When sigma is known, what is the formula for a 95% CI for the population mean?

When sigma is known, what is the formula for a 95% CI for the population mean?

• A. Xbar ± z0.975 * sigma / n
• B. Xbar ± z0.975 * sqrt(n) / sigma
• C. Xbar ± z0.975 * sigma / sqrt(n)
• D. Xbar ± z0.975 * sigma * sqrt(n)

Answer: (C)

When sigma is known, the sample mean X̄ is normally distributed around the true mean μ with standard deviation σ/√n. To capture the middle 95% of that distribution, you use the z critical value z_{0.975} (about 1.96). So the 95% confidence interval for μ is X̄ ± z_{0.975} × (σ/√n), i.e., X̄ ± 1.96 × (σ/√n). Why the others don’t fit: the standard error should shrink with √n in the denominator, not n, so X̄ ± z × (σ/n) is too narrow. Inverting the relationship gives √n in the numerator, which doesn’t match the standard error. Multiplying by √n instead of dividing by √n makes the interval far too wide and inappropriate for estimating μ.

When sigma is known, the sample mean X̄ is normally distributed around the true mean μ with standard deviation σ/√n. To capture the middle 95% of that distribution, you use the z critical value z_{0.975} (about 1.96). So the 95% confidence interval for μ is X̄ ± z_{0.975} × (σ/√n), i.e., X̄ ± 1.96 × (σ/√n).

Why the others don’t fit: the standard error should shrink with √n in the denominator, not n, so X̄ ± z × (σ/n) is too narrow. Inverting the relationship gives √n in the numerator, which doesn’t match the standard error. Multiplying by √n instead of dividing by √n makes the interval far too wide and inappropriate for estimating μ.