module six: outlier detection for two sample case
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Module Six: Outlier Detection for Two Sample Case. - PowerPoint PPT PresentationTRANSCRIPT
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Module Six: Outlier Detection for Two Sample Case
Two sample plot, also known as Youden’s Plot, is a scatter plot with a confidence region. Youden used it for detecting labs with unusual testing results when two samples are tested in n different lab. Youden plot is a special case of the bivariate control chart, and the idea behind is the Principal Component Analysis.
In this module, we will discuss Principal Component Analysis and how it is applied to construct bivariate control charts and discuss the interpretations of the plot.
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There are three types of laboratory testing, where two-sample plots can be applied:
• A group of labs participate in testing two similar materials using the same method. It is important to identify labs, if any, which perform extremely different from the rest in either one or both materials. A two-sample plot is used to detect the extreme labs.
This is the case studied by Youden (1954), and later extended by Mandel and Lashof (1974).
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2. A participated lab tests a variety of material using the same method as the standard lab, and the testing results are compared to the standardized lab to study if any particular test is extremely different from the standard lab. A paired bivariate control chart is used to determine how good the participated lab is when compared with the standard lab. Tracy, Young and Mason (1995) used a similar approach, bivariate control chart for studying paired measurements in quality control.
3. A lab test similar two or more materials using a standard procedure on a regular bases for many days. A bivariate or multivariate control chart is used for process control. In recent decades, multivariate control charts have been developed for process control of two or more quality characteristics simultaneously. Similar situation may occur in laboratory testing process control.
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The classical Youden’s Plot for Two-sample cases
The following inter-laboratory study about the percent of insoluble residue in cement reported by 29 Laboratories Row %residueA %residueB
1 0.31 0.22
2 0.08 0.12
3 0.24 0.14
4 0.14 0.07
5* 0.52 0.37
6 0.38 0.19
7 0.22 0.14
8* 0.46 0.23
9 0.26 0.05
10 0.28 0.14
11 0.10 0.18
12 0.20 0.09
13 0.26 0.10
14 0.28 0.14
15 0.25 0.13
Row %residueA %residueB
16 0.25 0.11
17 0.26 0.17
18 0.26 0.18
19 0.12 0.05
20 0.29 0.14
21 0.22 0.11
22 0.13 0.10
23* 0.56 0.42
24 0.30 0.30
25 0.24 0.06
26* 0.25 0.35
27 0.24 0.09
28 0.28 0.23
29 0.14 0.10
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Using One sample Box-Plot, we have
%residueA %residueB
0.0
0.1
0.2
0.3
0.4
0.5
0.6
%In
solu
abl
e r
esi
due
0.08
0.46
0.520.56
0.25
0.37
0.42
0.14
Side-By-Side Box Plot for Material A abd B -Insoluable Residue in Cement
Quick check suggests Lab #5, 8, and 23 are likely outliers. They are excluded in the two-sample plot.
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(from NIST website)
The vertical line passes through the median of the X-variable.
The Horizontal line passed through the median of the Y-variable.
The center of the graph is the intersection of the two median lines. This intersection is called the Manhattan Median.
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What is the 45o line for? What is the practical meaning of this line?
Why and How Youden Plot works?
Under the absolute perfect condition, the pair of data should be all equal when testing the same material twice using the same testing procedure. However, it is never the case. In fact, there are two major components of uncertainties: The systematic error and random error.
From common experience, systematic error of a given lab should be the same or very close, in theory, when performing the test under the same condition using the same procedure. And that each lab is different from others. This suggests that, if there is no random or unexpected error, under the perfect condition, but allowing lab differences, then, the pair of data should be equal (that x=y) for a given lab, but, may be different for different labs. Translating this situation, the pairs of data are located at the 45o line passing through the Manhattan Median.
Therefore, the distance from (x,y) pairs on the 45o line is the component of the systematic error for the given lab.
How can we identify the random error component from the plot?
In reality, a pair of data points are scattered on the graph. Only very few pairs will be located on the 45o line, and even so, x and y are really the same.
Consider a pair of data point which is away from center and are not on the 45o line. The distance from the point (x,y) to the center is the total error:
Total error = Systematic error + Random error
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(-7, -2)
Random Error Component(RaE): the distance from (-2,-7) to (-4.5, -4.5)
Systematic Error (SyE) component: the distance from (-4.5, 4.5) to (0,0)
The total error = 2 2(RaE) ( )SyE
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(x3 , y3)
How to determine (x3 , y3)?
For simplicity, assume center is (0,0). Then we see that x3 = (x2 + y2)/2. That is the coordinate for the systematic component is at ((x2 + y2)/2 , (x2 + y2)/2 ). Hence, systematic error component is the distance to from this point to (0,0):
If the center is at (x1 , y1), then the Systematic Error component is
22 2 2 22[( ) / 2] | | / 2x y x y
2 1 2 1(| ( ) ( ) |) / 2x x y y
The distance is allowed to be negative here for reflecting the quadrant. The magnitude is positive.
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NOTE: p in the formula is the Random Component, in our notation, we use: RaE.
If there is no systematic error, the uncertainty should involve only random error component. And therefore, the standard deviation of the random error components is an estimate of the random error variation. 95% coverage region is given by the radius = 2.45(s). The value 2.45 is based on the assumption the (X,Y) follows Bivariate Normal and are independent.
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Youden’s Calculation of the standard deviation and the radius
Youden plot is applied to identify labs with unusual high systematic error as well as labs have unusually high random errors.
•When a lab has both large total error, the data point will be far away from the center. They are the first group of labs that should be closely investigated.
•It may happen that a a material is less sensitive to different environment than the other. When this happens, the data points will tend to be parallel to X- axis or Y-axis, with a large variation due to a material. This can also be quickly identified.
•When a lab have an unusually high systematic error, but small random component, it will scattered along the 45o line. And that more data points are in upper right quadrant and lower left quadrant.
•When a lab has a large random component, it will be far from the 45o line, that is, will have higher probability to be in the upper-left and lower-right quadrants.
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In order to identify these unusual labs, labs with large systematic error, Youden suggested to draw a circle centered at the Manhattan median with the radius being a multiple of the variation due to random error.
The variation due to the random errors is the standard deviation of the random errors of labs, which is obtained by:
According to Youden (1959),
a 95% coverage probability of a circle is given by the circle with radius = 2.448(s)
A 99% coverage probability of a circle is given the cirvle with radius = 3.035(s)
The relationship between the coverage probability and the multiple b is given in Youden’s original paper in the Journal of Industrial Quality Control, 1959, p. 24-28:
Coverage probability = 1-exp(-b2/2)
2( )
1iRaE
sn
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An alternative approach to compute the systematic error components, random error components and the corresponding variations.
(-7, -2)
For a given (x,y) data point, its corresponding coordinate of the systematic component is ( (x+y)/2, (x+y)/2), and the difference between (x,y) and ((x+y)/2, (x+y)/2) along the X-axis is
X – (x+Y)/2 = (x-y)/2
The difference along the Y-axis = (y-x)/2
This suggests that the random error for the data point (x,y) is (x-y)/2.
For each lab,
1. compute the systematic component
[ (x+y)) – (x0+y0)]/2, where (x0 , y0) is the median origin.
2. Compute the random error component:
(x-y)/2
3. Compute variance and s.d. for each component.
sB measures the variation of between-lab systematic errors.
se measures the variation due to random errors.
4. To construct the circle with 95% coverage probability, the radius = 2.45(se)
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Activity: To construct a classical Youden’s Plot for Two-sample cases
The following inter-laboratory study about the percent of insoluble residue in cement reported by 29 Laboratories
Row %residueA %residueB
1 0.31 0.22
2 0.08 0.12
3 0.24 0.14
4 0.14 0.07
5* 0.52 0.37
6 0.38 0.19
7 0.22 0.14
8* 0.46 0.23
9 0.26 0.05
10 0.28 0.14
11 0.10 0.18
12 0.20 0.09
13 0.26 0.10
14 0.28 0.14
15 0.25 0.13
Row %residueA %residueB
16 0.25 0.11
17 0.26 0.17
18 0.26 0.18
19 0.12 0.05
20 0.29 0.14
21 0.22 0.11
22 0.13 0.10
23* 0.56 0.42
24 0.30 0.30
25 0.24 0.06
26* 0.25 0.35
27 0.24 0.09
28 0.28 0.23
29 0.14 0.10
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Variable N N* Mean StDev
(A-B)/2 25 4 0.04760 0.03473
(A+B)/2 25 4 0.1816 0.0570
The radius of the circle for the 95% coverage region is .03473 x 2.45 = .085
Hence the circle has the form:
(x-.25)2 + (y-.13)2 = (.85)2
Variable N Mean Median StDev
Material A 25 0.2292 0.2500 0.0731
Material B 25 0.1340 0.1300 0.0597
Hands-on activities using Mandel and Lashof’s data
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Principal Component Analysis -The concept Behind Two-Sample Plots
The idea behind the two-sample plot is the principal components and bivariate normal distribution. The following scatter plot illustrate the principal components for bivariate case.
70 80 90 100
80
90
100
X1, Tensile Strength-H
X2,
Ten
sile
Str
engt
h-E
Y1, Major PrincipalComponent
Y2, Minor PrincipalComponent
Scatter plot with Principal Compoment Coordinates
(79.81, 92.81)
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The scatter plot is from an inter-laboratory study in Mandel & Lashof (1974). The data are tensile strength of rubbers using two different materials, and testing in 16 laboratories. Laboratory Strength-E (X2) Strength-H (X1)
1 94 80
2 103 82
3 94 77
4 99 83
5 97 86
6 91 76
7 91 81
8 102 98
9 98 83
10 91 81
11 93 82
12 82 69
13 93 81
14 82 72
15 83 73
16 92 73
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Y1 and Y2 are new coordinates.
•Y1 represents the direction where the data values have the largest uncertainty.
•Y2 is perpendicular to Y1.
•They intersect at the sample averages = (79.81, 92.81) .
To find Y1 and Y2, we need to make transformation from X1 and X2. To simplify the discussion, we move the origin to and redefine the (X1,X2) coordinate as
x1 = X1 - , x2 = X2 - , so that the origin is (0,0).
The relationship is illustrated in the following graph. We would like to present the data of a given lab, p = (x1,x2) in terms of p = (y1,y2). From basic geometry relations, we see:
y1 = (cos) x1 + (sin) x2
y2 = (-sin) x1 + (cos) x2
1 2( , )x x
y2
x1
x2
1 2( , )x x
1x 2x
y1 p
The angle is determined so that the observations along the Y1 axis has the largest variability. But HOW?
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The transformation from (x1,x2) to (y1,y2) results several nice properties
1. The variability along y1 is largest.
2. Y1 and y2 are uncorrelated, that is, orthogonal.
3. The confidence region based on (y1,y2) is easy to construct, and provide useful interpretations of the two sample plots.
Questions remain unanswered are
1. How to determine the angle so that the variability of observations along the y1 axis is maximized?
2. How to construct the ellipse for confidence region with different levels of confidences?
3. How to interpret the two-sample plots?
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How to determine the y1 and y2 axis so that the variability of observations along the y1 axis is maximized and y2 is orthogonal to y1?Rewrite the linear relation between (y1,y2) and (x1,x2) in matrix notation:
y1 = (cos) x1 + (sin) x2
y2 = (-sin) x1 + (cos) x2'
1 1 2 1 1
'2 1 2 2 2
(cos ) (sin ) cos sin
( sin ) (cos ) sin cos
y x x x A XY AX
y x x x A X
NOTE: X is bivariate , so is Y, and
V(X) = , V(Y) = A’V(X)A =
and are called the eigen values. Which are the solutions of
And, V(Y1) = V(Y2) = Correlation between Y1 and Y2 = 0.
1
2
0
0
1 1 2
1 2 2
( ) ( , )
( , ) ( )
V x Cov x x
Cov x x V x
( ) 0V X I
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and are called the eigen values. Which are the solutions of
And, V(Y1) = V(Y2) = Correlation between Y1 and Y2 = 0.
The angle = if , when , = 45o
Note the angle depends on the correlation between X1 and X2 , as well as, on the variances of X1 and X2, respectively.
• When is close to zero, the angle is also close to zero. If V(X1) and V(X2) are close, then, the scatter plots are scattered like a circle. That is, there is no clear major principal component.
•When is close to zero and V(X1) is much larger than V(X2), then, the angle will be close to zero, and the data points are likely to be parallel to the X-axis. On the other hand, if V(X1) is much smaller than V(X2), the angle will be close to 900, and the data points will be more likely parallel to the Y-axis.
2
221
212arctan)5(.
21
21
211arctan
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Consider, now, we actually observe the following two sample data:
11 21
12 22
13 23
1 2n n
x x
x x
x x
x x
The sample means are given by
The sample variance-covariance matrix is given by 2
1 1 2
21 2 2
s rs s
rs s s
r is the Pearson’s correlation coefficient, and S2 is the sample variance. S is the sample standard deviation.
1
2
x
x
V(Y) is the solution of
The solutions for are given by
ˆ( )V X
ˆ( ) 0V X I
2 2 2 2 2 2 2 21 2 1 2 1 2( ) ( ) 4(1 )
2
s s s s r s s
NOTE: V(Y1) + V(Y2) = s12 + s2
2 = V(X1) + V(X2)
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2
221
212arctan)5(.
ss
srs
21 ss
Using the sample data, the angle is estimated by
21
211arctansrs
s
Case Example
We know use the Tensile strength data to demonstration the computation of principal components and related sample information.
For the Tensile Strength Example, X1 is the material H and X2 is the material E.
The number of labs, n= 16.
Using Minitab, we can obtain the following information:
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Variance-Covariances Matrix:
H E
H 46.4292 35.0292
E 35.0292 40.9625
Correlations = .8031
Principal Component Analysis: Tensile Strength-H, Tensile Strength-E
Eigen values are: 78.831 and 8.560, the solutions of
Linear Coefficients between (Y1, Y2) and (X1,X2)
Variable Y1 Y2
H 0.734 -0.679
E 0.679 0.734
These are the coefficients for y1 = (cos) x1 + (sin) x2
y2 = (-sin) x1 + (cos) x2
The angle =
= arctan[(78.831-46.4292)/35.0292] = 42.770
21 1 2
21 2 2
s rs s
rs s s
ˆ( )V X
ˆ( ) 0V X I
21
211arctansrs
s
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The sample means from the sample data are
Variable N Mean
H 16 79.81
E 16 92.81
In terms of (Y1, Y2), the means are
Variable N Mean
Y1 16 121.61
Y2 16 13.937
70 80 90 100
80
90
100
Tensile Strength-H
Te
nsile
Str
eng
th-E
Two-sample Scatter PLot for the Tensile Strength Data
Two sample scatter plot is
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Confidence Region for two-sample Plots
Each of the X1 and X2 can be treated as a univariate variable. In most cases, we consider each variable follows a normal distribution. The rules we introduced for one variable case do assume that each variable follows a normal distribution. We can apply outlier detecting methods for each variable. When we consider two variables simultaneously, [X1, X2] are bivariate, and the distribution for [X1,X2] is taken to be bivariate normal distribution. Because of this extension, we are able to construct ellipses that works similar to empirical rule. We can construct several ellipses so that the probability of having the pair of data inside the ellipse is .95 or .99 and so on.
The construction of the ellipse can be simplified when we use the principal components as described above. And the interpretations based on the principal components are very useful.
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Bivariate Normal Distribution and it’s application in two-sample plots
Because the ellipse region relies on bivariate normal distribution, we briefly give an introduction of the bivariate normal in the following.
The bivariate normal distribution of X1 and X2 has the form:
f(x1, x2) = (2exp(-Q/2)
Where ])())((2)(
[1
122
222
21
221121
211
2
XXXXQ
We usually use the notation :
),(~221
2121
2
1
2
1
BNX
X
2
1
is the mean vector.
221
2121
is the variance-covariance matrix.
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A ellipse Q = c, c>0 centered at can then be created in the (X1, X2) coordinate. .
The shape and the orientation of the ellipse is determined by the values of
and , and its size is determined by the choice of the constant c. The choice
of the constant c can be determined based on the level of confidence using the bivariate normal distribution.
When we collect two samples, the sample data provide sample means and sample variance-covariance matrix.
The sample means are given by
The sample variance-covariance matrix is given by 21 1 2
21 2 2
s rs s
rs s s
r is the Pearson’s correlation coefficient, and S2 is the sample variance. S is the sample standard deviation.
1
2
x
x
ˆ( )V X
2
1
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When replacing the population parameters by the corresponding sample information, we obtain the Hotelling’s T2:
2 22 1 1 1 1 2 2 2 2
2 2 21 1 2 2
1 ( ) 2 ( )( ) ( )[ ]
1
X x r X x X x X xT
r s s s s
T2 is distributed as
Which is a multiple of an F-distribution. The ellipse region is now given by
T2 = c*
The constant c* is determined using the F-distribution described above. The corresponding 100(1-)% percentile is from the F distribution with degrees of freedom (2, n-2).
For example, when participated labs, n = 16, then a 95% percentile of
F(2, n-2) = F(2,14) = 3.74. Therefore, c* = 2(255)/224 x (3.74) = 8.515
A 95% ellipse region can then be constructed using T2 = 8.515
)2,2(
2
)2(
)1(2
nF
nn
n
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How to construct a 100(1-)% region in a two-sample plot – the Youden’s Plot?
Under the (X1,X2) coordinate, a general form of an ellipse is given by :
Under the principal component coordinate and move the center to (0,0), the ellipse has the simple form in terms of (y1, y2):
1)())((2)(
22
22
12
2211
11
211
a
xX
a
xXxX
a
xX
122
22
11
21
b
y
b
y
The distance from the center, (0,0) to the ellipse curve along the major principal component is is , and the distance from the center (0,0) to the curve along the minor principal component is
22b
11b
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When matching the mathematical form with the statistical form of confidence ellipse region: T2 = c*, it is seen that, T2 = c* can be expressed in terms of the general form above.
Confidence ellipse region in terms of the principal components has a simple form:
1*
2
22
*1
21
c
y
c
y
*
1( ,0)c
*2(0, )c
y1
y2
*2(0, )c
*1( ,0)c
*1 2( ( ) ,0)c *
1 2( ( ) ,0)c
The sum of the distance from foci to curve is constant, and equal to
Hence, for any point on the ellipse (y1,y2), when y1 is given, y2 can be computed by
*12 c
* 222 2 1
1
y c y
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The sum of the distance from foci to curve is constant, and equal to
Hence, for any point on the ellipse (y1,y2), when y1 is given, y2 can be computed by
And, the original scale in terms of (x1,x2) at the center (0,0) is given by
x1 = (cosy1 – (sin)y2
x2 = (sin)y1 + (con)y2
Shifting the center from (0,0) back to , we have
*12 c
* 222 2 1
1
y c y
1 2( , )x x
1 1 1 2
2 2 1 2
21 1 1 2
(cos ) (sin )
(sin ) (cos )
where = arctan[( ) /( )
X x y y
X x y y
s rs s
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We are now ready to construct a 100(1-)% confidence ellipse region for the Tensile Strength data.
In terms of the principal component coordinate (y1, y2) at the center = (0,0).
A 95% coverage region has the region covered by the ellipse 1*
2
22
*1
21
c
y
c
y
Which is given by
2 21 2( ) ( )
178.831 8.515 8.56 8.515
Y Y
*1c *
2c= 25.908 = 8.537*
1 2( )c = 24.461
The vertices are (-25.908, 0) and (25.908, 0)
The foci are (-24.461, 0) and (24.461, 0)
The Vertical axis are intersected at (0, 8.537) and (0, -8.537)
The sum of the distances from two foci to the curve is 51.816
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140130120110
20
15
10
First Component
Seco
nd C
om
pone
nt
Score Plot of Tensile-Tensile
140130120110
20
15
10
First Component
Se
cond
Co
mp
one
nt
Score Plot of Tensile-Tensile
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To construct the ellipse curve, the point on the curve is given by:
* 2 221 2 1 2 1 1 1
1
( , ) ( , ) ( , 72.8884 .1086 )y y y c y y y
The original scale of the ellipse curve on the (X1, X2) is given by
1 1 1 2 1 2
2 2 1 2 1 2
(cos ) (sin ) 79.81 .734 .679
(sin ) (cos ) 92.81 .679 .734
X x y y y y
X x y y y y
Using the original scale of (X1, X2), the 95% coverage region is
2 21 1 2 2( 79.91) ( 79.81)( 92.81) ( 92.81)
1140.36 82.08 123.833
X X X X
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The two-sample plot with the 95% coverage probability for the tensile strength data is given by:
Raw Data
95% Ellipse
60 70 80 90 100
80
90
100
110
Material H
Ma
teri
al E
Two-Sample PLot for the Tensile Strength Data- Materal H Vs Material E
We notice that one lab is outside of the 95% coverage region. This is Lab 8 with two-sample results (98,102)
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Marginal Plots as an supplement for detecting outliers in two-sample case
The two-sample plot (Youden’s plot) takes the correlation between two-samples into account and are assumed to follow a bivariate normal distribution. This is a useful plot for detecting outliers. In addition to the two-sample plot, one have introduced box-plot for each sample. A two-dimensional marginal box-plot is a good addition for detecting outliers when we sue two-sample Youden’s plot. Using the Tensilt strength as an example, we use Minitab to construct the marginal box-plot in the following.
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65 70 75 80 85 90 95 100
80
85
90
95
100
105
Tensile Strength-H
Te
nsile
Str
eng
th-E
Scatter Plot between Material H and Ewith Marginal Box PLots
The horizontal Box plot is the box-plot for Tensile Strength of Material H. The vertical Box plot id for Material E. It is noticed that there is a clear outlier when testing Material H based on the marginal box plot. Lab 8 is the outlier lab when testing Material H. A similar finding was obtained using Two-sample plot.
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How to construct Two-sample Plot using Minitab?
Minitab does not have a procedure on the pull-down menu. However, a Minitab macro program, similar to what we call a subroutine, has been written for two-sample plots. I have added this macro into the Minitab Macro collection. The name of the macro is BCC (stands for Bivariate Control Chart). There are some restrictions when applying this macro.
Preparation of data:
1. Enter sample One in C1, which will be on the X-axis.
2. Enter sample Two in C2, which will be on the Y-axis.
3. Obtain the critical F-value of the F-distribution using any Statistics book or using Minitab: F(, 2, n-2), where n is the sample size with (1- being the coverage probability. For example, a 95% coverage region gives = .05.
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For the Tensile Strength inter-laboratory study, the number of participated labs = n = 16.
For, a 95% coverage region in the Two-sample plot, the critical value is F(.05, 2,14) = 3.74 from an F-table. The two degrees of freedom are ‘2’ for numerator d.f., and n-2 = 14 for the denominator d.f..
4. Use Minitab to determine F(.05,2,14):
• Go to Calc, choose Probability Distributions, then select ‘F’.
• In the F-distribution Dialog box, click on ‘Inverse cumulative probability.
• Enter d.f. 2 and 14, respectively.
• Click on Input Constant and enter .95.
5. Make the Session window an active window. Then, go to Editor Menu, and click on ‘Enable Commands’.
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6. In the Session window, you should see ‘MTB>’ appears. We will use Minitab commands to input the F(.05, 2,14) and run the BCC Macro.
7. In the Session window, enter the following two commands next to ‘MTB>’ :
MTB> let k11 = 3.74
MTB> %BCC
The ‘let’ statement defines a minitab constant: k11 = F(.05,2,14)
The ‘%’ before the macro name ‘BCC’ is to identify ‘BCC’ as a minitab macro.
The results from the %BCC macro are a Two-sample plot with both scatter plot and the 95% ellipse region, and some principal component computations stored in the worksheet.
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How to use Minitab to construct Marginal Box plot?
Marginal box plot is available in the Graph Menu.
1. Go to Graph, choose Marginal Plot.
2. In the Dialog box, enter Y-variable, and X-variable.
3. You can choose different types of marginal plots, including Histogram, Dot-lot and Box plot. Choose the one you would like to construct.
4. Click on the ‘Symbol’ selection, you can define the type of symbols for the scatter plot.
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How to use Minitab to conduct a Principal Component Analysis?
In my discussion of constructing principal components, and angle, the relationship between (X1, X2) and the principal components, I have used Minitab to obtain these results for the Tensile Strength example. Here is how Minitab does the analysis.
1. Go to Stat, choose Multivariate, then select Principal Components.
2. In the Dialog box, enter variables X1 and X2. Choose the Type of Matrix to be Covariance.
3. Click on the ‘Graph’ , then select ‘Score plot for first two components’.
4. Click on the ‘Storage’, you can choose to store the coefficients for computing (y1,y2) using (x1,x2)(that is cosand sin),and store the values of (Y1, Y2). For example, store Coefficients in C5 and Store (Y1,Y2) inC6, C7.
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How to Interpret Two-sample plots?
In a typical inter-laboratory testing study, the pairs of samples usually are of different, but similar materials are sent out periodically to a number of participating laboratories. All labs run a predetermined number of replicate measurements on each sample for a number of properties.
•For each property, the pair of two samples can be plotted as a Youden Plot or more generally, a bivariate control chart.
•Generally speaking, the points on the plot fall within an ellipse region, which may be defined as a 95% coverage region or 99%.
•The major axis (the major principal components) of the ellipse is approximately 45o.
•The length of the principal component ( based on previous discussion) is related to the ‘between laboratory variability’, and the length of the second principal component ( ) to the ‘material-laboratory interaction’.
*12 c
*22 c
49
A General Statistical Model for interpreting the possible source of variability
NOTE:
• If the between-lab variability is not homogeneous for two samples, the major principal component of the ellipse may not be 45o.
• In addition, the pattern and interpretation of the plot of the plot heavily depends on the model for the experiment and the source of variability in both samples and their inter-relationships.
• A general statistical model that includes important sources of variability is necessary in order to provide adequate interpretation of Youden’s two-sample plot under different assumptions of the sources of variability.
• Youden emphasizes that the two samples should be similar and reasonably close in the measurement of the property evaluated.However, this may not hold in inter-laboratory studies. A more general situation is to consider the following possibilities.
50
1. Material A and B have different population averages, and
2. Each lab has a systematic effect, denoted by LiA, and LiB, respectively.
3. The property is measured with an unexplainable random error from each laboratory, denoted by iA and iB, respectively.
The observed measurement from each lab , denoted by XiA and XiB can be expressed by the statistical model:
XiA = + LiA + iA , i = 1,2, 3, ----, n
XiB = + LiB + iB , I = 1,2,3, ---, n
NOTE that the systematic lab effects is assumed dependent only on the population of the material, not on the material itself. In other words, for different material with the same population average, the systematic lab effect remain the same, even though the material may be different.
It is important to realize that the interpretation of Youden plot depends critically on the specific assumptions on LiA, LiB, iA and iB.
When data are observed, and are estimated by the corresponding sample averages: , and the model is center to (0,0):
and A Bx x
51
xiA = XiA - = LiA + iA
xiB = XiB - = LiB + iB
AxBx
Expressing the variance-covariance of (x1, x2) in terms of the variance components of the systematic effects and random error, we obtain, under the assumption that systematic effects are independent of the random error, and the random errors are independent from lab to lab:
21 1 2
21 2 2
s rs s
rs s s
ˆ( )V X 2 2
2 2A A A B
A B B B
L R L L
L L L R
s s rs s
rs s s s
=
We can obtain the variance-covariance estimates from the data, and then, to estimate the variance component of each source of variation.
However, some assumptions are needed in order for us to do so, since we have three equations with five unknowns.
52
When we impose any assumptions, it is important to keep in mind that these assumptions should reflect the properties of populations underline the experiments. Considering an inter-laboratory experiment using two materials, A and B which can be described in terms of the model
XiA = + LiA + iA , i = 1,2, 3, ----, n
XiB = + LiB + iB , I = 1,2,3, ---, n
• It is often that the averages of two materials are not the same, that is
• The random error from each sample may have different variability (not homogeneous), that is, variance of random error for material A may not be the same as that for material B.
• There are situations where we can demonstrate the equal variance for the random errors of two material, but the systematic biases between two material may not be the same for different labs.
53
Based on the above scenarios, the following three types of model can be considered, and the variance components can be estimated for each type of model.
• One may consider the situation by assuming the systematic effects, the laboratory effect, are the same because of the standardized procedure is implemented, that is, LiA = LiB = L (additive model)
• A less restrictive assumption would be: The laboratory biases are in proportional relationships as so that LiA= K(LiB) (concurrent model).
• Foe each of above models, if we can demonstrate homogeneity of variances of random errors of two materials, we can assume: V(V(for all labs, (homogeneous variance).
54
Under the additive model: we obtain the following variance component estimate for each source of variation:
2122
221
21
221
2
221
2222
2221
, ,
: are components variance theTherefore,
, ,
srssssrssssrss
ssrsssssss
BA
BA
RRL
LRLRL
Under the concurrent model, the proportion is k = LA/LB. The variance component estimates are:
2122
2122
212
221
2222
22221
21 , ) ,
: are components variance theTherefore,
, ,
ssk
rssskrsssss
k
rs
kssrssssssks
BAB
BBBAB
RRL
LRLRL
55
Under the concurrent model, the proportion ‘k’ is estimated by
The average ratio of material A and material B.
.
)( , )( ,
: are components variance theTherefore,
, ,
1222
2112
212
221
2222
22221
sk
rssskrssssss
k
rs
kssrssssssks
BAB
BBBAB
RRL
LRLRL
n
xxk iBiA )/(
The variance component estimates are:
56
Under the homogeneous variance for additive model, the random error variance are assumed equal, that is:
And the variance component for each source is estimated by:
222RRR sss
BA
samples. twoof covariance theminus variancesample combined theiswhich
, 2/)( ,
: are components varianceThe
estimates. precised morefor components variancethe
estimate todata sample twocombinecan weTherefore,
, ,
2122
21
221
2
221
2222
2221
srsssssrss
ssrsssssss
RL
LRLRL
57
21
222
212
212
221
2222
22221
2
1
2 ,
: are components variance theTherefore,
, ,
srsk
ksssss
k
rs
kssrssssssks
RL
LRLRL
B
BBB
Under the homogeneous variance for concurrent model, the variance components are:
58
Case One: Two samples are similar, and the property is measured under the same standard procedure.
This is the most restricted situation. Under this situation, we are assuming:
1. The averages are approximately the same, that is
2. The lab systematic effects are equal for two samples, that is,
LiA = LiB = Li
3. The random error components follow the same normal distribution. and independent.
A Bx x
2 2 2 21
2 2 2 21
2 2 2 2 2 2 2 2
( 1)( ) / 2
( 1)( ) / 2
[( ) ( )] 4
A B A B
A B A B
A B A B A B
L L R R
L L R R
L L R R L L
n s s s s M
n s s s s M
M s s s s r s s
2 2 2 2tan [( ) ( ) ] /(2 )A B A B A BL L R R L Ls s s s M rs s
The principal components of the ellipse region for the general model is given by the following:
59
Under this case, the variance components are simplified to:
the principal components are reduced to
2 2 2 2 2 2 and A B A BL L L R R Rs s s s s s
2 21
22
( 1)(2 )
( 1)( )
and tan 1, that is, the angle 45
L R
R
o
n s s
n s
Under this situation, the Youden plot can be interpreted by:
1. The minor principal component, 2 , measures the random error.
2. The major principal component, 1, measures the systematic effect, if the random error is relatively small.
3. The ratio 1/ 2 measures the impact of the systematic effect with respect to the random error.
60
Case Two: Under this situation, we are assuming:
1. The averages are approximately, that is
2. The lab systematic effects are equal for two samples, that is,
LiA = LiB = Li
3. The random error components are independent, but follow two different normal distributions with different , but, small variances relative to .
Under these conditions, we have :
A Bx x
2Ls
.45ely approximat iscomponent principalemajor th of angle theis that ,1tan
2/))(1(
2/))(1()1(2
o
222
2221
BA
BA
RR
RRL
ssn
ssnsn
The same interpretation of Case One holds, here.
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Case Three: Under this situation, we are assuming:
1. The averages are approximatelyequal, that is
2. The lab systematic effects are equal for two samples, that is,
LiA = LiB = Li
3. The random error components are independent, but follow two different normal distributions with different variances:
Under these conditions, we have:
A Bx x
22222
222
2222
2221
)()2( where,
2/])[(tan
2/])(2)[1(
2/])(2)[1(
BA
BA
BA
BA
RRL
LRR
RRL
RRL
sssM
sMss
Msssn
Msssn
22 and BA RR ss
62
Case Four: When the model is a concurrent model
The systematic bias effects between material A and B is a retio relationship: LiA= k(LiB). The observed two-sample data are (XiA, XiB).
• An approach to determine if the model is concurrent or not
1. Compute the ratio of the variable with larger measurements to the smaller variable, so that the ratio is larger than one. Suppose XiA > XiB . Compute ki = XiA/ XiB for each lab.
2. Compute the average of ki :
3. If there is a consistent pattern of a similar ratios , then, the model is considered as concurrent model. (Note: there is no specific rule available.)
n
kk i
63
Interpretation of Youden’s Plot for case Three
For this case, the two variances of random errors of two materials are different, the angle of the ellipse depending on the magnitude of the variances.
1. If
Then, tan will be larger than one, and the angle is larger than 45o. When , then, the major principal component will approach 90o, and it will be almost vertical.
2. On the other hand, if
Then, Then, tan will be smaller than one, and the angle is smaller than 45o. When , then, the major principal component will approach 0o, and it will be almost horizontal.
22L
2 and s torelative small very is BA RR ss
22 n larger thamuch relatively is LR ssB
22L
2 and s torelative small very is AB RR ss
22 n larger thamuch relatively is LR ssA
64
Transforming concurrent model to additive model
And analyzing the concurrent model using an additive model based on the transformed data.
The complexity of concurrent model suggests for a transformation from concurrent model to additive model is preferred.
• Suppose XiA > XiB , and the average of ki is computed:
• Compute a new variable: X’iA = XiA/
• Analyze the two-sample data using the additive model approach based on (X’iA, XiB).
The analysis and the interpretation of Youden plots can be extended to this transformed two-sample data.
k
k
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Examples for the interpretation of Youden’s Plot
Consider the Tensile Strength Data. There are three rubber material, E,G and H. The data are: Row Laboratory E G H
1 1 94 45 80
2 2 103 43 82
3 3 94 42 77
4 4 99 50 83
5 5 97 47 86
6 6 91 66 76
7 7 91 96 81
8 8 102 128 98
9 9 98 73 83
10 10 91 54 81
11 11 93 81 82
12 12 82 45 69
13 13 93 127 81
14 14 82 150 72
15 15 83 70 73
16 16 92 89 73
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We conducted the (H,E) Youden plot for (H,E) before. In the following we will analyze the Youden plot using the variance components of an additive model.
1. A quick check of the data, the ratio between each pair of material is a little over one. Therefore, we will consider the additive model for this data.
2. Using Minitab, we obtain:
úûùêëéúûùêëé
9625.40
0292.354292.46
42 component principalmajor theof angle the,56.8 ,832.78
8.92 ,8.79
22
2121
o21
21
s
srss
XX
3. The variance components are:
44.2s 3.38, s ,92.5s :are deviations Standard
93.50292.359625.40
40.110292.354292.46s ,0292.35
A
A
RL
2
2R
2
B
B
R
R
L
s
s
67
Based on these information, the following conclusions are obtained:
1. The systematic effect dominates random errors. This indicates there are large laboratory differences.
2. The random errors from both material are small and about the same.
3. To see if there is any outlier labs, a 95% ellipse region is constructed, and showed one serious deficient lab. The two-sam ple plot with the 95% coverage probability for the
tensile strength data is given by:
Raw Data
95% Ellipse
60 70 80 90 100
80
90
100
110
Material H
Ma
teri
al E
Two-Sample PLot for the Tensile Strength Data- Materal H Vs Material E
We notice that one lab is outside of the 95% coverage region. This is Lab 8 with two-sample results (98,102)
4. The linear pattern indicates the systematic error is large. Since for a given, when comparing with others, a linear pattern suggests the data from both material are either both high or both low, which shows the existence of large systematic errors between labs.
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Youden plot for Material H and Material G
Results of descriptive summary and Principal Component analysis are given ine following
Covariances: H, G
H G
H 46.4292
G 33.8750 1178.7833
Variable N Mean Median TrMean StDev SE Mean
H 16 79.81 81.00 79.29 6.81 1.70
G 16 75.38 68.00 72.43 34.33 8.58
Eigenvalue 1179.8 45.4
Proportion 0.963 0.037
Variable PC1 PC2
H 0.030 1.000
G 1.000 -0.030
Cos = .03, hence the angle of the major principal component = 88o
The ratios of H to G are varied a lot. Therefore, concurrent model is not appropriate. We apply additive model for this data.
69
The variance components based on an additive model are:
8.33s 3.5, s ,8.5s :are deviations Standard
90.1144875.3378.1178
55.12875.334292.46s ,875.33
A
A
RL
2
2R
2
B
B
R
R
L
s
s
0 100 200
0
100
200
H
G
Scatter plot of material H and G with proper scale
Scatter plot indicates a huge variation among G material
70
Data95%Ellipse
60 70 80 90 100
0
100
200
Material H
Ma
teri
al G
Two Sample Youden Plot for Material H (X) Vs Material G (Y)
Youden plot indicates an potential outlier labs, Lab#8 with data (98,128) and Lab#14 with data (72,150).
65 70 75 80 85 90 95 100
30
50
70
90
110
130
150
HG
Marginal Plot for Material H and G
There is a clear outlier lab for material H. Material G is highly skewed and
large uncertainty among labs.
71
Conclusions:
1. Scatter plot indicates a huge variation among G testing material.
2. The major principal component is at 88o, almost vertical. This suggests labs testing material G are very inconsistent.
3. Further variance decomposition shows the the random error variance extremely dominates the uncertainty component among labs for material G.
4. The systematic bias effect is small. The random error for material H is also small.
5. There is an outlier lab: Lab #8 according to the 95% ellipse region.
6. Marginal box-plot indicates a clear outlier lab in material H. For material G, the variation is huge and stretched from 43 to 150. At least two labs are much larger than others. This result also indicates that the material G is much more sensitive to different environment than material H. Some extremely high strength deserve further investigation. In particular, one should try to locate special causes for such a high strength, and try to reproduce the results, if possible.
72
The issue about computation
In this module, we have used a variety of procedures in Minitab, I order to accomplish our analysis. The procedures we applied include:
•Graphs:
Plot, Marginal Plot, Box Plot, Time Series Plot, Histogram, Probability Plot.
•Statistical Procedures:
Descriptive statistics, Covariance, Correlation, Normality test, Regression, Individual control chart, Principal Component Analysis, and Bivariate control chart.
The combination of applying the right tool at the right time is the most important, yet, most difficult. It will take a good understanding of the statistical concepts as well as proficient usage of Minitab.
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Hands-On Activity#1:
Repeat the same analysis of the Tensile Data.
Hands-On Activity #2:
Conduct a thorough investigation of of the TAPPI Inter-laboratory study.
Hands-On activity #3:
Bring your own two-sample data and conduct a thorough investigation of your data.