Applying log-transformation when comparing two populations I have a survey data containing two samples drawing from two populations (or the same population but at different time series). Based on these two samples, I would like to estimate the total and its confidence intervals for each population. Since my sample distributions are highly skewed and both sample sizes are small. The recommended method is to first apply log-transform to the original data and compute CI in the log domain. 
However, since log-transform is nonlinear, i noticed that in the original domain sum(sample_1)>sum(sample_2), while in the log domain sum(log(sample_1)) is smaller than sum(log(sample_2)).
Can someone help me with this situation? Shall I still use the data in log domain. If the purpose of the study is to compare the total, or even further to use the data in prediction, using the original and log domain would give different results?
Thanks! 
 A: I guess you know the population size hence basically your goal is about comparing the means of the two populations.
But your observation about the skewness of the samples sounds like a little conflict with your goal: using the means to compare two populations is not pertinent if the data are skewed. 
More precisely, I think that using anyelse "centrality parameter" is not pertinenet too, but assuming the log-normal distribution is a reasonable assumption for your data, a common practice to statistically assess the comparison of the two populations is the following one. Assume $\log x_k \sim {\cal N}(\mu_1,\sigma_1^2)$ and $\log y_k \sim {\cal N}(\mu_2,\sigma_2^2)$.  Here $\mu_1$ is the population mean of the log-transformed variable in population 1 and $\mu_2$ is the population mean of the log-transformed variable in population 2. Using the classical two-samples model you get a confidence interval about the difference of means $\mu_1-\mu_2$. Transforming this interval with the exponential function gives a confidene interval about $\exp(\mu_1-\mu_2)=\frac{\exp \mu_1}{\exp \mu_2}$. Note that $\exp \mu_i$ is not the population mean of the nontransformed variable: it is the median (also considered as the geometric mean). 
Thus: 
1) In any case you get a confidence interval about the ratio of the medians (or geometric means) of the nontransformed variable.
And :
2) Under the additional assumption of equal variances ($\sigma_1=\sigma_2$), then $\frac{\exp \mu_1}{\exp \mu_2}$ is also the ratio of the means of the nontransformed variables. Indeed (see for instance en.wikipedia.org/wiki/Log-normal_distribution) the mean of the nontransformed variables in population $i=1,2$ is  $\exp(\mu_i+\frac12\sigma_i^2)$ and $\frac{\exp \mu_1+\frac12\sigma_1^2}{\exp \mu_2+\frac12\sigma_2^2}=\frac{\exp \mu_1}{\exp \mu_2}$ when $\sigma_1=\sigma_2$.
