Advanced Math/Stat Question
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?
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).