similar to: Trust p-values or not ?

R 2.7.2 Windows XP I am using apply to compute a series of Student's t-test from two matrices, sample1 and sample2. boo<-apply(sample1,1,t.test,sample2) I want to pick of the p-values from the tests, but can't seem to get it to work. I have tried several methods to get the values including: boo<-apply(sample1,1,t.test$t.test,sample2) boo<-apply(sample1,1,t.test,sample2)$t.test

David, Again you have my thanks!. You are correct. What I want is not technically a contrast. What I want is the estimate for "regional" and its SE. I don't mind if I get these on the log scale; I can get the anti-log. Can you suggest how I can get the point estimate and its SE for "regional"? The predict function will give the point estimate, but not (to my knowledge)

I have a model (run with glm) that has a factor, type. Type has two levels, "general" and "regional". I am trying to get estimates (and SEs) for the model with type="general" and type ="regional" using fit.contrast but I can't get the syntax of the coefficients to use in fit.contrast correct. I hope someone can show me how to use fit.contrast, or some

> On Oct 22, 2017, at 5:01 PM, Sorkin, John <jsorkin at som.umaryland.edu> wrote: > > David, > Again you have my thanks!. > You are correct. What I want is not technically a contrast. What I want is the estimate for "regional" and its SE. There needs to be a reference value for the contrast. Contrasts are differences. I gave you the choice of two references

David, predict.glm and se.fit were exactly what I was looking for. Many thanks! John John David Sorkin M.D., Ph.D. Professor of Medicine Chief, Biostatistics and Informatics University of Maryland School of Medicine Division of Gerontology and Geriatric Medicine Baltimore VA Medical Center 10 North Greene Street GRECC (BT/18/GR) Baltimore, MD 21201-1524 (Phone) 410-605-7119 (Fax)

David, Thank you for responding to my post. Please consider the following output (typeregional is a factor having two levels, "regional" vs. "general"): Call: glm(formula = events ~ type, family = poisson(link = log), data = data, offset = log(SS)) Deviance Residuals: Min 1Q Median 3Q Max -43.606 -17.295 -4.651 4.204 38.421 Coefficients:

Peter You question is difficult to answer without more information about the distribution of your residuals. Different residual patterns call for different transformations to stabilize the variance. One very common form of heterocedasticity is increasing variance with increasing values of an independent predictor, i.e. the variance of the residuals of y=x increase as x increases. In this case a

How can I get the p values from a glm ? I want to get the p values so I can add them to a custom report fitwean<- glm(data[,"JWean"]~data[,"Group"],data=data,family=binomial(link ="logit")) summary(fitwean) # This lists the coefficeints, SEs, z and p values, but I can't isolate the pvalues. names(summary(fitwean)) # I see the coefficients,

> On Oct 22, 2017, at 3:56 PM, Sorkin, John <jsorkin at som.umaryland.edu> wrote: > > David, > Thank you for responding to my post. > > Please consider the following output (typeregional is a factor having two levels, "regional" vs. "general"): > Call: > glm(formula = events ~ type, family = poisson(link = log), data = data, > offset =

R 2.8.1 Windows XP I am trying to analyze repeated measures data (the data are listed at the end of this Email message) and I need help to make sure that I have properly specified my model, and would like to know why lmer does not return a p value for Group, my fixed effect. My subjects are divided into two groups (variable GROUP), individual subjects are indicated by the variable SS, Value is

windows Vista R 2.10.1 Is it possible to get p values from gee? Summary(geemodel) does not appear to produce p values.: > fit4<- gee(y~time, id=Subject, data=data.frame(data)) Beginning Cgee S-function, @(#) geeformula.q 4.13 98/01/27 running glm to get initial regression estimate (Intercept) time 1.1215614 0.8504413 > summary(fit4) GEE: GENERALIZED LINEAR MODELS FOR

> On Oct 22, 2017, at 6:04 AM, Sorkin, John <jsorkin at som.umaryland.edu> wrote: > > I have a model (run with glm) that has a factor, type. Type has two levels, "general" and "regional". I am trying to get estimates (and SEs) for the model with type="general" and type ="regional" using fit.contrast ?fit.contrast No documentation for

An apparatus exists whereby a collection of balls is displaced to the top of a stack by suction. A top level (Level 1) each ball is shifted 1 unit to the left or 1 unit to the right at random with equal probability. The ball then drops down to level 2. At Level 2, each ball is again shifted 1 unit to the left or 1 unit to the right at random. The process continues for 15 levels and the balls are

I am using read.csv to read a CSV file (produced by saving an Excel file as a CSV file). The columns containing dates are being read as factors. Because of this, I can not compute follow-up time, i.e. Followup<-postDate-preDate. I would appreciate any suggestion that would help me read the dates as dates and thus allow me to calculate follow-up time. Thanks John John Sorkin M.D., Ph.D. Chief,

... of course, whether one **should** get them is questionable... http://www.nature.com/news/statisticians-issue-warning-over-misuse-of-p-values-1.19503#/ref-link-1 Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Fri, Apr

I am having trouble selecting rows of a dataframe that will be included in a regression. I am trying to select those rows for which the variable Meno equals PRE. I have used the code below: difffitPre<-lm(data[,"diff"]~data[,"Age"]+data[,"Race"],data=data[data[,"Meno"]=="PRE",]) summary(difffitPre) The output from the summary indicates that

Widows XP R 2.3.1 I have been trying to make a data structure that will contain both the coefficients from a linear regression along with column and row titles AND the call, i.e. myreg<-lm(y~x+y+z) whatIwant<-cbind(c(summary(myreg)$call,"",""),summary(myreg)$coefficients) Neither the statement above, nor any one of twenty variations I have tried work. I would appreciate

R 2.1.1 Win 2k Would someone suggest a method (or methods) that can be used to determine ntile cutpoints of a vector, i.e. to determine values that can be used to divide a vector into thirds such as 0-33 centile, 34-66 centile, 67-100 centile. If for example I had a vector: 1,2,3,4,5,6,7,8,9 and wanted to divide the vector into thirds I would have cut-points of 3, and 6. Thanks, John John

I want to place four plots on a page, and I would like to have all four plots share a common title. I have tried the following code, but the title is centered over the fourth graph and not centered across all four plots. Does anyone have any suggestions? R 2.1.1 windows xp oldpar<-par(mfcol =c(1,4),ask=TRUE) plot(p,varp) plot(p,SEp) plot(p,CVp) plot(p,ppval) title(paste("P and 95%CI

I am trying to model data in which subjects are followed through time to determine if they fall, or do not fall. Some of the subjects fall once, some fall several times. Follow-up time varies from subject to subject. I know how to model time to the first fall (e.g. Cox Proportional Hazards, Kaplan-Meir analyses, etc.) but I am not sure how I can model the data if I include the data for those