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MEASURES OF DISEASE FREQUENCY - EXERCISES

1. In a mass screening of 1,000 65-year old men, 100 were found to

have a certain disease. During the following 10-year period another

200 contracted this disease. Which measure (s) of disease frequency

can be calculated? Calculate this/these.

Answer: P = 100 / 1,000 = 0.10

CI Cumulative Incidence

= 200/ 1000-100 = 0.22 over a 10-year period

2. Among those admitted to a psychiatric treatment center there

were carriers of hepatitis B in some wards but not in others. To

investigate the extent to which this affected the occurrence of hepatitis

B among personnel, employees of the treatment center were examined

with regard to the presence of serological markers. Of 67 people

working on the wards with carriers, 14 had markers for hepatitis B. Of

72 people working on the other wards, 4 had these markers. Which

measure of occurrence of markers can be calculated? Calculate this for

each of the two personnel groups.

Answer: P1 = 14 / 67 = 0.21

P0 = 4 / 72 = 0.06

3. To enable early discovery and treatment of cervical cancer,1

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regular gynaecological check-ups were carried out on women aged

30-59 years. Follow-up of women who were found not to have the

disease at the initial examination covered 338,294 person-years at risk

and resulted in the identification of 123 new cases of carcinoma in

situ. Which measure of occurrence can be calculated? Calculate this.

Answer: I = 123 / 338,294 = 0.000363 per person-years

= 36.3 / 10 5 person-years

4. In a mass screening of 5,000 women, 25 of these were found to

have breast cancer. During the next five years 10 more of the

examined women developed breast cancer. Which measures of disease

occurrence can be calculated? Calculate these.

Answer: P = 25 / 5,000 = 0.005CI Cumulative Incidence

= 10/ 5000-25 = 0.002 over a 5-year period

5. During a 5-year period, 270 cases of duodenal ulcer occurred in

the male population of a city. The number of men in the city was18,500 at the beginning of the period and 21,500 at the end. Which

measure of disease occurrence can be calculated? Calculate this.

Answer: Number of cases = 270

Average population size = 18,500 + 21,500 / 2 = 20,000

Observation period = 5 years

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Total risk period = 20,000 x 5 = 100,000 person-years

I = 270 / 100,000 = 0.0027 per person-years

6. The figure below displays the follow-up experience of members

of a small study cohort between 1 January 1994 to 30 June 1999 from

entry (>) to follow-up until death (DC if due to disease C, DO for

other causes) or censoring (o). For those subjects contracting disease

C the time of diagnosis is also marked ( = onset of C).

= disease onset o= censoring or withdrawal

DC = death from disease C DO = death from other cause

Calculate the values of the following measures:

(a) incidence rate of disease C during the period from 1 Jan 1994 to31 Dec 1998,

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Answer: Prevalence of C on 30 September 1996 was 1/7 = 14 %

and Prevalence of C on 31 December 1998 was 3/5 = 60 %.

7. In the table below are given the size (in 1000s) of the male

population in Finland aged 0-14 years (the age range of childhood

in pediatrics!) on the 31 December in each year from 1991 to 2000.

The following numbers of cases describe the incidence and

mortality of acute leukaemia in this population for two calendar

periods: 5 years 1993 to 1997 (source: NORDCAN), and year 1999

only (source: Finnish Cancer Registry http://www.cancerregistry.fi/ ).

(a) Calculate the incidence rates of acute leukaemia in this

population for the two periods.(b) Calculate similarly the mortality rates of leukaemia.

1993-97 1999

person-years (in100000s)

1/2 (4.95+4.91)5=24.65 1/2 (4.85 + 4.81) 1 = 4.83

(a): incidence rate(per 10 5 y)

113/24.65 = 4.6 26/4.83 = 5.4

(b): mortality rate(per 10 5 y)

22/24.65 = 0.9 3/4.83 = 0.6

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http://www.cancerregistry.fi/http://www.cancerregistry.fi/

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(c) Is there evidence about any change in the incidence and/or

mortality between these two periods?

Answer: There are too few cases for any conclusions.(d) What would you conclude about the fatality of leukemia in

children?

Answer: It appears that most children survive, but this is not the

correct way of calculating survival or fatality measures!

8. The Alpha Tocopherol Beta Caroten (ATBC) Prevention Trial (N

Engl J Med 1994; 330:1029-35) addressed among other things the

possible benefits of daily intake of vitamin E supplements in reducingthe incidence of cancer among male smokers. The study population of

29133 regularly smoking 50-69 years old Finnish men were

randomized into two groups: active treatment (vitamin E

supplementation), and placebo (no supplementation). The following

results were obtained for cancer of the prostate after an averagefollow-up time of 6 years:

(a) Calculate the person-years at risk in the two study groups6

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separately.

Answer: Person-years in the two groups:

99(11.6/10000 )y= 85345 y,

151(17.8/10000) y= 84831 y

(b) Estimate the relative risk and excess risk measuring the

effect of daily supplementation with vitamin E on the risk prostate

cancer.

Answer: Relative risk:

incidence rate ratio = 11.6/17.8 = 0.652,

Excess risk:

rate difference = 11.6 17.8 = 6.2 per 10000 y,

(c) Estimate preventive fraction, whichever more appropriate, to

describe the proportional impact of vitamin E supplementation. Answer: Preventive fraction:

from rates: (17.8 11.6)/17.8 = 0.348 = 35%,

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