Reading the t-table and bracketing a tail

What the t-table lists

The t-table is laid out the other way round from the z-table. Its rows are degrees of freedom ($df=1,2,\dots ,30$ and $\infty$) and its columns are upper-tail areas $\alpha =0.10,0.05,0.025,0.01,0.005$. The cell $t_{\alpha ;df}$ is the value with area $\alpha$ to its right:

$$P(T_{df}>t_{\alpha ;df})=\alpha$$

So the table answers "given a tail area, what is $t$?" directly, and answers "given $t$, what is the tail area?" only approximately — by trapping the area between two columns.

Worked example — forwards: a critical value

A 95 % confidence interval on 12 degrees of freedom needs $t_{0.025;12}$ — 0.025 in each tail. Row 12, column 0.025: $2.179$. (R's qt(0.975, 12) prints 2.178813 — same number, more digits; the table rounds to three decimals and so does the marking scheme.)

A one-sided test at $\alpha =0.05$ on 25 df needs $t_{0.05;25}$: row 25, column 0.05 → $1.708$. Same table, different column: the sidedness of the question chooses the column, the sample size chooses the row.

Worked example — backwards: bracketing a p-value

A test on 9 df produced $t=2.5$. What is $P(T_9>2.5)$? Row 9 reads 1.383 · 1.833 · 2.262 · 2.821 · 3.250 under 0.10 · 0.05 · 0.025 · 0.01 · 0.005. The statistic 2.5 sits between 2.262 (area 0.025) and 2.821 (area 0.01), so the tail area sits between those two areas:

$$0.01<P(T_9>2.5)<0.025$$

That bracket is the answer on the paper. It is never interpolated — "about 0.017" earns nothing more, and a wrong interpolation earns less. For a two-sided test double both ends: $0.02<p<0.05$. If the statistic falls beyond the last column, write $p<0.005$; if before the first, $p>0.10$.

Negative t

The t distribution is symmetric like z: $P(T<-2.5)=P(T>2.5)$. A lower-tail test with $t=-2.00$ on 19 df reads row 19 exactly as an upper-tail test would read $+2.00$, then attaches the sign to the rejection region ($t<-1.729$).

The ∞ row

The last row of the table, $df=\infty$, reads 1.282 · 1.645 · 1.960 · 2.326 · 2.576 — the z critical values. As $df$ grows the t distribution converges to the standard normal, and by $df=30$ the difference is in the third decimal (2.042 against 1.960 at 0.025). The course rule: for $df>30$ read the ∞ row. A sample of 45 invoices with $s$ in the standard error is still a t problem — it reads $t_{0.025;44}$ as 1.96. R's qt(0.975, 44) prints 2.015368; the hand answer and R's differ by design, and Lesson 1.5 names that difference as one of three you must be able to explain.