I am trying to replicate results from a research paper that has the calculated 2.5 percentile and 97.5 percentile for a dataset, both with log10 transformed and untransformed versions. I can match their results for the untransformed data but not for the log10 transformed data. Is there a different approach needed when working with transformed data?

The dataset is:

data = c(1.0798,0.6047,1.2799,0.7581,0.6652,0.9692,1.1422)

They also define the 2.5 and 97.5 percentile as:

mean+- 1.96*sd

For their results they get the following:

| Stat          | Value  |
|Geomean        | -0.047 |
|s.d. (logmean) |  0.125 |
|Mean           |  0.897 |
|2.5 percentile |  0.510 |
|97.5 percentile|  1.578 |

Unfortunately, they do not provide more information on how they define each statistic in the table above.

My understanding of geomeans with log transformed data might be wrong, but when I calculate the mean and geomean I get the values around the other way (i.e. my geomean = the paper's mean).

I have done my calculations both in R and excel and I cannot get their percentiles using their formula. The dataset might be too small to really calculate these percentiles (?) but I would still like to replicate their results so I can make sure I am applying it appropriately to my work.

Am I missing something or is there a mistake in the paper? Any help would be greatly appreciated! Thank you.


1 Answer 1


this is how they did it: got the percentile of log transformed data, then applied exponential to it. what they call geomean is simply a mean of long transformed data.

  • your transformed variable's 97.5 percentile is $-0.047+1.96*0.125\approx 0.198$
  • exponentiate it to get the answer $10^{0.198}\approx 1.578$

logarithm is a monotonic function so if you obtain percentiles analytically then the order of operations (percentile vs log) doesn't matter. when you deal with empirical estimation, then the results depend on the order of operations, as you saw. in this case your source seems to think that the data comes from lognormal distribution. if that's the case, then what they did is appropriate. they are using parametric approach to percentiles. it's fairly common.

  • $\begingroup$ Thank you very much! That makes sense. It is not mentioned anywhere in the methods, might be standard practice? $\endgroup$
    – GPrice
    Jan 30, 2022 at 23:46
  • $\begingroup$ updated the answer to address your question $\endgroup$
    – Aksakal
    Jan 30, 2022 at 23:53

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