biometry

1. EXPERIMENTAL DESIGN for SBH

Prof. Dr. Md. Ruhul Amin

2. What ? Why ? How?

3. EXPERIMENTAL DESIGN
The preplanned
procedure by which
samples are drawn is
called
EXPERIMENTAL
DESIGN

4. Experimental Design
Experimental design is a set of rules used to
choose samples from populations. The rules are
defined by the researcher himself, and should
be determined in advance. In controlled
experiments, the experimental design describes
how to assign treatments to experimental
units, but within the frame of the design must
be an element of randomness of treatment
assignment. It is necessary to define

5. Experimental
error
Replication
Sample units
(observations)
Experimental
units
Size of
samples
Treatments
(population)
Experimental Design..

6. Basic Designs
1. Completely Randomized Design (CRD)
2. Randomized Block design (RBD)
3. Latin Square Design
CRD is known as “One-way design”

7. Designs commonly used in Animal
Science
i) One-way design (no interaction
effect)
a.
Fixed effects
b.
Random effects
ii) Factorial design (interaction effect)

8. Some important definitions
Treatments : Whose effect is to be determined. For
example
i)you are to study difference in lactation milk yield
in different breeds of cows. ….. Treatment is
breed of cows. Breed 1, Breed 2… are levels
ii) You intend to see the effect of 3 different diets on
the performance of broilers. ….. Treatment is
diet and diet1, diet2 and diet3 are levels (1,2,3)

9. …..definitions
Experimental units: Experimental material to which we
apply the treatments and on which we make
observations. In the previous two examples cow and
broilers are the experimental materials and each
individual is an experimental unit.
Experimental error: The uncontrolled variations in the
experiment is called experimental error. In each
observation of example(i) there are some extraneous
sources of variation (SV) other than breed of cow in
milk yield. If there is no uncontrolled SV then all cows
in a breed would give same amount of milk (!!!).

10. …..definitions
Replication: Repeated application of treatment
under investigation is known as replication. In
the example (i) no. of cows under each breed
(treatment) constitutes replication.
Randomization: Independence (unbiasedness)
in drawing sample.
Randomization, replication and error control
are three principles of experimental design.

11. Fixed Effects One-way ANOVA
1. Testing hypothesis to
examine differences between
two or more categorical
treatment groups.
3. Measurements are
described with dependent
variable, and the way of
grouping by an
independent variable
(factor).
2. Each treatment group
represents a population.

12. Fixed effects one-way ANOVA
• Consider an experiment with 15 steers and 3
treatments (T1, T2, T3)
• Following scheme describes a CRD
Steer
No
1
2
3
4
5
6
7
8
Treatm
ent
T2
T1
T3
T2
T3
T1
T3
T2
Steer
No
9
10
11
12
13
14
15
Treatm
ent
T1
T2
T3
T1
T3
T2
T1
NB: One treatment appeared 5 times. Equal no. of
replication/treatment – not necessary in one-way ANOVA

13. Fixed effects one-way ANOVA..
Data sorted by treatment for RANDOMIZATION
T1
T2
T3
Steer
Measure Steer
ment
Measure Steer
ment
Measure
ment
2
6
9
12
15
y11
y12
y13
y14
y15
y21
y22
y23
y24
y25
y31
y32
y33
y34
y35
1
4
8
10
14
3
5
7
11
13

14. Fixed effects one-way ANOVA..
In applying a CRD or when groups indicate a
natural way of classification, the objectives can
be
1. Estimating the mean
2. Testing the difference between
groups

15. Fixed effects one-way ANOVA..
Model
Y ij
ti
e ij
Where
Yij = Observation of ith treatment in jth replication
= Overall mean
ti = the fixed effect of treatment i (denotes an unknown
parameter)
eij = random error with mean ‘0’ and variance ‘ 2 ‘
The factor or treatment influences the value of observation

16. Fixed effects one-way ANOVA..
Treatment 1
Treatment 2
Look the difference

17. Fixed effects one-way ANOVA..
Problem 1:
An expt. was conducted to investigate the
effects of 3 different rations on post weaning
daily gains (g) in 3 different groups of beef calf.
The diets are denoted with T1, T2, and T3. Data,
sums and means are presented in the following
table.

18. Fixed effects one-way ANOVA..
T1
T2
T3
270
290
290
300
250
340
280
280
330
280
290
300
270
280
300
Total
1400
1390
1560
4350
n
5
5
5
15
y
280
278
312
290

19. One-way ANOVA: Hypothesis
Null hypothesis
Alternative hypothesis
Ho: There is no significant
difference between the
effect of rations on the daily
gains in beef calves ie
Effects of all treatments are
same.
Ha: There is significant
difference between the
effect of rations on the daily
gains in beef calves ie Effect
of all treatments are not
same.
Ho :
1
2
3
Ha :
1
2
3

20. Level of significance or confidence
interval
Commonly used level of significances
α=0.05
• True in 95% cases
• p<0.05
α=0.01
• True in 99% cases
• p<0.01
p< 0.05, conf. interval = 95% ; p< 0.01, conf. interval = 99%

21. Calculation of different Sum of
Squares(SS)
Total SS =
y
i
j
y
2
CF
T
ij
i
ij
N
2
y
Treatment SS =
, say Where CF
0
2
i.
n
i
CF
T , say
Error SS = Total SS – Treatment SS = T0-T= E say
CF stands for correction Factor

22. One-way ANOVA Table
Source of
variation
Degrees of
freedom
(df)
Treatment
Sum of squares
(SS)
k-1
T
Error
N-k
N-1
T0 =
i
T ' T /( k
n
T’/E’
i
CF
T0 –T = E
Total
F
2
y
i
Means square
(MS)
1)
E’ = E/(N-k)
y
2
CF
ij
If the calculated value of F with (k-1) and (N-k) df is greater than the tabulated
value of F with same df at 100α % level of significance, then the hypothesis
may be rejected ie the effects of all the treatments are not same. Otherwise
the hypothesis may be accepted. (N=Total no of observation, k=no of
treatments)

23. One-way ANOVA…
1. Grand Total (GT) =
2. CF =
y
( 4350 )
(
i
j
i
j
y
( 270
300
... ...
300 )
4350
ij
2
)2
ij
N
1261500
15
3. Total Corrected SS =
y
i
j
2
CF
( 270
ij
2
300
2
2
... ... 300 )
CF
= 1268700 – 1261500 = 7200
4. Treatment SS =
( y ij )
j
i
2
CF
n
i
1400
5
2
1390
5
2
1560
2
5
5. Error SS = Total SS – Treatment SS
= 7200-3640 = 3560
CF
1265140
1261500
3640

24. ANOVA for Problem 1.
Source
SS
df
MS
F
Treatment
3640
3-1=2
1820
6.13
Error (residual)
3560
15-3=12
296.67
Total
7200
15-1=14
The critical value of F for 2 and 12 df at α = 0.05 level of significance is F
0.05 (2,12 )= 3.89. Since the calculated F (6.13) > tabulated F or critical value
of F(3.89), Ho is rejected. It means the experiments concludes that there
is significant difference (p<0.05) between the effect different rations (at
least in two) of calves causing daily gain.
Now the question of difference between any two means will be solved by
MULTIPLE COMPARISON TEST(S).

25. Multiple Comparison among Group
Means (Mean separation)
There are many
tests such as
• Least significant
difference (LSD) test
• Tukey’s W-test
• Newman-Keul’s
sequential range test
• Duncan’s New
Multiple Range Test
(DMRT)
• Scheffe test

26. Multiple comparison: Least Significant
Difference(LSD) test
LSD compares treatment means to see whether
the difference of the observed means of
treatment pairs exceeds the LSD numerically.
LSD is calculated by
2 where t is the
s
t
r
value of Student’s t with error df at 100 %
level of significance, s2 is the MS of error and r
is the no. of replication of the treatment. For
unequal replications, r1 and r2 LSD= t s ( 1 1 )
r
1
r
2

27. Duncan’s Multiple Range Test(DMRT)
Duncan (1995) made , the level of
significance a variable from test to test. The
Least Significant Range (LSR) is defined by
k
LSR
SSR
s
r
The value of significant studentized range (SSR)
is given in Duncan (1955).
In case, a pair of means differs by more than its
LSR, they are declared to be significantly
different.

28. Random Effects One-way ANOVA: Difference between
fixed and random effect
Fixed effect
Random effect
Small number (finite)of groups or
treatment
Large number (even infinite) of groups or
treatments
Group represent distinct populations each The groups investigated are a random
with its own mean
sample drawn from a single population of
groups
Variability between groups is not
explained by some distribution
Effect of a particular group is a random
variable with some probability or density
distribution.
Example: Records of milk production in
Example: Records of first lactation milk
cows from 5 lactation order viz. Lac 1, Lac production of cows constituting a very
2, Lac 3, Lac 4, Lac 5.
large population.

29. One-way ANOVA, random effect
Source
SS
Between groups or
treatments
SSTRT
Residual (within groups or
treatments)
SSRES
df
MS=SS/df
a-1
MSTRT
N-a
For unbalanced cases n is replaced with
Expected Means
Square(EMS)
2
MSRES
1
a 1
2
n
N
n
i
N
i
2
2
T

30. Advantages of One-way analysis(CRD)
Popular design for
its simplicity,
flexibility and
validity
Can be applied with
moderate number
of treatments (<10)
Any number of
treatments and any
number of
replications can be
carried out
Analysis is straight
forward even one or
more observations
are missing

31. Two-way ANOVA
Suppose you intend to study the effectiveness of
3 different types of feed in 4 different strains of
hybrid broilers. You need to distribute your
treatments (3, feed) in a way so that birds of
each of the strains (4, blocks) receive each type
of feed. Randomization of the samples are to be
ensured in an efficient way. Total no. of records
= No. of treatments x No. of Blocks x No. of
replication (2 in this case) per treatment
(3x4x2=24)

32. You want to know
Why doing this
kind of expt. ?
1.Effect of type of feed on the
final live weight in broilers
(treatment effect)
2.Effect of strain on the final live
weight in broilers (block effect)
3.Joint effect of feed x strain on
the final live weight of broilers
( interaction effect)

33. Two-way ANOVA
B
L O
I
C K
II
S
III
IV
No. 1 (T3)
No. 7 (T3)
No. 13 (T3)
No. 19 (T1)
No. 2 (T1)
No. 8 (T2)
No. 14 (T1)
No. 20 (T2)
Broiler No.
No. 3 (T3)
No. 9 (T1)
No. 15 (T2)
No. 21 (T3)
(Treatment)
No. 4 (T1)
No. 10 (T1)
No. 16 (T1)
No. 22 (T3)
No. 5 (T2)
No. 11 (T2)
No. 17 (T3)
No. 23 (T2)
No. 6 (T2)
No. 12 (T3)
No. 18 (T2)
No. 24 (T1)

34. Two-way ANOVA
Observations can be shown sorted by treatments and blocks
T1
I
II
III
IV
y111
y121
y131
Y141
y112
y122
y132
y142
T2
y211
y212
y221
y222
y231
y232
y241
y242
T3
y311
y312
y 321
y322
y331
y332
y341
y342
treatment’ i ‘and block’ j ‘
Treatments
yijk indicates experimental unit ‘k’ in
Blocks

35. Statistical model in two-way ANOVA
y
ijk
t
i
j
t
ij
i = 1,…,a; j = 1,…,b; k = 1,….,n
Where
yijk= observation k in treatment i and block j
μ= overall mean
ti = effect of treatment i
βj = effect of block j
tβij = the interaction effect of treatment I and block j
eijk = random error with mean 0 and variance Ϭ2
a = no. of treatments; b= no. of blocks; n= no. of obs in each
treatment x block combination.
e
ijk

36. Sum of Squares, Degrees of Freedom
and Mean Squares in ANOVA
Source
SS
df
MS= SS/df
Block
SSBlk
b-1
MSBLK
Treatment
SSTRT
a-1
MSTRT
TreatmentxBlock
SSTRTXBLK
(a-1)(b-1)
MSTRTxBLK
Residual
SSRES
ab(n-1)
MSRES
Total
SSTOT
abn-1

37. Example: Two-way design
Recall that the objective of the experiment
previously described was to determine the
effect of 3 treatments (T1, T2, T3) on average
daily gain of steers, and 4 blocks were defined.
However, in this example 6 animals (3x2) are
assigned to each block. Therefore, a total of
4x3x2 = 24 steers were used. Treatments were
assigned randomly to steers within block.

38. Example: Two-way design
The data are as follows
Blocks
Treatments
I
II
III
IV
T1
826
806
864
834
795
810
850
845
T2
827
800
871
881
729
709
860
840
T3
753
773
801
821
736
740
820
835

39. Two-way: Computations
y (826 806
1. Grand Total =
2. Correction term for the mean =
i
y
(
i
C
j
k
SS
y
TOT
i
j
k
k
)2
ijk
835 ) 19426
2
19426
abn
3. Total SS=
j
......
ijk
15723728 . 17
24
2
C
ijk
826
2
806
2
.... ....
835
2
15775768
15723728 . 17
52039 . 83
4. Treatment SS=
y
(
SS
j
TRT
i
k
nb
)2
ijk
C
6630
8
2
6517
8
2
6279
8
2
15723728
. 17
8025 . 58

40. Two-way: Computations…
5. Block SS =
2
j k
j
y ijk
C
4785
2
5072
6
2
4519
6
2
5050
6
2
15723728 . 17
33816 . 83
6
na
6. Interaction SS
2
SS
( y ijk ) SS
k
TRTxBLK
i
j
(826 806 )
2
TRT
(864 871 )
2
SS
BLK
2
.... ....
C
(820 835 )
2
2
8025 . 58
33816 . 83
15723728
. 17
2
8087 . 42
7. Residual SS =
SS
RES
SS
TOT
SS
TRT
SS
BLK
SS
TRTxBLK
2110 . 00

41. ANOVA TABLE
Source
SS
df
MS
Block
33816.83
4-1 = 3
11272.28
Treatment
8025.58
3-1 = 2
4012.79
TreatmentxBlock
8087.42
2x3=6
1347.90
Residual
2110.00
3x4x(2-1)=12
175.83
Total
52039.83
23
F value for treatment : F = 4012.79/175.83 = 22.82
F value for interaction: F = 1347.90/175.83 = 7.67

42. Conclusion
The critical value for testing the interaction is
F0.05,6,12 = 3.00, and for testing treatments is
F0.05,2,12 = 3.89. So at p = 0.05 level of
significance, H0 is rejected for both treatments
and interaction.
Inference: There is an effect of treatments and
the treatment effects are different in different
blocks.

43. A practical example of one-way
ANOVA
Problem: Adjusted weaning weight (kg) of lambs
from 3 different breeds of sheep are furnished
below. Carry out analysis for i) descriptive
Statistics ii) breed difference.
Suffolk: 12.10, 10.50, 11.20, 12.00, 13.20,
10.90,10.00
Dorset: 11.50, 12.80, 13.00, 11.20, 12.70
Rambuillet: 14.20, 13.90, 12.60, 13.60, 15.10,
14.70, 13.90, 14.50

44. Analysis by using SPSS 14
Descriptive Statistics
N
minimum
maximum
mean
Std. dev
suff
7
10.00
13.20
11.4143
1.09153
dors
5
11.20
13.00
12.2400
.82644
ramb
8
12.60
15.10
14.0625
.76520
Valid N (list
wise)
5

45. ANOVA (F test)
a) ANOVA
Sum of
squares
df
Means
Squares
F
Sig.
Between
groups
27.473
2
13.736
16.705
.000
Within groups
13.979
17
.822
Total
41.452
19

46. Mean Separation
Post hoc tests
Homogenous subsets
Wean
Duncan
3
N
Subset for alpha =0.05
1
2
suff
7
11.414
dors
5
12.240
ramb
8
Sig.
14.063
.121
1.000

47. Interpretation of results
i) Null hypothesis (μ1=μ2=μ3) is rejected ie
there is significant (p<0.001) difference in
weaning wt. between breeds.
ii) Rambuillet has significantly (p<0.05)
highest weaning wt. among the 3 breeds and
there is no significant difference (p>0.05)
between weaning wt.s of Suffolk and Dorset.

48. Do you know
Statistics?????
You are going to be an Animal
Scientist!!!!
Booo----

49. Yes