Note: As for confidence intervals for m, we start by discussing the sampling distribution of sample proportions p.
Sampling Distributions – Proportion
For some population we are interested in the proportion
of items that belong to a particular group (e.g. we may be interested
in the proportion of y2k compliant modules in a software portfolio).
Let
p
denote this population proportion and let n denote sample size.
Theorem 3 Let n be large (i.e. n > 30). Consider all
possible samples of size n that may be selected from the
population. Compute the sample proportion (i.e. p) in each case. The
distribution of the p’s will be normal with mean mp= p and
standard deviation sp= sqrt(p(1-p)/n).
Problem: Consider the
population of modules in a software portfolio. 80% of these
modules are not y2k
compliant.
What proportion of samples of size 100 would you expect to
result in a sample proportion less than 70%. Solution: Since the population
proportion (expressed as a decimal) is
p=
0.8 then,
from Theorem 3,
mp= p=0.8
and sp=
sqrt(p(1-p)/n)
=sqrt(0.8(1-0.8)/100)=0.04.
Hence z=(0.7-0.8)/0.04=-2.5 and so the desired proportion is 0.62%
Note: We will only consider the case where n is large.