# Quants question

Not sure if most of us would find this question simple, I had a problem with this Schweser question… Q - Construct a 90 percent confidence interval for the starting salaries of 100 recently hired employees with average starting salaries of \$50,000 and a standard deviation of \$3,000 assuming the population has a normal distribution. A) 50000 +/- 1.65(3000). B) 50000 +/- 1.65(300). C) 30000 +/- 1.65(5000). D) 50000 +/- 1.65(30000). The correct answer according to Schweser is A. (90% confidence interval is X ± 1.65s = 50000 ± 1.65(3000) = \$45,050 to \$54,950) I thought the answer should have been B. Should we not divide the std deviation by sqrt(100) to get the right interval for a set of observations? Cheers Amit

you would’ve been right if the question was about average salary confidence interval. Confidence interval for average salary is 50,000 ± 1.65*3,000/sqrt(100), confidence interval for the starting salaries is 50,000± 1.65*3,000

We’ve had this before. A C.I. is an interval estimator of a parameter. The phrase “Construct a 90 percent confidence interval for the starting salaries of 100 recently hired employees” just doesn’t make sense, so no wonder the OP was confused.