In November 2018, 50,000 people living in the United States were asked whether they had a computer at home. 43,128 answered yes, they had a computer at home. There were 326,766,748 people living in the United States at the time of the survey. Determine an interval with 99% confidence for the number (how many) of people living in the United States who have a computer at home.

Question

Quizplus · Accepted Answer

We can use the formula for a confidence interval for a proportion:

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

where $\hat{p}$ is the sample proportion, $z_{\frac{\alpha}{2}}$ is the critical value from the standard normal distribution for the desired confidence level (in this case, 99% corresponds to $\alpha = 0.01$ and $z_{\frac{\alpha}{2}} = 2.576$), and $n$ is the sample size.

Plugging in the values, we get:

$\hat{p} = \frac{43{,}128}{50{,}000} = 0.86256$

$z_{\frac{\alpha}{2}} = 2.576$

$n = 50{,}000$

$	ext{CI} = 0.86256 \pm 2.576\sqrt{\frac{0.86256(1-0.86256)}{50{,}000}} = 0.85619 	ext{ to } 0.86893$

To find the interval for the number of people with a computer at home, we multiply the lower and upper bounds of this interval by the total US population:

Lower: $0.85619 * 326{,}766{,}748 = 280{,}561{,}930$

Upper: $0.86893 * 326{,}766{,}748 = 283{,}143{,}387$

Therefore, the 99% confidence interval for the number of people living in the United States who have a computer at home is 280,561,930 to 283,143,387. The correct choice is C.

In November 2018, 50,000 People Living in the United States