Marecek and colleagues developed a new electrochemical method for the rapid quantitative analysis of the antibiotic monensin in the fermentation vats used during its production.^9 The standard method for the analysis, which is based on a test for microbiological activity, is both difficult and

Question

Marecek and colleagues developed a new electrochemical method for the rapid quantitative analysis of the antibiotic monensin in the fermentation vats used during its production. $^9$ 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 α = 0.05.

*1 ppt is equivalent to 0.1%.

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 α = 0.05.

*1 ppt is equivalent to 0.1%.

Accepted Answer

This is an example of a paired data set since the acquisition of samples over an extended period introduces a substantial time-dependent change in the concentration of monensin. The comparison of the two methods must be done with the paired t-test, using the following null and two-tailed alternative hypotheses

[latex] H_0: \bar{d} =0 H_A: \bar{d} ≠0[/latex]

Defining the difference between the methods as

[latex]d = X_{elect} – X_{micro}[/latex]

we can calculate the difference for each sample

sample	1	2	3	4	5	6	7	8	9	10	11
d	2.8	1.4	-3.0	6.0	-6.6	-0.5	9.7	12.7	-1.6	4.0	-0.2

The mean and standard deviation for the differences are 2.25 and 5.63, respectively. The test statistic is

[latex]t_{exp}=\frac{\left|\bar{d} \right|\sqrt{n} }{s_d} =\frac{\left|2.25\right|\sqrt{11} }{5.63} =1.33[/latex]

which is smaller than the critical value of 2.23 for t(0.05, 10). Thus, the null hypothesis is retained, and there is no evidence that the two methods yield different results at the stated significance level.

Question 4.21: Marecek and colleagues developed a new electrochemical metho...

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