Question 1051022
p = 120/300 = .4
Find a 95% confidence interval estimate of population proportion of adults who support more gun control laws. 
x = 1.96
ME = {{{z*sqrt((p(1-p))/n)}}}
ME = {{{1.96*sqrt((.4*.6)/300)}}} = .055

.4 ± .055
We are 95% confident that the population proportion of adults who support more gun control laws is:
between .345 and .455 
0r
between
34.5%  and 45.5%