similar to: Ability estimates for partial credit model

Hola a todos, Os envío una consulta que considero sencilla pero me está resultando imposible de resolver. Si ejecutáis el siguiente código, obtendréis la gráfica que os adjunto: library(ltm) modelo <- rasch(LSAT) plot(modelo, main="Curva probabilidad pregunta 1",legend = TRUE, cx = "bottomright", items=1,xlab="Conocimiento",ylab="Probabilidad") Resulta

Hello All, I was wondering how I can get the overall Pearson chi^2 test of model fit with its df and p value in the LTM package for the 2PL models. Thanks, -- Davood Tofighi Department of Psychology Arizona State University [[alternative HTML version deleted]]

Dear R-List, to assess the model fit for 2PL-models, I tried to mimic the bootstrap-approach chosen in the GoF.rasch()-function. Not being a statistician, I was wondering whether the following simplification (omit the "chi-squared-expressed model fit-step") would be appropriate: GoF.ltm <- function(object, B = 50, ...){ liFits <- list() for(i in 1:B){ rndDat <-

In the near future I will release MiscPsycho, a package that contains various functions useful for applied psychometricians. I would like to include some data sets for distribution in the package, but have not created any of these on my own, but have used data distributed in other packages such as the LSAT data in the ltm package. Is it appropriate for me to distribute a data set in the package I

Hola a todos, Estoy usando la librería ltm, de teoría de respuesta al ítem, para crear gráficos. Resulta que añado unas líneas horizontales en los valores de 0.25, 0.50 y 0.75. Me gustaría que en el gráfico, desaparezcan las marcas de 0,.2, 0.4, 0.6 y 0.8 y aparezcan en 0.25, 0.50 y 0.75. He probado varias cosas que he visto por Googe pero no me han funcionado. ¿Alguna idea? library(ltm) # creo el

Dear R useRs, I have a vector with positive and negative numbers: A=c(0,1,2,3,0,4,5) Now if i-th element in vector A is > 0, then i-th element in vector B is a+1 else i-th element in vector b=a (or 0) vector A: 0 1 2 3 0 4 5 vector B: 0 2 3 4 0 5 6 What's the right way to do this. I still have some problems with for and if statements... Cheers, Andrej

Dear R Users, I am using the LLTM and the LRSM functions in the eRm package to do repeated measurements where there are 2 measurement points for a list of 10 items. I am trying to get ability estimates but am having trouble. I don't think that it is appropriate to use the pmat function since the person parameters are based on all 20 items. Rather, I think it would be more appropriate to

Dear useRs, I have two different length vectors: one column (1...m) and one row vector (1...n): 20 40 20 60 5 4 2 Now I have to calculate ratios between column vector elements and each row vector elements: 4 5 10 8 10 20 4 5 20 15 12 30 Thank's in advance for any suggestions, Andrej

Afternoon R help, I want to run Rasch/IRT analyses using the ltm package, however, I am using large scale survey data which requires weighting for accurate results. I attempted to create a weighted object to insert into the formulae of the ltm packages, however, the survey data only includes 30 replicate weights and a sampling weight. The svrepdesign requires additional information such as

I'm trying to kick off some of the rust and learn some of the R packages for Rasch modeling. When I tried using the eRm package, I get item difficulty estimates for all my items accept the first (in terms of order) item. #Begin code library(eRm) r.simulation <- sim.rasch(20,100) r.data <- r.simulation$items #eRm results erm.rasch <- RM(r.data) names(erm.rasch) erm.items <-

When I try to create a Rasch simulation of data using the sim.rasch function, I get more items than I intend #My code library(eRm) #Number of items k <- 20 #Number of participants n <- 100 #Create Rasch Data #sim.rasch(persons, items, seed = NULL, cutpoint = "randomized") r.simulation <- sim.rasch(n,k) I end up with 20 participants and 100 items, but the instructions say

Dear R useRs, I have big (23000 rows), vertical bar delimited file: e.g. A00001|Text a,Text b, Text c|345 A00002|Text bla|456 ... .. . Try using A <- read.table('filename.txt', header=FALSE,sep='\|') process stop at line 11975 with warning message: number of items read is not a multiple of the number of columns I have no problems with processing similar file, which is

Hi, What package would you recommend for analyzing the validity/reliability of multiple choice tests. Doing things such as classical test analysis, factor analysis, item response theory. I've used psychometric (item.exam), MiscPsycho (alpha.Summary), and ltm (rcor.test). MiscPsycho reported the numbers most similar to what I get in SPSS: corrected point biserial correlations,

Mando este mensaje de respuesta para dar las gracias tanto a Pedro como a Carlos por sus sugerencias. Ahora ya tengo los números que quiero en la gráfica, me falta por quitar los que no quiero. Teniendo en cuenta lo que dice Pedro, voy a preguntar al autor de la librería. saludos, Fernando En martes, 9 de octubre de 2018 13:23:51 CEST, Carlos Ortega <cof en qualityexcellence.es>

Dear all, Suppose I plot two normal distributions (A and B) side by side and add vertical line which hipotheticaly represent alpha value; e.g.: x <- seq(-3.5,5, length=1000) y <- dnorm(x) # Plot distribution A plot(y~x, type='l',axes=F,xlab="",ylab="",lwd=2) # Plot distribution B y2 <- dnorm(x-1.5) lines(y2~x,lwd=2) # Plot vertical line for alpha value

Dear all, Does anybody have an idea or suggestion how to construct (plot) 4-dimensional hypercube in R. Thanks in advance for any pointers. Regards, Andrej

Dear Listers: I used cronbach{psy} to evaluate the internal consistency and some set of variables gave me alpha=-1.1003, while other, alpha=-0.2; alpha=0.89; and so on. I am interested in knowing how to interpret 1. negative value 2. negative value less than -1. I also want to re-mention my previous question about how to evaluate the consistency of a set of variables and about the total

Dear R useRs, is there any simple, build in function to match specific regular expression in data file and write it to a vector. I have the following text file: *NEW RECORD *ID-001 *AB-text *NEW RECORD *ID-002 *AB-text etc. Now I have to match all ID fields and print them to a vector: 001 002 etc. I know that this is very simple with Perl or R-Perl interface, but if possible, I want to do

I have just submitted MiscPsycho to CRAN. MiscPsycho contains functions for miscellaneous psychometrics that may be useful for applied psychometricians. MML estimation already exists in the ltm package. Hence, a jml option is provided for users who prefer this method. The jml function gives back rasch difficulties and the same Infit and Outfit statistics as Winsteps. Also, jml is known to return

I have just submitted MiscPsycho to CRAN. MiscPsycho contains functions for miscellaneous psychometrics that may be useful for applied psychometricians. MML estimation already exists in the ltm package. Hence, a jml option is provided for users who prefer this method. The jml function gives back rasch difficulties and the same Infit and Outfit statistics as Winsteps. Also, jml is known to return