Sample size determination

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, though the results will become less accurate if there is a systematic error in the experiment. The result of increased accuracy through having a larger sample size is predicted 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 quantitative random variable (for example, the height of a person). Assuming that they have a random sample with independent observations, and also that 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 {n}}.

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

Implications of sample size

Central limit theorem

The central limit theorem 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). Typically B is generated in such a way that the range of values that are within a distance B of the estimated parameter value will be a 95% confidence interval, at least in an approximate sense.

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.^[1]

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

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

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=2{\sqrt {0.25/n}}=1/{\sqrt {n}}

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.

More exactly, it can be shown that the confidence interval (+/- the margin of error) is:

At 99% confidence: $({\hat {p}}-1.29/{\sqrt {n}},~~{\hat {p}}+1.29/{\sqrt {n}})$

At 95% confidence: $({\hat {p}}-0.98/{\sqrt {n}},~~{\hat {p}}+0.98/{\sqrt {n}})$ (Here 0.98 is often estimated as 1, as above, for convenience.)

At 90% confidence: $({\hat {p}}-0.82/{\sqrt {n}},~~{\hat {p}}+0.82/{\sqrt {n}})$

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

({\bar {x}}-B,{\bar {x}}+B),B=2\sigma /{\sqrt {n}}

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

4\sigma ^{2}/\varepsilon ^{2}\approx 4\sigma ^{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 unknown mean μ and known 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 H_a is true (i.e. a power of 1-β), and (2) reject H₀ with probability α when H₀ is true, then we need the following:

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

\Pr({\bar {x}}>z_{\alpha }\sigma /{\sqrt {n}}|H_{0}{\text{ true}})=\alpha

and so

'Reject H₀ if our sample average (

{\bar {x}}

) is more than

z_{\alpha }\sigma /{\sqrt {n}}

'

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 H_a is true. In this case, our sample average will come from a Normal distribution with mean μ^*. Therefore we require

\Pr({\bar {x}}>z_{\alpha }\sigma /{\sqrt {n}}|H_{a}{\text{ true}})\geq 1-\beta

Through careful manipulation, this can be shown to happen when

n\geq \left({\frac {\Phi ^{-1}(1-\beta )+z_{\alpha }}{\mu ^{*}/\sigma }}\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 n_i, i = 1, 2, ..., k. These n_i must conform to the rule that n₁ + n₂ + ... + n_k = n (i.e. that the total sample size is given by the sum of the sub-sample sizes). Selecting these n_i optimally can be done in various ways, using (for example) Neyman's optimal allocation.

According to Leslie Kish,^[2] 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 _{h=1}^{H}W_{h}{\bar {x}}_{h},

with

\operatorname {Var} ({\bar {x}}_{w})=\sum _{h=1}^{H}W_{h}^{2}\,\operatorname {Var} ({\bar {x}}_{h}).

^[3]

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 {Var} ({\bar {x}}_{w})=\sum _{h=1}^{H}W_{h}^{2}\,\operatorname {Var} (h)\left({\frac {1}{n_{h}}}-{\frac {1}{N_{h}}}\right),

^[4]

which can be made a minimum if the sampling rate within each stratum is made proportional to the standard deviation within each stratum: $n_{h}/N_{h}=kS_{h}$ .

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 {n(h)}{N(h)}}={\frac {KS(h)}{\sqrt {C(h)}}},

^[5]

or, more generally, when

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

^[6]

Notes

^ NIST/SEMATECH, "7.2.4.2. Sample sizes required", e-Handbook of Statistical Methods.
^ Kish (1965) ^{[page needed]}
^ Kish (1965), p.78.
^ Kish (1965), p.81.
^ Kish (1965), p.93.
^ Kish (1965), p.94.

References

Bartlett, J. E., II, Kotrlik, J. W., & Higgins, C. (2001). "Organizational research: Determining appropriate sample size for survey research", Information Technology, Learning, and Performance Journal, 19(1) 43-50.
Kish, L. (1965), Survey Sampling, Wiley. ISBN 047148900x