Agriculture Reference
In-Depth Information
Now the Bartlett test in this case reduces to a chi-
square (
number of degrees of freedom:
2
) test with
n
χ
−
1 degrees of freedom, where:
2
n
χ
1
=
/
M
C
−
2
2
σ
σ
df
ln
1/df
Consider
this
simple
example
involving
four
9
0.909
−
0.095
0.1111
2
) all based on 5 degrees of freedom:
σ
variances (
7
0.497
−
0.699
0.1429
9
0.076
−
2.577
0.1111
7
0.103
−
2.273
0.1429
2
2
σ
σ
d.f.
ln
5
0.146
−
1.942
0.2000
5
178
5.182
Total 37
0.7080
5
60
4.094
5
98
4.585
5
68
4.202
2
=
σ
/
=
/
=
S
df
df
13.79
37
0.3727
Total
404
18.081
(
)
(
)
=
(
)(
−
)
=−
df
ln
S
37
0.9870
36.519
2
M
=
(
df
)
ln
(
S
)
−
df ln
σ
=−
36.519
−
(
−
54.472
)
=
17.96
S
=
100.9;
ln
(
S
)
=
4.614
=
+[
/(
)(
)
]
(
−
)
C
1
1
3
4
0.7080
0.0270
=
M
=
(
5
)
[
(
4
)(
4.614
)
−
18.081
]=
1.88,
with 3d.f .
1.057
2
χ
4df
=
/
=
/
M
C
17.96
1.057
=
+
(
)/
[
(
)(
)(
)
]=
C
1
5
3
4
5
1.083
16.99
∗∗∗
,
=
with 3d.f .
2
χ
3df
=
/
=
1.88
1.083
1.74ns
As this value of chi-square exceed the value from sta-
tistical tables with three degrees of freedom we can say
that the five error variances show significant (
p
2
value in this case is smaller that the
corresponding value from tables with 3 degrees of free-
dom, we would say that the four variances were
not
significantly different
at the 5% level.
Consider now the case where each of the variances
under test is based on different degrees of freedom. Now
the Bartlett test is based on:
χ
As the
<
0.01)
heterogeneity
.
It should be noted that the Bartlett test is
over
sensitive
to deviations and it is usual under practical situations to
only consider heterogeneity of variance occurring where
we have significant at the 99.9% level, or higher.
2
M
(
df
)
ln
(
S
)
−
df ln
σ
×
Detecting significant treatment
environment
interactions
The most common method, by far, of detecting geno-
type
where
S
is equal to:
2
S
df
[
df .
σ
]
/(
df
)
environment interactions is to carry out an
appropriate analysis of variance.
Various methods have been proposed for the statis-
tical analysis of interactions in general and genotype
×
and
+{
(
)/
[
(
−
)
]}
[
(
/
df
i
)
−
/(
)
]
C
1
1
3
n
1
.
1
1
df
×
environment interactions in particular. Consider first a
simple example where a number of genotypes are each
grown in a number of different environments. Vari-
ance components can be used to separate the effects
of genotypes, environments and their interaction by
equating the observed mean squares in the analysis of
and
2
n
χ
=
/
M
C
−
1
To further examine this consider the simple exam-
ple involving five variances, each based on a different
