The one-sample t-test and bracketing p

s for σ, t for z

When $\sigma$ is unknown the four steps are unchanged and the statistic swaps in the sample SD:

$$t=\frac{\bar{x}-{\mu }_0}{s/\sqrt{n}},\mathrm{df}=n-1$$

Under $H_0$ this has Student's t-distribution (L9.3), so the critical value comes from the t-table — row $n-1$, column $\alpha$ (one-sided) or $\alpha /2$ (two-sided) — and the p-value comes from the same table as a bracket. The choice between z and t is made by whether $\sigma$ is given, never by $n$: a large $n$ with $s$ moves you to the $\infty$ row of the t-table, which is still a t-test.

Worked example — do files take longer than budgeted?

Cedar & Stone Accounting budgets 50 hours to complete a client file. The partner suspects files take longer. A sample of $n=16$ files: $\bar{x}=53.0$ hours, $s=6.4$ hours. Test at $\alpha =0.05$.

Step 1. $H_0:\mu =50$, $H_a:\mu >50$; $\mu$ is the mean completion time for all files. Upper-tailed, from longer.

Step 2. $\mathrm{SE}=6.4/\sqrt{16}=1.6$;

$$t=\frac{53.0-50}{1.6}=1.875,\mathrm{df}=15$$

Step 3 — critical value. Row 15, column 0.05: $t_{0.05;15}=1.753$. Rejection region $t>1.753$.

Step 3 — p-value, bracketed. Row 15 of the table reads

The statistic 1.875 sits between 1.753 and 2.131, so the tail area beyond it lies between the two column headings:

$$0.025<p<0.05$$

No interpolation: the bracket is the answer, and it is enough — the whole bracket lies below $\alpha =0.05$. (R's exact value is 0.0402, inside the bracket.)

Step 4. $1.875>1.753$, equivalently $p<0.05$: reject $H_0$ at the 5 % level. There is evidence that the mean time to complete a file exceeds the 50-hour budget. At $\alpha =0.01$ the bracket sits entirely above 0.01, so the same data would not reject.

Condition: $n=16<30$, so the test relies on completion times being roughly normal — check the sample for skew and outliers, and say so.

When the bracket straddles α

Sometimes the bracket contains $\alpha$: a $t_{15}$ of 2.3 sits between 2.131 (0.025) and 2.602 (0.01), giving $0.01<p<0.025$ — decisive at 5 %, decisive at 1 %, but at $\alpha =0.02$ the table cannot say. Then the critical-value route decides ($t_{0.02}$ is not tabulated either, so on the paper such a question is not set); in R, pt gives the exact value. In this course the bracket and the critical value always agree, because they read the same row.

Two-sided brackets

For $H_a:\mu \ne 50$ the same $t=1.875$ is compared with $t_{0.025;15}=2.131$ — do not reject — and the bracket is doubled at both ends: $0.05<p<0.10$. Doubling both ends is the one-line rule; L10.8 shows R printing 0.0804, inside it.