# doc ```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.
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