0
$\begingroup$

It seems like a basic questions, but I found so many answers on the internet... I have an experiment in which I observe 3 plants per modality. I have 6 different treatments; All the conditions are okay to run an ANOVA ( normality, homogeneity and independance). Is it correct to run an ANOVA with only 3 replications per treatment ? Knowing that I also ran a Kruskal Wallis, I obtained the same significant differences than the ANOVA, but only without the Bonferroni adjustment. With the adjustment, none of my results are significant.

Thanks

$\endgroup$
2
  • $\begingroup$ What is modality? You have 6 treatments, each treatment has 3 replicates, so 18 observations in total? Why do you think an ANOVA could be wrong? $\endgroup$ Aug 2 '19 at 12:45
  • $\begingroup$ Modality = treatment in my case sorry ; + 1 control, so 21 observations in total ; I don't really know the reasons... A professor told me that with less than 30 replications, I should run a Kruskal Wallis rather than an ANOVA ; and then I found contradictory afirmations on the internet so.... $\endgroup$
    – CM31
    Aug 2 '19 at 12:53
1
$\begingroup$

ANOVA doesn't have a minimum sample (or category) size requirement, as long as it can be estimated. The only difference between an ANOVA and Kruskal-Wallis is that KW is the non-parametric version of ANOVA, used when the assumptions of ANOVA do not hold.

The reason why someone could suggest using a KW instead of an ANOVA in such cases is if you cannot really be certain that the assumptions hold, since you have so few data points to check the assumptions, so it's better to err on the side of caution. You would however lose power this way.

Also, there are other better approaches than Boferroni for p-adjustment.

$\endgroup$
3
  • $\begingroup$ Thanks a lot for your help $\endgroup$
    – CM31
    Aug 2 '19 at 13:05
  • $\begingroup$ To see how low power can be, consider a two-group KW with 3 per group. Since the lowest achievable exact significance level is 10% (NB the chi-square approximation isn't much good in this case), you would never reject at 5% -- so power is actually 0 if you're looking for p<0.05. Adding more groups gives you more significance levels, but the Bonferroni will pretty much undo that, leaving you more or less back at having essentially no power. A partial solution would be a good parametric assumption (one you can defend on the basis of theory or previous studies, say) but even then ... ctd $\endgroup$
    – Glen_b
    Aug 3 '19 at 2:08
  • $\begingroup$ ctd... power is going to be a problem unless effect sizes are quite large. You need to compute power before collecting data so you don't waste experiments $\endgroup$
    – Glen_b
    Aug 3 '19 at 2:08
0
$\begingroup$

Yes, it's fine to do an ANOVA. A Kruskal-Wallis test is basically an ANOVA on the ranks instead of the values themselves. If you do the ANOVA and it turns out that the residual distributions are a problem (and cannot be dealt with via a transformation or generalized linear model), you can do a Kruskal-Wallis instead.

I would not do a Bonferroni, but I also recommend not paying too much regard to p-values in general.

$\endgroup$
1
  • $\begingroup$ Thanks !! Have a nice day $\endgroup$
    – CM31
    Aug 2 '19 at 13:13

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.