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hammer [34]
3 years ago
5

Because of staffing decisions, managers of the Gibson-Marion Hotel are interested in the variability in the number of rooms occu

pied 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.
Mathematics
1 answer:
olchik [2.2K]3 years ago
5 0

Answer:

a) s^2 =30^2 =900

b) \frac{(19)(30)^2}{30.144} \leq \sigma^2 \leq \frac{(19)(30)^2}{10.117}

567.28 \leq \sigma^2 \leq 1690.224

c) 23.818 \leq \sigma \leq 41.112

Step-by-step 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:

\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(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:

\frac{(19)(30)^2}{30.144} \leq \sigma^2 \leq \frac{(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

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