The t-Test

The $z$-test needs the population standard deviation $\sigma$. In practice it is almost never known — so it is estimated by $s$, and that extra uncertainty is exactly what the $t$-distribution accounts for.

When to use the t-distribution

Use $t$ rather than $z$ when:

• the population standard deviation $\sigma$ is unknown and estimated by the sample $s$, and
• the sample is small (roughly $n < 30$).

For large samples $s$ estimates $\sigma$ well and the two tests agree, which is why $t$ and $z$ become interchangeable as $n$ grows.

The $t$-distribution looks like the standard normal — symmetric, bell-shaped, centred on 0 — but has heavier tails. Extreme values are more likely, because $s$ itself varies from sample to sample.

As the degrees of freedom increase, the $t$-distribution converges to the standard normal. By $n = 30$ the difference is slight; by $n = 100$ it is negligible.

Degrees of freedom

For a one-sample $t$-test:

$$\nu = n - 1$$

One degree of freedom is lost because $\bar{x}$ was computed from the same data. Once the mean and $n-1$ of the values are fixed, the last value is determined — it is not free to vary.

For a paired test on $n$ pairs, $\nu = n - 1$ as well, since the test is applied to the $n$ differences.

The one-sample t-test

$$t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \qquad \nu = n-1$$

A sample of 16 has mean 105 and standard deviation 12. Test at the 5% level whether the population mean exceeds 100.

1. Hypotheses. $H_0: \mu = 100$; $H_1: \mu > 100$ — one-tailed, since "exceeds" gives a direction.

2. Critical value. $\nu = 15$, one-tailed 5%: $t_{\text{crit}} = 1.753$.

3. Test statistic.

$$\text{SE} = \frac{12}{\sqrt{16}} = 3 \qquad t = \frac{105 - 100}{3} = 1.67$$

4. Compare. $1.67 < 1.753$, so the statistic is not in the critical region.

5. Conclude. Do not reject $H_0$: at the 5% level there is insufficient evidence that the population mean exceeds 100.

Confidence intervals using t

$$\bar{x} \pm t\,\frac{s}{\sqrt{n}}$$

with $t$ read at $\nu = n-1$ for the required confidence level.

A 95% interval for the sample above:

$\nu = 15$, two-tailed 5%: $t = 2.131$.

$$105 \pm 2.131(3) = 105 \pm 6.39 = (98.6,\ 111.4)$$

Notice the interval contains 100 — consistent with the test's failure to reject $H_0$. The two methods must always agree.

The paired t-test

When the same subjects are measured twice — before and after a treatment, or by two methods — the observations are not independent, and a two-sample test would be wrong.

Instead, work with the differences $d = x_2 - x_1$ and apply a one-sample test to them:

$$t = \frac{\bar{d} - 0}{s_d/\sqrt{n}} \qquad \nu = n - 1$$

testing $H_0: \mu_d = 0$ — no average change.

Pairing is powerful because it removes the variation between subjects. Ten people's weights may differ enormously, yet their individual weight changes can be measured with precision.

Assumptions

Every $t$-test assumes:

1. The sample is random.
2. The population is approximately normally distributed. This matters most for small samples; the Central Limit Theorem protects larger ones.
3. The observations are independent — within the sample for a one-sample test, or between pairs for a paired test.

For a two-sample $t$-test using a pooled estimate, a fourth assumption is added: the two populations have equal variances.