Agriculture Reference
In-Depth Information
β
j
is the effect of
j
th level of variety
; j ¼
4
;
Doses
of N Variety 1 Variety 2 Variety 3 Variety 4
N1
13.0 17.0 15.5 20.5
13.4 17.5 15.6 20.8
13.5 17.4 15.5 20.4
N2
14.0 18.0 16.2 24.5
14.2 18.5 16.1 25.6
14.6 18.9 16.3 25.0
N3
25.0 22.0 18.5 32.0
24.8 22.5 18.6 32.5
24.3 22.3 18.7 32.0
Solution. The problem can be visualized as the
problem of two-way analysis of variance with
three observations per cell. The linear model is
γ
ij
is the interaction effect of
i
th level of nitrogen
and
j
th level of variety
:
We want to test the following null hypotheses:
H
01
: α
1
¼ α
2
¼ α
3
¼
0 against
H
11
: α
's are not
all equal,
H
02
: β
1
¼ β
2
¼ β
3
¼ β
4
¼
0 against
H
12
: β
's
are not all equal,
H
03
: γ
ij
's
¼
0 for all
i; j
against
H
13
: γ
ij
's are
not all equal
:
y
ijk
¼ μ þ α
i
þ β
j
þ γ
ij
þ e
ijk
;
where
α
i
Let the level of significance be
0.05.
Let us construct the following table of totals
α ¼
(
y
ij
):
is the effect of
i
th level of nitrogen
; i ¼
3
;
Nitrogen
Variety 1
Variety 2
Variety 3
Variety 4
Total (
y
i
00
)
Mean
N
1
39.90
51.90
46.60
61.70
200.10
16.675
N
2
42.80
55.40
48.60
75.10
221.90
18.492
N
3
74.10
66.80
55.80
96.50
293.20
24.433
Total(
y
0
j
0
)
156.800
174.100
151.000
233.300
715.200
Mean
17.422
19.344
16.778
25.922
19.867
Total number of observations
¼ mnl ¼
3
4
3
¼
36
¼ N
G ¼
13
:
0
þ
17
þ
15
:
5
þþ
22
:
3
þ
18
:
7
þ
32
: ¼
715
:
2
;
CF
¼
GT
2
= ¼
715
:
2
2
3
= ¼
14208
:
64,
0
2
17
2
5
2
3
2
7
2
32
2
TSS
¼
13
:
þ
þ
15
:
þþ
22
:
þ
18
:
þ
CF
¼
975
:
52
;
200
:
1
2
12
þ
221
:
9
2
12
þ
2
293
:
20
12
NSS
¼
CF
¼
395.182,
2
156
:
8
2
9
þ
174
:
10
2
9
151
:
00
9
VSS
¼
þ
CF
¼
472.131,
CF
NSS
VSS
¼
106
:
04,
SS
ð
N
V
Þ¼
1
=
339
:
9
2
þ
51
:
9
2
þ
46
:
6
2
þþ
66
:
80
2
þ
55
:
80
2
þ
96
:
5
2
ErSS
¼
TSS
NSS
VSS
SS N
V
ð
Þ
975
:
52
395
:
182
472
:
131
106
:
04
¼
2
:
17
:
