Sumber:
http://elsa.berkeley.edu/sst/regression.html
F.1. Hypothesis Testing
1. General Remarks
Let us assume U ~ N(0,2I). We had adopted the notation for our model Y = X + U and
the least squares residuals e = Y - X. The OLS estimator is = (X'X)-1X'Y. Making the
obvious substitutions allows us to write
e = Y - X(X'X)-1X'Y
e = (I - X(X'X)-1X')Y
e = (I - X(X'X)-1X')(X + U)
e = (I - X(X'X)-1X')X + (I - X(X'X)-1X')U
The first term is zero, so e = (I - X(X'X)-1X')U. Recall from the linear algebra chapter
that (I - X(X'X)-1X') is an idempotent matrix of order nxn. But what of its rank?
Rank(I - X(X'X)-1X') = Rank(I) - Rank(X(X'X)-1X')
The identity matrix is rank n. The rank of X is k so the rank of the second term is also k.
So we are left with the following conclusion
e'e =U'(I - X(X'X)-1X')'(I - X(X'X)-1X')U
e'e = U'(I - X(X'X)-1X')U ~ 22n-k
In what follows we will adopt the following naming and notation conventions:
e'e is often called the residual sum of squares, RSS
Y'Y is often called the total sum of squares, TSS
Y'Y - e'e is the estimated or explained sum of squares, ESS
Depending on the objective, the explained sum of squares is often
written in several other forms:
These are all found by substitution for the definition of Y and/or .
We now have sufficient machinery to test some hypotheses.
2. Student's t test
Define
In general we may wish to test the hypothesis
where
Some specific examples include the following
1. Choose a1 = 1 , ai = 0 for i=2,3,...,k, r=0. Then the null is that 1 = 0.
2. Choose ai = 1 , aj = 0 for j  i and r = 0. Then the null is that i = 0.
3. Choose ai = 1  i. Then the null is that i = r.
4. Choose ai = 1 , aj = -1 , all other al = 0. Then the null is that i - j = 0.
As a test statistic for the general form of the hypothesis test we propose