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.

A) 281,851,351 to 281,860,174
B) 280,856,020 to 282,849,297
C) 280,561,930 to 283,143,387
D) 281,852,005 to 281,861,808

To find the interval with 99% confidence, we will use the following formula:

$\text{Confidence Interval} = \text{Sample Estimate} \pm \text{Margin of Error}$

The sample estimate is the proportion of people who have a computer at home, which is:

$\hat{p} = \frac{\text{Number of people who have a computer at home}}{\text{Total population surveyed}} = \frac{43,128}{50,000} = 0.86256$

The margin of error can be found using the formula:

$ME = z^* \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

where $z^*$ is the critical value from the standard normal distribution corresponding to a 99% confidence level, which is 2.576, and $n$ is the sample size (50,000). Plugging in the values, we get:

$ME = 2.576 \times \sqrt{\frac{0.86256(1-0.86256)}{50,000}} \approx 0.007$

Therefore, the interval estimate is:

$0.86256 \pm 0.007$

To find the number of people living in the United States who have a computer at home, we multiply the interval estimate by the total population:

$326,766,748 \times (0.86256 \pm 0.007) \approx 280,561,930$ to $283,143,387$

Therefore, the best choice is C, which is the interval estimate calculated above.