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HOwdy I read R books about scale function for variable transformation. Acoording to this book scale function leads me to better regression results. Or am I worng? I hope somebody tell me about a scale function? Is it for variable transformation? -- Kum-Hoe Hwang, Phone : 82-31-250-3516Email : phdhwang@gmail.com [[alternative HTML version deleted]]

I am using R 2.1.1 under Mac OS 10.3.9. Two related problems (see notes 1. and 2. below) are illustrated by results of the following: y <- rnorm(10) x <- z <- 1:10 is.na(x[9]) <- TRUE lm0 <- lm(y ~ 1) lm1 <- lm(y ~ 1, weights = rep(1, 10)) add1(lm0, scope = ~ x) ## works ok add1(lm1, scope = ~ x) ## error lm2 <- lm(y ~ 1, offset = 1:10) add1(lm0, scope = ~ z) ## works ok add1(lm2, scope = ~ z) ## gives incorrect results (ignores the offset) glm0 <- glm(y ~ 1) glm...

...On Thu, 4 Aug 2005 d.firth at warwick.ac.uk wrote: > I am using R 2.1.1 under Mac OS 10.3.9. > > Two related problems (see notes 1. and 2. below) are illustrated by > results of the following: > > y <- rnorm(10) > x <- z <- 1:10 > is.na(x[9]) <- TRUE > > lm0 <- lm(y ~ 1) > lm1 <- lm(y ~ 1, weights = rep(1, 10)) > > add1(lm0, scope = ~ x) ## works ok > add1(lm1, scope = ~ x) ## error > > lm2 <- lm(y ~ 1, offset = 1:10) > > add1(lm0, scope = ~ z) ## works ok > add1(lm2, scope = ~ z) ## gives incorrect results (igno...

Hello guys, I am a total newbie on R, having downloaded it, read the documentation and started playing with it right now. My general question is what 'lr' model can be used for doing a linear regression on points that have a variance associated with them (ie. Monte Carlo simulation results). Actually my Data sets look like: Timestep Energy Variance_of_the_Energy 0.0005 -14.876840

Hi there, am trying to run a linear regression with a slope of 0. I have a dataset as follows t d 1 303 2 302 3 304 4 306 5 307 6 303 I would like to test the significance that these points would lie on a horizontal straight line. The standard regression lm(d~t) doesn't seem to allow the slope to be set. Any help very welcome. ed

...explanation of these different results? >> >> Since you didn't post your results and since your complaint was that they >> are "the same", it's hard to know what you are talking about. > >I think the question is why > >lm1 <- lm(y ~ factor(x)) >lm0 <- lm(y ~ 1) >anova(lm1) >anova(lm0, lm1) > >gives the same result, but > >lmR<-lmRob(y~factor(x)) >lmR0<-lmRob(y~1) >anova(lmR) >anova(lmR,lmR0) > >does not. > >I don't know the answer, but I think it is an interesting question. > >Best, &g...

Hi! Consider this code. setClass("Ctest" ,representation( test="character" ,bla="character" ,mod="lm" ) ) new("Ctest",test="bla") #This produces an error. #Error in validObject(.Object) : Invalid "Ctest" object: Invalid object for slot "mod" in class "Ctest": got class

Hi all, I am trying to bootstrap a small data set into 1000 "pseudodatasets" and then run an ANOVA on each one. Can anyone provide guidance on how I could do this? Thank you. -Dan Janes ************************************************ Dan Janes, Ph.D. Harvard University/OEB 26 Oxford St. Cambridge, MA 02138 Office: 617-496-2375 Fax: 617-495-5667 Email: djanes at oeb.harvard.edu

...ataset 1000 times. I received code that bootstrapped my dataset and ran an ANOVA on each pseudodataset, producing a histogram of F-values. Here is the code I received that produces this F-histogram. X <- data.frame(b=rnorm(2162),c =factor(rep(1:47,each=46))) ## linear fit with true values lm0 <- lm(b~c,data=X) Fvals <- numeric(1000) for (i in 1:1000) { X.boot <- X[sample(1:nrow(X),replace=TRUE),] Fvals[i] <- anova(lm(b~c,data=X.boot))$"F value"[1] } hist(Fvals) I am grateful for the assistance but the code does not provide what I need. I need the 1000 ps...

I'm running Samba 3.0.0 and seeing a lot of Oct 20 10:02:25 littleboy smbd[2062]: [2003/10/20 10:02:25, 0] lib/util_str.c:safe_strcpy_fn(577) Oct 20 10:02:25 littleboy smbd[2062]: ERROR: NULL dest in safe_strcpy messages. Most things (except the problem I posted earlier) seem to work perfectly.

...ckbc0 (kbd slot) pckbc0: using irq 1 for kbd slot wskbd0 at pckbd0: console keyboard, using wsdisplay0 pmsi0 at pckbc0 (aux slot) pckbc0: using irq 12 for aux slot wsmouse0 at pmsi0 mux 0 pcppi0 at isa0 port 0x61 midi0 at pcppi0: <PC speaker> sysbeep0 at pcppi0 lpt0 at isa0 port 0x378/4 irq 7 lm0 at isa0 port 0x290/8: W83627THF npx0 at isa0 port 0xf0/16: using exception 16 pccom0 at isa0 port 0x3f8/8 irq 4: ns16550a, 16 byte fifo pccom1 at isa0 port 0x2f8/8 irq 3: ns16550a, 16 byte fifo fdc0 at isa0 port 0x3f0/6 irq 6 drq 2 fd0 at fdc0 drive 0: 1.44MB 80 cyl, 2 head, 18 sec biomask c860 net...

Hello to the whole group. I am a newbie to R, but I got my way through and think it is a lot easier to handle than other software packages (far less clicks necessary). However, I have a problem with respect to the summary of regression results. The summary function gives sth like: Residuals: Min 1Q Median 3Q Max -0.46743 -0.09772 0.01810 0.11175 0.42252

...in numbers versions compared to trust in words versions. #Notice that this simply tests the difference. Of course, this #totally ignores cancer type. This works, at p<.05. But that is #still important because it rejects the view that the effect goes #the other way. verbal <- probtype==0 lm0 <- lm(trust ~ subj) lm1 <- lm(trust ~ subj+cancer) lm2 <- lm(trust ~ subj+cancer+verbal) #Then I compared successive models: anova(lm0,lm1) anova(lm1,lm2) #The first comparison shows that cancer matters. The second shows #that verbal is still significant. The result is slightly #differe...

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