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I have a set of values reported as the 'log mean'. I know from elsewhere in the text that it is the natural log that is being referred to. I'm trying to ascertain if there is a way I can derive the mean from this 'log mean', and also if there is a way to work back from the reported value of the '+/- 1 log standard deviation', thus allowing me to include the data from this study in a meta analysis I am trying to put together. Here's a copy of how the data is presented in the paper:

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    $\begingroup$ What's the log mean? It's not the same as the mean logarithm! If you mean the latter, then reverse whatever transformation you used in the first place, presumably either log base 10 or log base e (natural logarithms). Other bases are possible, but I doubt the question would arise in those contexts $\endgroup$
    – Nick Cox
    Jan 16 '18 at 17:11
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    $\begingroup$ More details and assumptions are needed for this question to be answerable. $\endgroup$
    – whuber
    Jan 16 '18 at 17:13
  • $\begingroup$ Thanks guys, apologies for the vague question. If you have the time, take a look at my edited version. Any and all help appreciated here. You will be included in the acknowledgments of any published paper that arises from this work. $\endgroup$
    – francis
    Jan 17 '18 at 14:18
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    $\begingroup$ There is nothing in the extract to explain what the authors did beyond a statement that the log mean is measured in the original units. My wild guess is that it is a geometric mean. You can't recover the mean from such results without extra assumptions. This is on all fours with this example. The geometric mean is 10. What is the original mean? A geometric mean of 10 is consistent with original values of 1 and 100 (mean 50.5) and with original value (and mean) 10 and with many more such possibilities. If you make a heroic assumption that the data are (e.g.) lognormal you can do more. $\endgroup$
    – Nick Cox
    Jan 17 '18 at 19:04
  • $\begingroup$ But then again the data summarized are clearly pretty wild with values up to 39911.67 mm hr$^{-1}$. Those are precisely the circumstances in which straight means do not work well. I realise this doesn't help your goals usefully, but there it goes. $\endgroup$
    – Nick Cox
    Jan 17 '18 at 19:07
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I will share some educated guesses, as a long-time consumer and interpreter of this kind of hydrogeologic information.

"Log mean" surely is the "geometric mean."

The relationships among the columns of the table are these:

  • Let $x_i$ be the "log mean" in row $i=1,2,3,4,5$. It is a statistical summary of data $x_{i1}, x_{i2}, \ldots, x_{in_i}$, individual measurements of hydraulic conductivity in mm/hr. The summary is obtained by taking the logarithms $$y_{ij} = \log(x_{ij}),\ j = 1, 2, \ldots, n_i.$$ Letting $$y_i = \frac{1}{n_i}\left(y_{i1} + y_{i2} + \cdots + y_{in_i}\right)$$ be the mean of the logs, we have $$x_i = \exp(y_i).$$ This is universally known as the geometric mean of the $x_{ij}$; the phrase "log mean" is unusual.

  • The "Antilog $S_1$" is the exponential of the standard deviations of the $y_{ij}.$ That is, $$\log(S_{1i}) = \sqrt{\frac{1}{n_i-1}\left((y_{i1}-y_i)^2 + \cdots (y_{in_i} - y_i)^2\right)}.$$ (It is possible the denominators of the fractions are the $n_i$. There's no way to tell from the information given.) $S_1$ is often called the "geometric standard deviation."

  • The "$\pm\text{s.d.}$" columns are the exponentials of $y_i \pm \log(S_{1i}).$ Equivalently, by virtue of properties of logarithms, the left column will be $x_i / S_{1i}$ and the right column will be $x_i S_{1i}$. I confirmed this by calculating the ratios of the right column to the "log mean" and of the "log mean" to the left column. In all cases they agree, to the stated precision, with the "Antilog $S_1$" values.

  • The "Range" reports the smallest and largest among the $x_{ij},j=1,2,\ldots, n_i$.

In other words, the authors were working with the logarithms of the hydraulic conductivities, the quantities I have named with "$y$". They computed the means and standard deviations of these logarithms, site by site. To report an interval of uncertainty--presumably because they wish to use these data to estimate average hydraulic conductivities at each site--they constructed a "one s.d. interval" of logarithms by subtracting and adding the standard deviation from each mean. Finally, to express these results in the original units (the $x$'s) rather than their logarithms, they exponentiated them all.

Incidentally, it doesn't matter what base of logarithms is used for these calculations. Indeed, we have no way of knowing, because this table displays all results in the original units: the logarithms are hidden.

Why do it this way instead of working with the hydraulic conductivities? Typically, they vary by orders of magnitude, even within relatively homogeneous units. Their logarithms, on the other hand, tend to have relatively symmetric distributions, with few outlying values. Basing statistical estimates on the logarithms therefore produces indications that are arguably more "typical" of the entire unit. Beware, though. When using hydraulic conductivities to estimate water speeds and related quantities, often the very largest conductivities can control the situation by offering preferential pathways of transport. That is why proper interpretation of this table requires not only a good understanding of how it was constructed, but also of hydrogeologic transport phenomena. To this end, it may be worthwhile to study the ranges of observed conductivities and watch for extremely high values compared to the "$\pm\text{ s.d.}$" intervals, such as at Site 3.

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