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]">news:9orn26$m80$[EMAIL PROTECTED]...
> 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 with the given confidence level (90% in
> this case)
> p = the proportion of successes (bruised apples in this case)
> e = the maximum error (4% in this case)
>
> Your problem is you don't know p since that is what you are trying to
> estimate. Ask yourself what value of p will make n a large as possible and
> then you can use this "worst case" solution.
>
> Another solution would be to estimate (roughly) the maximum value of p
(10%,
> 20% ...) and use it to find n. Whatever you do you should read up on
> "Calculating sample sizes for estimating proportions" in any good basic
> statistics text.
>
> GB
>
>
>
> "@Home" <[EMAIL PROTECTED]> wrote in message
> b25s7.46471$[EMAIL PROTECTED]">news:b25s7.46471$[EMAIL PROTECTED]...
> > 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?
> >
> >
> >
>
>
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