Suppose a 99 confidence interval for the mean salary of coll

Suppose a 99% confidence interval for the mean salary of college graduates in a town in Mississippi is given by [$37,469, $47,531]. The population standard deviation used for the analysis is known to be $15,600.

Solution

Confidence interval = [$37,469, $47,531

Point estimate of mean = mid point of confidence interval = 42,500

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Margin of error = 47,531-42,500 = 5,031

Margin of error can also be written as 2.58*std error as for 99% z alpha/2 =2.58

Hence 2.58 * std error = 5.031

std error = 1.95

Std error is also equal to sigma/rtn

So 15600/ rt n = 1.95

rt n = 8000

n = 64,000,000