Sample size determination

A relatively simple situation is estimation of a proportion. For example, we may wish to estimate the proportion of residents in a community who are at least 65 years old.

The estimator of a proportion is ${\displaystyle {\hat {p}}=X/n}$, where X is the number of 'positive' observations (e.g. the number of people out of the n sampled people who are at least 65 years old). When the observations are independent, this estimator has a (scaled) binomial distribution (and is also the sample mean of data from a Bernoulli distribution). The maximum variance of this distribution is 0.25n, which occurs when the true parameter is p = 0.5. In practice, since p is unknown, the maximum variance is often used for sample size assessments. If a reasonable estimate for p is known the quantity ${\displaystyle p(1-p)}$ may be used in place of 0.25.

For sufficiently large n, the distribution of ${\displaystyle {\hat {p}}}$ will be closely approximated by a normal distribution.[1] Using this and the Wald method for the binomial distribution, yields a confidence interval of the form

${\displaystyle \left({\widehat {p}}-Z{\sqrt {\frac {0.25}{n}}},\quad {\widehat {p}}+Z{\sqrt {\frac {0.25}{n}}}\right)}$ ,

where Z is a standard Z-score for the desired level of confidence (1.96 for a 95% confidence interval).

If we wish to have a confidence interval that is W units total in width (W/2 on each side of the sample mean), we would solve

${\displaystyle Z{\sqrt {\frac {0.25}{n}}}=W/2}$

for n, yielding the sample size

${\displaystyle n={\frac {Z^{2}}{W^{2}}}}$ , in the case of using .5 as the most conservative estimate of the proportion. (Note: W/2 = margin of error.)

Otherwise, the formula would be ${\displaystyle Z{\sqrt {\frac {p(1-p)}{n}}}=W/2}$ , which yields ${\displaystyle n={\frac {4Z^{2}p(1-p)}{W^{2}}}}$.

For example, if we are interested in estimating the proportion of the US population who supports a particular presidential candidate, and we want the width of 95% confidence interval to be at most 2 percentage points (0.02), then we would need a sample size of (1.96²)/(0.02²) = 9604. It is reasonable to use the 0.5 estimate for p in this case because the presidential races are often close to 50/50, and it is also prudent to use a conservative estimate. The margin of error in this case is 1 percentage point (half of 0.02).

The foregoing is commonly simplified...

${\displaystyle \left({\widehat {p}}-1.96{\sqrt {\frac {0.25}{n}}},{\widehat {p}}+1.96{\sqrt {\frac {0.25}{n}}}\right)}$

will form a 95% confidence interval for the true proportion. If this interval needs to be no more than W units wide, the equation

${\displaystyle 4{\sqrt {\frac {0.25}{n}}}=W}$

can be solved for n, yielding[2][3] n = 4/W² = 1/B² where B is the error bound on the estimate, i.e., the estimate is usually given as within ± B. So, for B = 10% one requires n = 100, for B = 5% one needs n = 400, for B = 3% the requirement approximates to n = 1000, while for B = 1% a sample size of n = 10000 is required. These numbers are quoted often in news reports of opinion polls and other sample surveys. However, always remember that the results reported may not be the exact value as numbers are preferably rounded up. Knowing that the value of the n is the minimum number of sample points needed to acquire the desired result, the number of respondents then must lie on or above the minimum.

A proportion is a special case of a mean. When estimating the population mean using an independent and identically distributed (iid) sample of size n, where each data value has variance σ², the standard error of the sample mean is:

${\displaystyle {\frac {\sigma }{\sqrt {n}}}.}$

This expression describes quantitatively how the estimate becomes more precise as the sample size increases. Using the central limit theorem to justify approximating the sample mean with a normal distribution yields a confidence interval of the form

${\displaystyle \left({\bar {x}}-{\frac {Z\sigma }{\sqrt {n}}},\quad {\bar {x}}+{\frac {Z\sigma }{\sqrt {n}}}\right)}$ ,

where Z is a standard Z-score for the desired level of confidence (1.96 for a 95% confidence interval).

If we wish to have a confidence interval that is W units total in width (W/2 on each side of the sample mean), we would solve

${\displaystyle {\frac {Z\sigma }{\sqrt {n}}}=W/2}$

for n, yielding the sample size

${\displaystyle n={\frac {4Z^{2}\sigma ^{2}}{W^{2}}}}$. (Note: W/2 = margin of error.)

For example, if we are interested in estimating the amount by which a drug lowers a subject's blood pressure with a 95% confidence interval that is six units wide, and we know that the standard deviation of blood pressure in the population is 15, then the required sample size is ${\displaystyle {\frac {4\times 1.96^{2}\times 15^{2}}{6^{2}}}=96.04}$, which would be rounded up to 97, because the obtained value is the minimum sample size, and sample sizes must be integers and must lie on or above the calculated minimum.

The table shown on the right can be used in a two-sample t-test to estimate the sample sizes of an experimental group and a control group that are of equal size, that is, the total number of individuals in the trial is twice that of the number given, and the desired significance level is 0.05.[4] The parameters used are:

• The desired statistical power of the trial, shown in column to the left.
• Cohen's d (= effect size), which is the expected difference between the means of the target values between the experimental group and the control group, divided by the expected standard deviation.

Mead's resource equation is often used for estimating sample sizes of laboratory animals, as well as in many other laboratory experiments. It may not be as accurate as using other methods in estimating sample size, but gives a hint of what is the appropriate sample size where parameters such as expected standard deviations or expected differences in values between groups are unknown or very hard to estimate.[5]

All the parameters in the equation are in fact the degrees of freedom of the number of their concepts, and hence, their numbers are subtracted by 1 before insertion into the equation.

The equation is:[5]

${\displaystyle E=N-B-T,}$

where:

• N is the total number of individuals or units in the study (minus 1)
• B is the blocking component, representing environmental effects allowed for in the design (minus 1)
• T is the treatment component, corresponding to the number of treatment groups (including control group) being used, or the number of questions being asked (minus 1)
• E is the degrees of freedom of the error component, and should be somewhere between 10 and 20.

For example, if a study using laboratory animals is planned with four treatment groups (T=3), with eight animals per group, making 32 animals total (N=31), without any further stratification (B=0), then E would equal 28, which is above the cutoff of 20, indicating that sample size may be a bit too large, and six animals per group might be more appropriate.[6]

Let X_i, i = 1, 2, ..., n be independent observations taken from a normal distribution with unknown mean μ and known variance σ². Consider two hypotheses, a null hypothesis:

${\displaystyle H_{0}:\mu =0}$

and an alternative hypothesis:

${\displaystyle H_{a}:\mu =\mu ^{*}}$

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

${\displaystyle \Pr({\bar {x}}>z_{\alpha }\sigma /{\sqrt {n}}\mid H_{0})=\alpha }$

and so

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

is a decision rule which satisfies (2). (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

${\displaystyle \Pr({\bar {x}}>z_{\alpha }\sigma /{\sqrt {n}}\mid H_{a})\geq 1-\beta }$

Through careful manipulation, this can be shown (see Statistical power#Example) to happen when

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

where ${\displaystyle \Phi }$ is the normal cumulative distribution function.