Question

A random sample of 13 hotels in a city had an average nighty room rate of $172.10 with a sample standard deviation of $23.90. What is the 99% confidence interval around this sample mean? O A. ($158.86, $185.34) O B. ($151.88, S192.32) OC. ($155.55, $168.65) D. ($165.48, 5178.72)

Answer 1

MAAN Given i a random sample of 13 hotels in a city average room nightly rate of with $142.10 Standard deviation of $ 23.90 . had an a sample ñ 12.10 SD . 23.90 n = 13 0 : 0.01 al = 00112 = 0.005 degrees of freedom = n-1 - 13-1 = 12 Hence, Critical value is to.005.12. - 3.05 C from Standard normal t-table] margin of error, Me t'obleme = 3.05 * 2390 V13 = 3.05 * 6.62866#3449 = 20.2174 : 20.22

99.4. confidence interval for the population mean = ã + ME = 172.10 + 20.22 = 172.10 – 20.22 172.10 + 20:22 = (151.88, 192.32) » Option "B" is correct