Agriculture Reference
In-Depth Information
Under
the given circumstances,
the test
Example
The following figures are
pertaining to lactation period (in months) for
two random samples of cows, drawn from two
normal populations fed with the same feed. Find
whether the variability in lactation period of
both the populations can be taken equal or not.
9.6.
statistic will be
n
1
i¼
1
ðx
i
μ
1
Þ
2
=n
1
F ¼
n
2
2
1
ðy
i
μ
1
Þ
=n
2
i¼
Lactation period (month)
Sample 1 4.5 3 5 7 6.5 5.5 6 8 7.5 4
Sample 2 6 7 6.5 5.5 7.5 6.5 8 10 8.5 9
with
n
2
d.f. the given alternative Under
hypothesis, the test is a two-tailed test, given that
n
1
¼
n
1
and
26 kg.
From the given information, we have
P
6
i¼
1
ðx
i
μ
1
Þ
6,
n
2
¼
7,
μ
1
¼
24 kg, and
μ
2
¼
2
Solution.
Given that (1) the populations are nor-
mal with unknown means.
(2) Sample sizes in each sample is 10, that is,
n
1
¼
10,
n
2
¼
10
To test
=n
1
¼
26
X
7
2
1
ðy
i
μ
1
Þ
=n
2
¼
22
:
57
; F ¼
1
:
152
:
i¼
2
.
Under the given conditions, the test statistic is
F ¼
1
2
against
1
H
0
:
σ
¼ σ
H
1
:
σ
6¼ σ
From the table, we have
F
0.025;6,7
¼
5.99
2
1
s
1
and
2
are the sample mean
where
s
s
and
F
0
:
975
;
6
;
7
¼
1
F
0
:
025
;
7
;
6
¼
16
=
:
98
¼
0
:
1754
:
s
2
2
Since 0.1754
5.12, so the null
hypothesis cannot be rejected, that means we
can conclude that both populations have the
same variance.
6.
<
cal
F <
squares.
Let the level of significance be
α ¼
0.05.
We have
Test of Equality of Two Population Variances
from Two Normal Populations with Unknown
Population Means
Let two independent random samples be
drawn from two normal populations (
2
1
X
10
1
n
1
2
s
1
¼
x
1
i
x
1
and
i¼
1
2
1
X
10
1
n
2
2
s
2
¼
x
2
i
x
2
and
x
1
,
x
2
,
i¼
1
x
3
,
...
,
x
n
1
) and (
y
1
,
y
2
,
y
3
,
...
,
y
n
2
). To test
H
0
:
n
1
X
n
2
X
10
i¼
1
ðx
1
i
Þ; x
2
¼
10
i¼
1
ðx
2
i
Þ:
1
1
2
2
σ
2
under the unknown means condition,
we have the test statistic
1
¼ σ
x
1
¼
2
1
S
F ¼
with (
n
1
1,
S
2
2
2
are the sample
mean square of the respective populations.
(a) When we are going to test
1
n
2
1) d.f. where
S
&S
From the given data, we have
1
2
H
0
: σ
¼ σ
x
1
¼
5
:
70
; x
2
¼
7
:
45
against
the
alternative
hypothesis
1
2
,wereject
2
1
2
2
H
1
: σ
6¼ σ
H
0
if cal
F >
tab
and
s
¼
2
:
638 and
s
¼
2
:
025
:
F
α=
2
;
ðn
1
; n
2
Þ
F < F
ð
1
α=
2
Þ;
1
1
or cal
Þ:
(b) When we are going to test
ðn
1
1
; n
2
1
2
1
s
F ¼
s
So,
with (9,9) d.f.
From the table, we have
2
¼
1
:
302
:
2
1
¼ σ
2
H
0
: σ
F
0.025;9,9
¼
4.03
against
the
alternative
hypothesis
and
:
Since 0.2481
<
cal
F <
4.03, so the test is non-
significant and the null hypothesis cannot be
rejected, that means we can conclude that both
populations have the same variance.
7.
F
0
:
975
;
9
;
9
¼
1
F
0
:
025
;
9
;
9
¼
14
=
:
03
¼
0
:
2481
2
2
H
1
: σ
1
> σ
2
, we reject
H
0
if cal
F >
tab
F
α=
2
;ðn
1
1
;n
2
1
Þ
.
(c) When we are going to test
1
¼ σ
2
H
0
: σ
against
the
alternative
hypothesis
2
2
H
1
: σ
1
< σ
2
, we reject
the
H
0
if cal
Test for Equality of Two Population Means
with Known Variances
F < F
ð
1
α=
2
Þ;ðn
1
1
;n
2
1
Þ
:
