Estimating a Proportion, Given a Small Sample
In this lesson, we explain how to estimate a confidence interval for
a proportion, when the sample size is small.
Confidence Interval: Proportion (Small Sample)
In the
previous lesson, we showed how
to estimate a confidence interval for a proportion when a simple random sample
includes at least 10 successes and 10 failures.
When the sample does not include at least 10 successes and 10
failures, the sample size will often be too small to justify the
estimation approach presented in the previous lesson.
This lesson describes how to construct a confidence interval for
a proportion when the sample has fewer than 10 successes and/or fewer than 10 failures. The key steps are:
Estimation Requirements
The approach described in this lesson is valid whenever the
following conditions are met:
 The sample includes at least 1 success and 1 failure.
The following examples illustrate how this works. The first example
involves a binomial experiment; and the second example, a hypergeometric
experiment.
Example 1: Find Confidence Interval When Sampling With Replacement
Suppose an urn contains 30 marbles. Some marbles are red, and the rest are
green. Five marbles are randomly selected,
with replacement, from the urn. Two of the selected marbles are red,
and three are green. Construct an 80% confidence interval for the proportion of
red marbles in the urn.
Solution: To solve this problem, we need to define the sampling
distribution of the proportion.

First, we assume that the population proportion is equal to the sample
proportion. Thus, since 2 of the 5 marbles were red, we assume the proportion
of red marbles is equal to 0.4.

Second, since we sampled with replacement, the sample proportion can be
considered an outcome of a binomial experiment.

Assuming that the population proportion is 0.4 and the sample proportion is the
outcome of a binomial experiment, the sampling distribution of the proportion
can be determined. It appears in the table below. (Previously,
we showed
how to compute binomial probabilities that form the
body of the table.)
Number of red marbles in sample

Sample proportion 
Probability 
Cumulative probability 
0

0.0 
0.07776 
0.07776 
1

0.2 
0.2592 
0.3396 
2

0.4 
0.3456 
0.68256 
3

0.6 
0.2304 
0.91296 
4

0.8 
0.0768 
0.98976 
5

1.0 
0.01024 
1.00 
We see that the probability of getting 0 red marbles in the sample is 0.07776;
the probability of getting 1 red marble is 0.2592; etc. Given the entries in
the above table, it is not possible to create an 80% confidence interval exactly.
However, we can come close. When the true population proportion is 0.4, the probability the
probability that a sample proportion falls between 0.2
and 0.6 is equal to 0.2592 + 0.3456 + 0.2304 or 0.8352. Thus, based on this
sample, we can say that an 83.52% confidence interval is described by
the range from 0.2 to 0.6.
Sample Planning Wizard
As you may have noticed, the steps required to construct a sampling distribution
based on a binomial experiment are not trivial.
They can be timeconsuming and complex. Stat Trek's
Sample Planning Wizard does this work for you  quickly, easily, and
errorfree. In addition to constructing a confidence interval, the Wizard
creates a summary report that lists key findings and documents analytical
techniques. Whenever you need to construct a confidence interval, consider
using the Sample Planning Wizard. The Sample Planning Wizard is a premium tool
available only to registered users. >
Learn more
Example 2: Find Confidence Interval When Sampling Without Replacement
Let's take another look at the problem from Example 1. This time, however, we
will assume that the marbles are sampled
without replacement. Suppose an urn contains 30 marbles. Some marbles
are red, and the rest are green. Five marbles are randomly selected, without
replacement, from the urn. Two of the selected marbles are red, and three are
green. Construct an 80% confidence interval for the proportion of red marbles
in the urn.
Solution: To solve this problem, we need to define the sampling
distribution of the proportion.

First, we assume that the population proportion is equal to the sample
proportion. Thus, since 2 of the 5 marbles were red, we assume the proportion
of red marbles is equal to 0.4.

Second, since we sampled without replacement, the sample proportion can be
considered an outcome of a hypergeometric experiment.

Assuming that the population proportion is 0.4 and the sample proportion is the
outcome of a hypergeometric experiment, the sampling distribution of the
proportion can be determined. It appears in the table below.
(Previously,
we showed
how to compute hypergeometric probabilities
that form the body of the table.)
Number of red marbles in sample

Sample proportion 
Probability 
Cumulative probability 
0

0.0 
0.0601 
0.0601 
1

0.2 
0.2577 
0.3178 
2

0.4 
0.3779 
0.6957 
3

0.6 
0.2362 
0.9319 
4

0.8 
0.0625 
0.9944 
5

1.0 
0.0056 
1.0000 
We see that the probability of getting 0 red marbles in the sample is 0.07776;
the probability of getting 1 red marble is 0.2592; etc. Given the entries in
the above table, it is not possible to create an 80% confidence interval exactly.
However, we can come close. When the true population proportion is 0.4, the probability the
probability that a sample proportion falls between 0.2
and 0.6 is equal to 0.2577 + 0.3779 + 0.2362 or 0.8718. Thus, based on this
sample, we can say that an 87.18% confidence interval is described by
the range from 0.2 to 0.6.
It is informative to compare the findings from Examples 1 and 2. In both
problems, the interval
estimate ranged from 0.2 to 0.6. However, the
confidence level was greater for Example 2 (which sampled without
replacement) than for Example 1 (which sampled with replacement). This
illustrates the fact that precision is greater when sampling without
replacement than when sampling with replacement.