Agriculture Reference
In-Depth Information
-ratio
corresponding to treatment and replication is
greater than the corresponding table value at the
α
If
the
calculated
values
of
F
Analyze the data and find out the best variety of
paddy.
Rep-1
Rep-2
Rep-3
Rep-4
level of significance with (
t
1), (
t
1)
V2 28.8
V4 14.5
V3 34.9
V1 27.8
(
1) degrees of
freedom, respectively, then the corresponding
null hypothesis is rejected. Otherwise, we con-
clude that there exist no significant differences
among the treatments/replications with respect to
the particular character under consideration; all
treatments/replications are statistically at par.
Upon rejection of the null hypothesis, corres-
ponding LSD/CD is calculated using the follow-
ing formula:
Replication:
r
1) and (
r
1), (
t
1)(
r
V3 22.7
V6 23.5
V5 17.7
V5 16.2
V5 17.0
V3 36.8
V2 31.5
V2 30.6
V6 22.5
V2 40.0
V4 15.0
V4 16.2
V4 16
V5 15.4
V1 28.5
V6 22.3
V1 27.3
V1 38.5
V6 22.8
V3 27.7
Solution. From the layout of the experiment, it
is clear that the experiment has been laid out in
randomized block design with six varieties in
four replications.
So the model for RBD is given by
y
ij
¼ μ þ α
i
þ β
j
þ e
ij
,
i ¼
6,
j ¼
4
r
2ErMS
t
where,
y
ij
¼
LSD
α
=ð
CD
α
Þ¼
t
2
;ðt
1
Þðr
1
Þ
and
effect due to the
i
th variety in jth replicates
μ ¼
general effect
Treatment:
α
i
¼
additional effect due to the
i
th variety
β
j
¼
additional effect due to the
j
th replicate
r
2ErMS
r
e
ij
¼
errors associated with
i
th variety in
j
th rep-
LSD
α
ð
CD
α
Þ¼
t
2
;ðt
1
Þðr
1
Þ
;
licate and are i.i.d.
2
)
The hypotheses to be tested are
N
(0,
σ
where
t
is
the number of
treatments and
H
0
: α
1
¼ α
2
¼ α
3
¼ α
4
¼ α
5
¼ α
6
against
H
1
: α
i
's are not all equal and
H
0
: β
1
¼ β
2
¼ β
3
¼ β
4
against
H
1
:
t
2
;ðt
1
Þðr
1
Þ
is the table value of
t
at
α
level of
significance
and (
t
1)(
r
1) degrees of
freedom.
By comparing the mean difference among the
replication/treatment mean with corresponding
LSD/CD value, appropriate decisions are taken.
All
β
j
's are not all equal
:
0.05.
We shall analyze the data in the following
steps:
Make the following table from the given
information.
Let the level of significance be
α ¼
Example 10.11.
An experiment was conducted
with six varieties of paddy. The following table
gives the layout and corresponding yield (q/ha).
y
0
j
Mean
Replication
V1
V2
V3
V4
V5
V6
Total (
y
0
j
)
R1
27.30
28.80
22.70
16.00
17.00
22.50
134.30
22.38
R2
38.50
40.00
36.80
14.50
15.40
23.50
168.70
28.12
R3
28.50
31.50
34.90
15.00
17.70
22.80
150.40
25.07
R4
27.80
30.60
27.70
16.20
16.20
22.30
140.80
23.47
Total
y
i
0
122.10
130.90
122.10
61.70
66.30
91.10
594.20
Mean
y
i
0
30.53
32.73
30.53
15.43
16.58
22.78
