t tests for a mean and the paired t-testEdexcel A-Level Further Maths: Revision notes

When $X\sim\mathrm{N}(\mu,\sigma^2)$ and $\sigma^2$ is unknown, replace $\sigma$ by the sample standard deviation $s$. The statistic$T=\frac{\bar X-\mu}{S/\sqrt n}$then follows a $t$ distribution with $\nu=n-1$ degrees of freedom. It is symmetrical about $0$, like the Normal distribution, but has heavier tails, more so for small $n$, and approaches $\mathrm{N}(0,1)$ as $n$ increases. Critical values come from $t$-tables or a calculator, e.g. the upper $5\%$ point of $t_9$ is $1.833$. The population must be Normal. With the variance known, use $z$ instead.

To test $H_0:\mu=\mu_0$: 1. calculate $\bar x$ and $s$ (use $s^2=\frac{1}{n-1}\left(\sum x^2-\frac{(\sum x)^2}{n}\right)$ if given summary data); 2. $t=\frac{\bar x-\mu_0}{s/\sqrt n}$; 3. compare with the critical value of $t_{n-1}$ (one-tailed for $\mu>\mu_0$ or $\mu<\mu_0$; two-tailed, with the level split, for $\mu\ne\mu_0$); 4. conclude in context. Example: $n=8$, $\sum x=99.4$, $\sum x^2=1236.48$, $H_0:\mu=12.0$ against $\mu\ne12.0$. $\bar x=12.425$, $s^2=0.205$, $t=\frac{0.425}{\sqrt{0.205/8}}=2.65$. The two-tailed $5\%$ critical value of $t_7$ is $2.365$, so reject $H_0$.

A confidence interval for $\mu$ uses $t$ in place of $z$:$\bar x\pm t_{n-1}\frac{s}{\sqrt n},$ where $t_{n-1}$ is the upper $\frac{\alpha}{2}$ point for a $(100-\alpha)\%$ interval. Example: $n=8$, $\bar x=24.3$, $s=1.9$, $95\%$: $t_7=2.365$, so $24.3\pm2.365\times\frac{1.9}{\sqrt8}=(22.71,\ 25.89)$. The interval is wider than the one using $z=1.96$ because $t$ allows for the extra uncertainty from estimating $\sigma$. As before, a $(100-\alpha)\%$ interval matches a two-tailed test at the $\alpha\%$ level.

Use a paired test when each observation in one sample is naturally linked to one in the other, such as the same people before and after, or matched pairs. Calculate the difference $d$ for each pair, then treat the differences as a single sample from $\mathrm{N}(\mu_d,\sigma_d^2)$. To test $H_0:\mu_d=0$:$t=\frac{\bar d}{s_d/\sqrt n},\qquad\nu=n-1,$where $n$ is the number of pairs. The assumption is that the differences are Normally distributed. Example: scores before and after give differences $5,3,-1,7,5,6,3,7$, so $\bar d=4.375$, $s_d=2.669$ and $t=\frac{4.375}{2.669/\sqrt8}=4.64$. Against the upper $5\%$ point of $t_7$, $1.895$, reject $H_0:\mu_d=0$ in favour of $\mu_d>0$. Paired data are not independent, so a two-sample test of means would be wrong. A paired test also removes variation between individuals. Careful: a significant difference shows an effect, but without a control group it does not prove the cause.