Student’s t and its table

Why a second table

L9.1's interval needed $\sigma$. In practice $\sigma$ is estimated by the sample SD $s$, and the standardised sample mean becomes

$$t=\frac{\bar{x}-\mu }{s/\sqrt{n}}$$

This is not a standard normal. The denominator now carries its own sampling variability — $s$ is sometimes too small, sometimes too large — and the extra randomness pushes more probability into the tails. The result is Student's t-distribution: symmetric, bell-shaped, centred at 0 like $z$, but with heavier tails, so a given tail probability sits further out than it does for $z$.

How much heavier depends on how well $s$ estimates $\sigma$, which depends on the sample size. The family is indexed by degrees of freedom,

$$\mathrm{df}=n-1$$

for a single sample — the $n-1$ of the sample variance. With small df the tails are fat; as df grows, $s$ pins down $\sigma$ and $t$ converges to $z$. The $\infty$ row of the t-table is the z-table's critical values.

Reading the table forward

The course's t-table lists upper-tail critical values $t_{\alpha ;\mathrm{df}}$: the value with area $\alpha$ to its right, for $\alpha =0.10,0.05,0.025,0.01,0.005$ and df $=1,\dots ,30$ and $\infty$. Row = df, column = tail area. Watch the 0.025 column shrink toward 1.960:

For a 95 % interval from $n=5$ the multiplier is 2.776, not 1.96 — 42 % wider, the price of estimating $\sigma$ from four degrees of freedom. Two more forward lookups: $t_{0.05;19}=1.729$ (row 19, column 0.05 — the 90 % interval multiplier for $n=20$), and $t_{0.005;9}=3.250$ (the 99 % multiplier for $n=10$).

By symmetry the lower-tail value is the negative: $P(T_{19}<-1.729)=0.05$. And a two-sided interval at level $1-\alpha$ uses the $\alpha /2$ column — 95 % reads the 0.025 column, exactly as $z_{0.025}=1.96$ served L9.1.

Reading the table backward — bracketing

Module 10 will hand you a computed $t$ and ask for its tail probability. The table cannot give it exactly; it gives a bracket. Suppose $t=2.1$ with df $=14$. Row 14 reads 1.761 in the 0.05 column and 2.145 in the 0.025 column; 2.1 sits between them, so

$$0.025<P(T_{14}>2.1)<0.05$$

The area to the right of 2.1 is less than the area to the right of 1.761 and more than the area to the right of 2.145. Never interpolate — the course brackets, and the two bounds are the answer. R's pt(2.1, 14, lower.tail = FALSE) would give 0.0272, safely inside.

What the heavy tails are not about

The tails are heavy because $s$ estimates $\sigma$ — not because $n$ is small in itself, and not because the population is skewed. A small $n$ makes the estimate worse, which is why df matters; a skewed population is a different problem, addressed by the robustness guideline in L9.4. Get the reason right: the paper asks it as a one-line question.