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 $38,300, with a sample standard deviation of $8,400. (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 = = 38300

sample standard deviation = s = 8400

sample size = n = 28

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

At 98% confidence level

= 1-0.98% =1-0.98 =0.02

/2 =0.02/ 2= 0.01

t _/2,df = t_0.01,27 = 2.47

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

= 2.47* ( 8400/ 28)

Margin of error = E =3921.003

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

- E < < + E

38300 - 3921.003 < < 38300 + 3921.003

34379 < < 42221

(34379,42221)

A 98% confidence interval for the mean account valuation is between $34379 and $42221.