Question

A survey of 25 randomly selected customers found the ages shown​ (in years). The mean is 32.32years and the standard deviation is 9.77years.

​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 10.0 years?

Answer 1

(A) Given that

sample size n = 25

mean (xbar) = 32.32

standard deviation (s) = 9.77

degree of freedom (df) = n-1

= 25-1

= 24

alpha level = 1-confidence level

= 1-0.80

= 0.20

t critical = 1.32 (using t distribution table for df = 24 and alpha = 0.20)

confidence interval =

(B) Margin of error =

= 2.58 (already calculated in confidence interval calculation)

(C) If the standard deviation was known to be 10.0 years, then we have to use z distribution instead of t distribution.

z critical value for 80% confidence level is 1.28

Therefore, confidence interval will be narrower because the margin of error will get decreased due to lower critical value