Which table, which df

◈ 7 cards

z for a known σ or any proportion; t on n − 1, min(n₁ − 1, n₂ − 1), n₁ + n₂ − 2, n_d − 1 or n − 2 otherwise; df above 30 read the ∞ row — and the critical-value and p-value routes always reach the same decision.

Six rules, one table

The $t$ -table has one dimension the $z$ -table lacks: a row. Choosing the row is a mark on its own, and the six rules cover every procedure in the course.

Procedure	Table	df	Example
one mean, σ known · any proportion	$z$	—	$p$ with $n = 500$
one mean, $s$	$t$	$n - 1$	$n = 16 \to 15$
two means, Welch (default)	$t$	$min (n_{1} - 1, n_{2} - 1)$	$12, 10 \to 9$
two means, pooled (equal variances stated)	$t$	$n_{1} + n_{2} - 2$	$12, 10 \to 20$
paired	$t$	$n_{d} - 1$	12 pairs $\to 11$
slope	$t$	$n - 2$	$n = 14 \to 12$

And the rule over all of them: df above 30 → the ∞ row, whose entries are the $z$ critical values (1.645, 1.960, 2.576). A one-sample test with $n = 40$ has 39 df and reads 1.960, not 2.042 — the handout’s own convention.

The Welch row is where the two conventions live side by side. By hand the course uses the minimum rule, which is conservative (a smaller df means a larger critical value). R computes the Satterthwaite fraction — 19.62 for the L11.3 samples — and its $p$ is correspondingly smaller. Both are correct; the paper marks the hand rule and expects you to explain the gap (L14.4).

Why the two routes cannot disagree

A test can end at step 3 either way: compare the statistic with the critical value, or compare the $p$ -value with $α$ . They are the same comparison in two currencies. The critical value $t_{α}$ is defined as the point with tail area $α$ ; the $p$ -value is the tail area beyond the observed $t$ . So $t_{obs} > t_{α}$ exactly when the area beyond $t_{obs}$ is smaller than the area beyond $t_{α}$ — that is, when $p < α$ . Same tail, same table, same decision.

Worked once. L10.5’s one-sample test: $t = 1.875$ on 15 df, upper-tailed, $α = 0.05$ . Critical route: $t_{0.05; 15} = 1.753 < 1.875$ → reject. $p$ -value route: 1.875 sits between 1.753 (0.05) and 2.131 (0.025), so $0.025 < p < 0.05$ , below $α$ → reject. Two routes, one verdict.

They only seem to disagree when something else differs: one route run two-sided and the other one-sided, or a $p$ bracketed from the wrong row, or a critical value read at $α$ instead of $α /2$ for a two-sided test. Every apparent conflict is one of those.

Which route to show

Show both when the question says "both routes"; otherwise the critical value is faster and the $p$ -value is more informative, and the marker gives step 3 either way. The $p$ -value route has one advantage on the paper: a bracket such as $0.025 < p < 0.05$ tells the reader what would happen at any $α$ , which is what "interpret your $p$ -value" is asking for.

The df rules. The row is a mark; df above 30 read the ∞ row, whose entries are the z critical values.