Confidence intervals from small samplesAQA A-Level Further Maths: Revision notes

For a normal population with unknown variance, replacing $\sigma$ by the unbiased estimate $s$ changes the distribution of the standardised mean. When the sample is small, $s$ can be well away from $\sigma$, and $\frac{\bar X-\mu}{s/\sqrt n}\sim t_{n-1},$ the t-distribution with $\nu=n-1$ degrees of freedom. It is symmetric about $0$ with heavier tails than the standard normal, so its critical values are larger. As $n$ grows it approaches the normal distribution, which is why the normal can be used for large samples.

A symmetric $C\%$ confidence interval for $\mu$ is $\bar x\pm t_{\nu}\,\frac{s}{\sqrt n},$ where $t_\nu$ is the critical value from the $t$-table with $\nu=n-1$ degrees of freedom and tail probability $\frac{100-C}{2}\%$. For $\nu=7$: $95\%$ uses the $2.5\%$ column ($2.365$) and $99\%$ uses the $0.5\%$ column ($3.499$). Calculate $\bar x$ and $s$ first, using $s^2=\frac{1}{n-1}\left(\sum x^2-\frac{(\sum x)^2}{n}\right).$

Chocolate bars: $n=8$, $\bar x=36.2$, $s=2.4$.

1. $\nu=7$, and the $2.5\%$ critical value is $2.365$.
2. Standard error $=\frac{2.4}{\sqrt8}=0.849$.
3. Margin $=2.365\times0.849=2.01$.
4. Interval: $(34.19,\ 38.21)$. With the normal value $1.96$ the interval would be $(34.54,\ 37.86)$, which is too narrow for a sample of $8$.

The interval is wider with a higher confidence level, a larger $s$ or a smaller $n$ (a smaller $n$ also means a larger $t$ value). The method assumes the population is normally distributed and the sample is random. For inference, check whether a claimed mean lies inside the interval: inside means the data are consistent with the claim; outside gives evidence at the corresponding level that the mean is different, and the interval shows whether it is higher or lower. Always answer in context.