Salaries of 40 college graduates who took a statistics course in college have a mean, x , of $ 68,200. Assuming a standard deviation, , of $18,561, construct a 95% confidence interval for estimating the population mean . $___ < < $___ Round to the nearest integer as needed. Solution Confidence Interval CI = x ± Z a/2 * (sd/ Sqrt(n)) Where, x = Mean sd = Standard Deviation a = 1 - (Confidence Level/100) Za/2 = Z-table value CI = Confidence Interval Mean(x)=68200 Standard deviation( sd )=18561 Sample Size(n)=40 Confidence Interval = [ 68200 ± Z a/2 ( 18561/ Sqrt ( 40) ) ] = [ 68200 - 1.96 * (2934.752) , 68200 + 1.96 * (2934.752) ] = [ 62447.887,73952.113 ] = [ 62448 ,73953 ].