1. Given a population with a stable age distribution, how can one explain that the prevalence of a disease is decreasing despite a constant incidence rate?
Prevalence is associated with both the incidence rate and the mean duration of the disease of the interest. With the incidence rate held constant, decrease in the mean disease duration can cause the prevalence to decrease. This can happen with improved treatment that cures disease faster or a more aggressive subset of disease that kills faster becoming more common.
2. Which is a better method for evaluating the success of a preventive program, comparing the change in disease incidence rates or comparing changes in disease prevalence? Explain.
Incidence rate is more appropriate. As explained in #1, prevalence a composite measure of the incidence rate and the mean disease duration. Prevention (presumed to be primary prevention here) by definition should decrease new occurrence of disease, which should result in decrease in the incidence rate. This does not necessarily result in prevalence as soon as incidence, as a large pool of people who already have the disease of interest can keep existing.
3. In a mass screening of 1,000 65-year-old men, 100 were found to have polyps of the colon. During the following 10-year period, another 200 men with colon polyps were discovered. Which measure(s) of disease occurrence can be calculated? Calculate this/these.
The prevalence of colonic polyps at the beginningof the study period is 100 / 1000 = 10%. The incidence rate of colonic polyps during this 10-year period is 200 / 1000 / 10 year = 2% per year = 2 / 100 person-years. The cumulative incidence or risk, which is the proportion newly affected during this period, is 200 / 1000 = 20% (10-year risk).
4. An investigator notices that children develop chronic bronchitis in the first year of life in 3 of 20 households where both parents are chronic bronchitics, as compared with 5 in 100 households nationally. What measure of disease occurrence is used above?
Number of people affected in the first year of life / number at risk at the beginning of the first year Thus, these figures are risks or cumulative incidences.
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 was 18,500 at the beginning of the period, and 21,500 at the end. Which measure of disease occurrence can be calculated? Calculate this.
The number at risk increased during this time, thus, not everyone has the same length of follow-up. This makes risk an inappropriate measure, thus, incidence rate is appropriate. The increase 21500 - 18500 = 3000 people are assumed to have contributed 2.5 years each. Thus, the total follow up duration is 18500 * 5 + 3000 * 2.5 = 100000. Therefore, the incidence rate is 270 / 100000 person-years = 27 / 10,000 person-years.
6. A sample of 10 people who suffer from migraine headaches are followed for various lengths of time, and the number of migraines are recorded. Using the data below, estimate the average rate of migraines in units of #/person-year.
part6 <- read.table(head = T, text = '
"Subject" "follow_up" "num_migraine"
1 3 0
2 6 2
3 2 1
4 4 2
5 4 1
6 8 2
7 4 1
8 5 3
9 5 2
10 3 0
')
average_rate <- sum(part6$num_migraine) / sum(part6$follow_up)
The average rate is calcuated by dividing the total number of events in the cohort with the total length of the follow-up. Thus,
(Total # of migraine headaches) / (Total length of follow-up in months) = 0.32 / person-month = 3.82 / person-year.
7. Assuming a constant incidence rate, the exact relationship between t-year cumulative incidence (Ct) and incidence rate (I) is given by Ct = [ 1 - exp(- I * t) ]. A good approximation, under certain circumstances, is: Ct \( \approx \) I * t.
incidence_rate <- c(1, 2, 5, 10, 20, 50, 100, 200, 500)
A <- incidence_rate / 1000
B1 <- 1 - exp(-1 * A * 1)
C1 <- 1 - exp(-1 * A * 10)
B2 <- A * 1
C2 <- A * 10
res.7 <- data.frame(incidence_rate, A, B1, C1, B2, C2)
res.7[,2:6] <- lapply(res.7[,2:6], function(x) sprintf("%.4f", x))
res.7
incidence_rate A B1 C1 B2 C2
1 1 0.0010 0.0010 0.0100 0.0010 0.0100
2 2 0.0020 0.0020 0.0198 0.0020 0.0200
3 5 0.0050 0.0050 0.0488 0.0050 0.0500
4 10 0.0100 0.0100 0.0952 0.0100 0.1000
5 20 0.0200 0.0198 0.1813 0.0200 0.2000
6 50 0.0500 0.0488 0.3935 0.0500 0.5000
7 100 0.1000 0.0952 0.6321 0.1000 1.0000
8 200 0.2000 0.1813 0.8647 0.2000 2.0000
9 500 0.5000 0.3935 0.9933 0.5000 5.0000
In the short duration case (B1 vs B2, t = 1 year), up to an incidence rate of 0.2 / person-year, corresponding risk B1 (0.1813) and risk B2 (0.200) are close enough. In the long duration case (C1 vs C2, t = 10 years), up to an incidence rate of 0.02 / person-year, corresponding risk C1 (0.1813) and risk C2 (0.200) are close enough. The rule of thumb is probably the approximate risk less than or equal to 0.2.
1. Of 2,872 individuals who had received radiation treatment in childhood because of enlarged thymus, 24 developed cancer of the thyroid and 52 developed benign thyroid tumors. A comparison group consisted of 5,055 children who had received no such treatment (brothers and sisters of those who had received radiotherapy. During the follow-up period, none in the comparison group developed thyroid cancer, but six developed benign thyroid tumors.
a. Calculate the appropriate ratio measure of association for thyroid cancer. Of which measure of disease occurrence is this a ratio?
The risk in two groups are compared. However, it follows: Risk ratio = (24 / 2872) / (0 / 5055) = Not defined. As the risk in the control group was zero, the risk ratio comparing the irradiated group to control is not well-defined.
b. Calculate the appropriate ratio measure of association for benign thyroid tumors.
The risk ratio is: (52 / 2872) / (6 / 5055) = 15.25 It means the rate of benign thyroid tumors are 15.25 times higher in the radiated group.
c. Suppose the study population was followed for approximately 50 years. Estimate the incidence rate for thyroid cancer among patients who received childhood irradiation for thymus enlargement, assuming constant incidence over time.
24 / (2872 * 50) = 0.0001671309 / 1 person-year = 1.67 / 10,000 person-years.
2. In a certain noisy factory, workers are given earplugs and are expected to wear them. An industrial hygienist inspecting the plant found that 100 of the 500 workers in the factory were not wearing the earplugs because they regarded them as uncomfortable and as a nuisance. When all the workers were given a hearing test, it was found that 16 of the earplug wearers and 40 of the nonwearers had developed significant hearing loss. All had had normal hearing at their preemployment exams 1 year earlier, when the plant opened.
a. What was the risk ratio of hearing loss in non-wearers as compared to wearers?
Risk ratio = (40 / 100) / (16 / 400) = 10 Thus, not wearing earplugs increase the risk of hearing loss 10-fold.
b. What was the risk of hearing loss associated with not wearing earplugs?
The risk difference (attributable risk), i.e., the risk in the exposed - the risk in the unexposed is the risk associated with exposure. Risk difference = (40 / 100) - (16 / 400) = 0.36 Thus, not wearing earplugs increase the risk of hearing loss by 36 percentage points.
c. What proportion of the hearing loss in the nonwearers was associated not wearing earplugs?
Attributable proportion = 0.36 / 0.40 = 0.90 = 90%. Among the 40 hearing loss among the non-wearers, 36 of them (90%) are attributed to their practice of not wearing earplugs.
d. If the associations above were causal and everyone wore their earplugs, what reduction in hearing loss would occur in the factory population as a whole?
The number would be reduced by 36, from 56 to 20. The reduction is the same as the number calculated in the attributable proportion part, and 20 / 500 = 4% is the same as the risk in the earplug wearers (16 / 400 = 4%).
e. What proportion of the hearing loss in all the workers was associated with not wearing earplugs? Among 56 cases of hearing loss, 36 were associated with not wearing earplugs. Thus, Population attributable risk = 36 / 56 = 0.643 = 64.3% Thus, 64.3% of the hearing loss was attributable to the practice of not wearing ear plugs among all the workers (population attributable risk percent).
Population attributable risk percent (PAR%): ((56/500) - (16/400)) / (56/500)