In a survey of 2255 randomly selected US adults (age 18 or older), 1787 of them use the internet...

Question:

In a survey of {eq}2255 {/eq} randomly selected US adults (age 18 or older), {eq}1787 {/eq} of them use the internet regularly. Of the internet users, {eq}1054 {/eq} use social networking sites. Find the {eq}95 \% {/eq} confidence interval of the proportion of all US adults who use a social networking site.

Confidence interval

The confidence interval is the estimation for the population parameters and it provides the lower and upper limits for the true value of population parameters. And the width of the confidence interval increases as the sample size increases.

Given Information

Number of internets using regularly; 1787

Number of social networking sites users; 1054

The value of sample proportion for social networking sites users of all US adults is;

{eq}\begin{align*} \hat P &= \dfrac{X}{n}\\ &= \dfrac{{1054}}{{2255}}\\ &= 0.4674 \end{align*}{/eq}

The value of confidence limit is given as;

{eq}\begin{align*} P\left( {\hat P - {Z_{\alpha /2}}\sqrt {\dfrac{{\hat P\left( {1 - \hat P} \right)}}{n}} < p < \hat P + {Z_{\alpha /2}}\sqrt {\dfrac{{\hat P\left( {1 - \hat P} \right)}}{n}} } \right) &= 0.95\\ P\left( {0.4674 - 1.95\sqrt {\dfrac{{0.4674\left( {1 - 0.4674} \right)}}{{2255}}} < p < 0.4674 - 1.95\sqrt {\dfrac{{0.4674\left( {1 - 0.4674} \right)}}{{2255}}} } \right) &= 0.95\\ P\left( {0.44692 < \mu < 0.48789} \right) &= 0.95 \end{align*}{/eq}

Therefore, the 95% confidence interval for the true proportion of all US adults who use a social networking site is (0.44692, 0.48789).