Agriculture Reference
In-Depth Information
calculate the critical difference (least significant
difference) value, which is given as
ANOVA table
SOV
d.f.
SS
MS
F
r
2ErMS
m
Feed
4
2.498
0.625
0.123
CD
=
LSD
¼
t
0
:
025
;
err
:
df
:
Breed
3
3324.934
1108.311
218.623
Error
12
60.834
5.069
r
2
5
:
07
Total
19
3388.27
¼
t
0
:
05
;
12
:
5
r
2
5
:
07
Let the level of significance be
α ¼
0.05.
¼
2
:
179
¼
3
:
103
:
5
3.26, that is,
F
cal
< F
tab
. Therefore, the test is nonsignificant
and we cannot reject the null hypothesis
The table value of
F
0.05;4,12
¼
Arrangement of the breeds according to their
milk production is given below:
H
01
.
We conclude that there exists no significant dif-
ference among the effects of feeds with respect to
milk production.
Next, let us check whether there exists any
significant difference between milk production
and different breeds of cows. The table value of
F
0.05;3,12
¼
B
10
C
46
D
27
6
) C > D > A > B
21
:
96
:
3
:
The difference between pairs of means is
greater than the CD/LSD value. So we conclude
that all the breeds are significantly different from
each other and that breed C has the highest milk
production. Hence, breed C is the best breed with
respect to milk production per day.
Using MS Excel program, in the following
slides, the above calculation has been presented:
3.49, that is,
F
cal
> F
tab
. Therefore,
the test
is significant and we reject
the null
hypothesis
H
02
, meaning there exists a significant
difference among the breeds of cows w.r.t and
milk production. Now our task is to find out
which breed is significantly different from other
and which breed is the best or worst with respect
to milk production. To compare the breeds we
Step 1: Select “Anova: Two-Factor Without
Replication,” as shown below.
Slide 10.4: Arrangement of two-way classified data in MS Excel work sheet
