![[t-distribution.png|300]] $ f(df) = \frac{\Gamma\paren{\frac{df+1}{2}}}{\sqrt{df\ \pi}\Gamma\paren{\frac{df}{2}}}\paren{1 + \frac{t^2}{df}}^{-\frac{df + 1}{2}} $ The $t$-distribution is a family of [[Probability Distribution|probability distributions]] used for [[Estimator|estimating]] the [[Mean|mean]] of a [[Normal Distribution|normal distribution]] with small [[Count|sample sizes]] and unknown [[Population|population]] [[Standard Deviation|standard deviation]]. The parameter $df$ is the *degrees of freedom* that corresponds to the sample size $df = n - 1$, because if $n - 1$ values and the [[Sample Mean|sample mean]] are known, the last value can be calculated, making only $n - 1$ values *free to change*. The $t$-distribution is approximately bell-shaped, with heavier tails under lower degrees of freedom, and approaches the normal distribution as $df \to \infty$. Let $t_\alpha (df)$ be a number such that $P(T\ge t_\alpha(df)) = \alpha$. The t-distribution can be used to create a confidence interval for a small sample with error $E = \frac{t_{\alpha/2}s}{\sqrt{n}}$: $ \begin{align*} P\paren{-t_{\alpha/2} < \frac{\bar{X} - \mu}{s/\sqrt{n}}<t_{\alpha/2}} &\approx 1 - \alpha \\ P\paren{\bar{X}-\frac{t_{\alpha/2}s}{\sqrt{n}} < \mu < \bar{X} + \frac{t_{\alpha/2}s}{\sqrt{n}}} &\approx 1 - \alpha \end{align*} $ Which means that $\bar{X}-\frac{t_{\alpha/2}s}{\sqrt{n}} < \mu < \bar{X} + \frac{t_{\alpha/2}s}{\sqrt{n}}$ gives a $(1 - \alpha)100\%$ confidence interval for $\mu$.