Question

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 294 rooms occupied per day and a sample standard deviation of 26 rooms.

What is the point estimate of the population variance? 676

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). ( , )

Answer 1

Solution :

Given that,

s = 26

s² = 676

n = 24

Degrees of freedom = df = n - 1 = 24 - 1 = 23

At 90% confidence level the ² is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

1 - / 2 = 1 - 0.05 = 0.95

²_L = ²_/2,df = 35.172

²_R = ²_{1 - /2,df} = 13.091

The 90% confidence interval for ² is,

(n - 1)s² / ²_/2 < ² < (n - 1)s² / ²_{1 - /2}

(24 - 1) * 676 / 35.172 < ² <  (24 - 1) * 676 / 13.091

442.0 < ² < 1187.7

(442.0 , 1187.7)

A 90% confidence interval estimate of the population standard deviation is,

21.0 < < 34.5

(21.0 , 34.5 )