Two Independent Proportions Power Analysis | Stata Data Analysis Examples

Introduction

Power analysis is the name given to the process for determining the sample size for a research study. The technical definition of power is that it is the probability of detecting a “true” effect when it exists. Many students think that there is a simple formula for determining sample size for every research situation. However, the reality it that there are many research situations that are so complex that they almost defy rational power analysis. In most cases, power analysis involves a number of simplifying assumptions, in order to make the problem tractable, and running the analyses numerous times with different variations to cover all of the contingencies.

In this unit we will try to illustrate how to do a power analysis for a test of two independent proportions, i.e., the response variable has two levels and the predictor variable also has two levels. Instead of analyzing these data using a test of independent proportions, we could compute a chi-square statistic in a 2×2 contingency table or run a simple logistic regression analysis. These three analyses yield the same results and would require the same sample sizes to test effects.

Description of the Experiment

It is known that a certain type of skin lesion will develop into cancer in 30% of patients if left untreated. There is a drug on the market that will reduce the probability of cancer developing by 10%. A pharmaceutical company is developing a new drug to treat skin lesions but it will only be worthwhile to do so if the new drug is 5% better than the existing drug.

The pharmaceutical company plans to do a study with patients randomly assigned to two groups, the control (untreated) group and the treatment group. The company wants to know how many subjects will be needed to test a difference in proportions of .15 with a power of .8 at alpha equal to .05.

The Power Analysis

We will make use of the built-in Stata program power which can be used to determine the sample size needed for tests of two independent proportions as well as for tests of means. The power command needs the following information in order to do the power analysis: 1) the keyword twoporportions, 2)the expected proportion of cancer the untreated group (p1 = .3), 3) the expected proportion of cancer in the treated group (p2 = .3 – .15 = .15), 4) the type of test to be run, and 5) the required level of power (power = .8 for this experiment).

power twoproportions .3 .15, test(chi2)(.8)

Performing iteration ...

Estimated sample sizes for a two-sample proportions test
Pearson's chi-squared test 
H0: p2 = p1  versus  Ha: p2 != p1

Study parameters:

        alpha =    0.0500
        power =    0.8000
        delta =   -0.1500  (difference)
           p1 =    0.3000
           p2 =    0.1500

Estimated sample sizes:

            N =       242
  N per group =       121

This is all well and good but a two-sided test doesn’t make much sense in this situation. We want to test for a drug that reduces the probability of cancer not for one that increases the probability. In this case we should be using one-tail test and we do this by using the onesided option in power.

power twoproportions .3 .15, test(chi2) onesided

Estimated sample sizes for a two-sample proportions test
Pearson's chi-squared test 
H0: p2 = p1  versus  Ha: p2 < p1

Study parameters:

        alpha =    0.0500
        power =    0.8000
        delta =   -0.1500  (difference)
           p1 =    0.3000
           p2 =    0.1500

Estimated sample sizes:

            N =       190
  N per group =        95

This is better. The output from the power command indicates that we need to use 95 subjects in each group to find a change in probability of .15 for a power of .8 when alpha equals .05.

Just as a check let’s run the analysis specifying each of the two sample sizes.

power twoproportions .3 .15, test(chi2) n(216) onesided
 
Estimated power for a two-sample proportions test
Pearson's chi-squared test 
H0: p2 = p1  versus  Ha: p2 < p1

Study parameters:

        alpha =    0.0500
            N =       216
  N per group =       108
        delta =   -0.1500  (difference)
           p1 =    0.3000
           p2 =    0.1500

Estimated power:

        power =    0.8440

Now because we believe that we know a lot about the incidence of cancer in the untreated group we would like to make the control group half as large as the treatment group. We can easily do this by including the nratio option.

power twoproportions .3 .15, test(chi2) nratio(2) onesided

Estimated sample sizes for a two-sample proportions test
Pearson's chi-squared test 
H0: p2 = p1  versus  Ha: p2 < p1

Study parameters:

        alpha =    0.0500
        power =    0.8000
        delta =   -0.1500  (difference)
           p1 =    0.3000
           p2 =    0.1500
        N2/N1 =    2.0000

Estimated sample sizes:

            N =       210
           N1 =        70
           N2 =       140

As you can see, we will need more subjects overall than for equal sized groups but we can have a much smaller untreated group.

In the end, the company has decided to use 75 patients in the control group and 150 in the treatment group. Let’s see what the power is.

power twoproportions .3 .15, test(chi2) n1(75) n2(150) onesided

Estimated power for a two-sample proportions test
Pearson's chi-squared test 
H0: p2 = p1  versus  Ha: p2 < p1

Study parameters:

        alpha =    0.0500
            N =       225
           N1 =        75
           N2 =       150
        N2/N1 =    2.0000
        delta =   -0.1500  (difference)
           p1 =    0.3000
           p2 =    0.1500

Estimated power:

        power =    0.8271

With this unbalanced design we have an estimated power of .8271, which the company deems acceptable.

Introduction

Description of the Experiment

The Power Analysis

See Also