One-sample t-testAQA A-Level Further Maths: Revision notes

To test a claim about the mean $\mu$ of a normal population when the population variance is unknown and the sample is small, replace $\sigma$ by the unbiased estimate $s$. The resulting statistic does not follow a normal distribution; it follows the t-distribution with $\nu=n-1$ degrees of freedom: $t=\frac{\bar x-\mu_0}{s/\sqrt n}\sim t_{n-1}.$ The $t$-distribution is symmetric about $0$ like the normal but has heavier tails, which allows for the extra uncertainty from estimating $\sigma$. As $\nu$ increases it approaches the standard normal.

Use the unbiased estimate of the population variance: $s^2=\frac{1}{n-1}\left(\sum x^2-\frac{(\sum x)^2}{n}\right).$ Example: $n=10$, $\sum x=1840$, $\sum x^2=339\,136$. Then $\bar x=184$ and $s^2=\frac{339136-338560}{9}=64$, so $s=8$. If the question gives the variance, take the square root to get $s$ before dividing by $\sqrt n$.

State hypotheses about the population mean: $H_0:\mu=\mu_0$ against

• $H_1:\mu>\mu_0$ or $H_1:\mu<\mu_0$ (one-tailed), with the whole significance level in one tail, or
• $H_1:\mu\neq\mu_0$ (two-tailed), with half the significance level in each tail. Look up the critical value in the $t$-table using $\nu=n-1$ and the tail probability. For a $5\%$ two-tailed test use the $2.5\%$ column. With $\nu=8$: one-tailed $5\%$ is $1.860$ and two-tailed $5\%$ is $\pm2.306$. The critical region is the set of $t$ values beyond the critical value(s).

1. State $H_0$ and $H_1$ in terms of $\mu$.
2. Find $\bar x$ and $s$, and calculate $t=\frac{\bar x-\mu_0}{s/\sqrt n}$.
3. Find $\nu=n-1$ and the critical value.
4. Compare: if $t$ is in the critical region, reject $H_0$.
5. Conclude in context, with suitably cautious wording. Example: café coffees, $n=9$, $\bar x=243.4$, $s=9$, $\mu_0=250$, $H_1:\mu<250$. $t=\frac{-6.6}{3}=-2.2$. The critical value for $\nu=8$ at $5\%$ is $-1.860$. Since $-2.2<-1.860$, reject $H_0$: there is evidence that the mean volume is below $250$ ml.

The test is valid only if the population is normally distributed and the sample is random. With small samples you cannot rely on the central limit theorem, so the normality assumption matters. The population variance is unknown, which is why $s$ and the $t$-distribution are used. For large samples the $t$ values are close to normal values, so the conclusions are the same.