Researchers are designing a randomized, double-blind, placebo-controlled trial to evaluate… — USMLE Step 2 CK

Researchers are designing a randomized, double-blind, placebo-controlled trial to evaluate the efficacy of a novel anti-obesity medication. Initially, the trial is designed to detect a mean weight loss difference of $5.0\text{ kg}$ between the treatment and placebo groups. Based on pilot data, the expected standard deviation of weight loss is estimated at $10.0\text{ kg}$ in both groups. Using a two-sided $\alpha$ of $0.05$ and $80\%$ power, the investigators calculate a required sample size of $64$ participants per group.

Prior to patient enrollment, a newly published meta-analysis indicates a larger population variability, suggesting the true standard deviation is $15.0\text{ kg}$ . In light of this, the investigators revise their protocol to ensure even more conservative clinical benefits can be detected, setting the new minimum detectable mean difference at $3.0\text{ kg}$ .

Assuming the type I error rate ( $\alpha$ ) and statistical power remain unchanged, what is the newly calculated required sample size per group?

Answer: 400 participants

Answer

The newly calculated required sample size is 400 participants per group.

The required sample size for a continuous outcome in a superiority trial is calculated using the formula

N = \frac{2(Z_{\alpha/2} + Z_{\beta})^2 \sigma^2}{\Delta^2}

, where

\sigma

is the standard deviation and

\Delta

is the minimum detectable difference. From this formula, it is clear that the sample size is directly proportional to the variance (

\sigma^2

) and inversely proportional to the squared difference (

\Delta^2

).

When the standard deviation increases from

10.0

15.0

(a factor of

1.5

3/2

), the variance increases by a factor of

(3/2)^2 = 9/4

.
When the minimum detectable difference decreases from

5.0

3.0

, the required sample size increases by the inverse square factor, which is

(5/3)^2 = 25/9

.

Multiplying the original sample size of

64

by these two adjustment factors yields:

64 \times \frac{9}{4} \times \frac{25}{9}

. The

9

s cancel out, leaving

64 \times \frac{25}{4} = 16 \times 25 = 400

Step-by-Step Solution

Identify the mathematical relationship between sample size, variance, and effect size.

Sample size

N \propto \frac{\sigma^2}{\Delta^2}

To determine how changes in standard deviation (

\sigma

) and minimum detectable difference (

\Delta

) mathematically scale the required sample size.

Calculate the multiplier for the change in variance.

Variance Multiplier =

(\frac{15.0}{10.0})^2 = 1.5^2 = \frac{9}{4}

Sample size is directly proportional to the variance (

\sigma^2

), which is the square of the standard deviation.

Calculate the multiplier for the change in the minimum detectable difference.

Difference Multiplier =

(\frac{5.0}{3.0})^2 = \frac{25}{9}

Sample size is inversely proportional to the square of the effect size difference (

\Delta^2

Calculate the new sample size by applying both multipliers to the original sample size.

New

N = 64 \times \frac{9}{4} \times \frac{25}{9} = 64 \times \frac{25}{4} = 16 \times 25 = 400

To accurately find the final required sample size after adjusting for both new parameters without needing complex Z-score recalculations.

Key Concept

In clinical trial design, the required sample size for a continuous outcome is directly proportional to the variance (the square of the standard deviation) and inversely proportional to the square of the minimum detectable difference.

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic