For a clinical endpoint that can be approximated by a normal distribution in an SE study, the \(100(1 - \alpha)\%\) confidence interval for the population mean, \(\mu\), is

\( \bar{Y} \pm \left [ t_{n-1, 1-\alpha/2}s/\sqrt{n} \right ] \)

where

\( \bar{Y}=\sum_{i=1}^{n}Y_i/n \) is the sample mean,

\( t_{n-1, 1-\alpha/2} \) is the appropriate percentile from the \(t_{n-1}\) distribution, and

\( s^2= \sum_{i=1}^{n}(Y_{i} - \bar{Y})^2 / (n-1) \) is the sample variance and estimates \(σ^{2}\).

If σ is known, then a z-percentile can replace the t-percentile in the \(100(1 - \alpha)\%\) confidence interval for the population mean, \(\mu\), that is,

\( \bar{Y} \pm \left( z_{1- \alpha /2}\sigma / \sqrt{n} \right) \)

If n is relatively large, say n ≥ 60, then \(z_{1 - \alpha/2} ≈ t_{n - 1,1 - \alpha/2}\).

If it is desired for the \(100(1 - \alpha)\%\) confidence interval to be

\( \bar{Y} \pm \delta \)

then

\( n= z_{1-\alpha/2}^{2}\sigma^2/\delta^2 \)

For example, the necessary sample size for estimating the mean reduction in diastolic blood pressure, where \(σ = 5\) mm Hg and \(δ = 1\) mm Hg, is \(n = \dfrac{1.96^{2} \times 5^{2}}{1^2} = 96\).