similar to: Servreg $loglik

I am a student (and very to new to R) working on a senior design project that is attempting to determine the demand distributions for single copy newspaper draws at individual sales outlet locations. Our sales data is right-censored, because sell-outs constitute a majority of the data, and we are also testing the relevance of including covariates (weather, seasonality, economic condition, etc.).

Hello, I would like to know if there is a function in R that will test for normality and handle censored data sets. Currently, I evaluate each censored data set by the extent to which a normal scores plot approximate a straight line. For complete data sets I use shapiro.test(). Below is an example of a censored data set. data1<-c(0.00, 0.00, 0.00, 5.86, 5.17, 8.17, 5.12, 4.92, 7.08,

Can survreg() handle interval-censored data like the documentation says? I ask because the command: survreg(Surv(start, stop, event) ~ 1, data = heart) fails with the error message Invalid survival type yet the documentation for Surv() states: "Presently, the only methods allowing interval censored data are the parametric models computed by 'survreg'"

Dear all, I am new to R and may make beginner mistakes. Sorry. I am learning using R to do survival analysis. As a start I used the example script code provided in the documentation of predict.survreg of the survival package: # Draw figure 1 from Escobar and Meeker fit <- survreg(Surv(time,status) ~ age + age^2, data=stanford2, dist='lognormal') plot(stanford2$age, stanford2$time,

I'm fitting a regression model for censored data with three categorical predictors, say A, B, C. My final model based on the survreg function is Surv(..) ~ A*(B+C). I know the three-way group mean estimates can be computed using the predict function. But is there any way to obtain two-way group mean estimates, say estimated group mean for (A1, B1)-group? The sample group means don't

Hi, If I used drop1 in a weighted lm fit, it seems to ignore the weights in the AIC calculation of the dropped terms, see the example below. Can this be right? Yan -------------------- library(car) > unweighted.model <- lm(trSex ~ (river+length +depth)^2- length:depth, dno2) > Anova(unweighted.model) Anova Table (Type II tests) Response: trSex Sum Sq Df F value

R2.3, WinXP Dear all, I am using the following functions: f1 = Phi1+(Phi2-Phi1)/(1+exp((log(Phi3)-log(x))/exp(log(Phi4))) f2 = Phi1+(Phi2-Phi1)/(1+exp((log(Phi3)-log(r)-log(x))/exp(log(Phi4))) subject to the residual weighting Var(e[i]) = sigma^2 * abs( E(y) )^(2*Delta) Here is my question, in steps: 1. Function f1 is separately fitted to two different datasets corresponding to

Hi, This is more of a general question than a pure R one, but I hope that is OK. I want to combine one-tailed independent p-values using the weighted version of fisher's inverse chi-square method. The unweighted version is pretty straightforward to implement. If x is a vector with p-values, then I guess that this will do for the unweighted version: statistic <- -2*sum(log(x)) comb.p <-

I've successfully plotted (in the plot and abline code below) a simple regression of Lambda1_2 on VV1_2. I then successfully regressed Lambda1_2 on VV1_2, VV1_22 and VV1_212 producing lm2.l. When I go to plot lm2.l using abline I get the warning: "1: In abline(lm2.l, col = "brown", lty = "dotted", lwd = 2) : only using the first two of 4 regression coefficients"

Greetings, I'm having trouble finding a simple way to calculate a weighted histogram where there may be zero raw counts in a given interval. Given equal-length vectors of data 'data' and weights 'w', and breaks (intervals) for the histogram, I calculate a weighted histogram as follows (see MASS's 'truehist' for an unweighted histogram): bin <- cut(data,

Hi, im little bit confused about Cohen's Kappa and i should be look into the Kappa function code. Is the easy formula really wrong? kappa=agreement-chance/(1-chance) many thanks christian ############################################################################### true-negativ:7445 false-positive:3410 false-negativ:347 true-positiv:772 classification-aggrement:68,6%

In the HistData package, I have a data frame, PearsonLee, containing observations on heights of parent and child, in weighted form: library(HistData) > str(PearsonLee) 'data.frame': 746 obs. of 6 variables: $ child : num 59.5 59.5 59.5 60.5 60.5 61.5 61.5 61.5 61.5 61.5 ... $ parent : num 62.5 63.5 64.5 62.5 66.5 59.5 60.5 62.5 63.5 64.5 ... $ frequency: num 0.5 0.5

I would like to apply network measures (such as betweenness centrality, upper boundedness, etc.) to a weighted graph with non-integer weights, defined by a euclidean distance matrix. The package sna provides the measures that I want to use, but seems only to operate on binary graphs. I have read work by Mark Newman (http://aps.arxiv.org/abs/cond-mat/0407503/), who suggests that a weighted graph

-----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1 Is it possible to use the instant messaging feature of Polycom phones in Asterisk? At the moment I'm seeing this in the SIP messaging when I try to send one from a Polycom 450. <--- SIP read from UDP:<CENSORED POLYCOM IP>:5060 ---> INVITE sip:0100@<CENSORED>:5060;user=phone SIP/2.0 Via: SIP/2.0/UDP <CENSORED POLYCOM

I am trying to use lm for a simple linear fit with weights. The results I get from IDL (which I am more familiar with) seem correct and intuitive, but the "lm" function in R gives outputs that seem strange to me. Unweighted case: > x<-1:4 > y<-(1:4)^2 > summary(lm(y~x)) Call: lm(formula = y ~ x) Residuals: 1 2 3 4 1 -1 -1 1 Coefficients:

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 <-

Hello all, I am using the function lm to do my weighted least square regression. model<-lm(Y~X1+X2, weight=w) What I am confused is the r.squared. It does not seem that the r.squared for the weighted case is an ordinary 1-RSS/TSS. What is that precisely? Is the r.squared measure comparable to that obtained by the ordinary least square? <I also notice that model$res is the unweighted

Hi, I need to find all minimum spanning trees of an unweighted graph. Is there a way in R to do that? Thanks -- View this message in context: http://n4.nabble.com/Minimum-Spanning-Trees-tp1569351p1569351.html Sent from the R help mailing list archive at Nabble.com.

Hi, is there any other routine for factor analysis in R then factanal? Basically I'am interested in another extraction method then the maximum likelihood method and looking for unweighted least squares. Thanks in advance Sigbert Klinke

Hello R folks, Reproducible code below - I'm trying to do a weighted mean by a factor and can't figure it out. Thanks in advance for your assistance. Mike data<-data.frame(c(5,5,1,1,1), c(10,8,9,5,3), c("A","A","A","B","B"))