Question

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

Step 1. Given information

Sample proportion $\left(\hat{p}\right)=49\%=0.49$

Sample size $\left(n\right)=2001$

Step 2. Formula used

The formula to compute the confidence interval for population proportion is:

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

Step 3. Calculation

From the standard normal table, the z-score at $95\%$confidence level is $1.96$.

The $95\%$confidence interval for the proportion of adult that play video games can be calculated as:

role="math" localid="1663923507517" $$\begin{gathered} CI=\hat{p}\pm z_{\alpha /2}\times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \\ =0.49\pm 1.96\times \sqrt{\frac{0.49(1-0.49)}{2001}} \\ =(0.468,0.512) \end{gathered}$$

Thus, the confidence interval is $(0.468,0.512)$

There is $95\%$ probability that the proportion of adult that play video games lies between $0.468$ and$0.512$.