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92 Modern Analytical Chemistry

methods. In general, paired data sets are used whenever the variation being investi-
gated is smaller than other potential sources of variation.
In a study involving paired data the difference, d i , between the paired values for
–
each sample is calculated. The average difference, d, and standard deviation of the
–
differences, s d , are then calculated. The null hypothesis is that d is 0, and that there
is no difference in the results for the two data sets. The alternative hypothesis is that
–
the results for the two sets of data are significantly different, and, therefore, d is not
equal to 0.
–
The test statistic, t exp , is derived from a confidence interval around d
ts d
0 = d ±
n
where n is the number of paired samples. Replacing t with t exp and rearranging gives
dn
t exp =
s d
The value of t exp is then compared with a critical value, t(a, n), which is determined
by the chosen significance level, a, the degrees of freedom for the sample, n, and
whether the significance test is one-tailed or two-tailed. For paired data, the degrees
of freedom is n – 1. If t exp is greater than t(a, n), then the null hypothesis is rejected
and the alternative hypothesis is accepted. If t exp is less than or equal to t(a, n), then
the null hypothesis is retained, and a significant difference has not been demon-
paired t-test strated at the stated significance level. This is known as the paired t-test.
Statistical test for comparing paired data
to determine if their difference is too
4
large to be explained by indeterminate EXAMPLE .21
error.
Marecek and colleagues developed a new electrochemical method for the rapid
quantitative analysis of the antibiotic monensin in the fermentation vats used
9
during its production. The standard method for the analysis, which is based
on a test for microbiological activity, is both difficult and time-consuming. As
part of the study, samples taken at different times from a fermentation
production vat were analyzed for the concentration of monensin using both
the electrochemical and microbiological procedures. The results, in parts per
thousand (ppt),* are reported in the following table.

Sample Microbiological Electrochemical
1 129.5 132.3
2 89.6 91.0
3 76.6 73.6
4 52.2 58.2
5 110.8 104.2
6 50.4 49.9
7 72.4 82.1
8 141.4 154.1
9 75.0 73.4
10 34.1 38.1
11 60.3 60.1
Determine whether there is a significant difference between the methods at
a= 0.05.
*1 ppt is equivalent to 0.1%.