Agriculture Reference
In-Depth Information
A has more vase life than variety
, that is,
H
1
:
μ
1
> μ
2
. Let us select the level of significance,
α ¼
B
¼
ðn
1
Þs
2
x
þðn
2
Þs
2
y
1
1
2
s
;
n
1
þ n
2
2
0.05. According to
H
1
, the test is a one-sided
2
2
2
test. We assume that
.
The test statistic, under the given null hypothesis
and unknown variance but equal, is
σ
A
¼ σ
B
¼ σ
ð
unknown
Þ
2
x
2
y
s
and
s
by the sample mean squares.
F ¼
s
2
x
s
2
2
B
First we test
H
0
: σ
A
¼ σ
by
with
2
y
2
2
ðn
1
1
; n
2
1
Þ
d
:
f
:
against the
H
1
: σ
A
6¼ σ
B
:
x y
t ¼
r
1
n
1
þ
;
Thus,
1.44 with (6,7) d.f.
From the table, we have
F ¼
36/25
¼
1
n
2
F
0.025;6,7
¼
5.12 and
s
F
0
:
975
;
6
;
7
¼
F
0
:
025
;
7
;
6
¼
=
:
¼
:
:
1
15
70
0
18
Since
2
2
B
with (
are the sample
mean vase life of two varieties and
n
1
+
n
2
2) d.f. where
x; y
0.18
<
cal
F <
5.12,
H
0
: σ
A
¼ σ
cannot be
2
is the
composite sample mean square and given by
rejected. So we can perform the
t
-test.
s
2
2
2
6
2
5
2
¼
ðn
1
1
Þs
1
þðn
2
1
Þs
¼
ð
7
1
Þ
þð
8
1
Þ
216
þ
175
2
s
¼
¼
391
=
13
¼
30
:
07
:
n
1
þ n
2
2
7
þ
8
2
13
64
72
8
s
s
t ¼
¼
¼
2
:
82
:
1
7
þ
1
8
15
56
30
:
07
30
:
07
From the table, we have
t
0.05,13
¼
1.771.
where
t
1
and
t
2
are the table values of
t
distri-
Since cal
1.771, the test is significant and
we reject the null hypothesis, that is, we accept
μ
1
> μ
2
. That means the vase life of variety A is
more than that of the vase life of variety B.
9.
t >
bution at (
1) degrees of
freedom, respectively, with upper
n
1
1) and (
n
2
level of
significance for acceptance or rejection of
α
H
0
.
Test for Equality of Two Population Means
with Unequal and Unknown Variances
The problem for test of significance of equal-
ity of two population means under unknown
and
Example 9.9.
Given below are the two samples
about the body weight (kg) of two breeds of cows.
Is it possible to draw inference that the body weight
of breed A is greater than that of breed B, assuming
that bodyweight behaves like a normal population?
unequal
population
variances
(i.e.,
2
2
σ
1
6¼ σ
2
) is known as
Fisher-Berhans prob-
Sample
size
Sample
mean
Sample mean
square
lem
. Under the null hypothesis
H
0
:
μ
1
¼ μ
2
Variety
against
H
1
: μ
1
>μ
2
, the test statistic
Breed A
7
185
100
Breed B
8
175
64
x y
s
t ¼
s
Solution.
We are to test the null hypothesis
H
0
:
μ
A
¼ μ
B
(under unknown and unequal popula-
tion variances) against alternative hypothesis
H
1
: μ
A
>μ
B
.
Let the level of significance
2
1
2
2
n
2
n
1
þ
s
does not follow
t
distribution, and as such
0.05.
Under the given conditions we apply
Cochran's approximation to Fisher-Berhans
problem. Thus, the test statistic is given by
α ¼
ordinary
-table will not be sufficient for com-
paring. According to Cochran's approxima-
tion, the calculated value of above
t
t
statistic
is compared with
x y
s
t ¼
s
;
2
2
t
¼
t
1
s
1
=n
1
þ t
2
s
2
=n
2
1
2
n
1
þ
s
;
s
=n þ s
=n
2
n
2
