DONE EASILY BY CREATING YOUR OWN MODELS -- Re: T-test on linear combination of means

[Joe Ward]

Paige Miller writes:   ----- Original Message ----- From: "Paige Miller" <[EMAIL PROTECTED]> To: <[EMAIL PROTECTED]> Sent: Thursday, June 27, 2002 2:04 PM Subject: T-test on linear combination of means   > Suppose I have three (or more) samples, from three (or more) different > populations. According to my subject matter expert, he wants to estimate > a linear combination of the means, say for example > >       0.5*mu1 + 0.5*mu2 - mu3 > > where mu1, mu2 and mu3 are the population means. I know how to compute > this estimate, it is done by simply replacing the population means with > the sample means. If I assume the original populations are normal and > that the population variances are equal, I can compute the variance of > this linear combination. Pretty straightforward stuff. > > However, I want to create a t-test to test the null hypothesis that this > linear combination of means is equal to zero, using an estimate of > variance derived from the data, rather than a population variance, which > is unknown. In doing so, I run into the mathematical difficulty that I > do not know the proper degrees of freedom for this test. (And yes, I > know that for the special case of estimating mu1 - mu2 there are > textbook formulas for this t-test, however I really am interested in > linear combinations of more than two means). > > I feel like I am missing something very obvious; however, if someone > knows, or can point me to the proper formula for a t-test on a linear > combination of means, it would be greatly appreciated. > > -- > Paige Miller > [EMAIL PROTECTED] > http://www.kodak.com > > "It's nothing until I call it!" -- Bill Klem, NL Umpire > "When you get the choice to sit it out or dance, I hope you dance" -- > Lee Ann Womack > > ================================================================= > Instructions for joining and leaving this list, remarks about the > problem of INAPPROPRIATE MESSAGES, and archives are available at: > .                  http://jse.stat.ncsu.edu/                    .   > ===JOE WARD WRITES ==============================================================   Hi, Page �   Yep, you can test any hypothesis of interest as long as it can be expressed as a LINEAR COMBINATION OF YOUR MODEL PARAMETERS.   You may get responses to your question that appear to be easier to implement than the one suggested here. The advantage to the process described here is its generality. For example some of the responders may point out that you are dealing with a contrast (the weights on the mu's sum to zero). The method described here does not require that the weights sum to zero. You can test any hypothesis that leads to a linear combination of the model parameters.equal to a specified constant. The constant does not have to be zero.   Consider what I think may be your ASSUMED MODEL   Y = mu1* X1 + mu2*X2 + mu3*X3 + E1   where        Y  = response variable       X1 = 1 if the corresponding element of Y is from Population #1, 0 otherwise       X2 = 1 if the corresponding element of Y is from Population #2, 0 otherwise       X3 = 1 if the corresponding element of Y is from Population #3, 0 otherwise       E1 = Error or Residual   I think that your Hypothesis is:     0.5*mu1 + 0.5*mu2 - mu3 = k ( it can be any value, but you suggested k=0)   So you can impose this restriction on your ASSUMED MODEL to obtain a RESTRICTED MODEL   (BUT IF YOU USE SAS YOU CAN USE THE 'TEST' STATEMENT THAT DOES ALL THE WORK.)   or in your special case    0.5*mu1 + 0.5*mu2 - mu3 = 0   or mu3 = .5*mu1 + .5*mu2  and substituting gives your RESTRICTED MODEL   Y = mu1* X1 + mu2*X2 + (.5*mu1 + .5*mu2)  *X3 + E2 Y = mu1*(X1 + .5X3) + mu2*(X2 + .5X3) + E2   Then you might let    Z1 = (X1 + .5X3)                                 Z2 = (X2 + .5X3) and your RESTRICTED MODEL is: Y = mu1*Z1 + mu2*Z2 + E2 Remember to use the �no intercept� or �pass through the origin�  option to solve this model.   Note: attempts to solve the restricted model without using the 'no intercept' option may give a message about �singularity.�   and if you obtain least-squares solutions to the ASSUMED MODEL 1 AND RESTRICTED MODEL 2   then F = (SSQE2-SSQE1)/ (3-2)   =     F(df1,df2) = F(1,N-3)= (SSQE2-SSQE1)/ (1)                   (SSQE1)/(N-3)                                                     (SSQE1)/(N-3) In order to get the t statistic that you desire, then you take the sqroot of F, since F( 1, df2) = t^2(df2)   then  t(N-3)  = sqroot F(1,N-3)   === HAVE FUN !!!!     --- Joe   ********************************** Joe H. Ward,  Jr. 167 East Arrowhead Dr. San Antonio, TX 78228-2402 Phone: 210-433-6575 Fax:     210-433-2828 Email:  [EMAIL PROTECTED] http://www.northside.isd.tenet.edu/healthww/biostatistics/wardindex ============================== Health Careers High School 4646 Hamilton Wolfe Road San Antonio, TX 78229 **********************************