I am trying to adjust a hierarchical multiple regression model and no matter which transformations I use (z-transformation, sqrt, cuberoot, inv, inv sqrt ...), I do not manage to get the residuals normally distributed. What can I do? Does anyone have any suggestions? I also checked out a related thread, the transformations suggested there did not help me. My residuals are normal according to D'Agostino Normality Test, but not according to Shapiro-Wilk (which is the crucial one according to my supervisor). I cannot use a non-parametric model. I would appreciate your help a lot! Thanks, brobdingnag!
1 Answer
Why can't you use non-parametric tests? Depending on what you are trying to learn there is probably a meaningful permutation test that would give meaningful results without needing the normality assumption (and is more meaningful that rank based tests).
Also note that the normality tests are really answering the wrong question. If you residuals are near normal (how near depends on sample size and other factors) then the normal based tests may give results that are close enough even though the normality tests may reject the "exact normality" hypothesis.
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$\begingroup$ Thanks for your reply! I want to figure out the independent effect of a cognitive ability on a psychological state whilst controlling for certain personality traits. i am using hierarchical multiple regression to figure out how much additional variance of the psychological state is explained by the cognitive ability on top of some traits in personality. I have to meet the assumption that the residuals are normal. $\endgroup$ Mar 14, 2014 at 21:51
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$\begingroup$ oh i forgot that the box cox transformation does not do the job... $\endgroup$ Mar 14, 2014 at 21:53
normalization
tag, in order that you see that "normalize" doesn't mean 'transform to normality'. There are many, many posts here discussing the issues with explicit tests of normality assumptions. Failure to reject doesn't mean your residuals are normal (that is D'Agostino test does NOT say your residuals are normal). On the other hand, a Shapiro-Wilk rejection on this data may be of no consequence whatever. $\endgroup$