### Math 225 Course Notes

### Section 6.6:
Confidence Interval for the Difference Between Two Population Proportions

This confidence interval should be used whenever the situation
involves two independent samples from different populations
and the sample proportions are compared.
As in all examples presented in this chapter,
the
central limit theorem
allows the use of the
normal distribution
to justify the formulas.
In this situation,
there are different formula
for the mean and standard error,
but the same logic and procedure for finding confidence intervals
remains.

Demonstration of these ideas is shown in an
example.

The sampling distribution of
is summarized by:
mean()
=

and
SE()
=

The shape will be approximately normal for sufficiently large samples.
This is reasonable to assume if each sample individually
meets the rule of thumb for the approximation.

We are interested in the different population rates
of hyperactivity
in two separate populations of retarded children.
Random samples of size 100 and 120 are respectively chosen,
with sample proportions of .38 and .40 respectively.
Give a 95% confidence interval
for the difference in population proportions.
The sampling distribution
for the difference in sample proportions
will be approximately normal
since each sample is sufficiently large.
(There are at least 5 children of each type in each sample.)

We may estimate the standard error by plugging in our sample proportions
instead of the unknown population proportions,
finding
= .0660.

Using our general formula for confidence intervals,

(.38 - .40) +/- (1.96)(.0660)

or
-.02 +/- .13

There is very little evidence that the true two population
proportions differ by a substantial amount.

Last modified: Feb 28, 1996

Bret Larget,
larget@mathcs.duq.edu