A common problem faced by statisticians is calculating the sample size required to yield a certain for a test, given a predetermined rate α. As follows, this can be estimated by tables for certain values, by Mead's resource equation, or, more generally, by the : Tables 0.2 0.5 0.8 0.25 84 14 6 0.50 193 32 13 0.60 246 40 16 0.70 310 50 20 0.80 393 64 26 0.90 526 85 34 0.95 651 105 42 0.99 920 148 58 The table shown at right can be used in a to estimate the sample sizes of an and a 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 is 0.05. The parameters used are: The desired of the trial, shown in column to the right. (=effect size), which is the expected difference between the of the target values between the experimental group and the , divided by the expected . Mead's resource equation Mead's resource equation is often used for estimating sample sizes of , 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. All the parameters in the equation are in fact the of the number of their concepts, and hence, their numbers are subtracted by 1 before insertion into the equation. The equation is: 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 (including ) 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 (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. Cumulative distribution function Let Xi, i = 1, 2, ..., n be independent observations taken from a with unknown mean μ and known variance σ2. Let us consider two hypotheses, a : and an alternative hypothesis: 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 when Ha is true (i.e. a of 1-β), and (2) reject H0 with probability α when H0 is true, then we need the following: If zα is the upper α percentage point of the standard normal distribution, then and so 'Reject H0 if our sample average () is more than ' is a 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 Through careful manipulation, this can be shown to happen when where is the normal . Stratified sample size With more complicated sampling techniques, such as , the sample can often be split up into sub-samples. Typically, if there are H such sub-samples (from H different strata) then each of them will have a sample size nh, h = 1, 2, ..., H. These nh must conform to the rule that n1 + β