<PRE><B>Question 203867</B>
If the inspection division of a county weights and measures department wants to estimate the mean amount of soft-drink fill in 2-liter bottles to within +/- 0.01 liter with 95% confidence and also assumes that the standard deviation is 0.05 liter, what sample size is needed?
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Since E = z*[s/sqrt(n)]
sqrt(n) = z*[s/E]
n = [z*s/E]^2
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Your Problem:
n = [1.96*0.05/0.01]^2
n = 96.04
Rounding up you get n = 97 (this is the required sample size)
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Cheers,
Stan H.
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