Question

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 January, a sample of 16 days has a sample mean of 48 rooms. This information is used to calculate an interval estimate for the population mean to be from 40 to 56 rooms. What is the level of confidence of this interval? Report the level of confidence as a percentage and use 2 decimal places.

Answer 1

Solution:

Given: The population of rooms rented daily is assumed to be normally distributed for each month with a standard deviation of 24 rooms.

Thus  

Sample Size = n 16

Confidence Interval estimate is: ( 40 , 56 )

We have to find the level of confidence of this interval.

First find E = Margin of Error

E = ( Upper Limit - :Lower Limit ) / 2 = ( 56 - 40 ) / 2 = 16 / 2 = 8

Now use following formula:

Look in z table for z = 1.3 and 0.03 and find corresponding area.

Area = 0.9082

Now use following formula to find c:

Area = ( 1 + c ) / 2

0.9082 = ( 1 + c) / 2

2*0.9082 = 1 +c

1.8164 = 1 +c

c = 1.8164 - 1

c = 0.8164

c = 81.64 %

Thus  the level of confidence of this interval is  c = 81.64 %