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 32.52 years and the standard deviation is 9.09 39 15 27 43 250 years. 30 15 39 38 32 a) Construct a 99% confidence interval for the mean age of all customers, assuming that the assumptions and conditions for the confidence 46 30 44 22 40 interval have been met. 40 22 24 35 39 b) How large is the margin of error? 27 44 34 24 39 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? O.O) (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. OB. The new confidence interval () is wider than the interval from part a.

Explanation / Answer

The statistical software output for this problem is:

One sample T summary confidence interval:
: Mean of population

99% confidence interval results:

Hence,

a) Confidence interval: (27.44, 37.60)

b) Margin of error = (37.60 - 27.44)/2 = 5.08

c) Option B; 27.37; 37.67

Mean Sample Mean Std. Err. DF L. Limit U. Limit 32.52 1.818 24 27.435164 37.604836

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.