What if the primary response variable is binary?
When the outcome in a CTE trial is a binary response and the objective is to compare the two groups with respect to the proportion of success, the results can be expressed in a 2 × 2 table as
Group # 1  Group # 2  
Success  \(r_1\)  \(r_2\) 
Failure  \(n_1  r_1\)  \(n_2  r_2\) 
There are a variety of methods for performing the statistical test of the null hypothesis \(H_0\colon p_1 = p_2\), such as a ztest using a normal approximation, a \(χ^2\) test (basically, a square of the ztest), a \(χ^2\) test with continuity correction, and Fisher's exact test.
The normal and \(χ^2\) approximations for comparing two proportions are relatively accurate when these conditions are met:
\( \dfrac{n_1(r_1+r_2)}{(n_1+n_2}\ge 5, \dfrac{n_2(r_1+r_2)}{(n_1+n_2}\ge 5, \dfrac{n_1(n_1+n_2r_1r_2)}{(n_1+n_2}\ge 5, \dfrac{n_2(n_1+n_2r_1r_2)}{(n_1+n_2}\ge 5 \)
Basically when the expected number in each cell is greater than 5, the normal or Chi Square approximation is useful.
Otherwise, Fisher's exact test is recommended. All of these tests are available in SAS PROC FREQ of SAS and will be discussed later in the course.
A sample size formula for comparing the proportions \(p_1\) and \(p_2\) using the normal approximation is given below:
\( n_2=\left( \dfrac{AR+1}{AR}\right)\left( z_{1\alpha/2}+z_{1\beta}\right)^2\bar{p}(1\bar{p})/(p_1p_2)^2 \)
where \(p_1  p_2\) represents the effect size and
\( \bar{p}= (AR \cdot p_1+p_2) / (AR+1) \)
is the weighted average of the proportions.
Example
Using PROC POWER to calculate sample size when comparing two binomial proportions Section
An investigator wants to compare an experimental therapy to placebo when the response is success/failure via a twosided, 0.05 significance level test and 90% statistical power. She knows from the medical literature that 25% of the untreated patients will experience success, so she decides that the experimental therapy is worthwhile if it can yield a 50% success rate. With equal allocation, \(n_2 = \dfrac{(2)(1.96 + 1.28)^2{0.375(10.375)}}{(0.25)^2} = 79\). Thus, the investigator should enroll \(n_1 = 79\) patients into treatment and \(n_2 = 79\) into placebo for a total of 158 patients.
With an unequal allocation ratio of \(AR = 3, n_1 = 168\) and \(n_2 = 56\). Again, notice that the allocation ratio of AR = 3 yields a total sample size larger than that for the allocation ratio of AR = 1 (224 vs. 158).
This is a program that illustrates the use of PROC POWER to calculate sample size when comparing two binomial proportions.
***********************************************************************
* This is a program that illustrates the use of PROC POWER to *
* calculate sample size when comparing two binomial proportions. *
***********************************************************************;
proc power;
twosamplefreq groupweights=(1 1) groupps=(0.25 0.50) alpha=0.05 power=0.9
test=Fisher sides=2 ntotal=.;
plot min=0.1 max=0.9;
title "Sample Size Calculation for Comparing Two Binomial Proportions (1:1 Allocation)";
run;
proc power;
twosamplefreq groupweights=(1 3) groupps=(0.25 0.50) alpha=0.05 power=0.9
test=Fisher sides=2 ntotal=.;
plot min=0.1 max=0.9;
title "Sample Size Calculation for Comparing Two Binomial Proportions (3:1 Allocation)";
run;
SAS PROC POWER for Fisher’s exact test yields \(n_1 = 85\) and \(n_2 = 85\) for \(AR = 1\), and \(n_1 = 171\) and \(n_2 = 57\) for \(AR = 3\).
Try it! Section
The answer is a total of 752 subjects.
Here is the output you should have obtained from SAS ...
The POWER Procedure 


Fixed Scenario Elements  
Distribution  Exact conditional 
Method  Walter's normal approximation 
Number of Sides  2 
Alpha  0.05 
Group 1 Proportion  0.3 
Group 2 Proportion  0.4 
Group 1 Weight  1 
Group 2 Weight  1 
Nominal Power  0.8 
Computed N Total  

Actual Power  N Total 
0.801  752 