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

Key Concepts

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.

Formula

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.

Example

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.