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STAT 430 (Fall 2017): Tutorial 8
Balanced Incomplete Block Design
Luyao Lin
November 7th/9th, 2017
Department Statistics and Actuarial Science, Simon Fraser University
Block Design
• Complete
• Random Complete Block Design
• Incomplete Block Design
• Balanced Incomplete Block Design (BIBD)
• Latin Square Design
1
Incomplete Block Design
• Why we need it?
• Advantage of it?
• Disadvantage of it?
• How to do it?
2
Why ‘Incomplete’ Block Design? (Motivation)
Block sizes
• Complete Block Design, every block can hold every treatment level
at least once.
• Incomplete Block Design, the block size is less than the treatment
level.
Suppose we have 4 different treatment levels, but each
block can only have 2 experiment units.
• Balanced Incomplete Block Design (BIBD)
3
Balanced Incomplete Block Design (BIBD)
• v treatment level
• r : each treatment appears in r blocks
• b blocks; b > r
• k each block size (# of units each block can have)
• λ: each pair of the treatment appears together in λ blocks (why we
care about this ?)
• total # of units: bk or vr
4
Balanced Incomplete Block Design (BIBD)
• v treatment level
• r each treatment appears in r
blocks
• b blocks; b > r
• k each block size (# of units
each block can have)
• λ: each pair of the treatment
appears together in λ blocks
• total # of units: bk or vr
5
Properties of BIBD
Requirements:
(i) Each treatment must appear the same number of times in the
design: vr = bk
(ii) each pair of treatments appears together in λ blocks
r(k − 1) = λ(v − 1)
Advantage:
• all treatment contrasts estimable
• all pairwise comparisons are estimated with the same variance
• tends to give the shortest CIs for contrasts
Disadvantage: for certain value of b, k , v , r , BIBD might not exist. If
b = v = 8, r = k = 3, λ =?
6
r(k − 1) = λ(v − 1)
• for treatment say A, it appears
in r blocks
• within those r blocks, there are
in total r(k − 1) treatments
other than A itself
• on the other hand, we have
v − 1 other treatment levels
• A is supposed to appear with
each of them in λ blocks
• so that λ(v − 1) = r(k − 1)
• v treatment level
• r each treatment appears in r
blocks
• b blocks; b > r
• k each block size (# of units
each block can have)
• λ: each pair of the treatment
appears together in λ blocks
• total # of units: bk or vr
7
How to analyze BIBD?
Yhi = µ+ θh + τi + εij
• εi.i.d.∼ N(0, σ2)
• h = 1, . . . , b, i = 1, . . . , v
• (h, i) in the design
•∑τi = 0
∑θh = 0
Assumptions for Yhi ∼ N(µ+ θh + τi , σ2)
• normality
• equal variance
• independence
• no interaction effect (hard to verify)
9
Parameter Estimation: least square
• unadjusted estimate: τi = Y.i
E (Y.1 − Y.2) = τ1 − τ2 +1
7
( 7∑h=4
θh −11∑h=8
θh
)6= τ1 − τ2
• adjusted estimate (section 11.3.2)
r(k − 1)τi − λ∑p 6=i
τp = kQi
Qi = Ti −1
k
∑h
nhiBh
(Intra-block equations)
• Ti is the total of response on treatment i
• Bh is the total of response in block h
• nhi is 1 if i is in block h or 0 otherwise
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adjusted estimate (section 11.3.2)
r(k − 1)τi − λ∑p 6=i
τp = kQi
along with∑
i τi = 0
r(k − 1)τi + λτi = kQi
τi =kQi
r(k − 1) + λ=
kQi
λv=
kTi −∑
h nhiBh
λv
because λ(v − 1) = r(k − 1)
τi =kQi
λv
11
Other Parameters Estimate
Yhi = µ+ θh + τi + εij
• τi = kQi
λv
• µ = Gbk ; G is the grand total
•
θh =Bh
k− G
bk−∑v
i=1 nhi τik
• σ2 = mse = SSEbk−b−v+1
12
ANOVA table
bk − b − v − 1 SSE =b∑
h=1
v∑i=1
nhi e2hi =
b∑h=1
v∑i=1
(yhi − µ− θh − τi )2
=b∑
h=1
v∑i=1
y 2hi −
1
k
b∑h=1
B2h −
v∑i=1
Qi τ2i
bk − 1 SStot =b∑
h=1
v∑i=1
y 2hi −
1
bkG 2
b − 1 SSθ =1
k
b∑h=1
B2h − 1
bkG 2
v − 1 SSTadj =v∑
i=1
Qi τ2i
F0 =MsTadj
MSE
13
CI for the contrasts
• τi = kλvQi
• Var(Qi ) = σ2
• Var(∑
ci τi ) =∑
c2ikλv σ
2
• CI for∑
ci τi : (k
λv
∑ciQi ± w
√∑c2i
k
λvσ2
)• wB = ttk−b−v+1,α/2m,wT = qv ,bk−b−v+1,α/
√2
Section 11.4.2 has a good example for BIBD page 355
14
Big picture
The Latin Square Design
1. ‘incomplete block design’
2. one treatment
3. two blocking variables
4. only one single treatment is applied within each combination of
blocking variables
5. the level of treatment = the level of two blockingsoperators
Material 1 2 3 4 5
1 A=24 B=20 C=19 D=24 E=24
2 B=17 C=24 D=30 E=27 A=36
3 C=18 D=38 E=26 A=27 B=21
4 D=26 E=31 A=26 B=23 C=22
5 E=22 A=30 B=20 C=29 D=31
16
why use a Latin Square?
1. impossible to use each treatment level for the same combination of
blocking
2. for example, consider an experiment with four diets, each to be
given to four cows in succession
3. each cow can only be given a single diet during a single time period.
4. one block factor is the cow, the other block factor is the time or the
ordertime
cow 1 2 3 4
1 C A D B
2 A B C D
3 D C B A
4 B D A C
5. Latin square is not unique; as long as each letter (treatment)
appears in each row and column exactly once.
17
R function
latin=
function (n, nrand = 20)
{
x = matrix(LETTERS [1:n], n, n)
x = t(x)
for (i in 2:n) x[i, ] = x[i, c(i:n, 1:(i -
1))]
if (nrand > 0) {
for (i in 1:nrand) {
x = x[sample(n), ]
x = x[, sample(n)]
}
}
x
}
latin (5)
[,1] [,2] [,3] [,4] [,5]
[1,] "E" "D" "C" "B" "A"
[2,] "D" "C" "B" "A" "E"
[3,] "C" "B" "A" "E" "D"
[4,] "A" "E" "D" "C" "B"
[5,] "B" "A" "E" "D" "C"
latin (4)
[,1] [,2] [,3] [,4]
[1,] "A" "C" "B" "D"
[2,] "B" "D" "C" "A"
[3,] "D" "B" "A" "C"
[4,] "C" "A" "D" "B"
18
Model
yijk = µ+ αi + τj + βk + εijk
• yijk : ith row, kth col, jth treatment; i , j , k = 1, 2, . . . , p
• µ overall mean, αi ith row effect; τj jth treatment effect; βk kth
column effect ∑αi =
∑τj =
∑βk = 0
• εijk ∼ iid N(0, σ2)
• N = p2
SST =SSrows + SScol + SStreatment + SSE
p2 − 1 =p − 1 + p − 1 + p − 1 + (p − 2)(p − 1)
19
ANOVA
SST =∑i
∑j
∑k
y2ijk −
y2•••N
SSTreatment =1
p
∑j
y2•j• −
y2•••N
SSRows =1
p
∑i
y2i•• −
y2•••N
SSCols =1
p
∑k
y2••k −
y2•••N
• y••• : total for all =∑
i
∑j
∑k yijk
• yi•• : total for row i
• y••k : total for row k
• y•j•k : total for treatment j
20