search for: 0.9998

[this message needed manual improvement by the mailing list administrator since it was `HTMLified' .. ``please do not''] Apologies for bothering you about a fairly trivial matter. I have been getting some inconsistencies with the display digits in R V1.5. I have been using the hypergeometric distribution function, and have found that when printing out the results from this

k=lm(y~x) summary(k) returns R^2=0.9994 lm(y~x) is supposed to find coef. a anb b in y=a*x+b l=lm(y~x+0) summary(l) returns R^2=0.9998 lm(y~x+0) is supposed to find coef. a in y=a*x+b while setting b=0 The question is why do I get better R^2, when it should be otherwise? Im sorry to use the word "MS exel" here, but I verified it in exel and it gives: R^2=0.9994 when y=a*x+b is used

Hi I am trying to read selected fields from a xml file with R using xml package. So far I have learned the basics of this package by going through the manual, examples, tutorial, and so on (www.omegahat.org/RSXML) . The problem is that I am getting stuck when it comes down to more complex xml files. I am a novice in R and xml, and was wondering if someone could help me out with here.

I got a set of data which has seasonal trend in form of sinx, cosx, I don't have any idea on how to deal with it. Can you give me a starting point? Thanks, Terry

Hello, I've encountered a very weird issue with the method subset(), or maybe this is something I don't know about said method that when you're subsetting based on the columns of a data frame you can only use constants (0.1, 2.3, 2.2) instead of variables? Here's a look at my data frame called 'ea.cad.pwr': *>ea.ca.pwr[1:5,] MAF OR POWER 1 0.02 0.01 0.9999 2 0.02

Hi all, I am running R1.5.0 under Unix. I am repeating my earlier question with a few details added. I have the following tree fitted as the tree object 'my.tree': node), split, n, deviance, yval * denotes terminal node 1) root 5807 0.9998 0.0001722 2) V604 < 0.5 5798 0.0000 0.0000000 * 3) V604 > 0.5 9 0.8889 0.1111000 * And I have a data.frame called

Stock returns and other financial data have often found to be heavy-tailed. Even Cauchy distributions (without even a first absolute moment) have been entertained as models. Your qq function subtracts numbers on the scale of a normal (0,1) distribution from the input data. When the input data are scaled so that they are insignificant compared to 1, say, then you get essentially the

I'm attempting to sort a 3 dimensional array that looks like this > x , , 1 [,1] [,2] [1,] 9 9 [2,] 7 9 , , 2 [,1] [,2] [1,] 6 5 [2,] 4 6 , , 3 [,1] [,2] [1,] 2 1 [2,] 3 2 Such that it ends up like this .... > y , , 1 [,1] [,2] [1,] 2 1 [2,] 3 2 , , 2 [,1] [,2] [1,] 6 5 [2,] 4 6 , , 3 [,1] [,2]

Dear R-Users, I have a dataframe (''forexdata'') of daily returns from the foreign exchange market for three currencies - British Pound (bp), Canadian Dollar(cd), Deustche Mark (dm) vis-a-vis the US Dollar and the Date Of Trade(yymmdd). For some dates the returns are missing (recorded as zero) as there were no trades in that currency for that date. My task is to substitute the

Hi, does the shapiro wilk test in R-1.3.0 work correctly? Maybe it does, but can anybody tell me why the following sample doesn't give "W = 1" and "p-value = 1": R> x<-1:9/10;x [1] 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 R> shapiro.test(qnorm(x)) Shapiro-Wilk normality test data: qnorm(x) W = 0.9925, p-value = 0.9986 I can't imagine a sample being