Confidence intervals can be used in place of the statistical tests. Reporting of confidence intervals is more informative because it indicates the magnitude of the treatment difference and how close it approaches the equivalence zone.

The \(100(1 - \alpha)\%\) confidence interval that corresponds to testing the null hypothesis of non-equivalence versus the alternative hypothesis of equivalence at the \(\alpha\)* *significance level has the following limits

\( \text{lower limit } = \text{min } \left [ 0, \left( \bar{Y}_{E} - \bar{Y}_{A} \right) - s \sqrt{\frac{1}{n_E}+\frac{1}{n_A}} t_{n_{E}+n_{A}-2, 1- \alpha} \right ] \)

\( \text{upper limit } = \text{max } \left [ 0, \left( \bar{Y}_{E} - \bar{Y}_{A} \right) + s \sqrt{\frac{1}{n_E}+\frac{1}{n_A}} t_{n_{E}+n_{A}-2, 1- \alpha} \right ] \)

This confidence interval does provide \(100(1 - \alpha)\%\) coverage - (see Berger RL, Hsu JC. Bioequivalence trials, intersection-union tests, and equivalence confidence sets. *Statistical Science* 1996, 11: 283-319).

Some researchers mistakenly believe that a \(100(1 - 2\alpha)\%\) confidence interval is consistent with testing the null hypothesis of non-equivalence versus the alternative hypothesis of equivalence at the \(\alpha\) significance level. Note that the Berger and Hsu \(100(1 - \alpha)\%\) confidence interval is similar to the \(100(1 - 2\alpha)\%\) confidence interval in its construction except that (1) the lower limit, if positive, is set to zero, and (2) the upper limit, if negative, is set to zero.

If the \(100(1 - \alpha)\%\) confidence interval lies entirely within \(\left(-\Psi, +\Psi \right)\), then the null hypothesis of non-equivalence is rejected in favor of the alternative hypothesis of equivalence at the \(\alpha\) significance level.

For a non-inferiority trial, the two-sample t statistic labeled \(t_{inf}\) , previously discussed,can be applied to test:

\( H_0: \left\{ \mu_E - \mu_A \le - \Psi \right\} \text{ vs. } H_1: \left\{ \mu_E - \mu_A > - \Psi \right\}\)

Because a non-inferiority design reflects a one-sided situation, only the \(100(1 - \alpha)\%\) lower confidence limit is of interest:

If the \(100(1 - \alpha)\%\) lower confidence limit lies within \(\left(-\Psi, +∞\right)\), then the null hypothesis of inferiority is rejected in favor of the alternative hypothesis of non-inferiority at the \(\alpha\) significance level.

The FDA typically is *more stringent than is required* in non-inferiority tests. The FDA typically requires companies to use \(\alpha = 0.025\) for a non-inferiority trial, so that the one-sided test or lower confidence limit is comparable to what would be used in a two-sided superiority trial.

**Equivalence**

**Non-Equivalence**

**Non-Inferiority**

**Inferiority**

##
Example
Section* *

As an example, suppose an investigator conducted an equivalence trial with 30 patients in each of the experimental therapy and active control groups \(\left(n_E = n_A = 30\right)\). He defines the zone of equivalence with \(\Psi = 4\). The sample means and the pooled sample standard deviation are

\( \bar{Y}_E = 17.4, \bar{Y}_A = 20.6, s = 6.5 \)

The t percentile, \(t_{58,0.95}\), can be found from the TINV function in SAS as TINV(0.95,58), which yields that \(t_{58,0.95} = 1.67\). Thus, using the formulas in the section above, the lower limit = min{0, -3.2 - 2.8} = min{0, -6.0} = -6.0; the upper limit = max{0, -3.2 + 2.8} = max{0, -0.4} = 0.0. This yields the 95% confidence interval for testing equivalence of \(\mu_E - \mu_A\) is (-6.0, 0.0). Because the 95% confidence interval for \(\mu_E - \mu_A\) does not lie entirely within \(\left(-\Psi, +\Psi \right) = \left(-4, +4 \right)\), the null hypothesis of non-equivalence is not rejected at the 0.05 significance level. Hence, the investigator cannot conclude that the experimental therapy is equivalent to the active control.

Suppose this had been conducted as a non-inferiority trial instead of an equivalence trial, and he defines the zone of non-inferiority with \(\Psi = 4\), i.e., \(\left(-4, +∞ \right)\). The 95% lower confidence limit for \(\mu_E - \mu_A\) is -6.0, which does not lie within \(\left(-4, +∞\right)\). Therefore, the investigator cannot claim non-inferiority of the experimental therapy to the active control.

A real example of a non-inferiority trial is the VALIANT trial in patients with myocardial infarction and heart failure. Patients were randomized to valsartan monotherapy \(\left(n_V = 4,909\right)\), captopril monotherapy \(\left(n_C = 4,909\right)\), or valsartan + captopril combination therapy \(\left(n_{VC} = 4,885\right)\). The primary outcome was death from any cause. One objective of the VALIANT trial was to determine if the combination therapy is superior to each of the monotherapies. Another objective of the trial was to determine if valsartan is non-inferior to captopril, defined by \(\Psi = 2.5\%\) in the overall death rate.

##
Switching Objectives
Section* *

Suppose that in a non-inferiority trial, the 95% lower confidence limit for \(\mu_E - \mu_A\) not only lies within \(\left(-\Psi, +∞\right)\) to establish non-inferiority, but also lies within \(\left(0, +\Psi\right)\). It is safe to claim the superiority of the experimental therapy to the active control in such a situation (without any statistical penalty).

**Non-Inferiority and Superiority**

In a superiority trial, suppose that the 95% lower confidence limit for \(\mu_E - \mu_A\) does not lie within \(\left(0, +\Psi\right)\), indicating that the experimental therapy is not superior to the active control. If the protocol had specified non-inferiority as a secondary objective and specified an appropriate value of \(\Psi\), then it is safe to claim non-inferiority if the 95% lower confidence limit for \(\mu_E - \mu_A\) lies within \(\left(-\Psi, +∞\right)\).

**Non-Inferiority**