Confidence Intervals

A confidence interval (CI) turns a single point estimate into an honest range of plausible values for an unknown population parameter, together with a stated level of reliability. Where a hypothesis test answers a yes/no question, a confidence interval reports the precision of an estimate — and the two are deeply connected. This lesson builds confidence intervals for a mean (with $\sigma$ known and unknown), for a difference of means, and for a proportion; analyses the width of an interval and how to control it; and stresses the correct interpretation, which is the most frequently misstated idea in the whole topic.

Where this sits in AQA 7367

This is Paper 3 Statistics (7367/3S) content (Paper 3: 2 h, 100 marks, AO1 40% / AO2 25% / AO3 35%). Confidence intervals are the estimation counterpart to hypothesis testing and sit naturally after the t-distribution, reusing its critical values. The construction is AO1 (substitute into $\bar x \pm z\,\sigma/\sqrt n$ and read tables), but the marks that separate candidates are AO2 — the interpretation of the interval and the link to significance tests. It builds directly on the sampling distribution $\bar X \sim N(\mu, \sigma^2/n)$ and on the t-distribution from the previous two lessons.

Core theory: the meaning of "confidence"

A $95\%$ confidence interval for a parameter $\theta$ is a procedure: if the sampling process were repeated many times, building an interval the same way each time, about $95\%$ of those intervals would contain the true $\theta$ . The confidence level describes the long-run success rate of the method, not the probability for one particular interval.

This distinction is essential. Once you have computed an interval such as $(500.0, 504.0)$ , the parameter $\theta$ is a fixed (if unknown) constant and the interval is fixed too — so $\theta$ either is or is not inside it; there is no " $95\%$ probability" about a settled fact. The randomness lives in the sampling, before the interval is drawn. The correct phrasing is therefore "we are $95\%$ confident that $\theta$ lies in $(a,b)$ ," meaning "this interval was produced by a method that works $95\%$ of the time."

The general shape of every interval below is

$\text{estimate} \;\pm\; (\text{critical value}) \times (\text{standard error}),$

where the critical value comes from the normal or t-distribution at the required level and the standard error measures the estimate's variability.

Confidence interval for the mean ( $\sigma$ known)

If $X_1, \ldots, X_n \sim N(\mu, \sigma^2)$ with $\sigma$ known, then $\bar X \sim N(\mu, \sigma^2/n)$ , so standardising and inverting the central $(1-\alpha)$ probability gives

$\bar x \;\pm\; z_{\alpha/2}\,\frac{\sigma}{\sqrt n},$

where $z_{\alpha/2}$ leaves $\alpha/2$ in each tail of $N(0,1)$ .

Deriving the interval from a pivot

It is worth seeing exactly where the $\pm z_{\alpha/2}\sigma/\sqrt n$ comes from, because the same argument produces every interval in the topic. Start from the standardised sample mean — a pivot, a quantity whose distribution is known and free of the unknown $\mu$ :

$Z = \frac{\bar X - \mu}{\sigma/\sqrt n} \sim N(0,1).$

By definition of $z_{\alpha/2}$ , the central probability is

$P\!\left(-z_{\alpha/2} \le \frac{\bar X - \mu}{\sigma/\sqrt n} \le z_{\alpha/2}\right) = 1 - \alpha.$

Now rearrange the inequality to isolate $\mu$ in the middle. Multiplying through by $\sigma/\sqrt n$ and then subtracting $\bar X$ and negating (which flips the inequalities) gives

$P\!\left(\bar X - z_{\alpha/2}\frac{\sigma}{\sqrt n} \le \mu \le \bar X + z_{\alpha/2}\frac{\sigma}{\sqrt n}\right) = 1 - \alpha.$

The random endpoints $\bar X \pm z_{\alpha/2}\sigma/\sqrt n$ bracket the fixed $\mu$ with probability $1-\alpha$ . Replacing the random $\bar X$ by its observed value $\bar x$ yields the computed interval — and shows precisely why " $95\%$ " attaches to the procedure (the random endpoints) and not to the fixed $\mu$ . This pivot-and-rearrange recipe is the single idea behind the mean, difference and (after the test-inversion of Going further) proportion intervals alike.

Confidence level	$z_{\alpha/2}$
90%	1.645
95%	1.960
99%	2.576

These three values are worth memorising — $1.645$ , $1.960$ , $2.576$ recur throughout inference.

Worked example — CI for a mean, $\sigma$ known

A machine fills bottles with mean volume $\mu$ and known standard deviation $\sigma = 5$ ml. A random sample of $25$ bottles has $\bar x = 502$ ml. Construct a $95\%$ CI for $\mu$ .

$\text{SE} = \frac{\sigma}{\sqrt n} = \frac{5}{\sqrt{25}} = 1. \quad (\text{M1 standard error})$ $502 \pm 1.960 \times 1 = 502 \pm 1.96. \quad (\text{M1 form; A1})$

So the $95\%$ CI is $(500.04,\ 503.96)$ ml. (M1 SE; M1 $\bar x \pm z\,\text{SE}$ with $z = 1.96$ ; A1 endpoints. Interpretation: we are $95\%$ confident the mean fill volume lies between $500.04$ and $503.96$ ml.)

Confidence interval for the mean ( $\sigma$ unknown)

When $\sigma$ is unknown, replace it with the sample standard deviation $s$ and use the t-distribution with $n-1$ degrees of freedom — exactly as in the t-test:

$\bar x \;\pm\; t_{n-1,\,\alpha/2}\,\frac{s}{\sqrt n}.$

Because the t critical value exceeds the matching $z$ , the interval is wider than the $\sigma$ -known case — the price of not knowing $\sigma$ .

Worked example — CI for a mean, $\sigma$ unknown

A sample of $10$ measurements gives $\bar x = 48.3$ and $s = 2.1$ . Construct a $95\%$ CI for $\mu$ .

$t_{9,\,0.025} = 2.262, \qquad \text{SE} = \frac{2.1}{\sqrt{10}} = 0.6641. \quad (\text{B1 critical value; M1 SE})$ $48.3 \pm 2.262 \times 0.6641 = 48.3 \pm 1.502. \quad (\text{M1; A1})$

So the $95\%$ CI is $(46.80,\ 49.80)$ . (B1 $t_{9,0.025}$ ; M1 SE; M1 $\bar x \pm t\,\text{SE}$ ; A1 endpoints. Using $z = 1.96$ here would wrongly give $\pm 1.30$ — too narrow.)

To see the cost of not knowing $\sigma$ concretely, compare the two cases for this sample. With $\sigma$ known and equal to $2.1$ , the half-width would be $1.96 \times 0.6641 = 1.302$ , giving $(47.00, 49.60)$ . With $\sigma$ unknown the t-multiplier $2.262$ widens this to $\pm 1.502$ . The interval is about $15\%$ wider — the penalty for having to estimate the spread from only ten observations. As $n$ grows, $t_{n-1,\,0.025}$ falls towards $1.96$ and the two intervals converge; for $n = 100$ the difference is negligible. This is the estimation counterpart of the heavier-tailed t-distribution from the previous lesson: less certainty about $\sigma$ means a wider net for $\mu$ .

Width of a confidence interval

The full width of a ( $\sigma$ -known) interval is

$w_{\text{full}} = 2\,z_{\alpha/2}\,\frac{\sigma}{\sqrt n},$

so the interval narrows if you increase $n$ (the only practical lever), lower the confidence level (smaller $z$ , but less reliable), or have a smaller $\sigma$ (rarely in your control). The dependence on $\sqrt n$ is important: to halve the width you must quadruple the sample, since $1/\sqrt n$ halves only when $n$ is multiplied by $4$ . This diminishing return is why precision is expensive — going from $n = 100$ to a half-width that is one-tenth as large demands $n = 10000$ , a hundredfold increase in data. The confidence level enters only through the multiplier $z_{\alpha/2}$ , so raising confidence from $95\%$ to $99\%$ widens every interval by the factor $2.576/1.960 = 1.31$ , a fixed $31\%$ inflation regardless of the data.

Sample size for a target precision

To achieve a half-width (margin of error) $w$ , rearrange $z_{\alpha/2}\,\sigma/\sqrt n \le w$ :

$n \ge \left(\frac{z_{\alpha/2}\,\sigma}{w}\right)^2.$

Worked example. How large a sample gives a $95\%$ CI with half-width $2$ ml when $\sigma = 5$ ?

$n \ge \left(\frac{1.96 \times 5}{2}\right)^2 = (4.9)^2 = 24.01 \;\Rightarrow\; n = 25.$

Always round up to the next integer — rounding down would leave the interval slightly too wide.

Confidence interval for a difference of two means

For independent samples from $N(\mu_1,\sigma_1^2)$ and $N(\mu_2,\sigma_2^2)$ , the estimator $\bar X_1 - \bar X_2$ has variance $\sigma_1^2/n_1 + \sigma_2^2/n_2$ (variances add), giving:

Known variances: $(\bar x_1 - \bar x_2) \;\pm\; z_{\alpha/2}\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}.$

Unknown but equal variances (pooled $s_p$ ): $(\bar x_1 - \bar x_2) \;\pm\; t_{n_1+n_2-2,\,\alpha/2}\,s_p\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}.$

A key reading: if this interval excludes $0$ , then $0$ is not a plausible value for $\mu_1 - \mu_2$ , which is evidence of a genuine difference (equivalent to rejecting $H_0:\mu_1 = \mu_2$ in a two-tailed test at level $\alpha$ ).

Worked example — CI for a difference of means

Two production lines have known standard deviations $\sigma_1 = 3$ , $\sigma_2 = 4$ (mm). Samples give line 1: $n_1 = 36$ , $\bar x_1 = 50.2$ ; line 2: $n_2 = 64$ , $\bar x_2 = 48.7$ . Construct a $95\%$ CI for $\mu_1 - \mu_2$ and state whether the lines differ.

$\bar x_1 - \bar x_2 = 50.2 - 48.7 = 1.5. \quad (\text{B1 point estimate})$ $\text{SE} = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} = \sqrt{\frac{9}{36} + \frac{16}{64}} = \sqrt{0.25 + 0.25} = \sqrt{0.5} = 0.7071. \quad (\text{M1 SE; variances added})$ $1.5 \pm 1.96 \times 0.7071 = 1.5 \pm 1.386. \quad (\text{M1; A1})$

So the $95\%$ CI is $(0.114,\ 2.886)$ mm. Since the interval excludes $0$ , there is evidence at the $5\%$ level that the line means differ (line 1 produces larger items). (B1 estimate; M1 SE adding the two variances; M1 $\pm z\,\text{SE}$ ; A1 endpoints and the "excludes 0" conclusion.)

Confidence interval for a proportion

For a large sample the sample proportion $\hat p$ is approximately normal,

$\hat p \sim N\!\left(p,\ \frac{p(1-p)}{n}\right),$

and estimating the standard error with $\hat p$ gives the $95\%$ (or general) interval

$\hat p \;\pm\; z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}},$

valid when $n\hat p > 5$ and $n(1-\hat p) > 5$ (so the normal approximation to the binomial holds).

Worked example — CI for a proportion

In a survey of $400$ voters, $220$ support a policy. Construct a $95\%$ CI for the true proportion $p$ .

$\hat p = \frac{220}{400} = 0.55, \qquad \text{SE} = \sqrt{\frac{0.55 \times 0.45}{400}} = \sqrt{0.00061875} = 0.024875. \quad (\text{M1 } \hat p;\ \text{M1 SE})$ $0.55 \pm 1.96 \times 0.024875 = 0.55 \pm 0.0488. \quad (\text{M1; A1})$

So the $95\%$ CI is $(0.501,\ 0.599)$ . (M1 $\hat p$ ; M1 SE with $\hat p(1-\hat p)/n$ ; M1 $\hat p \pm z\,\text{SE}$ ; A1 endpoints. Since the interval lies entirely above $0.5$ , there is evidence of majority support.)

Notice that the entire interval lies above $0.5$ , which is the proportion corresponding to "no majority." Because $0.5$ is not a plausible value, the data support a genuine majority — the same conclusion a two-tailed test of $H_0:p = 0.5$ would reach, illustrating the duality once more. Had the interval straddled $0.5$ (for example if the sample had been smaller and hence the interval wider), we could not claim a majority, because values below $0.5$ would remain plausible.

One-sided confidence intervals

Sometimes only one direction matters — a minimum guaranteed lifetime, or an upper bound on a contaminant. A one-sided interval places the whole error probability $\alpha$ in a single tail, using $z_\alpha$ (e.g. $1.645$ at $95\%$ ) rather than $z_{\alpha/2}$ :

$\text{lower bound: } \mu \ge \bar x - z_\alpha\frac{\sigma}{\sqrt n}, \qquad \text{upper bound: } \mu \le \bar x + z_\alpha\frac{\sigma}{\sqrt n}.$

For instance, with $\bar x = 1020$ hours, $\sigma = 40$ , $n = 25$ , a one-sided $95\%$ lower confidence bound for the mean lifetime is

$\mu \ge 1020 - 1.645 \times \frac{40}{\sqrt{25}} = 1020 - 1.645 \times 8 = 1020 - 13.16 = 1006.84 \text{ hours},$

Confidence Intervals

Confidence Intervals

Where this sits in AQA 7367

Core theory: the meaning of "confidence"

Confidence interval for the mean (σ\sigmaσ known)

Deriving the interval from a pivot

Worked example — CI for a mean, σ\sigmaσ known

Confidence interval for the mean (σ\sigmaσ unknown)

Worked example — CI for a mean, σ\sigmaσ unknown

Width of a confidence interval

Sample size for a target precision

Confidence interval for a difference of two means

Worked example — CI for a difference of means

Confidence interval for a proportion

Worked example — CI for a proportion

One-sided confidence intervals

More in Mathematics

Confidence interval for the mean ( $\sigma$ known)

Worked example — CI for a mean, $\sigma$ known

Confidence interval for the mean ( $\sigma$ unknown)

Worked example — CI for a mean, $\sigma$ unknown