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
SRS = simple random sample.
cheers
Michelle
Paul Bellamy [EMAIL PROTECTED] wrote in message
UD2t7.56886$[EMAIL PROTECTED]">news:UD2t7.56886$[EMAIL PROTECTED]...
Thanks alot - what does SRS mean?
Also what does frequentist mean - I have also seen that word?
Dennis Roberts [EMAIL PROTECTED]
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
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%
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
sampling error 2% 1
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
sampling
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
deviation value.
Donald - Thank you for your cogent explanation of a concept that is a bit
hard to grasp.
After researching it more, I determined that there is a gaping hole in my
knowldege relating to the area of inferences on a population proportion so I
am somethat admittedly in the dark and have to study up a
your formula is right on the money, but suppose your problem supplies no
SD - see my recent message in this thread.
Dennis Roberts [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
this is the typical margin of error formula for building a confidence
interval were
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:
this is the typical margin of error formula for building a confidence
interval were the sample mean is desired to be within a certain distance of
the population mean
n = sample size
z = z score from nd that will produce desired confidence level (usually
1.96 for 95% CI)
e = margin of error
? 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 dealing w/a normal distribution?
? How does thi
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 for?
could you express E as a % of a standard deviation .
What's E? The above formula doesn't have a (capital) E.
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
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 distribution?
well, let's see ... e
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