similar to: Problems with ks.test

Hello R-Users, I'm new to R so I apologize in advance for any big mistake I might be doing. I'm trying to fit a set of samples with some probabilistic curve, and I have an important question to ask; in particular I have some data, from which I calculate manually the CDF, and then I import them into R and try to fit: I have the x values (my original samples) and the y values

Hi list, I have a group of data. It looks like they follow a exponential distribution. In R, how can I esimate lamda, that is the rate in pexp, of the distribution and can I use Kolmogorov-Smirnov for hypothesis testing in such a situation? I have read the "8.2 Examing the distribution of a set of data" of "An Introduction to R" but I did not find any clues on this issue.

Does anyone know if R have any problems with the exponential random number generation (function rexp)? I comment it because I executed data<-sort(rexp(100)) plot(data,dexp(data)/(1-pexp(data)),type="l") and the graphic isn't constant. (Note: exponential distribution have a constant hazard failure rate). Thank you, Juan

Hi all, Using the exponential distribution to exemplify: The dexp function is the PDF (1) and pexp is the CDF (2), that is obtained integrating the PDF. How can I get the qexp and the rexp? Considering that I have the PDF, how this two are mathematically related to the PDF? (1) ke^{-kx} (2) 1-e^{kx} Thanks in advance.

Hi, I have a problem with Kolmogorov-Smirnov test fit. I try fit distribution to my data. Actualy I create two test: - # First Kolmogorov-Smirnov Tests fit - # Second Kolmogorov-Smirnov Tests fit see below. This two test return difrent result and i don't know which is properly. Which result is properly? The first test return lower D = 0.0234 and lower p-value = 0.00304. The lower 'D'

Is the kolmogorov-smirnov test valid on both continuous and discrete data? I don't think so, and the example below helped me understand why. A suggestion on testing the discrete data would be appreciated. Thanks, a <- rnorm(1000, 10, 1);a # normal distribution a b <- rnorm(1000, 12, 1.5);b # normal distribution b c <- rnorm(1000, 8, 1);c # normal distribution c d <- rnorm(1000,

I got a distribution function and a empirical distribution function. How do I make to Kolmogorov-Smirnov test in R. Lets call the empirical distribution function >Fn on [0,1] and the distribution function >F on [0,1] ks.test( ) thanks for the help -- View this message in context: http://www.nabble.com/Kolmogorov-Smirnov-test-tp23296096p23296096.html Sent

Dear All I am trying to replicate a numerical application (not computed on R) from an article. Using, ks.test() I computed the exact D value shown in the article but the p-values I obtain are quite different from the one shown in the article. The tests are performed on a sample of 37 values (please see "[0] DATA" below) for truncated Exponential, Pareto and truncated LogNormal

Hi, Interpretation problem ! so what i did is by using the: >fit1 <- fitdist(vectNorm,"beta") Warning messages: 1: In dbeta(x, shape1, shape2, log) : NaNs produced 2: In dbeta(x, shape1, shape2, log) : NaNs produced 3: In dbeta(x, shape1, shape2, log) : NaNs produced 4: In dbeta(x, shape1, shape2, log) : NaNs produced 5: In dbeta(x, shape1, shape2, log) : NaNs produced 6: In

Hi, why don't you try try ks.test(VeriSeti1, VeriSeti2)$p.value All the best Jenny >How can i print only the P-Value of the kolmogorov smirnov test? > > >> ks.test(VeriSeti1, VeriSeti2) > > Two-sample Kolmogorov-Smirnov test > >data: VeriSeti1 and VeriSeti2 >D = 0.5, p-value = 0.4413 >alternative hypothesis: two-sided > > >This expression

Although I saw this issue being discussed many times before, I still did not find the answer to: why does R can not calculate p-values for data with ties (i.e. - sample with two or more values the same)? Can anyone elaborate some details about how does R calculate the p- values for the Kolmogorov Smirnov test statistics? I can understand the theoretical problem that continuous distributions do

Hi r-users,   I would like to use Kolmogorov smirnov test but in my observed data(xobs) there are ties.  I got the warning message.  My question is can I do something about it?   ks.test(xobs, xsyn)           Two-sample Kolmogorov-Smirnov test data:  xobs and xsyn D = 0.0502, p-value = 0.924 alternative hypothesis: two-sided Warning message: In ks.test(xobs, xsyn) : cannot compute correct

dear list I use runif to generate a ramdom number between min and max runif(n, min=0, max=1) however , the syntaxe of rexp does not allow that rexp(n, rate = 1) and it generate a number with the corresponding rate. The question is: how to generate a number between min and max using rexp(). Regards -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000

Hi, I was wondering whether there is an implementation of the Kolmogorov-Smirnov goodness of fit test for discrete, ordinal scale data in R - I've only managed to find the test for continuous data. Thanks! Gila

I am hereby forwarding the data & method use to calculate the Kolmogorov-Smirnov goodness of fit test made manually by me in R launguage which deffers with the actual inbuilt formula as shown below. Further I have plot the graph in R. In that graph how to add trendline (i.e. straight line passing through maximum points in plot) to a Plot. R script is as follows please run this script to see

> x <- runif(100) > y <- runif(100) > ks.test(x,y) Two-sample Kolmogorov-Smirnov test data: x and y D = 0.11, p-value = 0.5806 alternative hypothesis: two-sided ok I expected that, but: > ks.test(runif(100), "runif") One-sample Kolmogorov-Smirnov test data: runif(100) D = 0.9106, p-value < 2.2e-16 alternative hypothesis: two-sided How

Hello, While doing test of normality under R and SAS, in order to prove the efficiency of R to my company, I notice that Anderson Darling, Cramer Van Mises and Shapiro-Wilk tests results are quite the same under the two environnements, but the Kolmogorov-smirnov p-value really is different. Here is what I do: > ks.test(w,pnorm,mean(w),sd(w)) One-sample Kolmogorov-Smirnov test data: w D

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

hello! i don't want to test my sample data for normality, but exponential- lognormal- or gammadistribution. as i've learnt the anderson-darling-test in R is only for normality and i am not supposed to use the kolmogorov-smirnov test of R for parameter estimates from sample data, is that true? can you help me, how to do this anyway! thank you very much! nadja

I'm trying to use modified Kolmogorov-Smirnov test with a Normal which I don't know it's parameters. Somebody told me about the lilifor function in R, but just can't find it. Does anybody know how I can test with the modified Kolmogorov-Smirnov test? Porqu? usar una base de datos relacional cualquiera, si pod?s usar PostgreSQL?