Question

Suppose that we randomly select 50 billing statements from each of the computer databases of the Hotel A, the Hotel B, and the Hotel C chains, and record the nightly room rates. The means and standard deviations for the data are given in the table.

	Hotel A	Hotel B	Hotel C
Sample Average ($)	140	180	120
Sample Standard Deviation	17.7	22.6	12.5

(a) Find a 95% confidence interval for the difference in the average room rates for the Hotel A and the Hotel B chains. (Use Hotel A − Hotel B. Round your answers to two decimal places.)

$______ to $______

(b)Find a 99% confidence interval for the difference in the average room rates for the Hotel B and the Hotel C chains. (Use Hotel B − Hotel C. Round your answers to two decimal places.)

$______ to $______

Answer 1

Here is this scenario we have to calculate the Confidence Interval for difference in population means based on given sample information.

To compute the Confidence Interval for (a) & (b) we need to use t distribution because here the population standard deviations is unknown.

Now,

a) The 95% confidence interval for the difference in the average room rates for the Hotel A and the Hotel B chains. (Use Hotel A − Hotel B) and

(b)Find a 99% confidence interval for the difference in the average room rates for the Hotel B and the Hotel C chains. (Use Hotel B − Hotel C.) Is calculated as below,

The t critical value is calculated using t table or using Excel at given degrees of freedom.

a) The 95% confidence interval for the difference in the average room rates for the Hotel A and the Hotel B chains. (Hotel A − Hotel B) is (-48.06, -31.94).

b) The 99% confidence interval for the difference in the average room rates for the Hotel B and the Hotel C chains. ( Hotel B − Hotel C) is (50.35, 69.64).

Thank you.