Agriculture Reference
In-Depth Information
the alternative hypothesis remain the same,
that is,
which is a standard normal variate with mean
zero and variance unity.
For acceptance or rejection of
H
0
, we have to
compare the calculated value of
H
0
:
Vase life of two types of flowers are same
:
with the
appropriate table value keeping in view the
alternative hypotheses.
τ
H
1
:
Vase life of type A
>
vase life of type B
;
that is
; H
0
: μ
1
¼ μ
2
H
1
: μ
1
>μ
2
:
Example 9.19.
To test the variability in vase life
of a cut flower, a sample of 100 cut flowers gives
s.d. 8.5 h. Can we conclude that the vase life of
the cut flower has the variability of 65 h
2
?
Let the level of significance be
0.05;
being a one-sided test, the critical value is
1.645 for standard normal variate
α ¼
τ
. The test
statistic is
Solution.
Given that the sample is large of size
(
x
1
x
2
s
n
1
2
) 100. The population variance is assumed to
be 65 h
2
, and s.d. of the sample is 8.5 h. Under
the given condition, the null hypothesis is the
population s.d.
n
τ ¼
s
Nð
0
;
1
Þ:
n
1
þ
s
n
2
2
n
2
6
p
h.
¼
H
1
: σ
0
Thus,
That is,
06.
The test statistic for the above null hypothesis is
H
0
: σ ¼
8
:
06 against
6¼
8
:
72
68
64
100
þ
4
145
100
4
1
τ ¼
r
¼
r ¼
p
τ ¼
s
n
σ
0
8
:
06
65
5
8
:
0
:
44
81
100
:
45
r ¼
σ
r
¼
57
¼
0
:
77
:
0
:
2
0
4
ð
Þ
2
100
2
n
¼
204
¼
3
:
322
:
1
:
Let the level of significance be
α ¼
0.05, and
Since the cal
645, so the test is
significant, we reject the null hypothesis
and accept the alternative hypothesis
τ >
1
:
the corresponding critical value of
(standard
normal variate) is 1.96. So the calculated value
of
τ
H
1
:
vase life of type A is more than that of type
B.
(1.96). The test is
nonsignificant and the null hypothesis cannot be
rejected. Hence, the population variance of vase
life of the cut flower under consideration can
be taken as 65 h
2
.
4.
jj<
tabulated value of
τ
3.
Test for Significance of Specified Population
Standard Deviation
Let (
Test of Significant Difference Between Two
Standard Deviations
Let us draw two independent samples
n
drawn randomly from a population with vari-
ance
x
1
,
x
2
,
...
,
x
n
) be a large sample of size
ð
x
11
;
2
. The sampling distribution of the sam-
ple s.d.
σ
x
12
; ...; x
1
n
1
Þ
and
ð
x
21
; x
2
2
; ...; x
2
n
2
Þ
of sizes
s
n
follows an approximately normal
distribution with mean
s
n
1
2
n
1
and
n
2
with means
x
1
; x
2
and variance
;
EsðÞ¼σ
and vari-
s
n
2
2
σ
1
2
from two populations with variances
2
2
n
VsðÞ¼
σ
ance
as the sample size
n
tends to
σ
2
2
, respectively.
For a large sample, both
and
.
2
2
n
s
n
1
and
s
n
2
are
n !1; s
n
N σ;
σ
infinity. Thus, as
and
s
n
1
N σ
1
,
σ
1
2
distributed as
s
n
2
2
n
1
.
Our objective is to test
H
0
: σ ¼ σ
0
:
N σ
2
,
σ
2
2
2
n
2
To test the above null hypothesis, we have the
following approximate test statistic
Now,
Es
n
1
s
n
2
ð
Þ Eðs
n
1
ÞEðs
n
2
Þ¼σ
1
σ
2
and
τ ¼
s
n
σ
0
s
σ
r
σ
p
Vs
n
1
s
n
2
1
2
n
1
þ
σ
0
2
n
ð
s
n
1
s
n
2
Þ ¼
ð
Þ
¼
:
SE
2
2
n
2
