To determine the number of replicates needed to detect a \u2018true\u2019 difference between sample means use the following formula (Sokal and Rohlf, 1981 Biometry: The Principles and Practice of Statistics in Biological Research. W.H. Freeman & Co., New York. p 262, Box 9.13):<\/p>\n\n\n\n
N = 2(s\/d)2<\/sup> * {t\u03b1<\/sub> [v] + t2(1-P)<\/sub>[v]}2<\/sup><\/p>\n\n\n\n where<\/p>\n\n\n\n To determine how many animals you will need (Nstable<\/sub>), you need some idea of what the variability is for the parameters you will be measuring. Nstable<\/sub> is obtained through an interative method. To calculate Nstable<\/sub>, you need the df (which is a function of N). Use an estimate of N (Nintitial<\/sub>) to obtain an estimate of df (dfinitial<\/sub> ) which is then used to calculate N2<\/sub>. This new estimate of N (N2<\/sub>) is then used to calculate a new df, (df2<\/sub>), which is used, in turn, for the calculation of a new estimate of N (N3<\/sub>). This method is repeated until a \u2018stable\u2019 N (Nstable<\/sub>) is found.<\/p>\n\n\n\n In the following example, we want to measure how chemicals affect CYP1A enzyme activity in fish. We look up the data we have (or the data of others) on the variability of this enzyme in fish:<\/p>\n\n\n\n Control fish: 269 \u00b1 49 pmol product\/min\/mg protein, n = 3 individual fish<\/p>\n\n\n\n Treated fish: 1,453 \u00b1 139 pmol product\/min\/mg protein, n =3 individual fish<\/p>\n\n\n\n If unknown, the term s\/d can be replaced by CV\/D were CV is the Coefficient of Variation (in %) and D is d in %. The coefficient of variation for these activities varied between ~ 10 \u2013 18% (that is, 49\/269 * 100 = 18.2%; 139\/1453*100 = 9.6%); in other studies we conducted, the CV for this enzyme varied between 34 \u2013 55%. We would like to detect at least a 50% difference between means. Using an average coefficient variation of 30%, an \u03b1 level = 0.05, a desired probability of P = 0.8 and 32 degrees of freedom1<\/sup>:<\/p>\n\n\n\n N2<\/sub> = 2 (30\/50)2<\/sup> * {t0.05<\/sub>[32] + t2(1-0.8)<\/sub>[32]}2<\/sup><\/p>\n\n\n\n = 0.72 * {2.037 + 0.853}2<\/sup><\/p>\n\n\n\n = 6<\/p>\n\n\n\n Note: Since the enzyme activity could go either up or down with treatment, look up the \u2018t\u2019 values in a Two-tailed Student\u2019s t Table. The value of \u2018t\u2019 for an \u03b1 of 0.05 and 32 df (t0.05<\/sub>[32]) = 2.037; the value of \u2018 t\u2032 for a P of 0.8 and 32 df (t2(1-0.8)<\/sub>[32]) = (t0.4<\/sub>[32]) = 0.853.<\/p>\n\n\n\n The second \u2018round\u2019 of calculations (using N2 <\/sub>= 6) leads to a df2<\/sub> of 40 and a new N3<\/sub> of 6, it is then stable and Nstable <\/sub>= 6. So, 6 fish per group is the number required to detect a significant difference of at least 50 % between the treatments (at a 0.05 level \u03b1), with a likelihood of detecting this difference 80% of the time, if this difference truly exists (this is the power of the test, P).<\/p>\n\n\n\n\n
Example<\/h3>\n\n\n\n