Question

Two different techniques for quantifying species diversity within a community are being compared in order to determine which technique yields the most consistent values. Higher consistency is indicated by lower variation among replicate estimates. Each technique involves employing a particular sampling protocol in the field and then using the results to calculate an index of diversity. Each sampling protocol is repeated six times on the same community, yielding six estimates of the species diversity of that community. The two indexes measure diversity on different scales (their means are different), so it is necessary to compare their coefficients of variation rather than their variances. If the two techniques yield the following results, which appears to be more consistent (less variable)?

Technique 1: 8.6, 9.7, 7.6, 8.5, 7.7, 9.9

Technique 2: 19.5, 18.0, 16.3, 17.6, 16.6, 19.3

Answer 1

Coefficient of variation = standard deviation / mean *100

$\large = \frac{s}{\bar{x}}*100$

Where s = sample standard deviation

1

N= Total number of observations = 6

$\large \bar{x} = \frac{ \sum x_i}{N}$ for all i

Thus, for technique 1.

i	$\large x_i$	$\large x_i^2$
1	8.6	73.96
2	9.7	94.09
3	7.6	57.76
4	8.5	72.25
5	7.7	59.29
6	9.9	98.01
Total	52	455.36

$\large s^2 = \frac{455.36- \frac{52^2} {6}}{6-1}$ = 0.939

So, s = 0.969

$\large \bar{x} = \frac{52}{6} = 8.67$

So, coefficient of variation = $\large \frac{0.969}{8.667} *100$ =11.2%

For technique 2:

i	$\large x_i$	$\large x_i^2$
1	19.5	380.25
2	18	324
3	16.3	265.69
4	17.6	309.76
5	16.6	275.56
6	19.3	372.49
Total	107.3	1927.75

$\large s^2 = \frac{1927.75- \frac{107.3^2} {6}}{6-1}$ = 1.774

So, s = 1.332

$\large \bar{x} = \frac{107.3}{6} = 17.88$

So, coefficient of variation $\large = \frac{1.332}{17.88}*100$ = 7.45%

Since, coefficient of variation of second technique is less than the coefficient of variation of first technique, the second technique is more consistent.