Agriculture Reference
In-Depth Information
Table 10.2
ANOVA table for one-way classification of
Variety 1
Variety 2
Variety 3
Variety 4
data
Sources of
variation (SOV)
12.5
13.2
11.5
14.5
13.4
13.2
11.6
14.2
d.f.
SS
MS
F
-ratio
12.6
13.4
11.7
14.3
Class/group
k
1
CSS MSC MSC/
ErMS
12.8
12.9
11.8
14.2
12.9
13.7
12.2
14.6
Error
n k
ErSS
ErMS
12
13.8
11
15.2
Total
n
1
TSS
12.2
11.9
15.6
13
15.5
12.4
e
ij
are independently and normally
distributed with zero mean and common variance
σ
Since
Solution. We are to test whether all the varieties
are equal or not w.r.t. the panicle length, that is,
to test the null hypothesis
2
, the test statistic under
H
0
is given by
CSS
σ
1
k
ErSS
σ
1
n k
CMS
ErMS
;
F ¼
¼
2
1
2
H
0
: α
1
¼ α
2
¼ α
3
¼ α
4
¼
0 against the
H
1
: α
0
s are not all equal,
where
α
i
which has an
F
distribution with (
k
1) and
is the effect of
i
th
ði ¼
1,2,3,4
Þ
variety
:
(
n k
) d.f.
If
the calculated value of
F >
F
α;k
1
;nk
, then
H
0
is rejected; otherwise, it is
This is a fixed effect model and can be
written as
accepted.
So we have Table
10.2
as the following
ANOVA table.
For all practical purposes, the various sums of
squares are calculated using the following short-
cut formulae:
y
ij
¼ μ þ α
i
þ e
ij
:
From the above information, we calculate the
following quantities:
Variety Variety 1 Variety 2 Variety 3 Variety 4
Total(
y
i
0
) 113.8
X
k
n
i
80.2
97.2
102.6
G ¼
Grand total
¼
1
y
ij
;
i¼
1
j¼
GT
¼
12
:
5
þ
13
:
4
þ
12
:
6
þþ
15
:
2
þ
15
:
6
;
where
n ¼
X
k
i¼
2
¼
G
2
Correction factor
ð
CF
Þ¼ny
1
n
i
;
þ
15
:
5
¼
393
:
8
n
GT
2
8
2
CF
¼
= ¼
393
:
= ¼
30
5169
:
281
Total sum of squares
ð
TSS
Þ
¼
X
X
2
ðy
ij
yÞ
5
2
4
2
6
2
2
2
TSS
¼
12
:
þ
13
:
þ
12
:
þþ
15
:
X
i
X
j
X
X
6
2
5
2
þ
15
:
þ
15
:
CF
¼
43
:
9387
2
ij
ny
2
2
ij
CF
¼
y
¼
y
80
2
9
20
2
6
þ
2
113
:
80
:
97
:
2
i
j
i
j
ð
Þ¼
þ
SS Variety
Class sum of squares
ð
CSS
Þ
¼
X
X
X
8
2
i
¼
2
2
i
ny
2
n
i
α
n
i
ðy
i
yÞ
n
i
y
¼
2
118
:
1
þ
CF
¼
26
:
46597
i
i
0
@
P
1
A
7
2
y
ij
X
X
k
y
2
i
0
n
i
CF
ErSS
2686
ANOVA for one-way analysis of variance
¼
TSS
SS Variety
ð
Þ¼
5
:
j
¼
n
i
CF
¼
n
i
i
i¼
1
where
y
i0
¼
X
j
y
ij
is the sum of observations
Source of variation
d.f. SS
MS
F
Variety (between)
3
26.4659
8.8219
13.1274
for
i
th class
;
Error (within)
26
17.4728
0.6720
Error sum of squares
ð
ErSS
Þ¼
TSS
CSS
:
;
Total
29
43.9387
Example 10.1.
The following figures give pani-
cle lengths (cm) of 30 hills of 4 local cultivars of
rice. Test (1) whether the panicle lengths of all
the four varieties are equal or not and (2) the
variety having the best panicle length.
2.975, that is,
F
cal
> F
tab.
So the test is significant at 5% level
of significance and we reject the null hypothesis.
Thus, we can conclude that the panicle lengths of
all the four varieties are not equal.
The table value of
F
0.05;3,26
¼
