search for: 0.328

Hi all, Using: R version 2.8.1 Patched (2009-03-07 r48068) on OSX (10.5.6) with survival version: Version: 2.35-3 Date: 2009-02-10 I get the following using the first example in ?summary.survfit: > summary( survfit( Surv(futime, fustat)~1, data=ovarian)) Call: survfit(formula = Surv(futime, fustat) ~ 1, data = ovarian) time n.risk n.event survival

Hi, I have been using lm in R to do a linear regression and find the slope coefficients and value for R-squared. The R-squared value reported by R (R^2 = 0.9558) is very different than the R-squared value when I use the same equation in Exce (R^2 = 0.328). I manually computed R-squared and the Excel value is correct. I show my code for the determination of R^2 in R. When I do not set 0 as the

Dear R users I have a big matrix like 6021 1188 790 290 1174 1015 1990 6613 6288 100714 6021 1 0.658 0.688 0.474 0.262 0.163 0.137 0.32 0.252 0.206 1188 0.658 1 0.917 0.245 0.331 0.122 0.148 0.194 0.168 0.171 790 0.688 0.917 1 0.243 0.31 0.122 0.15 0.19 0.171 0.174 290 0.474

Hi I am using R to fit a survival function to my data (with a weibull distribution). Data: Survival of individuals in relation to 4 treatments ('a','b','c','g') syntax: ---- > survreg(Surv(date2)~males2, dist='weibull') But I have some problems interpreting the outcome and getting the parameters for each curve. --------- Value Std.

Hello I was trying to estimate the weibull model using nls after putting OLS values as the initial inputs to NLS. I tried multiple times but still i m getting the same error of Error in nlsModel(formula, mf, start, wts) : singular gradient matrix at initial parameter estimates. The Program is as below > vel <- c(1,2,3,4,5,6,7,8,9,10,11,12,13,14) > df <- data.frame(conc, vel) >

I have a data frame, lets call it dat, with 3 columns ( mc, yr, ret) which represent market cap, year, and return. mc is a factor, mc, and ret are real numbers. I want to add a column to the data calculated as follows. For each year, I want to split the data by mc decile, then calculate the mean ret within that mc decile, and finally subtract that year's decile mean from the raw return. Then

I have the R code at the end. The last command gives me "1 observation deleted due to missingness". I don't understand what this error message. Could somebody help me understand it and how to fix the problem? > summary(afit) Df Sum Sq Mean Sq F value Pr(>F) A 2 0.328 0.16382 0.1899 0.82727 B 3 2.882 0.96057 1.1136 0.34644 C

Dear All, First of all I would like to say I do not have much knowledge about this subject, so most of you can find it really easy. I am doing a linear regression and I want to test if the slope of the curve is 0. R gives the summary statistics: Call: lm(formula = x ~ s) Residuals: Min 1Q Median 3Q Max -0.025096 -0.020316 -0.001203 0.011658 0.044970 Coefficients:

Dear R-helpers, I'm stuck with a little problem that surely has an easy solution but I can't think of a way to solve it. I'd really appreciate any help you can offer me! I'll provide a small example. Given a dataframe data.txt that looks like this: ID freq Var Var_mean Ratio_mean Var_median Ratio_median Var_sum Ratio_min Var_max Ratio_max Var_min

Hi all, I am using "lm" to fit some anova factor models with interactions. The default setting for my unordered factors is "treatment". I understand the resultant "lm" coefficients for one factors, but when it comes to the interaction term, I got confused. > options()$contrasts unordered ordered "contr.treatment"

Hi! Some background: I am trying to create an application that would encode video taken by USB camera using Theora and then send it to the client. I have almost succeeded, but I have one problem. When I grab video frames from the camera and encode them they form 4KB OGG pages, then I send them over TCP/IP to the client application. Since I want to achieve as small latency as possible I

Hi, I am analyzing a data set with greater than 1000 independent cases (collected in an unrestricted manner), where each case has 3 variables associated with it: one, a factor variable with 0/1 levels (called XX), another factor variable with 8 levels (X) and a third response variable with two levels (Y: 0/1). I am trying to see if X1 has an effect on the relationship between X2 and the

I have a data set called datastep4 with 211484 rows and 95 columns > dim(datastep4) [1] 211484 95 The first few column names are given below, note the first one is "RESPONDED" > names(datastep4)[1:5] [1] "RESPONDED" "VAR_30" "VAR_31" "VAR_32" "VAR_33" A table of RESPONDED shows mostly zeros >

Hi there, I am doing multiple comparisons for data that is not normally distributed. For this purpose I tried both functions kruskal{agricolae} and kruskalmc{pgirmess}. It confuses me that these functions do not yield the same results although they are doing the same thing, don't they? Can anyone tell my why this happens and which function I can trust? kruskalmc() tells me that there are no

Hello,.. Apologies for the newbie question but... I have a matrix R, and I know that *B %*% t(b) = R* *I'm trying to solve for B *(aka. 'factoring the correlation matrix' I think) Please help! I've read that 'to solve for B we define the eigenvalues of R and then apply the techniques of Principal Component Analysis' This made me reach for princomp() but now I'm

Hello. I'm a new user of R, and I have a question regarding the use of aov and lm-functions. I'm doing a fractional factorial experiment at our production site, and I need to familiarize myself with the analysis before I conduct the experiment. I've been working my way through the examples provided at http://www.itl.nist.gov/div898/handbook/pri/section4/pri472.htm

hello, i'd appreciate help with my glmer. i have a dependent which is an index (MH.index) ranging from 0-1. this index can also be considered as a propability. as i have a fixed factor (stage) and a nested random factor (site) i tried to model with glmer. i read that it's possible to use a quasibinomial distribution, for this kind of data, which i than actually did - but firstly (1)

http://bugs.freedesktop.org/show_bug.cgi?id=25319 Summary: KSnapshot in allocation mode crashes X Product: xorg Version: unspecified Platform: Other OS/Version: All Status: NEW Severity: normal Priority: medium Component: Driver/nouveau AssignedTo: nouveau at lists.freedesktop.org

The optim documentation states (second from last sentence of Details Section) that "Any names given to par will be copied to the vectors passed to \code{fn} and \code{gr}." This does not seem to be the case when the method argument is set to "Brent". Consider finding an optimum with the "Brent" method and a fn argument that does not rely on a named par argument,

Hello all- I am a relatively new user of R and am working through a graduate course in Statistics that uses Minitab, SAS and some Matlab. I like using R but am having some trouble lining up the output from lda() to that of the other programs' results. The dataset below is a modified set of wine data from the Pinot Noir data set as an illustration of the 2 group LDA scenario. Mo Ba