Confidence intervals for a Normal meanEdexcel International 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. Its confidence level (for example $95\%$) says how reliable the method is: if you took many random samples and built an interval from each in the same way, about $95\%$ of those intervals would contain the true parameter. The parameter $\mu$ is a fixed number; it is the interval that changes from sample to sample, because it is built from the sample mean $\bar X$. A higher confidence level gives a wider interval, and a larger sample gives a narrower one. A confidence interval gives more information than a single estimate such as $\bar x$, because it shows how precise that estimate is.

Let $X\sim N(\mu,\sigma^2)$ with $\sigma$ known, and take a random sample of size $n$. Then $\bar X\sim N\left(\mu,\frac{\sigma^2}{n}\right)$, so $\frac{\bar X-\mu}{\sigma/\sqrt n}\sim N(0,1)$. The confidence limits are $\bar x\pm z\,\frac{\sigma}{\sqrt n},$ where $\frac{\sigma}{\sqrt n}$ is the standard error and $z$ cuts off the required central area of $N(0,1)$:

• $90\%$: $z=1.645$
• $95\%$: $z=1.96$
• $98\%$: $z=2.326$
• $99\%$: $z=2.576$ Worked example: $\sigma=8$, $n=25$, $\bar x=1012$. The standard error is $1.6$, so the $95\%$ interval is $1012\pm1.96\times1.6=(1008.9,\ 1015.1)$.

The interval is symmetrical about $\bar x$, with half-width $z\frac{\sigma}{\sqrt n}$ and width $2z\frac{\sigma}{\sqrt n}$. The width falls if $n$ increases, and rises if the confidence level rises or $\sigma$ is larger. To find the sample size for a given width $w$: $2z\frac{\sigma}{\sqrt n}\leq w\ \Rightarrow\ \sqrt n\geq\frac{2z\sigma}{w}\ \Rightarrow\ n\geq\left(\frac{2z\sigma}{w}\right)^2.$ Always round $n$ up to the next whole number. For $\sigma=4.5$, $z=1.96$ and $w=2$: $\sqrt n\geq8.82$, so $n\geq77.8$ and the smallest sample is $78$. You can also work backwards: if a $95\%$ interval is $(49.2,\ 54.8)$ with $\sigma=10$, then $\bar x=\frac{49.2+54.8}{2}=52$ and the half-width $2.8=1.96\times\frac{10}{\sqrt n}$ gives $n=49$.

A confidence interval does not say that $95\%$ of the population values lie in it, and it does not give the probability that a fixed $\mu$ is in a particular interval. Valid statements are:

• if the method were repeated many times, about $95\%$ of the intervals would contain $\mu$;
• the interval is a set of plausible values for $\mu$ given this sample. Different random samples give different intervals, so two intervals that differ are not a sign that one is wrong, provided they overlap sensibly. Even a correct method fails to capture $\mu$ about $5\%$ of the time at the $95\%$ level. When commenting in context, refer to the actual quantity: 'we are $95\%$ confident that the mean journey time lies between $50.5$ and $54.3$ minutes' gains the mark that 'the mean is in the interval' might not.

A $c\%$ confidence interval corresponds to a two-tailed hypothesis test at the $(100-c)\%$ significance level. To test $H_0:\mu=\mu_0$ against $H_1:\mu\neq\mu_0$:

• if $\mu_0$ lies inside the interval, do not reject $H_0$;
• if $\mu_0$ lies outside the interval, reject $H_0$: there is evidence that $\mu$ differs from $\mu_0$. Example: a $95\%$ interval of $(49.2,\ 54.8)$ contains $50$, so at the $5\%$ level we do not reject $\mu=50$; but it does not contain $56$, so $\mu=56$ is rejected. A one-tailed test at the $5\%$ level needs a different interval (a $90\%$ interval matches), so use this link only for two-tailed tests. State conclusions in context and avoid saying that $H_0$ is proved.