A survey of 25 randomly selected customers found the ages shown (in years). The

Question

A survey of 25 randomly selected customers found the ages shown (in years). The mean is 33.44 years and the standard deviation is 9.99 years. a) Construct a 98% 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? a) What is the confidence interval? (OD) (Round to two decimal places as needed.)

Explanation / Answer

Mean is 33.44 and s is 9.99

if n is 25 then standard error is SE=s/sqrt(N)=9.99/sqrt(25)=1.998

for 98% confidece, the z is 2.33

a) lower limit of the confidence interval is mean-SE*z=33.44-1.998*2.33 =28.7847

upper limit is mean+SE*z=33.44+1.998*2.33 =38.095

b) margin of error is SE*z=1.998*2.33=4.655

c) the standard error for standard deviation of 10 would give standard error of 10/sqrt(25)=2

thus confidence interval 's lower limit is mean-SE*z =33.44-2*2.33=28.78

upper limit is mean+SE*z=33.44+2*2.33 =38.1

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