Adjusted R-squared formula I am studying linear regression lately and I notice this adjusted r-squared formula in a youtube video:
$$adj. R^2 = \frac{\frac{SSE}{n-k}}{\frac{SSTO}{n-1}}$$
While the formula that I know is this:
$$adj. R^2 = \frac{\frac{SSE}{n-k-1}}{\frac{SSTO}{n-1}}$$
I was about to comment in the yt vid that something is wrong but some people agree with it.

*

*How come that both formula seems to be applicable?

*Why does the other one has a n-k degrees of freedom while the other has n-k-1?

I'm so confused. I don't think of a way that you can equate the other formula to another formula.
 A: I now understand what I'm asking with the help of Dave and V.Vancak from math.stackexchange.com, but if my wording is wrong please correct me. So, the adjusted r-squared formula:
$$adj. R^2 = \frac{\frac{SSE}{n-k}}{\frac{SSTO}{n-1}}$$
Indicates that k includes the explanatory variables and the intercept term. Which implies that k = number of parameters.
While this formula:
$$adj. R^2 = \frac{\frac{SSE}{n-k-1}}{\frac{SSTO}{n-1}}$$
Indicates that k = explanatory variables only and the 1 is the intercept term.
All in all, these two formula are the same.
A: In the adjusted R squared the numerator should be the unbiased estimator of $\sigma^2$, namely the SSE divided by the degrees of freedom of the residuals, that is, the sample size minus the number of regression coefficients. Usually, $k$ denotes the number of explanatory variables, and thus you add $1$ for the intercept $\beta_0$ (hence $n - (k+1)$). However, sometimes $k$ denotes the number of coefficients including the constant (hence $n-k$).
