Agriculture Reference
In-Depth Information
Solution. Under the given condition, the null
hypothesis is H 0 : P 1 ¼ P 2 against the alternative
hypothesis
p 1 pð Þ
τ ¼
s
:
1
n 1 þ
1
n 2
^
^
P
1
P
H 1 : P 1 6¼ P 2 .
That means there exists no significant differ-
ence between the two proportions against the
existence of significant difference.
Let
Example 9.22.
From two large samples of 500
and 600 of electric bulbs, 30 and 25%, respec-
tively, are found to be defective. Can we con-
clude that the proportions of defective bulbs are
equal in both the lots?
the level of significance be
α ¼
0.05,
and the test
statistic for
the above null
hypothesis is
p 1 pð Þ
P ¼ n 1 p 1 þ n 2 p 2
500
0
:
3
þ
600
0
:
25
3
11
τ ¼
s
;
where
n 1 þ n 2 ¼
¼
500
þ
600
1
n 1 þ
1
n 2
1
ð Þ
3
11
0
:
3
0
:
25
05
0 : 02697 ¼
0
:
nτ ¼
s
¼
:
:
1
8539
8
11
1
500 þ
1
600
Since the calculated value of
jj
is less than
is distributed as
χ
2 with
k
d.f. as
n 1 ,
n 2 ,
n 3 ,
the tabulated value of
(1.96),wehaveto
accept the null hypothesis and conclude that
proportions of damaged bulbs are equal in both
the lots.
7.
τ
...
,
n k !1
.
To test
H 0 : P 1 ¼ P 2 ¼ P 3 ¼¼P k ¼
P ð
known
Þ
against all alternatives,
the
test statistic is
2
χ
Test
The
(
) 2
2 test is one of the most important
and widely used tests in testing of hypo-
thesis.Itisusedbothinparametricandin
nonparametric tests. Some of the impor-
tant uses of
χ
X k
x i n i P
n i
2
χ
¼
p
with
k
d
:
f
:;
1
1
2
2
and if cal
χ
> χ
α;k ;
we reject
H o :
2 test are testing equality of
proportions, testing homogeneity or signifi-
cance of population variance, test for good-
ness of fit, tests for association between
attributes, etc.
(a)
χ
In practice
will be unknown. The unbi-
ased estimate of
P
P
is
P ¼ x 1 þ x 2 þþx k
n 1 þ n 2 þþnk :
Then the statistic
2
χ
Test for Equality of k
(
2)
Population
0
@
1
A
2
Proportions
Let
X k
x i n i P
n i
X k be independent ran-
dom variables with
X 1 ,
X 2 ,
X 3 ,
...
,
2
χ
¼
q
X i ~
B
(
n i ,
P i ),
i ¼
1, 2,
1
1
3,
...
,
k
;
k
2. The random variable
(
) 2
2 with (
is asymptotically
χ
k
1) d.f.
X k
X i n i P i
n i P i ð 1 P i Þ
2
2
If cal
χ
α; k 1 , we reject
H 0 : P 1 ¼
p
P 2 ¼ P 3 ¼¼P k ¼ P ð
unknown
Þ
.
1
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