PROC POWER ;
twosamplemeans =equiv_diff
test= 91
lower = 101
upper =96
meandiff= 8
stddev = .
ntotal = 0.8
power = 0.05;
alpha RUN;
Sample Size for Non-Inferiority Trials in SAS
Introduction
PROC POWER1 can be used for sample size calculations for non-inferiority testing. See 2 for explanation of non-inferiority and how to perform Sample size in SAS (including comparing proportions). Below we give 2 sample size examples for the following types of studies:
two-sample comparison of means for Non-inferiority (i.e. testing if one treatment mean is non-inferior to the another treatment mean).
Paired-sample comparison of means (i.e. 2 treatment means recorded on 1 group of patients are equivalent within a set tolerance)
Two Sample Non-inferiority test: Comparing means for parallel design (unpaired)
This example is a sample size calculation for the following hypotheses: \(H_0:\mu2-\mu1\le -\theta\) versus \(H_1: \mu2-\mu1\gt -\theta\).
A client is interested in conducting a clinical trial to compare two cholesterol lowering agents for treatment of hypercholesterolemic patients through a parallel design. The primary efficacy parameter is a low-density lipidprotein cholesterol (LDL-C). We will consider the situation where the intended trial is for testing noninferiority. For establishing it, suppose the true mean difference is 0 and the noninferiority margin is chosen to be -0.05 (-5%). Assuming SD = 0.1, how many patients are required for an 80% power and an overall significance level of 5%?
As shown below, a total sample size of 102 is recommended, which equates to 51 in each group.
Comparing means for crossover design (paired)
Here we assume there is no carry-over effect and that the variance is known. For more information see3. There is no obvious way in SAS to do cross over non-inferiority, however, given the one sided test, you can half the alpha using the equiv_diff option to give you the non-inferiority sample size.
Estimating the within patient variance and correlation.
Let’s consider a standard two-sequence, two period crossover design. Suppose that the sponsor is interested in showing non-inferiority of the test drug against the reference with the non-inferiority margin -20%. Assume power of 80%. Based on the results from previous trials, it is estimated that the variance (of the difference) is 0.2 (20%). Suppose that the true mean difference is -0.1 (-10%). What is the required sample size, assuming significance level of 5%?
Alpha = 0.025 is used below, instead of 0.05 because you are doing non-inferiority (a one sided test). Note that this is still the sample size for alpha=0.05. The below shows a sample size of 13 patients is required.
pairedmeans =equiv_diff
test=-0.3
lower=0.1
upper= -0.1
meandiff =0.2
stddev= 0.5
corr = 0.025
alpha = .
npairs = 0.8; power
References
Version
{r} #| echo: false si <- sessioninfo::session_info(“sample_s_noninferiority”, dependencies = FALSE) si$external <- structure(list(“SAS” = “9.04.01M7P08062020”), class = c(“external_info”, “list”)) si