A study reports a sample mean serum creatinine of 1.2 mg/dL with a standard error of 0.1 mg/dL. If the sample size were increased from 100 to 400 while the standard deviation remained unchanged, what would be the new standard error?

• A 0.025 mg/dL
• B 0.05 mg/dL ✓
• C 0.1 mg/dL
• D 0.2 mg/dL

Explanation

Standard error = SD / sqrt(n). Since SE × sqrt(n) = SD, and SD remains constant, SE is inversely proportional to sqrt(n). Quadrupling n from 100 to 400 means sqrt(n) doubles (10 to 20), so SE halves from 0.1 to 0.05. Option A would require increasing n sixteenfold. Option C ignores the effect of sample size. Option D would occur if n were decreased.

Reference: Methods in Biostatistics by B. K. Mahajan, 7th ed.

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