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 $32,300, 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

Solution :

Given that,

= 32,300

s = 8,500

n = 28

Degrees of freedom = df = n - 1 = 28 - 1 = 27

At 98% confidence level the t is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

t _/2,df = t_0.01,27 = 2.473

Margin of error = E = t_/2,df * (s /n)

= 2.473 * (8,500 / 28)

= 3973

Margin of error = 3973

The 98% confidence interval estimate of the population mean is,

- E < < + E

32,300 - 3973 < < 32,300+ 3973

28327 < < 36273

(28327, 36273 )