[Math] Probability of having a disease given a symptom is reported

probability

A medical research team wishes to assess the usefulness of a certain symptom in the diagnosis of a particular disease. In a random sample of 775 patients with the disease, 744 reported having the symptom. In an independent random sample of 1380 subjects without the disease, 21 reported that they had the symptom. Find the predictive value positive and the predictive value negative for the syptom for the following hypothetical disease rates: 0.0001, 0.01, and 0.1

Best Answer

I'll give this a shot. I'm interpreting "predictive value" to mean the estimated probability that the disease is present, given that the symptom is (not) reported.

Let S be be event that the patient reports the symptom, and let D be the event that the patient has the disease.

Based on the data, we can estimate the probability of a patient reporting the symptom given that they have the disease as:

P(S | D) = 744/755

And we can estimate the probability of a patient reporting the symptom given that the patient does not have the disease as:

P(S | ¬D) = 21/1380

We know logically that P(S) = P(S ∧ D) + P(S ∧ ¬D), because these events are mutually exclusive. By Bayes' theorem we have:

P(S ∧ D) = P(D) * P(S | D)

P(S ∧ ¬D) = P(¬D) * P(S | ¬D) = (1 - P(D)) * P(S | ¬D)

And therefore:

P(S) = P(D) * 744/755 + (1 - P(D)) * 21/1380

P(S) = P(D) * 67391/69460 + 21/1380

Again by Bayes' theorem we have for the positive case:

P(D | S) = P(S | D) * P(D) / P(S)

P(D | S) = 68448 / (67391 + (1057 / P(D)))

And the negative case:

P(D | ¬S) = P(¬S | D) * P(D) / P(¬S)

P(D | ¬S) = (1 - P(S | D)) * P(D) / (1 - P(S))

P(D | ¬S) = 1012 / (-67391 + (68403 / P(D)))

Now for the given values of P(D), 0.0001, 0.01, and 0.01, we can compute the respective values of P(D | S): 22816/3656797, 22816/57697, and 22816/25987. Also we can compute the respective values of P(D | ¬S): 92/62178419, 92/615719, and 1012/616639.

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