2
$\begingroup$

I have done an one-way anova test for to compare among 4 groups. I did the homogeneity test for variance and the result showed the variance was not equal but there was a significant difference between group (p=0.013). Thus, I did the Dunnett's T3 test for post-hoc test. However, the p values between all the groups were more than 0.05. How would this happened? Does it mean there are no significant difference between the 4 groups?

Thank you for answering my question.

$\endgroup$
2
  • $\begingroup$ ANOVA assumes homogeneity of variances. Only under this assumption it is the test of that there is at list one difference. Dunnett's test is for nonhomogeneity case and therefore doesn't have to comply with ANOVA results. $\endgroup$
    – ttnphns
    Commented Apr 13, 2013 at 13:47
  • $\begingroup$ @ttnphns Oh, I see. This solved my doubt. Thank you very much. $\endgroup$
    – Lin
    Commented Apr 13, 2013 at 14:04

1 Answer 1

2
$\begingroup$

Be careful not to interpret ANOVA as a test of whether there is at least one different group among the bunch. It is a test with the null hypothesis that the between group variance is 0. A significant result doesn't guarantee some group will deviate significantly from some other group.

$\endgroup$
2
  • $\begingroup$ Thanks for your answer. As I know, if the ANOVA test showed significant result, there should be a significant difference between at least 1 pair of groups, am I right? However, the post-hoc test showed no difference between ALL groups. So, I wondered is this possible? $\endgroup$
    – Lin
    Commented Apr 13, 2013 at 6:10
  • 1
    $\begingroup$ I was hoping my answer would steer you away from that idea. ANOVA tests for significant variance among the set of groups. Significant variance can be present without any pair of groups being significantly different. Granted, significant variance should lead one to a stronger belief that some group dyad will have significantly different means, but no guarantees. $\endgroup$
    – ndoogan
    Commented Apr 13, 2013 at 22:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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