Advanced Problem: Clinical Trial Bias Analysis

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A pharmaceutical company is testing a new drug claimed to lower cholesterol levels. The lead researcher believes the drug is effective and designs a clinical trial. Here are the details:

1. 200 participants with high cholesterol are randomly divided into two groups of 100 each.
2. Group A receives the new drug, Group B receives a placebo.
3. Cholesterol levels are measured before and after a 30-day treatment period.
4. The researcher considers the treatment effective if there’s a statistically significant difference (p < 0.05) between the groups.

Results:
– Group A (Drug): Mean cholesterol reduction of 15 mg/dL, standard deviation 8 mg/dL
– Group B (Placebo): Mean cholesterol reduction of 12 mg/dL, standard deviation 7 mg/dL

The lead researcher concludes that the drug is effective in lowering cholesterol.

 

Drug Placebo 20 15 10 5 0 Mean Cholesterol Reduction (mg/dL) Error bars represent ± 1 standard deviation 15 ± 8 12 ± 7

Tasks

  1. Perform a t-test to check if the difference between the groups is statistically significant at the 0.05 level.
    Show your work.
  2. Calculate the 95% confidence interval for the difference in mean cholesterol reduction between the two groups.
  3. Identify potential sources of confirmation bias in this study design and analysis.
  4. Suggest improvements to the study design to minimize bias and increase reliability.
  5. Discuss the ethical implications of confirmation bias in medical research.

Solution

1. T-test for statistical significance:

Null hypothesis (H₀): There is no difference between the drug and placebo groups.
Alternative hypothesis (H₁): There is a difference between the drug and placebo groups.

t = (x̄₁ – x̄₂) / √(s₁²/n₁ + s₂²/n₂)

Where:
x̄₁ = 15, x̄₂ = 12 (mean reductions)
s₁ = 8, s₂ = 7 (standard deviations)
n₁ = n₂ = 100 (sample sizes)

t = (15 – 12) / √((8²/100) + (7²/100))
t = 3 / √(0.64 + 0.49)
t = 3 / √1.13
t = 3 / 1.063
t ≈ 2.82

Degrees of freedom: df = n₁ + n₂ – 2 = 198

The critical t-value for df=198 and α=0.05 (two-tailed) is approximately 1.97.

Since 2.82 > 1.97, we reject the null hypothesis. The difference is statistically significant at the 0.05 level.

2. 95% Confidence Interval:

CI = (x̄₁ – x̄₂) ± t₀.₀₂₅ * √(s₁²/n₁ + s₂²/n₂)

Where t₀.₀₂₅ for df=198 ≈ 1.97

CI = (15 – 12) ± 1.97 * √(8²/100 + 7²/100)
CI = 3 ± 1.97 * 1.063
CI = 3 ± 2.09
CI = (0.91, 5.09)

We can be 95% confident that the true difference in mean cholesterol reduction between the drug and placebo groups is between 0.91 and 5.09 mg/dL.

3. Potential sources of confirmation bias:
– The lead researcher’s prior belief in the drug’s effectiveness may influence study design or interpretation.
– Only positive results (cholesterol reduction) are being measured.
– The statistical threshold (p < 0.05) is arbitrary and may lead to over-interpretation of marginal results.
– Potential selective reporting of outcomes that confirm the hypothesis.

4. Improvements to study design:
– Use a double-blind design where neither participants nor researchers know who receives the drug or placebo.
– Pre-register the study protocol and analysis plan.
– Include multiple outcome measures beyond just cholesterol reduction.
– Use a larger sample size to increase statistical power.
– Consider a longer treatment period to assess long-term effects.
– Include an active control group with a known cholesterol-lowering drug.

5. Ethical implications:
– Confirmation bias can lead to overestimation of drug benefits and underestimation of risks.
– It may result in ineffective or potentially harmful treatments being approved.
– Biased research wastes resources and may misdirect future research efforts.
– It can erode public trust in medical research and healthcare recommendations.
– There’s an ethical obligation to design and conduct studies that prioritize patient safety and scientific integrity over desired outcomes.

Drug Placebo 20 15 10 5 0 Mean Cholesterol Reduction (mg/dL) Error bars represent ± 1 standard deviation 15 ± 8 12 ± 7