Confidence Intervals for a Mean

A sample mean of $50.3$ cm is a single best guess for the population mean, but on its own it says nothing about how far off it might be. A confidence interval replaces the single number with a range, such as $(49.99, 50.61)$, built so that intervals made this way capture the true mean a stated percentage of the time. Paper 6 asks you to calculate these intervals, to find a sample size or a confidence level from one, and, just as often, to interpret one correctly in words. The calculation is short; the interpretation is where most marks are lost.

From a probability statement to an interval

Suppose the population is normal with known standard deviation $\sigma$, so that the sample mean satisfies $\bar{X} \sim N\left(\mu, \dfrac{\sigma^2}{n}\right)$. Standardising, $Z = \dfrac{\bar{X} - \mu}{\sigma/\sqrt{n}}$ is standard normal, and $95\%$ of its probability lies between $-1.96$ and $1.96$:

So

$$P\left(-1.96 < \frac{\bar{X} - \mu}{\sigma/\sqrt{n}} < 1.96\right) = 0.95.$$

Rearrange the inequalities to put $\mu$ in the middle:

$$P\left(\bar{X} - 1.96\frac{\sigma}{\sqrt{n}} < \mu < \bar{X} + 1.96\frac{\sigma}{\sqrt{n}}\right) = 0.95.$$

Read this carefully. The random quantity is $\bar{X}$, so the interval is random and $\mu$ is fixed. Before the sample is taken, there is a $0.95$ probability that the interval you are about to calculate will contain $\mu$. Once you substitute the observed $\bar{x}$, you have one particular interval, called a $95\%$ confidence interval for $\mu$.

The quantity $z\dfrac{\sigma}{\sqrt{n}}$ is the margin of error, half the width of the interval. The interval is symmetrical about $\bar{x}$.

When the formula can be used

The syllabus covers two situations.

Situation	What to use for $\sigma$	Why $\bar{X}$ is normal
Population normal, variance known	$\sigma$	Exactly normal for any $n$
Large sample, any population	$\sigma$ if known, otherwise $s$ from the sample	Approximately normal by the Central Limit Theorem

In the second case the unbiased estimate $s^2$ (see unbiased estimates) replaces the unknown $\sigma^2$, and the interval is approximate. With a large sample, $s$ is close enough to $\sigma$ for this to make little difference.

A small sample from a normal population with unknown variance needs a different distribution (the $t$-distribution), which is not on the 9709 syllabus. Paper 6 will not ask for it.

Interpreting a confidence interval

Imagine taking many random samples of the same size and calculating a $95\%$ confidence interval from each. The intervals jump about because each sample has a different mean. About $95\%$ of them contain $\mu$; about $5\%$ miss it. The graph below shows ten such intervals for a population with $\mu = 50$ (the vertical line). Nine of them cross the line; the seventh one up misses it.

Once a particular interval has been calculated, it either contains $\mu$ or it does not; $\mu$ is not a random variable. Saying "there is a $95\%$ probability that $\mu$ lies in $(49.99, 50.61)$" is the classic misstatement. Examiners accept wording such as "$95\%$ of intervals constructed in this way would contain the population mean".

Because each interval contains $\mu$ with probability $0.95$ before it is calculated, the number of intervals that contain $\mu$, out of $k$ independent ones, has the distribution $B(k, 0.95)$. This is a popular final part to a question.

A confidence interval also gives a quick check on a claim. If a claimed value of $\mu$ lies outside a $95\%$ confidence interval, the sample is evidence against the claim; if it lies inside, the sample is consistent with it. This is closely related to a two-tailed hypothesis test at the $5\%$ level.

What controls the width

The width of the interval is $2z\dfrac{\sigma}{\sqrt{n}}$, so:

• A higher confidence level means a larger $z$ and a wider interval. To be more sure of catching $\mu$ you need a bigger net.
• A larger sample means a narrower interval, in proportion to $\dfrac{1}{\sqrt{n}}$. Quadrupling $n$ halves the width.
• A more variable population means a wider interval.

Practice questions

Answers

1. $20.5 \pm 2.326 \times \dfrac{4}{\sqrt{36}} = 20.5 \pm 1.5507$, giving $(18.95, 22.05)$.

2. $\bar{x} = 52.3$; $s^2 = \tfrac{1}{99}\left(274\,900 - \tfrac{5230^2}{100}\right) = \tfrac{1}{99}(274\,900 - 273\,529) = 13.848$; $s = 3.7214$. $52.3 \pm 1.96 \times \dfrac{3.7214}{10} = 52.3 \pm 0.7294$, giving $(51.57, 53.03)$. The sample is large, so the Central Limit Theorem applies.

3. Margin of error $= 1.4$, so $1.96 \times \dfrac{\sigma}{\sqrt{50}} = 1.4$ and $\sigma = \dfrac{1.4\sqrt{50}}{1.96} = 5.05$.

4. $2.576 \times \dfrac{10}{\sqrt{n}} \le 1.5 \Rightarrow \sqrt{n} \ge 17.173 \Rightarrow n \ge 294.9$. The smallest sample size is $295$.

5. $\bar{x} = 13$ and the margin is $0.64$. $z \times \dfrac{2.8}{8} = 0.64 \Rightarrow z = 1.829$. $\Phi(1.829) = 0.9663$, so each tail has $0.0337$ and the level is $1 - 2(0.0337) = 0.9326$, that is $93.3\%$.

6. (a) The population mean is a fixed value, not a random variable, so it either lies in $(4.2, 5.8)$ or it does not. The $95\%$ refers to the method: if many samples were taken, about $95\%$ of the intervals calculated would contain the population mean. (b) The number containing $\mu$ is $B(3, 0.95)$. $P(\text{exactly } 2) = 3(0.95)^2(0.05) = 0.135$.

7. (a) $10.22 \pm 2.576 \times \dfrac{0.5}{\sqrt{20}} = 10.22 \pm 0.2880$, giving $(9.932, 10.508)$ mm. (b) $9.9$ lies below the interval, so the sample gives evidence (at the $1\%$ level) that the mean length is not $9.9$ mm; the claim appears to be incorrect, and the mean appears to be higher.

8. (a) $\bar{x} = 43$ and the $90\%$ margin is $2.6 = 1.645 \times \dfrac{\sigma}{\sqrt{n}}$, so $\dfrac{\sigma}{\sqrt{n}} = 1.5805$. The $99\%$ margin is $2.576 \times 1.5805 = 4.0715$, giving $(38.93, 47.07)$. (b) $\dfrac{10}{\sqrt{n}} = 1.5805 \Rightarrow \sqrt{n} = 6.327 \Rightarrow n = 40.03$, so $n = 40$.