Question

As part of an annual review of its accounts, a discount brokerage selects a random sample of 28 customers. Their accounts are reviewed for total account valuation, which showed a mean of $36,700, with a sample standard deviation of $8,500. (Use t Distribution Table.) What is a 98% confidence interval for the mean account valuation of the population of customers? (Round your answers to the nearest dollar amount.) 98% confidence interval for the mean account valuation is between $ and $

Answer 1

DATE Answer: We are given that, As part of an annual review of its accounts, a discount brokerage selects a random sample of 28 customers Their allocants are reviewed for total account Ivalaation, which showed a •$36,700, ceith a sample standard deviation of $8,500 What is 98% confidence interval for the mean account valuation of the population of customers? s- $8,500 Here , n=28, P = $36,700, confidence devel 98 % We know that the Confidence interval for Population mean es with unknown population standard deviation is, X e t s where, t - sample mean t-value for given ven confidence level resing t-distribution table. - Sample standard deviation Sample size S Here t-value at 98% confidence level with (n-1)=(28-1)=27 degrees of freedom using t-distribution table is, 2.473

DATE .. Lower bound 10 X - t (m) 36,700 2.473 8,5oo √28 36,70o - (2.473 X 1606. 34901) 36, 700 3972.5011 32 727. 4989 32 72 8. oo. ... Lower bound - $32728.00 Upper bound X tt (5) = 36700 + 2.473 8,500 (28 36.700 + (2.473 X 1606. 34901) 36.700 + 3972.50II 40672.5011 40673.00 upper bound $40673.00 98% confidence interval for the mean a ccount valuation between $ 132728 and $ 40673 is