Agriculture Reference
In-Depth Information
To test the
x
n
1
)
and (
y
1
,
y
2
,
y
3
,
...
,
y
n
2
) two independent
random samples drawn from two normal
populations
H
0
:
μ
1
¼ μ
2
for (
x
1
,
x
2
,
x
3
,
...
,
that the varieties do not differ significantly with
respect to grain yield.
8.
Test for Equality of Two Population Means
with Equal but Unknown Variance
To test the
, respec-
tively, and under the condition of known
variances
and
2
1
2
2
N μ
1
; σ
N μ
2
; σ
H
0
:
μ
1
¼ μ
2
against
H
1
: μ
1
6¼ μ
2
for (
y
n
2
)
two independent random samples drawn
from two normal populations
x
1
,
x
2
,
x
3
,
...
,
x
n
1
) and (
y
1
,
y
2
,
y
3
,
...
,
2
2
σ
1
and
σ
2
, the test statistic is
and
2
1
N μ
1
; σ
x y
σ
, respectively, under the condition
τ ¼
s
;
N μ
2
; σ
2
2
1
2
2
n
2
n
1
þ
σ
σ
1
¼ σ
2
¼ σ
2
that
(unknown). The test statis-
tic under the given
H
0
is
when
τ
is in a standard normal variate.
x y
t ¼
r
1
n
1
þ
1
n
2
Example 9.7.
The following data are pertaining
to grain yield of two varieties of paddy under the
same management. If the grain yield follows a
normal distribution with known variances 36 and
64, respectively, for two varieties test whether
these two varieties differ significantly with
respect to grain yield (q/ha).
s
n
1
+
n
2
with (
2) degrees of freedom where
2
x
þðn
2
1
Þs
2
y
ðn
1
1
Þs
s
2
¼
and
s
2
x
and
s
2
y
are the
n
1
þn
2
2
sample mean squares
for
two samples,
respectively.
If the calculated absolute value of
t
is greater
(Grain yield q/ha)
Variety A: 25, 30, 31, 35, 24, 14, 32, 25, 29, 31, 30
Variety B: 30, 32, 28, 36, 42, 18 ,16 , 20 , 22, 40
than the table value of
t
at upper
α
=
2
level of
significance and at (
2) d.f., then the
test is significant and the null hypothesis is
rejected; that means the two population means
are unequal.
Note
n
1
þ n
2
Solution.
Assumption: grain yield follows a
normal distribution. Population variances are
known. Let the level of significance be
: This test statistic is known as two-
sample
α ¼
0.05.
statistic. Before
performing the test we are to test first
H
0
: σ
t
statistic or Fisher's
t
μ
A
¼ μ
B
(under
known population variances) against alternative
hypothesis
So the null hypothesis is
H
0
:
1
2
by
-test as already discussed.
If it is accepted, then we perform
¼ σ
F
.
Und
er
t
h
e given condition, the test statistic is
τ ¼
H
1
: μ
A
6¼ μ
B
-test statis-
tic; otherwise not. In that situation we perform
the following test.
t
s
σ
AB
which follows as a standard
n
1
þ
σ
2
A
2
B
n
2
Example 9.8.
Given below are the two samples
about the vase life of two varieties of tuberose. Is it
possible to draw inference that the vase life of
variety A is more than that of the variety B, assum-
ing that vase life behaves like a normal population?
normal variate.
From the samp
le
observations,
w
e have
n
A
¼
11
n
B
¼
A ¼
:
B ¼
:
and
10 and
27
82 and
28
4
27
:
82
28
:
4
58
3
0
:
¼
58
9
0
:
τ ¼
q
36
¼
p
p
Sample
size
Sample
mean
Sample mean
square
:
277
þ
6
:
4
:
677
64
10
11
þ
36 h
2
Variety A
7
72 h
¼
:
0
58
109
¼
0
:
186
:
25 h
2
Variety B
8
64 h
3
:
We know that at
α ¼
0.05,
the value of
Solution. H
0
: Both the varieties of tuberose
have the same vase life, that is,
τ
α=
2
¼
96. So we
cannot reject the null hypothesis. We conclude
1
:
96, as the calculated
jj<
1
:
H
0
:
μ
1
¼ μ
2
against the alternative hypothesis,
H
1
: Variety
