Question 1176083
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Hi
Using p = .5 as an example when mean not known.
99.9% confident ( z = 3.09 ) invNorm(.999)
 within 4% 
z = 3.29 (two tailed)
n = {{{(z/ME)^2 (p(1-p)))}}}
n = {{{(3.09/.04)^2 (.5(.5)))}}} = 1492  (Rounded up to nearest  whole number)

Wish You the Best in your Studies.
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