Degrees of freedom

Degrees of freedom, written $df$, count the independent pieces of information a statistic has left after some are spent estimating other quantities. Make that concrete. Four measurements with a sample mean of 10 must total 40. Pick the first three freely, say 7, 12, and 6, and the fourth is forced: $40 - 25 = 15$. Three values were free, so $df = 3 = n - 1$.

The count matters because it selects the curve that sets the multiplier. A one-sample $t$ interval at 95% uses $t^*$ at $n - 1$ degrees of freedom: 2.262 when $n = 10$, 2.064 when $n = 25$, and 1.984 when $n = 100$. Same level, three multipliers, differing only in how much information $s$ carries. All approach $z^* = 1.960$ from above without arriving.

$n - 1$ is not a universal rule. A paired $t$ uses one fewer than the number of pairs, not of measurements. A chi-square test on an $r$ by $c$ table uses $(r-1)(c-1)$, which depends on the shape of the table, not the sample size: a 2 by 3 table carries 2 degrees of freedom whether the counts total 60 or 6000.

Two habits cause most of the damage. One is reading the table at row $n$: a sample of 20 has 19 df and $t^* = 2.093$, while row 20 gives 2.086. The other is expecting a whole number: Welch's two-sample $t$ with $s_1 = 6$, $n_1 = 12$, $s_2 = 10$, and $n_2 = 15$ returns $df = 23.40$. Nothing is counted there: it is the $t$ curve that best approximates a statistic whose distribution is not $t$, and that density exists for any positive $df$.

The Welch value is penned in by $\min(n_1 - 1, n_2 - 1) \le df \le n_1 + n_2 - 2$, so 23.40 must land between 11 and 25, a fast data entry check. A calculator reports the decimal; a printed table forces a whole row, so round down. $t^*$ is 2.069 at 23 df against 2.067 at 23.40, so the interval comes out slightly wide, not falsely narrow.