How do you compute a 95% confidence interval for a population mean with unknown sigma?

How do you compute a 95% confidence interval for a population mean with unknown sigma?

• A. x-bar ± z_(0.975) * sigma / sqrt(n)
• B. x-bar ± z_(0.975) * s / sqrt(n)
• C. x-bar ± t_(n-1, 0.975) * s / sqrt(n)
• D. x-bar ± t_(n-1, 0.95) * s / sqrt(n)

Answer: (C)

For a 95% confidence interval for a population mean when the population standard deviation is unknown, you use the t distribution with n-1 degrees of freedom. The interval centers on the sample mean and uses the estimated standard error, s divided by the square root of n. The critical value comes from the t distribution at the 0.975 percentile (two-sided 95% interval), reflecting the added uncertainty from estimating sigma. So the correct form is: x-bar ± t_{n-1,0.975} × (s/√n). This uses s as the estimate of sigma and the t-quantile to account for variability due to estimating sigma. As n grows large, the t distribution approaches the normal distribution, but the exact construction for unknown sigma uses t, not z. Using z with sigma would assume you know the true population standard deviation, which isn’t the case here. Using a t-quantile with 0.95 would yield a narrower interval than 95% coverage, so it’s not correct for a 95% confidence level.

For a 95% confidence interval for a population mean when the population standard deviation is unknown, you use the t distribution with n-1 degrees of freedom. The interval centers on the sample mean and uses the estimated standard error, s divided by the square root of n. The critical value comes from the t distribution at the 0.975 percentile (two-sided 95% interval), reflecting the added uncertainty from estimating sigma.

So the correct form is: x-bar ± t_{n-1,0.975} × (s/√n). This uses s as the estimate of sigma and the t-quantile to account for variability due to estimating sigma. As n grows large, the t distribution approaches the normal distribution, but the exact construction for unknown sigma uses t, not z.

Using z with sigma would assume you know the true population standard deviation, which isn’t the case here. Using a t-quantile with 0.95 would yield a narrower interval than 95% coverage, so it’s not correct for a 95% confidence level.