Sampling and estimation

Populations, samples and randomness Paper 6

The population is the whole set of items of interest: every bag of rice a machine fills, every voter in a town. A sample is the part of the population that is actually observed. A survey of the whole population (a census) is usually impossible or wasteful: it costs too much, takes too long, or destroys what it tests (you cannot test the strength of every rope by breaking it).

A quantity describing the population, such as its mean $\mu$, is fixed but unknown. A quantity worked out from a sample, such as $\bar{x}$, is known but changes from one sample to the next. Statistics uses the sample to say something about the population, and that is only trustworthy if the sample is random.

In a random sample every member of the population has the same chance of being chosen, and the choices are independent. Randomness protects against bias that the person choosing may not even notice, and it is what makes the probability results of this topic valid.

Using random numbers. Number the members of the population, for example 0001 to 8000. Use a calculator or computer random number generator (or a table of random digits) to produce four-digit numbers. Take the member with each number produced, ignoring numbers above 8000 and any repeats, until the sample is the size you want.

Why a sampling method may be unsatisfactory Paper 6

When a question asks why a method is unsatisfactory, name the specific group that is over- or under-represented, in context, and say why that matters. Typical faults:

• Not everyone can be chosen. Interviewing shoppers at 10 am on a weekday misses people who work in the day.
• The sample chooses itself. Volunteers or people who reply to an online poll often have stronger views than everyone else.
• The choice is linked to what is measured. Asking people at a gym about exercise habits will overestimate how much the town exercises.
• Convenience or order. The first 30 names on an alphabetical list, or the top layer of a crate, may differ systematically from the rest.
• The sample is too small to give a reliable estimate.

Particular named methods such as quota or stratified sampling are not required.

The sample mean as a random variable Paper 6

Take a random sample $X_1,X_2,\dots ,X_n$ from a population with mean $\mu$ and variance ${\sigma }^2$. The sample mean
$$\bar{X}=\frac{X_1+X_2+\dots +X_n}{n}$$
is itself a random variable: a different sample gives a different value. Using the results of the previous topic, the total has mean $n\mu$ and variance $n{\sigma }^2$, and dividing by $n$ gives
$$\mathrm{E}(\bar{X})=\mu ,\mathrm{Var}(\bar{X})=\frac{{\sigma }^2}{n}\mathrm{.}$$
So sample means are centred on the population mean, and they vary less than single observations: the larger the sample, the closer $\bar{X}$ tends to be to $\mu$. The standard deviation of $\bar{X}$, $\frac{\sigma }{\sqrt{n}}$, is sometimes called the standard error.

Its distribution. If the population is normal, $X\sim \mathrm{N}(\mu ,{\sigma }^2)$, then $\bar{X}\sim \mathrm{N}(\mu ,\frac{{\sigma }^2}{n})$ exactly, for any $n$.

The Central Limit Theorem (CLT). If the population is not normal, or its distribution is unknown, then for a large sample $\bar{X}$ is still approximately $\mathrm{N}(\mu ,\frac{{\sigma }^2}{n})$. Only this informal statement is needed; a sample of about 30 or more is usually treated as large. Say "by the Central Limit Theorem" whenever you use a normal distribution for $\bar{X}$ from a population that is not known to be normal.

Masses of single eggs, $X\sim \mathrm{N}(48,6^2)$, and of the mean of 9 eggs, $\bar{X}\sim \mathrm{N}(48,2^2)$. The tall, narrow curve is the sample mean: it is centred on 48 like a single egg, but is far less spread out.

Unbiased estimates of the mean and variance Paper 6

The population mean and variance are usually unknown and must be estimated from a sample. The estimates in MF19 are
$$\bar{x}=\frac{\sum x}{n},s^2=\frac{\sum (x-\bar{x}{)}^2}{n-1}=\frac{1}{n-1}(\sum x^2-\frac{(\sum x{)}^2}{n})\mathrm{.}$$
They are unbiased: an individual estimate will usually miss the true value, but the method gives the right answer on average over all possible samples.

Why $n-1$? The spread of the sample is measured about $\bar{x}$, which is calculated from the same values and so sits in their middle. Deviations from $\bar{x}$ are therefore slightly smaller, on average, than deviations from the true $\mu$ would be. Dividing by $n-1$ instead of $n$ corrects exactly for this. Dividing by $n$ gives a biased estimate that is too small on average.

Summarised and coded data. If you are given $\sum x$ and $\sum x^2$, use the second form. If you are given $\sum (x-a)$ and $\sum (x-a{)}^2$ for some constant $a$, work with $y=x-a$: then $\bar{x}=a+\bar{y}$, and $s^2$ is the same for $x$ and $y$, because subtracting a constant does not change the spread.

Confidence intervals for a population mean Paper 6

A single estimate $\bar{x}$ gives no idea of how far from $\mu$ it might be. A confidence interval gives a range of plausible values. Because $\bar{X}\sim \mathrm{N}(\mu ,\frac{{\sigma }^2}{n})$, there is a probability 0.95 that $\bar{X}$ lies within $1.96\frac{\sigma }{\sqrt{n}}$ of $\mu$. Turning this round, the 95% confidence interval for $\mu$ is
$$\bar{x}\pm 1.96\frac{\sigma }{\sqrt{n}}\mathrm{.}$$
For other levels, replace 1.96 by the value $z$ with $\Phi (z)=\frac{1}{2}(1+\text{level})$, from the critical values in MF19: 1.645 for 90%, 1.96 for 95%, 2.326 for 98%, 2.576 for 99%. A higher level gives a wider interval; a larger sample gives a narrower one.

The formula needs one of two situations:

• the population is normal with known variance ${\sigma }^2$: then any $n$ will do;
• the sample is large: then $\bar{X}$ is approximately normal by the Central Limit Theorem, and the unknown ${\sigma }^2$ is replaced by its unbiased estimate $s^2$.

Interpreting it. $\mu$ is fixed; it is the interval that changes from sample to sample. A 95% confidence interval means: if many samples were taken and an interval worked out from each, about 95% of those intervals would contain $\mu$. So of 20 independent 95% intervals, on average one would miss $\mu$. A claimed value of $\mu$ outside the interval is doubtful; one inside it is consistent with the data.

Confidence intervals for a proportion Paper 6

If $x$ members of a large random sample of size $n$ have some property, the sample proportion is $\widehat{p}=\frac{x}{n}$. For large $n$, MF19 gives the approximate distribution of a sample proportion as $\mathrm{N}(p,\frac{p(1-p)}{n})$. The unknown $p$ in the variance is replaced by $\widehat{p}$, giving the approximate confidence interval
$$\widehat{p}\pm z\sqrt{\frac{\widehat{p}(1-\widehat{p})}{n}},$$
with the same values of $z$ as before. Give the limits as proportions (or as percentages if the question does), to 3 significant figures, and interpret them in context.