similar to: data manipulation

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

I'm using R for Mac OSX version 1.14 (2129) and the geoR package version 1.6-5 (the current version in the R repository). I'm running R in OS 10.4.6 on a Mac G4 iBook (933MHz, 640 MB DDR SDRAM). I searched the R archive and did not find a posting on this issue. I want to use the variog and variog4 functions of geoR to characterize the pattern of spatial autocorrelation of tree

Hi all, If we wanted to study the effect on the mean of the hourly data based on the hours within a day... and we wanted to do Anova analysis... We have two choices: Please see below: Why are these two approaches giving very different p-values? And which one shall I use? Thanks a lot! 1. treating the hours as double/floating numbers: anova(lm(hourlydata~as.double(hours_factors))) Df Sum

I tried to bootstrap the correlation between two variables x1 and x2. The resulting distribution has two distinct peaks, how should I interprete it? The original code is attached. Y. C. Tao ---------------- library(boot); my.correl<-function(d, i) cor(d[i,1], d[i,2]) x1<-c(-2.612,-0.7859,-0.5229,-1.246,1.647,1.647,0.1811,-0.07097,0.8711,0.4323,0.1721,2.143,

Dear All,   is there a way to set low and high limits to a simulation with rlnorm()?   as an example:  a <-rlnorm(500,0.7,1)     I get the summary of   Min. 1st Qu. Median Mean 3rd Qu. Max. 0.1175 1.0590 2.1270 3.4870 4.0260 45.3800 I would like to set limits so that the simulated values minimum would be greater then 0.5 and maximum of less than 30. If during simulation a

Buenas noches estimados colegas: Adjunto la matriz,de la que quiero obtener, los promedios de cada elemento de los vectores que la componen. Me explico, el primer vector, tiene como primer elemento 23 y la suma de todos los elementos es 134; por lo tanto el promedio será 0.1716, el segundo elemento de dicho vector es 20, entonces el promedio sería 0.1492, del tercero el promedio es 0.1343 y así

When I do ARMA(2,2) using one lag of LCPIH data This is eview result > > *Dependent Variable: DLCPIH > **Method: Least Squares > **Date: 08/12/11 Time: 12:44 > **Sample (adjusted): 1970Q2 2010Q2 > **Included observations: 161 after adjustments > **Convergence achieved after 14 iterations > **MA Backcast: 1969Q4 1970Q1 > ** > **Variable Coefficient Std.

Dear all, I have a question regarding the output of the coxph function. What does the 'max possible' exactly mean in the output below? Many thanks. coef exp(coef) se(coef) robust se z Pr(>|z|) smocc_zyban -0.4384 0.6451 0.8667 0.9473 -0.463 0.644 self 1.1857 3.2728 0.1405 0.1443 8.216 2.22e-16 *** smocc_zyban:self

Hello. I had a big collection of Web pages. Now I have this collection divided into clusters. Every page can be relevant or not. I made a table: relevant non relevant cluster1 1 20 cluster2 0 15 cluster3 3 35 . . . . . . . . . I cluster1 I have 21 Web pages, 1

Hi all Suppose I do the following: set.seed(1000) library(xtable) x <- runif( 10 ) y <- 1 + 2*x + rnorm( length(x) ) test.lm <- lm( y ~ x ) summary( test.lm ) xtable ( summary( test.lm ) ) The final xtable output follows: % latex table generated in R 1.8.1 by xtable 1.2-2 package % Tue Jun 22 09:56:36 2004 \begin{table}[ht] \begin{center} \begin{tabular}{rrrrr} \hline &

Dear friends, I am trying to understand and implement instrumental variables regression using R. I found a small (simple) example here which purportedly illustrates the mechanics (using 2-stage least-squares): http://www.r-bloggers.com/a-simple-instrumental-variables-problem/ Basically, here are the R commands (reproducible example) from that site: # ------ begin R library(AER)

Hello, I am using the Ljung Box test in R to compute if the resiudals of my fitted model is random or not. I am not sure though what the results mean, I have looked at various sources on the internet and have come up with contrasting explanations (mainly because these info deal with different program languages, like SAS, SPSS, etc). I know that my residuals should appropriate white noise( is

Dear Users, I ran factor analysis using R and SAS. However, I had different outputs from R and SAS. Why they provide different outputs? Especially, the factor loadings are different. I did real dataset(n=264), however, I had an extremely different from R and SAS. Why this things happened? Which software is correct on? Thanks in advance, - TY #R code with example data # A little

I had found the author's (Wolfgang Viechtbauer) earlier meta-analytic code in R, MiMa, useful. so I have been exploring metafor using an example dataset from MiMa. metafor provides a lot more. However, MiMa provided parameter estimates, standard errors, z values, etc. for individual moderators in the meta-analysis, but I don't see how to obtain these from metafor. Have you any help

Dear Sir or Madam? I am a doctor of urology,and I am engaged in developing a nomogram of bladder cancer. May I ask for your help on below issue? I set up a dataset which include 317 cases. I got the Binary Logistic Regression model by SPSS.And then I try to reconstruct the model ?lrm(RECU~Complication+T.Num+T.Grade+Year+TS)? by R-Project,and try to internal validate the model through

Dear Users, In case of ridge logistic regression, i want to calculate the optimum penalty using aic and bic criteria. Here is the sample code: fit <- lrm(RES ~CAT01+NUM01+NUM02+CAT02+CAT03+CAT04+NUM03+CAT05+CAT06+NUM04+ CAT07+CAT08+NUM05+NUM06, data = train.data, x = TRUE, y = TRUE) pentrace(fit, penalty = list(seq(.001, 5, by=.1))) output: Best penalty: penalty df 1.001

Hi All, I want to extract elements of elements in a list. Here's an example of what I mean: If I create a list: x <- as.list(100) for(loop in c(1:100)) { x[[loop]] <- summary(runif(100)) } > head(x) [[1]] Min. 1st Qu. Median Mean 3rd Qu. Max. 0.02271 0.25260 0.58130 0.52120 0.77270 0.99670 [[2]] Min. 1st Qu. Median Mean 3rd Qu. Max. 0.006796 0.259700

Dear R gurus I am analysing data from a study of behaviour and shade utilization of chimpanzees. I am using GLMs in R (version 2.13.0) to test whether shade/sun utilization is predicted by behaviour observed. I am thus interested in whether an interaction of behaviour (as a predictor) and presence in the sun/shade (also predictor) predicts the counts I have for the respective categories. I have

Hi, R-users, data is a matrix like this AMR BS GE HR MO UK SP500 1974 -0.3505 -0.1154 -0.4246 -0.2107 -0.0758 0.2331 -0.2647 1975 0.7083 0.2472 0.3719 0.2227 0.0213 0.3569 0.3720 1976 0.7329 0.3665 0.2550 0.5815 0.1276 0.0781 0.2384 1977 -0.2034 -0.4271 -0.0490 -0.0938 0.0712 -0.2721 -0.0718 1978 0.1663 -0.0452 -0.0573 0.2751 0.1372 -0.1346

Did something change with exception handling recently? A bunch of lit bots are showing slower compile times for many tests. Ciao, Duncan. On 20/06/12 07:53, llvm-testresults at cs.uiuc.edu wrote: > > lab-mini-03__O0-g__clang_DEV__x86_64 test results > <http://llvm.org/perf/db_default/v4/nts/1283?compare_to=1278&baseline=999> > > Run Order Start Time Duration >