David,
On 2019-03-20 12:38, David Winsemius wrote:
On 3/19/19 12:49 PM, Jeff Newmiller wrote:
Highly off topic. Try StackOverflow.
As it stands it's off-topic for SO. (You would just be making more
work for those of us who know the rules but need 4 close votes for
migration.) Better would b
On 3/19/19 12:49 PM, Jeff Newmiller wrote:
Highly off topic. Try StackOverflow.
As it stands it's off-topic for SO. (You would just be making more work
for those of us who know the rules but need 4 close votes for
migration.) Better would be immediately posting at CrossValidated.com
(i.e
Hi,
Since folks are taking the time to point out some subtle issues here, taking an
example from the UCLA Stats web site:
https://stats.idre.ucla.edu/other/mult-pkg/faq/general/faq-why-is-the-mann-whitney-significant-when-the-medians-are-equal/
Grp1 <- rep(c(-2, 0, 5), each = 20)
Grp2 <- rep(c(
No. Do not call plot.xmean.ordinaly() if the argument is not of class
xmean.ordinaly, because that function assumes that it is such an object.
That is one reason why it is better to call plot() than to be more specific.
On March 19, 2019 3:29:11 PM PDT, Kim Jacobsen wrote:
>Mailing list now inc
Hi Javed,
Easy.
A<-c(2000,2100,2300,2400,6900,7000,7040,7050,7060)
median(A)
[1] 6900
B<-c(3300,3350,3400,3450,3500,7000,7100,7200,7300)
median(B)
[1] 3500
wilcox.test(A,B,paired=FALSE)
Wilcoxon rank sum test with continuity correction
data: A and B
W = 26.5, p-value = 0.233
alternative
Highly off topic. Try StackOverflow.
On March 19, 2019 10:42:24 AM PDT, Philip Rhoades wrote:
>People,
>
>I have only a general statistics understanding and have never actually
>used Bayes' Theorem for any real-world problem. My interest lies in
>developing some statistical approach for addres
Actually the list is not moderated in the usual sense of the word. If
you subscribe, your posts are not moderated. Only your first posting
after subscription would be moderated, but for the purpose of preventing
persons with obvious spamming goals.
And there are several different moderators. I
> On Mar 19, 2019, at 2:06 PM, Evan Cooch wrote:
>
> Just curious -- if R-help is a moderated list (which in theory , it is -- my
> posts have been 'modertated', to the degree that they aren't released to the
> list until someone approves them), and if these 'statistics discussion'
> quest
Any reasonable test of whether two samples differ should be scale and
location invariant. E.g., if you measure temperature it should not matter
if you units are degrees Fahrenheit or micro-Kelvins. Thus saying the
medians are 3500 and 6200 is equivalent to saying they are 100.035 and
100.062: it
Just curious -- if R-help is a moderated list (which in theory , it is
-- my posts have been 'modertated', to the degree that they aren't
released to the list until someone approves them), and if these
'statistics discussion' questions are inappropriate to the mission (as
described), then...wh
Rhelp is not a forum for discussions of statistics. Instead it is for
persons who have specific questions about the use of R.
Please read the list info page where you started the subscription
process. And do read the Posting Guide. Both these are linked at the
bottom of this response.
There
People,
I have only a general statistics understanding and have never actually
used Bayes' Theorem for any real-world problem. My interest lies in
developing some statistical approach for addressing the subject above
and it seems to me that BT is what I should be looking at? However,
what I
> This is my function:
>
> wilcox.test(A,B, data = data, paired = FALSE)
>
> It gives me high p value, though the median of A column is 6900 and B
> column is 3500.
>
> Why it gives p value high if there is a difference in the median?
Perhaps becuase a) because you are testing the wrong data or
Good suggestion, and for my purposes, will solve the problem. Thanks!
On 3/18/2019 12:37 PM, Ben Tupper wrote:
> Hi,
>
> Might you replaced 'T' with a numeric value that signals the TRUE case
> without rumpling your matrix? 0 might be a good choice as it is never an
> index for a 1-based indexi
We've had this conversation.
A) This is off-topic for R-Help. Your question is about the statistical test,
not about the R coding.
B) A difference in sample statistics, whether or not it "looks" large, is not
sufficient for statistical significance.
On 3/19/19, 12:48 PM, "R-help on behalf of
Hi
This is my function:
wilcox.test(A,B, data = data, paired = FALSE)
It gives me high p value, though the median of A column is 6900 and B
column is 3500.
Why it gives p value high if there is a difference in the median?
Regards
[[alternative HTML version deleted]]
_
I want to obtain the individual biological parameters (ie age at takeoff,
final height velocity at takeoff) after a sitar model. (I do understand a
sitar model fits a mean curve for the population).
I have managed to obtain the parameters for one individual using the code
below,
library(sit
I want to obtain the individual biological parameters (ie age at takeoff,
final height velocity at takeoff) after a sitar model. (I do understand a
sitar model fits a mean curve for the population).
I have managed to obtain the parameters for one individual using the code
below,
library(sit
Wow! I've never come across this approach before. This will keep me reading for
a good few days.
Thanks Duncan!
-Original Message-
From: Duncan Murdoch [mailto:murdoch.dun...@gmail.com]
Sent: 18 March 2019 19:45
To: Jeff Newmiller ; r-help@r-project.org; William
Sones
Subject: Re: [R
Dear duncan,
Sorry to bother you with such a silly mistake I
didn,t notice it!
Sent: Tuesday, March 19, 2019 6:01 PM
To: akshay kulkarni; R help Mailing list
Subject: Re: [R] Fw: problem with nlsLM function
On 19/03/2019 8:26 a.m., akshay kulkarni wrote:
>
> de
On 19/03/2019 8:26 a.m., akshay kulkarni wrote:
dear members,
also,I can provide HM1,HM2 and HM3 if needed
From: R-help on behalf of akshay kulkarni
Sent: Tuesday, March 19, 2019 5:43 PM
To: R help Mailing list
Subjec
dear members,
I am getting the "singular gradient error" when I use nls
for a function of two variables:
> formulaDH5
HM1 ~ (a + (b * ((HM2 + 0.3)^(1/2 + (A * sin(w * HM3 + a) +
C)
HM1 is the response variable, and HM2 and HM3 are predictors.
The problem is I get the
Hello All,
wonder if you have thoughts on a clever solution for this code:
df <- data.frame(a = c(6,1), b = c(1000,1200), c =c(-1,3))
#the caveat here is that the number of rows for df can be anything from 1 row
to in the hundreds. I kept it to 2 to have minimal reproducible
t<-seq(-5
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