Pls help me... About probability calculation and question...?

Question

The sensitivity of a symptom is the probability that the symptom is present given that the person has a disease.

The specificity of a symptom is the probability that the symptom is not present given that the person does not have a disease.

Suppose the disease is lung cancer and the symptom is cigarette smoking. If we assume 87% of people with lung cancer and 20.9% of people without lung cancer are smokers. What are the sensitivity and specificity in this case? Can cigarette smoking be used as a diagnostic tool for predicting lung cancer? Why?

Answer 1

Sens = P(symptom + | disease +)
Spec = P(symptom - | disease -)

disease = lung cancer
symptom = cigarette smoking

we are told...

P(smoker|Lung cancer) = 0.87
P(smoker| no Lung cancer) = 0.209

Sens = P(smoker|Lung cancer) = 0.87
spec = P(not smoker| no Lung cancer) = 1 - P(smoker| no Lung cancer) = 1 - 0.209 = 0.791

To answer the last question it seems to me that we want to find:

P(Lung cancer | smoker)

This can be done using Bayes' Theorem.

For two event A and B (A' is the complement of A)

P(A|B) = P(B|A)*P(A)/(P(B|A)*P(A) + P(B|A')P(A'))

In your case, A = Lung Cancer and B = smoker. So

P(Lung Cancer | Smoker) = P(Smoker | Lung Cancer)*P(Lung Cancer) / (P(Smoker | Lung Cancer) * P(Lung Cancer) + P(Smoker | No Lung Cancer) * P(No Lung Cancer)) = 0.87p/(0.87p + 0.209(1 - p)) where p is the proportion of lung cancer cases in the population. According the Cancer Society, in 2006 p was about 0.25. Plugging this into the above equation give a value of

P(Lung Cancer | smoker) = 0.87*0.25/(0.87*0.25 + 0.209*0.75) = 0.58

Based on these numbers the evidence isn't too strong that smoking can be used a diagnostic tool for predicting lung cancer.

NOTE:

Philo's answer above makes the tacit assumption that 50% of the population has lung cancer. This is clearly not the case. However, if you took p in my formula to be 0.5, you would see that I get the same answer as Philo.

Answer 2

set up a table:

....... s .... ~s
----------------------|
.L . 87 ... 13 .....| . 100
~L . 20.9 . 79.1 .| . 100
-------------------------------
.....107.9 . 92.1 | . 200

sensitivity = P(s|L) = 87%
specificity = P(~s|~L) = 79.1%

P(L|s) = 87/107.9 = 80.63%

What you want to do with this contingency table is compute the chi square statistic. First compute the expected value for each cell in the table by multiplying row and column totals and dividing by overall total: I'll add these in parentheses to the existing table:

....... s .... ~s
----------------------|
.L . 87 ... 13 .....| . 100
.... (54) ...(46)
~L . 20.9 . 79.1 .| . 100
.... (54) ...(46)
-------------------------------
.....107.9 . 92.1 | . 200

Second step, compute differences between expected and observed values, square them, divide by the expected values, and sum them up:

(87-54)²/54 + (13-46)²/46 + (20.9-54)²/54 + (79.1-46)²/46 = 87.95

Finally, we have (2 rows - 1)(2 columns -1) = 1 degrees of freedom. We look up our result in a chi square table and find the probability smoking is unrelated to cancer is effectively 0.