In: Statistics and Probability
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 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? |
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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.
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.