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A surveyor wants to estimate the average income of daily users of a transit service. The surveyor would like to estimate this

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Answer #1

(1- α)*100% confidence interval the margin of error=z(α /2)*sd/sqrt(n)

90% confidence interval margin of error=z(0.1/2)*sd/sqrt(n)

or, 1000=1.645*30000/sqrt(n)

or, sqrt(n)=1.645*30000/1000=49.35

n=2435.4 (2436 , next whole number)

answer is 2436

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