Question

Even within a particular chain of hotels, lodging during the summer months can vary substantially depending on the type of room and the amenities offered. 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 50 billing statements from each of the computer databases of each of the three hotel chains are given in the table.

Hotel A Hotel B Hotel C

Sample average ($) 145 160. 125

Sample standard deviation 17.6 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 C chains. (Round your answers to two decimal places.)

Answer 1

Here we have given that

n1=number of billing statements for Hotel A = 50

n2 =number of billing statements for Hotel C = 50

=Mean of Hotel A billing statements = 145

= Mean od Hotel C billing statements = 125

S1=Sample standard deviation of Hotel A =17.6

S2=Sample standard deviation of Hotel C=12.5

Now we want to find

95% confidence interval for difference in the average room rate for Hotel A and Hotel C.

For finding this CI 1^st we want to check the two population variance equal or not. (To check the homogenicity of variance assupmption)

Formula is as follows

Claim: To check whether the two population variance equal or not.

The hypothesis is

v/s

Test statistics is

Fstat=

        =

         = 1.98

=level of significance=0.05

Degrees of freedom 1=n1-1=50-1=49

Degrees of freedom 2= n2-1= 50-1=49

F-critical = 1.61 Using Excel = (prob=0.05, d.f1=49, D.F2=49)

Decision:

Here F-tab > F-critical

Then we reject Null hypothesis that the population variances are not equal.

Now, We want to find he 95 % confidence interval for difference in the two population means

Formula is as follows

Now, first we can find the t critical value,

Degrees of freedom =n1+n2-2 =50+50-2=98

C=confidence level =0.95

=level of significance= 1-c = 1-0.95 =0.05

We get

Tc=1.984 (using excel =TINV(prob=0.05, d.f=98)

Now we get the 95% confidence interval for the difference in the two population mean

Interpretation:

This interval shows that we are 95% confident that the difference in the population mean will falls within that interval.