STATISTICS

1. 1
DESIGN OF EXPERIMENTS
BASIC PRINCIPLES OF EXPERIMENTAL DESIGNS
Prof. Ronald. A. Fisher pioneered the study of experimental designs with his classical
book “The Design of Experiments”. According to him, the basic principles of the design of
experiments are
1. Replication
2. Randomization
3. Local Control
Replication
Replication means “the repetition of treatments under an investigation”. The repetition of
treatments results in more reliable estimate than is possible with a single observation.
The following are the chief advantages of replication.
1. Replication serves to reduce experimental error and thus enables us to obtain more
precise estimate of treatment effects. Replication is an important but limited role in
increasing the efficiency of the designs.
2. From a statistical theory, we know that the standard error of the mean of a sample size ‘n’
is
n

, where ‘ ’ is population standard deviation. Thus, if a treatment replicated ‘r’
times, then the standard error of its mean effect is
r

where,
2
 , the variance of
individual plot is estimated from the error variance.
Thus, the precision of an experiment is inversely proportional to the square root of the
replication.
Randomization
“The process of assigning the treatments to various experimental units in a purely chance
manner is called randomization”.
In this we allocate the treatments to the experimental units randomly, so that each
treatment has an equal chance of showing its effect.
The following are the main objectives of randomization.
i. The validity of statistical tests of significance, for e.g., the t-test for the testing the
significance of the difference between two means or the ANOVA F-test for testing the

2. 2
homogeneity of several means, depends on the fact that the statistic under consideration
obeys some statistical distribution.
ii. Randomization eliminates the experimental error and also it ensures that different
treatments by their replication on the average are subject to equal environmental effects.
Local Control
“The process of reducing the experimental error by dividing the relatively heterogeneous
experimental area into homogeneous blocks is known as local control”.
In the experimental material say field for agricultural experimentation, the heterogeneous
area can be divided into number of homogeneous subgroups and then different treatments are
allocated to various units at random over the entire field.
The following are the main advantages of local control.
I. Local control, by reducing the experimental error increases the efficiency of the design.
II. Several designs are developed on the basis of local control principle. Some of the
important designs are Randomized Block Design and Latin Square Design.
COMPLETELY RANDOMIZED DESIGN (CRD)
The CRD is the simplest of all the designs based on principles of randomization and
replication. In this design all the treatments are allocated at random to the experimental units over
the entire experimental material. Let us suppose that we are having ‘k’ treatments, the ith
treatment being replicated times, i=1,2…..k. Then the whole experimental material is divided
into 

k
i
irN
1
experimental units and then the treatments are distributed completely at random
over the units subject to the condition that the ith
treatment occurs ir times. In particular case, if
rri  , i=1,2,……k i.e., if each treatment is repeated an equal no. of times, then krrN
k
i
 1
and randomization gives every group of ‘r’ units and equal chance of receiving the treatments.
Advantages:
1. CRD results in the maximum use of the experimental units since all the experimental
material can be used.
2. The design is very flexible i.e., any no. of treatments can be used and different treatments
can be applied an equal no. of times.
3. The statistical analysis remains simple even though some of the observations are rejected
or lost or missing.

3. 3
4. It provides the maximum no. of degrees of freedom for the estimation of the error
variance.
Disadvantages:
1. As the local control principle is not applied, then the experimental error is more in CRD.
2. In certain circumstances the design is less informative than the other designs.
Applications:
CRD is mostly used in agricultural experiments, laboratory techniques and
methodological studies like physics, chemistry, biology etc.
Statistical analysis:
Mathematical model:
In this case the linear mathematical model will be
iijiij rjkitx ,....,2,1;,....,2,1,  
Where ijx = The yield of j-th unit from i-th treatment.
µ = The general mean effect.
it = The treatment effect =  i .
ij = Error effects are independently and identically distributed
(i.i.d.) N(0,e
2
)
Also we have
0
0
)(
1
1
11











k
i
i
k
i
i
k
i
i
k
i
i
t
kk
k
t



Assumptions to the model:
1. All the observations ijx are independent.
2. The parent population must be normal.
3. ij ’s are independently and identically distributed (i.i.d.) N(0,e
2
)
Null hypothesis:
We want to test the equality of the population means i.e. the homogeneity of different
treatments. Hence the null hypothesis is given by

4. 4
  kH .....: 210
Or
0.....: 210  ktttH
And the alternative hypothesis is given by
kH   .....: 211
Now, in this classification, the linear mathematical model is given by,
iijiij rjkitx ,....,2,1;,....,2,1,   ---------- (1)
iijij tx   ------------ (2)
According to the principle of least squares, we have
.min)(
1 1
2
1 1
2
istx
E
k
i
r
j
iij
k
i
r
j
ij
i
i


 
 



According to the principle of maxima and minima, we have
0&0 





it
EE

2
TS = Total Sum of Squares =  

k
i
r
j
ij
i
xx
1 1
2
.. )(
2
TRS = Treatment Sum of Squares =  

k
i
r
j
i
i
xx
1 1
2
... )(
2
ES = Error Sum of Squares =  

k
i
r
j
iij
i
xx
1 1
2
. )(
Degrees of freedom:
The degrees of freedom for totals = N – 1
The degrees of freedom for treatments = k – 1
The degrees of freedom for errors = N – 1 – (k – 1)
= N – k
Mean Sum of Squares (M.S.S.):
Mean sum of squares is obtained by dividing the value of sum of squares with the
corresponding degrees of freedom.

5. 5
)(...
)(
1
...
2
2
2
2
says
kN
S
errorstodueSSM
says
k
S
treatmentstodueSSM
e
E
tr
TR






ANOVA TABLE
Source of
variation
Degrees
of
freedom
Sum of
squares
Mean Sum
of Squares
F-Ratio
Treatments 1k 2
TRS 2
trs
],1[~)( 22
2
2
kNkFss
s
s
F etr
e
tr
Errors kN  2
ES 2
es
Totals 1N
Conclusion: If the calculated value of F is less than the tabulated value of F at  % LOS then
we accept our null hypothesis H0, otherwise we reject H0.
PROBLEM
1. For testing the variety effect in a completely randomized experiment, the yields of Barley in
grams are shown in the following table. Perform an ANOVA.
320
1V
340
4V
398
5V
360
2V
350
3V
372
4V
455
2V
417
3V
420
1V
358
5V
400
3V
353
1V
334
5V
331
2V
358
4V
370
4V
340
1V
375
2V
320
5V
325
3V
430
5V
358
1V
378
3V
395
4V
328
2V
383
2V
275
3V
375
4V
308
5V
400
1V

6. 6
Solution:-
For the given data, our null hypothesis is given by,
0H : There is no significant difference between the varieties.
And the alternative hypothesis is given by,
1H : There is a significant difference between the varieties.
Varieties Observations Total ( .iT ) j
ijx2
1V 320 420 353 340 358 400 2191 807173
2V 360 455 331 375 328 383 2232 841084
3V 350 417 400 325 378 275 2145 780523
4V 340 372 358 370 395 375 2210 815698
5V 398 358 334 320 430 308 2148 780288
G = 10926
i j
ijx2
= 4024766
Here N = 30

7. 7
6666.44543
1334.9738.45516
1334.973
2.39792497689846667.8140165.7668378303041667.800080
30
10926
6
2148
6
2210
6
2145
6
2232
6
2191
8.45516
2.39792494024766
30
10926
4024766
222
222222
2
.
2
.2
2
2
22













TRTE
i i
i
TR
i j
ijT
SSS
N
G
n
T
S
N
G
xS
ANOVA TABLE
Source of
Variation
Degrees of
Freedom
Sum of
Squares
Mean Sum
of Squares
F-Ratio
Treatments
(varieties)
4 973.1334 243.2834
)4,25(~3237.7
2834.243
7467.1781
FF 
Errors 25 44543.6666 1781.7467
Totals 29
The tabulated value of F(25, 4) = 5.75 at 5% level of significance.
Since the calculated value of ‘F’ is greater than the tabulated value of ‘F’ at 5% level of
significance, hence we reject our null hypothesis 0H .
i.e., There is a significant difference between the varieties.
RANDOMIZED BLOCK DESIGN (RBD)
In field experimentation, if the whole experimental area which is not homogeneous and
the fertility gradient is only in one direction, then a simple method of controlling the variability of
the experimental material consists in stratifying (or) grouping the whole area into relatively
homogeneous strata (or) subgroups perpendicular to the direction of the fertility gradient. If the

8. 8
treatments are applied at random to relatively homogeneous units with in each strata (or) block
and replicated over all the blocks, the design is known as “Randomized Block Design”.
In CRD we do not resort to the grouping of the experimental area and allocate the
treatments at random to the experimental units. But in RBD, treatments are allocated at random
with in the units of each strata (or) block i.e., Randomization is restricted. Also variation among
blocks is removed from variation due to error. Hence if it is desired to control one source of
variation by stratification, the experimenter should select RBD rather than CRD.
Advantages:
1. Accuracy: This design has been shown to be more accurate than CRD.
2. Flexibility: In RBD no restrictions are placed on the no. of treatments or the no. of replicants.
3. Ease of Analysis: Statistical analysis is simple and rapid. Any number of treatments may be
omitted from the analysis without complicating it.
Disadvantage:
RBD is not suitable for large no. of treatments.
Statistical analysis:
Mathematical model:
In this case the linear mathematical model will be
hjkibtx ijjiij ,....,2,1;,....,2,1,  
Where ijx = The yield of j-th unit from i-th treatment.
µ = The general mean effect.
it = The i-th treatment effect.
jb = The j-th block effect.
ij = Error effects are independently and identically distributed
(i.i.d.) N(0,e
2
)
Also we have
0
11
  
h
j
j
k
i
i bt
Assumptions to the model:
1. All the observations ijx are independent.
2. The parent population must be normal.
3. ij ’s are independently and identically distributed (i.i.d.) N(0,e
2
)
Null hypothesis:

9. 9
We want to test the equality of the population means i.e. the homogeneity of different
treatments as well as blocks. Hence the null hypotheses are given by




hB
kT
H
H
.2.1.0
..2.10
.....:
.....:
And the alternative hypothesis is given by
hB
kT
H
H
.2.1.1
..2.11
.....:
.....:




In this classification, the linear mathematical model is given by,
hjkibtx ijjiij ,....,2,1;,....,2,1,   ---------- (1)
jiijij btx   ------------ (2)
According to the principle of least squares, we have
.min)(
1 1
2
1 1
2
isbtx
E
k
i
n
j
jiij
k
i
n
j
ij
i
i


 
 



According to the principle of maxima and minima, we have
0&0,0 








ji b
E
t
EE

Consider,
2
TS = Total Sum of Squares =  

k
i
h
j
ij xx
1 1
2
.. )(
2
TRS = Treatment Sum of Squares =  

k
i
h
j
i xx
1 1
2
... )(
2
BS = Block Sum of Squares =  

k
i
h
j
j xx
1 1
2
... )(
2
ES = Error Sum of Squares =  

k
i
h
j
jiij xxxx
1 1
2
.... )(
Degrees of freedom:
The degrees of freedom for totals = N – 1
The degrees of freedom for treatments = k – 1

10. 10
The degrees of freedom for blocks = h – 1
The degrees of freedom for errors = N – 1 – (k – 1) – (h – 1)
= N – k – h + 1
= kh – k – h + 1
= k(h – 1) – 1(h – 1)
= (h – 1)(k – 1)
Mean Sum of Squares (M.S.S.):
Mean sum of squares is obtained by dividing the value of sum of squares with the
corresponding degrees of freedom.
)(...
)(
1
...
)(
1
...
2
2
2
2
2
2
says
kN
S
errorstodueSSM
says
h
S
blockstodueSSM
says
k
S
treatmentstodueSSM
e
E
b
B
tr
TR









ANOVA TABLE
Source of
variation
Degrees of
freedom
Sum of
squares
Mean
Sum of
Squares
F-Ratio
Treatments 1k 2
TRS 2
trs
)]1)(1(,1[~)(
)]1)(1(,1[~)(
22
2
2
22
2
2


hkhFss
s
s
F
hkkFss
s
s
F
eb
e
b
B
etr
e
tr
TR
Blocks 1h 2
BS 2
bs
Errors )1)(1(  hk 2
ES 2
es
Totals 1N
Conclusion: If the calculated values of ‘F’ are less than the tabulated values of ‘F’ at  %
LOS then we accept our null hypotheses, otherwise we reject our null hypotheses.
PROBLEM
1. Consider the results given in the following table for an experiment involving six treatments in
four randomized blocks. The treatments are indicated by numbers within parentheses.
Blocks Treatment and Yield
1 (1)
24.7
(3)
27.7
(2)
20.6
(4)
16.2
(5)
16.2
(6)
24.9
2 (3)
22.7
(2)
28.8
(1)
27.3
(4)
15.0
(6)
22.5
(5)
17.0

11. 11
3 (6)
26.3
(4)
19.6
(1)
38.5
(3)
36.8
(2)
39.5
(5)
15.4
4 (5)
17.7
(2)
31.0
(1)
28.5
(4)
14.1
(3)
34.9
(6)
22.6
Test whether the treatments differ significantly.
Solution:- For the given data, our null hypothesis is given by
0H : There is no significant difference between the treatments as well as
blocks.
And the alternative hypothesis is given by
1H : There is a significant difference between the treatments as well as
blocks.
To evaluate various sums of squares, consider the following table.
Blocks
Treatments
1 2 3 4
Total
)( .iT
j
ijx2
1 24.7 27.3 38.5 28.5 119 3649.88
2 20.6 28.8 39.5 31 119.9 3775.05
3 27.7 22.7 36.8 34.9 122.1 3854.83
4 16.2 15 19.6 14.1 64.9 1070.41
5 16.2 17 15.4 17.7 66.3 1101.89
6 24.9 22.5 26.3 22.6 96.3 2328.71
Total )( . jT 130.3 133.3 176.1 148.8 G=588.5 i j
ijx2
=15780.77
Here N = 24, 64 ..  ji nandn
Now,
26.1350
51.1443077.15780
24
)5.588(
77.15780
2
2
22



  N
G
xS
i j
ijT

12. 12
1925.901
51.144307025.15331
24
)5.588(
4
3.96
4
3.66
4
9.64
4
1.122
4
9.119
4
119 2222222
2
.
2
.2



  N
G
n
T
S
i i
i
Tr
4283.219
51.144309383.14649
24
)5.588(
6
8.148
6
1.176
6
3.133
6
3.130 22222
2
.
2
.2



  N
G
n
T
S
j j
j
B
And
6392.229
4283.2191925.90126.1350
2222


 BTrTE SSSS
ANOVA TABLE
Source of
variation
Degrees of
freedom
Sum of
squares
Mean
Sum of
Squares
F-Ratio
Treatments 5 901.1925 180.2385
]15,3[~78.4
3093.15
1428.73
]15,5[~77.11
3093.15
2385.180
FF
FF
B
TR


Blocks 3 219.4283 73.1428
Errors 15 229.6392 15.3093
Totals 23
The tabulated values of F(5, 15) and F(3, 15) at 5% level of significance are 2.90 and
3.29 respectively.
Conclusion:- Since the calculated values of F are greater than the tabulated values of F at 5%
level of significance, hence we reject our null hypothesis 0H i.e., there is a significant
difference between the treatments as well as blocks.

13. 13
Unit-5
LATIN SQUARE DESIGN (LSD)
In this design the total no. of rows is equal to the total no. of columns i.e., total no. of
treatments is equal to the no. of replications.
Let us consider the total no. of treatments be ‘m’, the total experimental material of a
field is divided into
2
mmm  experimental units (or) plots and all these plots are arranged
into a square, so that each row as well as each column contains ‘m’ plots. The ‘m’ treatments are
the allocated at random to these rows and columns in such a way that every treatment occurs once
and only once in each row and in each column. Such a layout (or) design is known as Latin
Square Design. This design is extensively used in agricultural experiments.
Advantages:
1. With two-way grouping, LSD controls more of the variation than CRD and RBD.
2. The statistical analysis is very simple, but slightly complicated than RBD.
3. More than one factor can be investigated simultaneously.
Disadvantages:
1. The fundamental assumption that the interaction between different factors may not be
true in general.
2. In case of missing plots, when several units are missing, the statistical analysis becomes
quiet complex.
3. In field layout, RBD is much easy to manage than LSD.
Statistical analysis:
Mathematical model
In this case the linear mathematical model will be
mkjitcrx ijkkjiijk ,....,2,1,  
Where ijkx = The yield of k-th treatment effect on i-th row and j-th column.
µ = The general mean effect.
ir = The i-th row effect.
jc = The j-th column effect.
kt = The k-th treatment effect.
ijk = Error effects are independently and identically distributed
(i.i.d.) N(0,e
2
)

14. 14
Also we have
0
111
  
m
k
k
m
j
j
m
i
i tcr
Assumptions to the model:
1. All the observations ijkx are independent.
2. The parent population must be normal.
3. ij ’s are independently and identically distributed (i.i.d.) N(0,e
2
)
Null hypothesis:
We want to test the equality of the population means i.e. the homogeneity of different
treatments as well as blocks. Hence the null hypotheses are given by
0H : There is no significant difference between rows, columns as well as treatments.
And the alternative hypothesis is given by
1H : There is a significant difference between rows, columns as well as treatments.
In this classification, the linear mathematical model is given by,
mkjitcrx ijkkjiijk ,....,2,1,   ---------- (1)
kjiijkijk tcrx   ------------ (2)
According to the principle of least squares, we have
.min)(
1 1 1
2
1 1 1
2
istcrx
E
m
i
m
j
m
k
kjiijk
m
i
m
j
m
k
ijk


  
  



According to the principle of maxima and minima, we have
0&0,0,0 











kji t
E
c
E
r
EE

Consider,
also,
Similarly,

15. 15
2
TS = Total Sum of Squares =   

m
i
m
j
m
k
ijk xx
1 1 1
2
... )(
2
RS = Row Sum of Squares =   

m
i
m
j
m
k
i xx
1 1 1
2
..... )(
2
CS = Column Sum of Squares =   

m
i
m
j
m
k
j xx
1 1 1
2
..... )(
2
TRS = Treatment Sum of Squares =   

m
i
m
j
m
k
k xx
1 1 1
2
..... )(
2
ES = Error Sum of Squares =   

m
i
m
j
m
k
kjiijk xxxxx
1 1 1
2
......... )2(
Degrees of freedom:
The degrees of freedom for totals = 12
m
The degrees of freedom for rows = 1m
The degrees of freedom for columns = 1m
The degrees of freedom for treatments = 1m
The degrees of freedom for errors = )1(312
 mm
= 232
 mm
= )2)(1(  mm
Mean Sum of Squares (M.S.S.):
Mean sum of squares is obtained by dividing the value of sum of squares with the
corresponding degrees of freedom.
)(
)2)(1(
...
)(
1
...
)(
1
...
)(
1
...
2
2
2
2
2
2
2
2
says
mm
S
errorstodueSSM
says
m
S
treatmentstodueSSM
says
m
S
columnstodueSSM
says
m
S
rowstodueSSM
e
E
tr
TR
c
C
r
R













16. 16
ANOVA TABLE
Source of
variation
Degrees of
freedom
Sum of
squares
Mean
Sum of
Squares
F-Ratio
Rows 1m 2
RS 2
rs
)]2)(1(),1[(~
)]2)(1(),1[(~
)]2)(1(),1[(~
2
2
2
2
2
2



mmmF
s
s
F
mmmF
s
s
F
mmmF
s
s
F
e
tr
TR
e
c
C
e
r
R
Columns 1m 2
CS 2
cs
Treatments 1m 2
TRS 2
trs
Errors )2)(1(  mm 2
ES 2
es
Totals 12
m
Conclusion: If the calculated values of ‘F’ are less than the tabulated values of ‘F’ at  %
LOS then we accept our null hypotheses, otherwise we reject our null hypotheses.
PROBLEM
1. An experiment was carried out to determine the effect of claying the ground on the field of
barley grains; amount of clay used were as follows:
A: No Clay; B: Clay at 100 per acre; C: Clay at 200 per acre and D: Clay at 300 per acre.
The yields were in plots of 8 meters by 8 meters and layout was:
Column
Row
I II III IV
I
D
29.1
B
18.9
C
29.4
A
5.7
II
C
16.4
A
10.2
D
21.2
B
19.1
III
A
5.4
D
38.8
B
24.0
C
37.0
IV
B
24.9
C
41.7
A
9.5
D
28.9
Perform an ANOVA.
Sol.:- For the given data, our null hypothesis is
0H : There is no significant difference between rows, columns as well as
treatments.
And the alternative hypothesis is given by

17. 17
1H : There is a significant difference between rows, columns as well as
treatments.
Now, to compute 222
, CRT SandSS , consider the following table.
Column
Row
I II III IV
Total
)( ..iT
j k
ijkx2
I 29.1 18.9 29.4 5.7 83.1 2100.87
II 16.4 10.2 21.2 19.1 66.9 1187.25
III 5.4 38.8 24.0 37.0 105.2 3479.6
IV 24.9 41.7 9.5 28.9 105 3284.36
Total
)( .. jT
75.8 109.6 84.1 90.7 360.2 10052.08
 
0775.1943
0025.810908.10052
16
2.360
08.10052
2
2
22



  N
G
xS
i j k
ijkT
 
3125.259
0025.8109315.8368
16
2.360
4
)105(
4
)2.105(
4
)9.66(
4
)1.83(
22222
2
..
2
..2



  N
G
n
T
S
i i
i
R
 
2725.155
0025.8109275.8264
16
2.360
4
)7.90(
4
)1.84(
4
)6.109(
4
)8.75(
22222
2
..
2
..2



  N
G
n
T
S
j j
j
C

18. 18
To compute 2
TRS , consider the following table.
Treatment Observations Total )( ..kT
A 5.7 10.2 5.4 9.5 30.8
B 18.9 19.1 24.0 24.9 86.9
C 29.4 16.4 37.0 41.7 124.5
D 29.1 21.2 38.8 28.9 118
360.2
 
1225.1372
0025.8109125.9481
16
2.360
4
)118(
4
)5.124(
4
)9.86(
4
)8.30(
22222
2
..
2
..2



  N
G
n
T
S
k k
k
TR
And
37.156
1225.13722725.1553125.2590775.1943
22222


 TRCRTE SSSSS
ANOVA TABLE
Source of
variation
Degrees of
freedom
Sum of
squares
Mean
Sum of
Squares
F-Ratio
Rows 3 259.3125 86.4375
]6,3[~55.17
0617.26
3742.457
]6,3[~98.1
0617.26
7575.51
]6,3[~32.3
0617.26
4375.86
FF
FF
FF
TR
C
R



Columns 3 155.2725 51.7575
Treatments 3 1372.1225 457.3742
Errors 6 156.37 26.0617
Totals 15
The tabulated value of F(3, 6) at 5% level of significance is 4.76
Conclusion: Since the calculated values of F for rows and columns are less than the tabulated
value of F at 5% level of significance and the calculated value of F for treatments is greater than

19. 19
the tabulated value of F at 5% level of significance, hence we accept our null hypothesis for rows
and columns and we reject our null hypothesis for treatments.
i.e., There is no significant difference between rows and columns but there is a significant
difference between the treatments.
‘’………………………………………………………………………………………..

20. 20
ESTIMATION OF MISSING VALUES IN R.B.D
Let the observation )'(' sayxxij  in the j-th block and receiving the i-th treatment be
missing, then that value can be estimated by using the following formula.
)1)(1(
..



tr
GtTrT
x ij
Where jT. = Total of the known observation in the column
containing x .
.iT = Total of the known observation in the row
containing x .
G = Grand total of the known observations.
r = No. of Blocks
t = No. of treatments
For two missing values, say '''' yandx , let '''' 21 RandR be the total of the known
observations in the rows containing '''' yandx respectively. Let '''' 21 CandC be the totals of
the known observations in the columns containing '''' yandx respectively. And ‘S’ be the total
of all the known observations.
Then '''' yandx can be estimated by solving the following two equations.
xStCrRytr
yStCrRxtr


22
11
)1)(1(
)1)(1(
PROBLEM
1. In the table given below are the yields of 6 varieties in a 4 replicate experiment for which one
value is missing. Estimate the missing value and analyze the data.
Treatments
Blocks
1 2 3 4 5 6
Block
Totals
)( . jT
1 18.5 15.7 16.2 14.1 13.0 13.6 91.1
2 11.7 - 12.9 14.4 16.9 12.5 68.4
3 15.4 16.6 15.5 20.3 18.4 21.5 107.8
4 16.5 18.6 12.7 15.7 16.5 18.0 98.0

21. 21
Sol.: Estimate of missing value
Let ‘x’ be the missing value of the given data. We can estimate the missing value
of the given data by using the following formula.
)1)(1(
..



tr
GtTrT
x ij
Where,
jT. = 68.4, .iT = 50.9, G = 365.3, r = 4 and t = 6
Therefore,
25.14
15
7.213
15
3.3654.3056.273
53
3.365)9.50(6)4.68(4






x
Analysis of the data
Now, the given data becomes
Treatment
Totals )( .iT
62.1 50.9 57.3 64.5 64.8 65.7 365.3
Treatments
Blocks
1 2 3 4 5 6
Block
Totals
)( . jT
1 18.5 15.7 16.2 14.1 13.0 13.6 91.1

22. 22
Now,
2174.145
4251.60026425.6147
24
)55.379(
6425.6147
2
2
22



  N
G
xS
i j
ijT
5755.12
4251.60020006.6015
24
)55.379(
4
)7.65(
4
)8.64(
4
)5.64(
4
)3.57(
4
)15.65(
4
)1.62( 2222222
2
.
2
.2



  N
G
n
T
S
i i
i
TR
7537.56
4251.60021788.6059
24
)55.379(
6
)98(
6
)8.107(
6
)65.82(
6
)1.91( 22222
2
.
2
.2



  N
G
n
T
S
j j
j
B
8882.75
7537.565755.122174.145
2222


 BTRTE SSSS
ANOVA TABLE
Source of
variation
Degrees of
freedom
Sum of
squares
Mean
Sum of
Squares
F-Ratio
2 11.7 14.25 12.9 14.4 16.9 12.5 82.65
3 15.4 16.6 15.5 20.3 18.4 21.5 107.8
4 16.5 18.6 12.7 15.7 16.5 18.0 98.0
Treatment
Totals )( .iT
62.1 65.15 57.3 64.5 64.8 65.7 379.55
j
ijx2
988.55 1071.0725 830.39 1064.75 1065.42 1127.46 6147.6425

23. 23
Treatments 5 12.5755 2.5151
]15,3[~73.3
0592.5
9179.18
]5,15[~01.2
5151.2
0592.5
FF
FF
B
TR


Blocks 3 56.7537 18.9179
Errors 15 75.8882 5.0592
Totals 23
The tabulated values of F(15, 5) and F(3, 15) at 5% level of significance are 4.62 and 3.29
respectively.
Conclusion: Since the calculated value of F for treatments is less than the tabulated value of F
and the calculated value of F for blocks is greater than the tabulated value of F at 5 % level of
significance, hence we accept our null hypothesis for treatments and we reject our null hypothesis
for blocks.
ESTIMATION OF MISSING VALUES IN L.S.D
Let us suppose that in mm Latin Square, the observation in the i-th row, j-th column
and receiving the k-th treatment is missing. Let us assume that its value is x , i.e., xxijk  .
R = Total of the known observations in the i-th row, i.e., the row containing ‘ x ’
C = Total of the known observations in the j-th column, i.e., the row containing ‘ x ’
T = Total of the known observations receiving k-th treatment, i.e., total of all known
treatment values containing ‘ x ’
S = Total of known observations.
Now the formulae for various sum of squares in L.S.D. becomes

24. 24
  2
2
2222
2
E
2
C
2
R
2
T
2
E
2
22
2
TR
2
22
2
C
2
22
2
R
2
2
22
T
m
x)(S
2x)(Tx)C(x)(R
m
1
x
SSSSS
m
x)(S
-xw.r.t.constants
m
x)(T
S
m
x)(S
-xw.r.t.constants
m
x)(C
S
m
x)(S
-xw.r.t.constants
m
x)(R
S
m
x)(S
-xw.r.t.constantsxS

















We will choose ‘x’ so as to minimize E.
 
)2)(1(
2)(
2)()23(
)(
43
2
20
2
2








mm
STCRm
x
STCRmxmm
m
xS
xTCR
m
x
x
E
SHORT ANSWER QUESTIONS
1. Define design of experiments.
Ans.: “The logical construction of the experiment in which the degree of uncertainty with
which the inference is drawn may be we defined”.
The subject matter of the design of experiments includes
1. Planning of the experiment.
2. Obtaining relevant information from it regarding the statistical hypothesis under
study.
3. Making a statistical analysis of the data.
2. Define Experimental Error.
Ans.: Let us suppose that a large heterogeneous field is divided into different plots of equal shape
and size and different treatments are applied to these plots. If the yields from some of the
treatments are more than those of the others, the experimenter is faced with the problem of
deciding if the observed differences are really due to treatments effects (or) due to chance factors.
In field experimentation it is a common experience that the fertility gradient of the soil does not
follow any systematic pattern but behaves in an erratic fashion. Experience tells us that even if

25. 25
the same treatment is used on all the plots, the yields would still vary due to the differences in soil
fertility. Such a variation from plot to plot, which is due to random factors beyond human
control, is called experimental error.
The variations are due to
i. The inherent variability in the experimental material to which treatments are applied.
ii. The lack of uniformity in the methodology of conducting the experiments
iii. Lack of representative ness of the sample to the population under study.
3. Define Treatments.
Ans.: Various objects of comparison in a comparative experiment are termed as treatments. For
e.g., in field experimentation, different fertilizers or different varieties of crop are the treatments.
4. Define Experimental Units.
Ans.: The smallest division of the experimental material to which we apply the treatments and on
which we make observations on the variable under study is termed as experimental unit. For e.g.,
in field experiments, ‘the plot’ of land is an experimental unit and in other experiments, the
experimental unit may be a patient in a hospital etc.
5. Define Blocks.
Ans.: In agricultural experiments, most of the times, we divide the whole experimental area into
relatively homogeneous subgroups or strata. These strata which are more uniform amongst
themselves are known as blocks.
6. Define Replication.
Ans.: Replication means the repetition of treatments under investigation.
7. Define Precision.
Ans.: The reciprocal of the variance of the mean is termed as the precision (or) the amount of
information of a design is known as precision. Thus, for an experiment, a treatment replicated ‘r’
times, the precision is given by
22
1
)(
1

r
r
xVar

8. Define efficiency of design.
Ans.: Consider the designs D1 & D2 with error variances per unit  1
2
&  2
2
, and replications r1
& r2 respectively. The efficiency D1 w.r.t D2 is defined as the ratio of the precisions of D1 & D2.
i.e., 2
22
2
11


r
r
E 
If E = 1 then D1& D2 are said to be equally efficient.
If E > 1 then D1 is more efficient than D2.

26. 26
If E < 1 then D1 is less efficient than D2.
9. Give any two merits of C.R.D.
Ans.:
1. CRD results in the maximum use of the experimental units since all the experimental
material can be used.
2. The design is very flexible i.e., any no. of treatments can be used and different treatments
can be applied an equal no. of times.
3. The statistical analysis remains simple even though some of the observations are rejected
or lost or missing.
4. It provides the maximum no. of degrees of freedom for the estimation of the error
variance.
LIST OF ESSAY QUESTIONS
1. Explain the principles of experimental designs
2. Explain CRD
3. Explain RBD
4. Explain LSD.
5. Explain (i) Treatments (ii) Experimental units (iii) Experimental error (iv) Replication
and (v) Precision
LIST OF SHORT ANSWER QUESTIONS
1. Define Experiment.
2. Define Treatments
3. Define Efficiency of design
4. Define blocks
5. Define Yield
6. Give any two merits of CRD.
7. Distinguish between CRD and RBD.