Slide set 12
Stat402 (Spring 2016)
Last update: March 28, 2016
Stat 402 (Spring 2016): Slide Set 12
A 23 Factorial in a RCBD
•
Treatments: A, B, C each at 2 levels; 8 treatment combinations. Thus
we need blocks of size: 8.
•
Consider 2 blocks:
Block 1
Block 2
ab
c
(1)
bc
c
b
b
(1)
bc
abc
a
ab
ac
a
abc
ac
• The above table shows an example how the treatment combination
were randomly applied to experimental units/runs within each block.
1
Stat 402 (Spring 2016): Slide Set 12
A 23 Factorial in a RCBD (Cont’d 1)
•
Data:
Trt
(1)
a
b
ab
c
ac
bc
abc
Block Total
Block I
1
2
4
6
4
3
6
10
36
Block II
3
1
5
8
7
3
6
11
44
Trt. Total
4
3
9
14
11
6
12
21
80
2
Stat 402 (Spring 2016): Slide Set 12
A 23 Factorial in a RCBD (ANOVA)
•
•
Computations:
802
16
2
= 400 (SS for the Mean)
CF
=
Total SS
=
1 + 22 + · · · + 62 + 112 = 532
Total (Corrected) SS
=
532-400=132
Block SS
=
Trt SS
=
362
442
+
8
8 − CF = 404 − 400 = 4
32
212
42
+
+
·
·
·
+
2
2
2 − CF = 522 −
Error SS
=
6 (By Subtraction)
400 = 122
ANOVA Table:
SV
Block
Trt
Error
Total
DF
1
7
7
15
SS
4
122
6
132
3
Stat 402 (Spring 2016): Slide Set 12
•
Factorial Effects and their Sums of Squares:
Now the 7 d.f. for Treatment Sum of Squares is partitioned into 7 single degree of d.f.
sums of squares corresponding to the 7 factorial effects:
Treatment
Combination
(1)
a
b
ab
c
ac
bc
abc
Contrast
Divisor for Estimate
Estimate of Effect
Divisor for SS
SS of Effect
Observed
Total
4
3
9
14
11
6
12
21
I
+
+
+
+
+
+
+
+
80
16
5.0
16
400
A
+
+
+
+
8
8
1.0
16
4
B
+
+
+
+
32
8
4.0
16
64
Factorial Effect
AB
C
AC
+
+
+
+
+
+
+
+
+
+
+
+
20
20
0
8
8
8
2.5 2.5
0
16
16
16
25
25
0
BC
+
+
+
+
0
8
0
16
0
ABC
+
+
+
+
8
8
1
16
4
4
Stat 402 (Spring 2016): Slide Set 12
A 2k Factorial in Incomplete Block Experiments
•
In factorial experiments the number of treatment combinations increases
rapidly as the number of factors and/or the number of levels of each
factor increases.
•
For example, a 27 has 128 different treatment combinations and it is
very difficult if not impossible to find blocks of 128 experimental units
that are homogenous. (Why?)
•
Because variation between experimental units within blocks tend to
increase as the block size increase (which results in a larger experimental
error), it is desirable to keep the block size small.
5
Stat 402 (Spring 2016): Slide Set 12
A 22 Factorial in Blocks of Size 2
•
Suppose that we have a 22 factorial experiment but we can obtain blocks
of only 2 homogeneous units each. We have 4 treatment combinations
(1), a, b and ab, which can be allocated to the 2 blocks in 3 different
ways to obtain 3 designs.
•
We shall consider each of these designs separately:
Design 1:
Block I
(1)
ab
Block II
a
b
6
Stat 402 (Spring 2016): Slide Set 12
A 22 Factorial in Blocks of Size 2 (Cont’d design1 )
•
We can write the model for the design, assuming one replication of the
treatment combinations as follows:
y(1)
yab
ya
yb
•
=
=
=
=
µ(1) + β1 + (1)
µab + β1 + ab
µ a + β 2 + a
µ b + β 2 + b
Here β1 and β2 represent the respective effects of the two blocks. Using
this model, we shall look at the contrasts of the treatment combinations
which define the main effects and interactions to see the effect of putting
1/2 of the treatment combinations in one block and the other 1/2 in the
other block:
7
Stat 402 (Spring 2016): Slide Set 12
A 22 Factorial in Blocks of Size 2 (Cont’d design 1)
A: (a-1)(b+1)
 = 1/2[ab + a − b − (1)]
= 1/2[µab + β1 + ab + µa + β2 + a − µb − β2
−b − µ(1) − β1 − (1)]
= 1/2[µab + µa − µb − µ(1)] + 1/2[ab + a − b − (1)]
B: (a+1)(b-1)
B̂
= 1/2[ab + b − a − (1)]
= 1/2[µab + β1 + ab + µb + β2 + b − µa − β2
−a − µ(1) − β1 − (1)]
= 1/2[µab + µb − µa − µ(1)] + 1/2[ab + b − a − (1)]
8
Stat 402 (Spring 2016): Slide Set 12
AB: (a-1)(b-1):
d
AB
= 1/2[ab − a − b + (1)]
= 1/2[µab + β1 + ab + µ(1) + β1 + (1) − µa − β2
−a − µb − β2 − b]
= 1/2[µab − µa − µb + µ(1)] + (β1 − β2) + 1/2[ab − a − b + (1)]
•
We can see that if we use this design, then we cannot separate the AB
interaction effect from the block effect. The effect of AB interaction is
said to be confounded with the block effect.
•
From above analysis we can see that, if we use Design 1, the effects of
A and B are free of the block effect while the effect of AB is not.
9
Stat 402 (Spring 2016): Slide Set 12
A 22 Factorial in Blocks of Size 2 (Cont’d design 2)
Design 2:
Block I
(1)
b
Block II
a
ab
In this case, the model is given by
y(1)
yb
ya
yab
=
=
=
=
µ(1) + β1 + (1)
µ b + β 1 + b
µ a + β 2 + a
µab + β2 + ab
10
Stat 402 (Spring 2016): Slide Set 12
Hence,
 = 1/2[µab + β2 + ab + µa + β2 + a − µb − β1
−b − µ(1) − β1 − (1)]
= 1/2[µab + µa − µb − µ(1)] + (β2 − β1) + 1/2[ab + a − b − (1)]
B̂
d
AB
•
= 1/2[µab + β2 + ab + µb + β1 + b − µa − β2
−a − µ(1) − β1 − (1)]
= 1/2[µab + µb − µa − µ(1)] + 1/2[ab + b − a − (1)]
= 1/2[µab + β2 + ab + µ(1) + β1 + (1) − µa − β2
−a − µb − β1 − b]
= 1/2[µab − µa − µb + µ(1)] + 1/2[ab − a − b + (1)]
If we use this design, then we see that the B effect and the AB
interaction effect will be free of block effect and that the main effect A
will be confounded with blocks.
11
Stat 402 (Spring 2016): Slide Set 12
A 22 Factorial in Blocks of Size 2 (Cont’d design 3)
Design 3:
Block I
(1)
a
Block II
a
ab
The model in this case is
y(1)
ya
yb
yab
=
=
=
=
µ(1) + β1 + (1)
µ a + β 1 + a
µ b + β 2 + b
µab + β2 + ab
12
Stat 402 (Spring 2016): Slide Set 12
In a similar manner as above we can show that
 = 1/2[µab + µa − µb − µ(1)] + 1/2[ab + a − b − (1)]
B̂ = 1/2[µab + µb − µa − µ(1)] + (β2 − β1) + 1/2[ab + b − a − (1)]
ˆ = 1/2[µab − µb − µa − µ(1)] + 1/2[ab − b − a + (1)]
AB
•
In this design, A and AB effects are free of block effects and the main
effect B is confounded with block effect.
Remarks
1. Only one effect is confounded with blocks in each design
2. It is possible to confound a given effect by the choice of the design.
13
Stat 402 (Spring 2016): Slide Set 12
3. We usually try to confound the higher order interaction effects in a given
factorial experiment since these are the least important. When an effect
is confound it cannot be estimated; hence, no sum of squares exists to
test its effect.
4. Each of the above designs is an example of an incomplete block design
(as opposed to a complete block design)
5. Confounding can be thought of as a method available for increasing
the accuracy of factorial experiments by reducing the block size at the
expense of reduced accuracy on estimating some of the effects.
Randomization in Incomplete Blocks
1. Given an experimental plan, numbers are assigned to blocks at random.
2. Then treatment combinations are randomly applied to the units within
each block.
14