What is the general form of a 95% confidence interval for a population mean when sigma is known, and how does it change when sigma is unknown?

• A. xbar ± z*(sigma/sqrt(n)); If sigma unknown: xbar ± t_{0.975, n−1} (s/sqrt(n)).
• B. xbar ± z*(sigma/sqrt(n)) for both cases.
• C. xbar ± t_{0.975, n−1} (s/sqrt(n)) for both cases.
• D. xbar ± z*(s/sqrt(n)) in all cases.

Answer: (A)

The essential idea is how the uncertainty in estimating the mean is measured and which distribution you use to balance that uncertainty. If the population standard deviation sigma is known, the sample mean x̄ has a normal distribution with standard error sigma/√n. To capture the middle 95% of that distribution, you go out 1.96 standard errors on each side, giving the interval x̄ ± z0.975 · (sigma/√n), where z0.975 is the 97.5th percentile of the standard normal distribution (about 1.96).

If sigma is unknown, you don’t know the true variability of the population mean, so you replace sigma with the sample standard deviation s and use the t distribution with n−1 degrees of freedom to account for the extra uncertainty. The 95% interval then becomes x̄ ± t0.975, n−1 · (s/√n). The t critical value is slightly larger than 1.96 for small samples, making the interval a bit wider; as n grows, the t distribution approaches the normal, and the interval becomes closer to the known-sigma form.

In short, known sigma uses a normal-based interval with sigma in the standard error; unknown sigma uses a t-based interval with s and df = n−1 in the standard error.