Question

Social Networking. A Pew Internet de American Life Project examined Internet social networking. Among a sample of $929$online adults $18-29$years old, $836$said they use social networking sites. Determine a $95\%$ confidence interval for the percentage of all online adults $18-29$ years old who use social networking sites.

Short answer

The $95\%$confidence interval for the proportion of all online adults $18-29$years old who use social networking sites is $(0.879,0.917).$

The confidence range for the percentage of all online adults $18-29$years old who use social networking sites lies between $87.9\%$and $91.7\%$, according to confidence.

Step-by-step solution

Given information

Pew Internet and American life project examined internet social networking. Among a sample of $929$online adults$18-29$years old,$836$said they use social networking sites. Use the one-proportion plus-four$z-$interval procedure to find the required confidence interval. Interpret your results.

Explanation

From the given,

Sample proportion $p^{\text{'}}=\frac{x+2}{n+4}$

$\frac{836+2}{929+4}=0.898$

Here $x$and $n-x$are both $5$or greater, so the one-proportion $z-$interval procedure is appropriate. It has a $95\%$confidence interval, which means $\alpha =0.05$

To find

$Z_{\frac{a}{2}}=Z_{\frac{0.05}{2}}$

$=1.96$

Then the confidence interval is

$p^{\text{'}}-z_{\frac{a}{2}}\sqrt{\frac{p^{\text{'}}\left(1-p^{\text{'}}\right)}{n}}$to $p^{\text{'}}+z_{\alpha /2}\sqrt{\frac{p^{\text{'}}\left(1-p^{\text{'}}\right)}{n}}$

$0.898\pm 1.960\sqrt{\frac{0.898(1-0.898)}{50}}$

$=0.898\pm 1.960(0.009908)$

$=(0.879,0.917)$

As a result, the $95\%$confidence interval for the proportion of all online adults $18-29$years old who use social networking sites is $(0.879,0.917)$.

The confidence range for the percentage of all online adults $18-29$years old who use social networking sites lies between $87.9\%$and $91.7\%$, according to $95\%$confidence.