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

Answer:

(a) 900

(b) [567.35 , 1689.72]

(c) [23.82 , 41.11]

Step-by-step 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  $\frac{(n-1)s^{2} }{\sigma^{2} }$ ~ $\chi^{2} __n_-_1$

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 < $\frac{(n-1)s^{2} }{\sigma^{2} }$ < 30.14) = 0.90

P($\frac{10.12}{(n-1)s^{2} }$ < $\frac{1 }{\sigma^{2} }$ < $\frac{30.14}{(n-1)s^{2} }$ ) = 0.90

P($\frac{(n-1)s^{2} }{30.14}$ < $\sigma^{2}$ < $\frac{(n-1)s^{2} }{10.12}$ ) = 0.90

90% confidence interval for $\sigma^{2}$ = $[\frac{19s^{2} }{30.14} , \frac{19s^{2} }{10.12}]$

                                                   = $[\frac{19*900 }{30.14} , \frac{19*900 }{10.12}]$

                                                   = [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($\sqrt{\frac{(n-1)s^{2} }{30.14}}$ < $\sigma$ < $\sqrt{\frac{(n-1)s^{2} }{10.12}}$ ) = 0.90

90% confidence interval for $\sigma$ = $[\sqrt{\frac{19s^{2} }{30.14}} , \sqrt{\frac{19s^{2} }{10.12}} ]$

                                                 = [23.82 , 41.11]

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