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soccergal12 Group Title

A certain test is used to screen for Alzheimers disease. People with and without the disease were sampled and the following table resulted. Given the general Alzheimer disease rate is 0.001, what is the probability that a randomly selected individual will have an erroneous test result? [Hint: an error occurs if the individual has the disease and receives a negative test result or…?] Test Disease No Disease Total + 436 5 441 - 14 495 509 Total 450 500 950

  • one year ago
  • one year ago

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  1. kropot72 Group Title
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    Let P(A) = probability that the individual has the disease and receives a false negative result. \[P(A)=0.001\times \frac{5}{441}\] Let P(B) = probability that the individual does not have the disease and receives a false positive result. \[P(B)=(1-0.001)\times \frac{14}{509}=0.999\times \frac{14}{509}\] Event A and event B cannot happen together therefore the events are mutually exclusive. P(A or B) = P(A) + P(B) = probability of an erroneous test result.

    • one year ago
  2. soccergal12 Group Title
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    i don't get the right answer. the answer should be 0.010021

    • one year ago
  3. kropot72 Group Title
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    @Zarkon

    • one year ago
  4. kropot72 Group Title
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    @UnkleRhaukus

    • one year ago
  5. soccergal12 Group Title
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    ????????

    • one year ago
  6. kropot72 Group Title
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    There is only 1 chance in 1000 that a randomly selected individual has the disease. Assuming that the individual does not have the disease and is tested, the probability of a false positive (based on the sampling result) is 14/509 = 0.0275. The probability that a randomly selected selected individual has the disease and gives a false negative will add to the value of 0.0275. Assuming that the question has been stated with no errors, I must conclude that the answer 0.010021 is incorrect. Have you any other way of checking the answer?

    • one year ago
  7. soccergal12 Group Title
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    well, maybe. i was told the answer was 0.010021. that's okay. i will skip this question

    • one year ago
  8. UnkleRhaukus Group Title
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    @kropot72, your woking is great, i think you just read the table wrong ( its a trick table to read ) \[P(A)=0.001\times\frac{14}{450}\approx0.000031\] \[P(B)=(1-0.001)\times\frac5{500}\approx0.00999\]

    • one year ago
  9. kropot72 Group Title
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    @UnkleRhaukus Thank you for your help. Your reading of the table sure gets the given correct answer. With more thought I am sure I will get my head around the correct reading of the table :)

    • one year ago
  10. soccergal12 Group Title
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    thank you. that makes more sense

    • one year ago
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