I have conflicting web-based references for the following problem:

If a response variable y is normally distributed,
and so:
Beta(be) ~ N(Beta,sigma^2/Sxx)

Beta(be)-Beta / [sigma/sq(Sxx)] ~ N(0,1)

Where:
Beta = slope of least square fitted line
(be) = best estimate
Sxx = sum of squares of x

With:
sigma^2=var(y) will be unknown

can estimate sigma^2 using sigma(be)^2=RSS/(n-2)

Then
Beta(be)-Beta / se[Beta(be)] = Beta(be)-Beta / [sigma(be)/sq(Sxx)]

Has "Students" t-distribution with n-2 df

Where:
se = standard error


For t-test:

x1(avg)-x2(avg)/se(x)[pooled]


(Q1) is [sigma(be)/sq(Sxx)] the standard error or the standard
deviation?
(Q2) if [sigma(be)/sq(Sxx)] is the standard deviation then the
standard error should be [sigma(be)/sq(Sxx)]/sqrt(1/n)?


Any comment would be greatly appreciated.

Roy Carambula