Decision of Uncertainty
Darylisha Jones
QNT/561
February 14, 2011
Paul Thomasman
Decision of Uncertainty
Introduction
Decision: Extending automobile warranties or not? It comes a time when one has to make that decision of extending an automobile warranty when it has expired. Because auto warranties provide well needed protection, it does not come cheap be any means. Therefore, the decision to be made is the price of purchasing a warranty over the cost of repairs without a warranty. An extended warranty supplies the ability of possessing coverage of an automobile. It supplies coverage for repairs, parts, rentals, and even labor at a warranty rate rather paying out- of -pocket for every issue.
Research
In order to make the correct decision, I will research information on purchasing extended warranty of a 2005 Chevrolet Impala. In one year my warranty of my vehicle will expire, and I will have to decide on purchasing an extended warranty to protect my vehicle. After researching the effects of not extending a warranty can result in high auto repair bills. An average cost of an engine repair without protection for an Impala is $2,000. Therefore, information of whether or not major repairs are needed for this vehicle must be taken into consideration. After gaining information from many auto repair shops of their experience of servicing vehicles, it is wise to acquire extending a warranty. The chances of auto repairs being needed within five years on my vehicle will likely become a factor, therefore gaining information on extending my warranty will help in making the correct decision for my vehicle overall.
Interpretation of Data (using Bayes’ Theorem) To interpret my data, I have decided to utilize Bayes’ Theorem to assist in making the right decision. Bayes’ theorem emulates the process of logical inference by determining the degree of confidence in possible conclusions based on the available evidence (University of Phoenix, 2010). Bayesian statistics provides extension of a classical approach by using sampling data as well as considering all of the other reachable information (Cooper & Schindler, 2006). Therefore, Bayes’ theorem would be appropriate for predicting the likelihood of car repairs within five years as well as the probability of the warranty ensuring such repairs. In the next five years of owning my Chevy Impala, I establish that there is a highly chance of needing repairs that will cost well over $2,000 dollars that not covered without a warranty. These repairs, which are not covered, are at 5% chance of having repairs carried out on the vehicle. Finally the next step is to set up the variables to study the probability of engine repairs. Engine Repair (1) = ER1 = N/A coverage excess of $2,000 Engine Repair (2) = ER2 = Coverage with warranty Probability of repairs not being covered by a warranty (P (ER1) = .05) Probability of repairs being covered by a warranty (P (ER2) = .95

After researching within the next five years it is highly likely that my vehicle will need repairs with a 90% chance. P (B|ER1) = .90
Chevrolet mechanics supports that they produce quality vehicles, which are capable of holding up over time however, there is a chance that within five years I may need some kind of repairs rather they are small or big. Therefore, the probability of my vehicle not needing any type of service at all is .15 P (B|ER2) = .15
Posterior probability according to Bayes’ provides: P (ER1/B) = P (ER1) P (B|ER1) P (ER1) P (B|ER1) + P (ER2) P (B|ER2) = (.5)(.90) = .045 (.5) (.90) + (.95) (.15) .1875 = .24
According to this probability there is a chance that my vehicle will need repairs and not being covered with an extended warranty will be at .24. Therefore, if I decide to keep my vehicle for the next five years, the probability on having auto repairs will rise from .5 to .24.

Not Covered by Warranty = .05 .90 .0450 .0450 /.1875 = .24 Covered by Warranty = .95 .15 .1425 .1425 /.1875 = .76 P (B) = .1875 | = 1.00
Following with the information above, there is a .76 probability that my vehicle will indeed need some form of repairs within five years, and a .24 probability that my vehicle will need repairs, and will not be covered by an extended warranty.

Decision Because my warranty on my vehicle will soon become expired, I have to take in consideration what is the best decision collectively. Since car payments are not a factor in this vehicle, and this will be my vehicle for at least five more years. It would only be in my best interest to purchase the extended warranty for my vehicle with the help of the calculations from above. This decision will be feasible because in the long wrong money will be safe verses the money that would be lost with no protection and paying out- of-pocket for every repair. With the probability of having issues with my vehicle, and most likely being subjected to repairs it’s only right to purchase an extended warranty to protect my pockets as well as my vehicle. All decisions involves risk, it is just how high of a risk an individual is willing to take in making a decision. Using probability as a tool supplies ways to make sufficient decisions for an individual, including myself.

References
Cooper, D.R., & Schlinder, P.S. (2006). Business research methods (9th ed.). Boston,
MA: McGraw-Hill/Irwin