We have the following two groups of data for the concentration of the bacterium E. coli per 100 mL of water on each of 11 separate days: site 1-100, 82, 1198, 336, 5, 1071, 156, 31, 1120, 156, and 5 (average = 387.27); site 2-50, 20, 146, 738, 98, 1401, 98, 6488, 970, 98, and 50 (average = 923.36). The between-site averages for each day are therefore: 75, 51, 672, 537, 51.5, 1236, 127, 3259.5, 1045, 127, and 27.5 (average 655.32). But these averages are not representative of the central tendency of the data (as a bar graph of results clearly shows), so we could think of using the sample geometric means by taking averages ofthe logarithms of the data and then exponentiating the result. It is simplest to do this for the between-site averages.

In doing so we obtain the mean of the natural logarithms of these data as 5.4682 and so the sample geometric mean is e5.4682 = 237.03. (We get the same result if we take logarithms to base 10, giving an average value 2.3748, and then computing the sample geometric mean as 102.3748 = 237.03.) Is this the correct geometric mean?