Question 1127983

For the standard normal distribution there is a {{{95}}}% probability that a standard normal variable, {{{Z}}}, will fall between {{{-1.96}}} and {{{1.96}}}. 

In other words,

{{{P(-1.96 <Z< 1.96)=0.95}}}

{{{p-hat =0.23}}}
{{{n =455}}}

{{{p-hat+-Z*sqrt(p-hat(1-p-hat)/n)}}}

{{{0.23+-1.96*sqrt((0.23(1-0.23))/455)}}}

{{{0.23+-1.96*sqrt((0.23*0.77)/455)}}}

{{{0.23+-1.96*sqrt(0.1771/455)}}}

{{{0.23+-1.96*sqrt(0.00038923076923)}}}

{{{0.23+-1.96*0.019728932288}}}

{{{0.23+-0.03866870728448}}}

so,

{{{0.23+0.03866870728448= 0.2686679968}}}

{{{0.23-0.03866870728448= 0.1913320032}}}


Confidence Interval is:  ({{{0.1913320032}}},{{{ 0.2686679968}}})