Question

Video games A Pew Research Center report on gamers and gaming

estimated that $49\%$ of U.S. adults play video games on a computer, TV, game console, or portable device such as a cell phone. This estimate was based on a random sample of $2001$ U.S. adults. Construct and interpret a $95\%$ confidence interval for the proportion of all U.S. adults who play video games.

Short answer

The confidence interval is$(0.468,0.512)$

Step-by-step solution

Given Information

It is given that $\hat{p}=49\%=0.49$

$n=2001$

Concept Used

Formula to be used is$CI=\hat{p}\pm z_{a/2}\times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

Calculation

From normal standard table, at $95\%$confidence interval, $z$score is $1.96$

$95\%$confidence interval is calculated as

$CI=\hat{p}\pm z_{\alpha /2}\times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

$=0.49\pm 1.96\times \sqrt{\frac{0.49(1-0.49)}{2001}}$

$(0.468,0.512)$

Hence, probability is $95\%$that proportion of adults playing video games lies in range$(0.468,0.512)$