Aim: To Calculate Sample Size for Clinical trial: Non-inferiority design (Outcome variable - Ratio)

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What is non-inferirity margin (δ)?

It is the clinically acceptable difference between μ_C and μ_T to define non-inferiority.

In other words, even if μ_T is less than μ_C, but the difference is less than δ, then μ_T can be considered as non-inferior to μ_C.

Null hypothesis and alternate hypotheis in non-inferiority design are as follows.

H0: μ_C - μ_T > δ (implying μ_T is inferior)

The alternate hypothesis will be

μ_C - μ_T < = δ (implying μ_T is not inferior to μ_C; difference between μ_C and μ_T is less than the clinically acceptable non-inferiority margin )

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Please note that most of the times non-inferiority limit is positive. It implies that the direction of inferiority is towards lower side μ_C. In other words, inferiority of test arm is considered, if μ_T  < μ_C. Here non-inferiority margin (δ) has to be more than μ_C - μ_T. If δ is less than μ_C - μ_T, then test arm mean is already less than the control arm mean by more than acceptable margin. Here non-inferirity can not be established, even with infinite sample size.

In rare situations, inferiority of a treatment is considered when its mean is more than the control mean. In this situation, non-inferiority limit is negative. It implies that the direction of inferiority is towards higher side μ_C. In other words, inferiority of test arm is considered, if μ_T  > μ_C. Here non-inferiority margin (δ) must be negative and it must be less than μ_C - μ_T (considering signs).  If δ is more than μ_C - μ_T, then non-inferirity can not be established here, even with infinite sample size.

Mathematically it is possible to calculate the sample size using given formula, even if above conditions of (δ) are not met. However, this will give erroneous results.

 

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Example 1:

Mean reduction in SBP with a known antihypertensive drug is 30 ± 6 mm of Hg. A new antihypertensive drug is to be tested for non-inferiority with this known drug. The mean reduction in SBP with this test drug is expected to be 25 ± 6 mm of Hg. How much sample size shall be required, if non-inferiority margin is decided as 8 mm of Hg, confidence level of 95% and power of 80%.

Solution:

Here

μ_T = 25, μ_C = 30, SD_T = SD_C = 6, δ = 8, confidence level = 95%, power = 80%

Non-inferiority margin is positive, and it is more than μ_C - μ_T. So the sample size can be calculated.

After putting these values, we get required sample size in each group = 49.

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Suppose in above example, δ = 3, then δ < μ_C - μ_T(5). Hence, we can say that, test arm mean is less than control arm mean by the margin which is beyond the acceptable margin of 3. Here sample size can not be calculated.

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Example 2:

In pregnancies complicated with gestational diabetes, fetal macrosomia is a known risk. A non-inferiority trial is planned to establish non-inferiority of life-style modifications during pregnancy (test diet and prescribed exercise) to a known drug. It is expected that the mean birth weights in test therapy group and control therapy group are 4200 ± 100 gm and 4000 ± 100 gm. How much sample size shall be required, if non-inferiority margin is decided as - 250, confidence level of 95% and power of 80%.

Please note that, in above example, non-inferiority margin is negative, as inferiority is considered if mean is more. (more the mean birth weight, more inferior is the treatment)

Non-inferiority margin is negative, and it is less than μ_C - μ_T. (-250 < -200). So the sample size can be calculated.

After putting these values, we get required sample size in each group = 49.

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Suppose in above example, δ = -100, then δ (-100) > μ_C - μ_T (-200). Hence, we can say that, test arm mean is already "inferior"  to control arm mean by the margin which is beyond the acceptable margin of - 100. Here sample size can not be calculated.

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