#### Description:

Hello! In this video tutorial, we’ll demonstrate how to calculate sample size for an inequality Test of Two Independent Proportions in PASS. The hypotheses may be stated in terms of proportions, differences, ratios, or odds ratios, but all four hypotheses result in the same test statistic. This procedure computes power and sample size for a wide variety of two proportions tests.

As an example, suppose that we are designing a study to test the effectiveness of a new treatment. The standard treatment has a 60% cure rate, and want to compute the power for detecting differences of 5 and 10 percent in the cure rate with a two-sided Z-test at group sample sizes ranging from 50 to 650 and at a 0.05 significance level.

To perform this calculation in PASS, first load the Tests for Two Proportions procedure using the category tree or the search bar on the PASS Home Window.

For Solve For, select Power. This procedure also allows you to solve for sample size and effect size.
You have 2 choices for the power calculation method: Binomial Enumeration, which results in an exact power calculation, and the normal approximation. When the sample sizes are reasonably large and the proportions are between 0.2 and 0.8, the two methods will give similar results. For smaller sample sizes and more extreme proportions, the normal approximation is not as accurate so the binomial calculations may be more appropriate. We’ll use the normal approximation method for this example.

We’ll compute power for a two-sided Pooled Z-test. For alpha, enter 0.05. We’ll also assume equal group sample sizes and enter 50 to 650 by 100 for the sample size in each group.

To specify the effect size, we can enter proportions, differences, ratios, or odds ratios. For this example we’ll enter differences. For the difference to detect, D1, enter 0.05 and 0.10, and for P2, the standard group proportion, enter 0.6.
Now, click the Calculate button to perform the calculations and get the results. A separate line in the output is given for each combination of the input values. A sample size of 50 in each group has only 8% power to detect a difference of 0.05 and 18% power to detect a difference of 0.1 for the two-sided pooled z-test. A sample size of 650 in each group on the other hand has 46% power to detect a difference of 0.05 and nearly 97% power to detect a difference of 0.1. The plots give a graphical representation of the relationship between power and sample size for each difference.

To find the sample size necessary to detect these two differences with 80% power, go back to the procedure window and change Solve For to Sample Size and enter 0.80 for Power. Again, click calculate to get the result. The required group sample sizes to detect differences of 0.05 and 0.1 with 80% power are 1471 and 356, respectively.