similar to: How to make a figure plotting p-values by range of different adjustment values?

Hi Jim, Thanks for your help, I really appreciate it. Perhaps I'm misunderstanding, but does this formula run different ajustment values for this function? logit(p = doc$value, adjust = 0.025) I'm looking to plot the p-values of different adjustment values. Thanks so much, Kirsten On Wed, Jul 12, 2017 at 8:49 PM, Jim Lemon <drjimlemon at gmail.com> wrote: > Hi Kirsten,

Hi Kirsten, Perhaps this will help: set.seed(3) kmdf<-data.frame(group=rep(1:4,each=20), prop=c(runif(20,0.25,1),runif(20,0.2,0.92), runif(20,0.15,0.84),runif(20,0.1,0.77))) km.glm<-glm(prop~group,kmdf,family=quasibinomial(link="logit")) summary(km.glm) pval<-0.00845 padjs<-NA npadj<-1 # assume you have five comparisons in this family for(method in p.adjust.methods) {

Hi Ben, That's exactly right! Except for each set it's the sample population that is 400, 800 or 300. I want to take 3 samples, each of 100, where only the population differs. I can do this separately, but I'm having trouble putting them all on the same graph. I'd like to have sample on the x axis (1-300) and estimate on the y axis. I want to show how population affects the

Hi everyone, I'm having some trouble with my ifelse statements. I'm trying to put 12 conditions within 3 groups. Here is the code I have so far: dat$cond <- ifelse(test = dat$cond == "cond1" | dat$cond == "cond2" | dat$cond == "cond3" dat$cond == "cond4" yes = "Uniform" no = ifelse(test =

Hi! Thanks for taking the time to read this. The code below creates a graph that takes 100 samples that are between 5% and 15% of the population (400). What I'd like to do, however, is add two other sections to the graph. It would look something like this: from 1-100 samples take 100 samples that are between 5% and 15% of the population (400). From 101-200 take 100 samples that are between

Hi Kirsten, I can run your example code but I can't quite follow your division of sampling. Can you restate the the task? Below is what I think you are asking for, but I have the feeling I may be off the mark. Set A: 400 samples, draw 100 in range of 5 to 15 Set B: 800 samples, draw 100 in range of 5 to 15 Set C: 300 samples, draw 100 in range of 5 to 15 Ben > On Aug 5, 2017, at

James, I am running Windows 2008 R2 on Xen that is packaged with CentOS 5.4. I installed gplpv_Vista2008x64_0.11.0.238.msi using the falling steps: 1. Turn on testsigning 2. Reboot 3. Install gplpv_Vista2008x64_0.11.0.238.msi 4. Install process requests a reboot at the end. 5. At the end of the subsequent boot, there is a prompt for a reboot again. I

Not a reproducible example, so a bit of guessing here, but a) don't try to assign results to variables inside the ifelse. That is, remove all the single-equals signs and "test" variables. If you really need to conditionally assign variables then use "if"... but chances are good you don't need that. b) "closure" is effectively another word

Dear R experts, I want to plot a line chart with another secondary axis placed right to the standard secondary axis which one can access with the axis command, so that the data lines are seen in the same plot. Is there any way to do this in R? Many thanks, Kirsten.

Hello everyone, I have a .csv file with the following format: uniqueID SubjectID Distance_miles Tag 1 1001 5.5 3 2 1001 7 1 3 1001 6.5 1 4 1001 5 1 5 1002

I have a list of ICD9 (disease) codes with various formats - 3 digit, 4 digit, 5 digit. The first three digits of these codes are what I am most interested in. I would like to either add zeros to the 3 and 4 digit codes to make them 5 digit codes or add decimal points to put them all in the format ###.##. I did not see a function that allows me to do this in the formatting command. This seems

Hello, I have a problem with using replace() to convert a vector of dates from yyyy-mm-dd to julian date. For example, I type replace(x,2004-05-14,134) and I receive an error: Error in `[<-.data.frame`(`*tmp*`, list, value = 134) : new columns would leave holes after existing columns If I can successfully convert, I have a script that will convert all of the dates in

I have a researcher who is consistently get the warning message: In max(i) : no non-missing arguments to max; returning -Inf Best as I can tell the code is working properly and the output is as expected. I would like some help in understanding why he is getting this error message and what its implications are. I have his code. Sincerely, Kirsten Miles Support Specialist Research Computing Lab

Hello all, I have a data set where the response variable is the percent cover of a specific plant (represented in cover classes 0,1,2,3,4,5, or 6). This data set has a lot of zeros (plots where the plant was not present). I am trying to model cover class of the plant as a function of both total nitrogen and shrub cover. After quite a bit of research I have come across a conditional approach

Hello. I am looking for a function that will allow me to paste rows together without regard for the numbers of columns in the datasets to be joined. The only columns where it matters if they are aligned correctly are at the beginning - the rest of the columns represent differing numbers of ICD9 (disease) codes reported by each person(record) at a health visit. They are in no particular order.

Hey everyone, I'm working on a contour plot depicting asymptomatic prevalence at varying durations of infectiousness and force of infection. I've been able to work everything out except for this one - my legend title keeps getting cut off. Here's what I have: filled.contour(x=seq(2,30,length.out=nrow(asym_matrix)), y=seq(1,2,length.out=ncol(asym_matrix)), asym_matrix, color =

Hello! I have measurements (length and volume) of fish collected in two years. I want to know if the the relationship between length and volume is the same for both years. The number of fish measured is different for each year. I don't know whether lm or aov is more appropriate to use. Here are the two output options: Call: lm(formula = Volume ~ Length * Year) Residuals: Min 1Q

Hi all, though i know this is a simple question, i really hope someone could. I am just trying to compare percents respectively the difference in percents by a simple test. t.test usually compares means and is working like it is supposed to be. Now i wonder how i can use a test in R to test for significant difference between two percentages.

I would like to code records in a dataset with a 1 if any of the columns 9-67 contain a particular code, and zero if they don't. I've been working with "subset" and it seems that something like subset(data, data[9:67]--"12345") would work, but I have been unsuccessful so far. It seems like a simple problem - any help is appreciated!

I am interested in locating a script to implement a sampling scheme that would basically make it more likely that a particular observation is chosen based on a weight associated with the observation. I am trying to select a sample of ~30 census blocks from each ZIP code area based on the proportion of women in a ZCTA living in a particular block. I want to make it more likely that a block will