Exercise 1123 Algorithmic Because of staffing decisions mana

{Exercise 11.23 (Algorithmic)}

Because of staffing decisions, managers of the Gibson-Marimont Hotel are interested in the variability in the number of rooms occupied per day during a particular season of the year. A sample of 24 days of operation shows a sample mean of 280 rooms occupied per day and a sample standard deviation of 33 rooms.

What is the point estimate of the population variance?
______   

Provide a 90% confidence interval estimate of the population variance (to 1 decimal).
( ______ , ______ )

Provide a 90% confidence interval estimate of the population standard deviation (to 1 decimal).
( ______ , ______ )

Solution

What is the point estimate of the population variance?
______   

As s = 33

then

variance = s^2 = 33^2 = 1089 [answer]

**************************

Provide a 90% confidence interval estimate of the population variance (to 1 decimal).
( ______ , ______ )

As              
              
df = n - 1 =    23          
alpha = (1 - confidence level)/2 =    0.05          
              
Then the critical values for chi^2 are              
              
chi^2(alpha/2) =    35.17246163          
chi^2(alpha/2) =    13.09051419          
              
Thus, as              
              
lower bound = (n - 1) s^2 / chi^2(alpha/2) =    712.1196198          
upper bound = (n - 1) s^2 / chi^2(1 - alpha/2) =    1913.370219          
              
Thus, the confidence interval for the variance is              
              
(   712.1196198   ,   1913.370219   ) [answer]
              
*******************************************

Provide a 90% confidence interval estimate of the population standard deviation (to 1 decimal).
( ______ , ______ )

Also, for the standard deviation, getting the square root of the bounds,              
              
(   26.6855695   ,   43.7420875   ) [ANSWER]