In a survey of 3203 adults, 1447 say they have started paying bills online in the last year. Construct a​ 99% confidence interval for the population proportion. Interpret the results. A​ 99% confidence interval for the population proportion is ??

(0.4291, 0.4743)

Step-by-step explanation:

Using the relation :

p ± Zcritical * Sqrt[(p(1-p)) / n]

P = x / n =. 1447 / 3203 = 0.4517

1 - p = 0.5483

Zcritical at 99% = 2.575

Sqrt[(p(1-p)) / n] = sqrt(0.4517(0.5483)) / 3203) = 0.008793

p ± Zcritical * 0.008793

Lower boundary = 0.4517 - (2.575 * 0.008793) = 0.4291

Upper boundary = 0.4517 + (2.575 * 0.008793) = 0.4743

(0.4291, 0.4743)

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answer: i think it's a

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