Algebraically, E = [z(a/2) / SQRT(n)] x SD, so it must be that the margin of
error (maximum error as you called it) is a multiple of the population
standard deviation. Keep in mind what these values represent. E is the
margin of error of the estimate of mu, the population mean. SD is the
On Sun, 11 Nov 2001 01:30:27 +1100, "David Muir"
<[EMAIL PROTECTED]> wrote:
> Presently the Gaming Industry of Australia is attempting to define various
> new 'definitions of Standard Deviation'...in a concept to define infield
> metrics for the analysis of ma
i think you are asking the wrong question ... because, as far as i know ...
there is only really one standard deviation concept ... square root of the
variance (average of squared deviations around the mean in a set of data) ...
perhaps what you are really interested in is HOW should
Presently the Gaming Industry of Australia is attempting to define various
new 'definitions of Standard Deviation'...in a concept to define infield
metrics for the analysis of machines in terms which imply whether a machine
is being operated with respect to its defined percentage
Edward Dreyer wrote:
>
> A colleague of mine - not a subscriber to this helpful list - asked me if
> it is possible for the standard deviation
> to be larger than the mean. If so, under what conditions?
Of course - for example, if you analyse mean-corrected data...
It can even
possible for
the SD to be larger than the mean, but the distribution will then be not
symmetric.
Alan
Edward Dreyer wrote:
>
> A colleague of mine - not a subscriber to this helpful list - asked me if
> it is possible for the standard deviation
> to be larger than the mean. If so
Well, yes. the mean and standard deviation are not 'linked' for data with a
Normal distribution.
Dale Glaser asked:
Well, what about the standard normal distribution: N(0,1)?
The mean is 0, the standard deviation, 1.
If you add the restriction that the data not be less than 0, a
Title: RE: Mean and Standard Deviation
Well, what
about the standard normal distribution: N(0,1)?
Dale N. Glaser, Ph.D.
Pacific Science
& Engineering Group
6310 Greenwich
Drive; Suite 200
San Diego, CA
92122
Phone: (858)
535-1661 Fax: (858) 535-1665
http://www.pac
In article <[EMAIL PROTECTED]>, Edward
Dreyer <[EMAIL PROTECTED]> wrote:
> A colleague of mine - not a subscriber to this helpful list - asked me if
> it is possible for the standard deviation
> to be larger than the mean. If so, under what conditions?
>
E
At 04:32 PM 10/12/01 -0500, you wrote:
>A colleague of mine - not a subscriber to this helpful list - asked me if
>it is possible for the standard deviation
>to be larger than the mean. If so, under what conditions?
what about z scores??? mean = 0 and sd = 1
>At first blush I do
Title: RE: Mean and Standard Deviation
Edward Dreyer writes:
>A colleague of mine - not a subscriber to this helpful
>list - asked me if it is possible for the standard deviation
>to be larger than the mean. If so, under what conditions?
>
>At first blush I do not think so
A colleague of mine - not a subscriber to this helpful list - asked me if
it is possible for the standard deviation
to be larger than the mean. If so, under what conditions?
At first blush I do not think so - but then I believe I have seen
some research results in which standard
1.96 * (sigma / sqrt n)
> >
> > now, what n might it take to produce some e? we can rearrange the
formula
> ...
> >
> > sqrt n = (1.96 * sigma) / e
> >
> > but, we don't want sqrt n ... we WANT n!
> >
> > n = ((
On Sun, 30 Sep 2001 00:34:40 GMT, "John Jackson"
<[EMAIL PROTECTED]> wrote:
> Here is my solution using figures which are self-explanatory:
>
> Sample Size Determination
>
> pi = 50% central area 0.99
> confid level= 99%
Donald,
I totally agree w/your point about the stratification of the sample. My
facts were set up merely for simplicity's sake notwithstanding their clear
artificiality.
The only instances of multiple samples I have seen are in textbooks to prove
the CLT; that w/increasing numbers of sample mean
On Sun, 30 Sep 2001, John Jackson wrote:
> Here is my solution using figures which are self-explanatory:
>
> Sample Size Determination
>
> pi = 50% central area 0.99
> confid level= 99% 2 tail area 0.5
> sa
en my clarification, I would
> welcome your insights.
>
>
> "Donald Burrill" <[EMAIL PROTECTED]> wrote in message
> [EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> > On Fri, 28 Sep 2001, John Jackson wrote in part:
> >
> > > My formu
terval formula shown
> > below for ascertaining the maximum error.
> E = Z(a/2) x SD/SQRT N
> > The issue is you want to solve for N, but you have no standard
> > deviation value.
> Oh, but you do. In the problem you formulated, unless I
> misunderstood egregiously, you
On Fri, 28 Sep 2001, John Jackson wrote in part:
> My formula is a rearrangement of the confidence interval formula shown
> below for ascertaining the maximum error.
E = Z(a/2) x SD/SQRT N
> The issue is you want to solve for N, but you have no standard
> dev
/ sqrt n)
>
> now, what n might it take to produce some e? we can rearrange the formula
...
>
> sqrt n = (1.96 * sigma) / e
>
> but, we don't want sqrt n ... we WANT n!
>
> n = ((1.96 * sigma)/ e) ^2
>
> so, what if we wanted to be within 3 points o
Really sorry.
My formula is a rearrangement of the confidence interval formula shown below
for ascertaining the maximum error.
E = Z(a/2) x SD/SQRT N
The issue is you want to solve for N, but you have no standard deviation
value.
The formula then translates into n = (Z(a/2)*SD)/E)^2Note
gt;
> now, what n might it take to produce some e? we can rearrange the formula
...
>
> sqrt n = (1.96 * sigma) / e
>
> but, we don't want sqrt n ... we WANT n!
>
> n = ((1.96 * sigma)/ e) ^2
>
> so, what if we wanted to be within 3 po
? we can rearrange the formula ...
sqrt n = (1.96 * sigma) / e
but, we don't want sqrt n ... we WANT n!
n = ((1.96 * sigma)/ e) ^2
so, what if we wanted to be within 3 points of mu with our sample mean the
population standard deviation or sigma we
John Jackson wrote:
> the forumla I was using was n = (Z?/e)^2 and attempting to express .05 as a
> fraction of a std dev.
I think you posted that before, and it's still getting
garbled. We see a Z followed by a question mark, and
have no idea what was actually intended.
- Randy
===
OTECTED]">news:MGns7.49824$[EMAIL PROTECTED]...
> > re: the formula:
> >
> > n = (Z?/e)2
>
> This formula hasn't come over at all well. Please note that newsgroups
> work in ascii. What's it supposed to look like? What's it a formula fo
supposed to look like? What's it a formula for?
> could you express E as a % of a standard deviation .
What's E? The above formula doesn't have a (capital) E.
What is Z? n? e?
> In other words does a .02 error translate into .02/1 standard deviations,
> assuming you are
At 04:49 PM 9/26/01 +, John Jackson wrote:
>re: the formula:
>
> n = (Z?/e)2
>
>
>could you express E as a % of a standard deviation .
>
>In other words does a .02 error translate into .02/1 standard deviations,
>assuming you are dealing w/a normal dist
Thanks for the formula, but I was really interested in knowing what % of a
standard deviation corresponds to E.
In other words does a .02 error translate into .02/1 standard deviations?
"Graeme Byrne" <[EMAIL PROTECTED]> wrote in message
9orn26$m80$[EMAIL PROTECTED]"&g
re: the formula:
n = (Z?/e)2
could you express E as a % of a standard deviation .
In other words does a .02 error translate into .02/1 standard deviations,
assuming you are dealing w/a normal distribution
This sounds like homework but I will .
Anyway assume the normal approximation to the binomial can be used (is this
reasonable?) then the formula for estimating sample sizes based on a given
confidence level and a given maximum error is
n = z*Sqrt(p*(1-p))/e
where z = the z-scores associated wit
If you have a confidence level of 90% and an error estimate of 4% and don't
know the std deviation, is there a way to express the error estimate as a
fraction of a std deviation?
=
Instructions for joining and leaving this list a
Referring your example:
variance = 2_nd moment - (1_st moment),
that is:
2_nd moment = 0^2 * 0.2 + 1^2 * 0.3 + 2^2 * 0.2 + 3^2 * 0.2 + 4^2 * 0.1 =
4.5
1_st moment = 0 * 0.2 + 1 * 0.3 + 2 * 0.2 + 3 * 0.2 + 4 * 0.1 = 1.7
then
variance = 4.5 - (1.7)^2 = 1.61
then
standard deviation = sqrt(1.61
Chris Chiu wrote:
>
> Dear friends:
>
> Does anyone know / remember how to obtain the standard deviation of a set
> of numbers given only a frequency table?
>
> e.g.,
> xf(x)
> 00.2
> 10.3
> 20.2
> 30.2
> 40.1
(0
QUESTIONS
Dear friends:
Does anyone know / remember how to obtain the standard deviation of a set
of numbers given only a frequency table?
e.g.,
xf(x)
00.2
10.3
20.2
30.2
40.1
Many thanks.
Chris
ONE POSSIBLE ANSWER:
Here is a worked solution. I used the Windows
Chris Chiu wrote:
>
> Dear friends:
>
> Does anyone know / remember how to obtain the standard deviation of a set
> of numbers given only a frequency table?
>
> e.g.,
> xf(x)
> 00.2
> 10.3
> 20.2
> 30.2
> 40.1
>
> Many t
ng as the X values are fixed ... and the p values ... then you could
do it that way ... of course, without the X values .. you are lost
At 09:35 AM 1/6/01 -0500, Chris Chiu wrote:
>Dear friends:
>
>Does anyone know / remember how to obtain the standard deviation of a set
>of num
Dear friends:
Does anyone know / remember how to obtain the standard deviation of a set
of numbers given only a frequency table?
e.g.,
xf(x)
00.2
10.3
20.2
30.2
40.1
Many thanks.
Chris
=
Instructions
Mark Solberg <[EMAIL PROTECTED]> wrote:
>I've had some statistics coursework, probably just enough to be dangerous.
>
>Here's my problem. By the way this is an actual problem, not theoretical.
>I need to analyze the hold percentage on certain table games in the casino I
>work at.
I should think
exist, i.e. theft, bad game protection, etc..
>
>The data I have is for each day, I have the calculated hold percentage for
>each of the individual table games. There are multiple table games of each
>type, for example there are 7 blackjack tables.
>
>Q: I want to calculate the st
multiple table games of each
type, for example there are 7 blackjack tables.
Q: I want to calculate the standard deviation and confidence interval for
blackjack by week. Do I add the total win and total drop for the entire
week and establish a weekly hold percentage and use multiple weeks to
On Mon, 22 May 2000 13:24:25 +1000, "Glen Barnett"
<[EMAIL PROTECTED]> wrote:
>I assume you're talking about sample standard deviations,
>not population standard deviations (though interpretation
>of what it represents is similar).
>
> ...
>
>Note tha
Glen Barnett wrote:
>
> In article <[EMAIL PROTECTED]>,
> Neil <[EMAIL PROTECTED]> wrote:
> >I was wondering what the standard deviation means exactly?
> >
> >I've seen the equation, etc., but I don't really understand
> >what st dev is
In article <[EMAIL PROTECTED]>,
Neil <[EMAIL PROTECTED]> wrote:
>I was wondering what the standard deviation means exactly?
>
>I've seen the equation, etc., but I don't really understand
>what st dev is and what it is for.
I'm going to take a differ
There are a couple of (practical) features of the standard deviation that are
worth noting.
First, as a *descriptor* of the variation in a distribution, it is generally not
very good. I mean this is the sense that if you want to visualise the amount of
variation in a distribution the SD is only
In article <[EMAIL PROTECTED]>,
Neil <[EMAIL PROTECTED]> wrote:
>I was wondering what the standard deviation means exactly?
>I've seen the equation, etc., but I don't really understand
>what st dev is and what it is for.
>I am not a statistician as you can te
In article <mtAB4.8354$[EMAIL PROTECTED]>,
[EMAIL PROTECTED] says...
>
>My daughter has asked me if there are any tools / software programs that can
>resolve standard deviations, while Excel can determine a standard deviation
>of the Population, what formula is used for the
well, to be honest with you ... i have never heard of these terms before
... i am wondering if your daughter has the term standard deviation mixed
up with perhaps ... percentile rank?
At 01:43 AM 03/21/2000 +, Robert Meyer wrote:
>My daughter has asked me if there are any tools / softw
On Tue, 21 Mar 2000, Robert Meyer wrote:
> My daughter has asked me if there are any tools / software programs
> that can resolve standard deviations; while Excel can determine a
> standard deviation of the Population, what formula is used for the
> (A) 5th Standard Deviatio
Robert A. Meyer asked what is / which software calculates
>(A) 5th Standard Deviation
>(B) 10th Standard Deviation
>(C) 25th Standard Deviation
>(D) 40'th Standard Deviation
and T.S. Lim answered,
> I think you're looking for PERCENTILES.
I would say that there ac
In article <mtAB4.8354$[EMAIL PROTECTED]>,
[EMAIL PROTECTED] says...
>
>My daughter has asked me if there are any tools / software programs that can
>resolve standard deviations, while Excel can determine a standard deviation
>of the Population, what formula is used for the
T
My daughter has asked me if there are any tools / software programs that can
resolve standard deviations, while Excel can determine a standard deviation
of the Population, what formula is used for the
(A) 5th Standard Deviation
(B) 10th Standard Deviation
(C) 25th Standard Deviation
(D) 40&#x
n) from
an omnibus anova, and take the square root of it.
Just curious: What are you planning to do with a pooled standard deviation?
Anna Geyer wrote:
> How do I calculate pooled standard deviation? I have
> study with group of exercisers following forward over
> time. I wan
How do I calculate pooled standard deviation? I have
study with group of exercisers following forward over
time. I want look at weight by category of calorie
intake. I look at standard deviation for weight for
each calorie group but want one overall standard
deviation. Is this valid? Thank
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