After trying to read on this topic, I still have some clarifications remaining.
Context:
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).
Problem:
I am expressing the results as back-transformed mean ($\exp(\log(\text{mean}))$). Should I present the standard deviation as $\exp(\log(\text{sd}))$?
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 $\exp(\log(\text{mean}_A/\text{mean}_B))$?
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.
Example:
NON-TRANSFORMED, NATIVE VALUES
mean(GrpA) = 2.11 (SD 3.13) cfu/cm2, median(GrpA) = 1.01 (0.54 to 2.44) cfu/cm2
vs
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.
SO, USE TRANSFORMED LOG VALUES
exp(mean(log(GrpA))) = 1.107
exp(sd(log(GrpA))) = 3.094
vs
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?
EDIT #1
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.