The manager of an orchard expects about 70% of his apples to exceed the weight requirement for ""Grade A"" designation. At least how many apples must he sample to be 90% confident of estimating the true proportion within ±4%? A) 19 B) 23 C) 89 D) 356 E) 505

D) 356

Explanation:

ME = Z x √[ (P x Q) / N]

margin of error (ME) = 4% 90% confidence level (Z) = 1.645 (by convention) P = 70% of apples exceed Grade A Q = 30% of apples do not exceed Grade A N = sample size = ?

0.04 = 1.645 x √[ (0.7 x 0.3) / N]

0.04 = 1.645 x √ (0.21 / N)

0.04 = 1.645 x 0.458 / √N

0.04 = 0.7538 / √N

√N = 0.7538 / 0.04 = 18.84

N = 18.84² = 355.2 ≈ 356 (there is no 0.2 apples, you must round up)