В Wed, 15 Feb 2023 14:20:16 +
Nick Wray пишет:
> how to get the p values for changepoints in the "changepoint"
> package?
The documentation of this package is not always precise in describing
what the code actually does. It's possible to find out by reading the
source code that with default
Hello,
It seems to be right.
You have Chi-squared = 23.724, df = 8, p-value = 0.002549. So try the R
function ?pchisq:
pchisq(23.724, df = 8, lower = FALSE)
[1] 0.002549054
Hope this helps,
Rui Barradas
Em 23-05-2017 13:08, Dhivya Narayanasamy escreveu:
Hi,
I am working with bivariate tim
Hi Peter,
That was my first port of call before I posted this thread. Unfortunately,
it does not seem to explicitly state which test is used or how the p-value
is calculated.
Thanks,
Rob.
--
View this message in context:
http://r.789695.n4.nabble.com/P-value-from-Matching-tp4703309p4703345.h
On 16 Feb 2015, at 00:31 , Rob Wood wrote:
> Hi all,
>
> When using the match command from the matching package, the output reports
> the treatment effect, standard error, t-statistic and a p-value. Which test
> is used to generate this p-value, or how us it generated?
>
I would assume this
Thanks Michael.
I have read the original article and I guess whether the following formula
could resolve my question?
*for adjusted HR of each BMI category,*
fit<-coxph(Surv~factor(BMI-category)+..+covariates)
*for trend, *
fit<-coxph(Surv~as.numeric(BMI-category)+..+covariates)
*# p trend may be
On 1/4/2014 12:39 AM, zhu yao wrote:
Dear Sir
Many papers calculated the p value of trends for odds ratios of ordered
category variables. I have found the tabodds command in Stata. But how to
do it in R?
Thanks
*Yao Zhu*
If what you are looking at is a 2 x 2 x k table, where you want the odds
Thanks peter.
*Yao Zhu*
*Department of UrologyFudan University Shanghai Cancer CenterShanghai,
China*
2014/1/5 peter dalgaard
>
> On 04 Jan 2014, at 13:56 , zhu yao wrote:
>
> > Thanks for the suggestion.
> > The results is presented in following table. The authors calculated p
> value for
On 04 Jan 2014, at 13:56 , zhu yao wrote:
> Thanks for the suggestion.
> The results is presented in following table. The authors calculated p value
> for linearity and trend and they stated in the methods:
> " Linear and nonlinear trends of BMI associated with each mortality outcome
> were ob
On 04 Jan 2014, at 12:53 , Uwe Ligges wrote:
>
>
> On 04.01.2014 06:39, zhu yao wrote:
>> Dear Sir
>> Many papers calculated the p value of trends for odds ratios of ordered
>> category variables. I have found the tabodds command in Stata. But how to
>> do it in R?
>
> Depends on the method y
On 04.01.2014 06:39, zhu yao wrote:
Dear Sir
Many papers calculated the p value of trends for odds ratios of ordered
category variables. I have found the tabodds command in Stata. But how to
do it in R?
Depends on the method you want to use ... and most ladies and gents on
this list won't kn
Rosario Garcia Gil slu.se> writes:
>
> Hello
>
> I have run an anova analysis for
> the fallowing model: H_obs=mu+REGION+MANAGEMENT + e
>
> When I run it in ASRelm I get the p-value for mu, and,
> of course also for the two dependent variables (REGION
> and MANAGEMENT)
>
> When I run it in R
On 09/06/2012 05:00 AM, r-help-requ...@r-project.org wrote:
Hi, R experts
I am currently using lmekin() function in coxme package to fit a
mixed effect model for family based genetic data. How can I extract the p
value from a lmekin object? When I print the object in R console, I can
s
Dear All,
Thanks for all your explanations and sorry for the confusion. I will
need to consult some statistician for help.
Sincerely,
Li Sun
2012/6/24 S Ellison :
>
>
>>> But what if x is exact while y has some uncertainty Δy, in the
>>> relation y = k * x + b?
>>
>>Again, no: this is not a li
>> But what if x is exact while y has some uncertainty Δy, in the
>> relation y = k * x + b?
>
>Again, no: this is not a linear model. Assumption in a linear model is
>that the errors are identically distributed.
Surely not; errors in a linear model do not need to be homoscedastic. lm
handles h
Dear Li Sun:
You appear to have a good deal of statistical confusion. R-help is not a
statistical consulting service. You should consult a local statistician or,
if that's not possible,post on a statistical help list like
stats.stackexchange.com
-- Bert
On Sun, Jun 24, 2012 at 12:04 PM, Li SUN
2012/6/24 Uwe Ligges :
>
>
> On 24.06.2012 20:35, Li SUN wrote:
>>
>> Thanks David and Brian.
>>
>> But what if x is exact while y has some uncertainty Δy, in the
>> relation y = k * x + b?
>>
>> Now I need to fit some data like
>> x = 1, 2, 3, 4, 5
>> y±Δy
On 24.06.2012 20:35, Li SUN wrote:
Thanks David and Brian.
But what if x is exact while y has some uncertainty Δy, in the
relation y = k * x + b?
Now I need to fit some data like
x = 1, 2, 3, 4, 5
y±Δy = 1.1±0.1, 2.0±0.2, 3.1±0.2, 4.1±0.1, 5.0±0.2
Is
Thanks David and Brian.
But what if x is exact while y has some uncertainty Δy, in the
relation y = k * x + b?
Now I need to fit some data like
x = 1, 2, 3, 4, 5
y±Δy = 1.1±0.1, 2.0±0.2, 3.1±0.2, 4.1±0.1, 5.0±0.2
Is there any mechanism to pass x, y and Δ
On 24/06/2012 18:39, David Winsemius wrote:
On Jun 24, 2012, at 1:21 PM, Li SUN wrote:
Sorry for the confusion.
Let me state the question again. I missed something in my original
statement.
When using the linear model lm() to fit data of the form y = k * x +
b, where k, b are the coefficient
On Jun 24, 2012, at 1:21 PM, Li SUN wrote:
Sorry for the confusion.
Let me state the question again. I missed something in my original
statement.
When using the linear model lm() to fit data of the form y = k * x +
b, where k, b are the coefficients to be found, and x is the variable
and h
Sorry for the confusion.
Let me state the question again. I missed something in my original statement.
When using the linear model lm() to fit data of the form y = k * x +
b, where k, b are the coefficients to be found, and x is the variable
and has an error bar (uncertainty) Δx of the same lengt
On 24.06.2012 17:47, Li SUN wrote:
Hi All,
when using the linear model lm() to fit data of the form y = k * x +
b, is it possible to know the p-value for the parameters k and b? i.e.
can we find the result of the form (k, Δk; b, Δb)?
If you explain what Δk means in a linear model, we may be
On Wed, Mar 14, 2012 at 3:41 AM, cheba meier wrote:
> Hello,
>
> I have to compute the pooled z-value and I would like to know which way is
> more appropriate
>
>
> b <- c( -0.205,1.040,0.087)
> s <- c(0.449,0.167,0.241)
> n <- c(310, 342, 348)
> z <- b/s
>
> Z <- sum(z)/sqrt(length(n))
> P <- 2*(
Dear Peter,
Many thanks for the reply.
Sure, this is just an example and it makes zero sense!
Regards,
Dunia
2012/3/4 peter dalgaard
>
> On Mar 4, 2012, at 12:21 , Dunia Scheid wrote:
>
> > Dear all,
> >
> > I am fitting a GLM similar to
> >
> > library(MASS)
> > anorex.1 <- glm(Treat~Postwt+
On Mar 4, 2012, at 12:21 , Dunia Scheid wrote:
> Dear all,
>
> I am fitting a GLM similar to
>
> library(MASS)
> anorex.1 <- glm(Treat~Postwt+Prewt,family = binomial, data = anorexia)
I hope that's just for an example. The actual analysis makes zero sense to me...
>
> I have found two ways o
ame p-value?
>
> No, it is a nonlinear transformation and the p-values are not the same.
>
>>
>> I would like to apologize if these were trivial questions and I fear they
>> are not totally R-specific. Thus I understand if you don't have the time to
>> answ
f the hazard ratio in R (without hacking something
together like I did) seems reasonable to me at least to ask.
Cheers,
Josh
>
> Best,
> Thierry
>
>
>
> - Ursprüngliche Mail -
> Von: "Joshua Wiley"
> An: "Thierry Julian Panje"
> CC: r-he
Hi Thierry,
Could you give us an example of what exactly you are doing
(preferablly reproducible R code)? I may be misunderstanding you, but
if you are fitting cox proportional hazard models using the coxph()
function from the surival package, summary(yourmodel) should give the
SE, p-value based
Thanks for your help. I was not familiar with this form of notating the p
value.
--
View this message in context:
http://r.789695.n4.nabble.com/p-value-in-R-beginners-question-tp3915873p3918140.html
Sent from the R help mailing list archive at Nabble.com.
The p-value as I understand it, is the probability of seeing a phenomenon at
least as extreme as the one which you observed assuming the null hypothesis
is true. In other words, if you assume there is no difference (which is the
NULL for you here I believe), and there is an 81% chance of seeing dat
This is just scientific notation, so
8.15e-01 is the same as:
> 8.15*10^-1
[1] 0.815
niki wrote:
>
> Dear all,
>
> i have done some regression analyses but i do not understand the p value.
> These are the results
>
>t-value p value
> geno.1
Tatiana:
It sounds like you are in way over your head statistically. This is
not a statistics tutorial site (though sometimes good folks do help
witjh this). I suggest you try http://stats.stackexchange.com/ .
Better yet, work with your local statistician.
Cheers,
Bert
On Thu, Sep 15, 2011 at 6
On 11-02-07 19:21, Sascha Vieweg wrote:
Hello, knowing that some index y, with y(341)=2, SE=3, is t-distributed, I
(think I) can find an appropriate (left/one-sided) p-value with
R: y <- 2
R: yse <- 3
R: (p <- 1-pt(y/yse, df=341))
Now, some simulation resulted in the non-parametric distrib
Allright.. Appreciate the input on non-zero terminology (:-). What I wanted was:
rr<-data.frame(r1=rnorm(1000,10,5),r2=rnorm(1000,220,5))
with(rr,plot(r1,r2))
r3<-kde2d(r1,r2,lims=c(2,18,200,240))
filled.contour(r3)
//M
On 1. feb. 2011, at 21.26, David Winsemius wrote:
>
> On Feb 1, 2011, a
On Tue, Feb 01, 2011 at 08:31:57PM +0100, moleps wrote:
>
> My terminology is probably way off. I´ll try again in plain english.
>
> I´d like to generate a scatter plot of r1 & r2 and color code each pair
> according to the probability of observing the pair given that the two samples
> (r1 &
On Feb 1, 2011, at 2:31 PM, moleps wrote:
My terminology is probably way off. I´ll try again in plain english.
I´d like to generate a scatter plot of r1 & r2 and color code each
pair according to the probability of observing the pair given that
the two samples (r1 & r2) are drawn from tw
> Date: Tue, 1 Feb 2011 12:10:31 -0800
> From: nord...@dshs.wa.gov
> To: r-help@r-project.org
> Subject: Re: [R] p value for joint probability
>
> > -Original Message-
> > From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> > project.
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of moleps
> Sent: Tuesday, February 01, 2011 11:32 AM
> To: Peter Ehlers
> Cc: r-help@r-project.org
> Subject: Re: [R] p value for joint probability
>
My terminology is probably way off. I´ll try again in plain english.
I´d like to generate a scatter plot of r1 & r2 and color code each pair
according to the probability of observing the pair given that the two samples
(r1 & r2) are drawn from two independent normal distributions.
rr<-data.f
On Mon, Jan 31, 2011 at 09:42:27PM +0100, moleps wrote:
> Dear all,
>
> Given
>
> rr<-data.frame(r1<-rnorm(1000,10,5),r2<-rnorm(1000,220,5))
>
Hello.
There is already an answer to your question. However, i think
that the above command works in a different way than you
expect. The embedded assi
On 2011-01-31 12:42, moleps wrote:
Dear all,
Given
rr<-data.frame(r1<-rnorm(1000,10,5),r2<-rnorm(1000,220,5))
How can I add a column (rr$p) for the joint probability of each r1& r2 pair?
If you take the values in each pair to be observations
from two independent Normal distributions, it's
What hypothesis do you expect the p-values to be testing?
What you would get from regsubsets is unlikely to test any hypothesis of
interest.
If you really feel the need for p-values then the safest approach is probably
the function SnowsCorrectlySizedButOtherwiseUselessTestOfAnything in the
Is this the kind of thing you mean?
> wData <- within(wData, {
+ z <- (weigth - mean(weigth))/sd(weigth)
+ "p-value" <- 2*pnorm(-abs(z)) ## 2-sided
+ rm(z) })
> wData
employee_id weigth p-value
1 100150 0.3763641
2 101200 0.9081403
3 102300 0.1547139
4
What is your null hypothesis? What is your alternate hypothesis? What
is the test statistic? Why do you want a p-value?
Hadley
On Thu, Jul 22, 2010 at 5:40 PM, jd6688 wrote:
>
> Here is my dataframe with 1000 rows:
>
> employee_id weigth p-value
>
> 100 150
> 1
o: 'Soham'; r-help@r-project.org
> Subject: Re: [R] p value
>
> runif(1)
>
>
> Bert Gunter
> Genentech Nonclinical Statistics
>
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On
> Beh
runif(1)
Bert Gunter
Genentech Nonclinical Statistics
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Soham
Sent: Saturday, May 15, 2010 9:05 AM
To: r-help@r-project.org
Subject: [R] p value
How to compute the p-value of a stati
Hi Soham,
I don't feel your question is well defined.
But an equally ill defined answer would be:
Through a permutation test.
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Heb
I don't think that the p-value concept is as well defined for multivariate
distributions. Do you want the area under the curve corresponding to (x < t.x
& y < t.y) or (x < t.x | y < t.y) or ( t.x + t.y < C ) or all the area where
the height of the density is less than at t.x,t.y? or possibly o
Best wishes,
> Amor
>
>
> --- Peter Dalgaard schrieb am Fr, 13.11.2009:
>
> Von: Peter Dalgaard
> Betreff: Re: [R] p-value > 1
> An: "amor Gandhi"
> CC: r-h...@stat.math.ethz.ch
> Datum: Freitag, 13. November 2009, 15:24
>
> amor Gandhi wrote:
&g
the p-value of 0.004332?
Sorry for these questions.
Best wishes,
Amor
--- Peter Dalgaard schrieb am Fr, 13.11.2009:
Von: Peter Dalgaard
Betreff: Re: [R] p-value > 1
An: "amor Gandhi"
CC: r-h...@stat.math.ethz.ch
Datum: Freitag, 13. November 2009, 15:24
amor Gandhi wrot
amor Gandhi wrote:
Dear all,
I am trying to use SAMr-library(samr), it gives me p-value = 1.001, any idea
why?
Well, the package maintainer might be a better place to ask. (Bring an
example if you do.)
However, as a generic matter, the result is obviously nonsensical, and
could be due to
Don't throw out the baby with the bath water just yet. Note that even though
your first model is insignificant, the R-squared is very high. This is
because you fit the whole model with intercept and three coefficients on 1
degree of freedom. You need to first import the data, then run the model,
an
Lucas
This problem is very old --- older than keypunches. There are several
methods for selecting variables (forward, backwards, both, all subsets)
using a variety of criteria (p-values, R^2, adjusted R^2, Cp, AIC, BIC,
and more). Be sure you understand the methods, especially the tendency to
Hi Sascha,
Take a look at ?pf
HTH,
Jorge
On Wed, Sep 23, 2009 at 11:34 AM, Sascha Wolfer <> wrote:
> Dear List,
>
> is there an easy and fast way to compute the p value from a given F value
> and given degrees of freedom for an effect and the dfs for the residuals? I
> think of a function like
On Thu, 30 Oct 2008, GSt wrote:
Dear all,
I have a question concerning the p-value. When running coxph I get a p-value
= 0. :confused:
Can this be true? Why aren??t there decimal points? Is there a way to
find out the exact p-value?
The p-values are rounded, and of course they are not exac
Because your test statistics (z=-9.16) is too small. (well, z> 1.96 or
Z < -1.96 is p=0.05)
If you really want the exact p, you can try
pnorm(-9.16)
if I were you I will quote p<0.001 as well as the test statistics. (if
you are going to submit it to medical journals)
C
On Thu, Oct 30, 2008 at 1
On Fri, 2008-07-04 at 21:14 -0700, Michael Denslow wrote:
> Dear R-helpers,
>
> I am running metaMDS in the vegan package, which uses isoMDS in MASS,
> to perform Nonmetric Multidimentional Scaling (NMDS).
>
> I have seen some authors report a p-value for the NMDS ordination
> based on randomizat
You need to ask yourself a number of questions, e.g.
What is the hypothesis you wish to test?
What is the test statistic you wish to use to test it?
How can I get some information on where my value of that statistic sits
with respect to its null hypothesis distribution?
p-values do not exist inde
Not sure if this is what you are looking for but you can get
the p-value with something like this:
# Create a vector
mydata<-
c(132968364, 135945080, 156539568, 157817896, 162399496,
168344072, 173146584, 176302744, 182878168, 183946152,
18506
Something like. . .
> midpoint <- c(132968364, 135945080, 156539568, 157817896,
+ 162399496, 168344072, 173146584, 176302744,
+ 182878168, 183946152, 185068720, 190791232,
+ 84317660, 93708872, 106810172, 12684,
+ 148519056, 150945112, 155771432, 181069984,
+ 87104384
+ )
> shapiro.test(
Here's a sample:
> unif_rand_1 <- runif(1000);
> unif_rand_2 <- runif(1000);
> ks.test(unif_rand_1,unif_rand_2);
Two-sample Kolmogorov-Smirnov test
data: unif_rand_1 and unif_rand_2
D = 0.021, p-value = 0.9802
alternative hypothesis: two-sided
So in your case:
> ks.test( runif( length( you
Has your question been answered yet?
x=c(1,2,4,3,6,8)
y=c(3,2,5,7,4,6)
cor.test(x,y,method="spearman")
And that's how you "extract" the p-value:
cor.test(x,y,method="s")$p.value
Cheers,
Daniel
--
Abstrakthelfer helfen wenig
--
-Ursprün
On 2/28/08, Anne-Katrin Link <[EMAIL PROTECTED]> wrote:
> I would like to do a Spearman rank order test, and used the cor() function
> with the method "spearman".
> It gives me a number (correlation coefficient?) , but how can I get the
> p-value?
You're probably looking for rcorr() from Hmisc
check function ?cor.test().
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://med.kuleuven.be/biostat/
http
On Thu, 2008-02-28 at 16:17 +0100, Anne-Katrin Link wrote:
> Dear R-helpers,
>
> I would like to do a Spearman rank order test, and used the cor() function
> with the method "spearman".
> It gives me a number (correlation coefficient?) , but how can I get the
> p-value?
> Thank you for the help
The lme4 package is still in development. I can't guarantee the
evaluation of the deviance for the quasipoisson family. There is a
certain amount of mystery about the role of the dispersion parameter
and the null deviance in families like quasipoisson and quasibinomial.
For the time being don't
Try
str(ks.test(VeriSeti1, VeriSeti2)) # see ?str
then
ks.test(VeriSeti1, VeriSeti2)$p.value
Med venlig hilsen
Frede Aakmann Tøgersen
> -Oprindelig meddelelse-
> Fra: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] På vegne af Emre Unal
> Sendt: 3. oktober 2007 11:32
> Til: r-h
Hi,
why don't you try try
ks.test(VeriSeti1, VeriSeti2)$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
ks.test(x,y)[2]
On 10/3/07, Emre Unal <[EMAIL PROTECTED]> wrote:
> 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 hypo
69 matches
Mail list logo
