Question: Suppose we want a 90% confidence interval for the average amount spent on books by freshmen in their first year at a major university. The
Suppose we want 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, and the amount spent has a normal distribution, with a standard deviation =$30. The number of observations required is closest to?
I know the formula will be (1.645)(30/n)=2. I am having a hard time turning the formula around to figure "n".
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