search for: 0.038

Good day, All, Is there any way to maintain the number of decimal places in the type of situation below? I would like to maintain the number of decimal places in 0.667, despite the fact that its column-mates have a fourth decimal place. Thank you for your time. Jim dat.txt contents: MARKER ALLELES FREQ1 RSQR EFFECT2 STDERR CHISQ PVALUE rs6599753 C,T

Hi list, I have the following code to compute the acf of a time series acfresid <- acf(residfit), where residfit is the series when I type acfresid at the prompt the follwoing is displayed Autocorrelations of series ?residfit?, by lag 0.0000 0.0833 0.1667 0.2500 0.3333 0.4167 0.5000 0.5833 0.6667 0.7500 0.8333 1.000 -0.015 0.010 0.099 0.048 -0.014 -0.039 -0.019 0.040 0.018

En innebygd og tegnsett-uspesifisert tekst ble skilt ut... Navn: ikke tilgjengelig Nettadresse: https://stat.ethz.ch/pipermail/r-help/attachments/20060921/b479a8df/attachment.ksh

OK, Let's try this again! Here is the reproducible script; it is long because I had to copy the panel dataset here. My question is related to systemfit; I don't know how to get the result for the entire panel. #Reproducible script Empdata<- read.csv("/Users/ngwinuiazenui/Documents/UPLOADemp.csv") View(Empdata) install.packages("systemfit")

Sadly you failed to set your email program to send plain text and the data is corrupted at my end. I also think you need to reduce the size of the data set... the intent here is to increase your understanding, not debug your particular analysis. I will say that I am having a very challenging time understanding what you are trying to accomplish though. What are the equations that you think need

I can't get my simultaneous equations to work using system fit. Please help. #Reproducible script Empdata<- read.csv("/Users/ngwinuiazenui/Documents/UPLOADemp.csv") View(Empdata) str(Empdata) Empdata$gnipc<-as.numeric(Empdata$gnipc) install.packages("systemfit") library("systemfit") pdata <- plm.data(Empdata,

Full_Name: Adrian Custer Version: 0.90.0 OS: Linux on Thinkpad (pentium) and desktop (K6) Submission from: (NULL) (128.32.251.234) When I create a boxplot, and then try to overlay a lowess fit or just the points, the points do not appear in the highest level and the lowess curve does not reach the highest level. However, if I add one to each of the models, the problem is solved. I tried this

... and the mailing list is picky about attachments... whatever you attached did not conform to the stringent requirements mentioned in the Posting Guide. Pasting the code right into the email is usually safest, though you DO have to post using plain text (as the Posting Guide indicates) or your code may get mangled by the automatic html format removal. On May 15, 2018 7:04:31 AM PDT, Bert Gunter

Unless there is good reason not to, always cc the list -- there are lots of smarter folks than I on it who can help. I may or may not have time to look at this. Hopefully someone else will. -- Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip

Estimados Quisiera tener la sintaxis generar una serie de números entre 0 y 1 con distribución gamma de 25 puntos y que la suma de todos los puntos debe dar 1. Por ejemplo: c (0.000, 0.001, 0.012, 0.043, 0.078, 0.104, 0.117, 0.116, 0.108, 0.094, 0.078, 0.063, 0.049, 0.038, 0.028, 0.021, 0.015, 0.011, 0.008, 0.005, 0.004, 0.003, 0.002, 0.001, 0.001) No hemos podido

Dear all, I am so ashamed to pollute the list with a trivial question, but it is a long time I have not used R, and I need a result in the next one or two hour... I have a table which I have loaded with read.table, and I want to make the mean of its columns. > slides <- read.table("slides.txt") > slides [1:5,] V1 V2 V3 V4 V5 V6 V7 V8 1

Tengo un modelo de regresión de Cox y quiero obtener el plot ajustado por una covariable (sexo) con la función ?ggadjustedcurves?, pero me da el siguiente error: > cox2 <- coxph(os ~ imc_25 + sexo.1, data = datos) > cox2 Call: coxph(formula = os ~ imc_25 + sexo.1, data = datos) coef exp(coef) se(coef) z p imc_25 -0.621 0.537 0.299 -2.08 0.038 sexo.1M 0.714

Hi all, I can't find the error in the binomial GLM I have done. I want to use that because there are more than one explanatory variables (all categorical) and a binary response variable. This is how my data set looks like: > str(data) 'data.frame': 1004 obs. of 5 variables: $ site : int 0 0 0 0 0 0 0 0 0 0 ... $ sex : Factor w/ 2 levels "0","1": NA NA NA

I have a list of entrants (1-14 in this example) in a competitive event and corresponding win probabilities for each entrant. [(1, 0.049), (2, 0.129), (3, 0.043), (4, 0.013), (5, 0.015), (6, 0.040), (7, 0.066), (8, 0.038), (9, 0.204), (10, 0.022), (11, 0.234), (12, 0.044), (13, 0.068), (14, 0.035)] So, of course Sum(ps) = 1. In order to make some subsequent computations more

On Mon, Apr 13, 2009 at 12:41 PM, Dan Dube <ddube-at-advisen.com> wrote: > i use tapply and by often, but i always end up banging my head against > the wall with the output. The proposed solution of Dan's problem posted on R-help was: > do.call(rbind,a) When I use this 'solution' I get 'ERROR: second argument must be a list'. So head on wall continues. My

Estimado Jorge Perdóneme que lo moleste de nuevo, hay otra condición además de que sum(y)=1 y es que y[1] tiene que dar 0 en el ejemplo y<- c (0.0000000000, 0.6321985783, 0.2325728597, 0.0855587737, 0.0314753138, 0.0115791209, 0.0042597205, 0.0015670636, 0.0005764905, 0.0002120790) y[1]=0 sum(y)=1 esto se utiliza en el paquete EpiEstim donde se

Hi, How can I access the Values stored in kpssstat-class given by KPSS.test function and store it in a variable. For example: >x <- rnorm(1000) >test <- KPSS.test(ts(x)) >test ---- ---- KPSS test ---- ---- Null hypotheses: Level stationarity and stationarity around a linear trend. Alternative hypothesis: Unit root. ---- Statistic for the null

Dear All, I am trying to calculate a 95% confidence interval for the difference in two c statistics (or equivalently D statistics). In Stata I gather that this can be done using the lincom command. Is there anything similar in R? As you can see below I have two datasets (that are actually two independent subsets of the same data) and the respective c statistics for the variables in both cases.

Dear All, Apologies if you have a seen a question like this from me before. I am hoping that if I re-word my question more carefully someone may be able to offer more help than the last time I asked something similar. I am using R 2.9.2 and Windows XP. I am trying to determine if there is a statistically significant difference between two c-statistics (or equivalently D statistics). In Stata

I'm having a lot of problem with it recognizing "oh" for zero. I've tried both "o" and "oh". In one case, I just tried: $digit = o { out = "0"; } | fundamental {out = "2"; }; So I gave it a choice that was VERY far away from what I said. But when I said "o o o o o", more than 75% of the time, it had a bunch of