1
Chapter 4 Randomized Blocks, Latin
Squares, and Related Designs
2
4.1 The Randomized Complete
Block Design
•Nuisance factor: a design factor that probably has
an effect on the response, but we are not interested
in that factor.
•If the nuisance variable is knownand
controllable, we use blocking
•If the nuisance factor is knownand
uncontrollable, sometimes we can use the
analysis of covariance(see Chapter 14) to
remove the effect of the nuisance factor from the
analysis
3
•If the nuisance factor is unknownand
uncontrollable(a “lurking” variable), we hope
that randomizationbalances out its impact across
the experiment
•Sometimes several sources of variability are
combinedin a block, so the block becomes an
aggregate variable
4
•We wish to determine whether 4 different tips
produce different (mean) hardness reading on a
Rockwell hardness tester
•Assignment of the tips to an experimental unit;
that is, a test coupon
•Structure of a completely randomized experiment
•The test coupons are a source of nuisance
variability
•Alternatively, the experimenter may want to test
the tips across coupons of various hardness levels
•The need for blocking
5
•To conduct this experiment as a RCBD, assign all
4 tips to each coupon
•Each coupon is called a “block”; that is, it’s a
more homogenous experimental unit on which to
test the tips
•Variability betweenblocks can be large,
variability withina block should be relatively
small
•In general, a blockis a specific level of the
nuisance factor
•A complete replicate of the basic experiment is
conducted in each block
•A block represents a restriction on
randomization
•All runs withina block are randomized
6
•Suppose that we use b= 4 blocks:
•Once again, we are interested in testing the
equality of treatment means, but now we have to
remove the variability associated with the
nuisance factor (the blocks)
7
Statistical Analysis of the RCBD
•Suppose that there are a treatments (factor levels)
and bblocks
•A statistical model(effects model) for the RCBD
is
–is an overall mean,
i
is the effect of theith
treatment, and
j
is the effect of the jth block
–
ij
~ NID(0,
2
)
–1,2,...,
1,2,...,
ij i j ij
ia
y
jb
0,0
11
k
j
j
a
i
i
8
•Means model for the RCBD
•The relevant (fixed effects) hypotheses are
•An equivalent way for the above hypothesis
•Notations:jiijijijijy , 0 1 2
1
: where (1/ ) ( )
b
a i i j i
j
Hb
0:
210
aH Nyyayybyy
yyyy
bjyy
aiyy
jjii
a
i
i
b
j
j
a
i
b
j
ij
a
i
ijj
b
j
iji
/,/,/
,...,1,
,...,1,
1111
1
1
9
•ANOVA partitioning of total variability:2
.. . .. . ..
1 1 1 1
2
. . ..
22
. .. . ..
11
2
. . ..
11
( ) [( ) ( )
( )]
( ) ( )
()
a b a b
ij i j
i j i j
ij i j
ab
ij
ij
ab
ij i j
ij
T Treatments Blocks E
y y y y y y
y y y y
b y y a y y
y y y y
SS SS SS SS
10
SS
T
= SS
Treatment
+ SS
Blocks
+ SS
E
•Total N = abobservations, SS
T
has N –1degrees
of freedom.
•atreatments and bblocks, SS
Treatment
andSS
Blocks
have a –1 andb –1degrees of freedom.
•SS
E
has ab –1 –(a –1) –(b –1) = (a –1)(b –1)
degrees of freedom.
•From Theorem 3.1, SS
Treatment
/
2
, SS
Blocks
/
2
and
SS
E
/
2
are independently chi-square
distributions.
11
•The expected values of mean squares:
•For testing the equality of treatment means,2
1
2
2
1
2
2
)(
1
)(
1
)(
E
b
j
j
Blocks
a
i
i
Treatment
MSE
b
a
MSE
a
b
MSE )1)(1(,10
~
baa
E
Treatments
F
MS
MS
F
12
•The ANOVA table
•Another computing formulas:BlocksTreatmentsTE
b
j
jBlocks
a
i
iTreatments
a
i
b
j
ijT
SSSSSSSS
N
y
y
a
SS
N
y
y
b
SS
N
y
ySS
,
1
1
,
2
1
2
2
1
2
2
11
2
13
•Example 4.1
4.1.2 Model Adequacy Checking
•Residual Analysis
•Residual:
•Basic residual plots indicate that normality,
constant varianceassumptions are satisfied
•No obvious problems with randomization yyyyyye
jiijijijij
ˆ
14DESIGN-EXPERT Plot
Hardness
Residual
Norm al % probability
Normal plot of residuals
-1 -0.375 0.25 0.875 1.5
1
5
10
20
30
50
70
80
90
95
99
15DESIGN-EXPERT Plot
Hardness
2
2
2
2
Predicted
Residuals
Residuals vs. Predicted
-1
-0.375
0.25
0.875
1.5
-2.75 -0.31 2.13 4.56 7.00 DESIGN-EXPERT Plot
Hardness
Run Num ber
Residuals
Residuals vs. Run
-1
-0.375
0.25
0.875
1.5
1 4 7 10 13 16
16
•Can also plot residuals versus the type of tip
(residuals by factor) and versus the blocks. Also
plot residuals v.s. the fitted values. Figure 4.5 and
4.6 in Page 137
•These plots provide more information about the
constant variance assumption, possible outliers
4.1.3 Some Other Aspects of the Randomized
Complete Block Design
•The model for RCBD is complete additive. 1,2,...,
1,2,...,
ij i j ij
ia
y
jb
17
•Interactions?
•For example:
•The treatments and blocks are random.
•Choice of sample size:
–Number of blocks , the number of replicates
and the number of error degrees of freedom
–jiijjiij yEyE lnlnln)(ln)( 2
1
2
2
a
b
a
i
i
18
•Estimating miss values:
–Approximate analysis: estimate the missing
values and then do ANOVA.
–Assume the missing value isx. Minimize SS
E
to
find x
–Table 4.8
–Exact analysis )1)(1(
'''
ba
ybyay
x
ji
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4.1.4 Estimating Model Parameters and the General
Regression Significance Test
•The linear statistical model
•The normal equations1,2,...,
1,2,...,
ij i j ij
ia
y
jb
bba
a
aba
b
ba
yaa
yaa
ybb
ybb
yaabbab
ˆˆˆˆ
ˆˆˆ
ˆˆˆˆ
ˆˆˆˆ
ˆˆˆˆˆ
1
111
1
111
11
20
•Under the constraints,
the solution is
and the fitted values,
•The sum of squares for fitting the full model:
•The error sum of squares 0
ˆ
,0ˆ
11
b
j
j
a
i
i
yyyyy
jjii
ˆ
,ˆ,ˆ yyyy
jijiij
ˆˆˆˆ ab
y
a
y
b
y
yyyR
b
j
j
a
i
i
b
j
jj
a
i
ii
2
1
2
1
2
11
ˆˆˆ),,(
2
1111
2
),,(
a
i
b
j
jiij
a
i
b
j
ijE yyyyRySS
21
•The sum of squares due to treatments:ab
y
b
y
R
a
i
i
2
1
2
),|(
22
4.2 The Latin Square Design
•RCBD removes a known and controllable
nuisance variable.
•Example: the effects of five different formulations
of a rocket propellant used in aircrew escape
systems on the observed burning rate.
–Remove two nuisance factors: batches of raw
material and operators
•Latin square design: rows and columns are
orthogonal to treatments.
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•The Latin square design is used to eliminate two
nuisance sources, and allows blocking in two
directions (rows and columns)
•Usually Latin Square is a p psquares, and each
cell contains one of the pletters that corresponds
to the treatments, and each letter occurs once and
only once in each row and column.
•See Page 145
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•The statistical (effects) model is
–y
ijk
is the observation in the ith row and kth
column for the jth treatment, is the overall
mean,
i
is the ith row effect,
j
is the jth
treatment effect,
k
is the kth column effect and
ijk
is the random error.
–This model is completely additive.
–Only two of three subscripts are needed to
denote a particular observation.1,2,...,
1,2,...,
1,2,...,
ijk i j k ijk
ip
y j p
kp
25
•Sum of squares:
SS
T
= SS
Rows
+ SS
Columns
+ SS
Treatments
+ SS
E
•The degrees of freedom:
p
2
–1 = p –1 + p –1 + p –1 + (p –2)(p –1)
•The appropriate statistic for testing for no
differences in treatment means is
•ANOVA table (Table 4-10) (Page 146)
•Example 4.3)1)(2(,10
~
ppp
E
Treatments
F
MS
MS
F
26
•The residuals
•Table 4.13
•If one observation is missing,
•Replication of Latin Squares:
–Three different cases
–See Table 4.14, 4.15 and 4.16
•Crossover design: Pages 150 and 151 yyyyyyye
kjiijkijkijkijk 2ˆ )1)(2(
2)(
''''
pp
yyyyp
y
kji
ijk
27
4.3 The Graeco-Latin Square Design
•Graeco-Latin square:
–Two Latin Squares
–One is Greek letter and the other is Latin letter.
–Two Latin Squares are orthogonal
–Table 4.18
–Block in three directions
–Four factors (row, column, Latin letter and
Greek letter)
–Each factor has plevels. Total p
2
runs
28
•The statistical model:
–y
ijkl
is the observation in the ith row and lth
column for Latin letter j, and Greek letter k
–is the overall mean,
i
is the ith row effect,
j
is the effect of Latin letter treatment j,
k
is the
effect of Greek letter treatment k,
l
is the
effect of column l.
–ANOVA table (Table 4.19)
–Under H
0
, the testing statistic is F
p-1,(p-3)(p-1)
distribution.
•Example 4.4plkjiy
ijkllkjiijkl ,,1,,, ,
29
4.4 Balance Incomplete Block
Designs
•May not run all the treatment combinations in
each block.
•Randomized incomplete block design (BIBD)
•Any two treatments appear together an equal
number of times.
•There are atreatments and each block can hold
exactly k (k < a)treatments.
•For example: A chemical process is a function of
the type of catalyst employed. See Table 4.22
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4.4.1 Statistical Analysis of the BIBD
•atreatments and bblocks. Each block contains k
treatments, and each treatment occurs rtimes.
There are N = ar = bktotal observations. The
number of times each pairs of treatments appears
in the same block is
•The statistical model for the BIBD is)1(
)1(
a
kr
ijjiijy
31
•The sum of squaresBlocksadjustedTreatmentsTE
b
j
jijii
a
i
i
adjustedTreatments
b
j
jBlocks
ij
ijT
EBlocksTreatmentsT
SSSSSSSS
yn
k
yQ
a
Qk
SS
Nyy
k
SS
NyySS
SSSSSSSS
)(
1
1
2
)(
2
1
2
22
1
,
/
1
/
32
•The degree of freedom:
–Treatments(adjusted): a –1
–Error: N –a –b –1
•The testing statistic for testing equality of the
treatment effects:
•ANOVA table (see Table 4.23)
•Example 4.5E
adjustedTreatments
MS
MS
F
)(
0
33
4.4.2 Least Squares Estimation of the Parameters
•The least squares normal equations:
•Under the constrains,
we have jj
a
i
iijj
i
b
j
jijii
b
j
j
a
i
i
yknk
ynrr
ykrN
ˆˆˆ:
ˆˆˆ:
ˆˆˆ:
1
1
11 0
ˆ
,0ˆ
11
b
j
j
a
i
i
yˆ
34
•For the treatment effects,ai
a
kQ
kQkr
i
i
i
a
ipp
pi
,,2,1,ˆ
ˆˆ)1(
,1