For meta‐analysis studies and systematic reviews, it is important to pool the data from a set of similar clinical trials. To pool the data, one needs to know their SD. Many trial reports, however, contain only the median, the minimum and maximum values, and the sample size. It is therefore important to be able to estimate the SD S from the sample size n and range r. For small n ≤ 100, we improve existing estimators of r/S, the “divisor,” denoted by $\xi (n)$. This in turn yields improved estimators of the SD in the form $\widehat{S}=r/\widehat{\xi }(n)$ on simulated as well as real datasets. We provide numerical values of the proposed estimator as well as approximation by a simple formula $3\sqrt{\ln (n)}-1.4025$. Furthermore, for large n, we provide estimators $\widehat{\xi }(n)$ of the divisor $\xi (n)$ for the normal, exponential, and other bounded and unbounded distributions.

中文翻译：

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根据样本大小和范围估算样本方差。

对于荟萃分析研究和系统评价，重要的是从一组类似的临床试验中收集数据。为了汇总数据，需要知道其SD。但是，许多试验报告仅包含中位数，最小值和最大值以及样本量。因此，能够估计SD重要小号从样品尺寸Ñ和范围- [R 。对于小Ñ  ≤100，我们提高现有估计- [R /小号中，“除数，”记$\xi （ñ）$。这反过来以以下形式产生了SD的改进估计量$\widehat{\mathrm{小号}}=\mathrm{[R}/\widehat{\xi }（ñ）$在模拟数据集和真实数据集上。我们提供建议的估算器的数值以及通过简单公式得出的近似值$3\sqrt{\ln （ñ）}-\mathrm{1个}。4025$。此外，对于大n，我们提供估计量$\widehat{\xi }（ñ）$ 除数 $\xi （ñ）$ 用于正态分布，指数分布以及其他有界和无界分布。