A credit card company operates two customer service centers
\"A credit card company operates two customer service centers: one in Boise and one in Richmond. Callers to the service centers dial a single number, and a computer program routs callers to the center having the fewest calls waiting. As part of a customer service review program, the credit card center would like to determine whether the average length of a call (not including hold time) is different for the two centers. The managers of the customer service centers are willing to assume that the populations of interest are normally distributed with equal variances. Suppose a random sample of phone calls to the two centers is selected and the following results are reported:\"
\"a. Using the sample results, develop a 90% confidence interval estimate for the difference between the two population means.
b. Based on the confidence interval constructed in part a, what can be said about the difference between the average call times at the two centers?\"
Please explain how you did the equations. Thanks.
Solution
a)
Let the confidence interval be u(boise) - u(richmond).
Calculating the means of each group,
X1 = 195
X2 = 216
Calculating the standard deviations of each group,
s1 = 35.1
s2 = 37.8
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):
n1 = sample size of group 1 = 120
n2 = sample size of group 2 = 135
Thus, sD = 4.566262148
For the 0.9 confidence level, then
alpha/2 = (1 - confidence level)/2 = 0.05
z(alpha/2) = 1.644853627
lower bound = [X1 - X2] - z(alpha/2) * sD = -28.51083285
upper bound = [X1 - X2] + z(alpha/2) * sD = -13.48916715
Thus, the confidence interval is
( -28.51083285 , -13.48916715 ) [ANSWER]
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b)
As this whole interval is less than 0, then there is significant evidence that the mean call time at Richmond is greater than that in Boise. [CONCLUSION]
