Confidence intervals for a Normal meanEdexcel A-Level Further Maths: Revision notes

A confidence interval is a range of values, calculated from a sample, that is likely to contain the unknown population parameter. A $95\%$ interval is built by a method that, over many samples, produces intervals containing the true $\mu$ about $95\%$ of the time. The confidence level describes the method, not one interval. Once an interval is calculated, $\mu$ is a fixed number that is either inside it or not. So say: 'we are $95\%$ confident that $\mu$ lies in the interval', not 'there is a $95\%$ probability that $\mu$ is in the interval'.

If $X\sim\mathrm{N}(\mu,\sigma^2)$ with $\sigma$ known, then $\bar X\sim\mathrm{N}\left(\mu,\frac{\sigma^2}{n}\right)$, so the standard error is $\frac{\sigma}{\sqrt n}$. The confidence interval for $\mu$ is$\bar x\pm z\,\frac{\sigma}{\sqrt n}.$Common two-tailed values: $90\%$: $z=1.645$; $95\%$: $z=1.96$; $99\%$: $z=2.576$. Example: $\sigma=6$, $n=25$, $\bar x=503.2$. Standard error $=\frac{6}{5}=1.2$. A $95\%$ interval is $503.2\pm1.96\times1.2=(500.85,\ 505.55)$.

The interval is symmetric about $\bar x$, with width $2z\frac{\sigma}{\sqrt n}$. Width falls as $n$ increases (halving it needs four times the sample size) and rises with the confidence level or with $\sigma$. To find the sample size for a given maximum width $w$: solve $2z\frac{\sigma}{\sqrt n}\le w$, so $n\ge\left(\frac{2z\sigma}{w}\right)^2$, and round up to a whole number. Example: $\sigma=2.4$, $90\%$, $w=0.8$: $\sqrt n\ge\frac{2(1.645)(2.4)}{0.8}=9.87$, $n\ge97.4$, so $n=98$.

A $(100-\alpha)\%$ confidence interval contains the values of $\mu$ that a two-tailed test at the $\alpha\%$ level would not reject. So $95\%$ matches a $5\%$ two-tailed test, and $99\%$ matches a $1\%$ two-tailed test. If a claimed value $\mu_0$ lies outside the interval, reject $H_0:\mu=\mu_0$; if it lies inside, there is no evidence against it. This agrees with the test statistic $z=\frac{\bar x-\mu_0}{\sigma/\sqrt n}$ compared with the critical value. Example: $\bar x=12.84$, $\sigma=0.5$, $n=40$ gives a $99\%$ interval $(12.64,\ 13.04)$. The value $13.0$ is inside (not rejected), but $12.6$ is outside, and indeed $z=\frac{12.84-12.6}{0.0791}=3.04>2.576$.