solution:-
given that from the above information
margin of error E = 2% = 0.02
the value of 90% confidence from z table is z = 1.645
sample size n = (z/E)^2 * (p1(1-p1)+p2(1-p2))
(a) here given that p1 = 0.215 , p2 = 0.181
=> n = (1.645/0.02)^2 * (0.215*(1-0.215) + 0.181*(1-0.181))
=> n = 2145
(b) here estimate is not given
so we take p1 = 0.5 and p2 = 0.5
=> n = (1.645/0.02)^2 * (0.5*(1-0.5) + 0.5*(1-0.5))
=> n = 3383
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