Agriculture Reference
In-Depth Information
Thus, the analysis of covariance table can be
written as follows:
β
SOV
d.f.
SS(
x
)
SS(
y
)
SP(
x,y
)
SS(Reg.)
SS(
y
)
adj.
d.f.
adj.
Treatment
3
2.50
2365.50
44.75
Error
12
24.54
549.50
97.75
3.9898
390.0026
159.4974
11
2
y:x
¼
s
14
:
499
Treatment
+ Error
15
27.00
2915.00
53.00
1.9630
104.0370
2810.963
14
Treatment
(adjusted)
2651.4656
3
4.84; hence,
the test is significant, and the null hypothesis is
rejected.
The table value of
F
0.05;1,11
¼
For testing
H
01
: β ¼
0
;
we have test statistic
SS
ð
Reg
:Þ=
1
SS
ðyÞ
adj
:
=
11
¼
390
:
0026
=
1
159
:
4974
=
11
F ¼
¼
26
:
40 with
ð
1,11
Þ
d
:
f
:
For testing
H
02
: α
1
¼ α
2
¼ α
3
¼ α
4
¼
0
;
Tr
:
SS
ðyÞ
adj:
=
3
2651
:
4656
=
3
we have the test statistic
F ¼
11
¼
11
¼
60
:
95 with
ð
3,11
Þ
d
:
f
:
SS
ðyÞ
adj
:
=
159
:
4974
=
3.59; hence,
the test is significant, and the null hypothesis
is rejected. Thus, we can infer that the gain in
body weight due to different feeds varies and
also depends on the initial body weight of the
calves.
The table value of
F
0.05;3,11
¼
As both null hypotheses are rejected, we are
to work out the adjusted treatment means
and also the estimated adjusted variances for
different pairs of adjusted treatment means. We
have
β ¼
3
:
9898
; x ¼
10
:
75
; E
xx
¼
24
:
5
;
and
2
s
y:x
¼
:
We make the following table:
14
:
4998
2
ð
Þ
x
i
x
j
V y
0
i
y
0
j
¼ s
2
y:x
n
i
þ
1
n
j
þ
1
E
xx
2
ð
Þ
x
i
x
j
2
y:x
1
n
i
þ
1
n
j
þ
s
y
0
i
y
0
j
y
0
i
¼ y
i
β x
i
x
E
xx
x
i
y
i
x
i
x
ð
Þ
y
0
1
y
0
2
11.00
126.50
0.25
127.4975
7.2869
y
0
1
y
0
3
11.25
149.75
0.50
151.7449
7.3979
y
0
1
y
0
4
10.50
116.25
0.25
115.2526
7.5828
y
0
2
y
0
3
10.25
132.50
0.50
130.5051
7.5828
y
0
2
y
0
4
7.8417
y
0
3
y
0
4
7.2869
