search for: 75th

Hi List, Just wondering where the documentation exists for the statistics which makeup the bwplot. I'm guessing that if R is like similar products that the graph is constructed as The median is the filled circle. The box surrounding the filled circle depicts the 25th and 75th quartile. The range of values is given by the dotted lines (?whiskers?) outside of each box, and possible outliers are given by the open circles outside the box. Is this close? Thanks Kenneth E. Nussear Phone 775 784-1703 Ecology, Evolution and FAX 775 78...

How do I calculate the 95th percentile when I know the 25th, the median and the 75th?? Thanks, Harmony Tenney [[alternative HTML version deleted]]

...179974,3.599118565 102,2.716837723,3.316180099 I can regress log weight on log length simply enough, but how would I model the third quartile of log weights? In other words, rather than finding a 2nd quartile (or 50th percentile) regression line, e.g., mod=lm(logweight~loglength) can R find a 75th percentile line? Further, since my data is lake>1, is there a way to run 3rd quartile regressions on each lake? I would imagine that regressing each population would require some call of the subset function, but I cannot figure out how to call it. Thanks in advance, SR Steven H. Ranney Gr...

...ated summary tables that are being used to generate the charts? Realize I could use tapply or such to get something similar, but I have all the groupings and such already configured to generate the charts. Simply want to dump those values to a table so that I don't have to interpolate where the 75th percentile is on a visual chart. Appreciate any thoughts.. -- View this message in context: http://n4.nabble.com/Print-lattice-output-to-table-tp1375040p1375040.html Sent from the R help mailing list archive at Nabble.com.

...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 to a box which displays the 5th tothe 95th percentile instead of the standard 25th to 75th percentile. Is this possible and if so, how? I use both R and Splus, so I'd like to know for both flavours of S. Regards, Cornelis

Hi. I'm trying to create a 3d plot for a teaching example of finding a least-squares estimate of the parameters to fit a line to some data. I was hoping to get a nice plot with a clear, single minima where the derivative of the surface is zero. No matter how much I tinker, I can't seem to get a simple straightforward plot. Am I doing something wrong? Thanks in anticipation, Ross-c x

...(how many events per week over the past year for each of 4 values of some factor), some less straightforward. I've managed to do this by brute force, but I'd like to learn how to do it in more elegant, more R-like code. Consider for example the following, which graphs the 25th, 50th, and 75th percentile values per day of data$x perc <- function(code,data) { # select the part of the data with factor value slice <- data[data$factor == code,]; # calc quartiles for each day quarts <- tapply(slice$x, slice$day, functio...

I have 2 datafiles 'target' and 'observed' as shown below (I will gladly email these 2 small files to whomever). X25. And X75. Indicate the value of 25th and 75th-percentile of the target ('what should be') and the observed ('what is'). The i.value is simply the month. > target X i.value X25. X75. 1 one.month 1 10.845225 17.87237 2 one.month 2 12.235813 19.74490 3 one.month 3 14.611749 23.44810 4 on...

...) ) Now is where the problem starts. >From nonconfusing_regression , I get, say 2 added hot dogs per week for each centimeter of waistline along with a s.e. of 0.5 hot dogs per week, which I multiply by 1.96 to garner each side of the 95% c.i. If I want to show what the difference between the 75th percentile (say 100 cm) and 25th percentile (say 80 cm) waistlines are, I multiply 2 by 100-80=20 and get 40 hot dogs per week as the point estimate with a similar bumping of the s.e. to 10 hot dogs per week. What do I do to get the point estimate and 95% confidence interval for the difference bet...

Hello R-Help, Could anybody tell me how the boxplot-function calculates the upper and lower extremes of the notch contained in $conf which I assume is the confidence interval? Is it reliable for data which is not normally distributed? If not, how can I calculate and boxplot a specific confidence interval for not normally distributed data in R (increasing the sample size does not normalize the

Hi All: I am working with a dataset on Arsenic toxicity, and I am trying to calculate the 20th, 40th, 60th, 80th, and highest percentiles for a variable, dietary Moisture (variable name dMoist). The inbuilt function quantile(dMoist) would print 0, 25th, 50th, 75th, and 100th percentile. Does there exist a function that can calculate xth percentile (where x = 10th, 20th, ... etc) values? I looked for such functions in the documentation, but couldn't find one. Has anyone written a similar function that can be used? Is there a module or a function tha...

...help in calculating percentiles Hi All: I am working with a dataset on Arsenic toxicity, and I am trying to calculate the 20th, 40th, 60th, 80th, and highest percentiles for a variable, dietary Moisture (variable name dMoist). The inbuilt function quantile(dMoist) would print 0, 25th, 50th, 75th, and 100th percentile. Does there exist a function that can calculate xth percentile (where x = 10th, 20th, ... etc) values? I looked for such functions in the documentation, but couldn't find one. Has anyone written a similar function that can be used? Is there a module or a function tha...

Hi guys! I have to calculate the 95% CI for the IQR. I have 100samples with continuous numbers using a compute-intensive method. 1)I can calculate the 95%CI and the IQR, but I don't know how to calculate the CI of the IQR, does anybody have any idea, which method will be appropriate? 2)I have a 2x2 table with gender vs blue eyes or not. How do I calculate the expected two-by-two contingency

In the following case, library(splines) tt <- c(55, 251, 380, 289, 210, 385, 361, 669) nn <- rep(0:7,tt) ns(nn,4) ## knots are located at (0.25,0.5,0.75); quantiles = (2,5,7) we get the error Error in qr.default(t(const)) : NA/NaN/Inf in foreign function call (arg 1) because the 75th quantile (the location of the last "interior" knot) ends up on the boundary. As a result, the call to spline.des(...)$design within the bs() function call returns a design matrix that includes NaN values, which triggers the error when qr() is called. (This alternative works OK: ns(nn,4,...

Hi all, One quick question: How to put values of 25 and 75 percentile on boxplot? Thanks in advance. Nianqing Xiao, Ph.D NCI Center for Bioinformatics, NIH SAIC/Advanced Systems Group > 6116 EXECUTIVE BLVD 4026J > MSC 8335 > BETHESDA MD 20852 Phone: 301-451-6357 Fax: 301-480-4222 -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list --

Dear all, I am trying to add error bars on a boxplot but have encountered an error as indicated below. Is there a package I need to install or a library I have to load before this goes please. Thanks for any idea. Ogbos x<-replicate(20,rnorm(50)) boxplot(x,notch=TRUE,main="Notched boxplot with error bars") error.bars(x,add=TRUE) Error: could not find function "error.bars"

Hi, I'm using the rms package to do regression analysis using the lrm function. Retrieving odds ratios is possible using summary.rms. However, I could not find any information on how exactly the odds ratios for continuous variables are calculated. It doesn't appear to be the odds ratio at 1 unit increase, because the output of summary.rms did not match the coefficient's value. E.g.

Hello everybody, I need to know which samples (S1-S6) contain a value that is bigger than the median + five standard deviations of the column he is in. This is just an example. Command should be applied to a data frame wich is a lot bigger (over 100 columns). Any solutions? Thank you very much for your help!!! > s Samples A B C E 1 S1 1 2 3 7 2

Dear R-Helpers, I am a novice in survival analysis. I have the following code: for (i in 3:12) print(coxph(Surv(time, status)~a[,i], data=a)) I used it to fit the Cox Proportional Hazard models separately for every available parameter (columns 3:12) in my data set - with intention to compare the Hazard Ratios. However, some of my variables are in range 0.1 to 1.6, others in range 5000 to

Hi, i?m not part of the FLAC project, but i have a question regarding FLAC and Gapless support I hope, I get an answer from some of you ;-) We are currently try to add Gapless support on our device If we rip an CD with our device, we can find out, that one track follow after another so we can recognize, that the tracks are gapless or not. But how can we find that out on already existing