Page 1
Soo King Lim
- 1 -
1.0 Nested Factorial Design ................................................................... 3
1.1 Two-Factor Nested Design ......................................................................... 3
1.1.1 Analysis of Variance ........................................................................................ 4 Example 1 ................................................................................................................... 5 1.1.2 Staggered Nested Design for Equalizing Degree of Freedom ...................... 7
1.1.2 Three-Factor Nested Design ................................................................... 8
1.1.2.1 Analysis of Variance ..................................................................................... 9 Example 2 ................................................................................................................. 10
Page 2
Soo King Lim
- 2 -
Figure 1: Two-factor nested design ..................................................................... 3
Figure 2: Three-factor nested design ................................................................... 3
Figure 3: ANOVA table for two-factor nested design ......................................... 5
Figure 4: Data of two-factor nested design of example 1 .................................... 6
Figure 5: ANOVA results of example 1 .............................................................. 7
Figure 6: A staggered nested design .................................................................... 8
Figure 7: ANOVA table for three factor nested design ..................................... 10
Figure 8: Leakage current test results of the three factor nested design, mean
and sum of kth tester of ith facility, and mean and sum of ith device .. 11
Figure 9: The mean for ith device, jth facility, and kth tester ............................... 11
Figure 10: Mean and sum of ith device for jth facility .......................................... 11
Figure 11: ANOVA table for three factor nested experiment showing results of
F-test ................................................................................................... 13
Page 3
Soo King Lim
- 3 -
1.0 Nested Factorial Design
For standard factorial designs, where each level of every factor occurs with all levels
of the other factors and a design with more than one duplicate, all the interaction
effects can be studied. In a nested factor design, the levels of one factor like factor
B are similar but not identical for different levels of another factor like factor A.
These are also called hierarchical designs.
The two-factor and three-factor nested designs are shown in Fig. 1 and 2
respectively.
Figure 1: Two-factor nested design
In this example, machine is the fixed factor, while operator is a random factor.
Figure 2: Three-factor nested design
In this example, factor A is considered as fixed factor while, factor B and C is
considered as random factor.
1.1 Two-Factor Nested Design
The linear model for two factor nested design is
Page 4
Soo King Lim
- 4 -
yijk = μ + αi + βj(i) + ek(ij) (1)
where μ is the grand mean, αi is the effect of level i of factor A, βj(i) is the effect of
jth level of factor B nested within ith level of factor A, and ek(ij) is the error nested
within ith levels of factor A and jth level of factor B. k is the number of duplicate.
Replacing the parameters in equation (1) by their respective estimators yields
equation (2).
ijiijiijk yyyyyyy jkiy (2)
One assumes that the observations yijk are independent and follows normal
distribution with mean μij and variance σ2.
1.1.1 Analysis of Variance
If the number of levels of factor A is a, the number of levels of factor B nested under
each level of factor A is b, and the number of duplicate is n, the calculation of the
sums of squares and its degree of freedom are shown in equation (3) to (5).
The total sum of square SST is
abn
y
yyy
a
i
b
j
n
k
ijka
i
b
j
n
k
ijk
a
i
b
j
n
k
a
i
b
j
n
k
ijkijk
2
1 1 1
1 1 1
2
2
1 1 1
TSS
(3)
and it has (abn - 1) degree of freedom.
The sum of square due to factor A SSA is
abn
y
bn
y
yybn
a
1i
b
1j
n
1k
ijk
a
i
b
1j
n
1k
ijka
i
i
2
1
2
1
2
ASS
(4)
and it has (a - 1) degree of freedom. The sum of square due to B nested within A
SSB/A is
bn
y
n
y
yyn
a
i
n
k
ijk
a
i
b
j
n
k
ijka
i
b
j
iij
1
2
1 11 1
2
1
1 1
2
A/BSS
b
j (5)
and it has a(b - 1) degree of freedom. The sum of square due to error SSE is
calculated equal to
Page 5
Soo King Lim
- 5 -
SSE = SST - (SSA + SSB/A) (6)
which has ab(n - 1) degree of freedom.
The sum of square due to error SSE can also be calculated using ijij yky of
equation (7), which is
SSE =
a
i
b
j
n
k
iji yy1 1
2
1
jk (7)
The ANOVA table of two factor nested design showing their respective sum of
square, degree of freedom, mean square value, and calculated F-value is shown in
Fig. 3.
Factor Sum of
Square
Degree of
Freedom Mean Square
Calculated
F-value
F-value from
F-Table
A SSA (a - 1) SSA/(a - 1) MSA/MSE F[(a - 1), ab(n - 1)]
B SSB (b - 1) SSB(b - 1) MSB/MSE F[(b - 1), ab(n - 1)]
B/A SSB/A a(b - 1) SSAB/(a(b - 1)) MSB/A/MSE F[(a(b - 1)), ab(n - 1)]
Error SSE ab(n - 1) SSE/(ab(n - 1)) - -
Total SST (abn - 1) - - -
Figure 3: ANOVA table for two-factor nested design
Example 1
A company is interested to test if there is any difference among the percentage of
defect produced by three equipment and six operators in manufacturing floor. The
engineer uses a nested factor design with six operators 1, 2, 3, 4, 5, and 6, who
operate the machines two times. The percentage of defect produced and the associate
mean values are shown in Fig. 4. Note that in this design the equipment is a fixed,
whereas the operator is the random factor. The reason being the equipment is fixed
for two different selected groups of operators.
Equipment (i) 1 2 3
Operator (j) 1 2 3 4 5 6
% of Defect
yijk 5, 8 4, 2 4, 0 5, 3 2, 1 1, 4
2
1k
ijky 13 6 4 8 3 5
2
2
1
k
ijk
ij
y
y 6.5 3.0 2.0 4.0 1.5 2.5
Page 6
Soo King Lim
- 6 -
2
1
2
1j k
ijky 19 12 8
2x2
2
1
2
1
j k
ijk
i
y
y 4.75 3.00 2.00
3
1
2
1
2
1i j k
ijky 39
2x2x3
3
1
2
1
2
1
i j k
ijky
y 3.25
Figure 4: Data of two-factor nested design of example 1
Observation yijk is the kth replicate for k = 1 and 2 on equipment i, where i = 1, 2,
and 3, with operator j, where j = 1, 2, 3, 4, 5, and 6. This is a two-stage nested design.
If there are an equal number of levels of B within each level of A and an equal
number of duplicates then the design is a balanced nested design. The effects that
can be tested in this design are the effect due to equipment, which is factor A and
the effect of operator nested within the equipment B/A. The error term is nested
within levels of A and B. In this design, the interaction between A and B cannot be
tested because every level of B does not appear with every level of A.
Solution
For this experiment; a = 2, b = 2, and n = 2. The total sum of square is equal to
3x2x2
y
SS
23
1
2
1
2
13
1
2
1
2
1
2
T
i j k
ijk
i j k
ijky2x2x3
3941123542845
2222222222
= 54.25. Its degree of freedom is (abn - 1) = 12 - 1 = 11.
The sum of square of factor A is 3x2x22x2
SS
23 2 23
1
22 2
A
1i 1j 1k
ijk
i 1j 1k
ijk yy
= 12
39
4
81219 2222
= 15.5. Its degree of freedom is (a - 1) = 3 - 1 =2.
The sum of square of B/A is2x22
SS
3
1
2
1
2
1
3
1
2
1
22
1
A/B
i k
ijk
i j k
ijkyy
2
j
=4
81219
2
5384613 222222222
= 17.25. Its degree of freedom is a(b - 1) = 3(2
- 1) = 3.
Page 7
Soo King Lim
- 7 -
The sum of square due to error is SSE = 54.25 - 15.5 - 17.25 = 21.5. Its degree
of freedom is ab(n - 1) = 3x2(2 - 1) = 6.
The ANOVA results of the experiment for = 0.05 are shown in Fig. 5.
Factor Sum of
Square
Degree of
Freedom
Mean
Square
Calculated
F-value
F-value
from F-
Table for
= 0.05
p-value
Equipment 15.5 2 7.75 2.16
F0.05(2, 6) =
5.14 > 0.100
Operator
with
equipment
17.25 3 5.75 1.60 F0.05(3, 6) =
4.76 > 0.100
Error 21.5 6 3.58 - - -
Total 54.25 11 - - - -
Figure 5: ANOVA results of example 1
The results show that there is no difference among the number of defect produced
by equipment 1, 2, and 3 and there is no difference among six operator b1, b2, b3, b4,
b5, and b6 nested within each equipment.
In a nested experiment, the factors tested can be fixed or random or a
combination of both factors. The calculation of the sums of squares and the test
statistics do not change irrespective of whether these factors are fixed or random
types. However, the interpretation of the results depends upon the types of factors.
1.1.2 Staggered Nested Design for Equalizing Degree of Freedom
The nested factor design contains more information on factors at lower levels in the
hierarchy of the design than at higher levels because of the larger degree of freedom.
In higher level studies, the discrepancies in degrees of freedom among sources of
variation can be considerable. Staggered nested designs were developed to equalize
the degrees of freedom for the mean squares at each level of the hierarchy. The
staggered designs have unequal numbers of levels for factors that are nested within
other factors. The levels for factor B nested within factor A are varied from one level
of factor A to another in such a way that the degree of freedom for MSA and
MS(B/A) are almost equal.
A staggered nested design is given in Fig. 6. The degree of freedom for the
equipment is 2. For three operator/equipment combination, the degree of freedom
operator/equipment is 2 + 2 + 2 = 6. In this design, the degree of freedom for
operator/equipment is 2 + 1 + 1 = 4.
Page 8
Soo King Lim
- 8 -
Figure 6: A staggered nested design
1.1.2 Three-Factor Nested Design
The linear model for three factor nested design is
yijkl = μ + αi + βj + k(j) + (αβ)ij + (α)ik(j) + e(ijk)l (8)
where μ is the grand mean, αi is the effect of level i of factor A, βj is the effect of jth
level of factor B, k(j) is the effect of kth level of factor C nested with jth level of factor
B, (αβ)ij is the effect of ith level of factor interacts with jth level of factor B, (α)ik(j) is
the effect of ith level of factor A interacts with the kth level of factor C nested with jth
level of factor B, and ek(ijk) is the error nested within ith levels of factor A, jth level of
factor B, and kth level of factor C. l is number of duplicate. Replacing the parameters
in equation (3.61) by their respective estimators yields equation (9).
ijkijjkijijk
jiijjjkjii
yyyyyy
yyyyyyyyyyyy
jkl
jkl (9)
One assumes that the observations yijkl are independent and follows normal
distribution with mean μ and variance σ2.
Taking the summation for i = 1 to a, j = 1 to b, k = 1 to c, and l = 1 to n, it
yields the sum of square equation (10).
a
i
b
j
jiij
b
1j
c
k
jjk
b
j
j
a
i
i
a
i
b
j
c
k
n
l
ijkl
yyyycnyyan
yyacnyybcnyy
1 1
22
1
2
11
2
1 1 1 1
2
(10)
a
i
b
j
c
k
n
l
ijkijkl
a
i
b
j
c
k
jjkijijk yyyyyn1 1 1 1
2
1 1 1
2
y
Page 9
Soo King Lim
- 9 -
If we use the symbol A, B, and C factor, the simplified form of equation (11) can be
written as
SST = SSA + SSB + SSC/B + SSAB + SSAC/B + SSE (11)
1.1.2.1 Analysis of Variance
If the number of levels of factor A is a, the number of levels of factor B nested under
each level of factor A is b, and the number of replications is n, the calculation of the
sums of squares and the associated degrees of freedom are shown in equation (12)
to (3.70).
The total sum of square SST is
abcn
y
yyy
a
i
b
j
c
k
n
1l
ijkla
i
b
j
c
k
n
l
ijkl
a
i
b
j
n
k
a
i
b
j
c
k
n
1l
ijklijk
2
1 1 1
1 1 1 1
2
2
1 1 1
TSS
(12)
and it has (abcn - 1) degree of freedom.
The sum of square due to factor A SSA is
a
i
i yybcn1
2
ASS (13)
and it has (a - 1) degree of freedom. The sum of square due to factor B SSB is
2
1
BSS
b
j
j yyacn (14)
and it has (b - 1) degree of freedom. The sum of square due to C nested within B
SSC/B is
b
1j
c
k
jjk yyan
2
1
B/CSS (15)
and it has b(c - 1) degree of freedom. The sum of square due to interaction of factor
A and factor B is
a
i
b
j
jiij yyyycn1 1
2
ABSS (16)
Page 10
Soo King Lim
- 10 -
and its degree of freedom is SSAB is (a - 1)(b - 1). The sum of square due to factor
A interacts with factor C nested factor B is
a
i
b
j
c
k
jjkijijk yyyyn1 1 1
2
B/ACSS (17)
It has b(a - 1)(c - 1) degree of freedom.
The sum of square due to error SSE can be calculated using ijkyjkliy of
equation (9), which is
SSE = 2
1 1 1 1
i
a
i
b
j
c
k
n
l
ijkyjkly (18)
and its degree of freedom is abc(n - 1).
The ANOVA table of three factor nested design showing their respective sum
of square, degree of freedom, mean square value, calculated F-value, and F-test @
α = 0.05 is shown in Fig. 7.
Factor Sum of
Square
Degree of
Freedom
Mean
Square
Calculated
F-value
F-value from
F-Table
A SSA (a - 1) SSA/(a - 1) MSA/MSE F[(a - 1), abc(n - 1)]
B SSB (b - 1) SSB/(b - 1) MSB/MSE F[(b - 1), abc(n - 1)]
C/B SSC/B b(c - 1) SSC/B/
(b(c - 1)) MSC/B/MSE F[(b(c - 1)), abc(n - 1)]
AB SSAB (a - 1)(b - 1) SSAB/
(a - 1)(b - 1) MAB/MSE
F[((a - 1)(b - 1)),
abc(n - 1)]
AC/B SSAC/B b(a - 1)(c -1) SSAC/B/
b(a - 1)(c -1) MAC/B/MSE
F[(b(a - 1)(b - 1),
abc(n - 1)]
Error SSE abc(n - 1) SSE/
(abc(n - 1)) - -
Total SST (abcn - 1) - - -
Figure 7: ANOVA table for three factor nested design
Example 2
A test engineer has three correlation devices, where he wants to test them using four
testers randomly selected each from two facilities. The each correlation device is
test twice for its leakage current. Analyze the variance for all factors.
The results of leakage current test are shown in Fig. 8.
Page 11
Soo King Lim
- 11 -
Facility (j) 1 2 yi =
2
1
4
1
2
1j k l
ijkly iy Tester (k) 1 2 3 4 1 2 3 4
(i)
Device 1 2.2 2.0 2.1 2.3 2.5 2.7 2.8 2.5
38.0 2.38 2.4 2.4 2.2 1.8 2.8 2.5 2.5 2.3
Device 2 3.0 2.4 2.0 1.9 2.7 2.0 2.4 2.6
38.3 2.39 2.7 2.3 1.9 2.5 2.7 2.7 2.3 2.2
Device 3 2.5 2.2 1.8 2.6 2.7 2.6 2.4 2.7
37.6 2.35 2.1 2.2 1.7 2.3 2.5 2.4 2.4 2.5
Tester - Device
Total yjk 14.9 13.5 11.7 13.4 15.9 14.9 14.8 14.8
3
i j k l
ijkly1
2
1
4
1
2
1
= 113.9
y = 2.37
Mean of yik 2.48 2.25 1.95 2.23 2.65 2.48 2.47 2.47
yj =
3
1
4
1
2
1i k l
ijkly 53.5 60.4
Mean of total jy 2.23 2.52
Figure 8: Leakage current test results of the three factor nested design, mean and sum of kth
tester of ith facility, and mean and sum of ith device
Solution
The mean for ith device, jth facility and kth tester is n
yn
l
i
ijk
1y
jkl
is shown in Fig. 9,
while the mean and sum of ith device for jth facility is shown in Fig. 10.
B: Facility (j) 1 2
C :Tester (k) 1 2 3 4 1 2 3 4
Device 1 (i) 2.30 2.20 2.15 2.05 2.65 2.60 2.65 2.40
A Device 2 (i) 2.85 2.35 1.95 2.20 2.70 2.35 2.35 2.40
Device 3 (i) 2.30 2.20 1.75 2.45 2.60 2.50 2.40 2.60
Figure 9: The mean for ith device, jth facility, and kth tester
B Facility (j) 1 2 yi1 =
4
1
2
1
1
k l
kliy 1iy
yi2 =
4
1
2
1
2
k l
kliy 2iy C Tester (k) 1 2 3 4 1 2 3 4
A
Device 1 (i) 2.2 2.0 2.1 2.3 2.5 2.7 2.8 2.5
17.4 2.2 20.6 2.6 2.4 2.4 2.2 1.8 2.8 2.5 2.5 2.3
Device 2 (i) 3.0 2.4 2.0 1.9 2.7 2.0 2.4 2.6
18.4 2.3 19.6 2.5 2.7 2.3 1.9 2.5 2.7 2.7 2.3 2.2
Device 3 (i) 2.5 2.2 1.8 2.6 2.7 2.6 2.4 2.7
17.4 2.3 20.2 2.5 2.1 2.2 1.7 2.3 2.5 2.4 2.4 2.5
Figure 10: Mean and sum of ith device for jth facility
Based on equation (12) to (18), the sums of square and degrees of freedom of the
experiment are calculated.
Page 12
Soo King Lim
- 12 -
The value of a is 3, b is 2, c is 4, and n is 2.
The sum of square due to C nested within B SSC/B is
2 24
1
B/C 3x2SS1j k
jjk yy .
b
1j
c
k
jjk yyan
2
1
B/CSS
22
22222
)52.247.2(2)52.248.2(
)52.265.2()23.223.2()23.295.1()23.225.2()23.248.2(6
= 6x0.1668 = 1.001.
The sum of square due to factor A interacts with factor C nested factor B SSAC/B is
3
1
2
1
4
1
2
B/AC 2SSi j k
jjkijijk yyyy
222 )52.247.253.260.2(....)23.225.218.220.2(2.23)2.48-2.18-(2.302
= 2x0.291 = 0.783.
The sum of square due to interaction of factor A and factor B SSAB is
3
1
2
1
2
AB 4x2SSi j
jiij yyyy
222
222
)37.252.235.253.2()37.252.239.245.2()37.252.238.258.2(
)37.25.3.223.218.2()37.239.223.234.2(2.37)2.23-2.38-(2.188
= 8x0.024 = 0.189.
The sum of square for factor A SSA is
3
1
2
A 2x42xSSi
i yy
222 )37.235.2()37.238.2()37.238.2(16 = 0.015.
The sum of square for factor B SSB is
22
1
B 2x4x3SS
j
j yy
22 )37.252.2()37.223.2(24 = 0.992.
The total sum of square SST is 3x2x4x2
SS
23
1
2
1
4
1
2
3
1
2
1
4
1
2
1
2
T
i j k 1l
ijkl
i j k l
ijkl
y
y
48
0.1135.27.24.24.2....7.20.33.25.2.....4.20.24.22.2
2
222222222222
= 4.095
Page 13
Soo King Lim
- 13 -
The sum of square due to error SSE is equal to 4.095 - 0.015 - 0.992 - 1.001 - 0.189
- 0.783 = 1.115.
ANOVA table showing their respective sum of square, degree of freedom, mean
square value, calculated F-value, and F-table value @ α = 0.05 is shown in Fig. 11.
Factor Sum of
Square
Degree of
Freedom
Mean
Square
Calculated
F-value
F-value from
F-Table @ α = 0.05 p-value
A 0.015 2 0.0075 0.1616 F0.05(2, 24) = 3.40 > 0.100
B 0.992 1 0.9920 21.37 F0.05(1, 24) = 4.26 < 0.001
C/B 1.001 6 0.1668 3.595 F0.05( 6, 24) = 2.51 < 0.025
AB 0.189 2 0.0945 2.037 F0.05(2, 24) = 3.40 > 0.100
AC/B 0.783 12 0.0653 1.407 F0.05(12, 24) = 2.18 > 0.100
Error 1.115 24 0.0464 - - -
Total 4.095 48 - - - -
Figure 11: ANOVA table for three factor nested experiment showing results of F-test
The results show that there is no correlation between facility one and two and there
is difference for tester nested in facility. For other factors, the tests show that they
are not significant @ 95.0% confidence level.
Page 14
Soo King Lim
- 14 -
A
Analysis of variance ............................................. 3, 8
F
Fixed factor............................................................... 2
H
Hierarchical design ................................................... 2
N
Nested factor design ................................................. 2
P
p-value ................................................................ 6, 12
R
Random factor........................................................... 2
S
Staggered nested design ............................................ 6
T
Three-factor nested design ........................................ 7
Two-factor nested design .......................................... 2
V
Variance ................................................................ 3, 7