Q:

Should I use the mean difference or mean ratio for a t-test?

I have a quantitative variable (medication dose) that is paired: dose of medication before surgery and after surgery for 76 patients.
I was under the impression that I should calculate the mean of the differences of the before and after doses and then calculate the standard error of the mean difference, CI and t-test. In SPSS a paired-sample t-test would do this.
I was told instead to do the percentage of the second dose (after surgery, i.e., 600 mg) as a function of the first dose (before surgery, i.e., 1200 mg) for every patient and then calculate the mean, standard error, CI and t-test of the percentages.
Is this a valid analysis?

A:

A $t$-test assumes the relevant data (differences or ratios) are drawn from a normally distributed population.  Thus, at most, one of those possibilities affords a valid $t$-test.  In addition, a $t$-test of the mean ratio against $0$ makes no sense whatsoever, so you would need to specify some value that makes intrinsic sense given your data and the theoretical context (most commonly $1$).  Moreover, they ask subtly different questions—by asking about constant ratios, you wonder if those with larger starting values have a proportionally greater increase.  
I am a firm believer that the statistical analysis should match the substantive question that motivated the study.  Thus, your first job is to get as clear as possible on what you want to know, then pick the analysis that matches that.  Subsequently, you will need do determine that the assumptions of the analysis obtain.  You can check the normality of the differences or ratios with a qq-plot.  Ratios are not typically normally distributed.  A common strategy is to take the logarithm of every value, and compute the differences of the logs.  Those differences can be tested against $0$ with a one-sample $t$-test.  

