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18

Independent t-Test for Assessing the Difference

of Two Averages

KEY WORDS confidence interval, independent t-test, mercury.
Two methods, treatments, or conditions are to be compared. Chapter 17 dealt with the experimental design
that produces measurements from two treatments that were paired. Sometimes it is not possible to pair the
tests, and then the averages of the two treatments must be compared using the independent t-test.

Case Study: Mercury in Domestic Wastewater

Extremely low limits now exist for mercury in wastewater effluent limits. It is often thought that whenever
the concentration of heavy metals is too high, the problem can be corrected by forcing industries to stop
discharging the offending substance. It is possible, however, for target effluent concentrations to be so
low that they might be exceeded by the concentration in domestic sewage. Specimens of drinking water
were collected from two residential neighborhoods, one served by the city water supply and the other
served by private wells. The observed mercury concentrations are listed in Table 18.1. For future studies
on mercury concentrations in residential areas, it would be convenient to be able to sample in either
neighborhood without having to worry about the water supply affecting the outcome. Is there any
difference in the mercury content of the two residential areas?
The sample collection cannot be paired. Even if water specimens were collected on the same day,
there will be differences in storage time, distribution time, water use patterns, and other factors. Therefore,
the data analysis will be done using the independent t-test.

t-Test to Compare the Averages of Two Samples
Two independently distributed random variables y 1 and y 2 have, respectively, mean values η 1 and η 2 and
2 2
variances σ 1 and σ 2 . The usual statement of the problem is in terms of testing the null hypothesis that
the difference in the means is zero: η 1 − η 2 = 0, but we prefer viewing the problem in terms of the
confidence interval of the difference.
The expected value of the difference between the averages of the two treatments is:
(
Ey 1 – y 2 ) = η 1 η 2
–
are:
If the data are from random samples, the variances of the averages y 1 and y 2
()
()
Vy 1 = σ 1 /n 1 and Vy 2 = σ 2 /n 2
2
2
where n 1 and n 2 are the sample sizes. The variance of the difference is:
2 2
(
σ 1
------
Vy 1 – y 2 ) = ------ + σ 2
n 1 n 2
© 2002 By CRC Press LLC