Question

In a survey of 3105 ?adults, 1467 say they have started paying bills online in the...

In a survey of 3105 ?adults, 1467 say they have started paying bills online in the last year. Construct a? 99% confidence interval for the population proportion. Interpret the results.

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Answer #1

Both answer and interpretation to the question is below. I've also included the formuale and calculation as well. Please don't hesitate to give a "thumbs up" to the answer, in case you're satisfied with the answer.

A 99% Confidence interval will have a Z value of 2.575 (from the Z tables)

The sample proportion is, p^= x/n = 1467/3105 = .4725

Therefore, the 99% Interval should look like:

p^ +/- Z*sqrt(p^*p^'/n)

= .4725 +/- 1.96*sqrt( .4725*.5275/3105)

Answer to the question is as follows: 0.4550 to 0.4901

Interpertation: There is a 95% chance that the true population proportion will lie between .4550 and .4901

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