Question

A survey of 25 randomly selected customers found the ages shown​ (in years). The mean is 33.16 years and the standard deviation is 9.33 years.​ a) Construct a 95​% 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?

2727 3939 2626 3838 2424 4343 2121 1414 3737 4949 1313 3939 3434 3939 2626 4141 3030 2828 2828 4646 3232 4040 4343 3535 3737

​a) What is the confidence​ interval?

​(nothing​,nothing​)

​b) What is the margin of​ error?

The margin of error is (Nothing)

​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 ​(nothing​,nothing​)is narrower than the interval from part a.

B.The new confidence interval ​(nothing​,nothing​) is wider than the interval from part a.

Answer 1

a)

sample mean 'x̄=		33.160
sample size    n=		25
std deviation s=		9.330
std error ='sx=s/√n=9.33/√25=		1.8660

for 95% CI; and 24 df, value of t=		2.0639
margin of error E=t*std error    =		3.8512
lower bound=sample mean-E =		29.309
Upper bound=sample mean+E =		37.011
from above 95% confidence interval for population mean =(29.31 <μ<37.01

b)

margin of error E=t*std error    =

3.85

(please try 3.851 if required to 3 decimals)

c)

sample mean 'x̄=		33.160
sample size    n=		25
std deviation σ=		10.000

for 95 % CI value of z=		1.960
margin of error E=z*std error =		3.920
lower bound=sample mean-E=		29.24
Upper bound=sample mean+E=		37.08
from above 95% confidence interval for population mean =(29.24,37.08)

B.The new confidence interval ​ (29.24 , 37.08) is wider than the interval from part a.