# Orthogonal contrasts, ANOVA, why are there only as many contrasts there are degrees of freedom?

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.

• 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"?
– whuber
Dec 31, 2018 at 17:13
• "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. Jan 1, 2019 at 14:10
– baxx
Jan 1, 2019 at 22:55
• @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
– baxx
Jan 2, 2019 at 1:12
• @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
– baxx
Jan 2, 2019 at 1:13