6a.6 - Example: Comparative Treatment Efficacy Studies

An investigator wants to determine the sample size for comparing two asthma therapies with respect to the forced expiratory volume in one second \(\left(FEV_1\right)\). A two-sided, 0.05-significance level test with 90% statistical power is desired. The effect size is \(\Delta = 0.25 \text{ L}\) and the standard deviation reported in the literature for a similar population is \(\sigma = 0.75 \text{ L}\). The investigator plans to have equal allocation to the two treatment groups (AR = 1).

The first step is to identify the primary response variable. In this example, \(FEV_1\) is a continuous response variable. Assuming that \(FEV_1\) has an approximate normal distribution, the number of patients required for the second treatment group based on the z formula is

\(n_2 = \dfrac{(2)(1.96 + 1.28)^2(0.75)^2}{(0.25)^2 } = 189\)

Thus, the total sample size required is \(n_1 + n_2 = 189 + 189 = 378\). SAS Example: This is a program that illustrates the use of PROC POWER to calculate sample size when comparing two normal means.

***********************************************************************
* This is a program that illustrates the use of PROC POWER to         *
* calculate sample size when comparing two normal means.              *
***********************************************************************;

proc power;
twosamplemeans dist=normal groupweights=(1 1) alpha=0.05 power=0.9 stddev=0.75 
   meandiff=0.25 test=diff sides=2 ntotal=.;
plot min=0.1 max=0.9;
title "Sample Size Calculation for Comparing Two Normal Means (1:1 Allocation)"; 
run;

proc power;
twosamplemeans dist=normal groupweights=(2 1) alpha=0.05 power=0.9 stddev=0.75 
   meandiff=0.25 test=diff sides=2 ntotal=.;
plot min=0.1 max=0.9;
title "Sample Size Calculation for Comparing Two Normal Means (2:1 Allocation)"; 
run;

SAS PROC POWER, based on the t formula, yields \(n_1 + n_2 = 191 + 191 = 382\).

If the investigator had wanted an allocation ratio of \(AR = 2\) (twice as many subjects in the first group), then \(n_2 = \dfrac{(1.5)(1.96 + 1.28)^2(0.75)^2}{(0.25)^2} = 142\) and \(n_1 = 2 \times 142 = 284\).

The total sample size required is \(n_1 + n_2 = 142 + 284 = 426\).

SAS PROC POWER, based on the t formula, yields \(n_1 + n_2 = 143 + 286 = 429\).

Notice that the 2:1 allocation, when compared to the 1:1 allocation, requires an overall larger sample size (429 versus 382).

Now it is your turn to give it a try!