Loading in 2 Seconds...

Ch.08 Revising Judgments in the Light of New Information Dr. Ayham Jaaron

Loading in 2 Seconds...

- 265 Views
- Uploaded on

Ch.08 Revising Judgments in the Light of New Information Dr. Ayham Jaaron. Introduction.

Download Presentation
## Ch.08 Revising Judgments in the Light of New Information Dr. Ayham Jaaron

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

Introduction

- Suppose that you are a marketing manager working for an electronics company which is considering the launch of a new type of pocket calculator. On the basis of your knowledge of the market, you estimate that there is roughly an 80% probability that the sales of the calculator in its first year would exceed the break-even level. You then receive some new information from a market research survey. The results of this suggest that the sales of the calculator would be unlikely to reach the break-even level.
- How should you revise your probability estimate in the light of this new information?

Example....

- You then remember that it is possible to perform a ‘quick test on the component, though this test is not perfectly reliable. If the component is ‘OK’ then there is only an 80% chance it will pass the test and a 20% chance that it will wrongly fail. On the other hand, if the component is defective then there is a 10% chance that the test will wrongly indicate that it is ‘OK’ and a 90% chance that it will fail the test. When you perform the quick test the component fails. How should you revise your prior probability in the light of this result?

Example....

What is the new probabilities for the following cases:

- P(component Ok/Failed test)
- P(component defective/ failed test)
- P(component OK/pass test)
- P(component defective/pass test)

Example (2)

- An engineer makes a cursory inspection of a piece of equipment and estimates that there is a 75% chance that it is running at peak efficiency. Hethen receives a report that the operating temperature of the machine is exceeding 80◦C. Past records of operating performance suggest that there is only a 0.3 probability of this temperature being exceeded when the machine is working at peak efficiency. The probability of the temperature being exceeded if the machine is not working at peak efficiency is 0.8. What should be the engineer’s revised probability that the machine is operating at peak efficiency?

Solution...

- The probability tree for this problem is shown in Figure. It can be seen that the joint probabilities are:
- p(at peak efficiency and exceeds 80◦C) = 0.75 × 0.3 = 0.225
- p(not at peak efficiency and exceeds 80◦C) = 0.25 × 0.8 = 0.2
- so the sum of the joint probabilities is: 0.225 + 0.2 = 0.425
- and the required posterior probability is: 0.225/0.425 = 0.529

Continued..

- The tree for this problem is shown in Figure 8.4. The joint probabilities are:
- p(high sales occur and high sales forecast) = 0.2 × 0.9 = 0.18
- p(medium sales occur and high sales forecast) = 0.7 × 0.6 = 0.42
- p(low sales occur and high sales forecast) = 0.1 × 0.3 = 0.03
- so the sum of the joint probabilities is 0.63

Continued...

- which means that we obtain the following posterior probabilities:
- p(high sales) = 0.18/0.63 = 0.2857
- p(medium sales) = 0.42/0.63 = 0.6667
- p(low sales) = 0.03/0.63 = 0.0476

The effects of new information on the revision of probability judgments

- Suppose that a geologist is involved in a search for new sources of natural gas in southern England. In one particular location he is asked to estimate, on the basis of a preliminary survey, the probability that gas will be found in that location. Having made his estimate, he will receive new information from a test drilling.
- The ‘vaguest’ prior probability distribution that the geologist can put forward is to assign probabilities of 0.5 to the two events ‘gas exists at the location’ and ‘gas does not exist at the location’.

Continued...

- Clearly, if he went to the extreme of allocating a probability of 1 to one of the events this would imply that he was perfectly confident in his prediction. Suppose that having put forward the prior probabilities of 0.5 and 0.5, the result of the test drilling is received. This indicates that gas is present and the result can be regarded as 95% reliable. By this we mean that there is only a 0.05 probability that it will give a misleading indication.
- What are the posterior probabilities for both Gas existence and gas non existence?

Conclusion from previous problem

- It can be seen that these probabilities are identical to the probabilities of the test drilling giving a correct or misleading result. In other words, the posterior probabilities depend only upon the reliability of the new information. The ‘vague’ prior probabilities have had no influence on the result.

The effect of the reliability of information on the modification and Vagueness of prior probabilities

Conclusion (2)

- As we have seen in the previous problem and also in the figure, if the test drilling has only a 50% probability of giving a correct result then its result will not be of any interest and the posterior probability will equal the prior, as shown by the diagonal line on the graph. By considering the distance of the curves from the diagonal line, it can be seen that the more reliable the new information, the greater will be the modification of the prior probabilities.

Expected value of perfect information (EVPI)

- A year ago a major potato producer suffered serious losses when a virus affected the crop at the company’s North Holt farm. Since then, steps have been taken to eradicate the virus from the soil and the specialist who directed these operations estimates, on the basis of preliminary evidence, that there is a 70% chance that the eradication program has been successful. The manager of the farm now has to decide on his policy for the coming season and he has identified two options:

EVPI...continued

- (1) He could go ahead and plant a full crop of potatoes. If the virus is still present an estimated net loss of $20 000 will be incurred. However, if the virus is absent, an estimated net return of $90 000 will be earned.
- (2) He could avoid planting potatoes at all and turn the entire acreage over to the alternative crop. This would almost certainly lead to net returns of $30 000.
- The manager is now informed that Ceres Laboratories could carry out a test on the farm which will indicate whether or not the virus is still present in the soil. The manager has no idea as to how accurate the indication will be or the fee which Ceres will charge.

EVPI...continued

- However he decides initially to work on the assumption that the test is perfectly accurate. If this is the case, what is the maximum amount that it would be worth paying Ceres to carry out the test?

Expected Value of imperfect information (EVII)

- Suppose that, after making further enquiries, the farm manager receives new information that the Ceres test is not perfectly reliable. Manger has to decide whether to do the test or not. However, If the virus is still present in the soil the test has only a 90% chance of detecting it, while if the virus has been eliminated there is a 20% chance that the test will incorrectly indicate its presence. How much would it now be worth paying for the test?

Expected profit with imperfect

information = $62 155

Expected profit without the

information = $57 000

Expected value of imperfect

information (EVII) = $5 155

- So, it would not be worth paying Ceres more than $5155 for the test.

Download Presentation

Connecting to Server..