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 299 rooms occupied per day and a sample standard deviation of 32 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).

(____,_____)

Answer 1

Solution :

Given that,

s = 32

Point estimate = s² = 1024

n = 24

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

(a)

At 90% confidence level the ² value 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} = 13.090

The 95% confidence interval for ² is,

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

23 * 1024 / 35.172 < ² < 23*1024 / 13.090

669.6 < ² < 1799.2

(669.6 , 1799.2)

(b)

The 90% confidence interval for is,

25.9 < < 42.4

(25.9 , 42.4)