S501 (Section 3) EXAM 1 – SOLUTIONS Oct. 9, 2003

  1. (a) SEVERITY: qualitative, ordinal. RTIME: quantitative, continuous. HOSP: quantitative, discrete.

(b) m = 1*.6 + 2*.15 + 3*.1 +4*.1 +5*.05 = 1.85.

  1. s 2 = S (x - m )2 p(x) (sum over x in the table) = (1 – 1.85)2*.6 + (2 – 1.85)2*.15 + … remainging similar terms.
  2. P(HOSP > 2) = .1 + .1 + .05 = .25.
  3. Population: all emergency ambulance runs.
  4. statistic
  1. a) sample median = (0 + .7)/2 = .35 (average of the two middle terms).

    2. The median would not be affected because the two middle terms would remain the same.

    b) Estimate of population variance is s2 = (.2635)2 = .0649.

    c) s would not be affected: the spread in the data set would be the same. The whole data set would just be shifted 1.8 units to the right.

  2. a) binomial: B(n,p), with parameters n = 6, p = .9.

    3. b) P(4 detections) = C64 (.9)4(.1)2 = .098415 (note: C64 = 15).

    c) m = np = 6*.9 = 5.4, s = Ö npq = Ö 6*.9*.1 = Ö .54 = .735.

    d) 4 yrs = 208 weeks, so expect 208*.531441 = 110.5 weeks with 6 detections.

    e) P(1 detection) = .000054 so that, under the conditions of this problem, a week with only 1 detection is an extremely unlikely event. This suggests that the system is not functioning as stated – something may have changed, such as having an incompetent screener, defective x-ray machine, etc.

  3. a) P(type A, neither) = 2625/8766 = .2995.

    4. b) P(no peptic ulcer) = 1 – P(peptic ulcer) = 1 – 1796/8766 = 1 - .2049 = .7951. (There are other ways to organize this calculation.)

    c) P(none|type O) = 2892/4258 = .6792.

    d) Check the equation P(none) = P(none|type O): P(none) = 6087/8766 = .6944 which is not equal to P(none|type O) (see part (c)). Thus the events are not independent.

  4. a) C126 = 924

b) # of samples of 2 from remaining 8 = C82 = 28, so P(4 pre-treated in sample) = 28/924 = .03.

c) The treatment would tend to appear better than it actually is (if long-term steroids keep acting beneficially on back pain).

d) (i) To minimize placebo effect and to eliminate doctors treating the patients in the two groups differently. (ii) Both groups are taking some kind of pill, so are exposed to the same degree of placebo effect. Thus differences in the two groups can be ascribed to treatment effect rather than to placebo.