STATISTICS 402B
Spring 2016
Homework Set#7 Solution
1. Using 2 block generators to allocate runs to 4 blocks in a 24 factorial:
• Write contrasts for ABD and ACD for a 24 factorial:
Effect
ABD
ACD
Effect
ABD
ACD
(1)
d
+
+
a
+
+
ad
-
b
+
bd
+
ab
+
abd
+
-
c
+
cd
+
-
ac
+
acd
+
bc
+
+
bcd
-
abc
abcd
+
+
• Select treatment combinations with sign pairs (+, +), (+, −), (−, +) and (−, −) in the two contrasts
to divide the 16 runs into 4 blocks, as follows:
Block I
a
bc
d
abcd
Block II
b
ac
cd
abd
Block III
ab
c
acd
bd
Block IV
(1)
abc
ad
bcd
2. In a 27 factorial, 4 block generators are used to construct the plan. This implies that including the
generalized interactions of the 4 effects used as generators 4 + 6 + 4 + 1 = 15 effects will be confounded in
one replication.
(a) What is the block size?
Since 15 effects are confounded in a rep, their should be 16 blocks in a rep. Thus the block size is
= 27 /16 = 128/16 = 8
(b) What is the number of blocks per each replicate?
From part (i), the answer is 16.
(c) What is the total number of effects that will be confounded with blocks in each rep?
Clearly, as argued above the answer is 15.
(d) Select 4 block generators (with more than 2 letters each) so that none of thei generalized interactions
produce effects that have less than 3 letters. One such set is given in Table 7.9 of the text: ABCD,
EFG, CDE, ADG (or you can try to construct one by your self)
(e) This is also given in the Table 7.9 for this set: ABCD, EFG, CDE, ADG, ABCDEFG, ABE, BCG,
CDFG, ADEF, ACEG, ABFG, BCEF, BDEG, ACF, BDF
3. In a 26 factorial, if the block size is going to be 16, there will be 4 blocks per replicate.
• This means that there are 3 effects confounded in each replicate.
• We need to select 2 effects out of these 3 and the third effect confounded will be the generalized
interaction of those two selected.
• We attempt to select two effects arbitrarily so that no effects with less than 3 letters are confounded.
We try to this by ourselves or look at Table 7.9 to get an idea.
• They suggest using ABCF and CDEF with their generalized interaction is ABDE.
• By inspection, we see that another set can be constructed switching letters: ABCD, ADEF, and their
generalized interaction is BCEF
1
• Thus we get an idea of how to balanced confounding in 5 replications by selecting a sets such as:
ABCF
ABCD
ABCE
ABDF
ABEF
Here only
once in a
CDEF ⇒ ABDE
ADEF ⇒ BCEF
ACDF ⇒ BDEF
ACEF ⇒ BCDE
ACDE ⇒ BCDF
the 15 4-fi’s are confounded but we get balanced confounding since each 4-fi is confounded
rep, and no other effects are confounded.
• The abbreviated Anova table is:
SV
A, B, C, D, E, F
AB, AC , . . ., DF
ABC, ABD, . . .,DEF
ABCD, ABCE, . . .,BCDF
ABCDE, . . .,ABCDF
ABCDEF
Rep
Block(Rep)
Error
Total
d.f.
1 each (total=6)
1 each (total=15)
1 each (total=20)
1 each (total=15)
1 each (total=6)
1
4
15
237
319
%Inf.
100% each
100% each
100% each
80% each
100% each
100%
4. A single effect is confounded in each rep as there are only 2 blocks per rep.
(a) Find the effect confounded in each replication.
By inspection, it is easy determine that the following:
Rep No.
1
2
3
4
5
Effect Confounded
ABC
AC
AB
ABC
BC
(b) Provide a table of estimates of the factorial effects with their standard errors.
Term
A
B
A*B
C
A*C
B*C
A*B*C
Effect Estimate
1.6
1.9
0.25
0.4
0.3125
-1
-0.25
s.e.(effect)
0.3931
0.3931
0.4395
0.3931
0.4395
0.4395
0.5075
(c) Construct an analysis of variance table that inccludes F statistics and p-values. Add a column that
indicates relative percentage of information used in estmating each of the partially confounded effects.
2
Model
Error
C. Total
16
23
39
Effect Tests
Source
A
B
A*B
C
A*C
B*C
A*B*C
Rep
Block[Rep]
146.35625
35.54375
181.90000
DF
1
1
1
1
1
1
1
4
5
9.14727
1.54538
Sum of Squares
25.600000
36.100000
0.500000
1.600000
0.781250
8.000000
0.375000
34.150000
21.906250
5.9191
Prob > F
<.0001*
F Ratio
16.5655
23.3599
0.3235
1.0353
0.5055
5.1767
0.2427
5.5245
2.8351
Prob > F %Information
0.0005*
100%
<.0001*
100%
0.5750
80%
0.3195
100%
0.4842
80%
0.0325*
80%
0.6270
60%
(d) Interpret the results of this analysis. First, determine the effects that are significant. Then quantitatively explain the effect of each significant factor on the mean responses.
Main effects A and B and BC Interaction are significantly different from zero.
A
Least Squares Means Table
Level
Least Sq
Mean
6.1500000
7.7500000
-1
1
Std Error
Mean
0.27797306
0.27797306
6.15000
7.75000
LS Means Plot
12
Level
Least Sq
Mean
5.3000000
6.7000000
8.2000000
7.6000000
-1,-1
-1,1
1,-1
1,1
Std Error
0.40521188
0.40521188
0.40521188
0.40521188
LS Means Plot
10
-1
12
8
1
10
6
8
4
2
B*C
Least Squares Means Table
6
-1
1
4
A
2
-1
1
C
Effect of A is to increase mean yield by 1.6 units.
Effect of B is different at each level of C:
***** at the low level of C, effect of B is to increase mean yield by 3.25 units
***** at the high level of C, effect of B is to increase the mean yield only by 1.25 units.
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