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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