Agriculture Reference
In-Depth Information
if 4th, 5th, 6th, and 7th places are having
the same value, then each of these
identical values would be assigned ((4 +
5 + 6 + 7)/4
2
N
ða þ bÞða þ cÞðb þ dÞðc þ dÞ
ðad bcÞ
2
1
X
¼
24
24
) 5.5 rank).
(c) Find the sum of the ranks of both the
samples
¼
¼
12
¼
0
:
66
:
12
12
separately (say,
R
1
and
R
2
,
2 contingency table using the
formula, the calculated value of
From this 2
respectively, for 1st and 2nd sample).
(d) Calculate
2
χ
¼
0.66
<
the
U ¼ n
1
n
2
þ
n
1
ðn
1
þ
1
Þ
2
χ
2
¼
R
1
,
table value of
3.841 at 5% level of signifi-
cance. So we have no reason to reject the null
hypothesis that the two groups have come from
the same population, that is, the two groups do
not differ.
5.
n
2
are the sizes of the 1st and
2nd sample, respectively.
(e) If both
where
n
1
and
n
2
are sufficiently large
(
>
8), the sampling distribution of
U
can
be approximated to a normal distribution
and usual critical values could be used to
arrive at the conclusion. On the other
hand, if the sample sizes are small, then
the significance test can be worked out as
per the table values of the Wilcoxon's
unpaired distribution.
n
1
and
The Mann-Whitney U-Test
Analogous
to the parametric
t
-test,
the
Mann-Whitney
-test is aimed at determin-
ing whether two independent samples have
been drawn from the same population or not.
This test does not require any assumption
except that the continuity. Steps involved in
the process are given below:
(a) Rank the data of both the samples taking
them as one sample from low to high
(generally low ranking is given to low
value and so on).
(b) For tied ranks take the average of the
ranks for each of the tied values (e.g.,
U
Example 9.35.
The following table gives the
scores in mathematics of two groups of students.
Using Mann-Whitney
U
-test, can you say that
these two groups of students are the same at 5%
level of significance, that is, they have come
from the same population?
Students
1
Group
2
3
4
5
6
7
8
9
10
11
12
Group A
59
48
52
45
32
56
98
78
70
85
43
80
Group B
39
45
55
87
36
67
96
75
65
38
Solution.
Using the above information we get
the following table:
n
2
are greater than 8, so
normal approximation can be used with mean
μ
U
¼ n
1
n
2
=
Since both
n
1
and
ð
¼
Þ
2
60
and variance
Given that
n
1
¼
n
2
¼
12 and
10, so
U ¼ n
1
n
2
þ
n
1
ðn
1
þ
1
Þ
U
¼ n
1
n
2
þ
n
1
ðn
1
þ n
2
þ
1
Þ
2
R
1
σ
2
12
12
ð
12
þ
1
Þ
12
ð
12
þ
10
þ
1
Þ
¼
12
:
10
þ
143
:
5
¼
12
:
10
þ
¼
120
þ
23
2
12
¼
þ
:
¼
:
:
¼
σ
U
¼
:
:
120
78
143
5
54
5
143 and
11
958
