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 p.m. lunch period. The waiting time (operationally 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 stored in the file bank1.xls), 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 p.m. lunch period. A random sample of 15 customers is selected (and stored in the file bank2.xls), 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 Assuming that the population variances from both banks are not equal, is there evidence of a difference in the mean waiting time between the two branches? (Use 0.05 level of significance )

No there isn't.

To determine if there is evidence of a difference in the mean waiting time between the two branches, we can use a two-sample t-test. This test compares the means of two independent samples to see if they are significantly different from each other.

Here are the steps to perform the two-sample t-test:

Step 1: State the hypotheses.
- Null hypothesis (H0): The means of the two branches are equal.
- Alternative hypothesis (Ha): The means of the two branches are not equal.

Step 2: Set the significance level.
- The significance level (alpha) is given as 0.05 in this case.

Step 3: Calculate the test statistic.
- The test statistic for the two-sample t-test is calculated as:
t = (x1 - x2) / sqrt((s1^2 / n1) + (s2^2 / n2))
where x1 and x2 are the sample means, s1 and s2 are the sample standard deviations, and n1 and n2 are the sample sizes.

Step 4: Determine the critical value.
- The critical value is obtained from the t-distribution table or a statistical software, given the significance level and degrees of freedom.

Step 5: Compare the test statistic with the critical value.
- If the absolute value of the test statistic is greater than the critical value, we reject the null hypothesis and conclude that there is evidence of a difference in the means. Otherwise, we fail to reject the null hypothesis.

Now, let's perform the two-sample t-test using the given data:

Step 1: State the hypotheses:
- H0: μ1 = μ2 (The means of the two branches are equal)
- Ha: μ1 ≠ μ2 (The means of the two branches are not equal)

Step 2: Set the significance level:
- α = 0.05

Step 3: Calculate the test statistic:
- From the data provided, the sample means and sample standard deviations are:
x1 = 4.29 (from bank1.xls)
x2 = 7.07 (from bank2.xls)
s1 = 1.59 (from bank1.xls)
s2 = 2.11 (from bank2.xls)
n1 = n2 = 15 (sample sizes are both 15)

- Now, calculating the test statistic:
t = (x1 - x2) / sqrt((s1^2 / n1) + (s2^2 / n2))
= (4.29 - 7.07) / sqrt((1.59^2 / 15) + (2.11^2 / 15))
= -2.78 / sqrt(0.168 + 0.302)
= -2.78 / sqrt(0.47)
= -2.78 / 0.685
≈ -4.06

Step 4: Determine the critical value:
- Degrees of freedom (df) = n1 + n2 - 2 = 15 + 15 - 2 = 28
- Using a t-distribution table or statistical software, we find the critical value for a two-tailed test at α = 0.05 and df = 28 to be approximately ± 2.048.

Step 5: Compare the test statistic with the critical value:
- Since the absolute value of the test statistic (-4.06) is greater than the critical value (±2.048), we reject the null hypothesis.
- Therefore, we conclude that there is evidence of a difference in the mean waiting time between the two branches.

In conclusion, based on the given data and using a 0.05 level of significance, we have found evidence to suggest that there is a difference in the mean waiting time between the bank branch located in the commercial district and the one in the residential area.