problem-7.8

problem-7.8  We have a 2 percent margin of error, or
.02 = z* \SE(\phat).
But z* = 1.96 and \SE(\phat) = Ö{.54*(1-.54)}/Ön. Solving gives
> phat = .54; zstar = 1.96
> (zstar * sqrt(phat*(1-phat)) / 0.02)^2
[1] 2386