**550**wrote:

I have two sets of normalized expression data from a qPCR experiment. In order to perform my statistical analysis I've had to log transform the data from each set. I'd like to get a standard error associated with the mean of the log transformed set. Further, I'd like to back-transform (linearize) the data in order to get a meaningful fold-change (basically a ratio of the two means), along with an accompanying standard error.

This boils down to two questions:

How can I calculate a standard error for a back-transformed log mean? In other words, how can back-transform the standard error of a set of log-transformed values?

How can I incorporate the standard errors from two different back-transformed log means into a single standard error for the accompanying ratio of back-transformed log means?

I realize this might be a bit confusing, and I'm not sure where else to ask about it. Any help or pointers in the right direction would be greatly appreciated.

Are you sure that "back-transform" would linearize? I mean, if we use to "log-transform" data, it's to linearize exponentially distributed data no?

3.9kI'm not sure if this note helps: http://www.bmj.com/content/312/7038/1079.full

2.7kAre you sure that "back-transform" would linearize? I mean, usually, if we use to log transform data, it's to linearize exponentially distributed data no?

3.9kFold-change using your log-transformed values is equally as "meaningful" as that from the original values. For example if you used log base 2, then a difference in means of 1 = a mean fold-change of 2; difference of 2 = fold-change of 4 and so on.

48kYou should probably be doing everything in log space. If you get a standard error, you can always figure out the (now asymmetrical) confidence interval in linear space if needed.

25kYou shouldn't 'back-transform', the log transform was most likely done for a reason, e.g. to make the distribution symmetric and more similar to a normal distribution. The standard error makes much more sense when the error is normally distributed.

46k