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?


1 Answer 1


I think the equation is correct. v is computed from the two standard deviations but also from the sample sizes. The square root of v is the standard deviation of L. But the standard deviation of a computed parameter is a standard error, just like the standard deviation of a sample mean is the standard error of the mean (which is very different than the standard deviation of the data or the distribution itself).

When you look at a set of values, the standard deviation of those values is very different than the standard error of the mean of those values. But when you look at a estimated (via calculations) parameter, the "standard deviation" of that parameter and the "standard error" of that parameter are really the same. Or more precisely, the standard error is an estimate of the unknown standard deviation.

  • $\begingroup$ 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. $\endgroup$
    – StasK
    Commented Mar 6, 2013 at 14:36
  • $\begingroup$ Thank you for clarification Harvey. Does this mean that v is a good estimate of standard error if used together with the second equation, but if the equation is changes, for instance, if I replace z-distribution with t-distribution it may not be such a good estimate any longer? $\endgroup$
    – Mikko
    Commented Mar 7, 2013 at 6:49
  • $\begingroup$ @largh: I'm sure about SD vs. SEM as explained in my answer. Thanks for pointing out the issue of z vs. t. I'm not sure about that, but think that probably t is the right distribution to use. $\endgroup$ Commented Mar 8, 2013 at 0:30

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