This question cannot be answered without any further information. There are a lot of misconceptions, what should be normally distributed and when you can rely on your teststatistic although specific parameters are not normal. There are also discussions here on RG, e.g.

https://www.researchgate.net/post/In_case_of_larger_sample_size_more_than_700_sample_can_I_do_t-test_though_the_data_is_not_normally_distributed#view=5fcfe7b300e40a5b27345234

Just follow Bruce Weaver and some of his publications here on RG to get an impression and overview.

Hello Stamatia,

The glib answer is, "Of course you can!" The follow-up comment to that is, "But, doing so might not be wise."

As Rainer says, it depends on: (a) whether the normality assumption was correctly evaluated in the first place; (b) the degree to which the results depart from normality; and (c) the specific procedure and sample size. The only way to have tremendous confidence in how robust the anova might be to violations of the normality assumption would be to run simulations of its behavior (and recording false positive and false negative test results over many thousands of trials).

OR...you could just use a method that doesn't depend on distributional assumptions. Bootstrap/resampling methods are more commonly available in statistical software packages, and offer an easy option.

Good luck with your work.

If the assumption of normality is violated, you should use nonparametric tests, such as Kruskal- Wallis.

The hypothesis should not be rejected without a statistical test.

What's your samples size and what tool do you use to get to that conclusion ?

The only assumption that makes sense to me is that the expected value (i.e. the mean) is what you are interested in

The normality assumption was imposed more than a century ago in order to analytically come up with the distributional form of the statistics.

Data in real life are never normally distributed, but how far from normality and what tools you use to investigate it should be your focus

Hatim Al-salhi what you are suggesting is far from being correct, KW is an extension of the Wilcoxon test, which is known to be problematic because it is interpreted as a location test, while in fact it is a test of stochastic dominance.

Again what the test tells you is hard to interpret, aka that is why most people have a problem with frequentist methods.

Thank you all for your answers.

So i have 90 participants in total and those participants are separated in 3 groups.

And they are also separated in other two groups.

So 90 participants in 3 groups

And the same 90 participants in 2 groups.

I used the Shapiro wilk to see if the data are normally distributed.

So, Stamatia, you have two groupings making 3x2=6 subgroups (combinations), if I interpret you right. It is probably wise two consider the two groupings jointly in this way, which means you should consider a two-way ANOVA. This allows both groupings to have effect, and also allows interaction between the grouping factors.

Concerning you normality test, I'm afraid you might have done it by checking for normality of the whole dataset as one sample (this is quite common). But if you have done so, the result of the normality test is irrelevant, because if there are differences between groups, the total data should not look normally distributed. A check of normality should be done by looking at each group separately (each of the 6 groups). And personally I should not use a formal test, but only "looking".

A final point is that your partitioning of the variation to be shown in any ANOVA holds without normality assumptions. It's only when you want to go from description (of how the variation is explained by groupings) to inference (e.g. testing for statistical significance), that you more or less require normality.

I used spss to check the variables and I also added the histograms.

Before interpretating the results I did the split file command and choose the option organize output by groups.

Is it still possible that I might have done something wrong?

You say you used shapiro wilk to check if your data is normal

But still you don't say what the outcome of that test is or show a plot of your data 🤷🏼‍♂️

Depends what your null hypothesis is. If it is that the population residuals are normally distributed you can, if it is that the population means are the same, you cannot. I recommend deleting your question and writing one where you actually say what your null hypothesis is rather than letting people pretend that they know and potentially wasting their time.

The normality of the distribution is only a condition that must be met, but it is not the most important in the case of your sample. Much more important is the homogeneity of the variant at the level of the subgroups determined by the independent variables for the dependent variable. If this condition is not met there are important consequences on the analysis. This condition is verified by setting the Homogeneity tests option that calculates Levene statistics. If the value of p is less than 0.05, then there is an unequal variation between subgroups which determines the acceptance of the inhomogeneity hypothesis.

Best regards,