From: Greg Snow <Greg.Snow_at_imail.org>

Date: Sat, 26 Jun 2010 14:30:00 -0600

R-help_at_r-project.org<mailto:R-help_at_r-project.org> mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/postingguide. html

and provide commented, minimal, self-contained, reproducible code.

R-help_at_r-project.org mailing list

https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Received on Sat 26 Jun 2010 - 20:33:49 GMT

Date: Sat, 26 Jun 2010 14:30:00 -0600

No I mean something like this, assuming that the iris dataset contains the full population and we want to see if Setaso have a different mean than the population (the null would be that there is no difference in sepal width between species, or that species tells nothing about sepal width):

hist(out1, xlim=range(out1,obs1))

abline(v=obs1)

I don't have a reference (other than a text book that defines sampling distributions).

Atte Tenkanen kirjoitti 26.6.2010 kello 5.15:

Greg Snow kirjoitti 25.6.2010 kello 21.55:

I check, so you mean doing it this way:

t.test(sample(POPUL, length(SAMPLE), replace = FALSE), mu=mean(SAMPLE), alt = "less")

The values come from this kind of process: The musical composition is segmented into so-called 'pitch-class segments' and these segments are compared with one reference set with a distance function. Only some distance values are possible. These distance values can be averaged over music bars which produces smoother distribution and the 'comparison curve' that illustrates the distances according to the reference set through a musical piece result in more readable curve (see e.g. http://users.utu.fi/attenka/with6.jpg ), but I would prefer to use original values.

then, I want to pick only some regions from the piece and compare those values of those regions, whether they are higher than the mean of all values.

Atte

On Jun 24, 2010, at 6:58 PM, Atte Tenkanen wrote:

Is there anything for me?

There is a lot of data, n=2418, but there are also a lot of ties. My sample n—250-300

I do not understand why there should be so many ties. You have not described the measurement process or units. ( ... although you offer a

glipmse without much background later.)

i would like to test, whether the mean of the sample differ significantly from the population mean.

Why? What is the purpose of this investigation? Why should the mean of

a sample be that important?

The histogram of the population looks like in attached histogram, what test should I use? No choices?

This distribution comes from a musical piece and the values are 'tonal distances'.

http://users.utu.fi/attenka/Hist.png

That picture does not offer much insidght into the features of that measurement. It appears to have much more structure than I would expect for a sample from a smooth unimodal underlying population.

*--
*

David.

Atte

On 06/24/2010 12:40 PM, David Winsemius wrote:

On Jun 23, 2010, at 9:58 PM, Atte Tenkanen wrote:

Thanks. What I have had to ask is that

how do you test that the data is symmetric enough? If it is not, is it ok to use some data transformation?

when it is said:

"The Wilcoxon signed rank test does not assume that the data are sampled from a Gaussian distribution. However it does assume that

the

data are distributed symmetrically around the median. If the
distribution is asymmetrical, the P value will not tell you much

about

whether the median is different than the hypothetical value."

You are being misled. Simply finding a statement on a statistics
software website, even one as reputable as Graphpad (???), does
not

mean

that it is necessarily true. My understanding (confirmed
reviewing

"Nonparametric statistical methods for complete and censored
data"

by M.

M. Desu, Damaraju Raghavarao, is that the Wilcoxon signed-rank
test

does

not require that the underlying distributions be symmetric. The
above

quotation is highly inaccurate.

To add to what David and others have said, look at the kernel that

the

U-statistic associated with the WSR test uses: the indicator (0/1)
of

xi

+ xj > 0. So WSR tests H0:p=0.5 where p = the probability that
the

average of a randomly chosen pair of values is positive. [If
there

are

ties this probably needs to be worded as P[xi + xj > 0] = P[xi +
xj

<

0], i neq j.

Frank

*--
*

Frank E Harrell Jr Professor and Chairman School of
Medicine

Department of Biostatistics VanderbiltUniversity

R-help_at_r-project.org<mailto:R-help_at_r-project.org> mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/postingguide. html

and provide commented, minimal, self-contained, reproducible code.

[[alternative HTML version deleted]]

R-help_at_r-project.org mailing list

https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Received on Sat 26 Jun 2010 - 20:33:49 GMT

Archive maintained by Robert King, hosted by
the discipline of
statistics at the
University of Newcastle,
Australia.

Archive generated by hypermail 2.2.0, at Sat 26 Jun 2010 - 23:10:37 GMT.

*
Mailing list information is available at https://stat.ethz.ch/mailman/listinfo/r-help.
Please read the posting
guide before posting to the list.
*