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 (to nearest whole number)?

b. Provide a 90% confidence interval estimate of the population variance (to nearest whole number).

c. Provide a 90% confidence interval estimate of the population standard deviation (to 1 decimal).

Answer 1

a)
`\hat \sigma^2 =s^2 =30^2 = 900`

b)
`567.277 \leq \sigma^2 \leq 1690.224`

Rounded to the nearest number would be:

`567 \leq \sigma^2 \leq 1690`

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

And rounded :

`23.8 \leq \sigma \leq 41.1`

Explanation:

Data given and notation

s=30 represent the sample standard deviation

`\bar x=290` represent the sample mean

n=20 the sample size

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

The Chi Square distribution is the distribution of the sum of squared standard normal deviates .

Part a

The best point of estimate for the population variance is the sample variance, so on this case:

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

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

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

Since the Confidence is 0.90 or 90%, the value of
`\alpha=0.1` and
`\alpha/2 =0.05`, and we can use excel, a calculator or a table to find the critical values.

The excel commands would be: "=CHISQ.INV(0.05,19)" "=CHISQ.INV(0.95,19)". so for this case the critical values 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 \leq ((19)(30)^2)/(10.117)`

`567.277 \leq \sigma^2 \leq 1690.224`

Rounded to the nearest number would be:

`567 \leq \sigma^2 \leq 1690`

Part c

In order to find the confidence interval for the deviation we just need to take the square root for the interval of the variance, and we got:

`23.818 \leq \sigma \leq 41.112`

And rounded :

`23.8 \leq \sigma \leq 41.1`