Confidence Intervals for a Mean and a Proportion (9709 PS2) — Mathematics

Introduction

A confidence interval (CI) gives a range of plausible values for an unknown population parameter, estimated from sample data. Rather than a single point estimate, a CI captures the uncertainty of estimation. In 9709 PS2 you need to:

Construct a CI for a population mean $\mu$ when (i) the population is normal with known variance $\sigma^2$ , or (ii) the sample is large ( $n \geq 30$ is the working rule).
Construct an approximate CI for a population proportion $p$ from a large sample.
Interpret the meaning of a confidence level in context.

This topic draws directly on the sampling distribution of the mean $\bar{X} \sim \mathrm{N}\!\left(\mu,\, \dfrac{\sigma^2}{n}\right)$ from the prerequisite subtopic.

Core Concept

The Logic of a Confidence Interval

We know that the standardised sample mean $Z = \dfrac{\bar{X} - \mu}{\sigma/\sqrt{n}}$ follows $\mathrm{N}(0,1)$ . For a 95% CI, we want the middle 95% of this distribution, i.e. we find $z^*$ such that $\mathrm{P}(-z^* \leq Z \leq z^*) = 0.95$ . Rearranging for $\mu$ :

\mathrm{P}\!\left(\bar{X} - z^* \frac{\sigma}{\sqrt{n}} \leq \mu \leq \bar{X} + z^* \frac{\sigma}{\sqrt{n}}\right) = 0.95

Once $\bar{x}$ is observed, we obtain a fixed interval — $\mu$ is a fixed constant, not random, so we do not say "there is a 95% probability that $\mu$ lies in this interval." The correct interpretation is: "95% of all such intervals constructed by this method would contain $\mu$ ."

Critical Values $z^*$

Confidence level	$z^*$
90%	1.645
95%	1.960
99%	2.576

These are read from the normal distribution tables supplied in the exam ( $z^*$ is the value with $\Phi(z^*) = 1 - \tfrac{\alpha}{2}$ ).

Large Samples: Estimating $\sigma^2$

When $n$ is large and $\sigma^2$ is unknown, the sample variance $s^2$ is used as an estimate. By the Central Limit Theorem, $\bar{X}$ is approximately normal regardless of the population distribution, so the interval remains valid.

Confidence Interval for a Proportion

For a large sample of size $n$ with observed proportion $\hat{p} = x/n$ , the approximate distribution is:

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

Since $p$ is unknown, we substitute $\hat{p}$ for $p$ inside the standard error, giving an approximate CI.

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