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For example, if I have the data

$$ \begin{array}{l|l|l|l|l|l|l} A & low & & medium & & high & \\ \hline B & standard & new & standard & new & standard & new \\ \hline means & m1 & m2 & m3 & m4 & m5 & m6 \end{array} $$

And I want to create contrasts, then apparently there are 5 that can be created.

$$ \begin{align} C_1 &: \text{New v Standard} \\ C_2 &: \text{Low v High} \\ C_3 &: \text{(Low and High) v Medium} \\ C_4 &: \text{Interaction between $C_1$ and $C_2$} \\ C_5 &: \text{Interaction between $C_1$ and $C_3$} \\ \end{align} $$

If I create the additional contrast

$$ C_6 : \text{Medium v High} $$

What has been violated?

I don't really understand why this isn't an option.

Why there are only five

From my notes, with respect to the table provided, all I have to explain this is the following

Since the treatments occupy a six-dimensional vector space, it is only possible to find six orthogonal vectors. One of these dimensions is occupied by the overall mean:

$$ C_0 = \left[ 1 \quad 1 \quad 1 \quad 1 \quad 1 \quad 1 \right] $$

Hence there will only be 5 orthogonal contrasts - one for each treatment degree of freedom.

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    $\begingroup$ Nothing has been "violated:" $C_6$ is a perfectly fine contrast. In what sense is it "not an option"? Perhaps you are using an unusual definition of "contrast"? $\endgroup$
    – whuber
    Dec 31, 2018 at 17:13
  • $\begingroup$ "Apparently" is a bit mysterious. To whom is it apparent? You can create an infinite number of contrasts - most won't make much sense, but you can create them. $\endgroup$
    – Peter Flom
    Jan 1, 2019 at 14:10
  • $\begingroup$ sorry for the mystery - I've added some more information, hopefully that clears it up $\endgroup$
    – baxx
    Jan 1, 2019 at 22:55
  • $\begingroup$ @whuber but it couldn't be included as well as the rest presumably? I can't have c1 to c5 and c6, ? This is the impression that I have been given at least $\endgroup$
    – baxx
    Jan 2, 2019 at 1:12
  • $\begingroup$ @PeterFlom is there actually an infinite amount that can be created from an initially set of 5 means? I've added some more information, hopefully that clears it up some what $\endgroup$
    – baxx
    Jan 2, 2019 at 1:13

1 Answer 1

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There are only 5 that are all orthogonal to each other. But you can have different sets of 5. Your proposed 6th contrast is not orthogonal to the other 5, but if you drop one of those 5, then you can add it.

See this post from University of Southampton which gives examples of the different possible sets of contrasts for factors with different numbers of levels (they stop with 5 levels, but the idea is the same).

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