Agriculture Reference
In-Depth Information
for a rectangular distribution. We make the
following table:
or not. Actually in the median test, the null
hypothesis tested is that whether two sets of
scores differ significantly between them or
not. Generally it is used to test whether there
exists any significant difference between the
experimental group and the control group or
not. If the two groups have been drawn at
random from the same population, it is quite
obvious that half of the frequencies will lie
above and below the median. The whole pro-
cess is given stepwise:
(a) Arrange the ranks/scores of both the
group units taken together.
(b) A common median is worked out.
(c) Scores of each group are then divided into
two subgroups: (a) those above the com-
mon median and (b) those below the com-
mon median; generally the ranks equal to
the medians are ignored.
(d) These frequencies are then put into 2
x
F
0
(
x
)
F
n
(
x
)
j
F
n
ðxÞF
0
ðxÞ
j
0.089
0.089
0.1
0.011
0.217
0.217
0.2
0.017
0.278
0.278
0.3
0.022
0.358
0.358
0.4
0.042
0.394
0.394
0.5
0.106
0.404
0.404
0.6
0.196
0.486
0.486
0.7
0.214
0.524
0.524
0.8
0.276
0.572
0.572
0.9
0.328
0.942
0.942
1.0
0.058
α ¼
Let the level of significance
0.05.
From the table we get for
n ¼
10 the critical
value of K-S statistic
D
n
at 5% level of signifi-
cance is 0.409. Thus, the calculated value of
D
n
¼
Sup
x
½
j
F
n
ðxÞF
0
ðxÞ
j
¼
0
:
328
<
the table
2
value 0.409, so we cannot reject the null hypoth-
esis. That means we conclude that the given
sample is from the rectangular parent distribu-
tion. This example is due to Sahu (2007).
4.
contingency table.
2
value is obtained using the usual
norms.
(f) The decision rule is the same as that was
for
χ
(e)
χ
The Median Test
Synonymous to that of the parametric test to
test whether two samples have been taken
from the same population or not, the median
test is also used to test whether two groups
have been taken from the same population
2
test.
Example 9.34.
The following table gives the
scores of two groups of students. Test whether
the two groups could be taken as one or not.
Students
1
Group
2
3
4
5
6
7
8
9
10
11
12
Group 1
9
8
12
6
8
9
5
10
9
7
11
10
Group2
8
11
4
9
7
3
10
9
5
6
12
8
Solution.
Under the given condition, we can go
for median test to test the equality of two groups
with respect to their scores.
Taking both the groups as one can have the
following arrangement:
34556677888
89
99990001122
From the above arrangements, we have the
median score as the average of the 12th and 13th
ordered observation, that is, (8 + 9)/2
Below
Above
Group1
(a) 5
(b) 7
¼
8.5.
Group2
(c) 7
(d) 5
Now a 2
2 contingency table is prepared
