Definitions

Sample size

The sample size of a statistical sample is the number of observations that constitute it. It is typically denoted n, a positive integer (natural number).

Typically, all else being equal, a larger sample size leads to increased precision in estimates of various properties of the population. This can be seen in such statistical rules as the law of large numbers and the central limit theorem. Repeated measurements and replication of independent samples are often required in measurement and experiments to reach a desired precision.

A typical example would be when a statistician wishes to estimate the arithmetic mean of a continuous random variable (for example, the height of a person). Assuming that they have a random sample with independent observations, then if the variability of the population (as measured by the standard deviation σ) is known, then the standard error of the sample mean is given by the formula:

$sigma/sqrt\left\{n\right\}.$

It is easy to show that as n becomes large, this variability becomes very small. This yields to more sensitive hypothesis tests with greater statistical power and smaller confidence intervals.

Further examples

Central limit theorem

The central limit theorem is a significant result which depends on sample size. It states that as the size of a sample of independent observations approaches infinity, provided data come from a distribution with finite variance, that the sampling distribution of the sample mean approaches a normal distribution.

Estimating proportions

A typical statistical aim is to demonstrate with 95% certainty that the true value of a parameter is within a distance B of the estimate: B is an error range that decreases with increasing sample size (n). The value of B generated is referred to as the 95% confidence interval.

For example, a simple situation is estimating a proportion in a population. To do so, a statistician will estimate the bounds of a 95% confidence interval for an unknown proportion.

The rule of thumb for (a maximum or 'conservative') B for a proportion derives from the fact the estimator of a proportion, $hat p = X/n$, (where X is the number of 'positive' observations) has a (scaled) binomial distribution and is also a form of sample mean (from a Bernoulli distribution [0,1] which has a maximum variance of 0.25 for parameter p = 0.5). So, the sample mean X/n has maximum variance 0.25/n. For sufficiently large n (usually this means that we need to have observed at least 10 positive and 10 negative responses), this distribution will be closely approximated by a normal distribution with the same mean and variance.

Using this approximation, it can be shown that ~95% of this distribution's probability lies within 2 standard deviations of the mean. Because of this, an interval of the form

$\left(hat p -2sqrt\left\{0.25/n\right\}, hat p +2sqrt\left\{0.25/n\right\}\right)=\left(hat p -B, hat p+B\right)$

will form a 95% confidence interval for the true proportion.

If we require the sampling error ε to be no larger than some bound B, we can solve the equation

$varepsilon approx B=2sqrt\left\{0.25/n\right\}=1/sqrt\left\{n\right\}$

to give us

$1/varepsilon^2 approx 1/B^2=n$

So, n = 100 <=> B = 10%, n = 400 <=> B = 5%, n = 1000 <=> B = ~3%, and n = 10000 <=> B = 1%. One sees these numbers quoted often in news reports of opinion polls and other sample surveys.

Extension to other cases

In general, if a population mean is estimated using the sample mean from n observations from a distribution with variance σ², then if n is large enough (typically >30) the central limit theorem can be applied to obtain an approximate 95% confidence interval of the form
$\left(bar x - B,bar x + B\right), B=2sigma/sqrt\left\{n\right\}$

If the sampling error ε is required to be no larger than bound B, as above, then

$4sigma^2/varepsilon^2 approx 4sigma^2/B^2=n$

Note, if the mean is to be estimated using P parameters that must first be estimated themselves from the same sample, then to preserve sufficient "degrees of freedom," the sample size should be at least n + P.

Required sample sizes for hypothesis tests

A common problem facing statisticians is calculating the sample size required to yield a certain power for a test, given a predetermined Type I error rate α. A typical example for this is as follows:

Let X i , i = 1, 2, ..., n be independent observations taken from a normal distribution with mean μ and variance σ2 . Let us consider two hypotheses, a null hypothesis:

$H_0:mu=0$

and an alternative hypothesis:

$H_a:mu=mu^*$

for some 'smallest significant difference' μ* >0. This is the smallest value for which we care about observing a difference. Now, if we wish to (1) reject H0 with a probability of at least 1-β when Ha is true (i.e. a power of 1-β), and (2) reject H0 with probability α when Ha is true, then we need the following:

If zα is the upper α percentage point of the standard normal distribution, then

$Pr\left(bar x >z_\left\{alpha\right\}sigma/sqrt\left\{n\right\}|H_0 text\left\{ true\right\}\right)=alpha$

and so

'Reject H0 if our sample average ($bar x$) is more than $z_\left\{alpha\right\}sigma/sqrt\left\{n\right\}$

is a decision rule which satisfies (2). (Note, this is a 1-tailed test)

Now we wish for this to happen with a probability at least 1-β when Ha is true. In this case, our sample average will come from a Normal distribution with mean μ*. Therefore we require

$Pr\left(bar x >z_\left\{alpha\right\}sigma/sqrt\left\{n\right\}|H_a text\left\{ true\right\}\right)geq 1-beta$

Through careful manipulation, this can be shown to happen when

$n geq left\left(frac\left\{Phi^\left\{-1\right\}\left(1-beta\right)+z_\left\{alpha\right\}\right\}\left\{mu/sigma\right\}right\right)^2$

where $Phi$ is the normal cumulative distribution function.

Stratified sample size

With more complicated sampling techniques, such as Stratified sampling, the sample can often be split up into sub-samples. Typically, if there are k such sub-samples (from k different strata) then each of them will have a sample size ni, i = 1, 2, ..., k. These ni must conform to the rule that n1 + n2 + ... + nk = n (i.e. that the total sample size is given by the sum of the sub-sample sizes). Selecting these ni optimally can be done in various ways, using (for example) Neyman's optimal allocation.

According to Leslie Kish, there are many reasons to do this; that is to take sub-samples from distinct sub-populations or "strata" of the original population: to decrease variances of sample estimates, to use partly non-random methods, or to study strata individually. A useful, partly non-random method would be to sample individuals where easily accessible, but, where not, sample clusters to save travel costs.

In general, for H strata, a weighted sample mean is

$bar x_w = sum_\left\{h=1\right\}^H W_h bar x_h,$
with

$operatorname\left\{Var\right\}\left(bar x_w\right) = sum_\left\{h=1\right\}^H W_h^2 ,operatorname\left\{Var\right\}\left(bar x_h\right).$

The weights, W(h), frequently, but not always, represent the proportions of the population elements in the strata, and W(h)=N(h)/N. For a fixed sample size, that is n=Sum{n(h)},

$operatorname\left\{Var\right\}\left(bar x_w\right) = sum_\left\{h=1\right\}^H W_h^2 ,operatorname\left\{Var\right\}\left(h\right) left\left(frac\left\{1\right\}\left\{n_h\right\} - frac\left\{1\right\}\left\{N_h\right\}right\right),$

which can be made a minimum if the sampling rate within each stratum is made proportional to the standard deviation within each stratum: $n_h/N_h=k S_h$.

An "optimum allocation" is reached when the sampling rates within the strata are made directly proportional to the standard deviations within the strata and inversely proportional to the square roots of the costs per element within the strata:

$frac\left\{n\left(h\right)\right\}\left\{N\left(h\right)\right\} = frac\left\{K S\left(h\right)\right\}\left\{sqrt\left\{C\left(h\right)\right\}\right\},$

or, more generally, when

$n\left(h\right) = frac\left\{K\text{'} W\left(h\right) S\left(h\right)\right\}\left\{sqrt\left\{C\left(h\right)\right\}\right\}.$