similar to: 95% CI of a IQR and more

Dear R users, Something is strange in summary and IQR. Suppose, I have a data set and I would like to find the Q1, Q2, Q3 and IQR. x<-c(2,4,11,12,13,15,31,31,37,47) > summary(x) Min. 1st Qu. Median Mean 3rd Qu. Max. 2.00 11.25 14.00 20.30 31.00 47.00 > IQR(x) [1] 19.75 However, I test the same data set in SAS "proc univariate", and SAS shows that

This report follows the post http://tolstoy.newcastle.edu.au/R/e6/devel/09/03/0760.html where it is shown that quantile() and IQR() do not work as documented. In fact they do not check for numeric input even if the documentation says = : ?quantile x numeric vectors whose sample quantiles are wanted. Missing values are ignored. ?IQR x a numeric vector. > quantile(factor(1:9)) 0%

Dear all, from the help page of quantile: "x ??? numeric vectors whose sample quantiles are wanted. Missing values are ignored." from the help page of IQR: "x ??? a numeric vector." as a matter of facts it seems that both quantile() and IQR() do not check for the presence of a numeric input. See the following: set.seed(11) x <- rbinom(n=11,size=2,prob=.5) x <-

Hi all, I'm just getting started with R and I would appreciate some help. I'm having trouble creating a boxplot with whiskers at the 95th and 5th percentiles instead of at 1.5 * IQR. I have read the relevant documentation, and checked existing mails on this topic. I found a small modification that should work : https://stat.ethz.ch/pipermail/r-help/2001-November/016817.html and tried to

Dear R users & experts, I'd like to create a model using lm (or glm) under some constraints of how coefficients for each component could look like (sort of a range of coefficients that should be allowed). So let's go for an example : model=lm(age ~ eyecolor + height, data=inputdata) So let's suppose that R pops out a model with positive estimates for the coefficients eyecolor

Hi, I am trying to calculate confidence intervals using ci.numeric from epicalc package. If I generate a normal set of data and find the 99% and 95% CI, they seem too narrow to me. Am I doing something wrong?? The IQR goes from -0.62 to 0.62, so I thought the CI limits should be more extreme than these values. x<- rnorm(200,0,1) ci.numeric(x=mean(x),n=200,sds=sd(x),alpha=0.05) n

Hi i am doing an environmental research The equation is as follow: gam(y1 ~ x1 + s(x2) + s(x3) + s(x4), family = gaussian, fit = true) I would like to obtain the beta coefficient and 95CI of x4 (or s(x4)), what should I do? Thanks, Lung -- View this message in context: http://r.789695.n4.nabble.com/mgcv-beta-coefficient-and-95-CI-tp3320491p3320491.html Sent from the R help mailing list

Dear List, I'd like to make boxplots of a large number of observations (+/- 20.000), which are distributed log-normal and right skewed. The problem is that with standard boxplots a too large number of observations are displayed as outliers. I also tried to display the log of the observations, but even then there are to may outliers to my taste. So I'd like to change the standard IQR box

Dear all, I constructed this function called my.boxplot.stats by replacing fivnum() with quantile() in function boxplot.stats(). So I can try different quantile methods in bwplot(). The problem is I couldn't pass different values to the "type" argument to my.boxplot.stats, which in turn is an argument in bwplot(). Now I just have to manually change the "type" value in

Are you aware that there *already is* a function that does this? ?IQR (also your "function" iqr" is just a character string and would have to be parsed and evaluated to become a function. But this is a TERRIBLE way to do things in R as it completely circumvents R's central functional programming paradigm). Cheers, Bert Bert Gunter "The trouble with having an open mind

I noticed that boxplot computes a 95% CI for the median by using median +/- 1.58*IQR./sqrt(n) Where does the 1.58 constant come from? -- Rick White Statistical Consultant U.B.C. Vancouver B.C. Canada rick at stat.ubc.ca -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send

Dear list, I'm trying to implement the following function, but what I get is an error message and I don't understand where is the error: #outliers'identification: iqr=lapply(bb,function(){ inner_fencesl=quantile(x,0.25)-1.5*IQR(x) inner_fencesh=quantile(x,0.75)+1.5*IQR(x) outer_fencesl=quantile(x,0.25)-3*IQR(x) outer_fencesh=quantile(x,0.75)+3*IQR(x)}) where bb is a dataframe

??? IQR returns a single number. > IQR(rnorm(10)) [1] 1.090168 To your 2nd response: "I could have used average, min, max, they all would have returned the same thing., " I can only respond: huh?? Are all your values identical? You really need to provide a small reproducible example as requested by the posting guide -- I certainly don't get it, and I'm done guessing.

Dear R helpers, I have a problem with exporting a chart (to any format). The graphic device becomes inactive and I get the 'Error: invalid graphics state' error message. I searched the help, web and FAQ but couldn't find the solution. This is my code: I chart a histogram for differences in R2 by sample size (an extract from the data is below). Altogether I have n=2500 observations

The following is from version 1.6.2 of R under Windows, or 1.6.1 under Mac OSX/X11 > density(1) Error in if (!(lo <- min(hi, IQR(x)/1.34))) (lo <- hi) || (lo <- abs(x[1])) || : missing value where logical needed I am not sure how this should be handled. I encountered it in connection with densityplot(). In that connection, it might be enough to modify density() so that it

My box plot below is drawing its upper whisker all the way to the last point, instead of showing the point as an outlier. Am I misunderstanding, or is it a bug? Help(boxplot) states for the parameter ?range? that ?this determines how far the plot whiskers extend out from the box. If range is positive, the whiskers extend to the most extreme data point which is no more than range times the

Again, IQR returns two both a .25 and a .75 value and it failed, which is why I didn't use it before. Also, the first function just returns tha same value repeating. Since they are the same, before the second call, using the mode function is just a way to grab one value. I could have used average, min, max, they all would have returned the same thing. Mike On Tue, Apr 19, 2016 at 7:24 PM,

Hello R-Users, The following questions are not R-technical, but more of general statistical nature. 1. NORMALITY I built a normal linear regression model and now I want to check for the residual normality assumption. If I check the distribution graphically and look at the descriptive characteristics (skewness and kurtosis are below 1), I would confirm that the residuals are normally

> That didn't work Jim! It always helps to say how the suggestion did not work. Jim's function had a typo in it - was that the problem? Or did you not change the call to ddply to use that function. Here is something that might "work" for you: library(plyr) data <- data.frame(groupColumn=rep(1:5,1:5), col1=2^(0:14)) myIqr <- function(x) {

That didn't work Jim! Thanks anyway On Mon, Apr 18, 2016 at 9:02 PM, Jim Lemon <drjimlemon at gmail.com> wrote: > Hi Michael, > At a guess, try this: > > iqr<-function(x) { > return(paste(round(quantile(x,0.25),0),round(quantile(x,0.75),0),sep="-") > } > > .col3_Range=iqr(datat$tenure) > > Jim > > > > On Tue, Apr 19, 2016 at