Sampling — H2 Maths help

In statistics, we rarely have access to an entire population, so we draw a random sample of size $n$ and use it to make inferences. For a sample to be useful, every member of the population must have an equal chance of being selected; a biased sample (e.g. convenience sampling) can produce estimates that are systematically wrong.

What you need to know

A random sample $X_1, X_2, \ldots, X_n$ consists of independent, identically distributed observations from a population with mean $\mu$ and variance $\sigma^2$ .
The sample mean $\bar{X} = \dfrac{X_1 + X_2 + \cdots + X_n}{n}$ is itself a random variable with $\text{E}(\bar{X}) = \mu$ and $\text{Var}(\bar{X}) = \dfrac{\sigma^2}{n}$ .
If the population is normally distributed, then $\bar{X} \sim \text{N}\!\left(\mu, \dfrac{\sigma^2}{n}\right)$ exactly for any $n$ .
By the Central Limit Theorem (CLT), if $n$ is large (typically $n \geq 30$ ), then $\bar{X}$ is approximately normal even when the population distribution is unknown or non-normal.
To standardise: $Z = \dfrac{\bar{X} - \mu}{\sigma / \sqrt{n}} \sim \text{N}(0,1)$ (exactly or approximately as appropriate).
Unbiased estimates from a sample: population mean estimated by $\bar{x} = \dfrac{\sum x_i}{n}$ ; population variance estimated by $s^2 = \dfrac{1}{n-1}\sum(x_i - \bar{x})^2$ (divisor $n-1$ , not $n$ ).

Worked example 1 — exact normal population

The mass of a bag of rice is normally distributed with mean 1.02 kg and standard deviation 0.04 kg. A random sample of 16 bags is selected. Find the probability that the sample mean mass exceeds 1.03 kg.

The population is normal, so the result is exact:

\bar{X} \sim \text{N}\!\left(1.02,\ \frac{0.04^2}{16}\right) = \text{N}(1.02,\ 0.0001).

Standardise:

Z = \frac{\bar{X} - 1.02}{0.04/\sqrt{16}} = \frac{\bar{X} - 1.02}{0.01}.

P(\bar{X} > 1.03) = P\!\left(Z > \frac{1.03 - 1.02}{0.01}\right) = P(Z > 1) = 1 - \Phi(1) \approx 0.159.

Answer: 0.159 (3 s.f.)

Worked example 2 — CLT for a non-normal population

The number of minutes a student spends on social media per day follows an unknown distribution with mean 95 and variance 225. A random sample of 50 students is taken. Find the probability that the sample mean exceeds 100 minutes.

Since $n = 50$ is large, by the CLT:

\bar{X} \approx \text{N}\!\left(95,\ \frac{225}{50}\right) = \text{N}(95,\ 4.5).

P(\bar{X} > 100) = P\!\left(Z > \frac{100 - 95}{\sqrt{4.5}}\right) = P(Z > 2.357) \approx 1 - \Phi(2.357) \approx 0.00921.

Answer: 0.00921 (3 s.f.)

Worked example 3 — unbiased estimates

A random sample of 8 observations of a variable $X$ gives $\sum x = 56$ and $\sum x^2 = 420$ . Find unbiased estimates of the population mean and variance.

\bar{x} = \frac{56}{8} = 7.

s^2 = \frac{1}{n-1}\left(\sum x^2 - n\bar{x}^2\right) = \frac{1}{7}\!\left(420 - 8 \times 49\right) = \frac{420 - 392}{7} = \frac{28}{7} = 4.

Answer: unbiased estimate of mean $= 7$ ; unbiased estimate of variance $= 4$ .

Common mistakes

Using divisor $n$ instead of $n-1$ when computing the unbiased sample variance — this gives a biased estimate and is wrong in exam questions that ask for an “unbiased estimate”.
Forgetting to divide $\sigma^2$ by $n$ when writing the distribution of $\bar{X}$ — a common slip is writing $\bar{X} \sim \text{N}(\mu, \sigma^2)$ instead of $\text{N}(\mu, \sigma^2/n)$ .
Applying the exact normal result when the population is not stated to be normal and $n$ is small — the CLT only justifies approximate normality for large $n$ .
Confusing the standard deviation of $\bar{X}$ (i.e. $\sigma/\sqrt{n}$ ) with the population standard deviation $\sigma$ when standardising.