6.4 Sampling and estimation
- Syllabus
- 9709–2028–2029
- Topic
- 6.4
- Level
- A2
Learning objectives
6.4.1Samples and populations• understand the distinction between a sample and a population, and appreciate the necessity for randomness in choosing samples6.4.2Sampling methods• explain in simple terms why a given sampling method may be unsatisfactory Including an elementary understanding of the use of random numbers in producing random samples. Knowledge of particular sampling methods, such as quota or stratified sampling, is not required.6.4.3Sample mean• recognise that a sample mean can be regarded as a random variable, and use the facts that E X = n_ i and that Var X n 2 = v_ i6.4.4Normal distribution• use the fact that X_ i has a normal distribution if X has a normal distribution6.4.5Sample mean• use the Central Limit Theorem where appropriate Only an informal understanding of the Central Limit Theorem (CLT) is required; for large sample sizes, the distribution of a sample mean is approximately normal.6.4.6Unbiased estimates• calculate unbiased estimates of the population mean and variance from a sample, using either raw or summarised data Only a simple understanding of the term 'unbiased' is required, e.g. that although individual estimates will vary the process gives an accurate result 'on average'.6.4.7Mean confidence intervals• determine and interpret a confidence interval for a population mean in cases where the population is normally distributed with known variance or where a large sample is used6.4.8Proportion confidence intervals• determine, from a large sample, an approximate confidence interval for a population proportion.
A population is the target group while a sample is the observed subset
A population contains all units of interest; a sample is selected to estimate population characteristics. A parameter describes the population, while a statistic is calculated from the sample.
Define the target population before sampling and consider coverage, non-response and selection bias. Larger samples reduce random error but do not automatically remove systematic bias.
A survey of 500 randomly chosen voters is a sample; the proportion supporting a policy in all eligible voters is a population parameter.
A sample statistic is not the parameter itself, and a large biased sample can still mislead.
Random numbers select from a frame; they cannot repair a bad frame
| Random-number sampling step | Action |
|---|---|
| frame | list and number every covered population unit 1,…,N |
| generate | use a random-number generator/table with a fixed reading rule |
| screen | ignore values outside 1,…,N and repeated values when sampling without replacement |
| continue | select until the required sample size is reached |
| Unsatisfactory feature | Why results may mislead |
|---|---|
| convenience/voluntary participation | selected people may differ systematically from the target |
| incomplete frame | uncovered groups have no chance of selection |
| non-response | responders may differ from non-responders |
| leading question or faulty measurement | responses are shifted even if selection was random |
Choosing random numbers from a school email list gives random selection from that list, but it cannot represent pupils omitted from the list. Explain the missing group and the likely direction of distortion when context supports it.
The syllabus asks for simple criticism and elementary random-number use. Knowledge of named schemes such as quota or stratified sampling is not required.
‘Random’ is not a cure-all: it randomises selection within the sampling frame but does not remove frame, non-response or measurement bias.
The sample mean estimates a population mean and has its own sampling distribution
For independent observations Xᵢ with mean μ and variance σ², the sample mean X̄ has E(X̄)=μ and Var(X̄)=σ²/n. Its standard error is σ/√n (or estimated with s/√n).
Distinguish variability of individual observations from variability of means, and account for finite-population or dependence conditions when relevant.
If σ=12 and n=36, the standard error of X̄ is 2, even though individual values vary with standard deviation 12.
Increasing n reduces standard error by √n, not by n, and does not necessarily reduce measurement bias.
A normal population gives an exactly normal sample mean
IfindependentobservationscomefromX\sim N(\mu,\sigma^2),then for every sample size $n$,\bar X\sim N\left(\mu,\frac{\sigma^2}{n}\right).Thisisexact,notalarge−sampleapproximation.
For a sample-mean boundary $a$, standardise withZ=\frac{\bar X-\mu}{\sigma/\sqrt n}.
If individual values are N(100,152) and n=25, then Xˉ∼N(100,9). Thus P(Xˉ>106)=P(Z>3106−100)=P(Z>2).
Do not require large n when the population is normal. The mean stays μ, but the variance becomes σ2/n; use σ/n, not σ, in the z-score. CLT belongs to the next objective.
CLT makes a large-sample mean approximately normal
For a large independent random sample from a population with finite mean $\mu$ and variance $\sigma^2$, the Central Limit Theorem gives\bar X\approx N\left(\mu,\frac{\sigma^2}{n}\right).Onlyaninformalunderstandingofthisresultisrequired.
| Population | Distribution of Xˉ |
|---|---|
| normal | exactly normal for every n |
| not normal / shape unspecified | approximately normal when n is sufficiently large |
A more skewed or heavy-tailed population generally needs a larger sample before the approximation is convincing. Randomness and independence still matter; CLT does not remove sampling bias or dependence.
A non-normal population has mean 10 and standard deviation 4. For an independent sample of size 64, use Xˉ≈N(10,0.25), so P(Xˉ>11)≈P(Z>0.511−10)=P(Z>2).
CLT says the distribution of the sample mean becomes approximately normal; it does not say the individual observations or the population become normal.
Use $n-1$ for the unbiased variance estimate
| Data form | Unbiased mean estimate | Unbiased variance estimate |
|---|---|---|
| raw x1,…,xn | xˉ=∑x/n | s2={∑x2−(∑x)2/n}/(n−1) |
| frequencies f | xˉ=∑fx/∑f | s2={∑fx2−(∑fx)2/n}/(n−1), n=∑f |
Equivalently,s^2=\frac{\sum (x-\bar x)^2}{n-1}.The denominator $n-1$ corrects the downward bias caused by estimating $\mu$ with the same sample mean.
For data 2,4,6, xˉ=4 and ∑(x−xˉ)2=4+0+4=8, so the unbiased variance estimate is s2=8/(3−1)=4.
Unbiased means the estimation procedure gives the true population value on average over repeated random samples. One particular estimate can still be above or below the parameter.
Do not divide by n when the question asks for an unbiased population-variance estimate, and do not report s when variance s2 is requested.
A mean confidence interval is estimate plus/minus normal margin
| Allowed case | Standard error used |
|---|---|
| normal population, known variance σ2 | σ/n |
| large sample | s/n when σ is unknown, using the sample estimate |
A $100(1-\alpha)\%$ interval is\bar x\pm z_{1-\alpha/2}\times\operatorname{SE}(\bar X).Common two-sided values are $1.645$ for 90%, $1.96$ for 95% and $2.576$ for 99%.
A normal population has known σ=12. From n=36 observations, xˉ=50. A 95% interval is 50±1.96(612)=50±3.92, giving (46.08,53.92).
Interpret in context: we are 95% confident that the population mean lies between 46.08 and 53.92. Repeated use of this method captures the fixed population mean in about 95% of intervals.
Use the standard error, not the population spread itself. The Cambridge 9709 scope here uses the normal/large-sample cases stated above; do not introduce a t distribution unless another specification explicitly asks for it.
Estimate a population proportion with a large-sample normal interval
From $x$ successes in a large sample of size $n$,\hat p=\frac{x}{n},\qquad \widehat{\operatorname{SE}}(\hat p)=\sqrt{\frac{\hat p(1-\hat p)}{n}}.Checkthattheestimatedsuccessandfailurecountsarebothsufficientlylargeforthenormalapproximation.
A $100(1-\alpha)\%$ approximate interval is\hat p\pm z_{1-\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}.
If 40 of 100 sampled people support a proposal, p^=0.40 and SE≈0.4(0.6)/100=0.0490. A 95% interval is 0.40±1.96(0.0490)=0.40±0.096, giving approximately (0.304,0.496).
We are 95% confident that the population proportion supporting the proposal lies between about 30.4% and 49.6%, assuming the sample was random and the large-sample approximation is appropriate.
Do not stop after finding the standard error or treat the interval as a range for individual 0/1 responses. The interval estimates one population proportion.