Maximizing $R^2_{adj}$ is the same as minimizing ${SS_{res}^{(q)}\over n-q-1}$? When choosing q regressors from p regressors (q < p), show that choosing the subset model that maximizes the adj-R$^{2}$ value is equivalent to choosing the subset model that minimizes the ${SS_{res}^{(q)}\over n-q-1}$,
where R$^{2}$$_{adj}$ = 1 - ${n-1\over n-q-1}$(1-R$^{2}$$^{(q)}$), and R$^{2}$$^{(q)}$ is the usual R$^{2}$ value for the subset model.
The (q) superscripts refer to the fact that those values it is attached to are for the model with q regressors, not the full model with p regressors.
This is in regards to variable selection, and I think it is meant to be more of a thinking / rearrangement question.
Really I want to know if I am supposed to show that the max of adj-R$^{2}$ is the same as the min of ${SS_{res}^{(q)}\over n-q-1}$, and how I would go about doing that.
 A: The quantity $q$ in the formula for $R^{2}_{adj}$ is not a random variable.  It must be a pre-specified constant.  So if you "play around" with $q$ to optimize $R^{2}_{adj}$ you are not being honest about $q$ and the de-biasing property of the adjusted measure no longer holds.  If you want to use a data mining/data dredging procedure to arrive at a model it is better to de-bias regular $R^2$ (i.e., correct for overfitting) using the Efron-Gong optimism bootstrap (e.g., the validate function in the R rms package) or using 100 repeats of 10-fold cross-validation (also can be done using validate).
A: Okay, I feel like I should have looked at this for a couple more minutes before I posted my question, but I'll answer it now for anyone who needs help in the future.
So, if the question really is asking if max(R$^{2}$$_{adj}$$^{(q)}$) is the same as min(${SS_{res}^{(q)}\over n-q-1}$), then we will start by considering the max(R$^{2}$$_{adj}$$^{(q)}$).
Since R$^{2}$$_{adj}$$^{(q)}$ = 1 - ${n-1\over n-q-1}$(1-R$^{2}$$^{(q)}$), by saying we want to maximize it, we really want to make ${n-1\over n-q-1}$(1-R$^{2}$$^{(q)}$) as small as possible. To do this, R$^{2}$$^{(q)}$ must equal 1.
Okay, now let's look at min(${SS_{res}^{(q)}\over n-q-1}$). 
Another way to write SS$_{res}$$^{(q)}$ is (1-R$^{2}$$^{(q)}$)SS$_{tot}$$^{(q)}$ (this is easily seen by rearranging a few equations).
To make this as small as possible, R$^{2}$$^{(q)}$ must equal 1.
And that's our answer!
A: 
Maximizing $R^2_{adj}$ is the same as minimizing ${SS_{res}^{(q)}\over
 n-q-1}$?

Probably the fact that put you in the confusion is that $q$ is not precisely a free parameter/number here. You can maximize R$^{2}$$_{adj}$ but this mean that, given a full set of $K$ eligible predictors, you looking for a subset of them that maximize R$^{2}$$_{adj}$. Moreover note that maximize $R^2$ is a senseless goal.
At the end of the story you have a, before unknown, number of predictors, let me say $q<K$, that permit you to maximize
R$^{2}$$_{adj}$ = 1 - ${n-1\over n-q-1}$(1-R$^{2}$$^{(q)}$)
among all possible R$^{2}$$_{adj}$
But note that for any $q$ you can have several models (all combinations of $q$ elements in $K$). Then, said the number $q$ is not enough. You need a precise subset of $q$ elements.
