2
$\begingroup$

I am back-transforming y=log(x+1) data that represent individuals/m^2 of area. Data are in the form of X ± SE. I back-transformed the means, but I am wondering if it's possible to directly back-transform the standard errors and, if so, how? I need to back-transform these data because I am comparing to non-transformed data. Obtaining the raw data is not possible.

$\endgroup$
  • $\begingroup$ Is this a matter of confidence interval for the mean of a skewed variable with zeros? Or do you have a model for which that was the response or outcome? Why not use a generalised linear model with log link, either way? $\endgroup$ – Nick Cox Jul 4 '18 at 14:35
  • $\begingroup$ Hi Nick -- thanks for your comment. I am conducting a systematic review, and these data are from a paper that meets the review criteria. I am looking to back-transform the means and SE so that I can use the untransformed values to generate effect sizes to be used in meta-analysis. This is why I'm not analysing the data with a generalised linear model with log link -- I need the untransformed values to compare to other studies in the review. The authors originally transformed the data to accommodate for a variable with zeros. $\endgroup$ – pvichm Jul 4 '18 at 14:51
  • $\begingroup$ Could you please zoom-in on the exact issue. Why doesn't backtracking each step do the trick? $\endgroup$ – Jim Jul 4 '18 at 15:06
  • $\begingroup$ H Jim -- The paper discusses invertebrate response to fire. Sampling occurred in burnt and unburnt plots at six different times. Data were presented in a plot, wherein each point represented the mean number of individuals per m^2 of leaves with SE bars included, but these are log(x+1) transformed data. I lifted the means and SE off the plot using Plot Digitizer. I then back-transformed the means via y = 10^(x)-1, wherein y = untransformed mean and x = transformed mean. However, I am unsure how to back-transform the SE since they are dependent on variation among individual datapoints. $\endgroup$ – pvichm Jul 4 '18 at 15:16
  • $\begingroup$ was my answer clear? did it help? $\endgroup$ – Ben Bolker Jul 9 '18 at 1:12
4
$\begingroup$

In general, if you use a smooth nonlinear transformation $f(x)$ to transform a random variable $x$, its standard deviation will be approximately scaled (multiplied) by a factor $\phi=\left. \frac{\partial f}{\partial x} \right|_{x=\bar x}$, i.e. $\sigma(f(x)) = \phi \cdot \sigma(x)$, or $\sigma(x) = \sigma(f(x))/\phi$ (you can derive this using the first term in a Taylor expansion). In this case $\partial[\log(x+1)]/\partial x = 1/x$. So you can back-transform the value you have to get $x$, then multiply the standard error you have by that $x$ value.

This is a rather crude approach, but I think it's what I would recommend in the absence of any more information.

I might actually recommend that you transform your other (non-log-transformed) values in the other direction, to work on the log scale; if you are implicitly assuming Normally distributed values in your analysis, that's more likely to be true for abundances on the log scale than on the original (abundance/m$^2$) scale.

PS. If you are using $\log_{10}$ rather than natural log in your transformation, you should be aware that $\partial \log_{10}(x)/\partial x$ is $1/(\ln(10) \cdot x)$ (where $\ln$ is the natural logarithm; same as $\log$ above).

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.