Which elements determine the width of a 95% confidence interval for a mean when sigma is unknown?

Which elements determine the width of a 95% confidence interval for a mean when sigma is unknown?

• A. The standard error s/√n and the t critical value t_(n-1, 0.975).
• B. The mean and standard deviation
• C. The sample size only
• D. The variance of the population

Answer: (A)

When sigma is unknown, the precision of the estimated mean is captured by the standard error and the critical value from a t-distribution. The half-width of the 95% interval is t_(n-1, 0.975) times the standard error, where the standard error is s/√n (s is the sample standard deviation). Therefore the full width depends on s, n, and the t critical value, which itself depends on the degrees of freedom n-1. The mean’s value doesn’t set the width, and the population variance isn’t used directly because sigma is unknown. As n grows, s/√n shrinks and the t critical value approaches the usual z-quantile, so the interval narrows.

When sigma is unknown, the precision of the estimated mean is captured by the standard error and the critical value from a t-distribution. The half-width of the 95% interval is t_(n-1, 0.975) times the standard error, where the standard error is s/√n (s is the sample standard deviation). Therefore the full width depends on s, n, and the t critical value, which itself depends on the degrees of freedom n-1. The mean’s value doesn’t set the width, and the population variance isn’t used directly because sigma is unknown. As n grows, s/√n shrinks and the t critical value approaches the usual z-quantile, so the interval narrows.