Randomized Block Design
Caffeine and Endurance in 9 Bicyclists
W.J. Pasman, et al. (1995). “The Effect of Different Dosages of Caffeine on
Endurance Performance Time,” International Journal of Sports Medicine,
Vol. 16, pp225-230
Randomized Block Design (RBD)
• t > 2 Treatments (groups) to be compared
• b Blocks of homogeneous units are sampled. Blocks can
be individual subjects. Blocks are made up of t subunits
• Subunits within a block receive one treatment. When
subjects are blocks, receive treatments in random order.
• Outcome when Treatment i is assigned to Block j is
labeled Yij
• Effect of Trt i is labeled ai
• Effect of Block j is labeled bj
• Random error term is labeled eij
• Efficiency gain from removing block-to-block
variability from experimental error
Randomized Complete Block Designs
• Model (Block effects and random errors independent):
Yij    a i  b j  e ij  i  b j  e ij
t
a
i 1
i

 0 b j ~ N 0, 
2
b

e ij ~ N 0, 
2
e

• Test for differences among treatment effects:
• H0: a1  ...  at  0
• HA: Not all ai = 0
(1  ...  t )
(Not all i are equal)
Typically not interested in measuring block effects (although
sometimes wish to estimate their variance in the population of
blocks). Using Block designs increases efficiency in making
inferences on treatment effects
RBD - ANOVA F-Test (Normal Data)
• Data Structure: (t Treatments, b Subjects/Blocks)
• Mean for Treatment i:
y i.
• Mean for Subject (Block) j:
• Overall Mean:
y. j
y ..
• Overall sample size: N = bt
• ANOVA:Treatment, Block, and Error Sums of Squares

TSS  i 1  j 1 yij  y ..
t
b

SSB  t  y
SSE    y

2
df Total  bt  1

y 
SST  bi 1 y i .  y ..
2
df T  t  1
b
2
df B  b  1
t
j 1
.j
..

2
ij
 y i.  y . j  y ..
 TSS  SST  SSB
df E  (b  1)(t  1)
RBD - ANOVA F-Test (Normal Data)
• ANOVA Table:
Source
Treatments
Blocks
Error
Total
SS
SST
SSB
SSE
TSS
df
t-1
b-1
(b-1)(t-1)
bt-1
MS
MST = SST/(t-1)
MSB = SSB/(b-1)
MSE = SSE/[(b-1)(t-1)]
•H0: a1  ...  at  0 (1  ...  t )
• HA: Not all ai = 0
T .S . : Fobs
R.R. : Fobs
(Not all i are equal)
MST

MSE
 Fa ,t 1,( b 1)( t 1)
P  val : P ( F  Fobs )
F
F = MST/MSE
Comparing Treatment Means
b
1 b
1
y i.   yij  b    a i  b j  e ij  
b j 1
j 1
Mean for Treatment i :
b
b


 b b  ba i   b j   e ij 
j 1
j 1


b
b


1
Mean for Treatment i ': y i '.  b b  ba i '   b j   e i ' j 
j 1
j 1


Difference in Means :
1

y i.  y i '.  a i  a i '   e i.  e i '.
   
E e i.  E e i '.  0



   
V e i.  V e i '. 
 E y i.  y i '.  a i  a i ' 

 e2
b
V y i.  y i '.



COV e i. , e i '.  0
2 e2

b
^


V y i.  y i '. 
2MSE
b
Pairwise Comparison of Treatment Means
• Tukey’s Method- q from Studentized Range Distribution with n = (b-1)(t-1)
MSE
Wij  qa (t , v)
b
Conclude  i   j if y i.  y j .  Wij


Tukey' s Confidence Interval : y i.  y j .  Wij
• Bonferroni’s Method - t-values from table on class
website with n = (b-1)(t-1) and C=t(t-1)/2
Bij  ta / 2,C ,v
2 MSE
b
Conclude  i   j if y i.  y j .  Bij


Bonferroni ' s Confidence Interval : y i.  y j .  Bij
Expected Mean Squares / Relative Efficiency
• Expected Mean Squares: As with CRD, the Expected Mean
Squares for Treatment and Error are functions of the sample
sizes (b, the number of blocks), the true treatment effects
(a1,…,at) and the variance of the random error terms (2)
• By assigning all treatments to units within blocks, error
variance is (much) smaller for RBD than CRD (which
combines block variation&random error into error term)
• Relative Efficiency of RBD to CRD (how many times as
many replicates would be needed for CRD to have as
precise of estimates of treatment means as RBD does):
MSECR (b  1) MSB  b(t  1) MSE
RE ( RCB , CR) 

MSE RCB
(bt  1) MSE
Example - Caffeine and Endurance
•
•
•
•
Treatments: t=4 Doses of Caffeine: 0, 5, 9, 13 mg
Blocks: b=9 Well-conditioned cyclists
Response: yij=Minutes to exhaustion for cyclist j @ dose i
Data:
Dose \ Subject
0
5
9
13
1
36.05
42.47
51.50
37.55
2
52.47
85.15
65.00
59.30
3
56.55
63.20
73.10
79.12
4
45.20
52.10
64.40
58.33
5
35.25
66.20
57.45
70.54
6
66.38
73.25
76.49
69.47
7
40.57
44.50
40.55
46.48
8
57.15
57.17
66.47
66.35
9
28.34
35.05
33.17
36.20
Plot of Y versus Subject by Dose
90.00
80.00
70.00
Time to Exhaustion
60.00
50.00
0 mg
5 mg
9mg
40.00
13 mg
30.00
20.00
10.00
0.00
0
1
2
3
4
5
Cyclist
6
7
8
9
10
Example - Caffeine and Endurance
Subject\Dose
1
2
3
4
5
6
7
8
9
Dose Mean
Dose Dev
Squared Dev
TSS
0mg
36.05
52.47
56.55
45.20
35.25
66.38
40.57
57.15
28.34
46.44
-8.80
77.38
5mg
42.47
85.15
63.20
52.10
66.20
73.25
44.50
57.17
35.05
57.68
2.44
5.95
9mg
51.50
65.00
73.10
64.40
57.45
76.49
40.55
66.47
33.17
58.68
3.44
11.86
13mg
37.55
59.30
79.12
58.33
70.54
69.47
46.48
66.35
36.20
58.15
2.91
8.48
Subj MeanSubj Dev Sqr Dev
41.89
-13.34
178.07
65.48
10.24
104.93
67.99
12.76
162.71
55.01
-0.23
0.05
57.36
2.12
4.51
71.40
16.16
261.17
43.03
-12.21
149.12
61.79
6.55
42.88
33.19
-22.05
486.06
55.24
1389.50
103.68
7752.773
TSS  (36.05  55.24) 2    (36.20  55.24) 2  7752.773 dfTotal  4(9)  1  35

SSB  4(41.89  55.24)

   (33.19  55.24)   4(1389.50)  5558.00
SST  9 (46.44  55.24) 2    (58.15  55.24) 2  9(103.68)  933.12 dfT  4  1  3
2
2
df B  9  1  8
SSE  (36.05  41.89  46.44  55.24) 2    (36.20  33.19  58.15  55.24) 2 
 TSS  SST  SSB  7752.773  933.12  5558  1261.653 df E  (4  1)(9  1)  24
Example - Caffeine and Endurance
Source
Dose
Cyclist
Error
Total
df
3
8
24
35
SS
933.12
5558.00
1261.65
7752.77
MS
311.04
694.75
52.57
H 0 : No Caffeine Dose Effect (a1    a 4  0)
H A : Difference s Exist Among Doses
MST 311.04
T .S . : Fobs 

 5.92
MSE 52.57
R.R.(a  0.05) : Fobs  F.05,3, 24  3.01
P  value : P( F  5.92)  .0036 (From EXCEL)
Conclude that true means are not all equal
F
5.92
Example - Caffeine and Endurance
Tukey' s W : q.05, 4, 24
1
 3.90 W  3.90 52.57   9.43
9
Bonferroni ' s B : t.05 / 2, 6, 24
Doses
5mg vs 0mg
9mg vs 0mg
13mg vs 0mg
9mg vs 5mg
13mg vs 5mg
13mg vs 9mg
2
 2.875 B  2.875 52.57   9.83
9
High Mean
57.6767
58.6811
58.1489
58.6811
58.1489
58.1489
Low Mean Difference Conclude
46.4400
11.2367
5>0
46.4400
12.2411
9>0
46.4400
11.7089
13>0
57.6767
1.0044
NSD
57.6767
0.4722
NSD
58.6811
-0.5322
NSD
Example - Caffeine and Endurance
Relative Efficiency of Randomized Block to Completely Randomized Design :
t  4 b  9 MSB  694.75 MSE  52.57
(b  1) MSB  b(t  1) MSE 8(694.75)  9(3)(52.57) 6977.39
RE ( RCB , CR) 


 3.79
(bt  1) MSE
(9(4)  1)(52.57)
1839.95
Would have needed 3.79 times as many cyclists per dose to have the
same precision on the estimates of mean endurance time.
• 9(3.79)  35 cyclists per dose
• 4(35) = 140 total cyclists
RBD -- Non-Normal Data
Friedman’s Test
• When data are non-normal, test is based on ranks
• Procedure to obtain test statistic:
– Rank the k treatments within each block (1=smallest,
k=largest) adjusting for ties
– Compute rank sums for treatments (Ti) across blocks
– H0: The k populations are identical (1=...=k)
– HA: Differences exist among the k group means
12
k
2
T .S . : Fr 
T
 3b(k  1)

i 1 i
bk (k  1)
R.R. : Fr  a2 ,k 1
P  val : P(  2  Fr )
Example - Caffeine and Endurance
Subject\Dose
1
2
3
4
5
6
7
8
9
0mg
36.05
52.47
56.55
45.2
35.25
66.38
40.57
57.15
28.34
5mg
42.47
85.15
63.2
52.1
66.2
73.25
44.5
57.17
35.05
9mg
51.5
65
73.1
64.4
57.45
76.49
40.55
66.47
33.17
13mg
37.55
59.3
79.12
58.33
70.54
69.47
46.48
66.35
36.2
Ranks
Total
0mg
1
1
1
1
1
1
2
1
1
10
5mg
3
4
2
2
3
3
3
2
3
25
9mg
4
3
3
4
2
4
1
4
2
27
13mg
2
2
4
3
4
2
4
3
4
28
H 0 : No Dose Difference s
H a : Dose Difference s Exist


12
26856
2
2
T .S . : Fr 
(10)    (28)  3(9)( 4  1) 
 135  14.2
9(4)( 4  1)
180
R.R.(a  0.05) : Fr   .205, 41  7.815
P - value : P(  2  14.2)  .0026 (From EXCEL)
Conclude Means (Medians) are not all equal