Question

The national average cost of cable TV bills was reported as $99. You collect a random sample of 75 people in your town and find the average cable bill is $103 with a standard deviation of $13.50. You want to compute a 99% confidence interval estimate for the mean difference between the national average and your town average. What is 99% confidence interval?

Answer 1

Solution:

Confidence interval for Population mean is given as below:

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

From given data, we have

Xbar = 103

S = 13.5

n = 75

df = n – 1 = 74

Confidence level = 99%

Critical t value = 2.6439

(by using t-table)

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

Confidence interval = 103 ± 2.6439*13.5/sqrt(75)

Confidence interval = 103 ± 4.1215

Lower limit = 103 - 4.1215 = 98.88

Upper limit = 103 + 4.1215 = 107.12

Lower limit = 98.88

Upper limit = 107.12