View the step-by-step solution to:

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.

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.

Top Answer

a) => (30.11, 35.25) b) margin of error = 2.57 c)... View the full answer

Sign up to view the full answer

Other Answers

[ANSWERS] a. 80% CI = [ 30.033 , 35.327 ] b. ME = 2.647 c. 80%... View the full answer

Why Join Course Hero?

Course Hero has all the homework and study help you need to succeed! We’ve got course-specific notes, study guides, and practice tests along with expert tutors.

-

Educational Resources
  • -

    Study Documents

    Find the best study resources around, tagged to your specific courses. Share your own to gain free Course Hero access.

    Browse Documents
  • -

    Question & Answers

    Get one-on-one homework help from our expert tutors—available online 24/7. Ask your own questions or browse existing Q&A threads. Satisfaction guaranteed!

    Ask a Question
Ask a homework question - tutors are online