Question

In a random sample of 10 computers, the mean repair cost was $80 and the sample standard dev was $15. Use a 95% Confidence level.

What is the margin of error (E)?

What is the confidence interval?

Answer 1

sample size is n = 10

degree of freedom df= n-1 = 10 - 1 = 9

alpha = 1 - confidence level = 1 -0.95 = 0.05

using t distribution table for df = 9 and alpha = 0.05

t critical value = 2.262

Margin of error =

where t = 2.262, s= 15 and n = 10

Confidence interval =

where ME = 10.730 and x(bar) = 80

this implies

=(80-10.730, 80+10.730)

= (69.27, 90.73) (rounded to 2 decimals)