Z-Score vs T-Distribution

A z-score standardizes a value by measuring its distance from the mean in standard deviations. You compute it as $z = \frac{x - \mu}{\sigma}$, where $x$ is the value, $\mu$ (mu) is the mean, and $\sigma$ (sigma) is the standard deviation. For example, a score of 85 in a distribution with mean 70 and standard deviation 5 has $z = (85 - 70)/5 = 3$, so it sits 3 standard deviations above the mean. A negative z-score means the value is below the mean.

The $t$-distribution accounts for the extra uncertainty of estimating the population standard deviation with the sample standard deviation $s$. Its exact shape depends on the degrees of freedom: fewer degrees of freedom give fatter tails, and as they grow the curve approaches the standard normal. For example, a one-sample $t$ test with $n = 15$ uses the $t$ model with $15 - 1 = 14$ degrees of freedom. The test statistic is $t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}$, where $\bar{x}$ is the sample mean and $\mu_0$ the hypothesized mean.