Interpreting test results on log-transformed data I have data that is not normally distributed. I can log-transform it to be normally distributed, and then perform, for example, a t-test.
But how do I interpret the results of the t-test?
Do I have to transform back the p-value, before I can say that the difference between the two means is significant? Anything else I need to do? Or can I just take the results and interpret them as if the test was performed on the original data? I'm confused about this.
The data comes from a device that measures force in Newton and is ratio scaled.
 A: You're on the right track. You can definitely run a t-test on log-transformed data if it's well-behaved after transformation, and it definitely affects how you interpret your results.  In short, you can't make statements about the difference of the means because the mean(log(x)) is not the same as the log(mean(x)) -- the means don't transform well.  If you're okay with stating a difference in terms of log-transformed means (i.e. richter scale), then you can definitely do that and just state your results in log(mean) terms.  Otherwise, your t-test results will be in tearms of medians, because medians are preserved over a log transformation.  
If the data are very normal distributions after a log transform, mean(x) is roughly equal to median(x), and the median(log(x)) is the same as log(median(x)).
Thanks to the log property that allows you to coalesce differences of logs into a log of a ratio 
(log(X/Y) = log(X) - log(Y))

you can make statements about the difference in log(medians).  You can also can back-transform the CI values back into original units and make a statement about the ratio of the medians with no log() attached.  Median is a good measure of center in well-behaved, normally distributed data, so your statistical inference should stand.
Here are the steps:

1.  Transform the data (log(x) where x is an array of data in this case)
2.  Perform your t-test if your transformed data meet the assumptions of the t-test.  Check for:
  a. departure from normality
  b. significantly differing standard deviations
  c. lack of independence.
3.  Your CI values will now be in terms of ln(median(X)/median(Y)). 
    You can back-transform by taking the e-to-the-power-of(confidence interval values)... 
    This is sometimes written as EXP(<values>).  
4.  Now you've gotten rid of your ln(Mx/My) problem, 
    but your confidence interval is still in terms of the RATIO of medians.
5.  Your p-value will still stand without transformation.  
6.  State your conclusions in terms of ratio of medians.  
    Example:  "The median values of x and y are not the same.  
    We are 95% confident that X's median is between <lowerCI> and <higherCI> times that of Group B." 
    (because we're still talking median ratio).  

This is a bit clunky, so you can also talk in percentages, like "we are 95% confident that X's median is between  and  that of group Y"
note:  I was reading The Statistical Sleuth 3rd Ed and reviewing slides from SMU Data Science Masters course while I wrote this, so credit where credit is due.
A: If the logs of the data are really drawn from normally distributed populations with constant variance (but possibly different means), then the original data must have come from lognormal distributions with possibly different scales (due to differences in $\mu$, where adding something to the means on the log-scale has a multiplying effect on the original scale), and hence those populations will also differ in mean.
So a difference in population mean on the log scale will (if the other assumptions of the usual two-sample equal-variance t-test apply) imply a change in mean on the original (untransformed) scale.

You can convert the estimated size of difference on the log scale to a percentage increase on the original scale, and the ends of a confidence interval carry back as well.
So imagine you have an estimated difference in means of about 0.7 ($\hat{\delta}=\hat{\mu}_1-\hat{\mu}_2=0.7$) and (say) a 90% confidence interval for the difference in means of $(0.62, 0.78)$. then the ratio of means of the original populations is estimated to be $\exp(0.7)\approx 2.014$ (i.e. the first group has a mean about twice the size of the second group), and the 90% CI for the ratio of means would be $(e^{0.62}, e^{0.78})\approx (1.86,2.18)$.
Note that p-values carry over directly; we're still performing inference on the parameter $\mu$ (the mean on the log scale).
