Please help! <br><br><br> Much appreciated if anyone would help x

Use the graph to answer the question.

Sphinxa [80]

Answer:

The answer is B -5/18. you're welcome :)

Step-by-step explanation:

4 0

3 years ago

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

Anit [1.1K]

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.