4.16 Confidence intervalsIB Maths: Applications and Interpretation HL: Revision notes

A point estimate is one value that estimates a population parameter. The sample mean $\bar{x}$ estimates $\mu$. An unbiased estimate of the population variance is $s_{n-1}^2=\frac{\sum(x-\bar{x})^2}{n-1}$; the GDC gives $s_{n-1}$ (not $\sigma_n$, which divides by $n$). A confidence interval gives a range of plausible values for $\mu$: $\bar{x}\pm$ margin of error. The confidence level (for example $95\%$) says how often the method captures $\mu$: if many random samples were taken and an interval built from each, about $95\%$ of those intervals would contain the true mean. Any one interval either contains $\mu$ or it does not.

If the population is normal with known $\sigma$ (or $n$ is large, by the central limit theorem), then $\bar{X}\sim N\left(\mu,\frac{\sigma^2}{n}\right)$ and the interval is $\bar{x}\pm z\frac{\sigma}{\sqrt{n}}.$ The critical value $z$ leaves half the remaining probability in each tail: $1.645$ for $90\%$, $1.96$ for $95\%$ and $2.576$ for $99\%$ (use the inverse normal on the GDC, for example $z=\text{invNorm}(0.975)$). Example: $n=25$, $\bar{x}=4.20$, $\sigma=0.6$: margin $=1.96\times0.12=0.235$, interval $(3.96,\ 4.44)$. The GDC has a $Z$-interval option that takes $\sigma$, $\bar{x}$, $n$ and the level.

When $\sigma$ is unknown, replace it with $s_{n-1}$ and use the $t$-distribution with $\nu=n-1$ degrees of freedom, regardless of sample size (the population is assumed normal): $\bar{x}\pm t_{\nu}\frac{s_{n-1}}{\sqrt{n}}.$ The $t$-distribution is symmetric like the normal but has heavier tails, so $t_\nu>z$, especially for small $n$. Example: $n=12$, $\bar{x}=40.375$, $s_{n-1}=1.11$: $t_{11}=2.201$, margin $=2.201\times\frac{1.11}{\sqrt{12}}=0.708$, interval $(39.7,\ 41.1)$. On the GDC use the $T$-interval option with the data list or summary statistics.

The width is $2\times$ margin of error. It is smaller for a larger sample ($\sqrt{n}$ in the denominator), a smaller spread, and a lower confidence level (smaller critical value). A higher confidence level gives a wider interval: more certainty costs precision. Sample size. To get a margin of error at most $E$ with known $\sigma$, solve $z\frac{\sigma}{\sqrt{n}}\le E$, so $n\ge\left(\frac{z\sigma}{E}\right)^2$, and round up. For $z=2.576$, $\sigma=1.1$, $E=0.2$: $n\ge200.7$, so $n=201$.

Always conclude in the context of the question. Typical statements:

• 'We are $95\%$ confident that the mean mass of cats of this breed lies between $3.96$ kg and $4.44$ kg.'
• If a claimed value lies outside the interval, there is evidence that the true mean differs from the claim at that confidence level; if it lies inside, the data are consistent with the claim (this does not prove it). State the assumptions: a random sample, and a normal population (or a large sample).