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. Provide a 99% confidence interval for the proportion of people living in the United States who have a computer at home.

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Quizplus · Accepted Answer

To calculate a 99% confidence interval for the proportion of people with a computer at home, we use the formula for the confidence interval of a proportion: $p \pm z \sqrt{\frac{p(1-p)}{n}}$, where $p$ is the sample proportion, $z$ is the z-score corresponding to the confidence level (for 99%, $z \approx 2.576$), and $n$ is the sample size. Here, $p = 43128/50000 = 0.86256$, $n = 50000$, and plugging these values into the formula gives us a range, which is closest to the range provided in option B. This calculation accounts for the variability and the confidence level to estimate the true proportion in the population.

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