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HW-02

# HW-02 - HOMEWORK FOR EXST 7036(Chapter 2 Student Name Tanza...

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Unformatted text preview: HOMEWORK FOR EXST 7036 (Chapter 2) Student Name: Tanza Erlambang 2.2) For diagnostic testing, let X = true status (1 = diseases, 2 = no diseases) and Y = diagnosis (1 = positive, 2 = negative). Let π i = P (Y = 1| X = i), i = 1, 2 a. Explain why sensitivity = π 1 and specificity = 1 - π 2 Answer: Sensitivity = P (Y = 1| X = 1) = π 1 Specificity = P (Y = 2| X = 2) = 1 - P (Y = 1| X = 2) = 1 - π 2 b. Let γ denote the probability that a subject has a diseases. Given that the diagnosis is positive, use Baye’s theorem to show that the probability a subject truly has the diseases is : π 1 γ / [π 1 γ + π 2 (1 – γ)] Answer: Probability has diseases: P (X = 1 | Y = 1) = P(Y=1 | X=1)P(X= 1)/[P(Y=1 | X=1)P(X=1)+P(Y=1)P(X=2)] c. For mammograms for detecting breast cancer, suppose γ = 0.01, sensitivity= 0.86, and specifity = 0.88. Given a positive test result, find the probability that the woman truly has breast cancer. Answer: P (X = 1 | Y = 1) = P(Y=1 | X=1)P(X= 1)/[P(Y=1 | X=1)P(X=1)+P(Y=1)P(X=2)] =π 1 γ /[π 1 γ + π 2 (1– γ)] P(X=1) = 0.01; P(Y=1 | X=2) = 1 - π 2 = 1- 0.88 = 0.12 P(X=2) = 1 - γ = 1 – 0.01 = 0.99 So, probability the woman truly has breast cancer= 0.86 (0.01)/ [0.086(0.01) + 0.12 (0.99)] = 0.0675 1 d. To better understand the answer in ( c), find the joint probabilities for the 2 x 2 cross classification of X and Y. Discuss their relative sizes in the two cells that refer to a positive test result. Answer: Test Diagnosis +- Total-------------------------------------------------------------------- True Diseases 0.0086 0.0014 0.01 No Diseases 0.1188 0.8712 0.99 From Row, show that total “no diseases” is higher than diseases, it is mean that 99% women (subjects) do not have breast cancer. From Column corresponding to a positive test result that proportion “no diseases”= (0.1188) is very higher than the “diseases” = (0.0086) “diseases” category. 2.3) According to recent UN figures, the annual gun homicide rate is 62.4 per one million residents in the United States and 1.3 per one million residents in the UK a. Compare the proportion of residents killed annually by guns using the (i ) difference of proportion, (ii) relative risk Answer: P1 = 62.4/1 million= 0.0000624 P2 = 1.3/1 million = 0.0000013 (i ) Difference of proportion=P1 – P2= 0.0000624 - 0.0000013= 0.0000611 (ii) Relative Risk = P1/P2 = 0.0000624/0.0000013= 48 So, gun homicide is 48 times higher in the USA than UK 2 b . When both proportions are very close to 0, as here, which measure is more useful for describing the strength of association? Why? Answer: Relative risk is more useful, since different in proportion seems show there is no effect. 2.7) For adults who sailed on the Titanic on its fateful voyage, the odds ratio between gender (female, male) and survival (yes, no) was 11.4 (see R Dawson, J....
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HW-02 - HOMEWORK FOR EXST 7036(Chapter 2 Student Name Tanza...

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