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I was told that to calculate an approximate sample size, let n =[Z/2ME]^2; z=1.96 for a confidence level of .95.  Then n[1.96/2(.03)]^2 = 1,067.

My question is 1) where did we get ME from? 2) is this equation correct for calculating an approximate sample size?

thanks,
Wing

First, in words, this problem is using the "normalized z-value" to obtain the sample size needed to make sure that the calculated mean (based on the samples) is within a specified interval around the population mean at a given confidence interval and given a known population standard deviation.

It is easier to see what's going on if we write down the formula for the normalized z-value, also known as the z-score:

let σ = population standard deviation (assumed given)

then σ_m = σ/√n = standard deviation of the mean m calculated using n samples

∆m = m - m_pop = difference between the calculated mean and m_pop = the given population mean.

Then, we are given that ∆m has to be small enough so that 95% of the time t will be smaller than

z = ∆m/σ_m.

This corresponds to z = ±1.96.

Rearranging these expressions gives

n = [ (z･σ)/∆m ]^2.

Comparing with your formula,it looks like

σ/∆m = 1/(2ME) = 1/[ (2/(0.03) ]

This could mean that ME = ∆m = 0.03 and that σ = 1/2.

If this is true, then this is a correct method for calculating the sample size (to be correct, the method also requires a reasonably large sample size so that the statistics are approximately normal, which seems to be the case).

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#### randy patton

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college mathematics, applied math, advanced calculus, complex analysis, linear and abstract algebra, probability theory, signal processing, undergraduate physics, physical oceanography

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