Occupational Epidemiology

Part 1

You are a 0.5 FTE occupational epidemiologist at the state health department. Unfortunately with limited grant dollars, you aren’t able to work full time on occupational surveillance and program support – you split your time with the lead surveillance program. Dr. Zeldon is an occupational physician and an active member of the occupational health advisory board to the state. He has asked you to provide some epidemiologic expertise to a worksite wellness program at a large manufacturing plant where he provides health services. As per the company’s policy, all employees at the plant are routinely tested for alcoholism by lab testing and physician exam. (Lab testing and physician exam is the gold standard assessment of alcoholism.) This information is included in each of the employment records. The company (and Dr. Zeldon) are interested in the association between alcoholism and depression among the employees. You work with a small team to determine the methodology for the clinical depression screening. It is decided by the group that the company will offer clinical depression screening using a validated, gold standard questionnaire. (Assume that there is no misclassification of disease status.)

With the methods determined, you assist Dr. Zeldon in recruiting volunteers from the plant to participate in the depression screening. You recruitment was rather successful; 335 employees elect to participate. Of these, 98 test positive and 237 test negative for depression. In abstracting their alcoholic status (Exposure) from their employment records, you construct the following:

	Alcoholic	Non-Alcoholic
Depression Case	44	54
Not Depression Case	48	189

Concerned about employee productivity and morale, the company’s management team decides to set a new policy that requires mandatory depression screening for all remaining employees based on these results. The team conducts the testing as well as abstracts alcoholic status from the employment records for the remaining employees. Your results for the entire employee population (volunteers + remaining) follow:

	Alcoholic	Non-Alcoholic
Depression Case	221	136
Not Depression Case	121	472

1. Calculate the odds ratio for the study based on volunteers and its 95% confidence interval. For full credit, write out your step-by-step process for arriving at your answer.

OR = (44/48) / (54/189) = 3.21

ln(OR) = ln(3.21) = 1.17

V = 1/44 + 1/54 + 1/48 + 1/189 = 0.067

95%CI = e^(1.17 +/- 1.96SQRT(0.067))

= e^(0.66) to e^(1.68)

= 1.93 – 5.37

2. Calculate the odds ratio for the study based on the whole employee population and its 95% confidence interval. For full credit, write out your step-by-step process for arriving at your answer.

OR = (221/121) / (136/472) = 6.34

ln(OR) = ln(6.34) = 1.85

V = 1/221 + 1/136 + 1/121 + 1/472 = 0.022

95%CI = e^(1.85 +/- 1.96SQRT(0.022))

= e^(1.56) to e^(2.14)

= 4.76 – 8.50

3. Compare the measures of association for the volunteer population and the entire study population. Summarize the findings. Is selection bias affecting the volunteer study? If so, what is the direction of the bias?

The depressed individuals in the volunteer group were 3.21x more likely to be alcoholics. The depressed individuals in the population were 6.34x more likely to be alcoholics. The odds ratios alone seem to suggest a bias toward the null, however, the confidence intervals of both groups overlap. This information alone is not enough to evidence a bias, the overlapping confidence intervals imply that the differences in odds ratios could be explained by random sampling variance alone

4. Calculate the sampling fractions for the volunteer population relative to the entire study population. Does the relationship between sampling fractions support the direction of the bias determined in question 1? Why or why not? For full credit, write out your step-by-step process for arriving at your answer.

Depression No depression

Alcoholism exposure 44/221 = 0.20 48/121 = 0.40

No alcoholism exposure 54/136 = 0.40 189/472 = 0.40

Healthy controls were sampled from the population at the same rates, irrespective of alcoholism exposure status.

Cases of depression with a positive alcoholism exposure status were less likely to be sampled compared to cases without exposure status. This information does substantiate the bias towards the null determined above. Positive depression cases with alcoholism exposure were sampled at a lower rate than what’s observed in the total population, suggesting a weaker relationship between depression and alcoholism than what really exists

5. What is the healthy volunteer effect? Does this scenario illustrate an example of the healthy volunteer effect? Why or why not?

The volunteer bias is the specific selection bias associated with systematic differences between study volunteers and those who choose not to volunteer. In our case, those who were depressed but not alcoholic were sampled at a higher rate than those who were depressed and alcoholic. The healthier of those two groups is overrepresented, making this bias qualify to be considered a healthy volunteer bias.

Part 2 – A continuation of the scenario in Part 1

The large manufacturing plant continues to be interested in monitoring alcoholism and depression among their employees. However, with limited resources, they would like to use a method that can assess alcoholism that does not require a laboratory test and physician exam. They are hoping that there is a questionnaire screening tool that could be used instead of the more expensive methods. They ask you to investigate this possibility. You go to the literature and find a validity study for the Short Michigan Alcohol Screening Test (SMAST) which is a brief questionnaire to determine alcoholism. The results of the validity study demonstrate the SMAST to have low sensitivity and specificity (sensitivity of 70% and a specificity of 74%). You conduct some hypothetical analysis to demonstrate the impact of using the SMAST on the study findings.

6. Provide an interpretation of the sensitivity and specificity that you found from the validity study of the SMAST.

The validity study showed that the SMAST questionnaire would miss 30% of the individuals who are alcoholic. This is what a sensitivity of 70% means. 74% specificity means that 26% of the non-alcoholics would be falsely screened as being alcoholic.

7. If the employees’ alcoholic status had not been assessed by lab testing and physician exam, but by the SMAST questionnaire instead, you would have been concerned with misclassification of exposure status. Using data for the entire employee population and the information from the validity study for the SMAST, calculate the odds ratio and 95% confidence interval for this situation, where misclassification of exposure occurs non-differentially. For full credit, write out your step-by-step process for arriving at your answer.

Depressed Not depressed

Alcoholic (0.7x221)+(0.26x136) (0.7x121)+(0.26x472)

= 190 = 207

Not alcoholic (0.74x136)+(0.3x221) (0.74x472)+(0.3x121)

= 167 = 386

OR = (190/167)/(207/386) = 2.12

ln(OR) = ln(2.12) = 0.75

V(ln(OR)) = 1/190 + 1/207 + 1/167 + 1/386 = 0.019

95%CI = e^(0.75 +/- 1.96SQRT(0.019))

= e^(0.48) to e^(1.02)

= 1.62 – 2.77

8. Is a bias introduced from non-differential misclassification using the SMAST in your hypothetical analysis? If so, what is the direction of the bias?

Yes, the non-differential misclassification in the SMAST analysis introduced a significant bias toward the null

9. Now you are interested in conducting a hypothetical analysis assuming differential misclassification of exposure status had occurred using the SMAST questionnaire. For those diagnosed with depression, you estimate the SMAST score to have a sensitivity of 65% and specificity of 80%. For the non-cases, you estimate the SMAST to have a sensitivity was 80% and specificity was 80%. Calculate the odds ratio and 95% confidence interval for this scenario. For full credit, write out your step-by-step process for arriving at your answer.

Depressed Not depressed

Alcoholic (0.65x221)+(0.2x136) (0.8x121)+(0.2x472)

= 171 = 191

Not alcoholic (0.8x136)+(0.35x221) (0.8x472)+(0.2x121)

= 186 = 402

OR = (171/191)/(186/402) = 1.93

ln(OR) = ln(1.93) = 0.66

v = 1/171 + 1/191 + 1/186 + 1/402 = 0.019

95%CI = e^(0.66 +/- 1.96SQRT(0.019))

= e^(0.39) to e^(0.93)

= 1.48 – 2.53

10. Is a bias introduced from differential misclassification using the SMAST in your hypothetical analysis? If so, what is the direction of the bias?

This differential misclassification also biases the results of the analysis towards the null

11. What would you recommend to the company regarding the use of the SMAST instead of the laboratory testing and physician exam? Use information from the validity study (sensitivity of 70% and a specificity of 74%) and the results from your hypothetical analysis to justify your response. Make sure to describe the pros and cons of the method you recommend.

The company should screen all employees for alcoholism using the SMAST, on a periodic basis. Individuals who screen positive for alcoholism should then be referred for physician and lab testing to confirm the screener results.

This strategy saves the company money by using the expensive test only on the group of employees with 70% proportion of true alcoholics (based on sensitivity). The further testing will screen out the 30% who aren’t actually alcoholic, which saves those employees from receiving an inaccurate and possibly damaging mark of alcoholism in their employee record.

This strategy will fail to catch the 26% false negatives, however. To reduce the impact of that false negative rate, the company should administer the SMAST on a semi-regular interval.

The company should be cautious about using the SMAST to test for associations between alcoholism and other diseases, however. While the hypothetical analysis shown above seems to suggest that the screen is useful in showing some association between alcoholism and depression, the shown association’s strength is much lower than seen in the population. If this screener was used in the volunteer group, the combination of misclassification bias and healthy volunteer bias may lead inaccurately to no association being found between depression and alcoholism

Part 3

12. In what direction does non-differential misclassification usually bias the estimate of effect?

Nondifferential misclassification usually bias towards the null

13. In what direction does differential misclassification bias the estimate of effect?

Differential misclassification can bias either towards or away from the null, depending on the differentials of exposure status between cases and controls

14. Why is selection bias a greater concern for cross-sectional and case-control studies compared to cohort studies?

Selection bias refers to a systematic error in study participant recruitment resulting in a distortion of the measure of association between exposure and outcome. Due to the prospective nature of cohort studies, selection bias based on outcome status isn’t a concern because the outcome hasn’t occurred yet.

15. Self-reported data is often criticized as lacking validity, yet we so often rely on it for public health surveillance, research, and program evaluation activities. Describe how self-reported data may introduce bias. Use an example to illustrate your argument.

In this case study of employer-monitored alcoholism among employees, there is an incentive for individuals to under report-their alcoholism status. Maybe there is concern about discrimination from the mark in their employee record, or perhaps the employee is in denial about their alcoholism.

If the company relied on self-reported alcoholism exposure information in this instance, exposures would be underreported leading to a bias towards the null regarding the association between alcoholism and depression.