Aim: To calculate statistical power with given Sample Size for Clinical trial: Superiority design (Outcome variable - Ratio)

---

What is superiority margin?

It is the margin by which μ_T should be more than μ_C to define superiority.

In other words, only if μ_T is more than μ_C + δ, then μ_T is considered as superior to μ_C

Null hypothesis and alternate hypotheis in superiority design are as follows.

H0: μ_T - μ_C < δ (implying μ_T is not superior)

The alternate hypothesis will be
μ_T - μ_C < = δ (implying μ_T is superior to μ_C; difference between μ_T and μ_C is more than the superiority margin )

---

Example:

A superiority trial is conducted to test superiority of a new antihypertensive drug T compared with a known antihypertensive drug C. Mean reduction in SBP with a new and known antihypertensive drug is 35 ± 6 and 30 ± 6 mm of Hg. The trial is conducted by including 50 participants in each group. How much was the power of the trial to detect superiority, if superiority margin is decided as 4 mm of Hg, confidence level of 95%.

Solution:

Here

μ_T = 35, μ_C = 30, SD_T = SD_C = 6, δ = 4, confidence level = 95%, power = 80%, N₁=50, N₂=50.

After putting these values, we get power=20.85%.

---

Please note that the superiority (δ) has to be less than the difference between μ_T and μ_C

Mathematically, it is possible to calculate the sample size using given formula, even if δ is more than μ_T - μ_C . However, this will give erroneous results, because even with infinite sample size we can not prove superirity in this situation.

Because, if the difference between means of two treatents is less than the superiority margin, then it is already assumed that μ_T is less than μ_C + δ. In this situation, whatever sample size we take, it will not be possible to reject the null hypothesis. Following text will explain rhis conceptin detail.

---

Let us discuss why superiority margin needs to be between 0 and μ_T – μ_T. Suppose that an existing iron supplementation increases haemoglobin levels by 2.5 ± 0.5 mg /dl, and a new (test or experimental) supplementation is expected to increase it by 3 ± 0.5 mg /dl. In this situation, if the superiority margin is 0.6 or more, then it implies that the experimental drug shall be considered as superior, if its mean is equal to or more 3.1 (2.5 + 0.6) or more. In this situation, superiority can’t be determined as anticipated test mean (3 mg/dl) is already less than this limit of 3.1. We have to take the superiority margin of less than 0.5 (less than the anticipated difference between experimental (test) and existing (control) treatments). For example, if superiority margin is 0.3, it implies that superiority shall be considered, if test mean is more than 2.8 (2.5 + 0.3); provided that this superiority margin is also clinically acceptable margin to define superiority.

 

Similarly, suppose that we intent to decrease mean levels of serum bilirubin of new-borns by some intervention in certain pregnant women, in which risk of neonatal jaundice is high. With an existing intervention during ANC, mean bilirubin level at birth is 3.1 ± 0.8 mg /dl AND with new (experimental or test) intervention it is expected to be 2 ± 0.7 mg /dl. In this situation, we have to have superiority margin between 0 and – 1.1. Suppose superiority margin is -0.8, it implies that experimental intervention shall be considered superior to existing intervention, if it gives mean bilirubin levels of 2.3 (3.1 + (-0.8)) or less. We can conduct a clinical trial with superiority design in this situation. However, if superiority margin is selected as - (minus) 2, it implies that superiority shall be considered only when the experimental intervention mean is less than 1.1. But, the anticipated experimental intervention mean of 2 does not allow for superiority design.

Please choose superiority margin such that, its value is between 0 and μ_T – μ_T, and also which is clinically meaningful.