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Find an estimate of the sample size needed to obtain a margin of error of 33 for the 95% confidence interval of a population

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Answer #1

To find sample size (n) such that

Margin of error = 33

= 33 -* 21-0.95 n 2

33 = * 20.025 n

400 1.95996 33 /n

Vn- 400* 1.95996 33 23.7571

n 23.75712-564.399

n565

ans-> D) 588

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