# Determining Non-Inferiority with a 95% Confidence Interval

I have a sample of 1000 observations, divided into 2 groups: treatment and control (500 in each group). Each observation is dichotomous: 1 if an event occurred (bad event), and 0 if no event occurred.

I wish to test the non-inferiority hypothesis that P(treatment)-P(control)=0.01 vs. P(treatment)-P(control)<0.01, meaning, to show that the proportion of these bad events in the treatment group is not larger by more than 1% from the control group.

This is a one-sided test, and therefore the significance level for the test is 2.5% and not 5% (FDA requirement).

I wish to replace the test with a confidence interval. The idea is to calculate the 95% confidence interval to the proportions difference. If the upper limit of the CI is lower than 0.01, I declare non-inferiority, and otherwise, I fail to declare non-inferiority.

I wanted to ask you, if calculating a 95% CI for proportions difference, and using it's upper limit, is as identical to doing a one-sided test with a significance level of 2.5%? Is it just as conservative as the test?

I saw some references according to which this procedure is done, but I ain't sure.

To make things simpler I don't consider any correction here.

The confidence interval at level $1-\alpha$ corresponding to one-sided test is

$(-\infty ; (pT - pC) + z_{1-\alpha}s )$

where $s^2 = p(1-p)(1/n_1 + 1/n_2)$; $p$ being estimated as the pooled proportion.

The same CI($1-\alpha$) corresponding to the 2-sided test is

$( (pT - pC) - z_{1-\alpha/2} s ; (pT - pC) + z_{1-\alpha/2}s ).$

Thus to get the upper bound of the one sided CI($1-\alpha$) one should use a two sided CI($1 - 2\alpha$)

For a one-sided CI(97.5%) [i.e. $\alpha$=2.5%] one should use a two sided CI(95%).