Slide set 14
Stat402 (Spring 2016)
Last update: April 19, 2016
Stat 402 (Spring 2016): Slide set 14
Analysis of an Incomplete Block Experiment with Partial
Confounding
•
A 22 factorial experiment was run in blocks of size 2 with 4 replicates.
In Rep 1 the effect A is confounded, in Rep 2 the effect B and in Rep 3
and 4 the effect AB. The observations are:
Rep 1
Block 1
a
96
ab
112
Total 208
Rep 2
Block 2
(1)
90
b
102
Total 192
Block 3
b
84
ab
100
Total 184
Rep 3
Block 5
(1)
50
ab
80
Total 130
Block 6
a
48
b
60
Total 108
Block 4
(1)
72
a
90
Total 162
Rep 4
Block
(1)
ab
Total
7
31
45
76
Block
a
b
Total
8
36
24
60
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Stat 402 (Spring 2016): Slide set 14
Analysis of an Incomplete Block Experiment with
Partial Confounding (Cont’d 1)
X
(1120)2
= 78400 Total SS =
y 2−CF = 90266−78400 = 11866
CF =
16
(208)2 1922
602
178168
Block SS =
+
+· · ·+
−CF =
−78400 = 10684
2
2
2
2
•
The following table shows which effects are confounded in each replicate,
the confounded effects being denoted by a ‘∗0.
A
B
C
Rep 1
*
Rep 2
Rep 3
Rep 4
*
*
*
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Stat 402 (Spring 2016): Slide set 14
Analysis of an Incomplete Block Experiment with
Partial Confounding (Cont’d 2)
•
To begin the analysis, we have to form the totals for treatment
combination as in the case of a factorial in a CRD or a RCBD.
•
But here we have seperate sets totals to estimate each effect, because
different effects are confounded in different reps. For example, the
observations in Rep 3 and 4 cannot be used to estimate the AB effect.
For estimating
effect
I
A
B
AB
Totals for Estimating Each Effect
Treatment Combination
(1)
a
b
ab
243 270 270 337
(Omitting none)
153 174 168 225
(Omitting Rep 1)
171 180 176 237
(Omitting Rep 2)
162 186 186 212 (Omitting Rep 3,4)
Sample
Size
4
3
3
2
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Stat 402 (Spring 2016): Slide set 14
Using Contrasts for Estimating Confounded Effects
Now we can use the defining contrasts for each effect to evaluate the
estimates of effects using the corresponding set of totals.
Treatment
Combination
(1)
a
b
ab
Contrast
Divisor for Estimate
Estimate
Divisor for SS
SS
Factorial Effect
I
A
B AB
+
+
+
+
+
+
+
+
+
+
1120
78
72
2
16
6
6
4
70
13
12 0.5
16
12
12
8
78400 507 432 0.5
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Stat 402 (Spring 2016): Slide set 14
ANOVA Table
Source
d.f.
Treatments
3
A
1’
B
1’
AB
1’
Blocks
7
Error
5
Total
15
SS
939.5
507.0
432.0
0.5
10684.0
242.5
11866.0
MS
313.2
507.0
432.0
0.5
1526.3
48.5
F
6.46
10.45
8.91
0.01
So far we have considered the following experiments:
1. A 23 factorial in 2 blocks i.e., the block size used was 4. This is again
an example of a 2k factorial in 2p blocks. Here k=3 and p=1.
2. A 22 factorial in 2 blocks i.e., the block size used was 2. This is an
example of a 2k factorial in 2p blocks where k=2 and p=1.
3. In general, for 2p blocks the block size is 2k−p.
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Stat 402 (Spring 2016): Slide set 14
2k Factorial Experiments in 2p Blocks
Example 1: 23 Factorial Experiments in 22 Blocks We now consider an
example of a 2k experiment in 2p blocks where p takes the value 2.
Consider a 23 factorial experiment in 22=4 blocks, i.e., the block size is 2.
(k=3, p=2)
a) The treatment combinations, as before, are: (1), a, b, ab, c, ac, bc, abc.
b) Suppose that the following design was used:
Block 1
a
bc
Block 2
b
ac
Block 3
c
ab
Block 4
(1)
abc
c) Which of the effects are confounded in the experiment?
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Stat 402 (Spring 2016): Slide set 14
As before we can write out each of the effects using the appropriate
model and determine what effects are confounded. For example, using
A = 1/4[abc + ab + ac + a − bc − b − c − (1)]
we find that
 = 1/4[yabc,4 + yab,3 + yac,2 + ya,1 − ybc,1 − yb,2 − yc,3 − y(1),4]
= 1/4[µabc + µab + µac + µa − µbc − µb − µc − µ(1)]
+(β4 + β3 + β2 + β1 − β1 − β2 − β3 − β4)
+(contrast of error terms)
= A + (contrast of error terms)
which implies, as before, that A is not confounded with blocks and thus can
be estimated. If this procedure is repeated for all other effects, we would
find that AB, AC and BC, i.e., 3 effects in all, are confounded.
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Stat 402 (Spring 2016): Slide set 14
Example 1 (Cont’d 1)
d) The number of effects that are confounded will always be the number of
blocks in the replicate minus one.
e) How do we construct a factorial design where more than one effect is
confounded in a replicate?
f) In such a design, we would like to confound effects that are chosen by
the experimenter for reasons to be discussed later. The procedure for
doing this is illustrated in the following example.
Example 2: A 24 factorial in 4 blocks, i.e., 22 blocks of size 4 (k=4, p=2).
a) We have 16 treatment combinations : (1), a, b, ab, c, ac, ..., abcd.
b) Since the number of blocks is 4, we know that 3 effects will be confounded
in this design.
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Stat 402 (Spring 2016): Slide set 14
c) However, we select two (since p=2) effects to be used for constructing
the design.
d) These will be two of the effects that are going to be confounded. The
effects that are chosen for this purpose are called block generators.
e) In this example we choose ABC and BCD effects as block generators.
Now look at the defining contrasts for these effects:
Effect
ABC
BCD
Effect
ABC
BCD
(1)
d
+
a
+
ad
+
+
b
+
+
bd
+
-
ab
+
abd
-
c
+
+
cd
+
-
ac
+
acd
-
bc
bcd
+
abc
+
abcd
+
+
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Stat 402 (Spring 2016): Slide set 14
Example 2 (Cont’d 1)
f) The 16 treatment combinations are allocated to the 4 blocks as follows:
Block
Block
Block
Block
1:
2:
3:
4:
All
All
All
All
Combinations
Combinations
Combinations
Combinations
with
with
with
with
(+, +)
(+, −)
(−, +)
(−, −)
g) Thus the design is
Block 1
b
c
ad
abcd
Block 2
a
abc
bd
cd
Block 3
ab
ac
d
bcd
Block 4
(1)
bc
abd
acd
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Stat 402 (Spring 2016): Slide set 14
h) In this replicate there are 4 blocks, implying that 3 effects should be
confounded. Two of these are the ones used as block generators: ABC
and BCD.
i) The third confounded effect is found by the following rule: Pairs of
effects that are already confounded are multiplied together algebraically.
The odd-powered letters in the algebraic result are retained and the
even-powered letters are eliminated to form the other effects that are
confounded.
j) In the present example, the third effect confounded is , therefore, given
by
(ABC)(BCD) = AB 2C 2D = AD
Thus we see that AD is also confounded.
k) AD is called the generalized interaction of ABC anf BCD
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Stat 402 (Spring 2016): Slide set 14
Example 2 (Cont’d 4)
`) As before, we may check whether a given effect is confounded or not by
using the defining contrast. For example,
A = 1/8[−(1) + a − b + ab − c + ac − bc + abc − d + ad − bd + abd − cd + acd − bcd + abcd]
Thus,
 = A + 1/8[−β4 + β2 − β1 + β3 − β1 + β3 − β4 + β2 − β3
+β1 − β2 + β4 − β2 + β4 + −β2 + β4 − β3 + β1]
+contrast of error terms
= A + contrast of error terms
and therefore A is not confounded. Similarly,
d = AD + 1/8[β4 − β2 + β1 − β3 + β1 − β3 + β4 − β2
AD
−β3 + β1 − β2 + β4 − β2 + β4 − β3 + β1] + contrast of errors
Here we see that block effects do not cancel out showing that AD is
confounded with blocks.
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Stat 402 (Spring 2016): Slide set 14
Remarks
The usual approach for constructing a 2k factorial in 2p blocks is to use
a table such as Table 7.9) (p.316 of text). Once the block generators are
determined, the experimental plan may be constructed using the procedure
discussed above.
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Stat 402 (Spring 2016): Slide set 14
Suggested Blocking Arrangements
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