I am planning to do a 2-way mixed ANOVA with the within-subjects factor having four levels. Since the exact data do not matter, I hope to suffice by only providing a general question.

On my endeavours through the internet and different textbooks on statistics I have come to the point that I would like to compute this with ezANOVA from the package ez (as described by Andy Field in his "Discovering Statistics Using R") and since I will be interested especially in interactions between the 2-level between subjects factor and the 4-level within subjects factor I have decided to use type 3 sums of squares rather than the ezANOVA-standard type 2 sums of squares.

In order to calculate type 3 SSs, however, according to Field, I need to construct orthogonal contrasts, since in case I do not, R will not really return type 3 SSs to be (but something else instead, I suppose, on which the textbook does not elaborate. Any explanations of what R does in case will be highly appreciated.)

I have a couple of questions:

  1. Can I generally compute a mixed ANOVA using ezANOVA and type 3 SSs without specifying any contrasts at all and still get interpretable results? I am left confused since the intriguingly easy usage of ezANOVA appears to allow this. However, as stated above, Field writes that one needs to use orthogonal contrasts for type 3 SSs (Andy Field, Discovering Statistics Using R, 2013 reprint, p. 476).

IGNORE: 2. In case I truly need to construct orthogonal contrasts, how can I achieve this with a 4-level factor? I have tried to do this and I have come up with a Helmert contrast (as I have noted post hoc, when searching for a solution online). However, I am confused again, because as far as I know, the sum of products of the contrasts should add up to zero, which for Helmert contrasts does not hold, since the sum of products is not equal to 0:

    condition   C1      C2      C3      Cproduct
        1        3       0       0       0
        2       -1       2       0      -2
        3       -1      -1       1       1
        4       -1      -1      -1      -1
                                  sum = -2

The only explanation that I could come up with for myself is that possibly the "sum of products = 0" criterion is only one way of being sure that contrasts are orthogonal, but that they might also be orthogonal even if this criterion does not hold.

Or is there something special about Helmert contrasts that I am unaware of, some circumstance that qualifies them for orthogonality? STOP IGNORING

Thanks in advance for any help.

EDIT: Greg and Aniko have made me see the obvious mistake in basic arithmetics that I have made regarding the second question in my original post, so it might as well be ignored. Thanks for that one! However, maybe there is someone who could provide an answer to my first question regarding the behaviour of ezANOVA when using type 3 SSs without orthogonal contrasts. This would be great.

  • $\begingroup$ ?contr.sum() will help. $\endgroup$
    – Henrik
    Commented Aug 26, 2014 at 17:30
  • $\begingroup$ If Cproduct is supposed to be C1 * C2 * C3, then check your arithmetic: in condition 2 to product should be 0. $\endgroup$
    – Aniko
    Commented Aug 26, 2014 at 20:08

1 Answer 1


How did you compute the Cproduct column? It does not look like any combination of the other columns that makes sense to me.

The rule is that all products of pairs of contrasts sum to 0, so C1*C2 should sum to 0, C1*C3 should sum to 0, and C2*C3 should sum to 0, which they all do for Helmert contrasts in general, and your example in particular.

  • $\begingroup$ This, obviously was a very dumb mistake on my part. I wonder how I got up to asking this question anyway. Mea culpa, mea culpa! $\endgroup$
    – bunsenbaer
    Commented Aug 27, 2014 at 9:03
  • $\begingroup$ Ridiculous, really! $\endgroup$
    – bunsenbaer
    Commented Aug 27, 2014 at 9:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.