Question

The travel time between two cities is approximately normally distributed with a standard deviation of 2.25 minutes. You travel 12 times between the two cities, and it takes an average (mean) of 28 minutes. (a) Find a 99% confidence interval for the true mean travel time. (b) If you want for the confidence interval to be no wider than +0.95 minutes, what level of confidence would you need to use?

Answer 1

Part a

Confidence interval for Population mean is given as below:

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

From given data, we have

x̄ = 28

σ = 2.25

n = 12

Confidence level = 99%

Critical Z value = 2.5758

(by using z-table)

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

Confidence interval = 28± 2.5758*2.25/sqrt(12)

Confidence interval = 28 ± 2.5758*0.6495

Confidence interval = 28 ± 1.6731

Lower limit = 28 - 1.6731 = 26.33

Upper limit = 28 + 1.6731 = 29.67

Confidence interval = (26.33, 29.67)

Part b

We are given

E = 0.95

We have

E = Z*σ/sqrt(n)

0.95 = Z*2.25/sqrt(12)

0.95 = Z*0.6495

Z = 0.95/0.6495

Z = 1.462664

P(Z<-1.462664) + P(Z>1.462664) = 2*0.07178 = 0.14356

C = 1 - 0.14356 = 0.85644

Confidence level = 85.64%