Confidence Interval of a Proportion

This lesson describes how to construct a confidence interval for a sample proportion, p, when the sample size is large.

Key Considerations

The approach described in this lesson is valid whenever the following conditions are met:

• The sampling method is simple random sampling.
• Population size is at least 20 times larger than sample size.
• The sample includes at least 10 successes and 10 failures.

Note the implications of the third condition. If the population proportion were close to 0.5, the sample size required to produce at least 10 successes and at least 10 failures would be close to 20. But if the population proportion were extreme (i.e., close to 0 or 1), a much larger sample would be needed to produce at least 10 successes and 10 failures.

For example, imagine that the probability of success were 0.1, and the sample were selected using simple random sampling. In this situation, a sample size close to 100 might be needed to get 10 successes.

The Variability of the Sample Proportion

To construct a confidence interval for a sample proportion, we need to know the variability of the sample proportion. This means we need to know how to compute the standard deviation or the standard error of the sampling distribution.

• Suppose k possible samples of size n can be selected from the population. The standard deviation of the sampling distribution is the "average" deviation between the k sample proportions and the true population proportion, P. The standard deviation of the sample proportion σ_p is:

  σ_p = sqrt[ P * ( 1 - P ) / n ] * sqrt[ ( N - n ) / ( N - 1 ) ]

  where P is the population proportion, n is the sample size, and N is the population size. When the population size is much larger (at least 20 times larger) than the sample size, the standard deviation can be approximated by:

  σ_p = sqrt[ P * ( 1 - P ) / n ]

• When the true population proportion P is not known, the standard deviation of the sampling distribution cannot be calculated. Under these circumstances, use the standard error. The standard error (SE) can be calculated from the equation below.

  SE_p = sqrt[ p * ( 1 - p ) / n ] * sqrt[ ( N - n ) / ( N - 1 ) ]

  where p is the sample proportion, n is the sample size, and N is the population size. When the population size at least 20 times larger than the sample size, the standard error can be approximated by:

  SE_p = sqrt[ p * ( 1 - p ) / n ]

How to Find the Confidence Interval for a Proportion

Previously, we described how to construct confidence intervals. For convenience, we repeat the five steps below.

1. Choose the confidence level. The confidence level describes the uncertainty of a sampling plan. Often, researchers choose 90%, 95%, or 99% confidence levels; but any percentage can be used.
2. Compute the standard deviation or standard error. The standard deviation (σ_p) and the standard error (SE_p) can be computed from the following formulas:

   σ_p = sqrt[ P * ( 1 - P ) / n ]

   SE_p = sqrt[ p * ( 1 - p ) / n ]

3. Find the critical value. The critical value will be a z-score or a t-score. The value of the z-score or t-score depends on the confidence level. Common z-score critical values are 1.645 for a 90% confidence level, 1.96 for a 95% confidence level, and 2.576 for a 99% confidence level. Instructions for finding other z-score critical values and t-score critical values are provided in the lesson on margin of error.
4. Find the margin of error. You can compute the margin of error, based on one of the following equations.

   Margin of error = Critical value * Standard deviation of statistic

   Margin of error = Critical value * Standard error of statistic

5. Specify the confidence interval. The uncertainty is denoted by the confidence level. And the range of the confidence interval is defined by the following equation.

   Confidence interval = Sample statistic ± Margin of error

In the next section, we work through a problem that shows how to use this approach to construct a confidence interval for a proportion.

Test Your Understanding

Problem 1

A major metropolitan newspaper selected a simple random sample of 1,600 readers from their list of 100,000 subscribers. They asked whether the paper should increase its coverage of local news. Forty percent of the sample wanted more local news. What is the 99% confidence interval for the proportion of readers who would like more coverage of local news?

(A) 0.30 to 0.50
(B) 0.32 to 0.48
(C) 0.35 to 0.45
(D) 0.37 to 0.43
(E) 0.39 to 0.41

Solution

The answer is (D). The approach that we used to solve this problem is valid when the following conditions are met.

• The sampling method must be simple random sampling. This condition is satisfied; the problem statement says that we used simple random sampling.
• The sample should include at least 10 successes and 10 failures. Suppose we classify a "more local news" response as a success, and any other response as a failure. Then, we have 0.40 * 1600 = 640 successes, and 0.60 * 1600 = 960 failures - plenty of successes and failures.
• If the population size is much larger than the sample size, we can use an "approximate" formula for the standard deviation or the standard error. This condition is satisfied, so we will use one of the simpler "approximate" formulas.

Since the above requirements are satisfied, we can use the following five-step approach to construct a confidence interval.

1. Select a confidence level. In this analysis, the confidence level is defined for us in the problem. We are working with a 99% confidence level.
2. Compute the standard deviation or standard error. Since we do not know the population proportion, we cannot compute the standard deviation; instead, we compute the standard error. And since the population is more than 20 times larger than the sample, we can use the following formula to compute the standard error (SE) of the proportion:

   SE = sqrt [ p(1 - p) / n ]

   SE = sqrt [ (0.4)*(0.6) / 1600 ] = 0.012

3. Find critical value. The critical value is a factor used to compute the margin of error. Because the sampling distribution of a proportion is approximately normal when the sample size is large, we will express the critical value as a z-score.

   Common z-score critical values are 1.645 for a 90% confidence level, 1.96 for a 95% confidence level, and 2.576 for a 99% confidence level. So, the z-score for this problem will be 2.576.

4. Find the margin of error (ME):

   ME = critical value * standard error

   ME = 2.576 * 0.012 = 0.03

5. Specify the confidence interval. The range of the confidence interval is defined by the sample statistic + margin of error. And the uncertainty is denoted by the confidence level.

Therefore, the 99% confidence interval is 0.37 to 0.43. That is, the 99% confidence interval is the range defined by 0.4 + 0.03.