Question 1197341
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Hi  
 90% confident ⇒  z = 1.645, 
p = .50(Generally .5 is used if p is not given)  ME = .025
How large of a sample size is required?
n = {{{(z/ME)^2 (p(1-p)))}}}
n = {{{(1.645/.025)^2 (.5(.5)))}}} = 1082.41  0r sample Size = 1083 (always round Up)

 = CI	z = value
90%	z =1.645
92%	z = 1.751
95%	z = 1.96
98%	z = 2.326
99%	z = 2.576
Wish You the Best in your Studies.
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