similar to: weighted inverse chi-square method for combining p-values

Hi All, Can somebody tell me how to use R to combine p values using Fisher's method? thanks. Jiong The email message (and any attachments) is for the sole use of the intended recipient(s) and may contain confidential information. Any unauthorized review, use, disclosure or distribution is prohibited. If you are not the intended recipient, please contact the sender by reply email and

Hello all, I am using the function lm to do my weighted least square regression. model<-lm(Y~X1+X2, weight=w) What I am confused is the r.squared. It does not seem that the r.squared for the weighted case is an ordinary 1-RSS/TSS. What is that precisely? Is the r.squared measure comparable to that obtained by the ordinary least square? <I also notice that model$res is the unweighted

Greetings, I'm having trouble finding a simple way to calculate a weighted histogram where there may be zero raw counts in a given interval. Given equal-length vectors of data 'data' and weights 'w', and breaks (intervals) for the histogram, I calculate a weighted histogram as follows (see MASS's 'truehist' for an unweighted histogram): bin <- cut(data,

Hi, If I used drop1 in a weighted lm fit, it seems to ignore the weights in the AIC calculation of the dropped terms, see the example below. Can this be right? Yan -------------------- library(car) > unweighted.model <- lm(trSex ~ (river+length +depth)^2- length:depth, dno2) > Anova(unweighted.model) Anova Table (Type II tests) Response: trSex Sum Sq Df F value

In the HistData package, I have a data frame, PearsonLee, containing observations on heights of parent and child, in weighted form: library(HistData) > str(PearsonLee) 'data.frame': 746 obs. of 6 variables: $ child : num 59.5 59.5 59.5 60.5 60.5 61.5 61.5 61.5 61.5 61.5 ... $ parent : num 62.5 63.5 64.5 62.5 66.5 59.5 60.5 62.5 63.5 64.5 ... $ frequency: num 0.5 0.5

I would like to apply network measures (such as betweenness centrality, upper boundedness, etc.) to a weighted graph with non-integer weights, defined by a euclidean distance matrix. The package sna provides the measures that I want to use, but seems only to operate on binary graphs. I have read work by Mark Newman (http://aps.arxiv.org/abs/cond-mat/0407503/), who suggests that a weighted graph

Hello R folks, Reproducible code below - I'm trying to do a weighted mean by a factor and can't figure it out. Thanks in advance for your assistance. Mike data<-data.frame(c(5,5,1,1,1), c(10,8,9,5,3), c("A","A","A","B","B"))

I?m running some logistic regressions and I?ve been trying to include weights in the equation. However, when I run the model, I get this warning message: Here?s what it says: Warning message: In eval(expr, envir, enclos) : non-integer #successes in a binomial glm! I think it is because the weights are non-integer values. What is a good way to run logistic regressions in R when using

dear list, my question concerns the use of `eval' in defining the model formula for `nls' (version 2.6.2.). consider the following simple example, where the same model and data are used to perform unweighted and weighted fits. I intentionally used very uneven weights to guarantee large differences in the results #================================CUT=========================== ln

Hello I'm very much a beginner on meta-analysis, so apologies if this is a trivial posting. I've been sent a set data from separate experimental studies, Treatment and Control, but no measure of the variance of effect sizes, numbers of replicates etc. Instead, for each study, all I have is the mean value for the treatment and control (but not the SD). As far as I can tell, this forces

Dear R-experts: I am using survreg() to estimate the parameters of a Weibull density having right-censored observations. Some observations are weighted. To do that I regress the weighed observations against a column of ones. When I enter the data as 37 weighted observations, the parameter estimates are exactly the same as when I enter the data as the corresponding 70 unweighted observations.

Timothy, I believe you are mistaken. Fisher's exact test give the correct answer even in the face of small expected values for the cell counts. Pearson's Chi-square approximates Fisher's exact test and can give the wrong answer when expected cell counts are low. Chi-square was developed because it is computationally "simple". Fisher's exact test, particularly with tables

Dear all, sorry to bother you but I was looking for the "scale Inverse Chi square distribution" and I could not find it! If I remember well , when X~Scale-Inv-Chi-Square (a,b) then X~Inv-Gamma(a/2,a*b/2) but with the shape, scale and rate parameters I always get confused! Any suggestions? Thanks a lot for your prompt reply. Best regards, Giorgio Di Gessa

2 days ago I asked this on r-help, but no luck... since this is actually a programming question, I post it here again: my question concerns the use of `eval' in defining the model formula for `nls' when performing weighted fits. (I use version 2.6.2., but according to NEWS there were no changes to `nls' in 2.7.0, so the problem is still present). in this scenario their

R2.3, WinXP Dear all, I am using the following functions: f1 = Phi1+(Phi2-Phi1)/(1+exp((log(Phi3)-log(x))/exp(log(Phi4))) f2 = Phi1+(Phi2-Phi1)/(1+exp((log(Phi3)-log(r)-log(x))/exp(log(Phi4))) subject to the residual weighting Var(e[i]) = sigma^2 * abs( E(y) )^(2*Delta) Here is my question, in steps: 1. Function f1 is separately fitted to two different datasets corresponding to

Hi, How do we get the value of a chi square as we usually look up on the table on our text book? i.e. Chi-square(0.01, df=8), the text book table gives 20.090 > dchisq(0.01, df=8) [1] 1.036471e-08 > pchisq(0.01, df=8) [1] 2.593772e-11 > qchisq(0.01, df=8) [1] 1.646497 > nono of them give me 20.090 Thanks, cruz

Hi to all I found an formular of an ** ***p-Value Calculator for the Chi-Square test* *http://www.danielsoper.com/statcalc/calc11.aspx* *with the formula* *http://www.danielsoper.com/statkb/topic11.aspx* *what's the gamma function of this formula in r?* *df=5* *ch2=25.50878* *the following code does not give the result <0.001 for the values above * *p=

I've successfully plotted (in the plot and abline code below) a simple regression of Lambda1_2 on VV1_2. I then successfully regressed Lambda1_2 on VV1_2, VV1_22 and VV1_212 producing lm2.l. When I go to plot lm2.l using abline I get the warning: "1: In abline(lm2.l, col = "brown", lty = "dotted", lwd = 2) : only using the first two of 4 regression coefficients"

Dear all, I encounter some discrepancies when comparing the deviance of a weighted and unweigthed model with the AIC values. A general example (from 'nls'): DNase1 <- subset(DNase, Run == 1) fm1DNase1 <- nls(density ~ SSlogis(log(conc), Asym, xmid, scal), DNase1) This is the unweighted fit, in the code of 'nls' one can see that 'nls' generates a vector

Good morning. I am a competent (not sophisticated) Stata and SPSS user. Although I'm relatively new to R, I have learned to analyze unweighted data. But now I want to analyze the 2008 American National Election Study, which provides a single weight variable (v081010). In SPSS, I set the weight and proceed with the analysis. In Stata, I specify the weight within the command. How do I use the