conduct a study on support of ban on smartphones in class. 68/87 said yes. How large of a sample is needed to be sure the margin of error +/- 3.5% with 99% confidence. I know this is estimation...but I am not sure how to go about solving this problem.
99% confidence interval
z = 2.575
p hat = 68/87 =0.7816
E= Za/2 *sqrt((phat(1-phat)/n))
E = 2.575*sqrt(0.7816*0.2184/87)
E = 0.1141
phat-E < P < Phat+ E
0.7816-0.1141 <p< .7816+.1141
0.6675 < p < 0.8957