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!
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On 7 Oct 2006 21:55:51 -0700, "Hanz" <firstname.lastname@example.org> 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
>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
> model Response = Fert Column / ss3 solution;
>then A-(B+C) adjusted for Column is estimated as
>Parameter Estimate Error t Value Pr > |t|
>A vs.B+C -11.0000000 3.53225305 -3.11 0.0049
>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 /
>Easier to do by hand from the least squares means! Assuming zero
>covariances with standard error = sqrt(3*2.039**2).
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,
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