One is supplied here with a value for the sample proportion p (37% or 0.37) and a Margin Of Error E
(2% or 0.02) and can gain (1 − p) as 0.63 as well as a confidence level of z0.96 which corresponds to
2.05 in the Normal Distribution.
He can then follow the formula n = p(1 − p)zc2 ÷ E2
to reach n equal to 0.37(1 − 0.37)(2.05)2 divided by 0.022.
This gives the minimum sample size sought as 2449.006875 which is rounded forward to 2450.