T-interval

Estimating the spread from the sample adds uncertainty, which makes the reference curve heavier in the tails than the normal curve, so $t^*$ is always larger than the matching $z^*$. The interval is $\bar{x} \pm t^*\frac{s}{\sqrt{n}}$, where x-bar is the sample mean, s is the sample standard deviation, and the degrees of freedom are $n - 1$. For example, with x-bar = 50, s = 8, and n = 16, the standard error is $8/\sqrt{16} = 2$ and $t^* = 2.131$ at 95% confidence with 15 degrees of freedom, giving $50 \pm 4.262$, or 45.74 to 54.26. As n grows the t-distribution approaches the normal curve, so the two intervals nearly agree for large samples.