Just to add a note to the question asked by Bill Simpson.. I am confronted with a similar problem, but the model I try to fit on my data is nonlinear. I have a four-parameter, biexponential model describing the response times of a group of subjects performing a set of task. What I want to evaluate is whether the parameter values differ between (groups of) tasks. myfunc <-formula(x ~ a*exp(-b*age) + (c*exp(d*age)) ) both x (representing response time) and age are variables in the current scope. Part of the data frame is presented below. At the moment I fit the model for each task seperately. But now I cannot test the parameters for equality. A solution like in Bills case would be great. task type cor x var median stdev count pp nper nsw age 266 DS Repetition Correct 581.2427 108609.931 499.0 329.56021 103 007 22 17 19 267 TS Repetition Correct 508.1159 31364.839 461.0 177.10121 69 007 0 0 19 268 SRT Repetition Correct 215.8182 14352.028 188.0 119.79995 33 007 0 0 19 269 CRT-1 Repetition Correct 379.6250 3335.601 382.0 57.75466 48 007 0 0 19 270 CRT-2 Repetition Correct 488.4898 11872.963 461.0 108.96313 49 007 0 0 19 271 CRT-3 Repetition Correct 411.3600 7523.256 393.0 86.73670 50 007 0 0 19 272 D/CRT Repetition Correct 328.0313 1587.709 327.0 39.84606 32 007 0 0 19 273 DS Induction Correct 616.3548 74128.037 540.0 272.26464 31 007 22 17 19 274 TS Induction Correct 484.7867 54941.630 438.0 234.39631 75 007 0 0 19 282 DS Repetition Correct 681.6842 225197.191 519.0 474.54946 114 058 14 19 20 283 TS Repetition Correct 646.3538 106069.951 539.0 325.68382 65 058 0 0 20 284 SRT Repetition Correct 217.0833 1823.336 213.0 42.70054 36 058 0 0 20 -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To: r-help-request at stat.math.ethz.ch _._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._
> From: "Mark M. Span" <span at psy.uva.nl> > Date: Mon, 9 Oct 2000 16:27:35 +0200> Just to add a note to the question asked by Bill Simpson.. > > I am confronted with a similar problem, but the model I try to fit on my > data is nonlinear. > I have a four-parameter, biexponential model describing the response times > of a group of subjects performing a set of task. What I want to evaluate is > whether the parameter values differ between (groups of) tasks. > > myfunc <-formula(x ~ a*exp(-b*age) + (c*exp(d*age)) ) > > both x (representing response time) and age are variables in the current > scope. Part of the data frame is presented below. At the moment I fit the > model for each task seperately. But now I cannot test the parameters for > equality. A solution like in Bills case would be great.Use a model parametrizing a,b,c by groups and one without, and use anova to compare the models. A worked example is in V&R3, page 249. I'd try your example, but as you only have two age points (if age is the final column), it is non-identifiable (four parameters for two predictions only) so I assume you have much more data. -- Brian D. Ripley, ripley at stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272860 (secr) Oxford OX1 3TG, UK Fax: +44 1865 272595 -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To: r-help-request at stat.math.ethz.ch _._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._
Hi list members, I'm (gradually) coming up to speed on using R. SO far I am very much enjoying the learning process. I mentioned the software to a colleague at another research organisation, and he was interested. He had the following question: Can the following non-linear fit be carried out (I suppose by NLS), y~(x^a)*(w^b), where a and b are the parameters and x and w are variables? i.e. the formula has two independent variables. Does the software do this sort of thing, or will it only handle one independent variable? I gave it a go but did not get an answer out of it....the response being 2590.046 : 5.0 0.4 Error in qr(attr(rhs, "gradient")) : NA/NaN/Inf in foreign function call (arg 1) Before I persevere any further, does NLS have this capability, or am I on the wrong track? Thanks, Matt Redding ********************************DISCLAIMER**************************** The information contained in the above e-mail message or messages (which includes any attachments) is confidential and may be legally privileged. It is intended only for the use of the person or entity to which it is addressed. If you are not the addressee any form of disclosure, copying, modification, distribution or any action taken or omitted in reliance on the information is unauthorised. Opinions contained in the message(s) do not necessarily reflect the opinions of the Queensland Government and its authorities. If you received this communication in error, please notify the sender immediately and delete it from your computer system network. -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To: r-help-request at stat.math.ethz.ch _._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._._