Question

A​ new, non-evasive, test is to indicate the existance of colorectal cancers in patients who have colorectal cancer with a probability of 0.94. ​However, this test will also show the presense of colorectal cancer in patients who do not have colorectal cancer 3​% of the time.
The chance that a person taking this test has colorectal cancer is 0.08.
Answer the questions below.

a. Find the sensitivity of this test. Use two decimals.

b. Find the specificity of this test. Use two decimals.

c. A person taking this test has tested positive for colorectal cancer. What is the probability​ he/she has colorectal​ cancer? Use four decimals.

d. What is the probability that the person in part c does not have colorectal​ cancer? Use four decimals.

nothing

Answer 1

Answer)

Let say we have a sample size of 1000

Now, 0.08 has the cancer

So, 0.08*1000 = 80

Now when a person do have the cancer, then test detects it 0.94 of the times

So, 0.94*80 = 75.2 (true positive)

So, it detect negative for 80-75.2 = 4.8 (false negative)

So, 920 do not have cancer

And when the patients do not have the cancer

This test will show cancer 3% of the time

That is false positive = 3% of 920 = 27.6

and detect negative for 920-27.6 = 892.4 (true negative)

A)

Sensitivity is = true positive/(true positive + false negative)

Sensitivity = 0.94

B)

Specificity is = true negative/(false positive + true negative)

Specificity = 892.4/(27.6+892.4) = 0.97

C)

Probability is given by favorable/total

Total = total number tested positive = 75.2+27.6

Favorable = they actually has cancer = 75.2

Required probability is 75.2/(75.2+27.6) = 0.73151750972 = 0.7315

D)

As the sum of all the probabilities is = 1

So, p(does not have cancer) = 1 - 0.73151750972 =

0.26848249027

= 0.2685