# Normality test before testing the difference between two groups. Is it necessary?

I would like to know, when comparing the means in two groups, one with 15 patients and another with 70, if it is necessary to test for normality.

• Are you using a $t$-test? If so you may want to at least visually compare your distributions to the normal. But if you're worried about normality there are always non-parametric alternatives to the $t$-test. – dsaxton Nov 22 '16 at 18:46
• Using a nonparametric test is much better than testing for normality and assuming that such a test has reasonable power (if often doesn't). – Frank Harrell Nov 22 '16 at 18:51
• Normality tests show that my sample does not follow a normal distribution. If I do test t student the result is not significant. If we performed Mann-Whitney U the result is significant (<0.05). My question is about the size of my sample, if having more than 30 data in a single group, it is always assumed to be normal. Thanks – juanmeque Nov 22 '16 at 19:12
• None of that makes sense to me. It is not valid to do two tests. Prespecify one test. And I've seen a sample of size 50,000 where the t-test performed horribly. – Frank Harrell Nov 28 '16 at 18:41
• Related (possibly even duplicate): Is normality testing 'essentially useless'? – amoeba Nov 29 '16 at 9:05

While it is possible to test for normality, it is often not very useful to do so. Very few datasets come from an exactly normal distribution and many parametric statistical procedures work well even when the distribution is only "kind of normalish".

(I will note that the unequal sample size may mean that procedures might not be quite so robust to departures from normality as would be the case with equal samples.)

When the sample is small it contains little information about its underlying distribution and so the normal distribution test has low power and you get lots of false negatives. Conversely, when the sample is large and the test has high power, it starts to indicate significant departures in cases where the distribution is close enough to normal that there is no real problem.

Examine your data in a couple of normal distribution plots to get a feel for the shape of the distributions. If there is substantial deviation then you can either transform the data (log transformations are often appropriate) or use non-parametric methods. With sample sizes of 17 and 70 most non-parametric tests will have good power relative to the normal distribution based tests. For example, a permutations test will power equal to that of a Student's t-test.

Really you should provide a lot more information in your question, such as what the measurements are, what sort of tests you wish to perform, whether the research is exploratory or designed, what hypotheses you are interested in, and so on. That way the answers can be more specific and you will gain more assistance.

• Excellent answer. Thank you! My study is retrospective observational. I have a sample with 17 women with cancer and 70 women without cancer. I want to know if the average lesion size are different in both groups. Test Kolmogorov-Smirnov Cancer group p = 0.200 Non-cancer group p = 0.000 T-test p = 0.07 Mann-Whitney U p = 0.015 Is it correct in this case to use a non-parametric test like U Mann-Whitney? – juanmeque Nov 22 '16 at 21:31
• You want to make an inference about the lesion sizes, so you should look at the lesion sizes first, not the P-values from various tests. However, the tests agree quite well as the difference between P=0.007 and P=0.015 is trivial for almost all purposes. (And note that neither P-value is small enough to imply that the evidence for a difference between the groups is very strong.) – Michael Lew Nov 22 '16 at 22:29
• The mean in the cancer group is 23 and the mean in the free cancer group is 17. My doubt is that using statistical t-test the differences are not significant and using statistical mann Whitney the differences are significant – juanmeque Nov 22 '16 at 22:42
• "I want to prove" Sorry, you can't. "Statistically those differences are real" Sorry, a meaningless phrase. What you are probably after is a statistical justification for making a claim or inference. Your results seem to support prior assertions, and so gain some mutual corroboration from that. – Michael Lew Nov 22 '16 at 23:16
• Your results are at the margin for an asterisk by the arbitrary weak convention of taking less than 0.05 as sufficient. It's your responsibility to make the inference, not that of the statistical procedures. (The Mann-Witney U-test is probably a good choice.) – Michael Lew Nov 22 '16 at 23:19