similar to: kernel density to smooth plots

Hi, I have the binned data (observed and generated from model)  that I would like to test using the anderson-darling goodness of fit test.  But I'm not sure which package in R to use. I tried ad.test(...) but it does not recognise the test by Vito Ricci in FITTING DISTRIBUTIONS WITH R   > ad.test(hist_hume_beec[,1],hist_hume_beec[,2]) Error: could not find function "ad.test"

I'm getting an error message when using the new version of sqldf, > library(sqldf) > str(kdv) 'data.frame': 71 obs. of 3 variables: $ dpss: num 0.117 0.144 0.164 0.166 0.165 ... $ npdp: num 0.1264 0.0325 0.0109 0.0033 0.0055 ... $ logk: num 1.12 1.29 1.41 1.41 1.42 ... > test=sqldf("select * from kdv") Error in get("fun", env = this, inherits =

Dear WizaRds, A run of nls produces the following concise summary: > summary(cs.wt) Formula: 0 ~ wt.MM(conc, time, A1, a1, A2, a2) Parameters: Estimate Std. Error t value Pr(>|t|) A1 4.814e+02 2.240e+01 21.495 0.0296 * a1 7.401e-01 7.435e-02 9.956 0.0637 . A2 1.613e+02 1.738e+01 9.280 0.0683 . a2 1.770e-02 7.324e-03 2.417 0.2497

Hola, Manuel, No entiendo tu pregunta (la repito aqui para que sea mas explicito): hay alguna forma de comparar la probabilidad de supervivencias (en este caso anual) entre grupos sin utilizar un chi-cuadrado y un valor de P. Entiendo que lo que hace survdiff es comparar las curvas de supervivencia, pero yo quiero comparar la probabilidad de supervivencia entre grupos al final del estudio. Con

Hi, I have the following question about creating data frames. I want to create a data frame with 2 components: a vector and a matrix. Let me use a simple example: y <- rnorm(10) x <- matrix(rnorm(150), nrow=10) Now if I do dd <- data.frame(x=x, y=y) I get a data frame with 16 colums, but if, according to the documentation, I do dd <- data.frame(x=I(x), y=y) then str(dd)

Ted Harding said: > I can get the estimated RRs from > RRs <- exp(summary(GLM)$coef[,1]) > but do not see how to implement confidence intervals based > on "robust error variances" using the output in GLM. Thanks for the link to the data. Here's my best guess. If you use the following approach, with the HC0 type of robust standard errors in the

Hi, I am currently trying to do dose-response curves using weighted 4-parameter model (4PL). The weighting was based on 1/(expected variance) derived from historical data. I tried both drm() from drc package, and nls(), found very different results derived from drm() vs. nls() using "weights=" argument. d1<-read.table("d1.txt",sep='\t',header=T,row.names=1)

Hi, I've got a little problem using auto.arima. I run the following command auto.arima(drivers,ic="aic",d=1,D=1,max.order=10,max.p=5,max.q=5,max.P=5,max.Q=5,stepwise=FALSE,allowdrift=FALSE) and I get the following output : Series: drivers ARIMA(0,1,1)(5,1,1)[12] Coefficients: ma1 sar1 sar2 sar3 sar4 sar5 sma1 -0.6421

Hi all, today we had a performance issue transfering a big amount of data where one file was over 50GB. Rsync was tunneled over SSH and we expected the data to be synced within hours. However after over 10 hours the data is still not synced ... The sending box has rsync running with 60-80 % CPU load (2GHz Pentium 4) while the receiver is nearly idle. So far I had no acces to the poblematic

Hello, I would like to perform a Partial least square discriminate analysis (PLSDA) in R. To do this I use the package mixOmics. I could perform the PLSDA in R. however I would also like to perform a leave-one-out cross validation in order to assess the performance of my model. My supervisor told me that I should focus on the R2/Q2 ratios. However when I read the instruction for running the

Dear all, I'm using R 2.8.1 under Windows Vista on a dual core 2,4 GhZ with 4 GB of RAM. I'm trying to reproduce a result out of "Analysis of Financial Time Series" by Ruey Tsay. In R I'm using the fGarch library. After fitting a ar(3)-garch(1,1)-model > model<-garchFit(~arma(3,0)+garch(1,1), analyse) I'm saving the results via > result<-model

Hi I have data from 12 subjects. The measurement is log(expression) of a particular gene and can be assumed to be normally distributed. The 12 subjects are divided into the following groups: Infected, Vaccinated, Lesions - 3 measurements Infected, Vaccintaed, No Lesions - 2 measurements Infected, Not Vaccinated, Lesions - 4 measurements Uninfected, Not Vaccinated, No Lesions - 3 measurements

Full_Name: Ahmad Abu Hammour Version: rw0651 OS: windows 95 Submission from: (NULL) (63.23.128.44) Although I know that "ts package" is preliminary, I wanted to compare the results from R and SPSS. I ran ARIMA(2,1,2) in both softwares. I got NaN in standard errors of coefficients from R and real figures from SPSS. I changed "delta" in R to match that used by SPSS, I received

windows 7, R 2.12 I am trying to run a piecewise linear regression with a single knot, i.e. a regression composed of two straight lines where the two lines intersect at an x value given by the variable knot. I wish to estimate the slope of both lines, the value of knot, the x value where the two lines intersect, and an intercept. I am using the nls code below, and get the following error

Dear List, How do I extract the approximate Wald test for the frailty (in the following example 17.89 value)? What about the P-values, other Chisq, DF, se(coef) and se2? How can they be extracted? ######################################################> kfitm1 Call: coxph(formula = Surv(time, status) ~ age + sex + disease + frailty(id, dist = "gauss"), data = kidney)

Are these 225 compile time regressions real? It sure looks bad! Ciao, Duncan. On 01/12/11 09:39, llvm-testresults at cs.uiuc.edu wrote: > > bwilson__llvm-gcc_PROD__i386 nightly tester results > > URL http://llvm.org/perf/db_default/simple/nts/380/ > Nickname bwilson__llvm-gcc_PROD__i386:4 > Name curlew.apple.com > > Run ID Order Start Time End Time > Current 380