similar to: CI

Un texte encapsul? et encod? dans un jeu de caract?res inconnu a ?t? nettoy?... Nom : non disponible Url : https://stat.ethz.ch/pipermail/r-help/attachments/20061019/dc66a793/attachment.ksh

R-experts: A quick question, please. >From a lab exp, I got 12 positives out of 50. To get 90% CI for this , I think binom.test might be the one to be used. Is there a better way or function to calculate this? > binom.test(x=12, n=50, p=12/50, conf.level = 0.90) Exact binomial test data: 12 and 50 number of successes = 12, number of trials = 50, p-value = 1 alternative

I have the clinical study data. Year 0 Year 3 Retinol (nmol/L) N Mean +-sd Mean +-sd Vitamin A group 73 1.89+-0.36 2.06+-0.53 Trace group 57 1.83+-0.31 1.78+-0.30 where N is the number of male for the clinical study. I want to test if the mean serum retinol has increased over 3 years among subjects in the vitamin A group. > 1.89+0.36

What would be the R formulae for a two-sided test? I have a formula for a one-sided test: powertest <- function(a,m0,m1,n,s){ t1 = -qnorm(1-a) num = abs(m0-m1) * sqrt(n) t2 = num/s pow = pnorm(t1 + t2) } Would you pls let me know if you know of? Thank you, ej

I'm trying to replicate some of the examples from my textbook in R (my text uses Minitab). In this problem, I'm trying to construct a 95% confidence interval for these distance measurements [1]: > # Case Study 7.4.1, p. 483 > x <- scan() 1: 62 52 68 23 34 45 27 42 83 56 40 12: Read 11 items > alpha<-.95 > mean(x) + qt(c((1-alpha)/2, 1-((1-alpha)/2)), df=length(x)-1) *

A quick question, please! How do you extract a certain value of vector? i.e. x = c(2,5,3,6,21,3,6,24, ....) How do you get the 1st one (which is 2); the 5th one (which is 21); etc? thx much, ej [[alternative HTML version deleted]]

Hi, I have a glm gives summary as follows, Estimate Std. Error z value Pr(>|z|) (Intercept) -2.03693352 1.449574526 -1.405194 0.159963578 A 0.01093048 0.006446256 1.695633 0.089955471 N 0.41060119 0.224860819 1.826024 0.067846690 S -0.20651005 0.067698863 -3.050421 0.002285206 then I use confint(k.glm)

A quick question please! How do you rename column names? i.e. V1 --> Apple; V2 --> Orange, etc. thx much ej [[alternative HTML version deleted]]

It seems that confint.default returns an empty data.frame for objects of class mlm. For example: ``` nobs <- 20 set.seed(1234) # some fake data datf <- data.frame(x1=rnorm(nobs),x2=runif(nobs),y1=rnorm(nobs),y2=rnorm(nobs)) fitm <- lm(cbind(y1,y2) ~ x1 + x2,data=datf) confint(fitm) # returns: 2.5 % 97.5 % ``` I have seen proposed workarounds on stackoverflow and elsewhere, but

// R 2.3.1 Can someone please explain why this error returns? > y=numeric(100) > x=matrix(runif(16),4,4) > for(i in 2:100) + { + y[i]=which(rmultinom(1, size = 1, prob = x[y[i-1], ])==1) + } Error in rmultinom(n, size, prob) : too few positive probabilities thx much ej

>x=c(3.05176E-05,0.000457764,0.003204346,0.0138855,0.04165649,0.09164429,0.1527405,0.1963806,0.1963806,0.1527405,0.09164429,0.04165649,0.0138855,0.003204346,0.000457764,3.05176E-05) >y=c(0.0000306,0.0004566,0.0031985,0.0139083,0.0415539,0.0917678,0.1528134,0.1962831,0.1962994,0.1527996,0.0917336,0.0415497,0.0139308,0.0031917,0.0004529,0.0000301) I tried chisq.test, t-test, prop.test, etc,

Dear all, How to obtain the odds ratio (OR) and 95% confidence interval (CI) with 1 standard deviation (SD) change of a continuous variable in logistic regression? for example, to investigate the risk of obesity for stroke. I choose the happening of stroke (positive) as the dependent variable, and waist circumference as an independent variable. Then I wanna to obtain the OR and 95% CI with

An apparatus exists whereby a collection of balls is displaced to the top of a stack by suction. A top level (Level 1) each ball is shifted 1 unit to the left or 1 unit to the right at random with equal probability. The ball then drops down to level 2. At Level 2, each ball is again shifted 1 unit to the left or 1 unit to the right at random. The process continues for 15 levels and the balls are

Grandiosa Comunidad Saludos Quiero pedirles ayuda en los siguientes puntos; .- Pregunto si puedo sacar, aumentarle al summary también los siguientes puntos Intervalo de confianza, desviación estándar? .- como puedo obtener la correlación en variables dicotómicas ej. tau de kendall? .- Como puedo cambiar los títulos del ingles al español de los encabezados de un surfit(Surv(tiempo, estado))? --

Dear all, I was trying to see how the function 'confint' is defined. Doing > confint function (object, parm, level = 0.95, ...) UseMethod("confint") <environment: namespace:stats> does not really enlighten me. How can I get to see the implementation (I guess it should be possible according to the general philosophy of the R project)? Thanks in advance S??ren

I can't get confint.glm to work from within a function. Consider the following (using R 1.9.1, Windows 2000): # FIRST: SOMETHING THAT WORKS FROM A COMMAND PROMPT DF <- data.frame(y=.1, N=100) (fit <- glm(y~1, family=binomial, data=DF, weights=DF[,"N"])) Call: glm(formula = y ~ 1, family = binomial, data = DF, weights = DF[, "N"]) Coefficients:

the function confint uses the profiling method of the function of the package MASS confint.glm even after the package has been detached! 1: might this be the intenden behavior? 2. How does the function remember its 'MASS' functionality after detaching the package? R: 1.8.0; Windows 2000 Here is a sample program > set.seed(7882) > x<-rep(c(0,1),c(20,20)) >

Hello, I am trying to calculate and plot mean and confidence intervall for a set of data. This is the code that I am currently using: means <- sapply(data, mean, na.rm=TRUE) n <- sapply(data,length) stdev <- sqrt(sapply(data, var, na.rm=TRUE)) ciw <- qt(0.98, n) * stdev / sqrt(n) par(mgp=c(2,0.6,0), las=2, fin=c(7,3), mai=c(1,0.5,0.2,0.2), cex=0.8) plotCI(x=means, uiw=ciw,

help(qt) states that: "ncp non-centrality parameter delta; currently except for rt(), only for abs(ncp) <= 37.62" so I would expect that calling qt with non-centrality parameter exceeding 37.62 should fail, instead e.g. calling > mapply(function(x) qt(p = 0.9, df = 55, ncp = x),35:45) gives: [1] 40.21448 41.35293 42.49164 43.68862 44.82945 45.97048 47.11170 48.25310 [9]

Full_Name: Gerard Torrent Version: R version 2.8.0 (2008-10-20) OS: Linux 2.6.27.5-41.fc9.x86_64 #1 SMP Submission from: (NULL) (85.52.227.233) In some cases qt complains about NaNs and don't gives the correct result: > qt(0.1,, 0.1) [1] NaN Warning message: In qt(p, df, lower.tail, log.p) : NaNs produced But the result can be found: > pt(-1.60443e+06, 0.1) [1] 0.09999997 If I