In a study of 400 children, the mean serum calcium is 9.2 mg/dL with a standard deviation (SD) of 0.8 mg/dL. What is the 95% confidence interval for the true population mean?
- A 9.2 ± 1.57 mg/dL (7.63 to 10.77)
- B 9.2 ± 0.08 mg/dL (9.12 to 9.28) ✓
- C 9.2 ± 0.16 mg/dL (9.04 to 9.36)
- D 9.2 ± 0.31 mg/dL (8.89 to 9.51)
Correct answer: B. 9.2 ± 0.08 mg/dL (9.12 to 9.28)
Explanation
Standard Error of Mean (SEM) = SD/√n = 0.8/√400 = 0.8/20 = 0.04. 95% CI = mean ± 1.96 × SEM = 9.2 ± 1.96 × 0.04 = 9.2 ± 0.0784 ≈ 9.2 ± 0.08. Thus 95% CI ≈ 9.12 to 9.28 mg/dL. The ±1.96 SD interval would give the reference range for individuals (7.63 to 10.77), not the CI for the mean.
Reference: Park's Textbook of Preventive and Social Medicine, 27th ed.
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