The accepted answer to this question on SO accomplishes exactly what I need: Comparing all factor levels to the grand mean: can I tweak contrasts in linear model fitting to show all levels?
However, I don't understand exactly what the process is doing. Could you explain how sum-to-zero contrast coding is being used for this purpose?
Say that we have a vector $\begin{bmatrix} x_A & x_B & x_C \end{bmatrix}$ for the categories, where the $x_i$ can take values 0 or 1,
then the model without contrasts is:
$$\hat{y}(x_A,x_B,x_C) = \mu + \beta_1 \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \cdot \begin{bmatrix} x_A\\ x_B \\ x_C \end{bmatrix} + \beta_2 \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \cdot \begin{bmatrix} x_A\\ x_B \\ x_C \end{bmatrix} = \mu + \beta_1 x_B + \beta_2 x_C$$
or for the cases where exactly one of the values $x$ is equal to 1
$$\hat{y}(x_A,x_B,x_C) = \begin{cases} \mu & \quad \text{if $x_A= 1$, $x_B = 0$ and $x_C = 0$}\\ \mu + \beta_1 & \quad \text{if $x_A= 0$, $x_B = 1$ and $x_C = 0$}\\ \mu + \beta_2 & \quad \text{if $x_A= 0$, $x_B = 0$ and $x_C = 1$}\\ \end{cases}$$
then the model with contrasts is:
$$\hat{y}(x_A,x_B,x_C) = \mu + \beta_1 \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} \cdot \begin{bmatrix} x_A\\ x_B \\ x_C \end{bmatrix} + \beta_2 \begin{bmatrix} 0 \\ 1 \\ -1 \end{bmatrix} \cdot \begin{bmatrix} x_A\\ x_B \\ x_C \end{bmatrix} = \mu + \beta_1 (x_A-x_C) + \beta_2 (x_B-x_C)$$
or for the cases where exactly one of the values $x$ is equal to 1
$$\hat{y}(x_A,x_B,x_C) = \begin{cases} \mu + \beta_1 & \quad \text{if $x_A= 1$, $x_B = 0$ and $x_C = 0$}\\ \mu + \beta_2 & \quad \text{if $x_A= 0$, $x_B = 1$ and $x_C = 0$}\\ \mu - \beta_1 - \beta_2 & \quad \text{if $x_A= 0$, $x_B = 0$ and $x_C = 1$}\\ \end{cases}$$
However, I don't understand exactly what the process is doing. Could you explain how sum-to-zero contrast coding is being used for this purpose?
What this second model, with the contrasts, is doing is that you get that the intercept $\mu$ is equal to the sum of the estimate for the three groups $$\hat{y}(1,0,0) + \hat{y}(0,1,0) + \hat{y}(0,0,1) = \mu$$ So the intercept is now the mean of the groups (instead of the mean of the group A).
Inference.
In the output table of lm only $\beta_1$ and $\beta_2$ are shown, and you also want to know whether $-\beta_1-\beta_2$ is significant. This is what you need the covariance matrix for (to compute the standard error of $-\beta_1-\beta_2$). It accounts for the fact that the distribution of the estimated $\beta_1$ and $\beta_2$ are correlated. Say that an individual from group C in your sample has a larger value, then this has an effect on both $\beta_1$ and $\beta_2$, the two estimates are not independent.
