Question

A survey of 25 randomly selected customers found the ages shown​ (in years). The mean is 32.24 years and the standard deviation is 9.55 years.​ 30 41 48 36 20 36 37 36 26 27 17 23 39 35 21 10 31 42 32 41 43 28 47 25 35 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 10.0 ​years?

Answer 1

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.

Confidence interval for Population mean is given as below:

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

From given data, we have

Xbar = 32.24

S = 9.55

n = 25

df = n – 1 = 24

Confidence level = 90%

Critical t value = 1.7109

(by using t-table)

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

Confidence interval = 32.24 ± 1.7109*9.55/sqrt(25)

Confidence interval = 32.24 ± 3.2678

Lower limit = 32.24 - 3.2678 =28.9722

Upper limit = 32.24 + 3.2678 = 35.5078

Confidence interval = (28.97, 35.51)

​b) How large is the margin of​ error?

Formula for margin of error is given as below:

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

Margin of error = 1.7109*9.55/sqrt(25)

Margin of error =3.2678

c) How would the confidence interval change if you had assumed that the standard deviation was known to be 10.0 ​years?

When we know the value for the population standard deviation, then we need to use z or normal distribution instead of t-distribution for the calculation of confidence interval.

Confidence interval for Population mean is given as below:

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

From given data, we have

Xbar = 32.24

σ = 10

n = 25

Confidence level = 90%

Critical Z value = 1.6449

(by using z-table)

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

Confidence interval = 32.24 ± 1.6449*10/sqrt(25)

Confidence interval = 32.24 ± 3.2897

Lower limit = 32.24 - 3.2897 = 28.9503

Upper limit = 32.24 + 3.2897 = 35.5297

Confidence interval = (28.95, 35.53)

There is very small difference in the confidence intervals for the part (a) and (c).