Question

in a sample of 159,949 first year college students, the national survey of student engagement reported that 39% participated in community service or volunteer work.find the margin of error for 99% confidence.
...idk what formula to use.

Answer 1

how do I find the margin of error for the given values of c,s,and n

c=0.95, s=2.1,n=100

Answer 2

n=159,949
p̂ = 30%
H0: p = p0
Ha: p ≠ p0 or p >p0 or p<p0
C = 0.99, z* = 2.58
m = z* √p̂(1-p̂)/n = 0.0031
p̂ +- m = (0.3869, 0.3931)