advertisement

STA 6934 HW # 2 Chapter 6: Problems 8, 9, 13, 14, 15, 16 18, 19 8) Given: Probability of child’s gestation age < 37 weeks, P(A) = 0.142 Probability of birth weight less than 2500 gms, P( B) = 0.051 Probability of these two events occurring simultaneously, P(A ∏ B)= 0.031 (a) Venn Diagram: The overlapping region of A and B = P(A ∏ B) A B (b) No, A and B are not Independent. By definition, two events are said to be Independent if the outcome of one event doesn’t affect the other. Here, A and B are said to be dependent events. (c) P(AU B)= P(A) + P(B) – P(A ∏ B) = 0.142+0.051-0.031= 0.162 The Probability that A or B or both to occur is 0.162 (d) P(A| B)= P(A ∏ B) = 0.031 / 0.051 = 0.608 P(B) 9) (a) Probability that a woman who gave birth in 1992 was 24 years or younger, P(Age<= 24) = P(Age<15) + P(15-19) + P(20-24) = 0.003 + 0.124 + 0.263 = 0.39 (b) Probability that a woman who gave birth in 1992 was 40 years of age or older, P(Age>=40) = P(40-44) + P(45- 49) = 0.014 + 0.001 = 0.015 (c) Let P(A) be the probability of the mother who was not yet 20 years of age Let P(B) be the probability of the mother under 30 years of age Now, P(A| B) = P(A ∏ B) = 0.003 + 0.124 = 0.187 P(B) 0.003 + 0.124 + 0.263 + 0.290 (d) Let A be the probability of the mother under 40 years of age Let B be the probability of the mother under 35 years of age Now, P(A| B) = P(A ∏ B) = 0.085 P(B) = 0.85 0.085 + 0.014 + 0.001 13) Given: Sensitivity = 0.85, P(T+| D+) = 0.85 Specificity = 0.80, P(T-| D-) = 0.80 Let D+ be the event that an woman has breast cancer Let D- be the event that an woman does not have breast cancer Let T+ represent positive screening test Let T- represent negative screening test (a) Probability of a false negative test result, P(T-| D+) = 1 – sensitivity = 1 – 0.85 = 0.15 (b) Probability of a false positive test result, P(T+| D-) = 1 – specificity = 1 – 0.80 = 0.20 (c) Probability that she has cancer, given that the test is positive, P(D+ | T+) : P(D+ | T+) = P(D+) * P(T+| D+) P(D+) * P(T+| D+) + P(D-) * P(T+| D-) P(D+ | T+) = 0.85 * 0.0025 = 0.0105 0.85 * 0.0025+ 0.9975 * (1 – 0.80) Thus, we infer that given the mammogram is positive, there is a 1% probability that she has breast cancer. 14) Let D+ be the event that an individual has carpal tunnel syndrome Let D- be the event that an individual has carpal tunnel syndrome Let T+ represent positive test result Let T- represent negative test result Given: Sensitivity = 0.67, P(T+| D+) = 0.67 P(D+) = 0.15 Specificity = 0.58, P(T-| D-) = 0.58 P(D- ) = 0.85 (a) Predictive value of a positive test result P(D+ | T+): P(D+ | T+) = P(D+) * P(T+| D+) P(D+) * P(T+| D+) + P(D-) * P(T+| D-) P(D+ | T+) = 0.15 * 0.67 = 0.22 0.15 * 0.67 + 0.85 * (1 – 0.58) The predictive value of a positive result is 22% (b) (i)If P(D+) = 0.10, P(D-) = 0.9 P(D+ | T+) = 0.10 * 0.67 = 0.15 0.10 * 0.67 + 0.9 * (1 – 0.58) (ii) If P(D+) = 0.05, P(D-) = 0.95 P(D+ | T+) = 0.05 * 0.67 = 0.08 0.05 * 0.67 + 0.95 * (1 – 0.58) Thus we see from i and ii that for a 5% change in prevalence, there is a 7% change in predictive value. (c) 1, 000, 000 People Prevelance = 0.15 Prevelance = 0.85 CTS 150, 000 SENS.= 0.67 NON - CTS 850, 000 SENS.= 0.33 Test (+) 100, 500 Test (-) 49, 500 Test (+) 457, 500 Legend:CTS : Carpal Tunnel Syndrome SPEC= 0.42 Test (+) 357, 000 SPEC.= 0.67 Test (-) 493, 000 Test (-) 542, 500 15) Test Disease Positive Negative Total Present 302 179 481 Total Absent 80 372 452 382 551 933 (a) Sensitivity of radio nuclide ventriculography = P(T+ | disease) = 302 / 481 = 0.6278 Specificity of radio nuclide ventriculography = P(T- | No disease) = 372/452 = 0.8230 (b) Let D+ be the event that an individual has disease Let D- be the event that an individual doesn’t have disease Let T+ represent positive test result Let T- represent negative test result Sensitivity = 0.63, P(T+| D+) = 0.63 P(D+) = 0.10 Specificity = 0.82, P(T-| D-) = 0.82 P(D- ) = 0.9 P(D+ | T+) = P(D+) * P(T+| D+) P(D+) * P(T+| D+) + P(D-) * P(T+| D-) = 0.10 * 0.63 = 0.28 0.10 * 0.63+ 0.9 * (1 – 0.82) There is 28% probability that one has CAD given that he/ she tests positive (c) P(D- | T-) = P(D-) * P(T-| D-) P(D-) * P(T-| D-) + P(D+) * P(T-| D+) = 0.9 * 0.82 = 0.95 0.9 * 0.82+ 0.1 * (1 – 0.63) There is a 95% probability that one does not have CAD if he/she tests negative. 16) (a) Sensitivity : P(T+| D+) False positive result : P(T+| D-) Specificity : P(T-| D-) False negative result: P(T-| D+) As the cutoff point is raised, the probability of False positive result increases. Also, the probability of a false negative will decrease (b) ROC Curve 0.975 0.97 0.965 Selectivity 0.96 0.955 0.95 0.945 0.94 0.935 0.93 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 0.11 1- Specificity (c) The most appropriate cutoff level would be 9 ng/ml of cotinine, because this point appears the closest to the upper left hand corner of the above plot. This point maximizes sensitivity and specificity simultaneously. 18) Relative Risk = P( Pregnancy | Using safety methods) P( Pregnancy | without protection) Method None Diaphragm Condom IUD Pill Probability of Pregnancy 0.431 0.149 0.106 0.071 0.037 Relative Risk 0.3457 0.2459 0.1647 0.0858 Each contraception method offers a lower relative list than the previous one. Birth control pills offer the greatest protective effect from pregnancy in comparison with no contraception method. 19) (a) Probability of suffering = No. of symptoms No. of children SES Low Medium High No. of Children 79 122 192 .No. with Symptoms 31 29 27 Probability of Suffering 0.3924 0.2377 0.1406 (b) P(PRS – low, middle SES) = ( 29+31) / (79+122) = 0.299 Odds ratio= P(PRS- low, middle SES)/ 1 - P(PRS-low, middle SES) = 0.29 / (1 – 0.29) P(PRS- High SES)/ 1 - P(PRS-High SES) Odds Ratio = 0.14 / (1 – 0.14) 1.74 (c) Higher socioeconomic status appears to have a protective effect on the development of respiratory symptoms. Children with higher SES were less likely and those with lower SES were more likely to have respiratory systems.