Question

Refrigerators in a random sample of 55 refrigerators, the mean repair cost was $150.00 and the standard deviation was $15.50. Construct a 90% confidence interval for the population mean repair cost. Interpret the results.

Answer 1

146.5619<x<153.4381

Explanation:

The formula for calculating the confidence interval is expressed as:

CI = xbar±(z*s/√n)

xbar is the mean = 150

s is the standard deviation = 15.50

n is the sample size = 55

z is the z score at 90% CI = 1.645

Substitute and get CI

CI = 150±(1.645*15.50/√55)

CI = 150±(1.645*15.50/7.4162)

CI = 150±(1.645*2.09)

CI = 150±(3.4381)

CI = (150-3.4381, 150+3.4381)

CI = (146.5619, 153.4381)

CI = 146.5619<x<153.4381

Hence a 90% confidence interval for the population mean repair cost 146.5619<x<153.4381

The minimum cost is approximately $147 while the maximum cost of the refrigerator is $153