In: Statistics and Probability
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).
(____,_____)
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
2L
=
2
/2,df = 35.172
2R
=
21 -
/2,df = 13.090
The 95% confidence interval for
2 is,
(n - 1)s2 /
2
/2 <
2 < (n - 1)s2 /
21 -
/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)