similar to: Confidence intervals and polynomial fits

R-users E-mail: r-help@r-project.org My understanding is that package "mgcv" is based on "Generalized Additive Models: An Introduction with R (by Simon N. Wood)". On the page 126 of this book, eq(3.4) looks a quartic equation with respect to "x", not a cubic equation. I am wondering if all routines which uses cubic splines in mgcv are based on this quartic

Dear R users this may be a simple question - but i would appreciate any thoughts does anyone know how you would get one lower and one upper confidence interval for a set of data that consists of proportions. i.e. taking a usual confidence interval for normal data would result in the lower confidence interval being negative - which is not possible given the data (which is constrained between

I have the following binary data set: Sex Response 0 1 0 159 162 1 4 37 My commands library(MASS) sib.glm=glm(sib~sex,family=binomial,data=sib.data) summary(sib.glm) The coefficients in the output are Estimate Std. Error z value Pr(>|z|) (Intercept) -3.6826 0.5062 -7.274 3.48e-13

It seems that confint.default returns an empty data.frame for objects of class mlm. For example: ``` nobs <- 20 set.seed(1234) # some fake data datf <- data.frame(x1=rnorm(nobs),x2=runif(nobs),y1=rnorm(nobs),y2=rnorm(nobs)) fitm <- lm(cbind(y1,y2) ~ x1 + x2,data=datf) confint(fitm) # returns: 2.5 % 97.5 % ``` I have seen proposed workarounds on stackoverflow and elsewhere, but

Full_Name: Soren Feodor Nielsen Version: 2.5.0 OS: linux-gnu Submission from: (NULL) (130.225.103.21) The print-out of xtable in the following example is wrong; instead of yielding the correct ci's for the second model it repeats the ci's from the first model. require(xtable) require(MASS) data(cats) b1<-lm(Hwt~Sex,cats) b2<-lm(Hwt~Sex+Bwt,cats)

Hi, I am currently doing logistic regression analyses and I am trying to get confidence intervals for my partial logistic regression coefficients. Supposing I am right in assuming that the formula to estimate a 95% CI for a log odds coefficient is the following: log odds - 1.96*SE to log odds + 1.96*SE then I am not getting the right CI. For instance, this is a summary of my model:

Dear all, How to obtain the odds ratio (OR) and 95% confidence interval (CI) with 1 standard deviation (SD) change of a continuous variable in logistic regression? for example, to investigate the risk of obesity for stroke. I choose the happening of stroke (positive) as the dependent variable, and waist circumference as an independent variable. Then I wanna to obtain the OR and 95% CI with

Dear list, I just hit an error that stopped my make && make check-devel operation on my linux box (FC3, i686 P4 2GB RAM). Just to note that I've been building the development branch(?) for some time and this is the first hint of a problem. 1) updated the src tree using svn update 2) ran ../configure --with-recommended-package=no from my build directory 3) got: R is now configured

Greetings all, I was in the process of creating a function to generate profile likelihood confidence intervals for a proportion using a binomial glm. This is a component of a larger function to generate and plot confidence intervals for proportions using the above, along with bootstrap (BCa), Wilson and Exact to visually demonstrate the variation across the methods to some folks. I had initially

Full_Name: Christian Lajaunie Version: 2.5.1 OS: Fedora fc6 Submission from: (NULL) (193.251.63.39) confint() does not use the appropriate variance term when the design matrix contains a zero column (which of course should not happen). Example: A 10x2 matrix with trivial column 1: > junk <- data.frame(x=rep(0,10), u=factor(sample(c("Y", "N"), 10, replace=T))) The

Dear list, I am running some linear regressions through lapply, >lapply(c('EMAX','EC50','KOUT','GAMMA'),function(x)confint(lm(get(x)~RR0,dataset2))) I got results like [[1]] 2.5 % 97.5 % (Intercept) 0.6595789212 0.8821691261 RR0 -0.0001801771 0.0001489083 [[2]] 2.5 % 97.5 % (Intercept) -63.83694930

Dear R-helpers, I am using confint() within a function, and I want to turn off the message it prints: x <- rnorm(100) y <- x^1.1+rnorm(100) nlsfit <- nls(y ~ g0*x^g1, start=list(g0=1,g1=1)) > confint(nlsfit) Waiting for profiling to be done... 2.5% 97.5% g0 0.4484198 1.143761 g1 1.0380479 2.370057 I cannot find any way to turn off 'Waiting for. .." I tried

Hello, I am doing a simple linear regression analysis that includes few variables. I am using a bootstrap analysis to obtain the variation of my variables to replacement. I am trying to obtain the coefficients 95% confidence interval from the bootstrap procedure. Here is my script for the bootstrap: N = length (data_Pb[,1]) B = 10000 stor.r2 = rep(0,B) stor.r2 = rep(0,B) stor.inter =

I am hoping for some advice regarding limiting decimal points to 3. 'Round' produces the desired results except for the 97.5% confidence interval. Any advice as to how I modify the code to obtain output to 3 decimal points for all ouput is appreciated, regards Bob Green mod.multgran <-multinom(offence ~ grandiose * violent.convictions, data = kc, na.action = na.omit)

Dear R developers, I noticed a bug in the stats4 package, specifically in the confint method applied to ?mle? objects. In particular, when some ?fixed? parameters define the log likelihood, these parameters are stored within the mle object but they are not used by the ?confint" method, which retrieves their value from the global environment (whenever they still exist). Sample code: >

Hi, I have a problem that concerns combination of the package Maptools and Splancs I have 2 shapefiles that i want to manipulate (one of type point and one polygon).I import them in R using Maptools but then i can't estimate a quartic Kernel using Splancs. The package doesn't recognize the shapes (invalid points and poly argument).I don't know if this is an easy task but i have

Respected Sir I tried the strucchange My data is attached. However I tried the attached commands (last save.txt) to perform Bai Perron 2003... I t worked well but in the end it is giving warning that overlapping confidence interval... I am not sure how to proceed... Please Help Me Thanking You Ayanendu Sanyal -- Please have a look at our new mission and contribute into it (cut and paste the

Buenas, una pregunta. Si yo estoy calculando un modelo lineal, el caso más simple, 1 variable respuesta y una variable explicativa y creo un modelo, me da un R2 del 80% y quiero ver como es esa relacion entre las variables, para calcular el error de predicción del modelo, basta con ver el intervalo de confianza del modelo e irme a los extremos? Por si no me he expresado bien, un ejemplo tonto:

Enun ejemplo real estoy viendo como el intervalo de confianza usando lo que me comentas me ha salido mucho más pequeño de lo que la realidad luego refleja. ¿Cómo es esto posible?? Es decir, veo que para valores de 2,70 obtengo una respuesta de entre 2,69 y 2,90 y sin embargo luego en la realidad tengo valores entre 2,20 y 3 Gracias Jesús From: jorgeivanvelez en gmail.com Date: Thu, 21 Apr

Hi list, is there any direct way to obtain confidence intervals for the regression slope from lm, predict.lm or the like? (If not, is there any reason? This is also missing in some other statistics softwares, and I thought this would be quite a standard application.) I know that it's easy to implement but it's for explanation to people who faint if they have to do their own programming...