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 20 days of operation shows a sample mean of 290 rooms occupied per
day and a sample standard deviation of 30 rooms.
a. What is the point estimate of the population variance?
b. Provide a 90% confidence interval estimate of the population variance.
c. Provide a 90% confidence interval estimate of the population standard deviation.

Answer 1

(a) 900

(b) [567.35 , 1689.72]

(c) [23.82 , 41.11]

Explanation:

We are given that a sample of 20 days of operation shows a sample mean of 290 rooms occupied per day and a sample standard deviation of 30 rooms i.e.;

Sample mean,
`xbar` = 290 Sample standard deviation, s = 30 and Sample size, n = 20

(a) Point estimate of the population variance is equal to sample variance, which is the square of Sample standard deviation ;

`\sigma^(2)` =
`s^(2)` =
`30^(2)`

`\sigma^(2)` = 900

(b) 90% confidence interval estimate of the population variance is given by the pivotal quantity of
`((n-1)s^(2) )/(\sigma^(2) )` ~
`\chi^(2) __n_-_1`

P(10.12 <
`\chi^(2)__1_9` < 30.14) = 0.90 {At 10% significance level chi square has critical

values of 10.12 and 30.14 at 19 degree of freedom}

P(10.12 <
`((n-1)s^(2) )/(\sigma^(2) )` < 30.14) = 0.90

P(
`(10.12)/((n-1)s^(2) )` <
`(1 )/(\sigma^(2) )` <
`(30.14)/((n-1)s^(2) )` ) = 0.90

P(
`((n-1)s^(2) )/(30.14)` <
`\sigma^(2)` <
`((n-1)s^(2) )/(10.12)` ) = 0.90

90% confidence interval for
`\sigma^(2)` =
`[(19s^(2) )/(30.14) , (19s^(2) )/(10.12)]`

=
`[(19*900 )/(30.14) , (19*900 )/(10.12)]`

= [567.35 , 1689.72]

Therefore, 90% confidence interval estimate of the population variance is [567.35 , 1689.72] .

(c) 90% confidence interval estimate of the population standard deviation is given by ;

P(
`\sqrt{((n-1)s^(2) )/(30.14)}` <
`\sigma` <
`\sqrt{((n-1)s^(2) )/(10.12)}` ) = 0.90

90% confidence interval for
`\sigma` =
`[\sqrt{(19s^(2) )/(30.14)} , \sqrt{(19s^(2) )/(10.12)} ]`

= [23.82 , 41.11]

Therefore, 90% confidence interval estimate of the population standard deviation is [23.82 , 41.11] .