Question

As part of an annual review of its accounts, a discount brokerage selects a random sample of 27 customers. Their accounts are reviewed for total account valuation, which showed a mean of $39,900, with a sample standard deviation of $8,300. (Use t Distribution Table.) What is a 99% confidence interval for the mean account valuation of the population of customers? (Round your answers to the nearest dollar amount.)

Answer 1

Solution :

Given that,

= 39900

s = 8,300

n = 27

Degrees of freedom = df = n - 1 = 27 - 1 = 26

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t /2,df = t0.005,26 =2.779

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

= 2.779 * (8,300 / 27)

= 4439

Margin of error = 4439

The 99% confidence interval estimate of the population mean is,

- E < < + E

39900 - 4439 < < 39900 + 4439

35461 < < 44339

(35461, 44339)