# Confidence intervals using standard deviation - mistake or misunderstanding?

Hedges et al. 1999, Ecology 80: 1150-1156 reintroduces the old concept of using the natural logarithm of response ratio for ecologists as preferred statistics over p-values in comparative experiments and meta-analyses.

Response ratio, $R = \bar{X}_{E =treatment}/\bar{X}_{C=control}$; $L = ln(R) = ln(\bar{X}_{E}) - ln(\bar{X}_{C})$.

They indicate that taking a logarithm of the response ratio helps to tone down the violations against assumptions of normal distribution and homoscedasticity, which are evident in most experimental data sets.

Authors write: "If $\bar{X}_{E}$ and $\bar{X}_{C}$ are normally distributed and $\bar{X}_{C}$ is unlikely to be negative, then L is approximately normally distributed with mean approximately equal to the true log response ratio and variance, v, approximately equal to"

$\frac{(SD_{E})^2}{n_{E}\bar{X}_{E}^2}$ + $\frac{(SD_{C})^2}{n_{C}\bar{X}_{C}^2}$

They continue: "An approximate 100(1-$\alpha$)% confidence interval for the individual log response ratio parameter $\lambda$ is given by"

$L - z_{\alpha/2}\sqrt{v}\leq \lambda \leq L + z_{\alpha/2}\sqrt{v}$

"where $z_{\alpha/2}$ is the 100(1-$\alpha/2$)% point of the standard normal distribution, and the corresponding confidence interval for the (unlogged) response ratio $\rho$ is obtained by taking the antilogs of the confidence limits for the log response ratio."

In here, as I understand it, the authors indicate that standard deviation instead of standard error should be used to calculate the confidence intervals. Is this my misunderstanding or a mistake in the article?

• Well, that's a oversimplification. Standard deviation, or the square root of the variance, is a population parameter. Here, the distribution of $L$ is unknown, as it contains both unknown mean and unknown variance. Standard error is the estimate of this unknown standard deviation of the sample statistic. Commented Mar 6, 2013 at 14:36