An economist is interested in studying the incomes of consum

An economist is interested in studying the incomes of consumers in a particular region. The population standard deviation is known to be $1,000. A random sample of 50 individuals resulted in an average income of $15,000. What sample size would the economist need to use for a 95% confidence interval if the width of the interval should not be more than $100?

Solution

The formula for the confidence interval is xbar +- E, where E = z * sigma/sqrt(n) Because you both add and subtract E, the length of the confidence interval is 2E. So if you want the length of the confidence interval to be no more than 100, then 2E < 100 E < 50 z * sigma/sqrt(n) < 50 1.96 * 1,000/sqrt(n) < 50 1,000/sqrt(n) < 50/1.96 = 25.51 1/sqrt(n) < 25.51/1000 = 0.02551 1 < 0.02551sqrt(n) sqrt(n) > 1/0.02551 = 39.2 n > 39.2^2 = 1536.64, which should be rounded up to 1537. **************************************… The general formula for this is n = (z*sigma/E)^2 n = (1.96*1000/50)^2 = 1536.64, which again is rounded up to 1537.