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Difficulty: Very hardMeasures of Disease Frequency and Association

A public health department is evaluating a workplace intervention at a large chemical manufacturing plant. Plant records indicate that 20%20\% of the workers are stationed in the 'Zone A' processing area, where they are exposed to a specific volatile organic compound (VOC). The remaining 80%80\% of workers are stationed in other areas and are not exposed to this VOC.

An epidemiological analysis reveals that the overall incidence of a particular respiratory condition among all workers in the plant is 1818 cases per 1,0001,000 workers per year. The relative risk of developing this respiratory condition for workers in Zone A, compared to workers in other areas of the plant, is 5.05.0.

The plant management is considering a targeted intervention to upgrade ventilation strictly in Zone A, which would completely eliminate the VOC exposure and reduce the risk for Zone A workers to that of the unexposed workers.

Based on these data, how many workers need to benefit from this ventilation upgrade in Zone A to prevent one additional case of the respiratory condition per year?

  1. A
    20
  2. 25Answer
  3. C
    56
  4. D
    100
  5. E
    125

Answer

25
The question asks for the number of exposed workers who need to undergo a targeted intervention to prevent one case of the disease. This concept is equivalent to the Number Needed to Treat (NNT), which is calculated as 1/Absolute Risk Reduction1 / \text{Absolute Risk Reduction}. Because the intervention completely eliminates the exposure, the absolute risk reduction is equal to the Attributable Risk (AR) of the exposure. Attributable Risk is the absolute difference in incidence between the exposed and unexposed groups (Ie−IuI_e - I_u). First, we must derive the incidence in the unexposed group (IuI_u) using the formula for total incidence: Itotal=(Pexposed×Ie)+(Punexposed×Iu)I_{total} = (P_{exposed} \times I_e) + (P_{unexposed} \times I_u). We know that Itotal=18/1,000=0.018I_{total} = 18/1,000 = 0.018, Pexposed=0.2P_{exposed} = 0.2, Punexposed=0.8P_{unexposed} = 0.8, and RR=5.0RR = 5.0, meaning Ie=5IuI_e = 5 I_u. Substituting these yields: 0.018=(0.2×5Iu)+(0.8×Iu)=1.8Iu0.018 = (0.2 \times 5 I_u) + (0.8 \times I_u) = 1.8 I_u. This gives Iu=0.01I_u = 0.01. The incidence in the exposed group is therefore Ie=5×0.01=0.05I_e = 5 \times 0.01 = 0.05. The Attributable Risk is 0.05−0.01=0.040.05 - 0.01 = 0.04. The number of exposed workers who need the intervention is 1/0.04=251 / 0.04 = 25.

Step-by-Step Solution

1
Set up the equation for overall incidence based on the strata of exposed and unexposed workers.
Itotal=(Pe×Ie)+(Pu×Iu)I_{total} = (P_e \times I_e) + (P_u \times I_u), where PP is prevalence and II is incidence.
The total incidence of a disease in a population is the weighted average of the incidences in its component subpopulations.
2
Substitute known values into the equation to find the incidence in the unexposed group (IuI_u).
Given Pe=0.2P_e = 0.2, Pu=0.8P_u = 0.8, Itotal=0.018I_{total} = 0.018, and Ie=5×IuI_e = 5 \times I_u. Therefore, 0.018=(0.2×5Iu)+(0.8×Iu)=1.0Iu+0.8Iu=1.8Iu0.018 = (0.2 \times 5 I_u) + (0.8 \times I_u) = 1.0 I_u + 0.8 I_u = 1.8 I_u. Solving for IuI_u yields 0.010.01 (or 1010 per 1,0001,000).
To calculate the Attributable Risk, we must first establish the absolute incidence rates for both groups.
3
Calculate the incidence in the exposed group (IeI_e).
Ie=5×0.01=0.05I_e = 5 \times 0.01 = 0.05 (or 5050 per 1,0001,000).
The incidence in the exposed group is needed to find the excess risk due to the VOC.
4
Calculate the Attributable Risk (AR) for the exposed workers.
AR=Ie−Iu=0.05−0.01=0.04AR = I_e - I_u = 0.05 - 0.01 = 0.04 (or 4040 excess cases per 1,0001,000 exposed workers).
The Attributable Risk represents the excess risk of disease directly caused by the exposure, which is the exact amount of risk the intervention will eliminate.
5
Calculate the Number Needed to Treat (NNT).
NNT=1/AR=1/0.04=25NNT = 1 / AR = 1 / 0.04 = 25.
The number of workers who need the intervention to prevent one case is the reciprocal of the absolute risk reduction (which equals the Attributable Risk in this preventative context).

Key Concept

Attributable Risk (AR) and Number Needed to Treat (NNT) derivation from population incidence
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Topics

Measures of Disease Frequency and Association

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