HI listers,

Anyone could explain what does it mean to explain a
sampling scheme by induction...
I have the following scheme for sampling random sample
without replacement...
1-select first element with prob(n/N)
2-select second element with prob(n-1/N-1), if first
element was selected, otherwise with prob (n/N-1) if
first element was not selected
3-select the ith element with prob(n-u/(N-(i-1))) for
u<=n
In order to select n elements...
How could I show this by INDUCTION...

Thanks,

Ribeiro



       Abra sua conta no Yahoo! Mail, o �nico sem limite de espa�o para
armazenamento!
http://br.mail.yahoo.com/

This is venkateswara rao.
  i did my master of science in statistics from banaras hindu university
in the year 2007. i
learnt sas in hyderabad and the modules are BASE SAS, SAS SQL, SAS STAT.
Please any one can help me for getting a job as a statistical analyst.
please also give information about the openings for freshers. if
avilable then please send the information to my mail id
venkibhu@...

You should be able to use 'as.Date' on your data
and once they have the 'Date' class, then you can
just subtract them to get the difference in days.

Patrick Burns
patrick@...
+44 (0)20 8525 0696
http://www.burns-stat.com
(home of S Poetry and "A Guide for the Unwilling S User")

Deepak Manohar wrote:

> Dear All
> Can anyone guide in finding the difference between two dates which has
> its data set in Excel. The question here is I need to find the
> difference between two dates using R package from the Excel data.
>
>
> Thanks in Advance
>
> Thanking you
>
> Yours Sincerely
>
> Deepak M
>
> ------------------------------------------------------------------------
> Get the freedom to save as many mails as you wish. Click here to know
> how.
>
<http://in.rd.yahoo.com/tagline_mail_5/*http://help.yahoo.com/l/in/yahoo/mail/ya\
hoomail/tools/tools-08.html/>
>
>

Gaetan, I can't give you a precise answer. It appears that the method of
calculating the % overlap
measure is described in Cohen's 1988 book. Unfortunately the 'hyperlinks' on the
web page just take
you to Becker's index, not the source.

When I read your query I wondered whether overlap might be a simple rescaling of
the measure theta
= U/mn used in the spreadsheet I sent you last month. In case I didn't send you
the 2 papers in
Stats in Medicine that describe this, here they are. Anyway, theta is a kind of
measure of
non-overlap, which takes the value 0.5 if the two distributions are identical
and 1 or 0 if there's
no overlap. The correspondence between the standardised difference (i.e. Cohen's
d) and theta
assuming Gaussian distributions is shown in Table 1 of the first paper. The
percentage overlap
measure shown in Becker's table is similar, but NOT identical, to 2*theta-1.
Thus for d = 1, theta =
0.760 and  2*theta-1 = 0.520, which is close to Becker's 55.4%, but isn't
exactly the same as it.

Personally, I'd recommend simply reporting U/mn, which is easily obtained from
your data using
SPSS's ROC procedure, then use my spreadsheet to get a good CI - and reference
my papers.

Hope this helps.


Robert G. Newcombe PhD CStat FFPH
Professor of Medical Statistics
Department of Primary Care and Public Health
Centre for Health Sciences Research
Cardiff University
4th floor, Neuadd Meirionnydd
Heath Park, Cardiff CF14 4YS

Tel: 029 2068 7260
Fax: 029 2068 7236

Home page
http://www.cardiff.ac.uk/medicine/epidemiology_statistics/research/statistics/ne\
wcombe
For location see http://www.cardiff.ac.uk/locations/maps/heathpark/index.html


>>> "gaetanlion" <gaetanlion@...> 21/01/08 23:35:36 >>>

Within an unpaired hypothesis testing framework, the p value gives
you the probability that two samples come from population with
identical distribution.  But, even a p value of 0% does not tell you
how different two samples are.  To assess the magnitude of the
difference of two samples, I have been studying the Effect Size.

Regarding the Effect Size, I am struggling to calculate correctly the
Overlap and Nonoverlap areas of two samples' respective
distributions.  Someone sent me this helpful link on the subject:
http://web.uccs.edu/lbecker/Psy590/es.htm.  On page 2, of this
article Professor Lee A. Becker calculates the surface of the
Nonoverlap area.  Unfortunately, I calculate very different
Nonoverlap areas.

Example: Effect Size = 2.  I take the midpoint of 2 which is 1.  This
value of '1' is a Z value that I simply put into the NORMSDIST()
function in Excel to calculate the Area under the curve of one sample
which is equal to 84.1%.  The area beyond the curve (the right tail
of the first sample) = 1 – 84.1% = 15.9%.  The Overlap area = 2 x
15.9% = 31.7%.  And, the Nonoverlap area = 1 – 31.7% = 68.3%.

My example makes good sense to me.  But, in this case my Nonoverlap
area of 68.3% is very different than Lee A. Becker's of 81.1%.  See
page 2 of his paper (link shown above).


To Gaetan Lion 1013 27/12/2007

The key point here is that when you dichotomise a continuous variable (or an
ordinal one) you lose
a
lot of power - even when the cut point is close to the median - much more so
when it's out in the
tail. The main justification for dichotomising a variable in this way is for
concise presentation
of
results.

Also, we should be shifting away from using p-values as the main expression of
uncertainty due to
limited sample size, towards using appropriate measures of effect size with
confidence intervals.
An
appropriate effect size measure corresponding to the M-W test is U/mn, obtained
by dividing the M-W

U statistic by the product of the 2 sample sizes. This is the same as the area
under the ROC curve.

It is 0 or 1 if two samples don't overlap, and 0.5 if they are identical. The
AUROC is easily
obtained using SPSS's ROC procudure, but the CI meth
od isn't always reliable, it goes wrong when
U/mn is at or close to 0 or 1. A better method is obtained using the attached
Excel spreadsheet. It

comes in 2 versions, one starting with m, n and U, the other starting with m, n
and U/mn from SPSS.

The spreadsheet also displays 2 Gaussian curves corresponding to this degree of
separation, with a

confidence region.

Hope this helps.


>>> "gaetanlion" <gaetanlion@...> 26/12/07 00:43 >>>

I use the nonparametric Mann-Whitney test all the time to test if two samples
coming from
a population that is not normal are statistically different.  But, I find
dealing with average
rank (as required by Mann-Whitney test) not transparent or intuitive.  In such a
situation, I
would like to use instead the Chi Square test testing for Medians.  It be a lot
more
transparent to convey to the audience that the samples' medians are different
instead of
their average rank.

Can you think of any reason why I should not use the Chi Square test testing for
median as
described?

Thank you in advance for your answers.  Please describe your background briefly
to
attribute credibility to your answer.

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I am kishore here,There is a oppenings for freshers in CTS .

I am kishore here,There is a oppenings for freshers in CTS .

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Date of posting January 23, 2008

Thanks and regards,
Archan Bhattacharya
Statistician
Sciformix Technologies Pvt. Ltd.
Mumbai, INDIA.

I'm not sure if i want to talk directly about what is bothering me,
but I feel that creating a sort of mystery about this may be of help
for calling onto myself the help i may need to find the solution for
what is bothering me. I'm not sure if there is much more i want to
say about that right now. Because of my qualification in Applied
Statistics at Masters Level, those around me believe that I have a
great understanding of statistics and can help them in their
statistical analyses. I completed my masters in applied statistics
in 2002 in very unusual circumstances, within health difficulties,
difficulties to meet deadlines etc. I had to seek help at that time
from the university's student support services etc. In any case, I
succeeded somehow in acquiring an MSc in Applied Statistics in
circumstances which are almost unbelievable to me. I feel that i
have acquired an MSc without feeling the understanding of Applied
Statistics in my head. I do not consider that I sufficiently master
the subject nor do i feel very confident when questioned about it.
Some of those who are aware of my qualification have come to seek my
help in the past and while attempting to help them to some extent, i
needed to make them understand somehow that i'm uncertain if i can
help them as they expect. Right now, someone doing an MBA with
specialisation in Information Systems, has brought to me some
statistical analyses that he did, requesting me to verify whether
certain things are correct or incorrect in his analyses. He left
print-outs with me etc, and this is what bothering me right now, as
i attempt to seek inspiration concerning how to approach this
situation. I thought that venting a little bit overhere could
possibly be of some help. Thanking u in advance for your support.
Peace and blessings...

I think the first part is correct.

but for the second part... though it doesnt have much specific conditions, but it has the conditions that: 1 ball drawn from each jar and both will b black.so for minimizing the prob. each jar should contain atleast 1 black ball. so in  my view, in 1 jar there will 1 black ball, and 99 red balls, and the remaining part will b in another jar. so the minimum prob will b 0.0099.

Nishant Jain <nishantjain18@gmail.com> wrote:

without much thinking...... i guess......

to maximize the prob.... put 50-50 balls of each color in each jar.... that shd make the prob 0.25, i dont think prob could be higher that this.....

however, since no conditoin is specified... the min. prob can be 0.... for that put 100 black balls in one jar.... and put 100 red in other one.....

 

does this makes a sense?????

 

Nishant

 

On 1/22/08, venkat suresh <suresh_medabala@yahoo.co.in> wrote:

There are 100 red balls and 100 black balls, 2 jars, you have to find
the number of red and
      black balls to put in the 2 jars such that probability of
getting 2 black balls
(draw 1 ball from each jar) is maximized, again it was asked to find
the same for minimizing the probability.

---

Get the freedom to save as many mails as you wish. Click here to know how.

 

Regards....

 

Jitadeepa Nayak 
Statistical Trainee 
Indian Statistical Institute           
203,BT Road
Kolkata-700108

---

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without much thinking...... i guess......

to maximize the prob.... put 50-50 balls of each color in each jar.... that shd make the prob 0.25, i dont think prob could be higher that this.....

however, since no conditoin is specified... the min. prob can be 0.... for that put 100 black balls in one jar.... and put 100 red in other one.....

 

does this makes a sense?????

 

Nishant

 

On 1/22/08, venkat suresh <suresh_medabala@yahoo.co.in> wrote:

There are 100 red balls and 100 black balls, 2 jars, you have to find
the number of red and
      black balls to put in the 2 jars such that probability of
getting 2 black balls
(draw 1 ball from each jar) is maximized, again it was asked to find
the same for minimizing the probability.

---

Get the freedom to save as many mails as you wish. Click here to know how.

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Within an unpaired hypothesis testing framework, the p value gives
you the probability that two samples come from population with
identical distribution.  But, even a p value of 0% does not tell you
how different two samples are.  To assess the magnitude of the
difference of two samples, I have been studying the Effect Size.

Regarding the Effect Size, I am struggling to calculate correctly the
Overlap and Nonoverlap areas of two samples' respective
distributions.  Someone sent me this helpful link on the subject:
http://web.uccs.edu/lbecker/Psy590/es.htm.  On page 2, of this
article Professor Lee A. Becker calculates the surface of the
Nonoverlap area.  Unfortunately, I calculate very different
Nonoverlap areas.

Example: Effect Size = 2.  I take the midpoint of 2 which is 1.  This
value of '1' is a Z value that I simply put into the NORMSDIST()
function in Excel to calculate the Area under the curve of one sample
which is equal to 84.1%.  The area beyond the curve (the right tail
of the first sample) = 1 � 84.1% = 15.9%.  The Overlap area = 2 x
15.9% = 31.7%.  And, the Nonoverlap area = 1 � 31.7% = 68.3%.

My example makes good sense to me.  But, in this case my Nonoverlap
area of 68.3% is very different than Lee A. Becker's of 81.1%.  See
page 2 of his paper (link shown above).

Any assistance is greatly appreciated.

Guy

hai

         iam a statistics student i want to join the SAS training
course so pls send the institution address in chennai

Dear Statisticians,

If you are in the clinical research field (not finance) I have a
projects I need help with please contact me. If you know data
management, site management, or sas programming I want to speak with
you as well.

Regards,
Amy

Pls provide me a website regarding acturial statistics.
regards,
gowri

Dear Statisticians,

If you are in the clinical research field (not finance) I have a
projects I need help with please contact me. If you know data
management, site management, or sas programming I want to speak with
you as well.

Regards,
Amy