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 33.52 years and the standard deviation is 10.13 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 11.0 years? 34 44 13 37 49 32 35 45 38 43 38 29 47 26 29 14 22 26 34 34 39 48 39 23 20 a) What is the confidence interval? OD Round to two decimal places as needed.) b) What is the margin of error? The margin of error is 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.) O A. The new confidence interval ) is narrower than the interval from part a ) is wider than the interval from part a B. The new confidence interval (-

Explanation / Answer

Solution:- Given values :- n = 25 , mean = 35.52 , sd = 10.13 df = n-1 = 24 , t(0.01,24) = 2.492

a)  95% confidence interval for the population mean is X +/- t * s/sqrt(n)
=> 35.52 +/- 2.492*10.13/sqrt(25)
= 35.52 +/- 5.0488
= (30.47 , 40.57)

b) The margin of error is 5.05

c) => 95% confidence interval for the population mean is X +/- Z * s/sqrt(n)
=> 35.52 +/- 1.96*10.13/sqrt(25)
= (31.55 , 39.49)

option A. the new confidence interval (31.55 , 39.49) is narrower than the interval from part a.

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