Find the mean of the distribution of differences in sample proportions

Question

Consider taking random samples of size 50 from Population A with proportion 0.88 and

          random samples of size 70 from Population B with proportion 0.81. Find the mean of the

          distribution of differences in sample proportions, p^_A-P^_B

                  A) - .07

                   B)   .07

                   C) 1.07

                   D)  .87

Answer 1

The problem asks for the mean of the distribution of differences in sample proportions, denoted as μp^​A​−p^​B​​.

According to the central limit theorem for proportions, the mean of the distribution of sample proportions (μp^​​) is equal to the true population proportion (p).

Therefore, for two independent populations, the mean of the distribution of the differences in sample proportions is simply the difference between the true population proportions.

In this case:

Population A proportion (pA​) = 0.88

Population B proportion (pB​) = 0.81

The mean of the distribution of differences in sample proportions is:

μp^​A​−p^​B​​=pA​−pB​

μp^​A​−p^​B​​=0.88−0.81

μp^​A​−p^​B​​=0.07

The sample sizes (50 from Population A and 70 from Population B) are not needed to calculate the mean of the distribution of differences in sample proportions. They would be used to calculate the standard deviation of the distribution.

Looking at the options:

A) -0.07

B) 0.07

C) 1.07

D) 0.87

The correct answer is B) 0.07.