geometric mean, because the logarithm of zero does not exist. In this instance one has to be added to all the results (+1) before their logarithmic transformation and then after the average of the logarithms is calculated and the antilog has been taken, the added value has to be subtracted. The calculation of the measures of dispersion are given in Box 8.2.

## Box 8.2 Calculation of the measures of dispersion

The logarithmic standard deviation, s_{l }is calculated by entering the log values, x, into a scientific calculator, programmed to calculate the sample standard deviation. This can also calculated manually as s_{l }= √{[Σx^{2 }- (Σx)^{2}/n]/(n - 1)}. The 95 per cent confidence intervals of the geometric mean are calculated in two stages by the method below:

•

The standard error (se) of the logarithmic mean “m”, is calculated as s

_{l}/√n.

•

The 95 per cent confidence intervals are defined as m ± t

_{(0.025) }× se, where t_{(0.025) }

Student's t for α = 0.025 and for n - 1 degrees of freedom (from statistical tables).

is the value of

•

For bathing area A the antilog of the log mean is antilog 2.7758 = 597 cfu per 100 ml (Tables

8.8

and 8.9), the standard error (se) of the logarithmic mean 2.7758 is 0.6934/√12 = 0.2001 and

the 95 per cent confidence intervals of the log mean where t_{(0.025) }is the value of Student's t for α = 0.025 and for n - 1 degrees of freedom (from statistical tables; for 11 degrees of freedom =

2.201)

is 2.7758 ± 2.201 × 0.2001, giving 2.7758 - 0.4405 = 2.3353 and 2.7758 + 0.4405 =

3.2163

respectively. The antilogs of these values are 216 and 1,646 respectively.

•

For bathing area B three counts are below the limit of detection, so the transformation count +1

had been applied to all counts before taking logarithms. The antilog of the log average 1.1933 is 15.6; and the estimated geometric mean is obtained by subtracting 1 from this, giving 14.6, rounded-off to 15. The standard error is 0.9883/√12 = 0.2853 and the 95 per cent confidence intervals of the log mean are thus 1.1933 ± 2.201 × 0.2853, giving results of 1.1933 - 0.6279 = 0.5654 and 1.1933 + 0.6279 = 1.8212. The antilogs of these are 4 and 66 respectively.

# Log-normal distribution method

The log-normal distribution method involves the ranking of results and the transformation of data into logarithms to determine the log-normal distribution that fits most closely the experimental results. This can be done by hand fitting the data directly onto log-normal probability paper together with their corresponding cumulative frequencies. The concentration of micro-organisms corresponding to certain specified percentile points on the frequency distribution cumulative frequencies (50 per cent or 90 per cent) can be deduced by the graphic representation (WHO/UNEP, 1994b). This approach is quite similar to the calculation of percentile points on a continuous distribution of an infinite number of samples.

Percentile points, “p”, on the distribution of the n data values from the mean “m”, (or log mean) and the standard deviation “s” (or log standard deviation), as p = m + zs, where z is the standard normal variable for the desired percentile, obtained from tables of the quantiles (percentage points) of the standard normal distribution. The values of z for the 80-, 90- and 95-percentage points are 0.8416, 1.2816 and 1.6449 respectively. The