Hello, I am analyzing a data set based on the trait values under drought condition. The experiment was carried out on three separate drought environments. First I have done a single experiment analysis for each environment to extract the entry means. I have also extracted the degrees of freedom and standard error from single analysis. Now, I have combined the entry means from each environment to run a combine anova. The model is: Y=meu+Genotype+Environment+Genotype*Environment+Error. I have done the analysis. My F-test for the main effect genotype is non-significant, the interaction the term is significant at 10% and main effect environment is significant at 5%. Looking at the means for grain yield of each genotype I have found some differences which are higher than my LSD value at 5%. Then I performed a pairwise comparison test and found significance differences at 5%. Now, I am wondering how can that happen. Is it possible that the F-test is non significant and pairwise comparisons are significant? I would be very grateful if someone can explain me this issue with references.
Thank you.
If the omnibus test for some effect is non-significant, you shouldn't be proceeding to make pairwise comparisons anyway.
Also, the fact that you got a significant Genotype*Environment error means that there's not much point looking at the main effect of Genotype (or pairwise comparisons within it) anyway. The interaction tells you that the effects of Genotype are different in different environments, so those more broken-down effects are what you should be looking at in your follow-up comparisons.
Ultimately, your research question and hypotheses will dictate what comparisons you need to be making. There is no single automatic answer.
Well I want to why and how this happened. What could be the reason that I have non significant genotypic main effect but significant differences between the means of the genotypes?
I can't be sure without seeing the data, but I suspect it could be because you have an interaction. It is possible to have, for instance, opposite effects of genotype at different levels of environment, thus causing you to observe significant simple effects but not a significant main effect.
Dear Anil. If the F-test is not significant, we generally do not proceed to to calculate LSD and pairwise comparison. LSD test is not meant to replace F-test. LSD test is second step once the F-test is significant. I also get this sort of data, sometimes.
Hi Anik,
The fact that main effects can disappear in the face of an interaction with simple effects in the opposite direction follows straightforwardly from the basics of what an interaction means. For instance, a situation like that shown in the attached figure might cause you to observe an interaction, a simple effect of 1 vs. 2 at each level of a or b, but no main effect of a vs. b or of 1 vs. 2.
Hello Babu and Stephen,
Thank you for the reply. Is it possible to get a non significant main effect for genotype but significant differences in the mean of the genotypes? I have attached a document containing the anova for grain yield and the mean grain yield for all the genotypes. Please look at the LSD and there are differences higher than the LSD. My supervisor is telling me that if the main effect is not significant than why there are differences bigger than the LSD. Can you please see this and explain to me?.I will be very grateful. The + is significant at 10% alpha level
Hi Anik
I looked into your data. Looking at big LSD value, I calculated CV approx. 20% (based on Error variance and mean). A high CV might be indicative of heterogeneity of variance. Did you check homogeneity of variance? You can run Bartlett's test in SAS or in Genstat. If significant, go for appropriate transformation. This will hopefully solve the problem.
Good luck
Hi Anik,
Do you have missing values? If number of repetitions is not the same for all genotypes, you need to account for this when running pairwise comparisons. A single LSD is fine when balanced data.
Hello Arellano,
Yes I have missing data and different number of replications as well.
Hello Babu,
do you think it is a good idea to transform the yield values? Because then you will loose the information about the actual yield values. When you backtransform the results you will report the medians not the means. Isn´t it?
Hi Anik, You are not going to loose info on original yield values. You are still going to report mean values which will be back-transformed from log or other transformation. That will still be mean not median. The median values are compared only when you do non-parametric test. But, the means values back-transformed will not be the same as the original ones. however, it makes your comparison valid.
One thing more, if you had different number of replications I do not think you can use a single LSD value to compare all pair of means. You will have different values of LSD to compare different pair of means.
Like Babu mentioned, heterogeneity of variance (or non-circularity, in a repeated measures context) could cause the omnibus test to be non-significant in univariate ANOVA. There's an example in Franz & Loftus 2012 (http://faculty.washington.edu/gloftus/MiscMissingArticles/Franz.Loftus.PBR.2012.pdf)
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