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

Answer:

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 :

$23.8 \leq \sigma \leq 41.1$

Step-by-step 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:

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

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:

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

$567.277 \leq \sigma^2 \leq 1690.224$

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 :

$23.8 \leq \sigma \leq 41.1$