Slide set 7
Stat402 (Spring 2016)
Last update: February 26, 2016
Stat 402 (Spring 2016): Slide Set 7
Balanced Incomplete Block Design (BIBD)
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Recall that in a RCBD, every treatment must appear once in each block;
i.e, block size must equal a, the number of treatments.
Sometimes this is not possible because limitations on the availability
of enough material, people, time etc. in a block to accomodate all a
treatments.
For example, in the vascular graft experiment, a batch may not have
enough raw material for testing 4 extrusion pressures as required by the
RCBD that was used.
In another experiment where runs performed in a day formed a block
(i.e, Day is blocking factor) it may not possible to run a complete set of
the treatments in a day because of insufficient time.
A balanced incomplete block design allows the block size k to be smaller
that a (i.e k < a as long as every pair of treatments appear together the
same number of times in the experiment.
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Stat 402 (Spring 2016): Slide Set 7
Model
iid
yij = µ + τi + βj + ij , ij ∼ N (0, σ 2)
i = 1, . . . , a(treatments)
a
b
k
n
λ
j = 1, . . . , b (blocks)
no. of treatments
no. of blocks
block size (no. of exp. units or runs per block), k < a
no. of replications of each treatment
no. of times each pair of treatments appear together in the same block
N = an = bk=total number of observations. Also λ =
n(k−1)
a−1
In this design it is necessary to adjust the treatment means to allow for the
fact that not every treatment occurs in each block.
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Stat 402 (Spring 2016): Slide Set 7
Balanced Incomplete Block Design (BIBD) Cont’d
SST
=
SSBlocks =
Ti
Bi
Qi
Pa
Pb
2
i=1
j=1 (yij − ȳ.. )
Pb
k j=1(ȳ.j − ȳ..)2
= yi. = Total for treatment i
Pb
=
j=1 nij yij ; nij = 1 if trt i appears in block j;
nij = 0, otherwise
= sum of block totals of blocks containing treatment i
= Ti − (1/k)Bi
Estimate of τi = τ̂i = KQi/aλ=Adjusted Treatment effect
ȳi(adjusted) = τ̂i + ȳ..=Adjusted Treatment Mean
Pa
SST rt(adjusted) = k i=1 Q2i /aλ
SSE = SST − SST rt(adj.) − SSBlks
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Stat 402 (Spring 2016): Slide Set 7
ANOVA Table
SV
DF
T rt(adj.)
a−1
Blks
b−1
Error
N −a−b+1
T otal
N −1
SS
MS
F
SST rt(adj.) M ST rt(adj.)
M ST rt(adj.)/M SE
SSBlk
SSE
M SE = (s2E )
SST
A 100(1 − α)% CI for τp − τq is
p
(τ̂p − τ̂q ) ∓ tα/2,ν sE 2k/aλ
Note that τ̂p − τ̂q is identical to ȳp(adj.) − ȳq(adj.)
A single degree of freedom sum of
to test a contrast of treatment
Psquares
a
effects equal to zero is (i.e. H0 : i=1 ciτi = 0)
SSc =
k(
=1a ci Qi )2
iP
2
λa a
i=1 ci
P
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Stat 402 (Spring 2016): Slide Set 7
Computations for the BIBD
Example: (Catalyst Experiment: Table 4.22)
Catalyst
1
2
3
4
y.j
Blocks
1
2
3
4
73 74
−
71
−
75 67 72
73 75 68
−
76
−
72 75
221 224 207 218
yi.
218
214
216
222
870
The experimenter is comparing time of reaction yij for a chemical process
under 4 different types of catalysts. It is planned to use batches of raw
materials as blocks; however, a batch of raw materials only large enough to
do 3 runs. Thus they used a BIBD with a = 4 b = 4 k = 3 n = 3 λ =
2 N = 12 as above.
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Stat 402 (Spring 2016): Slide Set 7
CF
SST
SSBlk
2
= 870
12
P
P4
4
2
=
y
− CF
ij
i=1
j=1
P4 yij2
=
j=1 k − CF
= 63075
= 63156 − 63075 = 81.0
=
2212 +2072 +2242 +2182
3
− 63075 = 55.0
Treatment i Ti
BI
Qi
τ̂i
ȳi(adj)
1
218 (221 + 224 + 218) = 663
−3
−1.125 71.375
2
214 (207 + 224 + 218) = 649 −2.33 −0.875 71.625
3
216 (221 + 207 + 224) = 652 −1.33 −0.5
72.0
4
222 (221 + 207 + 218) = 646 +6.66
2.5
75.0
P4
SST rt(Adj) = k i=1 Q2i /λa
= 2[(−3)2 + (−2.33)2 + (−1.33)2 + (6.66)2]/2(4) = 22.75
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Stat 402 (Spring 2016): Slide Set 7
ANOVA Table (extracted from Montgomery: Table 4.24)
SV
d.f.
SS M S
F p-value
Catalyst(adj.
3 22.75 7.58 11.66
.0107
Batch
3
55
Error
5 3.25 0.65
Total
11 81.0
Thus reject H0 : τ1 = τ2 = τ3 = τ4 at α = .05 as p-value¡.05.
A 95% CI for τ1 − τ2 is:
√
(71.375 − 71.625) ± t.025,5 · 0.65
s
(2)(3)
(2)(4)
i.e.,(−0.25 ± (2.571)(0.7) or (−2.05, 1.55)
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