Agriculture Reference
In-Depth Information
Calculate the following quantities:
2
n
¼
20
2
¼
G
594
:
¼
:
;
CF
14711
40167
6
4
X
Obs
2
3
2
8
2
2
2
3
2
TSS
¼
:
CF
¼
27
:
þ
28
:
þþ
16
:
þ
22
:
14711
:
40167
¼
16156
:
72
14711
:
40167
¼
1445
:
318
;
6
X
4
j¼
1
y
1
1
6
2
3
2
7
2
4
2
8
2
RSS
¼
0
j
CF
¼
134
:
þ
168
:
þ
150
:
þ
140
:
14711
:
40167
88940
:
98
¼
14711
:
40167
¼
112
:
095
;
6
4
X
6
i¼
1
y
1
1
4
2
i
1
2
9
2
1
2
7
2
3
2
1
2
TrSS
¼
CF
¼
122
:
þ
130
:
þ
122
:
þ
61
:
þ
66
:
þ
91
:
14711
:
40167
0
63453
:
42
¼
:
¼
:
;
14711
40167
1151
953
4
ErSS
¼
TSS
RSS
TrSS
¼
1445
:
318
112
:
095
1151
:
953
¼
181
:
27
;
r
2MSE
r
where TSS, RSS, TrSS, and ErSS are the total,
replication, treatment, and error sum of squares,
respectively.
Construct the ANOVA table as given below.
CD
0
:
05
ð
Variety
Þ¼
t
0
:
025
;
err
:
d
:
f
:
r
2
12
:
085
¼
t
0
:
025
;
15
4
r
2
12
:
085
4
¼
2
:
131
¼
5
:
238
:
ANOVA
SOV
d.f.
SS
MS
F
Arrange the varietal mean values in descending
order and compare the difference between any
two treatment mean differences with that of the
critical difference value. If the critical difference
value be greater than the difference of two varietal
means, then the treatments are statistically at par;
there exists no significant difference among the
means under comparison.
Replication
3
112.095
37.365
3.091935
Variety
5
1151.953
230.3907
19.06471
Error
15
181.270
12.08467
Total
23
1445.318
The table value of
F
0.05;3,15
¼
3.29 and
F
0.05;5,15
¼
2.90. Thus, we find that the test
corresponding to replication is not significant,
but the test corresponding to the effect of
varieties is significant. So the null hypothesis of
equality of replication effects cannot be rejected;
that means there is no significant difference
among the effects of the replications; they are
statistically at par. On the other hand, the null
hypothesis of equality of varietal effect is
rejected; that means there exist significant
differences among the varieties. So we are to
identify the varieties, which varieties are signifi-
cantly different from each other, and the best
variety.
Calculate critical difference value at
Variety
Mean yield (bushels/acre)
V2
32.725
V1
30.525
difference
<
CD(0.05)
V3
30.525
V6
22.775
V5
16.575
V4
15.425
difference
<
CD(0.05)
Variety 2 is the best variety having highest
yield, but variety 1 and variety 3 are also at par
with that of variety 2. Variety 6 is significantly
different from all other varieties. Variety 5 and
variety 4 are statistically at par and are the lowest
yielders among the varieties.
α ¼
0.05
using the following formula:
