Question

A survey of 25 randomly selected customers found the ages shown​ (in years). The mean is 32.88 years and the standard deviation is 8.96 years.

​a) Construct a 90​% 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 9.0 ​years?

Answer 1

Part a

Here, we have to find the 90% confidence interval for the population mean. The confidence interval formula is given as below:

Confidence interval = Xbar ± t*S/sqrt(n)

We are given

Sample mean = Xbar = 32.88

Sample standard deviation = S = 8.96

Sample size = n = 25

Confidence level = C = 90% = 0.90

Degrees of freedom = n – 1 = 25 – 1 = 24

Critical t-value = 1.7109 (by using t-table)

Confidence interval = 32.88 ± 1.7109*8.96/sqrt(25)

Confidence interval = 32.88 ± 1.7109* 1.792

Confidence interval = 32.88 ± 3.0659

Lower limit = 32.88 - 3.0659 = 29.81

Upper limit = 32.88 + 3.0659 = 35.95

Confidence interval = (29.81, 35.95)

Part b

Margin of error = t*S/sqrt(n)

Margin of error = 1.7109*8.96/sqrt(25)

Margin of error = 1.7109*1.792

Margin of error = 3.0659

Part c

Here, we have to find the 90% confidence interval for the population mean. Here, we know the population standard deviation as σ = 9

Confidence interval = Xbar ± Z*σ/sqrt(n)

We are given

Sample mean = Xbar = 32.88

Population standard deviation = σ = 9

Sample size = n = 25

Confidence level = C = 90% = 0.90

Critical Z-value = 1.6449 (by using Z-table)

Confidence interval = 32.88 ± 1.6449*9/sqrt(25)

Confidence interval = 32.88 ± 1.6449*1.8

Confidence interval = 32.88 ± 2.9607

Lower limit = 32.88 – 2.9607 = 29.92

Upper limit = 32.88 + 2.9607 = 35.84

Confidence interval = (29.92, 35.84)