Sampling and estimation

6.4 Sampling and estimation

Samples and populations

A population is the whole group being studied. A sample is the part you collect data from. Sampling is cheaper and quicker, and it is the only option if testing destroys the item.

A random sample gives every possible sample of that size an equal chance of being chosen. It avoids bias, and it allows the probability results below to be used. For example, number the members and use random numbers to choose them.

Why a sampling method can be unsatisfactory

Give the specific reason, linked to the context.

The distribution of the sample mean

Different samples give different means, so the sample mean $\bar{X}$ is itself a random variable.

Larger samples vary less. If $X$ is normal, $\bar{X}$ is exactly normal for any $n$:

$\bar{X} \sim \mathrm{N}\left(\mu, \frac{\sigma^2}{n}\right)$

$\bar{X} \sim \mathrm{N}(50, 0.64)$, with a standard deviation of 0.8:

$z = \frac{51 - 50}{0.8} = 1.25 \qquad \mathrm{P}(\bar{X} > 51) = 1 - 0.8944 = 0.1056$

The Central Limit Theorem

Unbiased estimates

An unbiased estimate is one whose long-run average equals the true value. Dividing by $n - 1$ for the variance corrects the bias from using the sample's own mean.

$\bar{x} = 12 \qquad s^2 = \frac{1}{9}\left(1640 - \frac{120^2}{10}\right) = \frac{200}{9} = 22.2$

Confidence intervals for a mean

A confidence interval gives a range of likely values of the population mean.

$12 \pm 1.96 \times \frac{4}{8} = 12 \pm 0.98$

The interval is $(11.02, 12.98)$.

For a population that is not known to be normal, use a large sample and the Central Limit Theorem. If $\sigma$ is unknown, a large sample lets you use $s$ in its place.

What the interval means

The true mean does not vary, and the interval does. Avoid saying "the probability that $\mu$ is in this interval is 95%".

$0.95^5 = 0.774$

The size of the sample and the width

The width is $2z\dfrac{\sigma}{\sqrt{n}}$. Multiplying the sample size by 4 halves the width. A higher confidence level gives a wider interval.

$2 \times 1.96 \times \frac{4}{\sqrt{n}} \leq 1 \quad\Rightarrow\quad \sqrt{n} \geq 15.68 \quad\Rightarrow\quad n \geq 245.9$

So the sample needs at least 246. Always round up.

Confidence intervals for a proportion

For a large sample, with $\hat{p} = \dfrac{x}{n}$:

$\hat{p} = 0.2$, so

$0.2 \pm 1.96\sqrt{\frac{0.2 \times 0.8}{200}} = 0.2 \pm 0.0554$

The interval is $(0.145, 0.255)$.