Sample Size Calculator: Two Parallel-Sample Means

Hypothesis: Non-Inferiority / Superiority

 

Data Input: (Help) (Example)

Input

 

Results

α

 

 

 

β

 

 

 

Allowable difference

 

n

Expected variance

 

 

 

δ

 

 

 

 

Note:

Variables

Descriptions

α

One-sided significance level

1-β

Power of the test

Allowable difference

Acceptable mean difference between sample two and sample one (µ21)

Expected variance

Population variance

δ

True difference between the mean and reference value, δ>0, the superiority margin or value of δ<0, the non-inferiority margin

n

Sample size of each group


Help Aids Top

 

Application: This procedure is used to test non-inferiority and superiority that can be unified by the following hypotheses:

 

Procedure:

  1. Enter

a)      value of α, the probability of type I error

b)      value of β, the probability of type II error

c)      value of allowable difference

d)     value of expected variance, the population variance

e)      value of δ>0, the superiority margin or value of δ<0, the non-inferiority margin.

  1. Click the button “Calculate” to obtain result sample size of each group n.

 

Formula:                                                        (*)

 

Notations:

 

α:               The probability of type I error (significance level) is the probability of rejecting the true null hypothesis.

 

β:               The probability of type II error (1 – power of the test) is the probability of not rejecting the false null hypothesis.

δ:               The true difference between the two mean values at which the power is calculated.

μ2 – μ1:      Margin of equivalence is the largest change from the reference value (baseline) that is considered to be trivial.

 


Example: Consider the difference of 5% is a difference of clinical importance, thus the non-inferiority margin is chosen to be 5% (i.e., δ=-0.05). Also, suppose the true difference in mean of low density lipidproteins (LDLs) between treatment groups is 0% (i.e., μ2(test) – μ1(control)=0). Thus, by using (*), with the standard deviation is 10% (i.e., expected variance is 0.01), the required sample size to achieve an 80% power (β=0.2) at α=0.05 for correctly detecting such difference of 0.5 change obtained by normal approximation as n=50.

 

 

Reference: Chow, Shao and Wang, Sample Size Calculations In Clinical Research, Taylor & Francis, NY. (2003) Pages 57-59.

 

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