IV. Randomized Complete Block
Design (RCBD)
IV.A Design of an RCBD
IV.B Indicator-variable models and estimation
for an RCBD
IV.C Hypothesis testing using the ANOVA
method for an RCBD
IV.D Diagnostic checking
IV.E Treatment differences
IV.F Fixed versus random effects
IV.G Generalized randomized complete block
design
Statistical Modelling Chapter IV
1
IV.A Design of an RCBD
Definition II.6: A randomized complete block
design is one in which the number of
experimental units per block is equal to the
number of treatments and every treatment
occurs once and only once in each block, the
order of treatments within a block being
randomized.
–
–
–
•
b denotes no. of blocks
t denotes both no. of units in each block and no. of
treatments.
n = bt denotes total no. of observations.
In RCBD group units into blocks such that the
units in a block are as similar as possible.
Statistical Modelling Chapter IV
2
Forming blocks in field experiments
• Place plots parallel to
the trend and blocks
perpendicular to it.
L e ss ston y e nd o f
fie ld
I
II
• Suppose trend not as I
thought — went across
the field.
L e ss ston y s id e
o f fie ld
...
...
B lo ck
III
...
IV
...






I
. . .
II
. . .
III
. . .
IV
. . .
S to n y s id e of
fie ld
B lo ck
s to n ie r en d o f
fie ld
• Clearly, Blocks would be similar and plots different.
• In fact this experiment can be less sensitive than a CRD
— getting it wrong can be costly.
Statistical Modelling Chapter IV
3
a) Obtaining a layout for an RCBD
in R
• General set of expressions for obtaining
RCBD layout is given in Appendix B,
Randomized layouts and sample size
computations in R.
• To generate a layout for particular case,
need to substitute
– actual values for b, t and n
– actual names for Blocks, Units, Treats and the
data frame to contain them.
Statistical Modelling Chapter IV
4
Example IV.1 Penicillin yield
• In this example the effects of four treatments (A, B, C and D) on the
yield of penicillin are to be investigated.
• Corn steep liquor, an important raw material in producing penicillin, is
highly variable from one blending to another.
• To ensure that the results of the experiment apply to > 1 blend,
several blends to be used in experiment.
• The trial was conducted using the same blend in 4 flasks and
randomizing treatments to these 4.
• Altogether five blends were utilized.
• Crucial feature, making RCBD different from CRD, is that there are
– 2 unrandomized factors indexing the units: Blends, Flasks
– there is nesting between these factors: Flasks are nested within Blends
because randomize treatments to Flasks within Blends.
• Names to be used for the blocks, units and treatments for this
example are Blends, Flask and Treat, respectively.
• Also, b = 5 and t = 4 so that n = 20.
• Assigning these values and substituting these names into the general
expressions, yields the following output for this case.
Statistical Modelling Chapter IV
5
R
>
>
>
>
>
•
Flask is a nested
factor;
b <- 5
t <- 4
n <- b*t
• Nested within
RCBDPen.unit <- list(Blend=b, Flask=t) Blend
RCBDPen.nest <- list(Flask = "Blend")
> Treat <- factor(rep(1:t, times=b), labels=c("A","B","C","D"))
> data.frame(fac.gen(RCBDPen.unit), Treat) #basic systematic arrangement
Blend Flask Treat
Blend Flask Treat
Systematic
11
3
3
C
1
1
1
A
12
3
4
D
2
1
2
B
arrangement on
13
4
1
A
3
1
3
C
which
14
4
2
B
4
1
4
D
randomization
15
4
3
C
5
2
1
A
16
4
4
D
6
2
2
B
based
17
5
1
A
7
2
3
C
18
5
2
B
8
2
4
D
Blend & Flask order
19
5
3
C
9
3
1
A
determined by order
20
5
4
D
10
3
2
B
in RCBDPen.unit
> RCBDPen.lay <- fac.layout(unrandomized = RCBDPen.unit,
+
nested.factors = RCBDPen.nest,
+
randomized = Treat, seed = 311)
Statistical Modelling Chapter IV
6
Layout
> RCBDPen.lay
Units Permutation Blend Flask Treat
1
1
11
1
1
C
2
2
12
1
2
B
3
3
10
1
3
D
4
4
9
1
4
A
5
5
13
2
1
C
6
6
15
2
2
D
7
7
16
2
3
B
8
8
14
2
4
A
9
9
8
3
1
D
10
10
7
3
2
C
11
11
5
3
3
A
12
12
6
3
4
B
13
13
17
4
1
A
14
14
19
4
2
D
15
15
20
4
3
B
16
16
18
4
4
C
17
17
4
5
1
A
18
18
2
5
2
D
19
19
1
5
3
B
20
20
3
5
4
C
This layout is said to be
in standard order for
Blend then Flask:
In general the first factor
changes slowest and the
last fastest.
• So with the first blend, the Treatments are to be done in the order C,
B, D, A.
Statistical Modelling Chapter IV
7
IV.B Indicator-variable models
and estimation for an RCBD
a)Maximal model
• The maximal model used for an RCBD is:
 B +T = E  Y  = X B   X T  a n d v a r  Y  =  I n
2
where
Y is the n-vector of random variables for the response
variable observations,
 is the b-vector of parameters specifying a different mean
response for each block,
XB is the nb matrix indicating the block from which an
observation came,
 is the t-vector of parameters specifying a different mean
response for each treatment,
XT is the nt matrix indicating the observations that received
each of the treatments.
Statistical Modelling Chapter IV
8
Example IV.1 Penicillin yield (continued)
• The yields of penicillin, in nonrandom order
T re a tm e n t
A
89
84
81
87
79
1
2
3
4
5
B le n d
B
88
77
87
92
81
C
97
92
87
89
80
D
94
79
85
84
88
80
80
85
85
Yield
Yield
90
90
95
95
• initial exploration of the data — differences?
1
Statistical Modelling Chapter IV
2
3
Blend
4
5
A
B
C
Treatment
D
9
Yields in a vector in standard order
for Blend then Treatment
• Same order as systematic layout i.e. prerandomization layout
89 
88 
97 
94 
84 
77 
92 
79 
 8 1
87 
y =  ,
87
85 
 
87
 
92
 
89
 
84 
79 
 8 1
80 
 8 8 
XB
Statistical Modelling Chapter IV
1
1
1
1
0
0
0
0
0
0
= 
0
0

0

0

0

0
0
0
0
 0
0
0
0
0
1
1
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
1
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
1
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
,
0
0

0

0

0

0
1
1
1
1 
 1 
2
 = 3 ,
 
4
 
 5
XT
1
0
0
0
1
0
0
0
1
0
= 
0
0

1

0

0

0
1
0
0
 0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
0
1
0
0
0
1
0
0
,
0
1

0

0

0

1
0
0
0
1 
 1 
 
 = 2
 3 
 
 4
10
Estimator of expected values
• Our model also assumes Y ~ N(B+T, V)
• The model for the expectation is still of the form
E[Y] = Xq with X = [XB XT] and q = [ ].
• It can be shown that ˆ B +T = B  T  G
where B , T a n d G are the n-vectors of block,
treatment and grand means, respectively.
N ote that B = M B Y , T = M T Y and G = M G Y
where MB, MT and MG are the block, treatment
and grand mean operators, respectively.
• So once again the estimator of the expected
values are functions of means.
Statistical Modelling Chapter IV
11
Mean operators
• Suppose data arranged in the vector Y in
nonrandomized order with all the observations for
a block placed together.
– Standard order for blocks then treatments.
• Then the mean operators are:
MG = n
MB = t
1
Jb  Jt = n
1
Jn
1
MT = b
Ib  J t
1
J b  It
where  is called the direct product operator and,
• if Ar and Bc are square matrices of order r and c
A r  Bc
 a1 1B
= 
 a r 1B
a1r B 

a rr B 
Mean operators simpler than for CRD — divisors
factored out leaving matrices with 0s & 1s.
Statistical Modelling Chapter IV
12
Grand
mean
operator
for
standard
order
MG = 20
Statistical Modelling Chapter IV
1
J5  J 4
J4
J4
1
J4
=
20 J
4
J
 4
1
1
1
1
1
1
1
1
1
1 1
=

2 0 1
1

1

1

1

1
1
1
1
1
J4
J4
J4
J4
J4
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
J4
J4
J4
J4
J4
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
J4
J4
J4
J4
J4
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
J4 
J4 
J4 
J4 
J 4 
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1

1

1

1

1
1
1
1
1
13
1
M B = 4 I5  J 4
Block
 J
mean = 1  00
4 0
operator  0
1
for
1
1
standar  1
0
d order  00
4
4 4
4 4
4 4
4 4
0
0
1 0
= 
4 0
0

0

0

0

0
0
0
0
 0
Statistical Modelling Chapter IV
1
1
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0 4 4
J4
0 4 4
0 4 4
0 4 4
1
1
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0 4 4
0 4 4
J4
0 4 4
0 4 4
1
1
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
1
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0 4 4
0 4 4
0 4 4
J4
0 44
0
0
0
0
1
1
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
1
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0 44 
0 4 4 
0 4 4 
0 4 4 
J 4 
0
0
0
0
1
1
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
1
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
1
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
1
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
1
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
1
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
1
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
1
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
1
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
1
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
1
1
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
1
1
1
0
0
0
0
0
0
0
0
0
0
0
0

0

0

0

0
1
1
1
1 
14
Treatment
mean
operator
for
standard
order
MT = 5
1
J 5  I4
 I4
I
1 4
=  I4
5 I
4
I
4
I4
I4
I4
I4
I4
1
0
0
0
1
0
0
0
1
1 0
= 
5 0
0

1

0

0

0
1
0
0
 0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
Statistical Modelling Chapter IV
I4
I4
I4
I4
I4
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
0
I4 
I4 
I4 
I4 
I 4 
I4
I4
I4
I4
I4
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
0
1
0
0
0
1
0
0
0
1

0

0

0

1
0
0
0
1 
15
Estimators for example
Statistical Modelling Chapter IV
 B1 
 B1 
B 
1
B 
 1
B
 2
B2 
B2 
B2 
B3 
B 
B =  3 ,
B
 3
B
 3
B4 
B4 
B4 
B4 
B5 
B 
5
B 
 5
B5 
T A 
G 
 TB 
G 
T 
G 
C
T 
G 
 D
G 
T
 A
 
G
 TB 


G
T C 
 
TD 
G 
T A 
G 
T 
G
T =  B  and G =  
T
G 
 C
G 
T
 D
G 
T
 A
G 
 TB 
G 
T C 
G 
TD 
 
G
T A 
 
G
T 
 
B
T 
G 
C
 
G 
TD 
16
Estimates for the example
• The means are in the following table:
1
2
Blend
3
4
5
Means
A
89
84
81
87
79
84
Treatment
B
C
88
97
77
92
87
87
92
89
81
80
85
89
Statistical Modelling Chapter IV
D
94
79
85
84
88
86
Means
92
83
85
88
82
86
 92 
 92 
 92 
 92 
 83 
 83 
 83 
 83 
 85 
 85 
b = 
,
85 
 85 
 88 
888 
 88 
 88 
 82 
 82 
 82 
 82 


84 
85 
89 
86 
84 
85 
89 
86 
84 
85 
t =  ,
89
86 
84 
85 
89 
86 
84 
85 
89 
86 
 
8 6 
8 6 
8 6 
8 6 
8 6 
8 6 
8 6 
8 6 
8 6 
8 6 
g =  
86
8 6 
8 6 
86 
86 
86 
86 
86 
86 
86 
 
17
Estimates for the example
 92 
 92 
 92 
 92 
 83 
 83 
 83 
 83 
 85 
 85 
b = 
,
85 
 85 
 88 
888 
 88 
 88 
 82 
 82 
 82 
 82 


84 
85 
89 
86 
84 
85 
89 
86 
84 
85 
t =  ,
89
86 
84 
85 
89 
86 
84 
85 
89 
86 
 
8 6 
90 
8 6 
 9 1
8 6 
95 
8 6 
92 
8 6 
 8 1
8 6 
82 
8 6 
86 
8 6 
83 
8 6 
83 
8 6 
84 
g =   a n d ψ B+T = b  t  g =  
86
88
8 6 
85 
8 6 
86 
86 
87 
86 
 9 1
86 
88 
86 
80 
86 
 8 1
86 
85 
86 
82 
 
 
• These fitted value are different for each block-treatment
combination but display an additive pattern.
Statistical Modelling Chapter IV
18
Additivity
• The fitted value are those for a model that is additive in
Block and Treatment parameters:
–
B+T = E[Y] = XB  XT.
• So its fitted values display an additive pattern:
ˆ Β +Τ = B  T  G
• Hope an adequate
description of the data.
• In one direction, same trend
as means:
Statistical Modelling Chapter IV
19
b) Alternative expectation models
• There are 4 possible different models for the expectation
that we consider:
ψG = XG
no
ψ Β = X B
 b lo ck
ψΤ = X T
 tre a tm e n t
ψ Β+Τ = X B  X T 
• Note that:
tre a tm e n t o r b lo ck d iffe re n ce s 
 b lo ck
d iffe re n ce s o n ly 
d iffe re n ce s o n ly 
a n d tre a tm e n t d iffe re n ce s 
C  X G   C  X B   C  X B
C  X G   C  X T   C  X B
X T 
X T 
• Consequently: ψ G  ψ Β , ψ Τ , ψ Β + Τ a n d ψ Β , ψ Τ  ψ Β + Τ
• Also note that, like the CRD,
 the models B and T can be obtained from B+T by
setting either  or  equal to zero and
 G can be obtained from B and T by setting  = 1
and  = 1, respectively.
Statistical Modelling Chapter IV
20
Estimators of expected values
• Estimators of the expected values under the different
models:
ψˆ G = G
no
ψˆ Β = B
 b lo c k
ψˆ Τ = T
 tre a tm e n t
ψˆ Β + Τ = B  T  G
Statistical Modelling Chapter IV
tre a tm e n t o r b lo c k d iffe re n c e s 
 b lo c k
d iffe re n c e s o n ly 
d iffe re n c e s o n ly 
a n d tre a tm e n t d iffe re n c e s 
21
IV.C Hypothesis testing using the
ANOVA method for an RCBD
• An ANOVA will be used to choose between the 4
alternative expectation models for an RCBD.
a) Analysis of the penicillin example
Example IV.1 Penicillin yield (continued)
• The hypothesis test for the example RCBD is as follows:
Step 1:
Set up hypotheses
a) H0: 1 = 2 = 3 = 4 (or XT not required in model)
H1: not all population Treatment means are equal
b) H0: 1 = 2 = 3 = 4 = 5
(or XB not required in model)
H1: not all population Blend means are equal
Set a = 0.05.
Statistical Modelling Chapter IV
22
Hypothesis test
Step 2:
Calculate test statistics
• The analysis of variance table for a RCBD is:
S o u rce
B le n d s
F la sks [B le n d s]
T re a tm e n ts
R e sid ua l
T o tal
df
4
SSq
264
MSq
6 6 .0
F
3 .5 0
P ro b
0 .0 4 1
15
3
12
19
296
70
226
560
2 3 .3
1 8 .8
1 .2 4
0 .3 3 9
• Note that Flasks[Blends] in this table means "Flasks
within Blends".
Step 3:
Decide between hypotheses
• It would appear that there are significant differences
between the blends but not between the treatments so
that the expectation model that best describes the
response appears to be B = XB.
Statistical Modelling Chapter IV
23
Blocking effectiveness
• In our RCBD example there were significant
differences between the blends so that the
blocking based on blends has been effective.
• Turns out that, if the units within a block are as
similar as possible, there will be block
differences.
• If a CRD had been used,
– that is 4 treatments randomized to 20 flasks
irrespective of blends, then
– Residual SSq  Blend SSq + RCBD Residual SSq
– viz. 264 + 226 = 490 and the mean square 490/16 =
30.625.
– That is, residual MSq would have been twice (30.6 vs
18.8) as large and the experiment much less sensitive.
Statistical Modelling Chapter IV
24
b) Sums of squares for the analysis
of variance
• In this section we will use the generic names of
Blocks, Units and Treatments for the factors in an
RCBD.
• The estimators of the SSqs for the RCBD
ANOVA are the SSqs of the following vectors:
T o ta l o r U n its S S q :
DG = Y  G
B lo ck s S S q :
Be = B  G
U n its[B lo ck s] S S q :
DB = Y  B
T re a tm e n ts S S q :
Te = T  G
R e sid u a l S S q :
D B+T = Y  B  T  G
Statistical Modelling Chapter IV
= Y  B e  Te  G
25
SSq (continued)
• From section IV.B, Models and estimation for an
RCBD, we have that
G = MGY = n
B = MBY = t
1
1
T = MTY = b
 Jb
 Ib
1
 Jt  Y
 Jt  Y
 Jb
 It  Y
a n d le t Y = M U Y =  I b  It  Y .
Statistical Modelling Chapter IV
26
SSq (continued)
• It can be shown that the SSqs for the ANOVA are
given by
D G D G =  Y  G   Y  G  = Y Q U Y
w ith Q U = M U  M G
B e B e =  B  G   B  G  = Y Q B Y
w ith Q B = M B  M G
D B D B =  Y  B   Y  B  = Y Q B U Y
w ith Q B U = M U  M B
Te Te =  T  G   T  G  = Y Q T Y
w ith Q T = M T  M G
D B + T D B + T =  Y  B  T  G   Y  B  T  G  = Y Q B U
w ith Q B U
Res
Res
Y
= MU  M T  MB  MG
• All the Ms and Qs are symmetric and idempotent.
Statistical Modelling Chapter IV
27
ANOVA table is constructed as
follows:
S o u rc e
B lo cks
U n its [B loc ks]
T re a tm en ts
R e s id ua l
df
b1
SSq
Y Q B Y
b (t 1 )
Y Q B U Y
t1
Y Q T Y
(b  1)(t 1)
Y Q B U
Res
Y
M Sq
Y Q B Y
2
= sB
b 1
sB sB U
Y Q T Y
s T sB U
t 1
Y Q B U Y
Res
 b  1  t
T o ta l
b t 1
Statistical Modelling Chapter IV
=
 1
F
2
2
2
sT
p
pB
2
Res
pT
2
Res
2
= sB U
Res
Y Q U Y
28
Geometrical
interpretation
• The matrix Q orthogonally projects the
U
•
data vector into the bt-1 dimensional
part of the bt-dimensional data space
that is orthogonal to equiangular line.
This is partitioned, by QB and QBU, into
two subspaces:
a) the b1 dimensional part of the bdimensional Block space that is
orthogonal to equiangular line and
b) b(t1) dimensional Units[Blocks] space.
•
The latter space is then partitioned, by
QT and Q B U , into two subspaces:
R es
•
a) the t1 dimensional part of the tdimensional Treatment space that is
orthogonal to equiangular line and
b) the (b1)(t1) Residual subspace.
S o u rce
B lo cks
b1
SSq
Y Q B Y
U n its[B locks]
b (t 1 )
Y Q B U Y
T re a tm e n ts
t 1
Y Q T Y
R e sid u a l
(b  1 )(t 1 )
T o ta l
df
b t 1
Y Q B U
Res
Y
Y Q U Y
That is, the Units space is divided into
the three orthogonal subspaces:
– the Blocks subspace,
– Treatments subspace,
– Residual subspace.
•
Here Block and Treatment spaces are
column spaces of the matrices XB and
XT, respectively.
Statistical Modelling Chapter IV
29
Example IV.1 Penicillin yield (continued)
• The effects needed for the analysis have
been added to the means in the following
table:
1
2
B le n d
3
4
5
Means
E ffe c ts
Statistical Modelling Chapter IV
A
89
84
81
87
79
84
-2
T re a tm e n t
B
C
88
97
77
92
87
87
92
89
81
80
85
89
-1
3
D
94
79
85
84
88
86
0
M e a n s E ffe c ts
92
6
83
-3
85
-1
88
2
82
-4
86
0
30
Vectors
for SSQ
Total
Flask
Blend Flask[Blend] Treat
Yield deviations Effects deviations effects
dG = Q Uy
be = Q By
dB = Q BF y
te = Q Ty
y
= yg
Treat
A
B
C
D
A
B
C
D
A
B
Units SSq is YQUY = 560, C
Blend SSq is YQBY = 264, D
A
B
Flask[Blend] SSq is
C
YQBFY = 296,
D
A
Treatments SSq is
B
C
YQTY = 70 and
D
Residual SSq is 226.
SSq
89
88
97
94
84
77
92
79
81
87
87
85
87
92
89
84
79
81
80
88
3
2
11
8
-2
-9
6
-7
-5
1
1
-1
1
6
3
-2
-7
-5
-6
2
560
=bg
6
6
6
6
-3
-3
-3
-3
-1
-1
-1
-1
2
2
2
2
-4
-4
-4
-4
264
= yb
= t g
-3
-4
5
2
1
-6
9
-4
-4
2
2
0
-1
4
1
-4
-3
-1
-2
6
296
-2
-1
3
0
-2
-1
3
0
-2
-1
3
0
-2
-1
3
0
-2
-1
3
0
70
Residual
Flask[Blend]
deviations
d B +T = Q B F
Res
y
= y t bg
-1
-3
2
2
3
-5
6
-4
-2
3
-1
0
1
5
-2
-4
-1
0
-5
6
226
N o te o rth o g o n a l d e co m p o sitio n o f y = g  b e  t e  d B +T
Statistical Modelling Chapter IV
31
c) Expected mean squares
• To justify choice of test statistic, want to work out the
E[MSq]s under the 4 alternative expectation models.
• E[MSq]s under maximal model
Source
df
MSq
E[MSq]
ψ B+T
Blocks
b1
Y Q B Y
b 1
Units[Blocks]
Treatments
Residual
b(t1)
t1
(b1)(t1)
Y Q T Y
t 1
Y Q B U Y
Res
 2  qB  ψ 
 2  qT ψ 

2
 b  1   t  1
Total
bt1
• Residual MSq estimates the uncontrolled variation,
– that is the variation arising from uncontrolled differences between
units within the same block, both treatment and block differences
having been eliminated.
Statistical Modelling Chapter IV
32
E[MSq]s under the 4 alternative
expectation models
Source
df
MSq
E[MSq]
ψ B+T
Blocks
Y Q B Y
b1
b 1
Units[Blocks] b(t1)
Treatments t1
Residual
Y Q T Y
(b1)(t1)
t 1
Y Q B U Y
Res
ψT
 2  qB  ψ 

 2  qT ψ 

2
ψB
ψG
 2  qB  ψ 

2
 2  qT ψ 

2

2


2

2
2
2
 b  1   t  1
Total
qB  ψ  =
bt1
ψ Q Bψ
b 1
b
=
 t   i   .   b  1 and
2
qT ψ  =
i =1
ψ Q Tψ
t 1
t
=
 b  j   . 
2
 t  1
j =1
• Once again numerator of:
– qB(Y) is SSq of QBY = (MB – MG)Y
– qT(Y) is SSq of QTY = (MT – MG)Y
– where Y depends on model.
• Expressions qB(Y) and qT(Y) above are under maximal models
• To obtain those for reduced models set is and js to 0 or to .
• Could compute population means of MSqs if knew is, js and 2.
Statistical Modelling Chapter IV
33
Justifying the F ratios
• Clear from these E[MSq]s that if the Treatments F
is not significant then a model not involving XT is
required
– as those models are the ones for which qT(Y) = 0.
• Similarly, if the Blocks F is not significant then a
model not involving XB is required.
• In the case where both are not significant, then
the minimal model adequately describes the data.
• Generally, will only present the E[MSq]s under the
maximal model, realizing that q(Y) = 0 under the
H0 that removes the term from the model.
Statistical Modelling Chapter IV
34
Potential contributers to block and
treatment mean differences
1
2
B le n d
3
4
5
Means
A
89
84
81
87
79
84
T re a tm e n t
B
C
88
97
77
92
87
87
92
89
81
80
85
89
D
94
79
85
84
88
86
Means
92
83
85
88
82
86
• Two treatment means will differ because of the different treatments
involved and because of the different runs (the units in this example)
involved in the observations from which the means are calculated;
• but block differences will not contribute to treatment mean differences
as all treatments involve the same set of blocks.
• E[MSq]s reflect this fact.
• The Treatment F again involves the question:
– "Is the variance of the treatment means greater than can be expected
from uncontrolled differences between the runs?"
Statistical Modelling Chapter IV
35
d) Summary of the hypothesis test
• See notes
Statistical Modelling Chapter IV
36
e) Comparison with traditional twoway ANOVA
• As for the analysis for the CRD, the above and the
traditional two-way ANOVA tables are essentially the
same —the values of the F-statistics are exactly the
same.
As indented,
see
Treatments
confounded
with
Units[Blocks]
S o u rc e
B lo c k s
U n its [B lo c k s ]
T re a tm e n ts
R e s id u a l
df
b1
b (t 1 )
t 1
(b  1 )(t 1 )
T o ta l
b t 1
S o u rc e in tw o -w a y A N O V A
B e tw e e n B lo c k s
B e tw e e n T re a tm e n ts
E rro r
T o ta l
Residual is inherent variability of Units; Error?
– the two tables have in common 3 sources that are labelled
differently
– but the tables differ in that our table includes the line
Units[Blocks] — this source is partitioned.
Statistical Modelling Chapter IV
37
f) Computation of the ANOVA in R
• The expressions for analyzing a
randomized complete block design are
summarized in Appendix C, Analysis of
designed experiments in R.
Statistical Modelling Chapter IV
38
Example IV.1 Penicillin yield (continued)
• First the data is
entered into a data
frame so that it
contains
– the factors Blend,
Flask and Treat and
– the numeric vector
Yield
Here data is in
nonrandom order.
Statistical Modelling Chapter IV
> RCBDPen.dat
Blend Flask Treat Yield
1
1
1
A
89
2
1
2
B
88
3
1
3
C
97
4
1
4
D
94
5
2
1
A
84
6
2
2
B
77
7
2
3
C
92
8
2
4
D
79
9
3
1
A
81
10
3
2
B
87
11
3
3
C
87
12
3
4
D
85
13
4
1
A
87
14
4
2
B
92
15
4
3
C
89
16
4
4
D
84
17
5
1
A
79
18
5
2
B
81
19
5
3
C
80
20
5
4
D
88
39
Model formula for aov function
• As for CRD, use the aov function, either
with or without the Error as part of the
model.
• In this case the uncontrolled variation is:
– Blend differences
– differences between Flasks within Blends (we
denote Flasks[Blends]).
• R shorthand for this: Blend/Flask
– expands to Blend + Blend:Flask.
Statistical Modelling Chapter IV
40
Output
> RCBDPen.aov <- aov(Yield ~ Blend + Treat +
+
Error(Blend/Flask), RCBDPen.dat)
> summary(RCBDPen.aov)
Blend occurs outside and
Error: Blend
inside the Error function —
Df Sum Sq Mean Sq
necessary to get correct fitted
Blend 4
264
66
values for diagnostic checking.
Error: Blend:Flask
Df Sum Sq Mean Sq F value Pr(>F)
Treat
3 70.000 23.333 1.2389 0.3387
Residuals 12 226.000 18.833
>
>
>
>
#Compute Blend F and p
Blend.F <- 66/18.833
Blend.p <- 1-pf(Blend.F, 4, 12)
data.frame(Blend.F,Blend.p)
Blend.F
Blend.p
1 3.504487 0.0407441
Statistical Modelling Chapter IV
Computation of
Blend F and p.
41
Output
> RCBDPen.NoError.aov <- aov(Yield ~ Blend + Treat,
RCBDPen.dat)
> summary(RCBDPen.NoError.aov)
Df Sum Sq Mean Sq F value Pr(>F)
Blend
4 264.000 66.000 3.5044 0.04075
Treat
3 70.000 23.333 1.2389 0.33866
Residuals
12 226.000 18.833
F and p for
Blend, but
controversial
• ANOVA table from the expression that
– includes Error in model resembles our table — prefer
– without is like the traditional ANOVA table.
Statistical Modelling Chapter IV
42
IV.D Diagnostic checking
•
•
Again, we have assumed Y ~ N(, 2I) where, for the
maximal model, B+T = E[Y] = XB  XT
For this model to be appropriate requires a similar set of
behaviours as for the CRD:
a) response is operating additively (see section IV.B,
Indicator variable models and estimation for an
RCBD) as specified by the maximal model: a
treatment has about the same additive effect on each
unit;
b) variability of the units within a block are the same for
each block;
c) each observation displays the covariance implied by
the model (independence for Blocks fixed and equal
correlation within blocks for Blocks random); and
d) that the response of the units is normally distributed.
Statistical Modelling Chapter IV
43
Diagnostic plots
• Same set of diagnostic plots as for the CRD can be used.
– Residual-versus-fitted-values
– Normal probability plots.
• A particular pattern to look out for in the Residual-versusfitted-values plot for this type of design is evidence of a
curvilinear relationship
– indicates nonadditivity between the blocks and treatments
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
*
_________________________
systematic trend in residuals
Statistical Modelling Chapter IV
46
Nonadditivity
• Such nonadditivity may be transformable by take logs,
square root or reciprocals of the data and analyzing
these.
• Another type of block-treatment interaction would occur
where say a particular blend had a poison in it that
affected only process B.
– Then only the observation corresponding to that particular
combination of blend and treatment would be affected.
– It would be extremely low leading to an extreme residual.
• Possible to test for transformable nonadditivity using
Tukey's one-degree-of-freedom-for-nonadditivity,
• Can be used with any design with an additive expectation
model ( 2 terms), including regression (not CRD).
• Involves detecting whether or not there is a curvilinear
relationship between the residuals and fitted values.
• For this, and subsequent designs, diagnostic checking
should be based on the two plots and this one degree-offreedom.
Statistical Modelling Chapter IV
47
An R function from dae,
tukey.1df
• tukey.1df(aov.obj, data,
error.term="within")
• where
– aov.obj is an aov object or aovlist object
created from a call to aov,
– data is optional and is a data.frame
containing the original response variable and
factors used in the call to aov, and
– error.term is the error.term whose
residuals are to be tested for nonadditivity.
Statistical Modelling Chapter IV
48
Statistical Modelling Chapter IV
4
2
res
0
-2
-4
80
85
90
95
fit
-2
0
2
4
6
Normal Q-Q Plot
-4
#
# Diagnostic checking
#
res <- resid.errors(RCBDPen.aov)
fit <- fitted.errors(RCBDPen.aov)
data.frame(Blend,Flask,Treat,Yield,res,fit)
Blend Flask Treat Yield
res fit
1
1
1
A
89 -1.000000e+00 90
2
1
2
B
88 -3.000000e+00 91
3
1
3
C
97 2.000000e+00 95
4
1
4
D
94 2.000000e+00 92
5
2
1
A
84 3.000000e+00 81
6
2
2
B
77 -5.000000e+00 82
7
2
3
C
92 6.000000e+00 86
8
2
4
D
79 -4.000000e+00 83
9
3
1
A
81 -2.000000e+00 83
10
3
2
B
87 3.000000e+00 84
11
3
3
C
87 -1.000000e+00 88
12
3
4
D
85 -2.392617e-15 85
13
4
1
A
87 1.000000e+00 86
14
4
2
B
92 5.000000e+00 87
15
4
3
C
89 -2.000000e+00 91
16
4
4
D
84 -4.000000e+00 88
17
5
1
A
79 -1.000000e+00 80
18
5
2
B
81 -2.614662e-15 81
19
5
3
C
80 -5.000000e+00 85
20
5
4
D
88 6.000000e+00 82
> plot(fit, res, pch=16)
> qqnorm(res, pch = 16)
> qqline(res)
Sample Quantiles
>
>
>
>
>
>
6
Example IV.1 Penicillin yield (continued)
-2
-1
0
1
2
Theoretical Quantiles
From plots, no serious departures
from the assumptions apparent 49
Example IV.1 Penicillin yield (continued)
> tukey.1df(RCBDPen.aov, RCBDPen.dat,
+ error.term="Blend:Flask")
$Tukey.SS
[1] 2.001082
S o u rc e
df
SSq
M Sq
B le n ds
$Tukey.F
[1] 0.0982679
$Tukey.p
[1] 0.7597822
$Devn.SS
[1] 223.9989
F la sk s[B le n ds]
T re a tm en ts
R e s id ua l
N o n a d d itiv ity
D ev iatio n
T o ta l
E [M S q]
 
4
264
6 6 .0

2
 qB 
15
3
296
70
2 3 .3

2
 qT 
12
226
1 8 .8
1
11
19
2 .0
224
560
2 .0
2 0 .4

 
F
3 .5 0
P ro b
0 .0 41
1 .2 4
0 .3 39
0 .1 0
0 .7 60
2
The hypotheses for the one-degree-of-freedom is:
H0:
Blends and Treatments are additive
H1:
Blends and Treatments are nonadditive
H0 cannot be rejected — no evidence of transformable
nonadditivity.
Statistical Modelling Chapter IV
50
IV.E Treatment differences
• For the purposes of the scientist the effect
of the blocks are not of primary interest
• Rather, attention is likely to be focused on
treatment differences which can be
investigated using the treatment means.
• The discussion of multiple comparisons
and submodels for the analysis of a CRD
applies here also.
Statistical Modelling Chapter IV
51
Example IV.1 Penicillin yield (continued)
• The treatment means
are:
A
84
T re a tm e n t
B
C
85
89
D
86
• As the treatment levels are qualitative a multiple
comparison procedure would be used to examine
the differences.
• However they are not significantly different so that
we shall not apply such a procedure.
Statistical Modelling Chapter IV
52
Example IV.1 Penicillin yield (continued)
• Bar chart illustrates:
Fitted values for Yield
80
Yield (%)
60
40
20
A
B
C
D
Treatment
Statistical Modelling Chapter IV
53
IV.F Fixed versus random effects
a) Another maximal model for the RCBD
• Two alternative maximal models for RCBD:
E  Y  = X B  X T  a n d v a r  Y  =  I
2
E  Y  = X T  a n d V =  In   B  Ib  J t 
2
2
• Difference is that  dropped from 2nd expectation
model and covariance of observations from different
units in the same block is  B2 , rather than being
zero.
Statistical Modelling Chapter IV
54
Variance matrices for RCBD for b=3, t=4
B lo c k s fix e d
2
3
4
1
B lo ck
II
2
3
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
I
U n it
1
2
3
4
1
2
1

2
0
0
0
0
0

2
0
0
0
0

2
0
0
0

2
0
0

2
0

2
2
III
4
1
2
3
4
3
0
0
0
0
0
0

4
0
0
0
0
0
0
0

1
0
0
0
0
0
0
0
0

2
0
0
0
0
0
0
0
0
0

3
0
0
0
0
0
0
0
0
0
0

4
0
0
0
0
0
0
0
0
0
0
0
2
2
2
2
B lo c k s ra n d o m
Statistical Modelling Chapter IV
3
4
1
 B2
 B2
 B2
0
0
0
0
0
0
0
0


2
B
0
0
0
0
0
0
0
0

2
B
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
I
U n it
1
III
4
1
2
3
4
0

2
1
2
Notice that, for Blocks
random, the covariance
between units from the
same block is non-zero
and is equal for all blocks.
2
B lo ck
II
2
3

2
 B
2

2
B
3

2
B
4

2
B
1
2
3
4
1
2
0
0
0
0
0
0

2
 B
2

2
B

2
B
0
0
0
0
0
0

2
B
2


2
B
0
0
0
0
0
0
2
B

2

0
0
0
0
0
0
2
B
0

2


2
B

2
B

2

2
B

2
B
 B2

2
B

0
0
2
B
0
0
2
B


2
B


2
B
 B2
2

B
2
0
0
2
B
 B2
2
B

2

0
0
3
0
0
0
0
0
0
0
0
4
0
0
0
0
0
0
0
0
2
B
0

2

 B2
2
B

0
0
0
 B2
 B2
 B2
 B2
 B2
2
 B
2
 B2
 B2
 B2
 B2

2
 B2
 B
2
 B2

2
 B
2
55
Fixed versus random factors
• Definition IV.2: A factor will be designated as random if
it is considered appropriate to use a probability
distribution function to describe the distribution of effects
associated with the population set of levels.
• Definition IV.3: A factor will be designated as fixed if it
is considered appropriate to have the effects associated
with the population set of levels for the factor differ in an
arbitrary manner, rather than being distributed according
to a regularly-shaped probability distribution function.
• As far as the model is concerned,
– random effects are modelled using terms in the variation model
– fixed effects are modelled using terms in the expectation model.
• So when we are deciding whether a factor is random or
fixed, we are choosing which mathematical model best
describes the population distribution for the response
variable.
Statistical Modelling Chapter IV
56
Making the choice
• Need to consider the population set of levels and how the set of
response variable effects corresponding to this set of levels behaves.
• To be classified as
– random, we require that
• the set of population levels is large in number and
• the effects are “well-behaved” so that a regularly-shaped probability
distribution function with some variance is appropriate for describing
them.
– fixed, the effects do not have the restrictions that are placed on
random effects.
• There might be a small or a large number of levels in the population
and
• their effects do not have to conform to a regularly-shaped probability
distribution function because the model allows for arbitrary
differences between them.
• For example, effects from factor modelled in expectation model
– If they display a systematic trend (perhaps involving polynomial
submodels)
– If factor for a small set of treatments that are to be compared.
• In both cases, seems inappropriate to model the effects as being, say
normally distributed, with some variance.
– Pattern in the treatment effects may well be quite irregular — no
interest in the form of this distribution.
Statistical Modelling Chapter IV
57
Summary
In practice
– Random if
i. large number of population levels and
ii. random behaviour
– Fixed if
i. small or large number of population
levels and
ii. systematic behaviour
Statistical Modelling Chapter IV
58
Units & Blocks — fixed or random?
• Effects from individual units treated alike (for example, animals, plots
of land, runs of a chemical reactor) are anticipated to arise randomly
and the effects could well follow a probability distribution, say a
normal distribution.
– Hence appropriate to model them via a term in the variation model.
• Must always model terms to which other terms have been
randomized as random effects
– because Treatments are randomized to Units[Block] in an RCBD,
Units[Block] must be random.
• What about Block effects in the RCBD?
– It could be either depending on the anticipated effects of the blocks.
• Suppose the blocks are groups of plots and are contiguous and a
systematic trend is anticipated:
– The distribution of block effects cannot be regarded as a random sample
— they display a systematic pattern.
– The factor Blocks should be designated as fixed.
• However, suppose each block is in a separate location to other
blocks and could be regarded as a random sample of all blocks
obtained by dividing up the whole area under study.
– It seems likely that the population block effects could be described by a
probability distribution such as the normal distribution and the factor
Blocks could be designated as random.
• If there is some doubt, safest to not make the assumption of some
probability distribution and to designate the factor as fixed.
Statistical Modelling Chapter IV
59
Example IV.1 Penicillin yield (continued)
• Should Blends be designated as fixed or random?
– It was said at the outset that it was expected that there would be a
lot of variability from blend to blend — that is why the RCBD was
employed.
– However, a systematic pattern in the average yields of the blends
cannot be anticipated.
– Rather, it seems reasonable that the effects of the population set
of blends can be described by a probability distribution.
– So Blends should be a random factor.
• Analysis needs to be revised, using a call to aov in which
Blends is not included outside the Error function.
RCBDPen.aov <- aov(Yield ~ Treat +
Error(Blend/Flask), RCBDPen.dat)
• This will change the fitted values and Tukey's one-degreeof-freedom-for-nonadditivity.
Statistical Modelling Chapter IV
60
b) Estimation and analysis of
variance for Blocks random
• Estimator of expected values under the model
2
2
E  Y  = X T  a n d V =  In   B  Ib  J t 
are
ˆ T = T = MTY
ψ
the same as for the model
E  Y  = X T  a n d V =  In
2
• Block hypotheses become
H0:  B = 0
2
H 1:  B  0
2
• That is, can  B2 be dropped from V?
• Also, as expectation model no longer involves the sum
of two terms, Tukey’s one-degree-of-freedom for
nonadditivity is no longer applicable.
Statistical Modelling Chapter IV
61
ANOVA table for the RCBD
• Form same irrespective of whether Blocks fixed or random
E [M S q]
S o u rc e
df
B lo cks
b -1
U n its [B loc ks]
b (t-1 )
B lo cks F ixe d

2
 t B
 

2
 qT 
2
 qB 
2
 qT 
t-1

R e s id ua l
(b -1 )(t-1)

T o ta l
 

T re a tm en ts
2
B lo cks R a n d om

2
 
2
b t-1
• However, E[MSq]s differ — qB(Y) becomes t  B2
• The F-statistic for testing this hypothesis is again the
ratio of the Block and Residual mean squares.
• Thus the test for both fixed and random block effects
are the same —not always the case.
Statistical Modelling Chapter IV
62
IV.G Generalized randomized
complete block design
• Difference between generalized and ordinary RCBDs is that in
GRCBD each treatment occurs > 1 in a block.
• As before we let b be no. of blocks and t no. of treatments.
• In addition let
– k denote no. of units per block and
– g no. of times a treatment occurs in a block
that is, k = t  g and n = b  k.
• The R expressions for obtaining a layout for this design is given in
Appendix B, Randomized layouts and sample size computations in
R.
• Advantages of this design
– more df for the Residual compared to the standard RCBD.
– Also, you can test for Block:Treatment interaction, as is discussed in
chapter VI, Determining the analysis of variance table.
• Disadvantage of the design
– it has larger blocks
– so it is likely that the units within a block will be less homogeneous than
would be the case if a standard RCBD with smaller blocks were
employed.
Statistical Modelling Chapter IV
63
Analysis of GRCBD
• The model for the generalized RCBD, without the
Block:Treatment interaction, is virtually the same as that
for the RCBD so that, in this case, the analyses of
variance are similar.
• Thus, depending on whether Blocks are fixed or random
the maximal model, would be chosen from the two given
for the RCBD.
• For Blocks and Plots random, the ANOVA table is
S o u rc e
df
SSq
E [M S q]
B lo cks
b 1
Y Q B Y
U n its [B loc ks]
b  k  1
Y Q B U Y
T re a tm en ts
t 1
Y Q T Y
R e s id ua l
b  k  1   t  1
T o ta l
Y Q B U
Res
 B2 U  k  B2
 B2 U  q T   
Y
 B2 U
bk  1
• R expressions same as for the standard RCBD.
Statistical Modelling Chapter IV
64
Example IV.2 Design for a wheat experiment
• For example, suppose 4 treatments are to be compared
when applied to a new variety of wheat.
• The researcher wants to employ a generalized RCBD
with 12 plots in each of 2 blocks so that each treatment is
replicated 3 times in each block.
• Hence, b = 2, t = 4 and g = 3.
so that k = 4  3 = 12 and n = 2  12 = 24.
Layout for a generalized randomized complete
block experiment
Plots
Blocks
I
II
1
C
D
2
D
A
3
D
D
4
C
C
5
B
A
6
B
D
7
A
B
8
A
A
9
D
B
10
A
B
11
B
C
12
C
C
• The yield of wheat from each plot was measured.
Statistical Modelling Chapter IV
65
Analysis with Blocks and Plots random
• The model for the example:
E  Y  = X T  a n d V =  I2 4   B  I2  J1 2  =  M U  1 2 B M B
2
2
2
2
• The corresponding ANOVA table:
S o u rc e
B lo cks
P lo ts[B locks ]
df
T o ta l
E [M S q]
1
Y Q B Y
22
Y Q B P Y
T re a tm en ts
R e s id ua l
SSq
3
19
 B2 P  1 2 B2
Y Q T Y
Y Q B P
Res
 B2 P  q T   
Y
 B2 P
23
• Note that a RCDB b = 6, t = 4 and
– would also have n = 6  4 = 24,
– but would have (b  1)(t – 1) = 5  3 = 15 Residual df.
Statistical Modelling Chapter IV
66
IV.I Exercises
• Ex. IV.1-2 looks at quadratic forms for SSq
• Ex. IV.3 requires a design of an RCBD and
then analysis of data
• EX. IV.4 asks for the complete analysis of
an RCBD with a quantitative treatment
factor
• EX. IV.5 asks for the complete analysis of
an RCBD with a qualitative treatment factor
Statistical Modelling Chapter IV
67