After trying to read on this topic, I still have some clarifications remaining.


Comparing between 2 arms (categorical), measuring microbiological plate-counted bacteria concentration (colony forming units/cm2). Results are positive skewed, log transformation gives me a normal distribution which I am interested in as it loses less data points than a non-parametric test (EDIT #1: I read often that non-parametric tests is inefficient, and 'wastes data' compared to parametric tests).


  1. I am expressing the results as back-transformed mean ($\exp(\log(\text{mean}))$). Should I present the standard deviation as $\exp(\log(\text{sd}))$?

  2. If I am interested in mean difference (since I'm using a student t test), is it a $\exp(\log(\text{mean}_A)) - \exp(\log(\text{mean}_B))$, or $\exp(\log(\text{mean}_A/\text{mean}_B))$?

  3. How best to express confidence interval for this t-test comparison of log-transformed data? I read differing opinions - some suggest to back-transform the CI generated by the t-test(log data), accepting that the CI will not contain the back-transformed mean. Others say that back-transformed CI has no meaning and hence should not be presented.



mean(GrpA) = 2.11 (SD 3.13) cfu/cm2, median(GrpA) = 1.01 (0.54 to 2.44) cfu/cm2


mean(GrpB) = 1.80 (SD 3.24) cfu/cm2, median(GrpB) = 0.61 (0.29 to 1.48) cfu/cm2

Comparing GrpA vs GrpB will require non-parametric test (Mann-Whitney) since they are not distributed normally.


exp(mean(log(GrpA))) = 1.107

exp(sd(log(GrpA))) = 3.094


exp(mean(log(GrpB))) = 0.738

exp(sd(log(GrpB))) = 3.585

Comparing log(GrpA) vs log(GrpB) can be done with t-test since log(GrpA) and log(GrpB) are normally distributed

T-test of log(GrpA) vs log(GrpB):

95% CI -0.08288 to 0.89421

exp(-0.08288)=0.92047, exp(0.89421)=2.44534

Can I then express this statement on a scientific journal (stating in methods that data was log transformed to allow for parametric statistical testing): Mean Grp A was 1.107 (sd=3.094). Mean Grp B was 0.738 (sd=3.585). Mean difference was 1.107-0.738=0.369, 95% CI 0.92047 to 2.44534?


Thank you for your replies! (Will read more on the suggested generalised linear model with logarithmic link) I included above a specific example, given my untrained eyes and mind thought maybe this will express my dilemma and help my understanding of your pointers.

  • $\begingroup$ "as it loses less data points than a non-parametric test": what does this mean? $\endgroup$
    – Nick Cox
    Commented Feb 25, 2020 at 18:11
  • 1
    $\begingroup$ I would just use a generalised linear model with logarithmic link here. Switch to 1970s technology from older technique. $\endgroup$
    – Nick Cox
    Commented Feb 25, 2020 at 18:12
  • $\begingroup$ Do you have access to the raw individual values, or only to the arithmetic (regular) mean and SD? $\endgroup$ Commented Feb 27, 2020 at 15:50
  • $\begingroup$ I do have access to the individual values $\endgroup$
    – ltw
    Commented Feb 28, 2020 at 17:35

1 Answer 1


If you compute the SD of the log-transformed data and then take the antilog of that, exp(SD), you have computed the geometric SD factor. It has no units. You cannot add or subtract it from the geometric mean, but could multiply or divide the geo mean by it to get a sense of variation.

Wikipedia has an article on the geometric SD. The term was coined by Kirkwood. Here is a summary I wrote.

Regarding the CI, I don't understand your comments. If you compute the 95% CI of the mean of the logarithms of your values, and then antilog both confidence limits, you have computed the CI of the geometric mean. The geomean will be within that interval (but not at its center).

  • $\begingroup$ Hi Harvey. I edited my original post to include an example to try to express what I mean about the CI of mean difference between log(values). Thanks! $\endgroup$
    – ltw
    Commented Feb 27, 2020 at 2:24

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