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Suppose we wish to calculate a 90% confidence interval for the average amount spent on books...

Suppose we wish to calculate a 90% confidence interval for the average amount spent on books by freshmen in their first year at a major university. The interval is to have a margin of error of $2. Assume that the amount spent on books by freshmen has a Normal distribution with a standard deviation of U = $30.How many observations are required to achieve this margin of error?

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Answer #1

sample size = ( Z * \sigma / E )2

= ( 1.645 * 30 / 2 )2

= 608.86

Sample size = 609 (Rounded up to nearest integer)

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