Question 438280: A utility requires service operators to answer telephone calls from customers in an average of 0.1 minutes or less, and either respond to them or refer the customer to the proper department within 0.5 minute. The manager is interested in estimating the actual overall time for both components, in total. A pilot study of 30 actual operator times was drawn, and the results are given in the following table. If the service manager wants to 95% confident that the overall time is correctly estimated, with a 3 percent probability of error, what sample size should be taken.
Component Mean time standard deviation
Answer 0.1023 0.0183
Service 0.5290 0.0902
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A utility requires service operators to answer telephone calls from customers in an average of 0.1 minutes or less, and either respond to them or refer the customer to the proper department within 0.5 minute. The manager is interested in estimating the actual overall time for both components, in total. A pilot study of 30 actual operator times was drawn, and the results are given in the following table. If the service manager wants to 95% confident that the overall time is correctly estimated, with a 3 percent probability of error, what sample size should be taken.
Component Mean time standard deviation
Answer 0.1023 0.0183
Service 0.5290 0.0902
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Combining Answer and Service times you get:
mean(sum) = sum of means = 0.6313
std(sum) = sqrt[0.0183^2+0.0902^2) = 0.0920
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Since: E = zs/sqrt(n)
n = [zs/E]^2
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n = [1.96*0.0920/0.03]^2 = 37 when rounded up
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cheers,
Stan H.
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