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becomes useful (McBride et al. 1993; McDonald and Erickson 1994; McBride 1999). In the pharmaceutical area, a new drug is considered to be bioequivalent to a standard drug if the potency of the new drug is, say, at least 80% of the potency of the standard drug (Kirkwood 1981; Westlake 1988). In a similar way, a damaged site might be considered to be bioequivalent to a control site in terms of vegetation biomass if the mean biomass per unit area on the damaged site, μt, is at least 80% of the mean on the control site, μc. In that case, bioequivalence can be examined by testing the null hypothesis

H0: μt ≤ 0.8μc

against the alternative hypothesis

H1: μt > 0.8μc

Example 7.1:  Native Shrubs at Reclaimed and Reference Sites

As an example of how the concept of bioequivalence might be used to assess reclamation, consider the following hypothetical situation described by McDonald and Erickson (1994), noting that the analysis here is simpler than the one that they used. It is imagined that a mining company has paid a bond to a government agency to guarantee the successful reclamation of a strip mining site. Having carried out the necessary work, the company wants the bond released. However, the agency requires the company to provide evidence that the mined site is equivalent to an untouched control site with respect to the density of native shrubs.

A consultant has designed and carried out a study that involved randomly selecting eight plots from the treated site and matching them up on the basis of slope, aspect, and soil type with eight plots from the control site. The densities of native shrubs that were obtained are shown in Table 7.1. The (control − mined) site differences are also shown with their means and sample standard deviations.

A conventional approach for analyzing these results involves using a t-test to see whether the mean difference of d = 0.041 is significantly greater than zero. The null hypothesis is then that the mean density of

Table 7.1

Comparison between the Vegetation Density on Eight Paired Plots from an Undamaged Control Site and a Site Where Mining Has Occurred

Note: The difference is for the (control − mined) sites. Mean difference = 0.041; standard deviation of difference = 0.171.

This second analysis seems more realistic than the first one because the acceptance of the null hypothesis, possibly because of the small sample size, will result in the mined site being considered to need further remediation: The mined site is guilty until proved innocent, rather than innocent until proved guilty. The definition of equivalence in terms of the mined site having more than 80% of the shrub density of the control site would, of course, have been the subject of negotiations between the mining company and the government agency. Another percentage could be used equally well in the test.

7.4  Two-Sided Tests of Bioequivalence

The example just considered was quite straightforward because the test was one-sided, and the data were paired. A more complicated situation is where a previously damaged site is considered to be equivalent to an undamaged reference site, provided that the mean of a relevant variable at the first site is sufficiently close to the mean at the reference site.

Here the null hypothesis can be that the two sites are not equivalent (following the precautionary principle) or that they are equivalent. In the first case, the null hypothesis becomes that μd < μdL or μd > μdH, where the two sites are considered to be equivalent if μd, the true difference between them (damaged − reference), is within the range from μdL to μdH. In the second case the null hypothesis is that μdL ≤ μd ≤ μdH. It may be very important which of these null hypotheses is chosen, because with the first, a significant result leads to the conclusion that the two sites are equivalent, whereas with the second, a significant result leads to the conclusion that the sites are not equivalent.

The simplest way to test the null hypothesis that the two sites are not equivalent is to run the two one-sided test (TOST) developed by Schuirmann (1987) and Westlake (1988). Assuming normally distributed data, with equal variances for the potentially damaged site and the reference site, this proceeds as follows for a 5% level of significance:

1.Calculate the mean difference d between the potentially damaged site and the reference site, and the estimated standard error of this difference

SE(d) = sp√(1/n1 + 1/n2)

where n1 is the sample size for the damaged site, and n2 is the sample size for the reference site,

sp2 = [(n1 − 1)s12 + (n2 − 1)s22]/(n1 + n2 − 2)

172 Statistics for Environmental Science and Management, Second Edition

is the pooled-sample estimate of variance, s12 is the sample variance for the damaged site, and s22 is the sample variance for the reference site.

2.Use a t-test to see whether d is significantly higher than μdL at the 5% level, which involves seeing whether (d − μdL)/SE(d) is greater than or equal to the upper 5% point of the t-distribution with n1 + n2 − 2 df.

3.Use a t-test to see whether d is significantly lower than μdH at the 5% level, which involves seeing whether (d − μdH)/SE(d) is less than or equal to the lower 5% point of the t-distribution with n1 + n2 − 2 df.

4.If the tests at steps 2 and 3 are both significant, then declare that there is evidence for the equivalence of the two sites. The logic here is that if the observed difference is both significantly higher than the lowest allowed difference, and also significantly lower than the highest allowed difference, then there is certainly evidence that it is within the allowed range.

Of course, this test can be carried out using a different significance level if necessary, and it should be noted that, although it includes two t-tests, there is no need to allow for multiple testing because the probability of declaring the two sites to be equivalent when they are not is no more than α if the two t-tests are each carried out at the 100α% level (Berger and Hsu 1996).

If the null hypothesis is that the sites are equivalent (μdL ≤ μd ≤ μdH), then the two tests that are part of the TOST procedure must be modified. Part 2 of the above procedure changes to:

2.Use a t-test to see whether d is significantly lower than μdL at the 5% level, which involves seeing whether (d – μdL)/SE(d) is less than or equal to the lower 5% point of the t-distribution with n1 + n2 − 2 df.

This reveals whether there is any evidence that the true mean difference is lower than μdL. Similarly, part 3 of the procedure changes to:

3.Use a t-test to see whether d is significantly higher than μdH at the 5% level, which involves seeing whether (d − μdH)/SE(d) is greater than or equal to the upper 5% point of the t-distribution with n1 + n2 − 2 df.

Now, if either of these tests gives a significant result, then there is evidence that the two sites are not equivalent.

The test of the nonequivalence null hypothesis is more stringent than the test of the equivalence null hypothesis because evidence is required before sites are declared to be equivalent, rather than the other way round. With the nonequivalence null hypothesis, the TOST procedure carried out with a

5% level of significance can be shown to give evidence of equivalence if the sample mean difference falls within the interval

where t0.05,ν is the value that is exceeded with probability 0.05 for the t-distribution with ν = n1 + n2 − 2 df. On the other hand, with the equivalence

null hypothesis carried out with the same level of significance, there is no evidence against the null hypothesis if

The second interval may be much wider than the first one. This is demonstrated in Figure 7.1, which is for a hypothetical situation where two sites are considered to be equivalent if the mean difference is between −1 and +1.

There are procedures other than TOST for carrying out two-sided tests of bioequivalence, as reviewed by McBride (1999). Apparently the general view in the pharmaceutical literature, where most applications have been in the past, is that the TOST approach is best.

In Example 7.1, bioequivalence was expressed in terms of a ratio, with the equivalence of a damaged and a reference site being defined as the biomass per unit area of native plants in the damaged site being at least 80% of the value for the reference site. The two-sided version for this might then be that two sites are considered as equivalent providing that the ratio R = (density of native plants in an impacted area)/(density of native plants in a control area)

Figure 7.1

Bioequivalence intervals for a situation where two sites are considered to be equivalent if their true mean difference is between −1 and +1. It is assumed that a random sample of size 10 is taken from each of the two sites and gives a sample mean difference of d = −0.5 with an estimated standard error of SE(d) = 0.3. The top interval is the 95% confidence interval for the true mean difference between the sites, d ± 2.10 SE(d); the middle interval is the range of sample means that give evidence for equivalence calculated from equation (7.1); and the bottom interval is the range of sample means that give no evidence against the hypothesis of equivalence calculated from equation (7.2).

174 Statistics for Environmental Science and Management, Second Edition

should be within the range 0.8 to 1.2. McDonald and Erickson (1994) discuss procedures for use with this ratio-type of approach.

Specialized computer programs are now available to carry out bioequivalence tests. One is EquivTest/PK from Statistical Solutions (http://www. statsolusa.com), and another is Power and Sample Size Analysis (PASS) from Number Cruncher Statistical Systems (http://www.ncss.com).

Example 7.2:  PCB at the Armagh Compressor Station

For an example of a comparison between a reference site and a potentially contaminated site, some data were extracted from a much larger set described by Gore and Patil (1994). Their study involved two phases of sampling of polychlorinated biphenyl (PCB) at the site of the Armagh compressor station in Indiana County, Pennsylvania. The phase 1 sampling was in areas close to sources of PCB, while the phase 2 sampling was away from these areas. For the present purpose, a random sample of 30 observations was extracted from the phase 2 sampling results to represent a sample from a reference area, and a random sample of 20 observations was extracted from the phase 1 sample results to represent a sample from a possibly contaminated area.

The values for the PCB concentrations in parts per million (ppm) are shown in the left-hand side of Table 7.3 and plotted on the left-hand side of Figure 7.2. Clearly, the possibly contaminated sample has much more variable results than the reference sample, which complicates the comparison of the means. However, for data of this type, it is common to find that distributions are approximately lognormal (Section 3.3), suggesting that the comparison between samples is best made on the logarithms of the original results, which should be approximately normally distributed, with the variation being more similar in different samples. This turns out to be the case here, as shown by the right-hand sides of Figure 7.2 and Table 7.3.

It is in fact convenient to work with logarithms if it is desirable to define the equivalence between the two areas in terms of the ratio of their means. Thus, suppose that it is decided that the two areas are equivalent in practical terms, provided that the ratio of the mean PCB concentration in the possibly contaminated area to the mean in the reference area is between 0.5 and 1.0/0.5 = 2.0. Then this corresponds to a difference between the logarithms of means between log(0.5) = −0.301 and log(2.0) = +0.301, using logarithms to base 10. Then, for the tests of nonequivalence and equivalence described above, μdL = −0.301 and μdH = +0.301. These tests will be carried out here using the 5% level of significance.

From the logarithmic data in Table 7.3, the observed mean difference between the samples is d = 0.630, with estimated standard error SE(d) = 0.297. For the test for nonequivalence, it is first necessary to see whether d is significantly higher than −0.301, at the 5% level of significance. The t-statistic is t = (d − μdL)/SE(d) = 3.137, with 48 df. The probability of a value this large or larger is 0.001, so there is evidence that the observed mean is higher than the lowest value allowed. Next, it is necessary to test whether d is significantly lower than +0.301, at the 5% level of significance. As d exceeds 0.301, this is clearly not true. This nonsignificant