Thanks. I appreciate this isn't strictly an R question and will
pursue on another list.
The procedure I followed was inspired from
@article{
Author = {Baayen, R. Harald and Milin, Petar},
Title = {Analysing Reaction Times},
Journal = {International Journal of Psychological Research},
Cecile De Cat leeds.ac.uk> writes:
> I'm analysing reaction time data from a linguistic experiment (a variant of
> a lexical decision task). To ascertain that the data was normally
> distributed, I used *shapiro.test *for each participant (see commands
> below), but only one out of 21 returns a
Cecile:
On Wed, Apr 18, 2012 at 8:21 AM, Cecile De Cat wrote:
> Hello,
>
> I'm analysing reaction time data from a linguistic experiment (a variant of
> a lexical decision task). To ascertain that the data was normally
> distributed, I used *shapiro.test *for each participant (see commands
> be
Hello,
I'm analysing reaction time data from a linguistic experiment (a variant of
a lexical decision task). To ascertain that the data was normally
distributed, I used *shapiro.test *for each participant (see commands
below), but only one out of 21 returns a p value above p.0 05.
> f = functio
On 29/03/2011 5:33 PM, Patrizio Frederic wrote:
dear all,
here's a couple of questions that puzzled me in these last hours:
# issue 1 qnorm(1-10e-100)!=qnorm(10e-100)
qnorm(1-1e-10) == -qnorm(1e-10)
# turns on to be FALSE. Ok I'm not a computer scientist but,
# but I had a look at the R in
On Wed, Mar 30, 2011 at 10:33 AM, Patrizio Frederic
wrote:
> dear all,
> here's a couple of questions that puzzled me in these last hours:
>
> # issue 1 qnorm(1-10e-100)!=qnorm(10e-100)
>
> qnorm(1-1e-10) == -qnorm(1e-10)
>
> # turns on to be FALSE. Ok I'm not a computer scientist but,
> # but
dear all,
here's a couple of questions that puzzled me in these last hours:
# issue 1 qnorm(1-10e-100)!=qnorm(10e-100)
qnorm(1-1e-10) == -qnorm(1e-10)
# turns on to be FALSE. Ok I'm not a computer scientist but,
# but I had a look at the R inferno so I write:
all.equal(qnorm(1-1e-10) , -qno
Altogether I got five more or less silly solutions (not my judgment!), some of
them further discussed in private mail, for a problem where my expectation was
to get a simple one-liner back: "Check ?clt" or so...
Fortunately, with all of them I seem to arrive at a result that is consistent
with
Just to add to the silly solutions, here's how I would have done it...
mu <- 40
sdev <- 10
days <- 100:120 # range to explore
p <- 0.8
days[ match(TRUE, qnorm(0.2, mu*days, sqrt(sdev * sdev * days)) >= 4000) ]
Michael
On 9 January 2011 08:48, Bert Gunter wrote:
> If I understand what you have
If I understand what you have said below, it looks like you do NOT
have the problem solved manually. You CAN use qnorm , and when you do
so, your equation yields a simple quadratic which, of course, has an
exact solution that you can calculate in R.
Of course, one can use uniroot or whatever to so
On Sat, Jan 8, 2011 at 6:25 AM, Rainer Schuermann
wrote:
> It is _from_ a homework but I have the solution already (explicitly got that
> done first!) - this was the pasted Latex code (apologies for that, but in
> plain text it looks unreadable[1], and I thought everybody here has his / her
> f
On Sat, 8 Jan 2011, Rainer Schuermann wrote:
Sounds like homework, which is not an encouraged use of the Rhelp
list. You can either do it in theory...
It is _from_ a homework but I have the solution already (explicitly got
that done first!) - this was the pasted Latex code (apologies for th
On Jan 8, 2011, at 9:25 AM, Rainer Schuermann wrote:
Sounds like homework, which is not an encouraged use of the Rhelp
list. You can either do it in theory...
It is _from_ a homework but I have the solution already (explicitly
got that done first!) - this was the pasted Latex code (apologie
> Sounds like homework, which is not an encouraged use of the Rhelp
> list. You can either do it in theory...
It is _from_ a homework but I have the solution already (explicitly got that
done first!) - this was the pasted Latex code (apologies for that, but in plain
text it looks unreadable[1]
On Jan 8, 2011, at 6:56 AM, Rainer Schuermann wrote:
This is probably embarrassingly basic, but I have spent quite a few
hours in Google and RSeek without getting a clue - probably I'm
asking the wrong questions...
There is this guy who has decided to walk through Australia, a total
dist
This is probably embarrassingly basic, but I have spent quite a few hours in
Google and RSeek without getting a clue - probably I'm asking the wrong
questions...
There is this guy who has decided to walk through Australia, a total distance
of 4000 km. His daily portion (mean) is 40km with an sd
Hi Greg,
I'm making NoiseGenerators with different noise sources and components, the
meaning of my tests with R is to know which NoiseGenerator approached most
the Normal distribution function...
Thanks, for all the reactions.
Bosken
Greg Snow-2 wrote:
>
>
> Do your NoiseGenerotors need to
On Feb 25, 2010, at 12:20 PM, Bosken wrote:
Hi,
Thanks for your reaction.
The purpose of my test is to check if my NoiseGenerators really are
Normal
Distributed en witch circuit is the best!
So I need some good test to do this.
But what with: Fortune(), can't find anything about it..
Hi,
Thanks for your reaction.
The purpose of my test is to check if my NoiseGenerators really are Normal
Distributed en witch circuit is the best!
So I need some good test to do this.
But what with: Fortune(117) and fortune(234), can't find anything about it..
Thanks for the help!
Bosken
--
dhan.h
tml
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Bert Gunter
Sent: Thursday, February 25, 2010 12:24 PM
To: 'Greg Snow'; 'Bosken'; r-help@r-project.org
Subject: Re: [R] Normal distribution (Lillie.test())
...
tical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of Bosken
> Sent: Thursday, February 25, 2010 10:21 AM
> To: r-help@r-project.org
>
gt; From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of Bosken
> Sent: Tuesday, February 23, 2010 4:13 AM
> To: r-help@r-project.org
> Subject: Re: [R] Normal distribution (Lillie.test())
>
>
> Hi,
>
> Thanks for your reaction;
>
Hi,
Thanks for your reaction;
How do you come to the decision that my data not is normal distributed?
With the 69-95-99.7 test and Q-Q plot seems it ok! But these test are not
exact, they only give you an image.
Gr. Bosken
--
View this message in context:
http://n4.nabble.com/Normal-distrib
wrote:
> From: Bosken
> Subject: [R] Normal distribution (Lillie.test())
> To: r-help@r-project.org
> Received: Tuesday, 23 February, 2010, 7:22 AM
>
> Hi all,
>
> I have a dataset of 2000 numbers ( it's noise measured with
> a scoop )
>
> Now i want to
Hi all,
I have a dataset of 2000 numbers ( it's noise measured with a scoop )
Now i want to know of my data is normal distributed (Gaussian distribution).
I did already:
- 68-95-99.7 test
- Q-Q-plot
and now i used "nortest library" and the Lilli.test()
However i don't understad the output?
Thank you for all the answers!
Kjetil,
I am not sure if we are talking about the same thing. I only have a two
dimensional normal distribution which leads to three dimensional data.
You mean with "reject" I should not do such a test?
My data files contain about 26 points which I can reduce
On 11/17/2009 10:07 AM, rkevinbur...@charter.net wrote:
This is probably an even more basic question but shapiro.test return both the
statistic (w) and the significance (pw) of the statistic. For this test the
null-hypothesis is that the distirbution is not normal so very small values of
pw wo
This is probably an even more basic question but shapiro.test return both the
statistic (w) and the significance (pw) of the statistic. For this test the
null-hypothesis is that the distirbution is not normal so very small values of
pw would mean that there is very little chance that the distiri
On Tue, Nov 17, 2009 at 11:17 AM, Markus Mehrwald wrote:
> Hi all,
>
> I am completely new to R and my knowledge of statistics is quite small so I
> hope you can help my.
> I have three dimensional point data which represents (and this is what I do
> not know for sure) a normal distribution. Now I
Markus Mehrwald wrote:
> Hi all,
>
> I am completely new to R and my knowledge of statistics is quite small
> so I hope you can help my.
> I have three dimensional point data which represents (and this is what I
> do not know for sure) a normal distribution. Now I want to test if this
> is true o
Hi all,
I am completely new to R and my knowledge of statistics is quite small
so I hope you can help my.
I have three dimensional point data which represents (and this is what I
do not know for sure) a normal distribution. Now I want to test if this
is true or not and as I can remember from s
Normal distribution check within R can be done with functions available in
nortest package. This package consists of several normality tests. In order
to install package type install.packages('nortest'). Afterwards, you should
consider running ks.test() only if mu and sigma parameters are known (t
, October 01, 2009 10:47 AM
> To: r-help@r-project.org
> Subject: [R] Normal distribution
>
> Hi,
>
> I am dealing with how to check in R if some data that I have belongs to
> a
> normal distribution or not. I am not interested in obtaining the
> theoreticall frequencie
project.org [mailto:r-help-boun...@r-project.org] On
>> Behalf Of Noela Sánchez
>> Sent: Thursday, October 01, 2009 12:47 PM
>> To: r-help@r-project.org
>> Subject: [R] Normal distribution
>>
>> Hi,
>>
>> I am dealing with how to check in R if some data
t 1, 2009 at 12:56 PM, Richardson, Patrick
wrote:
?shapiro.test
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Noela Sánchez
Sent: Thursday, October 01, 2009 12:47 PM
To: r-help@r-project.org
Subject: [R] Normal distribution
Hi
Noela,
Package nortest has 5 tests of normality.
A word of caution: the wording of your question suggests that
you believe that, if the test you use does not reject the
normality hypothesis, you then have proof that your data come
from a Normal distribution. That's *NOT* true. Normality will
sti
Sent: Thursday, October 01, 2009 1:18 PM
To: Richardson, Patrick
Cc: Noela Sánchez; r-help@r-project.org
Subject: Re: [R] Normal distribution
Hi,
I think you can also use a qq-plot to do the same, no? You won't get a
statistic score + p.value, but perhaps you're more of a visual person?
:-)
t; From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf Of Noela Sánchez
> Sent: Thursday, October 01, 2009 12:47 PM
> To: r-help@r-project.org
> Subject: [R] Normal distribution
>
> Hi,
>
> I am dealing with how to check in R if some data that
?shapiro.test
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Noela Sánchez
Sent: Thursday, October 01, 2009 12:47 PM
To: r-help@r-project.org
Subject: [R] Normal distribution
Hi,
I am dealing with how to check in R if some data
Hi,
I am dealing with how to check in R if some data that I have belongs to a
normal distribution or not. I am not interested in obtaining the
theoreticall frequencies. I am only interested in determing if (by means of
a test as Kolmogorov, or whatever), if my data are normal or not.
But I have t
Thank You
It really works fine and i do understand it. Now i can sleep, its 0200 hours
here.
Why don't you just do this:
H<-rnorm(100, mean=5, sd=3000)
par(las=1)
hist(H, breaks=seq(4, 6, 1000), freq=F)
f<- function(x) exp(-(x-5000)^2/1800)/sqrt(1800*pi)
x<- seq(4, 60
Why don't you just do this:
H<-rnorm(100, mean=5, sd=3000)
par(las=1)
hist(H, breaks=seq(4, 6, 1000), freq=F)
f<- function(x) exp(-(x-5000)^2/1800)/sqrt(1800*pi)
x<- seq(4, 6, 100)
lines(x, dnorm(x, 5, 3000))
On Sat, Sep 12, 2009 at 5:47 PM, KABELI MEFANE wrote:
Dear All
let me go one step further by asking you if you could help me show that the
distribution of this data in normal. have a little idea (by trial and
error) but i seem to not fully understand how its done.
H<-rnorm(100, mean=5, sd=3000)
par(las=1)
hist(H, breaks=seq(4, 6,
Hi James,
(qnorm(.9)-qnorm(.1))
is the width of the 80% CI of a standard normal distribution. So its a
kind of standardization the other way round calculating sd from this.
The same holds for arbitrary p1 and p2. Without loss of generality
(since the normal distribution is symmetric), assume
Hi James
its just simple calculus, since with
q90<-qnorm(.9,me,sd)
q10<-qnorm(.1,me,sd)
mean<-(q90+q10)/2# the normal distribution is symmetric around the mean
sd<-(q90-q10)/ (qnorm(.9)-qnorm(.1)) # between 10th and 90th are
qnorm(.9)-qnorm(.1)=2.563103sds
hth.
g...@ucalgary.ca schrieb:
If we knew two pth quantiles for a normal distribution,
is it possible that we can find mean and sigma for the normal distribution
using R?
Let x ~ norm(mean, sigma).
Suppose that qnorm(0.9,mean,sigma) and qnorm(0.1,mean,sigma) are known.
Can we find mean and sigma using R?
Thanks,
-james
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