Question

Because of staffing decisions, managers of the Gibson-Marion Hotel are interested in the variability in the number of rooms occupied per day during a particular season of the year. A sample of days of operation shows a sample mean of rooms occupied per day and a sample standard deviation of rooms.

Required:
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)
`s^2 =30^2 =900`

b)
`((19)(30)^2)/(30.144) \leq \sigma^2 \leq ((19)(30)^2)/(10.117)`

`567.28 \leq \sigma^2 \leq 1690.224`

c)
`23.818 \leq \sigma \leq 41.112`

Explanation:

Assuming the following 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

Part a

For this case the best point of estimate for the population variance would be:

`s^2 =30^2 =900`

Part b

The confidence interval for the population variance is given by the following formula:

`((n-1)s^2)/(\chi^2_(\alpha/2)) \leq \sigma^2 \leq ((n-1)s^2)/(\chi^2_(1-\alpha/2))`

The degrees of freedom are given by:

`df=n-1=20-1=19`

Since the Confidence is 0.90 or 90%, the significance
`\alpha=0.1` and
`\alpha/2 =0.05`, the critical values for this case are:

`\chi^2_(\alpha/2)=30.144`

`\chi^2_(1- \alpha/2)=10.117`

And replacing into the formula for the interval we got:

`((19)(30)^2)/(30.144) \leq \sigma^2 \leq ((19)(30)^2)/(10.117)`

`567.28 \leq \sigma^2 \leq 1690.224`

Part c

Now we just take square root on both sides of the interval and we got:

`23.818 \leq \sigma \leq 41.112`