# 6a.4 - Confidence Intervals for Means

6a.4 - Confidence Intervals for Means

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 $$n = \dfrac{1.96^{2} \times 5^{2}}{1^2} = 96$$.

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