A hotel chain wants to estimate the average number of rooms rented daily in each month. The population of rooms rented daily is assumed to be normally distributed for each month with a standard deviation of 24 rooms. During June, a sample of 16 days has a sample mean of 48 rooms. This information is used to calculated an interval estimate for the population mean to be from 40 to 56 rooms. What is the level of confidence of this interval?

Answer:81.64%Step-by-step explanation:Data provided in the question: Standard deviation, s = 24 roomsSample size, n = 16 daysMean = 48 roomscalculated an interval = 40 to 56Now,Level of confidence = 100(1 - α)%Confidence interval = Mean ± [tex]z\frac{s}{\sqrt{n}}[/tex]Thus,Mean - [tex]z\frac{s}{\sqrt{n}}[/tex] = 40and,Mean + [tex]z\frac{s}{\sqrt{n}}[/tex] = 56or48 + [tex]z\times\frac{24}{\sqrt{16}}[/tex] = 56or[tex]z\times\frac{24}{4}[/tex] = 8orz = 1.33 for z = 1.33 , we have [tex]\frac{\alpha}{2}[/tex] = 0.918  [from standard z table]orα = 0.1836Therefore,Level of confidence = 100(1 - 0.1836)%orLevel of confidence = 81.64%