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A reputable polling organization in a certain country surveyed 106,300 ​adults, and 18​% of those polled...

A reputable polling organization in a certain country surveyed 106,300 ​adults, and 18​% of those polled reported that they smoked.

1. Calculate the margin of error for the proportion of all adults who smoke with 99​% confidence.

ME = [ ]? Round to four decimals as needed

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Answer #1

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given : n=106300 p = 0.18 q= 1-p = 0.82

Margin of error = Z * Sqrt ( p ( 1 - p) / n)

Z for 99% confidence is 2.576

Margin of error = 2.576 *sqrt(0.18*0.82 / 106300 )

= 0.00303

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