I have a data set that has 3 groups and a single variable. Under normal circumstances, a one-way ANOVA would be used for comparisons between my three groups. HOWEVER I ran a Shapiro-Wilk test for normality and Levine's test for homoegeinity and turns out my data is both: not normally distributed and heterogeneous.
What statistics should I run instead of a One-way ANOVA and what post-hoc test may be better?
You could use the non-parametric Kruskal-Wallis test for your unrelated 3 groups and you end up with a H-value which you check in a table if it is significant. I do not know anything about your group sizes.
What is your response variable? If it is a count, use a count-model (e.g. negative binomial), if it is a proportion, use a binomial model (or a beta-model, if the denominator is unknown), if it is a concentration-like measure, check if the logarithms are approximately normal distributed.
NB: Real data is never normally distributed, and variances are never identical in different groups. Tests like Shapiro-Wilk and Levene are useful ONLY to check if the data at hand is already sufficient to "see" that. They are NOT useful to judge whether or not any assumption is reasonable or tolerable. This depends on the kind and amount of deviation (of the real data from the assumptions) and on the sample size (and on the complexity of the model). This is both not adressed by these tests. Even worse: the more data you have, the less relevant and more tolerable are deviations, but the higher is the sensitivity of these tests to even irrelevant or negligible deviations!
Think about your data and if the assumptions you want to make actually make some sense or if they are not tolerable. If you think that they should make some sense, have a look at residual diagnostic plots to see if the data screams at you, saying that your assumptions are drastically stupid. If so, reconsider other distributional models making assumptions that seem to be more resonable in view of your data (I mentioned some above). If you think that there is no sensible assumption about the distribution, you can possibly bootstrap the sampling distribution of the statistic you are interested in (e.g. a mean difference).
PS: Note that ANOVA does not compare the means of the three groups. It compares a model including "group" as a predictor against a model not including "group" as a predictor. This does not seem to be the question you want to adress. Comparisons between means are always only between two groups, and this is usually some kind of t-test or z-test. There exist methods that perform t-tests on all possible pairs of groups using a pooled variance estimate for the standard errors and adjust the resulting p-values for multiple testing to control a family-wise error-rate. The vast majority of these "multiple comparison procedures" (e.g. Dunnett, Tukey, Scheffe, Duncan, ...) have nothing to do with ANOVA (exceptions are so-called ANOVA-protected procedures like Fisher's LSD, that are usually not recommended anyway, because they fail to control the FWER in a strong sense).
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1w-ANOVA is somewhat robust to non normality although strong skewness might hurt (classical transforms such as Box-Cox may help)
http://diposit.ub.edu/dspace/bitstream/2445/122126/1/671797.pdf
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heterogeneous variances is a no-go for classical 1w-ANOVA ; fortunately, you can resort to Welch ANOVA
https://statisticsbyjim.com/anova/welchs-anova-compared-to-classic-one-way-anova/
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Kriskall-Wallis is always an option ; just remember it testst for the medians
if your data is strongly skewed, medians may be significantly different from means
EDIT
as noted by Bruce Weaver below, the above is a sloppy and incorrect formulation : KW tests for (absence of) stochastic dominance, not for equality of medians.
the take-away message stays the same : KW and 1w-ANOVA do not test for the same thing and you have to decide if you can indeed use KW as a substitute for 1w-ANOVA, depending on your goals in the analysis
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Hello Ana Guadalupe Hernandez Reynoso. Many questions have been put to you in some other posts. Let me try to boil things down to the key pieces of information you need to provide.
Thanks,
Bruce
EDIT: Point 3 about software added.
You should use the kruskal-wallis test (Non-Parametric test) for three group and you could investigate/check the made results using the relative table (considering that significance level). Finally, the computational H index should be compared with its significance level due to the corresponding table.
Another poster said that the Kruskal-Wallis tests "tests for the medians". I assume that meant it tests a null hypothesis stating that all of the population medians are equal. If so, I disagree. To see why, read the following data into whatever statistical software you use, and perform a Kruskal-Wallis test on it. Then compute the group medians.
Group y
0 5
0 5
0 5
0 5
0 5
0 5
0 7
0 8
0 9
0 10
1 1
1 2
1 3
1 4
1 5
1 5
1 5
1 5
1 5
1 5
2 0
2 1
2 2
2 3
2 5
2 5
2 5
2 5
2 5
2 5
I did that using Stata's -kwallis- command, and got this result:
chi-squared with ties = 10.155 with 2 d.f.
probability = 0.0062
But, for all 3 groups, the median = 5.
It seems clear, therefore, that the null hypothesis being tested by the Kruskal-Wallis test does not state that all population medians are equal.
PS- Credit where credit is due: I generated the data used here by taking a dataset Ronán Michael Conroy used in his Stata Journal article on the Wilcoxon-Mann-Whitney test (see below), made a copy of his group 1 to use as my group 2, and reduced the first 4 values in group 2 by 1 point. ;-)
Article What Hypotheses do “Nonparametric” Two-Group Tests Actually Test?
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thanks, Bruce Weaver , for correcting my sloppy formulation
i edited my answer accordingly
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