Question

PLEASE ANSWER PARTS A, C, AND D. Thank you!

2.2 For diagnostic testing, let X = true status (1 = disease, 2 = no disease) and Y = diagnosis (1 = positive, 2 = negative). Let πi = P (Y = 1|X = i), i = 1, 2.

a. Explain why sensitivity = π1 and specificity = 1 − π2.

c. For mammograms for detecting breast cancer, suppose γ = 0.01, sensitivity = 0.86, and specificity = 0.88. Given a positive test result, find the probability that the woman truly has breast cancer.

d. To better understand the answer in (c), find the joint probabilities for the 2 × 2 cross-classification of X and Y . Discuss their relative sizes in the two cells that refer to a positive test result.

Answer 1

A)

Sensitivity is the probability that the diagnostic test is positive given that a subject has the disease, that is, P(Y = 1 | X = 1). so its probability is given by π1.

Specificity is the probability that the test is negative given that the subject does not have the disease, that is, P(Y = 2 | X = 2). so, its probability is given by 1 − π2.

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c) γ = 0.01, sensitivity = 0.86, and specificity = 0.88

probability that the woman truly has breast cancer P(X = 1 | Y = 1) is given by

= π1γ /(π1γ + π2(1 − γ))

so, putting value in above formula we, get

π1γ /(π1γ + π2(1 − γ)) =( 0.86 × 0.01)/(0.86 × 0.01 + 0.12(1 − 0.01))

= 0.0675.

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d)P(Y = 1 ∩ X = 1) = P(Y = 1 | X = 1) · P(X = 1) = 0.86 × 0.01 = 0.0086

P(Y = 2 ∩ X = 1) = 0.01 − 0.0086 = 0.0014

P(X = 2) = 1 − 0.01 = 0.99

P(Y = 1 ∩ X = 2) = P(Y = 1 | X = 2) · P(X = 2) = 0.12 × 0.99 = 0.1188

P(Y = 2 ∩ X = 2) = 0.99 − 0.1188 = 0.8712.

status	y=1	y=2	total
x=1	0.0086	0.0014	0.01
x=2	0.1188	0.8712	0.99
total	0.1274	0.8726	1

The probability of women who have breast cancer and tested positive for it is lower than the probability of women who do not have breast cancer but tested positive for it