cannot compute correct p-values with ties in: ks.test(x, pgev,
fit$mle[1], fit$mle[2], fit$mle[3])
You may want to use the ks.boot function in the Matching package which
implements a bootstrap ks-test which provides consistent pvalues
(achieved significance levels) when there are ties.
Hi,
I am trying the following:
library(ismev)
library(evd)
fit - gev.fit(x,show=FALSE)
ks.test(x,pgev,fit$mle[1],fit$mle[2],fit$mle[3])
but I am getting:
Warning message:
cannot compute correct p-values with ties in: ks.test(x, pgev,
fit$mle[1], fit$mle[2], fit$mle[3])
where x is:
[1] 239
Benjamin Dickgiesser wrote:
Hi,
I am trying the following:
library(ismev)
library(evd)
fit - gev.fit(x,show=FALSE)
ks.test(x,pgev,fit$mle[1],fit$mle[2],fit$mle[3])
1. The test *is* working. It simply warns (and does not report an error)
that you have ties in your data and the
Prof Brian Ripley schrieb:
On Sat, 16 Dec 2006, R. Villegas wrote:
2006/12/15, Carmen Meier [EMAIL PROTECTED]:
Hello r-group
I have a question to the ks.test.
I would expect different values for less and greater between data1 and
data2.
Does anybody could explain my point of
R. Villegas schrieb:
2
data1-c(8,12,43,70)
data2- c(70,43,12,8)
is the same for ks.test, isn't it?
Yes, it's the same. Wich version of R have you?.
2.4.0
Carmen
__
R-help@stat.math.ethz.ch mailing list
On Sat, 16 Dec 2006, R. Villegas wrote:
2006/12/15, Carmen Meier [EMAIL PROTECTED]:
Hello r-group
I have a question to the ks.test.
I would expect different values for less and greater between data1 and
data2.
Does anybody could explain my point of misunderstanding the function?
The help
Hello r-group
I have a question to the ks.test.
I would expect different values for less and greater between data1 and
data2.
Does anybody could explain my point of misunderstanding the function?
data1-c(8,12,43,70)
data2- c(70,43,12,8)
ks.test(data1,pnorm)
ks.test(data1,pnorm,alternative
2006/12/15, Carmen Meier [EMAIL PROTECTED]:
Hello r-group
I have a question to the ks.test.
I would expect different values for less and greater between data1 and
data2.
Does anybody could explain my point of misunderstanding the function?
data1-c(8,12,43,70)
data2- c(70,43,12,8)
R enthusiasts!
I have been simulating daily in-stream bacteria concentrations using a variety
of scenarios. I am using the ks.test (two sample,two-sided) for analysis. My
data sets are both of equal size (n=64).
My question is this: For the two sample, two sided ks.test, how is the exact
Dear all,
One can use ks.test(x,y) for a one-sample kolmogorov-smirnov test:
x being the data sample
y being a string specifying a distribution
I notice the help on ks.test does not tell you how to get such a list. Is
this a hole in my R knowledge?
Where can I get a list of the strings
On 5/9/06, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote:
Dear all,
One can use ks.test(x,y) for a one-sample kolmogorov-smirnov test:
x being the data sample
y being a string specifying a distribution
I notice the help on ks.test does not tell you how to get such a list. Is
this a hole in
For the ones in the stats package:
help.search(distribution$, package = stats)
seems to get the right set.
On 5/9/06, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote:
Dear all,
One can use ks.test(x,y) for a one-sample kolmogorov-smirnov test:
x being the data sample
y being a string
Dear Gabor
Thanks very much. I knew about help(), but didn't know there was a
help.search().
I have found what I was looking for by using it.
Desmond Campbell
Room C0.29, SGDP Centre
Institute of Psychiatry, De Crespigny Park, London SE5 8AF
Tel: +44 (0) 20 7848 0236
Fax: +44 (0) 20 7848
Bates [EMAIL PROTECTED]
Sent by: [EMAIL PROTECTED]
Date: 09.05.2006 07:23PM
cc: r-help@stat.math.ethz.ch
Subject: Re: [R] ks.test one-sample - where can I get a list of the
strings specifying the distribution?
On 5/9/06, [EMAIL PROTECTED] [EMAIL PROTECTED]
wrote
Hi
I would recommend graphical methods to compare two samples from
possible different distributions. See ?qqplot
Since the Kolmogorov-Smirnov test has in many cases very small
power, you can not conclude that two sample come from the same
distribution only because the ks.test is not significant.
I'm using ks.test() to compare two different
measurement methods. I don't really know how to
interpret the output in the absence of critical value
table of the D statistic. I guess I could use the
p-value when available. But I also get the message
cannot compute correct p-values with ties ... does
Dear experts,
Is it possible to use ks.test function to check the goodness of fit of the
conditional distribution Y|X=x?
For example, I would like to check that my data (Y,X) come from Norm(0.5+x,1)
using KS.
Thank you in advance,
Victoria Landsman.
[[alternative HTML version
Landsman
Sent: Thursday, April 07, 2005 3:16 PM
To: R-help list
Subject: [R] ks.test for conditional distribution Y|x
Dear experts,
Is it possible to use ks.test function to check the goodness of fit of the
conditional distribution Y|X=x?
For example, I would like to check that my data (Y,X) come
Hi Angela,
I believe you should introduce only df as parameters;
t distribution as by default mean=0; see this example.
x-rt(100,10)
ks.test(x, pt,10)
One-sample Kolmogorov-Smirnov test
data: x
D = 0.1414, p-value = 0.03671
alternative hypothesis: two.sided
Ciao
Vito
you wrote:
On 28 Aug 2003 at 8:06, Roger Koenker wrote:
But is it worth it to modify Kolmogorov-Smirnof fot estimated
parameters? It has very low power anyhow. If the null hypothesis is
exponential distributio (which is a scale family), what about using
the quantile transformation twice
new -
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
You appear to be applying the KS test after estimating parameters. The
distribution theory is for an iid sample from a known continuous
distribution (and does not otherwise depend on the distribution). Since
your H_0 is not pre-specified, that distribution theory is not correct.
(Some
On Thu, 28 Aug 2003, Prof Brian Ripley wrote:
You appear to be applying the KS test after estimating parameters. The
distribution theory is for an iid sample from a known continuous
distribution (and does not otherwise depend on the distribution). Since
your H_0 is not pre-specified, that
With the Shifted Exponential test, H_0 is data is a sample coming from a
Shifted Exponential distribution with shift=30 and lambda = 0.001566907
You appear to be applying the KS test after estimating parameters.
I do in order to define H_0 as explained above
The distribution theory is for an
On Thu, 28 Aug 2003, franck allaire wrote:
With the Shifted Exponential test, H_0 is data is a sample coming from a
Shifted Exponential distribution with shift=30 and lambda = 0.001566907
You got that after looking at the data. H_0 has to be specified before
looking at the data.
You appear
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