### Math 225 Course Notes

### Section 6.5:
Confidence Interval for a Population Proportion

It is appropriate to form confidence intervals
for a single population proportion
when sample is drawn and the statistic of interest is counting the proportion
in the sample that fall into a given category.
The methodology presented here is only appropriate for large samples.
As in all examples presented in this chapter,
the
central limit theorem
allows us to conclude that the sampling distribution
is approximately normal.
In this situation,
there are different formula
for the mean and standard error,
but the same logic and procedure for constructing confidence intervals
remains the same.

Another example shows how to apply these ideas
to an estimation problem.

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

and
SE()
=

The shape will be approximately normal for sufficiently large samples.
A general rule of thumb is that if np > 5
and n(1-p) > 5, then the distribution will be approximately normal.
*The techniques of this section should not be used to construct
confidence intervals if the sample size is too small*.

Suppose that 1000 American women
aged 50--54 are randomly selected,
and that 18 are found to have breast cancer.
Construct a 95% confidence interval for the proportion of
American women in this age group with breast cancer.
Our sample proportion is 18/1000 = .018.
We need to check that np is at least 5.

1000 * .018 = 18 > 5

(This will hold
whenever there are at least 5 successes and failures in the sample.)
We may conclude that
the sampling distribution
for the sample proportion
will be approximately normal,
by the central limit theorem.

We estimate the SE to by replacing p by
in the SE formula
.

sqrt( (.018)(.982)/1000 ) = .00420

From our large sample size,
we may use a reliability coefficient from the normal distribution.
.018 +/- (1.96)(.00420)

or
.018 +/- .008

Last modified: April 15, 1996

Bret Larget,
larget@mathcs.duq.edu