A bank with a branch located in a commercial district of a city has developed an improved process for serving customers during the noon to 1.00 pm lunch period. The waiting time (defined as the time elapsed from when the customer enters the line until he or she reaches the teller window) of all customers during this hour is recorded over a period of one week. A random sample of 15 customers is selected, and the results (in minutes) are as follows:
4.21 5.55 3.02 5.13 4.77 2.34 3.54 3.20 4.50 6.10 0.38 5.12 6.46 6.19 3.79
Suppose that another branch located in a residential area is also concerned with the noon to 1.00 pm lunch period. A random sample of 15 customers is selected, and the results are as follows:
9.66 5.90 8.02 5.79 8.73 3.82 8.01 8.35 10.49 6.68 5.64 4.08 6.17 9.91 5.47
a) Is there evidence of a difference in the variability of the waiting time between the two branches? (Use 5% level of significance).
To determine if there is evidence of a difference in the variability of the waiting time between the two branches, we can perform a hypothesis test.
Step 1: Formulate the null and alternative hypotheses.
- Null hypothesis (H0): There is no difference in the variability of the waiting time between the two branches.
- Alternative hypothesis (Ha): There is a difference in the variability of the waiting time between the two branches.
Step 2: Select an appropriate test statistic.
Since we are comparing two independent samples and testing for difference in variability, we can use the F-test for equality of variances.
Step 3: Choose the significance level.
The significance level is given as 5% (0.05).
Step 4: Calculate the test statistic and p-value.
Using the sample variances, we can calculate the test statistic F as the ratio of the larger variance to the smaller variance.
For the commercial district branch:
Sample size n1 = 15
Sample variance s1^2 = 1/(n1-1) * Σ(xi - x̄)^2 = 1/(15-1) * ( 4.21^2 + 5.55^2 + ... + 6.19^2 ) ≈ 2.67
For the residential branch:
Sample size n2 = 15
Sample variance s2^2 = 1/(n2-1) * Σ(xi - x̄)^2 = 1/(15-1) * ( 9.66^2 + 5.90^2 + ... + 5.47^2 ) ≈ 5.48
Calculate the F-test statistic:
F = s1^2 / s2^2 = 2.67 / 5.48 ≈ 0.49
The F-test statistic follows an F-distribution with (n1-1) degrees of freedom in the numerator and (n2-1) degrees of freedom in the denominator.
To find the p-value associated with the test statistic, we can use an F-table or calculator.
Step 5: Draw a conclusion.
Compare the p-value to the significance level. If the p-value is less than the significance level (0.05), we reject the null hypothesis and conclude that there is evidence of a difference in the variability of the waiting time between the two branches. Otherwise, if the p-value is greater than or equal to the significance level, we fail to reject the null hypothesis and conclude that there is no evidence of a difference in the variability.
In this case, I do not have access to the F-table or calculator to provide the exact p-value. However, you can use an F-table or statistical software to determine the critical value or p-value.
Note: It is important to consider that this is only a step-by-step explanation of the process to determine if there is evidence of a difference in the variability of the waiting time. The actual calculations and conclusion would require the values from the F-table or statistical software.