A survey of 25 randomly selected customers found the ages shown (in years). The mean is 32.68 years and the standard deviation is 10.04 years.
18 33 49 31 22
49 30 27 27 12
33 36 31 44 39
42 37 26 23 25
45 39 48 21 30
a) Construct a 80% confidence interval for the mean age of all customers, assuming that the assumptions and conditions for the confidence interval have been met.
b) How large is the margin of error?
c) How would the confidence interval change if you had assumed that the standard deviation was known to be 11.0 years?
a) What is the confidence interval? (Round to two decimal places as needed.)
b) What is the margin of error? (Round to two decimal places as needed.)
c) What is the confidence interval using the given population standard deviation? Select the correct choice below and fill in the answer boxes within your choice.
(Round to two decimal places as needed.)
A. The new confidence interval is ? is wider than the interval from part a.
B. The new confidence interval is ? is narrower than the interval from part a.
a) => (30.11, 35.25) b) margin of error = 2.57 c)... View the full answer