Date:         Sun, 8 Oct 2006 10:51:49 -0400
Reply-To:     Jay Weedon <jweedon@EARTHLINK.NET>
Sender:       "SAS(r) Discussion" <SAS-L@LISTSERV.UGA.EDU>
From:         Jay Weedon <jweedon@EARTHLINK.NET>
Organization: NewsGuy - Unlimited Usenet $19.95
Subject:      Re: NOTE: A vs.B+C is not estimable. BUT IT IS!
Comments: To: sas-l@uga.edu
Content-Type: text/plain; charset=us-ascii

On 7 Oct 2006 21:55:51 -0700, "Hanz" <hans_hockey@yahoo.com> wrote:

> Is that the only solution to getting estimates that are functions of
> treatment that are not contrasts? Use /noint and turn all other model
> terms into indicator variables and drop the last level so as to fool
> SAS?
>
>No, it is more complicated than that!
>
>Given that the LS means are:
>
>            Response        Standard                  LSMEAN
>Fert          LSMEAN           Error    Pr > |t|      Number
>
>A         15.1666667       2.0393472      <.0001           1
>B         15.0000000       2.0393472      <.0001           2
>C         11.1666667       2.0393472      <.0001           3
>D         13.8333333       2.0393472      <.0001           4
>E          9.8333333       2.0393472      <.0001           5
>
>from
>  model Response = Fert Column / ss3 solution;
>
>then A-(B+C) adjusted for Column is estimated as
>
>                                     Standard
>Parameter            Estimate           Error    t Value    Pr > |t|
>A vs.B+C          -11.0000000      3.53225305      -3.11      0.0049
>
>using
>
>model Response = F1 F2 F3 F4 C1 C2 / ss3 solution;
>estimate "A vs.B+C" intercept -3 F1 3 F2 -3 F3 -3 F4 0 C1 -1 C2 -1 /
>divisor=3 e;
>
>Easier to do by hand from the least squares means! Assuming zero
>covariances with standard error = sqrt(3*2.039**2).
>
>HaNZ

There's a whole section of SAS documentation on estimability that you
may want to refer to.

To get an intuitive feel for it, think about this:

GLM's "over-parameterized" model is

E(y) = b0 + b1*F1 + b2*F2 + b3*F3 + b4*F4 + b5*F5,

so that

E(y|A) = b0 + b1
E(y|B) = b0 + b2
E(y|C) = b0 + b3.

The mean of A minus the sum of means of B & C is thus
(b0 + b1) - (b0 + b2) - (b0 + b3), which reduces to
-b0 + b1 - b2 - b3. So you end up with 4 parameters to estimate 3
means, which means there's no unique solution of this quantity.

Contrast this with 2*mean of A minus sum of means of B & C, which is
2(b0 + b1) - (b0 + b2) - (b0 - b3), which reduces to
2b1 - b2 - b3, so there are 3 parameters to estimate 3 means, and
there's a unique solution.

When you remove the intercept from the model you get

E(y) = b1*F1 + b2*F2 + b3*F3 + b4*F4 + b5*F5,

so you have instead

E(y|A) = b1
E(y|B) = b2
E(y|C) = b3.

In this case the mean of A minus the sum of means of B & C is
b1 - b2 - b3, and 2*mean of A minus sum of means of B & C is 2*b1 - b2
-b3. In both cases 3 parameters are involved, so estimates have unique
solutions.

JW