I want to test for a difference in means between two sets of data. The data is standardized using $(X-\mu)/\sigma$. Can I perform two-sample t-test for a difference in means? In general, since I have standardized through mean shift, which means I've set the means at 0, should I expect mean difference at all?


No, you should not. Once the distribution is standardized, it has a mean of zero and a standard deviation of one. What you should do is go back to the original data and t-test that, that is, if you are sure it is at least approximately normally distributed.

  • $\begingroup$ Many thanks. But my issue is that each observation in my data is the sum of ten standardized values. I've done this to build an index and now I'd like to see whether different groups have different means or not. Any suggestion? @Carl $\endgroup$ – msmazh Sep 14 '16 at 5:24
  • $\begingroup$ If they do not have the same means, then they are not normally distributed. All you will be doing is testing your assumptions. $\endgroup$ – Carl Sep 14 '16 at 5:29
  • $\begingroup$ @Carl I don't understand how you conclude that they're not normally distributed if they don't have the same means. If they're internallly standardized, don't they all necessarily have the same means no matter what the distribution? $\endgroup$ – Glen_b -Reinstate Monica Sep 14 '16 at 6:28
  • $\begingroup$ @Glen_b The mean, median and mode of a skewed distribution are different. If one normalizes non-normal data to have equal mean and standard deviation, and we then discover that there is a significant bias between several different attempts at doing this, then all we are doing is quantifying how bad our assumption of normality was. $\endgroup$ – Carl Sep 15 '16 at 0:12
  • $\begingroup$ @Glen_b Maybe this will help. There is a difference between the expected value of a Beta distribution with a peak, and its best measure of location. $\endgroup$ – Carl Sep 15 '16 at 0:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.