Distribution of difference between two proportions

What is a sampling distribution?

The probability distribution of a given statistic based on a random sample of size n is called as a sampling distribution.

Sampling distribution of difference between two proportions:

  Let the two independent proportions be p₁ and p₂. Then, the mean of the sampling distribution of the difference between the two independent proportions (p₁ -p₂) is given by:

μ _p1-p2 = π₁ - π₂

Therefore, the standard error (S.E) of p1 -p2 is given by:

σ_p1-p2 = √π₁ ( 1 - π₁ ) / n1 + π₂ (1 - π₂ ) / n₂

As long as the proportions are not too close to 1 or 0 and the sample sizes n1 and n2 are not too small, the sampling distribution p1 -p2 is approximately equal to normal. For most purposes, the approximation is satisfactory if n1 and n2 are both at least 10 and neither lies within 0.10 of 0 or 1. An alternative rule is that the approximation is good if both Np and N (1 -π) are greater than 10 for both π₁ and π₂

Example:

Assume that 0.6 of middle school students but only 0.3 of middle school drop outs are able to pass a basic literacy test. If 10 students are sampled from the population of middle school pass outs and 15 students are sampled from the population of middle school drop outs, what is the probability that the proportion of drop outs that pass will be as high as the proportion of pass outs?

Solution:

The mean of the sampling distribution p1 -p2 is given by:

μ= π₁ -π₂ = 0.6 -0.3 = 0.3

The standard error is given by:

σ_p1-p2 = √π₁ ( 1 - π₁ ) / n1 + π₂ (1 - π₂ ) / n₂

=√0.6 ( 1 - 0.6 ) / 10 + 0.3 (1 - 0.3 ) / 15

=√0.024 + 0.014

=√0.038

=0.2

The solution of the problem is the probability that p1 -p2 is less than or equal to 0. The number of standard deviations above the mean associated with a difference in proportion of 0 is:

Z = p1 - p2 - μ _p1-p2 / σ_p1-p2

= 0 - 0.3 / 0.2

= - 1.5

Finally, from z-table it can be determined the number of standard deviations below the mean