Agriculture Reference
In-Depth Information
Table 9.6
Frequency distribution table for two-sample
On the other hand, if the fixed marginal
frequencies are moderately large, we can use
the
median test
2
χ
statistic for a 2
2 contingency table
Number of observations
using the formula
<θ
θ
Total
Sample 1
n
n
n
11
12
1
2
¼
ðn
11
n
22
n
12
n
21
Þ
N
n
1
n
2
n
1
n
2
:
Sample 2
n
21
n
22
n
2
2
χ
Total
n
1
n
2
N
If
n
1
,
n
2
,
n
1
, and
n
2
are small, we can get the
exact probability of the above
Example 9.41.
The following table gives the
distribution of nuts per palm for two different
samples of coconut. Test whether the samples
have been drawn from the same coconut popula-
tion or not.
table
with fixed
marginal frequencies as follows:
n
1
!n
2
!n
1
!n
2
!
n
11
!n
12
!n
21
!n
22
!N!
:
P ¼
Sample1
80
85
90
110
78
86
92
115
120
118
Sample2
110
82
89
114
122
128
130
126
128
Solution.
78, 80, 82, 85, 86, 89, 90, 110, 110,
114, 115, 118, 120, 122, 126, 128, 128, 130
From the arranged data, we get the median of
combined samples as 110 and we construct the
following table:
0.10216. As 0.10216
0.05, we cannot reject
the null hypothesis that the two distributions are
identical against both-sided alternatives that they
are not identical.
4.
>
Kolmogorov-Smirnov Two-Sample Test
In
2
test, we have come across with the test
procedure to test the homogeneity of two
distributions in the previous sections of this
chapter. But we know that
χ
Sample
<
110
110
Total
1
6
4
10
2
2
7
9
2
test is valid
under certain assumptions like a large sample
size. The parallel test to the above-mentioned
χ
χ
Total
8
11
19
The exact probability of getting likely distri-
bution of the above table is given by
2
tests which can also be used under small
sample conditions is Kolmogorov-Smirnov
two-sample test. Kolmogorov-Smirnov two-
sample test is the test for homogeneity of two
populations. We draw two random indepen-
dent samples (
n
1
!n
2
!n
1
!n
2
!
n
11
!n
12
!n
21
!n
22
!N!
¼
10
!
9
!
8
!
11
!
P ¼
!
¼
0
:
3
:
6
!
4
!
2
!
7
!
19
The probability of getting less likely distribu-
tion than the observed one is obtained from the
following table:
x
1
,
x
2
,
x
3
,
...
,
x
m
) and (
y
1
,
y
2
,
y
3
,
y
n
) from two continuous cumulative
distribution functions
...
,
, respectively.
The empirical distribution functions of the
variable are given by
F
and
G
Sample
<
112
112
Total
1
8
2
10
2
1
8
9
Total
9
10
19
F
m
ðxÞ¼
x < x
ð
1
Þ
0if
0.00438
Thus, the probability of getting the observed
table or more extreme table in one direction
is
Probability
¼
¼ i=m
x
ðiÞ
x < x
ðiþ
1
Þ
if
¼
x x
ðmÞ
and
1if
G
n
ðxÞ¼
x < y
ð
1
Þ
0if
0.30438.
By symmetry, we have the exact probability
of getting the observed distribution or less likely
¼
0.3 + 0.00438
¼
¼ i=m
y
ðiÞ
x < y
ðiþ
1
Þ
if
