•
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 2 Factorial in a RCBD (Cont’d 1)
3
Last update: March 28, 2016
Stat402 (Spring 2016)
Slide set 12
•
•
•
•
Stat 402 (Spring 2016): Slide Set 12
ab
c
(1)
bc
c
b
b
(1)
bc
abc
a
ab
ac
a
abc
ac
=
=
=
=
=
Total SS
Total (Corrected) SS
Block SS
Trt SS
Error SS
ANOVA Table:
=
CF
Computations:
= 400 (SS for the Mean)
SV
Block
Trt
Error
Total
DF
1
7
7
15
SS
4
122
6
132
6 (By Subtraction)
362
442
8 + 8 − CF = 404 − 400 = 4
42
32
212
2 + 2 + · · · + 2 − CF = 522 −
532-400=132
1 + 22 + · · · + 62 + 112 = 532
802
16
2
400 = 122
3
Stat 402 (Spring 2016): Slide Set 12
A 2 Factorial in a RCBD (ANOVA)
3
1
• The above table shows an example how the treatment combination
were randomly applied to experimental units/runs within each block.
Block 1
Block 2
Consider 2 blocks:
Treatments: A, B, C each at 2 levels; 8 treatment combinations. Thus
we need blocks of size: 8.
A 2 Factorial in a RCBD
3
Stat 402 (Spring 2016): Slide Set 12
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
Block I
(1)
ab
Block II
a
b
6
We shall consider each of these designs separately:
•
Design 1:
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.
•
•
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.
•
=
=
=
=
μ(1) + β1 + (1)
μab + β1 + ab
μa + β2 + a
μb + β2 + b
7
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:
y(1)
yab
ya
yb
We can write the model for the design, assuming one replication of the
treatment combinations as follows:
A 2 Factorial in Blocks of Size 2 (Cont’d design1 )
Stat 402 (Spring 2016): Slide Set 12
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?)
•
2
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.
•
Stat 402 (Spring 2016): Slide Set 12
ABC
+
+
+
+
8
8
1
16
4
5
BC
+
+
+
+
0
8
0
16
0
Stat 402 (Spring 2016): Slide Set 12
A 2 Factorial in Incomplete Block Experiments
k
4
Factorial Effect
AB
C
AC
+
+
+
+
+
+
+
+
+
+
+
+
20
20
0
8
8
8
2.5 2.5
0
16
16
16
25
25
0
A 2 Factorial in Blocks of Size 2
2
Treatment
Combination
(1)
a
b
ab
c
ac
bc
abc
Contrast
Divisor for Estimate
Estimate of Effect
Divisor for SS
SS of Effect
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:
Factorial Effects and their Sums of Squares:
•
•
Stat 402 (Spring 2016): Slide Set 12
Block I
(1)
b
y(1)
yb
ya
yab
=
=
=
=
Block II
a
ab
μ(1) + β1 + (1)
μb + β1 + b
μa + β2 + a
μab + β2 + ab
In this case, the model is given by
Design 2:
10
•
= 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)]
AB
11
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.
= 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)]
B̂
 = 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)]
Stat 402 (Spring 2016): Slide Set 12
Hence,
Stat 402 (Spring 2016): Slide Set 12
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
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.
= 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)]
Stat 402 (Spring 2016): Slide Set 12
•
AB
AB: (a-1)(b-1):
8
= 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)]
A 2 Factorial in Blocks of Size 2 (Cont’d design 2)
2
B̂
B: (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)]
A: (a-1)(b+1)
A 2 Factorial in Blocks of Size 2 (Cont’d design 1)
2
=
=
=
=
Block II
a
ab
μ(1) + β1 + (1)
μa + β1 + a
μb + β2 + b
μab + β2 + ab
Block I
(1)
a
Stat 402 (Spring 2016): Slide Set 12
14
2. Then treatment combinations are randomly applied to the units within
each block.
1. Given an experimental plan, numbers are assigned to blocks at random.
Randomization in Incomplete Blocks
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.
y(1)
ya
yb
yab
The model in this case is
Design 3:
12
Stat 402 (Spring 2016): Slide Set 12
A 2 Factorial in Blocks of Size 2 (Cont’d design 3)
2
2. It is possible to confound a given effect by the choice of the design.
1. Only one effect is confounded with blocks in each design
13
In this design, A and AB effects are free of block effects and the main
effect B is confounded with block effect.
Remarks
•
Stat 402 (Spring 2016): Slide Set 12
 = 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 a similar manner as above we can show that