search for: dof

Hi all, ########################################## dof=c(1,2,4,8,16,32) Q5=matrix(rt(100,dof),100,6,T,dimnames=list(NULL,dof)) par(mfrow=c(2,6)) apply(Q5,2,hist) myf=function(x){ qqnorm(x);qqline(x) } apply(Q5,2,myf) ########################################## These looks ok. However, I would like to achieve more. Apart from using a loop, is there are...

...logins do not work anymore (the session is immediately closed after a successful login). Interestingly, pipe and spool2dir are working (that is, the session won't be closed), dspam-exec and crm114-exec are not. If this happens, mail_debug output is as follows: imap-login: Login: user=<dof>, method=PLAIN, rip=127.0.0.1, lip=127.0.0.1, mpid=10420, secured imap: Debug: Loading modules from directory: /usr/lib/dovecot/modules imap: Debug: Module loaded: /usr/lib/dovecot/modules/lib20_autocreate_plugin.so imap: Debug: Module loaded: /usr/lib/dovecot/modules/lib20_fts_plugi...

...and the bayesm libraries allow us to make draws from the Inverse Wishart distribution. But I wanted to find out how exactly is the Inverse Wishart distribution parameterized in these libraries. The reason I ask is the following: Now its generally standard to express Inverse Wishart as IW(0.5 * DOF,0.5* Scale). (DOF-> Degree of freedom, Scale -> Scale parameter). If we follow standard usage when we refer to the Degree of Freedom of the above IW distribution it is = DOF (and not 0.5* DOF). Similarly the Scale parameters of the above IW it is= Scale (and not 0.5*Scale). For the MCM...

I need to fit a t copula with fixed degree of freedom let's say 4. I do not want to estimate the dof together with correlation matrix optimally. Instead fix the dof to 4 and only estimate the correlation matrix in the optimization routine. Is anyone aware of such estimation method in R. The packages and functions that I know of can't do this estimation. I searched online but couldn...

I'm running an ANOVA on some data for respiration in a forest. I am having a problem with my degrees of freedom. For one of my variables I get one fewer degrees of freedom than I should. I have 12 plots and I therefore expected 11 degrees of freedom, but instead I got 10. Any ideas? I have some code and output below: > class(Combined.Plot) [1] "character" >

...; is of unknown type") scores <- match.arg(scores) if(scores != "none" && !have.x) stop("requested scores without an 'x' matrix") p <- ncol(cv) if(p < 3) stop("factor analysis requires at least three variables") dof <- 0.5 * ((p - factors)^2 - p - factors) if(dof < 0) stop(gettextf("%d factors is too many for %d variables", factors, p), domain = NA) sds <- sqrt(diag(cv)) cv <- cv/(sds %o% sds) cn <- list(nstart = 1, trace = FALSE, lower = 0.005)...

Hello! I have a question about the factanal function. This function returns at the end test statistics like this: Test of the hypothesis that 4 factors are sufficient. The chi square statistic is 4.63 on 2 degrees of freedom. The p-value is 0.0988 Is it possible to get the chi square statistic and the p-value as variables, not the text on the screen? An object of class "factanal"

Hi there, Does anyone know how to explicitly refer to the p-value of thet test that the chosen number of factors is significant in a factor analysis. It's not in the list of values for the factanal command output yet it is printed out with the results. Thanks in advance. Wayne Dr Wayne R. Jones Statistician / Research Analyst KSS Group plc St James's Buildings 79 Oxford Street

...; is of unknown type") scores <- match.arg(scores) if(scores != "none" && !have.x) stop("requested scores without an 'x' matrix") p <- ncol(cv) if(p < 3) stop("factor analysis requires at least three variables") dof <- 0.5 * ((p - factors)^2 - p - factors) if(dof < 0) stop(gettextf("%d factors is too many for %d variables", factors, p), domain = NA) sds <- sqrt(diag(cv)) cv <- cv/(sds %o% sds) cn <- list(nstart = 1, trace = FALSE, lower = 0.005)...

...r by the drti ioctl. So whenever I run a program with an USDT probe, no tracepoint is installed. Only after I run the dtrace command the tracepoint is actually installed on the victim process. My question is about how Solaris discovers the correct instruction pointer (PC) on the target victim. The DOF only has information relative to the offset inside the function where the probe is defined and the address of the function in the _relocatable_ object. So, whenever I''m building a program with an usdt probe I get: libdtrace DEBUG: defined probe database:::query-start main() +0x13 (main)...

...to SAS results it shouldn't happen. If I try to use another working correlation (like AR-M or stat_M_dep), R just exits without giving any error message. Another doubt is how R estimates the scale parameter for Poisson? I get 1.36 in R, 1.25 in SAS (estimation by the square root of DEVIANCE/DOF) and 1.17 (estimation by the square root of Pearson's Chi-Square/DOF). I'm using R 1.4.1 but all the problems apply for S-Plus 2000 too. I hope anybody can help me. Sincerely, -- Frederico Zanqueta Poleto fred at poleto.com - ICQ# 4129787 -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-...

...0.283,0.613),dim=3,dispstr="un",df=6,df.fixed = TRUE) where c(0.785,0.283,0.613) is the correlation pattern of my data with 0.785 pearson correlation between variable 1-2, 0.283 correlation between 1-3 and 0.613 is the correlation between variable 2-3. In given command degree of freedom (dof) is fixed at 6 and i'm checking p-value of the estimate using gof copula using gofCopula(t.cop,x,500) for 500 iterations, where x is my data vector. I'm checking p-values of my each run by varying dofs from 2,3,...,6. But in every run the value of cramer von-mises is changing but p-value i...

...emade version I only use the "test" database. I am confused if the cv.glm generates new glm models for each simulation of if it uses the one provided? (3) is the cv.glm sampling using replacement = TRUE or not? Thanks in advance. LOT ***** cv.glm method glm0.dmi<-glm(DMI_kg~Sex+DOF+Avg_Nem+In_Wt) # Simulation for 50 re-samplings... perr1.vect<-vector() for (j in 1:50) { print(j) cv.dmi<-cv.glm(data.dmi, glm0.dmi, K = 10) perr1<-cv.dmi$delta[2] perr1.vect<-c(perr1.vect,perr1) } x11() hist(perr1.vect) mean(perr1.vect) sd(perr1.vect) ***** homem...

Hi, I want to compute the quantiles of Chi^2 distributions with different degrees of freedom like x<-cbind(0.005, 0.010, 0.025, 0.05, 0.1, 0.5, 0.9, 0.95, 0.975, 0.99, 0.995) df<-rbind(1:100) m<-qchisq(x,df) and hoped to get back a length(df) times length(x) matrix with the quantiles. Since this does not work, I use x<-c(0.005, 0.010, 0.025, 0.05, 0.1, 0.5, 0.9, 0.95, 0.975,

Dear all, I am having a (really) hard time getting pyears to work together with a ratetable to give me the number of expected events (deaths). I have the following data: dos, date of surgery, as.Date dof, date of last follow-up, as.Date dos, date of surgery, as.Date sex, gender, as.factor (female,male) ev, event(death), 0= censored at time point dof, 1=death at time point dof Could someone please help. I am searching the web 5 days now and I cannot find any simple example that I can reproduce or f...

Hi, I have Postfix + OpenLdap + DoveCot configuration, and it's running succesfuly, i wantto convert users pop3 password NTPassword and LMPassword, so i ne plain passwor dof users, how can i do that. (Normaly using perl's ntlmgen function i convert password , but in plain) thanks in advance

...= yes wins support = true server string = back from hell, asking for more! encrypt passwords = yes ssl CA certDir = /etc/ssl/certs update encrypted = yes If there is any other info you might need, then feel free to ask, I'll do my best. thanks dof. -- David Goodwin | dof@codepoets.co.uk | http://www.codepoets.co.uk Software Engineering, Uni. Of Wales Aberystwyth. Do not meddle in the affairs of wizards, for they are subtle and quick to anger. Do not go to the elves for counsel, for they will say both no and yes.

Hi, I have a faulty dtrace script (most probably too big bufsize). My machine hangs at boot (amd64 5.10 Version Generic_118855-10 64-bit). NOTICE: enabling probe 11 (dtrace:::ERROR) WARNING: /etc/svc/volatile: File system full, swap space limit exceeded WARNING: Sorry, no swap space to grow stack for pid 5 (autopush) WARNING: /etc/svc/volatile: File system full, swap space limit exceeded How to

If I wanted to fit a logit model and account for clustering of observations, I would do something like: library(Design) f <- lrm(Y1 ~ X1 + X2, x=TRUE, y=TRUE, data=d) g <- robcov(f, d$st.year) What would I do if I wanted to do the same thing with a probit model? ?robcov says the input model must come from the Design package, but the Design package appears not to do probit? Thanks very

..., total.df=NA, rest.df=NA, model.df=NA, LL=NA, AIC=NA, AICc=NA, BIC=NA) }else{ n <- object$n.obs p <- length(object$uniquenesses) m <- object$factors model.df <- (p*m) + (m*(m+1))/2 + p - m^2 total.df <- p*(p+1)/2 rest.df <- total.df - model.df # = object$dof LL <- -as.vector(object$criteria["objective"]) k <- model.df aic <- 2*k - 2*LL aicc <- aic + (2*k*(k+1))/(n-k-1) bic <- k*log(n) - 2*LL c(n=n, items=p, factors=m, total.df=total.df, rest.df=rest.df, model.df=model.df, LL=LL, AIC=aic, AICc=aicc, BIC=...