similar to: Model Description

Dear sir/madame, I am currently writing a meta-analysis on the complication and reoperation rates of 5 different treatment modalities after a distal radius fracture. I was able to pool the rates of the 5 different rates using R. Now I have to compare the pooled rates of the 4 treatment modalities with the golden standard separately. I though the chi squared test would be the best method. How do

> On Jun 23, 2017, at 5:53 AM, Jay Zola <jayjay.1988 at hotmail.nl> wrote: > > Dear sir/madame, > > > I am currently writing a meta-analysis on the complication and reoperation rates of 5 different treatment modalities after a distal radius fracture. I was able to pool the rates of the 5 different rates using R. Now I have to compare the pooled rates of the 4 treatment

Hello folks, I am on learning phase of R. I have developed Regression Model over six predictor variables. while development, i found my all data are not very linear. So, may because of this the prediction of my model is not exact. Here is the summary of model : Call: lm(formula = y ~ x_1 + x_2 + x_3 + x_4 + x_5 + x_6) Residuals: Min 1Q Median 3Q Max -125.302

Hi all, I'm using the medical risk package to determine the risk stratification Index based on the ICD9 codes. Although, I have been successful in using it, I'm unable to interpret the output. here is the sample code # Calculate RSI for each patient ("id") in dataframe cases <- data.frame(id=c(1,1,1,2,2,2,2,2), +

I have the following code: ``` rm(list=ls()) N = 30000 xvar <- runif(N, -10, 10) e <- rnorm(N, mean=0, sd=1) yvar <- 1 + 2*xvar + e plot(xvar,yvar) lmMod <- lm(yvar~xvar) print(summary(lmMod)) domain <- seq(min(xvar), max(xvar))??? # define a vector of x values to feed into model lines(domain, predict(lmMod, newdata = data.frame(xvar=domain)))??? # add regression line, using

But of course the whole point of additivity is to decompose the combined effect as the sum of individual effects. "Mislead" is a subjective judgment, so no comment. The explanation I provided is standard. I used it for decades when I taught in industry. Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into

Please help getting the R squared value in asymptotic regression model I use the code below model1<-nls(GN1~SSasymp (nrate,a,b,c), data = data.1 ) and R produced the modell coefficients without the R squared value? -- Ahmed M. Attia Research Assistant Dept. Of Soil&Crop Sciences Texas A&M University ahmed <ahmedatia@zu.edu.eg>.attia@ag.tamu.edu Cell phone:

David: I believe your response on SO is incorrect. This is a standard OFAT (one factor at a time) design, so that assuming additivity (no interactions), the effects of drugA and drugB can be determined via the model you rejected: For example, if baseline control (no drugs) has a response of 0, drugA has an effect of 1, drugB has an effect of 2, and the effects are additive, with no noise we

> On Mar 5, 2018, at 3:04 PM, Bert Gunter <bgunter.4567 at gmail.com> wrote: > > But of course the whole point of additivity is to decompose the combined effect as the sum of individual effects. Agreed. Furthermore your encoding of the treatment assignments has the advantage that the default treatment contrast for A+B will have a statistical estimate associated with it. That was a

Hi everyone, I am doing panel data analysis using the 'plm' package. However, I have noticed that the plm() function reports a different value of R-squared from the R-squared of the lm() function with time-demeaned data. To be clear, I have tried to compute the within model both manually (run an OLS regression with time-demeaned data using lm()) and by using plm(). The two methods give me

> On Aug 14, 2017, at 5:17 AM, peter dalgaard <pdalgd at gmail.com> wrote: > > >> On 14 Aug 2017, at 13:43 , Spencer Graves <spencer.graves at effectivedefense.org> wrote: >> >> >> >> On 2017-08-14 5:53 AM, peter dalgaard wrote: >>>> On 14 Aug 2017, at 10:13 , Troels Ring <tring at gvdnet.dk> wrote: >>>>

Dear R users, how can I extrapolate values listed in the summary of an lm model but not directly available between object values such as the the standard errors of the calculated parameters? for example I got a model: mod <- lm(Crd ~ 1 + Week, data=data) and its summary: > summary(mod) Call: lm(formula = Crd ~ 1 + Week, data = data, model = TRUE, y = TRUE) Residuals: Min

> On Mar 5, 2018, at 2:27 PM, Bert Gunter <bgunter.4567 at gmail.com> wrote: > > David: > > I believe your response on SO is incorrect. This is a standard OFAT (one factor at a time) design, so that assuming additivity (no interactions), the effects of drugA and drugB can be determined via the model you rejected: >> three groups, no drugA/no drugB, yes drugA/no drugB,

Thank you Charles Berry for your kind reply. I don't see anything wrong with the word "struggling". I have spent several hours trying various R packages like kernlab and GPfit to use GP to create a binary classification model which produces a prediction interval for each sample. I have been struggling because with all of them you may create a GP classification model but it only

Dear useRs, Lately I noticed a strange warning in the summary of a lm-object. Any idea what this warning is about? I'm using R 2.5.1 on Win XP pro. > x <- rnorm(100) > y <- rnorm(100) > summary(lm(y~x)) Call: lm(formula = y ~ x) Residuals: Min 1Q Median 3Q Max -1,77809 -0,68438 -0,04409 0,63891 2,30863 Coefficients: Estimate Std.

Hi I am trying to find the relationship between two variables. First I fitted a linear model between two variables and I found the following results: Residual standard error: 0.03253 on 2498 degrees of freedom Multiple R-squared: 0.5551, Adjusted R-squared: 0.5549 F-statistic: 3116 on 1 and 2498 DF, p-value: < 2.2e-16 Then I used the cor function to see the correlation between two variable

I have spatial data on a sphere (the Earth) for which I would like to run an gls model assuming that the errors are autcorrelated, i.e. including a corSpatial correlation in the model specification. In this case the distance metric should be calculated on the sphere, therefore metric = "euclidean" in (for example) corSpher would be incorrect. I would be grateful for help on how to

Hi folks, I'm trying to estimate bias-corrected percentile (BCP) confidence intervals on a vector from a simple for loop used for resampling. I am attempting to follow steps in Manly, B. 1998. Randomization, bootstrap and monte carlo methods in biology. 2nd edition., p. 48. PDF of the approach/steps should be available here: https://wyocoopunit.box.com/s/9vm4vgmbx5h7um809bvg6u7wr392v6i9 If

Note though that this has been put on hold on stats.stackexchange.com as off-topic. On 23/06/2017 19:33, Bert Gunter wrote: > Probably the wrong list. R-help is concerned with R programming, not > statistics methodology questions, although the intersection can be > nonempty. > > I suggest you post on stats.stackexchange.com instead, which *is* > concerned with statistics

While a plea about struggling may seem appropriate to you, it is just as content-free as a reply telling you to use Google... and like it or not, that tit-for-tat arises due to frustration with lack of specificity as detailed by Charles. That is, if you are constructive about documenting your issue with a reproducible example and mentioning what you have tried and how it failed, you won't