An electric utility tracks the response time to customer-reported outages. The a

Question

An electric utility tracks the response time to customer-reported outages. The accompanying data are a random sample of 40 response time (in minutes) for one operating division of this utility during a single month.

a. Use Proc Capability in SAS to compute the appropriate capability index when the utility believes the response time should be below two hours (120 minutes).

b. Based on your capability index value, how well do you think the process is performing?

c. Suppose the new goal is to achieve a 90% response rate in less than two hours (That is, 90% of all response times are under 2 hours, whereas, 10% can be above.) Explain how you might adapt a capability index to measure capability in this situation and calculate the new value.

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Explanation / Answer

a) Average observed response time, µ = 98.775

Standard deviation of observed response time, = 12.271

Upper specification limit, USL = 120 minutes

Lower specification limit, LSL = 0

Capability index, Cpk = MIN((µ-LSL)/3, (USL-µ)/3)

= MIN((98.775-0)/(3*12.271), (120-98.775)/(3*12.271)) = 0.577

b) Based on the capability index calculated above, the process is not performing well. For process to be capable of meeting the speciifcation limits of below 120 minutes, Cpk value should be at least 1.33, but Cpk at 0.577 this is far lower than that. Therefore, the process is not capable.

c) Z value for 90% confidence level = NORMSINV(0.90) = 1.28

The revised capability capability index is as follows

Cpk = MIN((98.775-0)/(1.28*12.271), (120-98.775)/(1.28*12.271)) = 1.35

New value of Capability index = 1.35

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