Agriculture Reference
In-Depth Information
these two values. So the null hypothesis cannot
be rejected. That means we can conclude that the
variance of no. of flowers per plant in rose can
be taken as 6.
4.
we can accept that the variance of number of
fruits per tree may be 200.
5.
Test of Equality of Two Population Variances
from Two Normal Populations with Known
Population Means
Let two independent random samples be
drawn from two normal populations (
Test of Significance for Hypothetical Popula-
tion Variance with Unknown Population
Mean from a Normal Population
If
x
1
,
x
2
,
x
n
be a random sample drawn
from a normal population with unknown
mean
x
1
,
x
2
,
x
3
,
...
,
x
3
,
...
,
x
n
1
) and (
y
1
,
y
2
,
y
3
,
...
,
y
n
2
). To test
H
0
:
2
under the condition, we have the of
known means test statistic
1
¼ σ
σ
2
2
0
μ
and variance
σ
to test
H
0
: σ
¼ σ
2
0
where
σ
is any specified value for the popu-
n
1
i¼
1
ðx
i
μ
1
Þ
lation
variance. Against
the
alternative
2
=n
1
hypotheses
H
1
:
ð
1
Þ σ
2
6¼ σ
2
0
; ð
2
Þ σ
2
> σ
2
0
;
and
F ¼
n
2
i¼
1
ðy
i
μ
2
Þ
2
2
ð
3
Þ σ
< σ
0
, the test statistic,
2
=n
2
P
n
i¼
1
ðx
i
xÞ
2
with
n
2
d.f.
(a) When we are going to test
n
1
and
2
χ
ðn
1
Þ
¼
2
2
2
H
0
: σ
1
¼ σ
2
0
σ
against
the
alternative
hypothesis
2
2
H
1
: σ
1
6¼ σ
2
, we reject
H
0
if cal
F >
tab
with (
n
1) degrees of freedom where
x
is
the sample mean
P
i¼
1
x
i
. The decision
rule is as usual for the test mentioned in 3.
F < F
ð
1
α=
2
Þ;ðn
1
;n
2
Þ
:
(b) When we are going to test
F
α=
2
;ðn
1
;n
2
Þ
or cal
1
n
¼
2
2
2
H
0
: σ
1
¼ σ
against
the
alternative
hypothesis
2
2
H
1
: σ
1
> σ
2
, we reject
H
0
if cal
F >
tab
Example 9.4.
A random sample of 25 trees of
jackfruit gives average fruits per plant as 40 with
variance 220 from a normal population. Test
whether the variance of no. of fruits per plant of
jackfruit can be taken as 200 given that the no. of
fruits per plant follows a normal distribution.
F
α=
2
;ðn
1
;n
2
Þ
.
(c) When we are going to test
2
2
2
H
0
: σ
1
¼ σ
against
the
alternative
hypothesis
2
2
H
1
: σ
1
< σ
2
, we reject
the
H
0
if cal
F < F
ð
1
α=
2
Þ;ðn
1
;n
2
Þ
:
Solution.
Given that
(
1) sample size
n ¼
25.
Example 9.5.
The following figures are the
weights of two batches of primary students
assumed to follow a normal distribution with
(
(2) Sample mean
ð
x ¼
40
Þ
.
220.
(4) The sample has been drawn from a normal
population with unknown mean.
To test
(3) Sample variance
¼
with respective mean
weights of 24 and 26 kg, respectively. Can we
assume from the data that both the populations
have the same variance?
and
2
1
2
2
N μ
1
; σ
N μ
2
; σ
2
2
200,
the test is a both-sided test, and under the null
hypothesis, the test statistic is
H
0
:
σ
¼
200 against
H
1
:
σ
6¼
Yield in t/ha
Batch 1: 16,18,20,22,24,30
Batch 2: 19,22,23,27,29,31,33
P
n
i¼
1
x
i
x
2
ð
Þ
2
¼
nS
25
220
200
¼
2
χ
24
¼
2
¼
27
:
5
;
σ
2
σ
2
is the sample variance.
From the table, we have
Solution.
Null hypothesis
H
0
: σ
1
¼ σ
2
against
where
S
2
2
2
χ
¼
9.89
the alternative hypothesis
2
, given
that the populations are normal with mean
24 t/ha and 26 t/ha, respectively.
H
1
: σ
1
6¼ σ
0.995,24
2
and
χ
¼
45.558.
Since
9.89
<
cal
0.005,24
2
χ
<
45.56,
H
0
cannot be rejected. That means
