(10 points) Starting salaries of 64 college graduates who have taken a statistics course have a mean of $42,500 with a standard deviation of $6,800. Find an 90% confidence interval for ????. (NOTE: Do not use commas or dollar signs in your answers. Round each bound to three decimal places.) Lower-bound: 41101.87 Upper-bound: 43898.13

Answer:41101.750 to 43898.250Step-by-step explanation:Using this formula X ± Z (s/√n)WhereX = 42500 --------------------------MeanS = 6800----------------------------- Standard Deviationn = 64 ----------------------------------Number of observationZ = 1.645 ------------------------------The chosen Z-value from the confidence table belowConfidence Interval Z80%. 1.28285% 1.44090%. 1.64595%. 1.96099%. 2.57699.5%. 2.80799.9%. 3.291Substituting these values in the formulaConfidence Interval (CI) = 42500 ± 1.645(6800/√64)CI = 42500 ± 1.645(6800/8)CI = 42500 ± 1.645(850)CI = 42500 ± 1398.25CI = 42500+1398.25 ~. 42500-1398.25CI = 43898.25 ~ 41101.75In other words the confidence interval is from 41101.750 to 43898.250