Question

Last year, Ashton Accounting conducted 214

corporate audits. For the annual report, Julie takes

a random sample of 20 of these audits and finds the

mean billable hours for the sample was 1680.4 with

a standard deviation of 235 hours. What is a 90%

confidence interval for the true mean billable hours

for all corporate audits last year?

Answer 1

Given that,

= 1680.4

s =235

n = 20

Degrees of freedom = df = n - 1 =20 - 1 = 19

At 90% confidence level the t is ,

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

t _/2,df = t_0.05,19 =   1.729    ( using student t table)

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

= 1.729* (235 / 20) = 90.8548

The 90% confidence interval estimate of the population mean is,

- E < < + E

1680.4 -90.8548 < < 1680.4+ 90.8548

1589.5452< < 1771.2548

( 1589.5452, 1771.2548 )