Section 5.6: Distribution of the Difference Between Two Sample Proportions

Key Concepts

This sampling distribution 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 find probabilities.

In this situation, there are different formula for the mean and standard error, but the same logic and procedure for solving problems remains the same.

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

Suppose that the proportion of hyperactive children in two separate populations of retarded children is .40 for each. Random samples of size 100 and 120 are respectively chosen. What is the probability that the difference in the sample proportions, ( ) is more than 16%?

The sampling distribution for the difference in sample proportions will be approximately normal with a mean of 0 and a standard deviation of = .0663. The z-score is

z = (.16 - 0) / .0663 = 2.41

Since the area to the right of 2.41 under the standard normal curve is .0080, the probability is less than 1% that the difference would be this large.