Calculating the degrees of freedom

Question

Hi I wonder if there is a standard way to calculate the degrees of freedom. I am having a hard time understanding why these sums of squares has only one degree of freedom.

1. In one-way ANOVA (one factor of $a$ levels), the sum of squares for a contrast $\sum_{i=1}^ac_i\tau_i$ for $\sum_{i=1}^ac_i = 0$ has one degree of freedom. $SS_C = \frac{\left(\sum_{i=1}^ac_i\bar{y}_{i\cdot}\right)^2}{\frac{1}{n}\sum_{i=1}^ac_i^2}$

2. In two-way ANOVA (factor A of $a$ levels and factor B of $b$ levels) and the model is $y_{ij} = \mu + \tau_i + \beta_j + \epsilon_{ij}$, $i = 1, 2, ..., a$, $j=1, 2, ..., b$. Suppose that there is only a single replicate and we assume $(\tau\beta)_{ij} = \gamma\tau_i\beta_j$ for a constant $\gamma$. The Tukey test for significance of interaction says that the following $SS_N = \frac{\left[\sum_{i=1}^a\sum_{j=1}^by_{ij}y_{i\cdot}y_{\cdot j} - y_{\cdot\cdot}\left(SS_A + SS_B + \frac{y_{\cdot\cdot}^2}{ab}\right)\right]^2}{abSS_ASS_B}$ has one degree of freedom. There is a paper talking about this test.

I am very confused about these cases and I wonder if there is any standard way that we can calculate the degrees of freedom.

Thank you very much for your help!

Answer 1

Every time you write the "equals" sign =, you spend one degree of freedom (unless you can derive this equal sign from others supplied before).

You wrote down one contrast (+1, -1, 0, 0)? That's one degree of freedom.

You wrote down two contrasts (+1, -1, 0, 0) and (-0.5, 0, -0.5, +1)? That's two degrees of freedom.

You wrote down three contrasts (+1, -1, 0, 0), (+1, 0, -1, 0) and (0, +1, -1, 0)? Well you can deduce the third one by subtracting the first one from the second one:

$$ \begin{pmatrix}+1 \\ 0 \\ -1 \\ 0 \end{pmatrix} - \begin{pmatrix} +1 \\ -1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ +1 \\ -1 \\ 0 \end{pmatrix} $$

So there are only two equalities you are testing, so that's two degrees of freedom.

You wrote down that you think $\gamma = 0$? That's one degree of freedom.

Answer 2

I think I have found the answer here. The degrees of freedom are "the rank of a quadratic form".