Agriculture Reference
In-Depth Information
other kind. Let
the number of
elements of one kind and
n
2
be the number
of element of other kind. That is,
n
1
¼
This formula is exact when the
n
is even
and the
is normally distributed as the
number of observations
r
n
1
might
n
increases. That
be the number of heads and
n
2
be the
number of tails, or n
1
might be the number
of pluses and
τ ¼
rEðrÞ
p
Var
means
~
Nð
0
;
1
Þ
and we can
ðrÞ
conclude accordingly.
n
2
might be the number of
minuses.
total number of observed
events, that is,
n ¼
Example 9.31.
In an admission counter, the boys
(B) and girls (G) were queued as follows:
B, G, B, G, B, B, B, G, G, B, G, B, B, G, B, B, B.
Test the hypothesis that the order of males and
females in the queue was random.
n
2
. To use the
one-sample run test, first we observe the
n
1
and
n ¼ n
1
+
n
2
events in the sequence in which
they occurred and determine the value of
r
, the number of runs.
Let
... x
n
be a sample drawn
from a single population. At first we find
the median of the sample and we denote
observations below the median by a minus
sign and observations above the median
by plus signs. We discard the value equal
to the median. Then we count the number
of runs (
x
1
,
x
2
,
x
3
,
Solution. H
0
: The order of boys and girls in the
queue was random against
H
1
: The order of boys and girls in the queue
was not random.
In this problem there are
n
1
¼
11 boys and
n
2
¼
6 girls. The data is
BGBGBBBGGBG
BBGBBB.
There are 11 runs in this series, that is,
r
) of plus and minus signs. If both
11.
Table
A.12
given in the
Appendix
shows that for
n
1
¼
r ¼
n
1
and
n
2
are equal to or less than 20,
then
Table
A.12
gives the critical value of
r
6, a random sample would be
expected to contain more than 4 runs but less
than 13. The observed
11and
n
2
¼
under
These are critical
values from the sampling distribution of
H
0
for
α ¼
0
:
05
:
r
< r ¼
11
< r ¼
13 fall
under
falls
between the critical values, we accept
H
0
. If the observed value of
r
in the region of acceptance for
α ¼
0
:
05. So we
H
0
.
accept
H
0
, that is, the order of boys and girls in
the queue was random.
For large
n
2
or both the number of
runs below and above, the sample median
value is a random variable with mean
EðrÞ¼
2
þ
n
1
and
Example 9.32a.
The following figures give the
production (million tons) of rice in certain state
of India. To test whether the production of rice
has changed randomly or followed a definite
pattern.
1 and variance Var
ðrÞ
¼
nðn
2
Þ
Þ
:
4
ðn
1
Year
1971
1972
1973
1974
1975
1976
1977
1978
1979
1980
Production
13.12
13
13.3
14.36
10.01
12.22
13.99
8.98
13.99
11.86
Year
1981
1982
1983
1984
1985
1986
1987
1988
1989
1990
Production
13.55
13.28
15.8
12.69
14.45
14.18
13.03
13.81
13.48
13.1
Year
1991
1992
1993
1994
1995
1996
1997
1998
1999
2000
Production
15.91
15.52
15.44
14.6
18.2
17.22
15.45
19.64
21.12
19.78
Year
2001
2002
2003
2004
2005
2006
2007
2008
2009
2010
Production
20
19.51
21.1
21.03
19.79
21.79
21.18
21.18
22.27
21.81
Solution.
We are to test the null hypothesis
H
0
:
median value and minus signs to those values
which are less than the median value; the infor-
mation are provided in the table given below.
Median
The series is random against
the alternative
hypothesis that
H
1
: The series is not random.
Let us first calculate the median and put plus
signs to those values which are greater than the
¼
15.02(m.t).
