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.

Answer 1

Using the z-distribution, it is found that the 95% confidence interval for the proportion of all U.S. adults who play video games is (0.4681, 0.5119). It means that we are 95% sure that the true proportion of all U.S. adults who play video games is between 0.4681 and 0.5119.

What is a confidence interval of proportions?

A confidence interval of proportions is given by:

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which:

• $\pi$ is the sample proportion.

• z is the critical value.

• n is the sample size.

In this problem, we have that:

• 95% confidence level, hence$\alpha = 0.95$, z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$, so $z = 1.96$.

• 49% out of 2001 U.S. adults play video games, hence $\pi = 0.49, n = 2001$.

The lower bound of the interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.49 - 1.96\sqrt{\frac{0.49(0.51)}{2001}} = 0.4681$

The upper bound of the interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.49 + 1.96\sqrt{\frac{0.49(0.51)}{2001}} = 0.5119$

The 95% confidence interval for the proportion of all U.S. adults who play video games is (0.4681, 0.5119). It means that we are 95% sure that the true proportion of all U.S. adults who play video games is between 0.4681 and 0.5119.

To learn more about the z-distribution, you can take a look at brainly.com/question/25730047