search for: svymeans

R-help, Sorry for this question (I guess it has been addressed before but I could not find it in the archives) but how can I see a function code when the following comes up: > svymean function (x, design, na.rm = FALSE, ...) { .svycheck(design) UseMethod("svymean", design) } <environment: namespace:survey> Thanks in advance

...3.3188 However, when I do the following: svymean(~W524787, clus1, deff='replace', na.rm=TRUE) mean SE DEff W524787 0.855547 0.011365 4.1158 Compare this to the value in the row 9 up from the bottom to see it is different. Computing the mean of the item by itself with svymeans agrees with the sample mean > mean(lower_dat$W524787, na.rm=T) [1] 0.8555471 Now, I know that there is a covariance between the variables, but I was under the impression that the sample mean was still of pragmatic utility, but to account for sample design only the standard error is affected....

Dear WizaRds, Working through sampling theory, I tried to comprehend the concept of stratification and apply it with Survey to a small example. My question is more of theoretic nature, so I apologize if this does not fully fit this board's intention, but I have come to a complete stop in my efforts and need an expert to help me along. Please help: age<-matrix(c(rep(1,5), rep(2,3),

Dear List: I am working to understand some differences between the results of the svymean() function in the survey package and from code I have written myself. The results from svymean() also agree with results I get from SAS proc surveymeans, so, this suggests I am misunderstanding something. I am never comfortable with "I did what the software" does mentality, so I am working to

Hello, Although my R code for the svymean () and svyquantile () functions works fine, I am stuck with the svyby () and make.formula () functions. I got the following error messages. - Error: object of type 'closure' is not subsettable # svyby () - Error in xx[[1]] : subscript out of bounds # make.formula () A reproducible example is appended below. I would appreciate if

Hi, I have been experimenting with the new Survey package. Specifically, I was trying to use some of the functions on the public-use survey data from NHIS (2000 Sample Adult file). Error 1): The first error I get is when I try to specify the complex survey design. nhis.design<-svydesign(ids=~psu, probs=~probs, strata=~strata, data=nhis.df, check.strata=TRUE) Error in svydesign(ids =

Hi, is there a way to analyze subpopulations (e.g. women over 50, those who answered "yes" to a particular question) in a survey using Survey package? Other packages (e.g. Stata, SUDAAN) do this with a subpopulation option to identify the subpopulation for which the analysis shoud be done. I did not see this option in the Survey package. Is there another way to do this?

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.

Hi R users, I am trying to use the aggregate function with a survey design object and survey functions, but get the following error. I think I am incorrectly using the syntax somehow, and it may not be possible to access variables directly by name in a survey-design object. Am I right? How do I fix this problem? I have used aggregate with "mean" and "weighted.mean", and

Hello I'm trying to get the proportion "true" for dichotomous variable for various subgroups in a survey. This works fine, but obviously doesn't give proportions directly: svytable(~SurvYear+problem.vandal, seh.dsn, round=TRUE) problem.vandal SurvYear FALSE TRUE 1995 8906 786 1997 17164 2494 1998 17890 1921 1999 18322 1669 2001 17623 2122 ...

Hello, I still apologise if the question i ask is naive. I would like to use R instead of usual softs (SAS, Stata) or at list to do things that the others do not. As a consequence I am trying to get familiar with R, and i am astonished because it does not seem to do things that are easy to do on other softs. I have got data with missing values (I am working on a questionnaire filled by students)

Dear WizaRds, dear Thomas, First of all, I want to tell you how grateful I am for all your support. I wish I will be able to help others along one day the same way you do. Thank you so much. I am struggling with a multistage sampling design: library(survey) multi3 <- data.frame(cluster=c(1,1,1,1 ,2,2,2, 3,3), id=c(1,2,3,4, 1,2,3, 1,2), nl=c(4,4,4,4, 3,3,3, 2,2), Nl=c(100,100,100,100,

Hi everyone, my apologies in advance if I'm overlooking something simple in this question. I am trying to use R's survey package to make a direct method age-adjustment to some complex survey data. I have played with postStratify, calibrate, rake, and simply multiplying the base weights by the correct proportions - nothing seems to hit the published numbers on the nose. I am trying to

Hi all, When I try the following with pkg Survey it returns the error below: ftable(svyby(~INCOME, ~AGECL+RACECL, svymean, design=q50), rownames=list(AGECL=c("<35", "35-44", "45-54", "55-64", "65-74", ">=75"), RACECL=c("white non hispanic", "non white or

How do I retrieve the replicates estimates from a crosstab done using svyby? Here is an example from the help page for svyby in the package: > data(api) > dclus1<-svydesign(id=~dnum, weights=~pw, data=apiclus1, fpc=~fpc) > rclus1<-as.svrepdesign(dclus1) > > a <- svyby(~api99, ~stype, rclus1, svymean, return.replicates=TRUE) > a$replicates NULL But, compare to > b

Dear list, I got some surprising results when using the svytotal routine from the survey package with data containing missing values. Some example code demonstrating the behaviour is included below. I have a stratified sampling design where I want to estimate the total income. In some strata some of the incomes are missing. I want to ignore these missing incomes. I would have expected that

Good morning! I have 4 questions which trouble me: 1. I want to test the hypothesis that the 2 proportions (the mean of a binomial) which come from 2 different samples are equal. I want to use the following function z= (p1-p2)/ sqrt((p1(1-p1)/n1)+(p2(1-p2)/n2)) which is one of the standard formulas for this case. Is there such a function in R? p1=the proportion from the first sample n1=the

Dear WizaRds, I am struggling to compute correctly a cluster sampling design. I want to do one stage clustering with different parametric changes: Let M be the total number of clusters in the population, and m the number sampled. Let N be the total of elements in the population and n the number sampled. y are the values sampled. This is my example data: clus1 <-

try resetting your factor levels and re-run? q50 <- update( q50 , INCOME = factor( INCOME ) , AGECL = factor( AGECL ) , RACECL = factor( RACECL ) ) On Sun, Jul 9, 2017 at 2:59 PM, Orsola Costantini via R-help < r-help at r-project.org> wrote: > Hi all, > > When I try the following with pkg Survey it returns the error below: > > ftable(svyby(~INCOME, ~AGECL+RACECL,

Hi everyone, apologies if the answer to this is in an obvious place. I've been searching for about a day and haven't found anything.. I'm trying to replicate Stata's confidence intervals in R with the survey package, and the numbers are very very close but not exact. My ultimate goal is to replicate Berkeley's SDA website with R (http://sda.berkeley.edu/), which seems to