HW 3 [PG] Pages 155-157:

Exercises: 7, 8, 9, 10, 11, 13, 14

7. a. (A intersection B) is the event that the individual is exposed to 
high levels of both carbon monoxide and nitrogen dioxide.
  b. (A union B) is the event that the individual is exposed to either
     carbon monoxide or nitrogen dioxide or both.
  c. The complement of A is the event that the individual is not exposed
     to high levels of carbon monoxide.
  d. The events A and B are not mutually exclusive.

8. b. Since P(A) x P(B) = 0.142 x 0.051 = 0.0072
           P(A intersection B) = 0.031,
     so P(A) x P(B) not equal P(A intersection B), these two events
     are not independent.
   c. P(A union B) = 0.141+0.051-0.031 = 0.162
   d. P(A|B) = P(A intersection B)/P(B) = 0.031/0.051=0.608.

9. a. P(<= 24) = P(<15 or 15-19 or 20-24)
            = P(<15) + P(15-19) + P(20-24)
            = 0.003 + 0.124 + 0.263
            =0.390
  b. P(=> 40) = 0.014+0.001=0.015
  c. P(<20|<30) = P(<20 and <30)/P(<30) = P(<20)/P(<30) 
                = (0.003+0.124)/(0.003+0.124+0.263+.290)
                = 0.187
  d. P(<40|=>35)= 0.085/(0.085+0.014+0.001)
                =0.850

10. a. 0.387
    b. P(Medicare or Medicaid or other)=0.345+0.116+0.033
                                       =0.495 
    c. P(Medicare|goverment program)=0.345/0.494 = 0.698

11. a. P(both uninsured) = 0.123 x 0.123 = 0.015
    b. P(both insured) = (1-0.123) x (1-0.123)=0.769
    c. P(5 adults uninsured) = 0.123 x 0.123 x 0.123 x 0.123 x 0.123
                             =0.000028

13. a. False negative
       P(-test|ca) = 1-sensitivity=1-0.85=0.15
    b. False positive
       P(+test|no ca) = 1-specificity=1-0.80=0.20
    c. P(ca)=0.0025, P(no ca)=0.9975, so

 P(ca|+test)=P(+test) P(ca) /[P(+test) P(ca) + P(+test|no ca)P(no ca)]
            = (0.0025 x 0.85)/[0.85 x 0.0025 + 0.20 x 0.9975]
            =0.0105

14. a.   P(+test|cts)=sensitivity=0.67
         P(+test|cts)=1-specificity=1-0.58=0.42
   If the prevalence of carpal tunnel syndrome is 15%, then
   P(cts)=0.15 and P(no cts)=0.85. The predictive value of a 
  positive test is 
                 P(cts|+test) = 0.22
   b. If the prevalence is 10%, then 
                 P(cts|+test) = 0.15
      If the prevalence is 5%, then
                 P(cts|+test) = 0.08.
Therefore, as the prevalence of cts decreases, the predictive value
of a positive test decreases as well.