x (number of levels – 1)
(3.14)
The degrees of freedom of the remaining factors (or interactions) are similarly calculated.
3.14 The Error Sum of Squares
Since more than one measurement is made in each experiment, it is necessary to calculate
the error sum of squares, Se.
S2 = ST – (SA + SB + SC + SD + SE + SF + SG)
(3.15)
The degrees of freedom for ve is:
ve = vT – (vA + vB + vC + vD + vE + vF + vG)
4. Illustration.
(3.16)
Defects known as streaks occur during the polishing operation on optical lenses. An
experiment was designed to minimize these streaks. Seven factors were studied at two levels
each. The lenses were graded as Good, Fair and Bad, where only the Bad lenses were
unacceptable to the customer. Hence, the objective is to maximize Good and Fair lenses.
From the data below:
Figure 4.1 Orthogonal Array and Result
Exp A
B
1
1
1
2
1
1
3
1
2
4
1
2
5
2
1
6
2
1
7
2
2
8
2
2
Column Total
C
1
1
2
2
2
2
1
1
D
1
2
1
2
1
2
1
2
E
1
2
1
2
2
1
2
1
F
1
2
2
1
1
2
2
1
G
1
2
2
1
2
1
1
2
I
1
1
4
8
11
1
3
14
43
II
9
10
10
7
3
4
10
1
54
III
5
4
1
0
1
10
2
0
23
(I)
1
1
4
8
11
1
3
14
43
(II)
10
11
14
15
14
5
13
15
97
(III)
15
15
15
15
15
15
15
15
120
This is best done by first calculating the response table for the factors (or interactions) using
the frequencies.
Figure 4.2 Response Table of Factor Effect
(I)
(II)
Level 1
Level 2
Level 1
Level 2
A
14
29
36
18
B
14
29
26
28
C
19
24
30
24
D
19
24
32
22
E
20
23
24
30
F
34
9
20
34
G
13
30
30
24
Figure 4.3 Analysis of Variance
Source
A
B
C
D
E
F
G
e
Pool
Y
Y
Y
Y
Y
Sq
0.29
15.70
0.03
41.03
0.28
13.50
0.77
249.5
v
2
2
2
2
2
2
2
304
Mq
0.15
7.88
0.01
20.56
0.15
6.75
0.33
0.82
F-ratio
-
Sq’
10.96
31.224
11.88
-
rho %
4.41
12.36
3.72
-
Pooled e
ST
249.7
240
249.7
238
0.80
-
254.6
320.0
79.5
100
5. References
[1]
N. Belavendram, Quality by Design, Prentice Hall, 1995.
[2]
D.C.Montgomery, Design and Analysis of Experiments, John Wiley & Sons,
Singapore,1991
[3]
J.S. Simonoff,. Analyzing Categorical , Springer-Verlag, New York, 2003