similar to: glm, poisson and negative binomial distribution and confidence interval

Hi, 1. How do I construct 95% prediction interval for new x values, for example - x = 30000? 2. How do I construct 95% confidence interval? my dataframe is as follows : >dt structure(list(y = c(26100000, 60500000, 16200000, 30700000, 70100000, 57700000, 46700000, 8600000, 10000000, 61800000, 30200000, 52200000, 71900000, 55000000, 12700000 ), x = c(108000, 136000,

fit lwr upr 1 218.4332 90.51019 346.3561 2 218.3906 90.46133 346.3198 3 218.3906 90.46133 346.3198 4 161.3982 44.85702 277.9394 5 192.4450 68.39903 316.4909 6 179.8056 56.49540 303.1158 7 219.5406 91.52707 347.5542 8 162.6761 46.65760 278.6945 9 193.8506 70.59838 317.1029 10 181.3816 58.11305 304.6502 11 221.2871 92.14366 350.4305 12 164.2947 47.91081 280.6785 13

Dear All, When I plot the values and linear regression line for one data set, it is fine. But for another one I see zigzags, when I plot the confidence interval >cd Depth CHAOsep12RNA 9,94 804 25,06 1476,833333 40,04 1540,561404 50,11 1575,166667 52,46 349,222222 54,92 1941,5 57,29 1053,507042 60,11 1535,1 70,04 2244,963303 79,97 1954,507042 100,31 2679,140625 >

With this data x <- c(0,0,1,1,2,2) y <- c(5,6,4,3,2,6) lwr <- y-1 upr <- y+1 xlab <- c("Low","Low","Med","Med","High","High") mydata <- data.frame(x,xlab,y,lwr,upr) I would like to make a dot plot and use lwr and upr as error bars. Above 0=Low. I would like there to be some space between the 5 and the 6 corresponding

I would like to try a simple parametric bootstrap, but unfortunately (stupidly?) my models are "overdispersed" gams & glms. I'm hoping for a function that generates overdispersed poisson or negative binomial data with a given mean, scale (& shape parameter). The loose definition I'm using is overdispersed poisson produces integer values with variance=const*mean &

Hi everyone, I am trying to build a table putting standard errors horizontally. I haven't been able to do it. library(memisc) berkeley <- aggregate(Table(Admit,Freq)~.,data=UCBAdmissions) berk0 <- glm(cbind(Admitted,Rejected)~1,data=berkeley,family="binomial") berk1 <- glm(cbind(Admitted,Rejected)~Gender,data=berkeley,family="binomial") berk2 <-

For a given linear regression, I wish to find the 2-tailed t-dist probability that Y-hat <= newly observed values. I generate prediction intervals in predict() for plotting, but when I calculate my t-dist probabilities, they don't agree. I have researched the issues with variance of individual predictions and been advised to use the variance formula below (in the code). I presume my

Enun ejemplo real estoy viendo como el intervalo de confianza usando lo que me comentas me ha salido mucho más pequeño de lo que la realidad luego refleja. ¿Cómo es esto posible?? Es decir, veo que para valores de 2,70 obtengo una respuesta de entre 2,69 y 2,90 y sin embargo luego en la realidad tengo valores entre 2,20 y 3 Gracias Jesús From: jorgeivanvelez en gmail.com Date: Thu, 21 Apr

Como usas la función image puedes consultar la ayuda ?image o help(image) y encontrarás el siguiente ejemplo donde se usa un diferente color Palette (mencionada por pepeceb en su respuesta). x <- 10*(1:nrow(volcano)) y <- 10*(1:ncol(volcano)) image(x, y, volcano, col = terrain.colors(100), axes = FALSE) # O puedes usar directamente el número para indicar el color image(x, y, volcano, col =

Hello, I have a data frame with the following columns: "date","name","value" the name is the same for each date I would like to get TukeyHSD p-value for the differences of "value" between "name"s in each "date" separately I tried different ANOVA (aov()) but can only get either tukey by "name" or by "data" but not

Hello I'm having trouble figuring out how to use the output of "segmented()" with a new set of predictor values. Using the example of the help file: ??set.seed(12) xx<-1:100 zz<-runif(100) yy<-2+1.5*pmax(xx-35,0)-1.5*pmax(xx-70,0)+15*pmax(zz-.5,0)+rnorm(100,0,2) dati<-data.frame(x=xx,y=yy,z=zz) out.lm<-lm(y~x,data=dati) o<-## S3

Buenas, una pregunta. Si yo estoy calculando un modelo lineal, el caso más simple, 1 variable respuesta y una variable explicativa y creo un modelo, me da un R2 del 80% y quiero ver como es esa relacion entre las variables, para calcular el error de predicción del modelo, basta con ver el intervalo de confianza del modelo e irme a los extremos? Por si no me he expresado bien, un ejemplo tonto:

In modelling functions some people like to use a weight of 0 to drop an observation instead of using a subset value of FALSE. E.g., weights=c(0,1,1,...) instead of subset=c(FALSE, TRUE, TRUE, ...) to drop the first observation. lm() and summary.lm() appear to treat these in the same way, decrementing the number of degrees of freedom for each dropped observation. However, predict.lm() does

Dear all,? I hope that you can help me on this. I have been struggling to figure this out but I haven't found any solution. I am running a generalised mixed effect model, gamm4, for an ecology project. Below is the code for the model: model<-gamm4(LIFE.OE_spring~s(Q95, by=super.end.group)+Year+Hms_Rsctned+Hms_Poaching+X.broadleaved_woodland? ? ? ? ? ? ?+X.urban.suburban+X.CapWks,

I am studying statistics using R and a book "Understandable Statistics", by Brase and Brase. The book has two worked examples for calculating a confidence interval around a predicted value from a linear model. The answers to the two examples in the book differ from those I get from R. The regression line, the standard error, and the predicted value in R and the book all agree for the

Dear all, I have a quite basic questions about anova analysis in R, sorry for this, but I have no clue how to explain this result. I have two datasets which are named: nmda123, nmda456. Each dataset has three samples which were measured three times. And I would like to compare means of them with Posthoc test using R, following please see the output: (CREB, mCREB and No virus are the name of

Dear all, I am studying a dataset using the aov() function. The independant variable 'cds' is a factor() with 8 levels and here is the result in studying the dependant variable 'rta' with aov() : > summary(aov(rta ~ cds)) Df Sum Sq Mean Sq F value Pr(>F) cds 7 0.34713 0.04959 2.3807 0.02777 Residuals 92 1.91635 0.02083 The dependant variable

Craig, Thanks for your follow-up note on using the asypow package. My problem was not only constructing the "constraints" vector but, for my particular situation (Poisson regression, two groups, sample sizes of (1081,3180), I get very different results using asypow package compared to my other (home grown) approaches. library(asypow) pois.mean<-c(0.0065,0.0003) info.pois <-

I am a new R user but a long time SAS user. I searched for a response to this question but no luck, so forgive me if this topic has been covered before. I am running a TukeyHSD post hoc test after running an ANOVA. I get the results of all pairwise comparisons, no problem. However, the output table is a little "busy", and I'd like to make the output easier to read. Specifically, I

I am attempting to run a Poisson EWMA model using Patrick Brandt's source code. I get the following error when I run the code: Error in dimnames(x) <- dn : length of 'dimnames' [1] not equal to array extent Dimnames(x) looks like this: [[1]] NULL [[2]] [1] "mip" "div" "nom" "unity" "mood"