MEASURES OF DISEASE - ANSWERS

Measures of Disease - Problem Bank Answers

Question 1
The percentage of freshman girls who become pregnant over the course of their high school year. Cumulative incidence

The percentage of senior boys who are fathers at the time of graduation. Prevalence

The number of live born babies who die of sudden infant death syndrome during the first year of life per 100,000 baby-years of follow-up. Incidence rate

The percentage of infants weighing less than 2,500 grams at birth. Prevalence

The lifetime risk of breast cancer. Cumulative incidence


Question 2
Incidence measures new cases of disease, which reflects a change in disease status. Prevalence captures both new and existing cases of disease; temporality of risk factor and outcome is easier to establish when the case occurs during the study (Kleinbaum pg 99).


Question 3
Design a) because this design allows for better detection of new cases of disease. (Kleinbaum pg 99). A study at one point in time tends to assess prevalent cases, given that temporality is difficult to establish.


Question 4
A. Calculate the prevalence of gastroenteritis in the class on October 1.

5/100

B. Calculate the prevalence of gastroenteritis in the class on October 30.

0/100

C. Calculate the cumulative incidence of gastroenteritis in the class during the month of October.

3/(100-5) = 3/95. Remember that only the population at risk at the beginning of October is eligible for the denominator.


Question 5
A. Compute the cumulative incidence of Legionnaires' disease among delegates and nondelegates.

Cumulative incidence (CI) among the delegates equals a/(a + b) =125/1,849 = 0.068 (in 41 days). 

CI among the nondelegates equals c/ (c + d) = 3/762 = 0.004 (in 41 days).


 B. Calculate the cumulative incidence ratio of Legionnaires' disease among delegates compared with nondelegates.

Cumulative incidence ratio equals CI-Delegates /CI-Non-delegates = 0.068/0.004 =17.0.

C. State in words the meaning of this measure.

The risk of Legionnaires' disease was 17 times greater among delegates than among nondelegates.

Or, a delegate had 17 times the risk of contracting Legionnaires' disease as did a nondelegate.

D. Calculate the cumulative incidence difference of Legionnaires' disease among delegates compared with nondelegates.

Cumulative incidence difference = CI_nondelegates-CI_delegates=0.068-0.004 = 0.064 (in 41 days).

E. State in words the meaning of this measure.

If delegate status is a "cause" of Legionnaires' disease, then 64 cases per 1,000 delegates would be eliminated if the delegates had been nondelegates.

Note that delegate status is not the actual cause of Legionnaire's disease but reflects which part of the hotel that the individual entered during the convention. Further investigation found that the organism that causes Legionnaires' disease was present in certain air conditioners.

F. Calculate the attributable proportion of Legionnaires' disease among the delegates.

Attributable proportion = [(CI_delegates-CI_nondelegates)/CI_delegates] x 100% = [(0.068 - 0.004)/0.068] x 100% = 94%.

G. State in words the meaning of this measure.

This means that 94% of the cases of Legionnaires' disease among the delegates can be attributed to their delegate status.


Question 6
a. What is the point prevalence on May 1, 1992?

4 cases / 8 people alive in the population = 0.5

Please note that the denominator is the population of interest. This includes those who have developed the disease (otherwise it would not be a proportion) and excludes those who have died.

b. What is the cumulative incidence for March 1, 1992 to December 31, 1992?  Interpret this measure.

2 incident cases in time period / 5 at risk at beginning of time period = 0.4

Only 5 at risk because person 2 had died and persons 4, 7, 8, and 10 already had the disease.

Interpretation: The probability of getting the disease between Mar 1 and Dec 31 among the at-risk population on Mar 1 was 0.4

c. What is the incidence density from March 1, 1992 to December 31, 1992? Interpret this measure. Calculate the person-time at risk for the 5 people at risk at the beginning of the time period.

Person 1 was at risk from January through March = 1 month Person 3 was at risk for the whole time period = 10 months Person 5 was at risk from January through August = 6 months Person 6 was at risk for the whole time period = 10 months Person 9 was at risk for the whole time period = 10 months

Total person-time at risk = 37 months

Number of incident cases / person-time at risk = 2/ 37 person-months = 0.054 cases per person-month

Interpretation: The rate of incidence of disease in the population at risk from Mar 1 1992 to Dec 31 1992 was 0.054 cases per person-month

d. Which measure estimates risk and which estimates rate?   What is the difference?  Which measure can be applied to individuals?

CI estimates risk, which is the average individual risk of getting the disease over the time period. ID estimates rate or the average rate of disease occurrence over the time period in the population.

Question 7
A. Prevalence of the disease on January I, 2011; July I, 2011; and December 31, 2011

i. 4/1,000.

ii. 6/996. Remember that the population size has decreased by 4 because of death.

iii. 4/994. Again, the population is smaller.

B. Cumulative incidence of disease during 2011

6/996. Only those who do not have the disease of interest are "at risk."

C. Cumulative incidence of death during 2011

6/1,000. Everyone in the population is at risk of dying.


Question 8
The incidence rate was 12/100,000 person-years, which is equivalent to person-years. Solving for "x", we arrive at 30 new cases.


Question 9
Group A

Group B

Age

(years)

Cases of illness

Population

Rate (A)

Cases of illness

Population

Rate (B)

0‑4

6

100



12

200



5‑19

60

200



30

400



Total

66

300

66/300=0.22

42

600

42/600=0.07



The incidence rate for Group A is 0.22 and for B is 0.07. Group B appears healthier than Group A.


Question 10
No. It is also necessary to know the size of the population and the amount of follow-up time in each state.


Question 11
The following calculation assumes that everyone is followed for the entire two years and that all cases occur at the end of the second year: 60/(100,000 persons x 2 years) = 60/200,000 person -years or 30/100,000 person-years. Different assumptions can be made regarding the length of follow up for cases and noncases and the time of case occurrence


Question 12
Incidence measures how often new cases occur over a period of follow-up whereas prevalence measures how often existing cases occur at a moment or short period of time.


Question 13
Prevalence and incidence are related through the formula P = I x D where P denotes prevalence, I denotes incidence rate, and D denotes the average duration of the illness in the population under study.


Question 14
a. Number of campers who developed gastroenteritis within 24 hours after eating potato salad at the dining hall. Incidence
b. Number of persons who reported having hypertension at some point during the following year as part of National Health Interview survey. Prevalence
c. Occurrence of acute myocardial infarction in participants during the first 10 years of follow up in Framingham heart study. Incidence


Question 15
Prevalence is useful for quantifying the general burden of disease, especially for chronic diseases that people live with for a long time. It is also useful to help decide how to allocate resources.
Incidence is useful for acute diseases that people tend to have for a shorter period of time, or when you’re interested in following an outbreak. It is useful for understanding disease causation.

Note: Sometimes, it can be appropriate to use one measure rather than the other. For example, in the 1970s and 1980s, it made more sense to calculate the incidence of AIDS because new cases came quickly and people died quickly; now that HIV is a disease that people live with for longer periods of time, it could make more sense to calculate the prevalence depending on your research question. For Zika virus, we’re interested in incident cases of microcephaly associated with the virus, but we might also be interested in the prevalence of maternal antibodies to Zika.

Another example has to do with smoking. Among youth, we are typically concerned with onset of smoking because most smokers initiate between the ages of 14 and 17. Therefore, we would want to measure incident cases of smoking among youth. With established adult smokers, however, we often measure the prevalence of smoking in order to estimate the burden of smoking in a given population. For example, the CDC reports the current cigarette smoking among adults 18 and older in the United States annually. This statistic is the population prevalence, and in 2014 the prevalence of smoking among US adults was 16.8%.


Question 16
c. Disease is rare


Question 17
a. In a steady state population, what condition is required for the following formula to hold?

Prevalence (P) ≈ Incidence density (ID) * duration of disease (D)

While incidence measures the frequency with which new disease develops, prevalence is defined as the total proportion of the population that is diseased. 

The disease is rare

b. Which equation can be used to link prevalence and incidence density in a steady state population when the condition from part a) is not met?

P =  ID * D

(ID * D) + 1

c. If the cumulative incidence of H1N1 (a specific strain of influenza) is 20.5/100,000 cases per year, and the duration of the flu is 2 weeks, what is the annual prevalence? (hint: Calculate the prevalence, incidence, or duration using the equations.)

Since the prevalence is <10% either equation is ok. Using either equation gives nearly the same result. Furthermore, since the disease is rare, CI approximates the ID*delta t (change in time); in this case the relevant time is 1 year, so CI/1=ID and we could use the CI in the above equations instead of the ID. We will calculate this both using CI to approximate ID and using ID directly. Either way is an acceptable answer to this question.

Estimate P using CI

Equation 1: (20.5/100,000)*(14/365) = prevalence of 7.8 *10^-6 

Equation 2: [(20.5/100,000)*(14/365)]/ {[(20.5/100,000)*(14/365)]+1} = prevalence of 7.8 *10^-6

Estimate P using ID

First we will convert the CI to the ID using the equation: R=CI=1-e^(-ID*Ät)

Using this formula requires us to assume that the ID is constant over the period of time for which we would like to calculate ID and prevalence. We think this is a reasonable assumption to make because we are interested in estimating the prevalence over a one-year period, which we can consider a yearly average of point prevalences at different points in the year.

20.5/100,000 year=1-e^(-ID*1 year)

0.000205-1=-e^(-ID) ln(0.999795)=-ID ID=0.000205=20.5/100,000


So we see that the CI and the ID are identical in this case. You can continue the problem as laid out above under “Estimate P using CI”


Question 18
a. Based on the above information, calculate the incidence rate of HIV infection among prisoners in the Nevada prisons.

The incidence rate of HIV infection among prisoners in the Nevada prisons is 2/1207 = .001657 per person-year.

b. Express the incidence rate calculated in part a in terms of cases per 1000 person-years.

1.7 cases per 1000 person years

c. Why can’t you obtain an estimate of risk based on the information provided?

Not all prisoners were followed for the same amount of disease-free follow-up time.

d. Why would estimating risk likely be inappropriate for these data?

Same reason as in c.

e. Calculate the prevalence of HIV infection among incoming prisoners in the Nevada prison under study.

The prevalence of HIV infection among incoming prisoners in the Nevada prisoners under study is 36/1105 = .0326


Question 19
Since the disease is rare (prevalence is <10%) you can use the formula in #1. (20.5/100,000)*(14/365) = prevalence of 7.8 *10^-6


Question 20
5/970 person-years = 5.15 per 1,000 person-years. 95 subjects each contributed 10 years of disease-free follow-up (950 person-years total), and the other 5 subjects who developed disease contributed an additional 20 person-years of follow-up before they developed disease. So, there were 5 new cancers per 970 person-years of follow-up.


Question 21
There were 5 women living with ovarian cancer on December 31, 1999 and none of the subjects left the study population. So, the prevalence was 5/100 or 5%.


Question 22
When the population is in steady state, P = IR x D, where P is prevalence, IR is incidence rate, and D is average duration. Thus, P = 500/100,000 person-years x 3 years or 1,500/100,000


Question 23
The main similarity is that they both quantify the number of new cases of disease that develop in a population at risk during a specified period of time. The main difference is that the incidence rate is a true rate that directly integrates person-time of observation into the denominator, and cumulative incidence is a proportion whose denominator is the population at risk at the start of the observation period. Time in cumulative incidence is expressed only by words that go along with the proportion.


Question 24
2. Cumulative incidence


Question 25
1. Prevalence


Question 26
3. Incidence rate


Question 27
1. Prevalence


Question 28
Crude heart disease death rates and developing country equals (0.30 x 2/100,000 person-years) + (0.40 x 20/100,000 person-years) + (0.30 x 40/100,000 person-years) = 20.6/100,000 person-years. The age-adjusted rate is better because the age structures of the two populations are different.


Question 29


Question 30


Question 31


Question 32


Question 33
D. All of the above


Question 34
C. Both A and B


Question 35
B. It would decrease