6.1 See text.
6.2 Answer appears on p. 140 is correct. Additional notes follow:
· There is a misprint in text. The question should read “the remaining 950.” This misprint is documented in the errata listing: http://www.sjsu.edu/faculty/gerstman/eks/errata.htm
· The answer to part C (p. 140) includes a life-table adjustment of the denominator. This assumes that the 64 cases have onsets half-way through the 5-year follow-up period. Thus, the 64 cases contribute half the potential person-time at risk, or 2.5 person-years each, to the denominator. The 886 non-cases contribute the full 5 person-years each. Thus, the person-time in the cohort is estimated as [(64 persons × 2.5 years] + (886 persons × 5 years)] = 4590 person-years. The IR = 64 / 4590 = .0139 years-1 = 13.9 per 1000 person-years.
· Without this life-table adjustment, the estimate of person-time in the cohort = 950 persons × 5 years = 4750 person-years and the IR = 64 / 4750 = 0.0135 years-1 = 13.5 per 1000 person-years. Note that the life-table adjustment has only a minor effect on the IR estimate because the disease is rare (i.e., the person-time denominator was only slightly down-adjusted).
6.3 - 6.9 See text.
6.10 Mortality rate and life expectancy.
(A) IR = 2 / 150 years = .01333 / year
(B) Life expectancy = 1 / .01333 year = 75 years
(C) Life expectancy is the reciprocal of mortality rate
6.11 Mortality rate and life expectancy.
(A) IR = 2 / 175 years = .01143 / year.
(B) Life expectancy = 1 / .01143 year = 87.5 years.
6.12 Person-time and a survival curve.
(A) Person-time = 1 + 2 + 4 + 4 = 11 years
(B) 50% survived two years; 50% survived four years
(C) Between years 0 and 1, the area = 4 persons × 1 year = 4
person-years.
Between years 1 and 2, there are 3 persons × 1 year = 3 person-years.
Between years 2 and 4, there are 2 persons × 2 years = 4 person-years.
Total, there are 4 + 3 + 4 = 11 person-years.
6.13 Breast cancer.
(A) IP = 250/ 9500 = .02632 = 26.3 per 1000
(B) Without the life-table adjustment of the denominator: IR = 250 / (9500 × 5) = 250 / 47,500 years = 0.005263 / year = 5.26 per 1000 person-years. With the life-table adjustment, subtract half of the follow-up time in the cases (250 × 2.5 = 625 person-years) 250 / (47,500 - 625) years = 250 / 46,875 years = 0.00533 / year = 5.33 per 1000 person-years.
6.14 Cohort study of CHD.
(A) 850 – 50 = 800 @ risk; IP = 100 / 800 = .125 = 12.5
per 100
(B) Person-years = (800 – 100) ∙ 10 + (100 ∙ 5) = 7500; thus
IR = 100 / 7500 person-years = .0133 per year = 1.33 per 100 person-years
6.15 Rates of driving
errors.
(A) one mistaken observation / 2 miles = half a mistake per mile
(B) 1 mistaken observation / 400 observations = 1 / 400
(C) “A” is a rate because the numerator and denominator are in different units;
the dimension of the rate is “mistaken observations per mile.” “B is a
proportion because the numerator is a subset of the denominator.
(D) 1 / 500 miles : 1 / 61,000 miles = 122:1
6.16 Open population.
(A) N = 6
(B) person-time = 4 + 4 + 2 + 3 + 9 + 6 = 28
(C) Average number of persons (N-bar) = 28 / 10 = 2.8
(D) IR = 4 / 28 years = .143 / year = 14.3 / 100 years
6.17 Another cohort.
(A) P = 10 / 150 = .0667
(B) P = (10 + 16) / 150 = .1733
(C) IP = 16 / (150 – 10) = .1143
(D) person-years = (140 – 16) ∙
5 + (16 ∙ 2.5) = 620 + 40 = 660
IR = 16 / 660 years = .0242 = 24.2 per 1000 years
6.18 Just like Exercise 6.1. (A) 1 in 7 (B) 2 in 7 (C) 2 in 6
6.19 Just like Exercise 6.3. (A) 12 (B) 10 (C) 10 (D) 30
6.20 (A) 1 in 3 (33.3%) (B) 1 in 12 person-years = 8.33 per 100 person-years
6.21 Twice the prevalence with the same risk. No, you cannot conclude population A has twice the risk because prevalence depends on both incidence and duration of disease. For example, population A and B can have the same risk but the cases in population A may survive twice as long, accounting for double the prevalence.