Question 1139079
assume random sample, and numbers and probability are sufficiently large to use normal approximation.
The point estimate is 1073/2383=0.450
the half-interval is z*sqrt(p*1-p)/n)=1.96*sqrt(0.45*0.55/2383)
0.45+/-1.96(1.01), the half-interval is 0.02
the whole interval is (0.43, 0.47)



The half-interval is z0.995*sqrt (p*(1-p)/n)
z is 2.576
p and 1-p are 0.5
the half-interval equals the error of 0.03
square both sides and z^2*p(1-p)/n=0.0009
so 6.64*0.25=0.0009n
n=1844.44 or 1845