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 $33,400, with a sample standard deviation of $8,650. (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,

Point estimate = sample mean =   = $33400

sample standard deviation = s = $8650

sample size = 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 * (8650 / 28)

= 4043

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

- E < <   + E

33400 - 4043 < < 33400 + 4043

29357 < < 37443

(29357,37443)

98% confidence interval for the mean account valuation is between $29357 and $37443