Question

A branch located in a commercial district of a city has the business objective of developing an improved process for serving customers during the noon-to-1:00 pm lunch period. Management decides to first study the waiting time in the current process. The waiting time is defined as the number of minutes that elapses from when the customer enters the line to when he or she reaches the teller window. Data are collected from a random sample of 15 customers. Suppose a competitor's branch located in a residential area is also concerned with the same business objective. Data are collected from a random sample of 15 customers.

1a. Is there evidence of a difference in the variability of the waiting time between the two branches (use α = 0.05)?

1b. Identify the appropriate test to use, explain why you use that test, and interpret your results.

2a. Is there any evidence of a difference in the mean waiting time between the two branches?

2b. Identify the appropriate test to use, explain why you use that test, and interpret your results.

Please show your work in excel using the Data Analysis function.

Commercial Bank	Residential Bank
4.21	9.66
5.55	5.9
3.02	8.02
5.13	5.79
4.77	8.73
2.34	3.82
3.54	8.01
3.2	8.35
4.5	10.49
6.1	6.68
0.38	5.64
5.12	4.08
6.46	6.17
6.19	9.91
3.79	5.47

Answer 1

-----------------------------------------------Answer No: 1(a) ---------------------------------------------------------------

The variability of the waiting time can be tested for the two branches with the help of F-test. The F-test can be conducted in Excel by using the following steps:

1. Enter the data in Excel and go to the ‘Data’ tab in the menu bar and select ‘Data Analysis’ as shown:
   
   
2. Make the selection in the resulting pop-up box as shown below and click OK.  
   
3. Make the following settings in the resulting window and click OK :
   
4. The following results are obtained:
   
5. The p-value of 0.1900 suggests that the null hypothesis of equality of variances can not be rejected. Hence this concludes to the fact that the variability in the two samples is not much different from each other.

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---------------------------------------------------------------------Answer 1.(b) -----------------------------------------------------------------------------

The problem involves comparison of the variances of two populations without any information of population variances. Therefore, the F-test comes into action assuming the two samples from two normal populations. F-test is carried out in the part (a) of this solution

The F-test concludes to the fact that there is not much difference between the variability between two samples.

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------------------------------------------------------------------Answer 2(a) ---------------------------------------------------------------------------------

The t-test for difference of means for populations with equal variances can be used. The variances are taken to equal because of the result of F-test conducted above. The steps to apply t-test in excel are as follows:

1. Enter the data in Excel and go to the ‘Data’ tab in the menu bar and select ‘Data Analysis’ as shown:
   
2. Make the following sections in the resulting pop-up box and click OK:
   
3. Make the following selections and click OK:
   
4. The following results are obtained:
   
5. The p-value 0.00029 is very much less than 0.05 , therefore null hypothesis of no difference between the two sample means, should be rejected.

Thus it can be concluded that the two samples differ significantly in their mean waiting times .

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-------------------------------------------------------------Answer 2(b)----------------------------------------------------------------------------

The t-test for equal variances has been used because of the fact that the F-test in part (a) of this question suggests that the two samples have no significant difference between the variances.

The t-test shows that the mean waiting time for the two samples differ significantly.