Why does this formula divide by n - 2 when in other forumulas n - 1 has been used?
you have sample mean - when you now estimate sample standard deviation - since you are reusing the sample mean - 1 degree of freedom is lost. So you divide by (n-1) – -1 because sample mean also came from the sample’s n values.
you have sample mean and sample standard deviation (2 degrees of freedom are gone). now you are getting to the std. error of the estimate. which is sqrt(sample variance / (n-2))
numerator is sample variance - which means n - 1 (mean) - 1 (std deviation).
I dug out one of my old undergrad stats texts trying to find this. There’s a nasty looking exercise at the end of the chapter asking the interested reader to prove that a given expression involving the SEE has a t-distribution with (n-2) degrees of freedom.
I think cpk123’s explanation is far less stressful.
variance use (n-1)
correlation use (n-2)
does that work?
Computation of SEE - is that something we need to worry about?
@CMLSML - Based on practice tests and mocks it appears that SEE will most likely always be given, not that the formula is super hard. I would be more concerned with learning the relation between SEE, MSS, RSS, and R^2
As for the N-1 vs N-2 debate, I think gazhoo has the best response. Remember, this is a degrees of freedom issue and I suppose I think of the intercept in regressions as the reason why you subtract 2 instead of 1.
I’m going to stick with this, hope I don’t need to understand why further than that and hope for the best. There are plenty of other areas I am focusing on learning the “why” behind things.