Question

A survey of 25 randomly selected customers found the ages shown​ (in years). The mean is 32.04 years and the standard deviation is 9.87 years.

​a) Construct a 80% 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?

Answer 1

Solution :

Given that,

= 32.04

a ) s = 9.87

n = 25

Degrees of freedom = df = n - 1 = 25 - 1 = 24

At 80% confidence level the t is ,

= 1 - 80% = 1 - 0.80 = 0.20

/ 2 = 0.20 / 2 = 0.10

t _/2,df = t_0.10,24 =1.318

b ) Margin of error = E = t_/2,df * (s /n)

= 1.318 * (9.87 / 25)

= 2.60

Margin of error = 2.60

The 80% confidence interval estimate of the population mean is,

- E < < + E

32.04 - 2.60 < < 32.04 + 2.60

29.44 < < 34.64

(29.44, 34.64 )

c ) s = 10.0

n = 25

Degrees of freedom = df = n - 1 = 25 - 1 = 24

At 80% confidence level the t is ,

= 1 - 80% = 1 - 0.80 = 0.20

/ 2 = 0.20 / 2 = 0.10

t _/2,df = t_0.10,24 =1.318

Margin of error = E = t_/2,df * (s /n)

= 1.318 * (10.0 / 25)

= 2.63

Margin of error = 2.63

The 80% confidence interval estimate of the population mean is,

- E < < + E

32.04 - 2.63 < < 32.04 + 2.63

29.40 < < 34.67

(29.40, 34.67 )