Soo King Lim

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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

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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

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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

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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

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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

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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.

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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.

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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

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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)

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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.

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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.

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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

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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.

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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