Hi,

In my opinion,

You should use two-way ANOVA.

Cheers,

Ramin

If you compare tests of hypothesis ANOVA TEST always be preferred. ..

when you want to use an ANOVA, your dependend variable would be the satisfaction which seems not to be metric, but ordinal. so it would not be possible to use an ANOVA (you could use some test for ordinal data, for example the krukal wallis H-test).

if you want to test if there are more researchers in one university who are highly satisfied than in another university (I guess this is your question) then you are comparing frequencies and so you should use the chi²-test.

Your response variable (satisfaction) seems to be ordinal, and you seem to have two predictors (faculty membership and university), so I would suggest an ordered logistic model. If the three universities are only three "random" universities you may model "university" as a random factor (-> ordered logistic mixed model).

Anova would require to translate the categories of "satisfaction" to numerical values. Hiw this is done will impact the results obtained by the Anova. So if you have any good reason for a particuar numerical coding, Anova could give sensible answers. If you don't think about the numerical coding, Anova can give you answers you will likely misinterpret. Instead of calling this "Anova" I would call it a linear model. Again, the predictor "university" can be modelled as a random factor (-> linear mixed model).

With a Chi-squared test you can test the hypothesis that some frequency distributions were all equal. This can surely be done with your data, but I doubt that the question (or the answer to such a question) is of any help. If you find that the distributions are "significantly different" you will still need to find which and where - but these questions are specifically adressed in the ordered logistic model anyway.

Thanks all.