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

If a population is normal with mean $\mu$ and variance $\sigma^2$, the mean $\bar X$ of a random sample of size $n$ is normal with mean $\mu$ and variance $\frac{\sigma^2}{n}$: $\bar X\sim N\left(\mu,\ \frac{\sigma^2}{n}\right).$ The standard deviation of $\bar X$, $\frac{\sigma}{\sqrt n}$, is called the standard error. It shrinks as $n$ grows, so larger samples pin down $\mu$ more precisely. A confidence interval uses this to give a range of plausible values for $\mu$.

A symmetric $C\%$ confidence interval for $\mu$ is $\bar x\pm z\,\frac{\sigma}{\sqrt n},$ where $z$ is the value with $C\%$ of the standard normal distribution between $-z$ and $z$. The common values are

• $90\%$: $z=1.645$
• $95\%$: $z=1.96$
• $99\%$: $z=2.576$. Example: $\sigma=12$, $n=36$, $\bar x=504$. Standard error $=2$, so the $95\%$ interval is $504\pm1.96\times2=(500.08,\ 507.92)$.

Often $\sigma$ is not known. If the sample is large, the sample standard deviation $s$ is a reliable estimate of $\sigma$, and the interval is still $\bar x\pm z\,\frac{s}{\sqrt n},$ using the same $z$ values. Use the unbiased estimate of the variance: $s^2=\frac{\sum x^2-n\bar x^2}{n-1}=\frac{1}{n-1}\left(\sum x^2-\frac{(\sum x)^2}{n}\right).$ Example: $n=50$, $\bar x=1172$, $s=96$: the $95\%$ interval is $1172\pm1.96\times\frac{96}{\sqrt{50}}=(1145.4,\ 1198.6)$.

A confidence interval tells you which values of $\mu$ are plausible at that level.

• If a claimed value of $\mu$ lies inside the interval, the data are consistent with the claim.
• If it lies outside, there is evidence at the corresponding level (for example $5\%$ for a $95\%$ interval) that the mean is different, and the interval shows in which direction. Always answer in context. The interval is wider when the confidence level is higher (larger $z$), when $\sigma$ is larger, and when $n$ is smaller. For a $95\%$ interval to have width $w$, solve $2\times1.96\times\frac{\sigma}{\sqrt n}\le w$ for $n$ and round up.