Agriculture Reference
In-Depth Information
2
χ
2
1.
χ
Test for Goodness of Fit when Pop-
ulation Is Completely Specified
The problem is to test
H
0
: P
1
¼ P
value of
is less than the tabulated value
0
of
4
, hence the test is nonsignificant and
the null hypothesis cannot be rejected. We
conclude that the results obtained in different
places are homogeneous in nature.
χ
:
05
;
1
;
0
0
0
P
2
¼P
2
; P
3
¼P
3
; ...; P
k
¼P
k
;
where
P
0
i
ði ¼
;
;
; ...kÞ
are specified
values. The test statistic under
1
2
3
H
0
is
(d)
χ
Test for Goodness of Fit
In biological, agricultural, and other
applied sciences, there are several laws;
verification of these theoretical laws is
taken up in practical field.
2
2
X
k
X
k
0
i
2
O
i
nP
ð
O
i
e
i
Þ
¼
nP
0
i
e
i
i¼
1
i¼
1
2
test for good-
ness of fit is a most widely used test. It is to
test the agreement between the theory and
practical. Now the agreement or disagree-
ment between the theoretical values
(expected) and observed values is being
judged through
χ
X
k
2
OðÞ
¼
i
n
nP
0
i¼
1
where
e
i
¼
expected frequency of the
0
i
th class
¼ nP
i
ði ¼
1
;
2
;
3
; ...; kÞ
and
which is
(in the limiting for as
2
test for goodness of fit.
For example, to test whether the different
types of frequencies are in agreement with
Mendel's respective theoretical
frequencies or not, different observed
frequencies are in agreement with the
expected frequencies in accordance with
different probability distributions or not.
Suppose a population is classified into
χ
2
n !1
)
distributed
as
χ
with
(
k
1) d.f. If
α
be the level of signifi-
cance,
then we reject
H
0
if cal
2
2
χ
χ
α;k
1
; otherwise we accept it.
Example 9.25.
Two different types of tomato
(red and yellow) varieties were crossed, and in
F
1
generation, 35, red; 40, yellow; and 70, mixed
(neither red nor yellow) types of plants are
obtained. Do the data agree with the theoretical
expectation of 1:2:1 ratio at 5% level of
significance?
k
mutually exclusive nominal like “male” or
“female” or may be interval of the domain
of some measured variable like height
groups. To test whether the cell
frequencies are in agreement with the
expected frequencies or not. In all these
cases,
Solution.
Total frequency 35 + 40 + 70
145.
The expected frequency for both red and yellow
145
¼
2
2
χ
known as frequency
χ
or
36, and mixed
14
4
4
1
¼
36
:
25
~
2
¼
72
:
5
~
73.
Pearsonian chi-square is most useful.
1
4
; P
2
¼
2
4
; P
3
¼
1
4
H
0
:
The observed frequencies support the theoretical frequencies
;
that is
; P
1
¼
against
1
4
; P
2
4
;
0
1
0
2
H
1
:
;
; P
6¼
6¼
The observed frequencies do not follow the theoretical frequencies
that is
1
4
:
0
3
P
6¼
H
0
the test statistic is
2
Under
From the table, we have
χ
0
:
05
;
2
¼
5
:
991. So
2
the calculated value of
χ
is less than the table
X
k
2
ð
O
i
e
i
Þ
2
χ
¼
with
k
1d
:
f
:
value of
2
. Hence, we cannot reject the null
hypothesis. That means the data agree with the
theoretical ratio.
2.
χ
e
i
i¼
1
2
2
2
ð
35
36
Þ
ð
70
73
Þ
ð
40
36
Þ
¼
þ
þ
χ
Test for Goodness of Fit When Some
Parameters of the Hypothetical Population
Are Unspecified
2
36
73
36
1
9
16
¼
36
þ
73
þ
36
¼
0
:
596 with 2 d
:
f
:
