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  = $30.How many observations are required to achieve this margin of error?

A) 25
B) 608
C) 609
D) 865

Question

Ahmed Soliman · Accepted Answer

Final Answer : C, Explanation :  The margin of error is given by:Margin of Error = z-value * (standard deviation / sqrt(sample size))We are given that the margin of error is $2, and we want a 90% confidence interval. From the Normal distribution, we know that a 90% confidence interval corresponds to a z-value of 1.645.Plugging in the given values, we get:2 = 1.645 * (30 / sqrt(n))Solving for n, we get:n = (1.645^2 * 30^2) / 2^2 = 608.71Since we cannot have a fractional number of observations, we must round up to the nearest whole number. Therefore, we need at least n = 609 observations. Therefore, the correct answer is C.

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