```Date: Sun, 8 Oct 2006 10:51:49 -0400 Reply-To: Jay Weedon Sender: "SAS(r) Discussion" From: Jay Weedon 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" 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 ```

Back to: Top of message | Previous page | Main SAS-L page