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I want to clarify I do know how to compute Bayes Theorem and conditional Probability questions. I just want to make sure I am doing this problem correctly because my teacher doesn't think he made a mistake on grading my homework.

QUESTION:

1 in 3000 High School students use steroids. A test they take has a 97.5% chance of being positive when the athlete has steroids, and a 98.5% of showing negative when they really aren't using steroids.

a) What is the chance the test is negative when the athlete is using steroids?

b) An athlete tests positive. What is the chance he is really using steroids?

For part "a" I have 1-.985, which is 1 - the probability of a True Negative.

For part "b" I have .0003(.975)/.0003(.975)+.9997(.025), which is the probability of a True Positive/ The sum of the chance of getting a True Positive and False Positive.

Any Help? If I need it, or reassurance if I am correct? Thank You!

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  • $\begingroup$ Could you explain how you computed the chance of a false positive result? $\endgroup$
    – whuber
    Sep 26, 2018 at 20:42
  • $\begingroup$ A True Positive has a probability of .975. A False Positive has a Probability of 1-.975. $\endgroup$ Sep 26, 2018 at 20:45
  • $\begingroup$ I didn't write part a correctly on her but I did on my paper. I just noticed this sorry! $\endgroup$ Sep 26, 2018 at 20:46
  • $\begingroup$ Wouldn't a false negative be 1-.985 though? Because A true negative is .985? $\endgroup$ Sep 26, 2018 at 20:49
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    $\begingroup$ It looks like you ought to review the definitions. $\endgroup$
    – whuber
    Sep 26, 2018 at 20:54

2 Answers 2

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Some comments.

First, $1/3000 = 0.0003333333$ and maybe shouldn't be rounded quite so much.

Second, terminology such as 'false positive' has been so badly abused that one can no longer be sure whether it is intended to mean "Non-user and Positive", "Non-user given Positive", or "Postive given Non-user". These phrases give rise to three different probabilities. So I have stopped using 'false positive'.

Third, the last factor in your denominator should be $$P(Pos|S^c) = 1 - P(Neg|S^c) = 1 - 0.985 = 0.015$$ (not 0.025).


You are given $P(S) = 1/3000$ (sometimes called prevalence of steroid use, $P(Pos | S) = 0.975$ (sometimes called sensitivity of the test), $P(Neg | S^c) = 0.985$ (sometimes called specificity of the test).

You seek $P(S | Pos)$ (sometimes called predictive power of a positive test). Because $P(S|Pos) = \frac{P(S \cap Pos)}{P(Pos)},$ we begin by using the 'Law of Total Probability' to find the denominator.

$$P(Pos) = P(Pos \cap S) + P(Pos \cap S^c) = P(S)P(Pos|S) + P(S^c)P(Pos|S^c)\\ = \frac{1}{3000}0.975 + \frac{2999}{3000}(1-0.985) = 0.0153.$$

(1/3000)*.975 
[1] 0.000325
(2999/3000)*(1-.985)
[1] 0.014995
(1/3000)*.975 + (2999/3000)*(1-.985)
[1] 0.01532

Then $$P(S|Pos) = \frac{P(S \cap Pos)}{P(Pos)} =\frac{0.000325}{0.01532} = 0.0212141.$$


Notes: Your answer, with extra parentheses for the denominator, is

.0003*(.975)/(.0003*(.975)+.9997*(.025))
[1] 0.01156812

Your answer to (a) should be $P(Neg|S) = 1 - P(Pos|S) = 1 - .975,$ and (a) is not directly related to (b).

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I am sorry for the confusion, here is the solution.

A) 1 - P(True -) = 1- .985 = .015

B) .0003(.975)/.0003(.975) + .9997(.025) = .02

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  • $\begingroup$ Your answer to (b) lacks parentheses in the denominator. With or without parentheses, it does not compute to .02 (which would be nearly correct). Also, do you mean .9997(.015) in the denominator? I make enough typos that I'm the last person to downvote because of typos. But perhaps you want to make changes to your Answer. (And maybe proofread mine.) $\endgroup$
    – BruceET
    Sep 27, 2018 at 0:34

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