D Some theory on equivalence trials

D.1 Type I error in equivalence trials

Assume we have a continuous outcome with variance $\sigma^2$ and two treatment groups of size $n_A$ and $n_B$, respectively. Now consider the difference $\bar d$ of the mean outcomes as treatment effect, so $\bar d \sim \mathop{\mathrm{N}}(\Delta, \sigma^2 \, \lambda^2)$ where the true treatment difference is $\Delta$ and $\lambda = \sqrt{1/n_A+1/n_B}$.

Equivalence is established if the $\gamma \cdot 100$ % confidence interval for the treatment difference is within the pre-specified interval of equivalence $(-\delta, \delta)$, i.e. $$\bar d \pm z_{(1+\gamma)/2} \, \sigma \, \lambda \subset (-\delta, \delta).$$ With $z = z_{(1+\gamma)/2}$ we obtain the equivalent requirement $$\bar d \in (-\xi, \xi) \, \mbox{ where } \xi = \delta - z_{(1+\gamma)/2} \, \sigma \, \lambda$$

to establish equivalence. This happens with probability

$$\begin{eqnarray} \tag{D.1} \operatorname{\mathsf{Pr}}(\bar d \in (-\xi, \xi) ) &=& \Phi\left(\frac{\xi - \Delta }{\lambda \, \sigma}\right) - \Phi\left(\frac{-\xi - \Delta }{\lambda \, \sigma}\right) \nonumber \\ &=& \Phi\left(\frac{\delta - \Delta }{\lambda \, \sigma} - z_{(1+\gamma)/2}\right) - \Phi\left(\frac{-\delta - \Delta }{\lambda \, \sigma} + z_{(1+\gamma)/2}\right). \end{eqnarray}$$

Now evaluate (D.1) at $\Delta = \delta$ to obtain the Type I error rate (more precisely an upper bound on the Type I error rate)

$$\begin{eqnarray} \tag{D.2} \alpha = \Phi(-z_{(1+\gamma)/2}) - \Phi\left(z_{(1+\gamma)/2} - \frac{2 \delta}{\lambda \, \sigma}\right). \end{eqnarray}$$

Likewise, the power $1 - \beta$ is derived from (D.1) with $\Delta = 0$:

$$1 - \beta = 2 \Phi\left(\frac{\delta}{\lambda \, \sigma} - z_{(1+\gamma)/2}\right) - 1$$ so $$1 - \beta/2 = \Phi\left(\frac{\delta}{\lambda \, \sigma} - z_{(1+\gamma)/2}\right).$$

Note that $1 - \beta/2 = \Phi(z_{1-\beta/2})$, and therefore

$$\begin{eqnarray} \tag{D.3} \frac{\delta}{\lambda \, \sigma} = z_{(1+\gamma)/2} + z_{1-\beta/2}. \end{eqnarray}$$

With (D.2) we obtain

$$\begin{eqnarray} \tag{D.3} \alpha \approx \Phi(-z_{(1+\gamma)/2}) = 1-(1+\gamma)/2 = (1-\gamma)/2 \end{eqnarray}$$

for relatively small $\beta$ (say $\beta \leq 30$ %), since the second term in (D.2)

$$\Phi\left(z_{(1+\gamma)/2} - \frac{2 \delta}{\lambda \, \sigma}\right) = \Phi(z_{(1+\gamma)/2} - 2 \, (z_{(1+\gamma)/2} + z_{1-\beta/2})) = \Phi(-z_{(1+\gamma)/2} - 2 \, z_{1-\beta/2})$$

is then very close to zero. Note that (D.3) differs from superiority trials, where $\alpha = 1-\gamma$.

D.2 Sample size calulations in equivalence trials

Assume equal group sizes, i.e. $n=n_A=n_B$, then $\lambda=\sqrt{2/n}$.

With (D.3) we then obtain

$$\begin{eqnarray*} n &=& \frac{2 \sigma^2 (z_{(1+\gamma)/2} + z_{1-\beta/2})^2}{\delta^2} \\ &=& \frac{2 \sigma^2 (z_{1-\alpha} + z_{1-\beta/2})^2}{\delta^2} \end{eqnarray*}$$

as the required sample size in each group for Type I error rate $\alpha$ and power $1-\beta$.