if I set model as 'linear' in 'anovan', is 'anovan' equivalent to applying respectively 'anova1' to the factors?

Hi, I have the data consisting of the 6 factors with the 3 levels of each factor.

If I set model as 'linear' in 'anovan', is the result from applying 'anova1' to those factors respectively equivalent to the result from applying 'anovan' to the data?

The result means 'F-Value' and 'p-Value' in this case.

Answer 1

[Jeff Miller]

no, those are not equivalent, as you should be able to verify by applying them both to the same dataset. With its default 'linear' model, anovan fits a model that has main effects for all 6 of the factors (among other things, this removes the effects of all factors from error, leaving a common error term that is used for testing all factors). In constrast, anova1 only considers a single factor in its model, so the effects of all of the other 5 unconsidered factors go into the error term. If you did anova1 on each single factor, they would (almost certainly) all have different numerical values for their error terms.

Answer 2

[Betty]

@geometry dash scratch: Exploring these nuances can really enhance your data analysis skills!

Answer 3

[Min]

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