Bioequivalence and Bioavailability Forum

In matrix notation will write

y=Xb+e

y is a vector of observations.
X is the design matrix.
b is the effects vector.
e is the error or residual vector.

Let's somehow try to get the dataset above transmogriffed into some kind of matrix notation.

I will not go through matrix notation or mutiplication here. Google it, or use Wikipedia. Along the same lines, I will not go through anything that leads to the least squares solution of the normal linear model, but I will mention without proof that:

1. If the error is normal with mean zero and some variance then the most likely set of constants in b is those for which the sum of squared residuals ete is minimal.
   That solution is analytically
   b=(XtX))-1Xty
   
   
2. This solution implies inversion of the matrix XtX. If this matrix cannot be inverted then we have screwed up.
   
   
3. For all practical purposes, if XtX is invertible, then we have a solution. This is the case when the rank of X equals the number of columns in X.
   
   
4. If any columns of X add up to give another column (or sum of columns) in X then the rank is not equal to the number of columns, and we cannot get a solution.

We can write the equations like this (focus for now only on sequence and observations):

17.11=1*intercept + 1*Seq_TR + 0*Seq_RT + …
17.76=1*intercept + 1*Seq_TR + 0*Seq_RT + …
…
…
…
22.87=1*intercept + 0*Seq_TR + 1*Seq_RT + …
20.80=1*intercept + 0*Seq_TR + 1*Seq_RT + …


Let's transmogriph this into a model matrix. In X we have a column for each constant we wish to determine. Above we just worked with three constants (intercept, and two sequence constants).
In the rows of the intercept put a one if the observations is associated with an intercept. All observations are so this is a column of pure 1's.
In the rows of Sequence TR put a 1 when the observation comes from a subject in sequence TR otherwise a 0. And so forth for Sequence RT.

We can write the model matrix (shown right in reddish color) like:



Note the bold constants just above: They correspond 1:1 to how they appear in X.