STAT 512 Balanced Incomplete Block Designs
BALANCED INCOMPLETE BLOCK DESIGNS (BIBD) & RELATED
• As with CBD, b blocks, but now each with k units, k < t
• N = bk
• NOW all treatments cannot fit in any one block, so blocks are called “ incomplete ”
• Examples: b = 4 , k = 3 , t = 4
Usually a bad idea block treatment
3
4
1
2
1
1
1
4
2
2
2
4
3
3
3
4
Usually a better idea block treatment
3
4
1
2
1
1
1
2
2
2
3
3
3
4
4
4
1

STAT 512 Balanced Incomplete Block Designs
Rules for BALANCED Incomplete Block Designs (BIBD’s)
• “First-order balance” : Each treatment must appear the same number of times in the (entire) design: r = N/t = bk/t
• “Second-order balance” : Each PAIR of treatments must appear together in the same number of blocks:
λ = r ( k − 1) / ( t − 1)
– pick any specific treatment, say “1”
– numerator is number of “non-1” assignments in blocks containing a “1”
– denominator is number of “non-1” treatments
2

STAT 512 Balanced Incomplete Block Designs
• BIBD’s don’t exist for every t , b , and k < t
• Necessary, but not Sufficient, Condition: r = bk/t and λ = r ( k − 1) / ( t − 1) must be integers
• Example: t = 7 , k = 5
– r = bk/t = b [5 / 7] = integer
– λ = [ r ][( k − 1) / ( t − 1)] = b [5 / 7][4 / 6] = b [10 / 21] = integer
– smallest integer solution is b = 21 → λ = 10 , r = 15
– EXIST? Yes, in this case use all possible combinations of 7 treatments in groups of size 5:
 
7
 
= 21
5
3

STAT 512 Balanced Incomplete Block Designs
• Easy Case:
– any k < t , one block for each unique subset of k treatments
  t
∗ b =
  k
 t − 1
∗ r = b [ k/t ] = [ t !
/ (( t − k )!
k !)][ k/t ] =
 k − 1


 t − 1
∗ λ = r [( k − 1) / ( t − 1)] =
 k − 1
  t − 2
 k − 1 t − 1
=
 k − 2


– the second design on the first slide is an example of this:
∗ k = t − 1 , b = t , r = t − 1 , λ = t − 2
• In General: combinatorics, tables
4

STAT 512 Balanced Incomplete Block Designs
Model
• y i,j
= α + β i
+ τ j
+ i,j or β i
+ τ j
+ i,j
– as with CBD, but not all ( i, j ) combinations are included
• y = X
1
β + X
2
τ +
• [ bk × 1] = [ bk × b ][ b × 1] + [ bk × t ][ t × 1] + [ bk × 1]
Estimable Functions ... need to characterize ( I − H
1
) X
2
• H
1
= X
1
( X
0
1
X
1
)
− 1
X
0
1
=
1 k diag ( J k × k
...
J k × k
)
– like CBD, but now k = t
5

STAT 512 Balanced Incomplete Block Designs
• H
1
X
2
=
1 k








← columns indexed by j →

# of times trt j appears in blk 1
# of times trt j appears in blk 2
.
..







# of times trt j appears in blk b
( k rows )
( k rows )
.
..
( k rows )
• Not the same as a CRD with r units/trt group

# of times trt j appears in this row of X
2
-
• ( I − H
1
) X
2
=









1 k no. of times trt j appears in blk 1
.
..
# of times trt j appears in this row of X
2
-
1 k no. of times trt j appears in blk b










6

STAT 512 Balanced Incomplete Block Designs
• e.g. for k = 4 , t = 6 , treatments 1 - 4 in first block:

X
2 | 1
=









− 1
4
3
4
− 1
4
− 1
4
...
−
− 1
4
− 1
4
1
4
3
4
−
− 1
4
−
3
4
1
4
1
4
−
−
−
1
4
1
4
1
4
3
4
0 0
0
0
0
0
0
0










• Note all rows sum to zero, so all linear combinations of rows add to zero, so only contrasts between treatments are estimable
(again)
7

STAT 512 Balanced Incomplete Block Designs
Reduced Normal Equations: (Need X
0
2 | 1
X
2 | 1
= X
0
2
( I − H
1
) X
2
)
• X
0
2
X
2
= r I
• ( X
0
1
X
1
)
− 1
= k
− 1
I
• { X
0
1
X
2
} ij
= N ( i, j ) , an “incidence matrix”
– number of times treatment j appears in block i
– always 0 or 1
• { ( X
0
1
X
2
)
0
( X
0
1
X
2
) } j,j 0
= P i
N ( i, j ) N ( i, j
0
)
– number of blocks containing treatments j AND j
0
– always r for j = j
0
– always λ for j = j
0
– ( r − λ ) I + λ J
• ( X
0
2
X
1
)( X
0
1
X
1
)
− 1
( X
0
1
X
2
) = r − λ
I + k
λ
J k
8

STAT 512 Balanced Incomplete Block Designs
• Left side of R.N.E.:
– [ X
0
2
X
2
− X
0
2
H
1
X
2
]ˆ
– [ r I − ( r − λ ) /k I − λ/k J ]ˆ
–
λ k
[ t I − J ]ˆ = I
2 | 1
• Right side of R.N.E.:
– X
0
2
( I − X
1
( X
0
1
X
1
)
− 1 X
0
1
) y
– X
0
2 y − 1 k
X
0
2
X
1
X
0
1 y
– T − k
− 1
( X
0
2
X
1
) B
∗ T = t -element vector of treatment totals
∗ B = b -element vector of block totals
∗ { ( X
0
2
X
1
) B } j
= sum of block totals containing trt j
9

STAT 512 Balanced Incomplete Block Designs
• Multiply both Right and Left expressions above by k/λ
– [ t I − J ]ˆ = k/λ [ T − k
− 1
( X
0
2
X
1
) B ]
– Now divide both sides of this by t
– j th row: ˆ j
− τ
¯
.
= k/ ( λt ) [ y
.j
− k
− 1 P i
N ( i, j ) y i.
]
– Second term in brackets is the sum of block averages for
(only!) blocks that contain a unit with treatment j .
– Many books use Q j for [-], “adjusted treatment j total”
– If c
0
τ is estimable,
– c 0 τ = k/ ( λt ) P j c j
Q j
(since
¯ drops out)
Note that k
λt
“plays the role of”
1 r
... but they aren’t equal
10

STAT 512 Balanced Incomplete Block Designs
Variance of Estimable Functions:
• 0 τ = k/ ( λt ) c
0
[ X
0
2
− X
0
2
H
1
] y
• V ar [ c 0 τ ]
= k
2
σ
2
/ ( λt )
2 c
0
[ X
0
2
− X
0
2
H
1
][ X
2
− H
1
X
2
] c
= k 2 σ 2 / ( λt ) 2 c
0
[ X
0
2
X
2
− X
0
2
H
1
X
2
] c left-side matrix !
= k
2
σ
2
/ ( λt )
2 c
0
[ λt/k I − λ/k J ] c
J term drops out for estimable functions
= [ kσ
2
/ ( λt )] c
0 c
– again, k
λt
“plays the role of”
1 r
...
11

STAT 512 Balanced Incomplete Block Designs
When is it worth using?
• V ar
BIBD
[ 0 τ ] = [ kσ 2
BIBD
/ ( λt )] c
0 c
• V ar
CRD
[ 0 τ ] = [ σ
2
CRD
/ ( N/t )] c
0 c
• Use r = N/t and λ = ( N/t )[( k − 1) / ( t − 1)] ...
• V ar
BIBD
/V ar
CRD
= [ k/ ( k − 1)] / [ t/ ( t − 1)] [ σ
2
BIBD
/σ
2
CRD
]
• “Multiplying ratio” is greater than one if k < t
• Compare to earlier results, e.g.
V ar
LSD
/V ar
CRD
= σ
2
LSD
/σ
2
CRD
12

STAT 512 Balanced Incomplete Block Designs
• Multiplying ratio increases as k decreases relative to t , e.g.
t k ratio
10 10 1/1
10 8 72/70
10 6 54/50
10 4 36/30
10 2 18/10
• Expected squared CI length criterion becomes: v u u t k/ ( k − 1) t/ ( t − 1)
σ
BIBD
/σ
CRD
< t (1 t
−
(1 −
α/ 2;
α/
N
2;
−
N b
−
− t t )
+ 1)
13

STAT 512 Balanced Incomplete Block Designs
ANOVA: SS’s for blocks and treatments aren’t orthogonal ...
• SST( blocks
∗
) = y
0
H
1 y = P b i =1 k (¯ i.
− ¯
..
)
2
• SST( blocks
∗
, treatments ) = y
0
Hy
SST( treatments | blocks
∗
) = SST( blocks
∗
, treatments ) - SST( blocks
∗
)
=
=
= k
λt
P j
Q 2 j
(slide 10) k
λt
P j
[ y
.j
−
( t − 1) k
( k − 1) t
P
1 k
P i
N ( i, j ) y i.
]
2 j r [¯
.j
− y
[ j ]
..
]
2 where ¯
[ j ]
..
is the average of all observations include a unit assigned to treatment j from blocks that
∗ : “ blocks ” includes the intercept here, which would ordinarily be removed first in “correction for the mean”
14

STAT 512 Balanced Incomplete Block Designs
Similar result for Test Power
• Hyp
0
: τ
1
= τ
2
= ...
= τ t
• Compare CRD with r = N/t , BIBD with same N
• For hypothetical τ , CRD Q ( τ ) = r P i
( τ i
− ¯ )
2
• For CRD, the distribution of the test statistic is:
F
0
( t − 1 , N − t,
Q (
τ
)
σ 2
CRD
)
• For BIBD, the distribution of the test statistic is:
F
0
( t − 1 , N − b − t + 1 , t/ ( t − 1) k/ ( k − 1)
Q (
τ
)
σ 2
BIBD
)
• i.e. noncentrality is decreased , and estimation variances increased by the same factor
15

STAT 512 Balanced Incomplete Block Designs
YOUDEN “SQUARE” DESIGNS
• Recall that in LSD, #(rows) = #(columns) = #(treatments)
• Now consider row-column designs in which
– ignoring rows leads to a CBD in columns
– ignoring columns leads to a BIBD in rows
• For example:
1 2 3
2 4 1
3 1 4
4 3 2
• Generally: t treatments, t rows, k < t columns
16

STAT 512 Balanced Incomplete Block Designs
• Structure:
– Columns are orthogonal to Treatments (this matters)
– Rows are orthogonal to Columns (this DOESN’T matter)
– Rows are not orthogonal to Treatments (this matters)
• H
1 is more complicated than it would be for a CBD or a BIBD, but
– H
1
X
2 is the same as for a BIBD with t blocks of size k .
– RNE’s are as with BIBD with row blocks (only)
– Point estimates, variance formulae, and non-centrality parameters can be computed ignoring row blocks
• Can replicate Youden Designs in any of the 3 ways discussed with
LSDs
17

STAT 512 Balanced Incomplete Block Designs
PARTIALLY BALANCED INCOMPLETE BLOCK DESIGNS
• For two associate classes ...
• t treatments, b blocks, k < t units/block
• Every treatment is applied to r units overall (as in BIBD’s)
• Everay pair of treatments are either:
– First associates , applied together in λ
1 blocks, or
– Second associates , applied together in λ
2 blocks
• Each treatment has the same number of 1st associates and the same number of 2nd associates
• For any two treatments that are i th associates, there are:
– p i
1 other treatments that are 1st associates of each
– p i
2
– q i other treatments that are 2nd associates of each other treatments that are 1st associates of one and 2nd associates of the other
18

STAT 512 Balanced Incomplete Block Designs
• For example:
1 2 4 5
2 3 5 6
1 3 4 6
– t = 6 , b = 3 , k = 4
– r = 2
– Treatments can be divided into two groups: (1,2,3) and (4,5,6)
– Any two treatments within a group are 1st associates, with
λ
1
= 1
– Any two treatments in different groups are 2nd associates, with
λ
2
= 2
• Analysis is somewhat messier than with BIBD’s, but there is far more flexibility in size for designs for given t and k
19

STAT 512 Balanced Incomplete Block Designs
BIBD’s: INTER- and INTRA-BLOCK INFORMATION
• Fixed-Effects analyses we’ve considered derive treatment information only from within-block comparisons (INTRA-block information)
• INTER-block comparisons are considered “sacrificed”; confounded with unknown block effects.
• Recall 0 τ = k
λt
P j c j
[1 × y
.j
− 1 k
P i
N ( i, j ) y i.
]
– for every block containing treatment j , weights for each data value in the estimate of 0 τ are
− 1 k
, − 1 k
, ..., 1 − k
λt
× c j times:
1 k
, − 1 k
, − 1 k
, ...
– for every block not containing treatment j , weights are all zero.
– in either case, 0 τ = P ij w ij y ij where P j w ij
= 0 for all i .
20

STAT 512 Balanced Incomplete Block Designs
• First, note that some of the INTRA-block info is not so obvious ...
for example: block treatment
3
4
1
2
1
1
1
2
2
2
3
3
3
4
4
4
• Is there trt 1-vs-trt 2 INTRA-block information contributed from blocks 3 and 4?
– Overall, V ar [ τ
1
− d
τ
2
] = c
0 c σ
2 k
λt
= 2 σ
2 k
λt
=
3
4
σ
2
– From just blocks 1 and 2,
V ar [ τ
1 d
τ
2
] = V ar [
1
2
[( y
11
− y
12
) + ( y
21
− y
22
)] ] = σ 2
– So the answer is “Yes, but indirect”, e.g.
[1-3 diff from block 3] - [2-3 diff from block 4] estimates 1-2
21

STAT 512 Balanced Incomplete Block Designs
“Recovery” of INTER-block information
• For example design above, if y ij
= β i
+ τ j
+ ij
,
+block
2
3
4
Expected Block Total Differences
-block
1 2 3
τ
4
− τ
3
+ 3 β
2
− 3 β
1
τ
4
− τ
2
+ 3 β
3
− 3 β
1
τ
3
− τ
2
+ 3 β
3
− 3 β
2
τ
4
− τ
1
+ 3 β
4
− 3 β
1
τ
3
− τ
1
+ 3 β
4
− 3 β
2
τ
2
− τ
1
+ 3 β
4
− 3 β
3
• Suppose block effects are assumed to be random :
β i
∼ i.i.d.
N (0 , δ 2 ) + µ
β
• Then differences between blocks also contain information about treatment differences
(but ... some of these differences are correlated ... common β ’s)
22

STAT 512 Balanced Incomplete Block Designs
• Practical, Realistic Assumption?
• Sometimes not:
– machinists ... perhaps some of those available were trained by same instructor, and that these are “systematically different” from others
– litters of animals ... perhaps some of those available were bred/raised in same facility, ...
• Sometimes maybe:
– autos actually selected randomly from a year’s production
– fields actually selected randomly from Iowa farms
23

STAT 512 Balanced Incomplete Block Designs
• y ij
= µ
β
+ τ j
+ ( β i
+ ij
) = µ
β
+ τ j
+
– E [
∗ ij
] = 0
– V ar [
∗ ij
] = σ 2 + δ 2

δ
2
, i = i
0
, j = j
0
– Cov [
∗ ij
,
∗ i 0 j 0
] =

0 , i = i
0
, j = j
0
∗ ij
• For block-ordered y :
– E [ y ] = µ
β
1 + X
2
τ
– V ar [ y ] = σ
2
I + δ
2 diag ( J k
, J k
, ..., J k
) = Σ
– Full, general treatment would involve generalized least squares with unknown Σ
24

STAT 512 Balanced Incomplete Block Designs
Neat Trick for BIBD
• Recall the INTRA-block estimator: c 0 τ = k/ ( λt ) P j c j
[ y
.j
− k
− 1 P i
N ( i, j ) y i.
]
• For random blocks, we can summarize INTER-block information using block totals:
– Let { z } i
= y i.
, i.e. the vector of block totals
– z
– b × 1
= kµ
β
Elements of
1 b × 1
+ U b × t
τ t × 1
+
∗ b × 1
∗ are i.i.d. w/ variance k
2
δ
2
+ kσ
2
– U = X
0
1
X
2
!
25

STAT 512 Balanced Incomplete Block Designs
– U = X
0
1
X
2
(again)
– U has:
∗ k 1’s in each row
∗ r 1’s in each column
∗ inner product of any two columns = λ
– One implication is that the sum of all columns in U is k 1
∗ ( 1 | U ) is of rank t (rather than t + 1 )
∗ only linear contrasts in τ ’s are estimable ...
– RNE’s: U
0
( I − 1 b
J ) U ˆ = U
0
( I − 1 b
J ) z
– 0 τ = k/ ( r − λ ) P j c j
[ k
− 1 P i
N ( i, j ) y i.
− r y ¯
..
]
– V ar [ 0 τ ] = [( k
2
δ
2
+ kσ
2
) / ( r − λ )] c
0 c
26

STAT 512 Balanced Incomplete Block Designs
Bottom Line:
• P d j c j
τ j and P j c j
τ j are independent statistics
• The weighted l.s. estimator (BLUE, given Σ ) is a weighted average of the two, with weights inversely proportional to their respective variances.
P d d j c j
τ j
= w
1 w
1
+ w
2
P d j c j
τ j
+ w
2 w
1
+ w
2
P g j c j
τ j
• w
1
= [ k
2
δ
2
+ kσ
2
] / ( r − λ ) (generally larger)
• w
2
= [ kσ 2 ] / ( tλ )

σ
2
• E [MSE] =
 k
2
δ
2
+ kσ
2
INTRA: fit of individual data
INTER: fit of block-total data
• Since, we ordinarily don’t know the weights, substitute:
– ˆ
1
= MSE
IN T ER
/ ( r − λ )
– ˆ
2
= k MSE
IN T RA
/ ( tλ )
27

STAT 512 Balanced Incomplete Block Designs
• If we knew σ and δ , we could write:
V ar [ τ ] = ( w
1 )
2
V ar [ 0 τ ] + (
= c
0 w
1
+ w
2 h c ( w
1 w
1
+ w
2
) 2 w
2
+ ( w
1
+ w
2 w
2 w
2 w
1
+ w
2
) 2
)
2 w
V ar [ 0 τ ]
1 i
= c
0 c w
1 w
2 w
1
+ w
2
• Again, can substitute estimates for σ
2 and k
2
δ
2
+ kσ
2
, although this does not necessarily lead to an estimator that is better than c 0 τ
28