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.

A) .862546 to .862573
B) .8586 to .8665
C) .8595 to .8656
D) .862548 to .862578

Question

Donia Bedewi · Accepted Answer

Final Answer : B, Explanation :   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 ± z p ( 1 − p ) n p \pm z\sqrt{\frac{p(1-p)}{n}} p ± z n p ( 1 − p ) ​ ​  , where   p p p   is the sample proportion,   z z z   is the z-score corresponding to the confidence level (for 99%,   z ≈ 2.576 z \approx 2.576 z ≈ 2.576  ), and   n n n   is the sample size. Here,   p = 43128 50000 = 0.86256 p = \frac{43128}{50000} = 0.86256 p = 50000 43128 ​ = 0.86256  ,   n = 50000 n = 50000 n = 50000  , and plugging these values into the formula gives us the interval   [ 0.8586 , 0.8665 ] [0.8586, 0.8665] [ 0.8586 , 0.8665 ]  , which corresponds to choice B.

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