Question

a random sample of 100 homes in the Bakersfield area showed an average value of $253,00 with a standard deviation of $11,725. We can say with 99% confidence that the mean proce of a home in Bakersfield, CA, is between _____ and ____ with a margin of error ____.

Answer 1

Sample size (n) = 100

Mean = $25300

Standard deviation = $11725

Confidence level = 99%

Z value for 99% confidence level = 2.58

Required confidence interval = Mean + z value × St. Dev/sqrt n

= 25300 + 2.58 × 11725/√100

= 25300 + 3025

Required confidence interval = ($22275, $28325)

Standard error = z value × standard deviation /sqrt n

= 2.58 × 11725/ √100

= 2.58 × 11725/10 = $3025