0

I am running an anova in SAS on the average hours worked for a country. With those countries grouped into 3 regions. When I run the anova I get a F value of 3.64 and P-value 0.0379. With these results I thought you would reject the null stating that the means for each region is the same. But when I run a post hoc test like Tukey, SNK or scheffe they tell me the means are not significantly different. Does anyone have an idea where I am going wrong?

statsstudent718
  • 61
  • 1
  • 2
  • 5
  • Why do you think anything is wrong? Clearly you'd *like* there to be a significant difference in a post hoc which you could try to claim 'caused' the significant result in the ANOVA, but there's generally no guarantee that will happen, since an ANOVA can reject due to a combination of relatively small effects, none of which will be significant on the post hoc tests you mention. – Glen_b Mar 15 '15 at 22:25
  • The section of [this answer](http://stats.stackexchange.com/questions/83030/can-anova-be-significant-when-none-of-the-pairwise-t-tests-is/83083#83083) labelled "Multiple comparisons t-tests" indicates an example where all the multiple comparisons would not reject *even when carried out at the original significance level that the ANOVA rejected at* --- that is, it can even happen with no adjustment to type I error rates for multiple comparisons. If that can happen, you can appreciate that if you are using a much lower significance level in the multiple comparisons, it can much more easily occur – Glen_b Mar 15 '15 at 23:46
  • 1
    Adding to @Glen_b's point, a significant ANOVA $F$ is equivalent to saying that there is *some* contrast that is significant at the same level, using the Scheffe critical value. That contrast may well not be a comparison of two means. – Russ Lenth Mar 16 '15 at 00:36

0 Answers0