WarrenB1

Way to confuse. Throw in a quant question, and stare at the blank faces of candidates.

By the way i'd think the answer to your question is NO.

SEE = std deviation of error terms.
SEE = sqrt(variance of error)
SEE = sqrt(SSE/n-k-1)

where as MSE = SSE/ n-k-1 <-- there is no square root here.
SSE = squared sum of all errors, or residual sum of errors.

SSE/n-k-1 is not equal to SEE.

By the way what is RMSE? seeing it for the first time.