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I am trying to understand why testing for non-inferiority requires pretty much the same sample size as testing for superiority (I assume the latter is the same as a one-sided test for a given MDE).

I was asked how big a sample we need to test that a certain change to our website's backend has no (negative) effect on visitor conversion. I said that it should be easier than testing for a lift in conversion. But that doesn't seem to be the case.

Running a one-sided test with 95% confidence and 90% power and assuming a 9% conversion rate and a 5% effect (0.45% lift) requires some 70k examples: http://powerandsamplesize.com/Calculators/Compare-2-Proportions/2-Sample-1-Sided

At the same time, running a non-inferiority test with a 0.45% margin requires 69k samples http://powerandsamplesize.com/Calculators/Compare-2-Proportions/2-Sample-Non-Inferiority-or-Superiority

Is that right or am I missing something?

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What alternative scenario are you assuming? If you assume the same positive true effect, then superiority will require a smaller sample size for the same power. If you assume no effect, then superiority testing does not even make sense.

On the other hand, if you look at superiority assuming a true improvement by x or non-inferiority with a non-inferiority margin of x assuming there is in truth no difference, then you get about the same probability of a significant result under the alternative hypothesis with the same sample size.

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  • $\begingroup$ I guess I am still confused why the following post says you need a lot less data for a non-inferiority test: blog.analytics-toolkit.com/2017/… There is a table in the middle. $\endgroup$
    – iggy
    Apr 2, 2018 at 4:26
  • $\begingroup$ The table in the blog assumes that there is in truth an improvement of 5 percentage points and under that assumption non-inferiority with a margin of 2 percentage points is easier to show than superiorty. Very approximately non-inferiority is as easy to show as superiorty assuming that the true difference is 5+2=7 percentage points (and sample size is about proportional to effect size squared i.e. $5^2/7^2\approx 0.5$). $\endgroup$
    – Björn
    Apr 2, 2018 at 5:21

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