similar to: relative risk regression with survey data

Howdy, Referencing the below exchange: https://stat.ethz.ch/pipermail/r-help/2006-April/103862.html I am still getting the same warning ("non-integer #successes in a binomial glm!") when using svyglm:::survey. Using the API data: library(survey) data(api) #stratified sample dstrat<-svydesign(id=~1,strata=~stype, weights=~pw, data=apistrat, fpc=~fpc)

Estimados usuarios: Estoy intentando reproducir el ejemplo 6.4 de Thomas Lumley. Complex Survey. Editorial Wiley. 2010 (ver la página en google:

Using the survey package I find it is convenient and easy to get estimated proportions using svymean, and their corresponding estimated standard errors. But is there any elegant/simple way to calculate corresponding confidence intervals for those proportions? Of course +/- 1.96 s.e. is a reasonable approximation for a 95% CI, but (incorrectly) assumes symmetrical distribution for a proportion.

Hello, I'm running R 2.14.1 on OS X (x86_64-apple-darwin9.8.0/x86_64 (64-bit)), with version 3.28 of Thomas Lumley's survey package. I was using predict() from svyglm(). E.g.: data(api) dstrat<-svydesign(id=~1,strata=~stype, weights=~pw, data=apistrat, fpc=~fpc) out <- svyglm(sch.wide~ell+mobility, design=dstrat, family=quasibinomial()) pred.df <-

I have survey data that I am working on. I need to make some multi-way tables and regression analyses on the data. After attaching the data, this is the code I use for tables for four variables (sweight is the weight variable): > a <- xtabs(sweight~research.area + gender + a2n2 + age) > tmp <- ftable(a) Is this correct? I don't think I need to use the strata and cluster

Greetings, I am revisiting code from several different files I have saved from the past and all used to run flawlessly; now when I run any of the svyglm related functions, I am coming up with an error: Error in model.frame.default(formula = F3ATTAINB ~ F1PARED, data = data, : the ... list does not contain 4 elements The following is a minimal reproducible example: library(RCurl)

hello I want to know why when I use the function "svyglm" for a logistic regression I get the AIC: NA. The code and the result is mestran below: mod2<-svyglm(APES_DICOT~Nivel_Educativo+Ocupacion_principal+Afiliacion_salud+Tiene_cuidador+Presencia_enfer_cronica+ Consumo_tabaco+Consumo_alcohol+Presencia_Dolor+ABC_fis+ABC_instr+Anergia+Actividad_fisica_ultimo_año+

Hello I want to know if the summary of the logistic model with survey Pr (> | t |) to test if the coefficient of the model is significant, ie is the p_valor wald test for the model coefficients, for I am interested to know if the three levels of the variable educational level are significant to the model (significance of handling 0.2), I present below the results of my model

summary(result) Call: svyglm(Injury ~ seat, sD, family = quasibinomial(link = "logit")) Survey design: svydesign(~1, prob = NULL, strata = Data[, 1], weights = Data[, 4], data = Data, fpc = ~fPc) Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) -4.256875 0.001421 -2996.7 <2e-16 *** seatbad 0.681504 0.001689 403.4 <2e-16 *** ---

I?m running some logistic regressions and I?ve been trying to include weights in the equation. However, when I run the model, I get this warning message: Here?s what it says: Warning message: In eval(expr, envir, enclos) : non-integer #successes in a binomial glm! I think it is because the weights are non-integer values. What is a good way to run logistic regressions in R when using

Dear Daniel and Jorge, Thank you very much and it does help. If I have a string "ABCD", how can I access the second element of the string "B"? Thanks, Miao 2012/7/27 Daniel Nordlund <djnordlund@frontier.com> > > -----Original Message----- > > From: r-help-bounces@r-project.org [mailto:r-help-bounces@r-project.org] > > On Behalf Of jpm miao

hi, i am not hitting an error when i copy and paste your code into a fresh console. maybe compare your sessionInfo() to mine? > sessionInfo() R version 3.4.1 (2017-06-30) Platform: x86_64-w64-mingw32/x64 (64-bit) Running under: Windows Server 2008 R2 x64 (build 7601) Service Pack 1 Matrix products: default locale: [1] LC_COLLATE=English_United States.1252

Hello, I want to specify a linear regression model in which the metric outcome is predicted by two factors and their interaction. glm() computes effects for each factor level and the levels of the interaction. In the case of singularities glm() displays "NA" for the corresponding coefficients. However, svyglm() aborts with an error message. Is there a possibility that svyglm()

Hi, I am trying to fit a model with an ordered response variable (3 levels) and 13 predictor variables. The sample has complex survey design and I've used 'svydesign' command from the survey package to specify the sampling design. After reading the manual of 'svyglm' command, I've found that you can fit a logistic regression (binary response variable) by specifying the

Say I have a dataframe like this: df <- data.frame(cbind(c(1,0,0,1),c(0,1,0,0),c(0,0,1,0))) names(df) <- c('a','b','c') I would like to create a factor in a new column, where the factor values are taken from the column names, like this: > df2 a b c f 1 1 0 0 a 2 0 1 0 b 3 0 0 1 c 4 1 0 0 a How would I do this? Thanks, Dan Daniel Nordlund Bothell, WA USA

¡Muchas gracias Olivier! Un saludo. El 16 de abril de 2015, 10:44, Olivier Nuñez <onunez en unex.es> escribió: > Mira el paquete survey. > Un saludo. Olivier > > ----- Mensaje original ----- > De: "Víctor Nalda Castellet" <victor.nalda.castellet en gmail.com> > Para: "r-help-es" <r-help-es en r-project.org> > Enviados: Miércoles, 15 de

I'm trying to use the survey package to calculate a risk difference with confidence interval for binge drinking between sexes. Variables are X_RFBING2 (Yes, No) and SEX. Both are factors. I can get the group prevalences easily enough with result <- svyby(~X_RFBING2, ~SEX, la04.svy, svymean, na.rm = TRUE) and then extract components from the svyby object with SE() and coef() to do the

Has anyone heard whether Revolution Analytics is going to release this capability to the R community? http://www.businesswire.com/news/home/20110201005852/en/Revolution-Analytics-Unlocks-SAS-Data Dan Daniel Nordlund Bothell, WA USA

Dear list, I am trying to set up a propensity-weighted regression using the survey package. Most of my population is sampled with a sampling probability of one (that is, I have the full population). However, for a subset of the data I have only a 50% sample of the full population. In previous work on the data, I analyzed these data using SAS and STATA. In those packages I used a propensity weight

If I have two matrices like x <- matrix(rep(c(1,2,3),3),3) y <- matrix(rep(c(4,5,6),3),3) How can I combine them to get ? 1 1 1 4 4 4 1 1 1 5 5 5 1 1 1 6 6 6 2 2 2 4 4 4 2 2 2 5 5 5 2 2 2 6 6 6 3 3 3 4 4 4 3 3 3 5 5 5 3 3 3 6 6 6 The number of rows and the actual numbers above are unimportant, they are given so as to illustrate how I want to combine the matrices. I.e., I am looking for