Agriculture Reference
In-Depth Information
the claim that the pumps manufactured by the
company fill a 1,000-l water tank in 8 min.
2.
cut flowers he took 100 flower sticks from each
type and found that the average vase life of two
types of flowers sticks were 72 and 68 h with
8 and 9 h standard deviations, respectively.
Assuming that (1) both the types of flowers
have the same variability and (2) different
variabilities, test whether the vase life of type A
is better than type B or not.
Test of Significance Between Two Means
Two large samples
ð
x
11
; x
12
; ...; x
1
n
1
Þ
and
ð
x
21
; x
22
; ...; x
2
n
2
Þ
of
sizes
n
1
and
n
2
2
n
1
with means
x
1
and
x
2
and variances
s
and
2
n
2
s
are drawn independently at random from
two populations with means
μ
1
and
μ
2
and
Solution.
Given that:
σ
2
2
, respectively.
As two large samples are drawn indepen-
dently
σ
1
2
and
variances
Type
Mean
S.D.
Sample size
from two
population
so
x
1
and
.
A
72 h
8 h
100
N μ
1
;
σ
1
2
x
2
N μ
2
;
σ
2
2
B
68 h
9 h
100
n
1
n
2
So to test
H
0
: μ
1
¼ μ
2
, the test statistic is
the given condition σ
1
2
¼ σ
2
2
(a)
Under
¼
σ
H
0
:
vase life of two type of flowers are equal
against
2
ð
unknown
Þ
and the null hypothesis
x
1
xð Þ
σ
1
2
τ ¼
s
;
n
1
þ
σ
2
2
H
1
:
vase life of type A
>
vase life
n
2
of type B. So the test statistic under
H
0
is
x
1
x
2
σ
which follows the distribution of a standard
normal variate with mean zero and unit
variance.
When population variance are unknown,
then these are replaced by respective sample
variances
τ ¼
^
q
1
Nð
0
;
1
Þ;
1
n
2
n
1
þ
where
2
n
2
n
2
. If both the population
variances are equal and unknown, then esti-
mate of common population variance is
given by
s
1
and
s
^
¼
n
1
s
n
1
2
þ n
2
s
n
2
2
n
1
þ n
2
8
2
9
2
100
þ
100
σ
¼
100
þ
100
100
ð
64
þ
81
Þ
145
2
:
¼
¼
200
¼
n
1
s
n
1
2
þ n
2
s
n
2
2
n
1
þ n
2
2
σ
:
So,
68
145
2
72
4
1
σ
1
2
¼ σ
2
2
¼ σ
2
Thus, the test statistic under
τ ¼
s
¼
p
¼
3
:
322
:
(unknown) comes out to be
:
45
1
100
þ
1
100
x
1
x
2
τ ¼
s
1
n
1
þ
1
n
2
Let the level of significance be
0.05.
This is a one-tailed test. Since the cal
τ > τ
0
:
05
¼
α ¼
2
σ
1
:
645, the test is significant, we
reject
the null hypothesis and accept
the
which follows
N
(0, 1).
alternative hypothesis
Vase life of
type A is more than that of type B.
H
1
:
Example 9.18.
A flower exporter is to select
large quantity of cut flowers from two lots of
flowers type A and B. To test the vase life of
Under the given condition of σ
1
2
6¼ σ
2
2
and
(b)
both being unknown
. The null hypothesis and
