So in our statistical test, we are using a dependent t-test because we are comparing the means between pre and post conditions in a physiology lab. Since there is a lot of variability between two subjects. For instance, one patient has pre/post values of 15.3 and 19.55% respectively while another has 2.74 and 3.02. My question is rather than find the means of both, can I simply compare the DIFFERENCE by making the pre condition always 1 and the post condition the difference between the two (i.e 19.55-15.3) meaning that when I run the test, the mean of pre condition will be exactly 1 and the post condition will be the difference between the two values. Is that appropriate?


1 Answer 1


One can use a paired t-test. As you can see in the link, the paired t-test is a comparison of differences, and that is indeed the appropriate test to perform, under one condition. That is, one should check to make sure that the difference between data pairs is not non-normally distributed. The paired t-test calculates the average difference between samples (which is also the difference of averages, but that is actually not as relevant.) Then, that average difference is divided by the standard error to make the t-statistic. That t-statistic is then the deviation from zero difference, so looking up what that statistic means in terms of probability, gives the likelihood of no difference.

If the paired differences are not normally distributed, then one can attempt to transform the data, by taking logs, square-roots, reciprocals and so forth of the original data sets to attempt to find a fairly normal difference histogram, or, one can just use a Wilcoxon signed-rank test, which, as it creates normal conditions by ranking the data, does not need normal conditions in the data itself.

  • $\begingroup$ How does one make sure that the difference is normally distributed? I don't know any way to be sure of that. Ranking the data does not produce normality; the ranks are distinctly non-normal; under continuity they only take integer values from 1 to n, each of which appears equally often (where n is the number of pairs; and they're dependent, since we're 'sampling' ranks without replacement). $\endgroup$
    – Glen_b
    Dec 4, 2016 at 22:19
  • $\begingroup$ @Glen_b True enough, loose language. One must avoid significant non-normality for the difference function of a paired t-test. In practice, known or not, it fairly easy to see when that is. Typically, significant t-tests for non-normality will be less significant than Wilcoxon S-RT, and, significant or not, it is also telling that non-significant t-test are then less non-significant. $\endgroup$
    – Carl
    Dec 5, 2016 at 2:35
  • $\begingroup$ @Glen_b Changed to "not non-normally distributed," which although more correct, is also confusing for the Newbie. $\endgroup$
    – Carl
    Dec 5, 2016 at 2:38

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