Research Methodology - Dr. Krishan K. Pandey

Testing of Hypotheses I 223
Thus, q 0 = 1 – p 0 = .2727
The test statistic z can be worked out as under:
z =
p$ −p$
pq
n
0 0
1
1 2
pq
+
n
0 0
2
=
400 400
−
500 600
b. 7273gb. 2727g
b. 7273gb. 2727g
+
500
600
0133 .
= = 4. 926
0. 027
As the H is one-sided we shall determine the rejection region applying one-tailed test (in the right tail
a
because H is of greater than type) at 5 per cent level and the same works out to as under, using
a
normal curve area table:
R : z > 1.645
The observed value of z is 4.926 which is in the rejection region and so we reject H in favour of H 0 a
and conclude that the proportion of smokers after tax has decreased significantly.
Testing the difference between proportion based on the sample and the proportion given
for the whole population: In such a situation we work out the standard error of difference between
proportion of persons possessing an attribute in a sample and the proportion given for the population
as under:
Standard error of difference between sample proportion and
N − n
population proportion or S. E. diff . p$ − p = p ⋅ q
nN
where p = population proportion
q = 1 – p
n = number of items in the sample
N = number of items in population
and the test statistic z can be worked out as under:
p$ − p
z =
N − n
p ⋅ q
nN
All other steps remain the same as explained above in the context of testing of proportions. We
take an example to illustrate the same.
Illustration 18
There are 100 students in a university college and in the whole university, inclusive of this college, the
number of students is 2000. In a random sample study 20 were found smokers in the college and the
proportion of smokers in the university is 0.05. Is there a significant difference between the proportion
of smokers in the college and university? Test at 5 per cent level.