I'm new to statistics and the only software i know how to use is SPSS. I need help....

I have 3 independent groups with sizes: 24, 27, 37. I wanted to test the difference between the means of their ages using ANOVA.

                                *mean age ± SD*    
                       *Group1:   40.7297 ± 6.01225  
                       *Group2:   31.5926 ± 4.93231 
                       *Group3:   32.125  ± 4.4557

I have no problem with variances because Levene's test says they have equal. When I tested the NORMALITY of each group (with Shapiro-Wilk), one of them is NOT (p=0.022). I tried transforming my data using natural log and checked for normality of each.This time the other group gave p=0.031. Is it correct if i still use ANOVA? or is there any other test I can use with this situation?

Any help would be much appreciated.


The ANOVA's parametric F-Test is fairly robust to the normality-assumption violation (Maxwell and Delaney, 2004), and you have one group barely reaching the 5% significance level, so it should be OK to perform the F-test. You could run modified-parametric F-test like Welch's or Brown/Forsythe Test, if SPSS still has these options. If the results differ between parametric and modified-parametric tests, you may want to stick with the result based on the latter. If you are trying to testing the equality of medians across the groups, I would not recommend the Kruskal-Wallis test unless the three groups' distributions are similarly shaped. This should not be the case, since only one group was deemed non-normal by the SW test.

By the way, you should not rely on Levene's test for testing heterogeneity of variance because the test is not powerful (i.e., tends to produce more type II errors). Regardless, your data seem fine since the sample sizes and SDs are not too far off of each other: Max(SS)/Min(SS) < 4 (Maxwell and Delaney, 2004).

  • $\begingroup$ The Kruskal-Wallis test does not assume similar distribution shapes, and is therefore widely applicable. In it's most general sense it is a test for stochastic dominance among $k$ groups. The question of similarity of distributions only arises when one wants to interpret the results of the Kruskal-Wallis test (and say, post hoc multiple comparisons' i.e. using Dunn's test) in terms of median difference. $\endgroup$
    – Alexis
    Aug 5 '14 at 18:54
  • 1
    $\begingroup$ I have edited the text for clarification. I am assuming that the poster wants test the equality of location. $\endgroup$ Aug 5 '14 at 19:17
  • $\begingroup$ So, i may use Kruskal-Wallis test if I just wanted to find out that one group dominates the other 2? $\endgroup$
    – don
    Aug 6 '14 at 5:43
  • $\begingroup$ @MasatoNakazawa, I just read about Welch's and Brown-Forsythe tests. Correct me if I'm wrong, (the way I understand it) those are alternative test when the assumption of homogeneity of variances has been violated, not the normality. here. Regarding what you said about the robustness of ANOVA against normality violation, thanks for your thoughts, maybe i'll go for it. $\endgroup$
    – don
    Aug 6 '14 at 7:43

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.