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The difference in the averages of the measurements is y c – y p = 0.157 − 0.151 = 0.006 µg/L. The
2 2
variances s c and s p of the city and private samples are nearly equal, so they can be pooled by weighting
in proportion to their degrees of freedom:
(
12 0.0071) + 9 0.0076)
(
s pool = ---------------------------------------------------------- = 0.00734 µg/L( ) 2
2
12 + 9
The estimated variance of the difference between averages is:



2 
1
1
1
1
2
Vy c –( y p ) = s y c + s y p = s pool ---- + ----- = 0.00734 ------ + ------ = 0.0013 µg/L( ) 2
2



10
13
n c
n p
and the standard error of y c – y p = 0.006 µg/L is s y c −y p = 0.0013 = 0.036 µg/L.
The variance of the difference is estimated with ν = 12 + 9 = 21 degrees of freedom. The 95%
confidence interval is calculated using α/2 = 0.025 and t 21,0.025 = 2.080:
( y c – y p ) ± t 21,0.025 s y c −y p = 0.006 ± 2.080 0.036) = 0.006 ± 0.075 µg/L
(
It can be stated with 95% confidence that the true difference between the city and private water supplies
falls in the interval of −0.069 µg/L and 0.081 µg/L. This confidence interval includes zero so there is
no persuasive evidence in these data that the mercury contents are different in the two residential areas.
Future sampling can be done in either area without worrying that the water supply will affect the outcome.
Comments
The case study example showed that one could be highly confident that there is no statistical difference
between the average mercury concentrations in the two residential neighborhoods. In planning future
sampling, therefore, one might proceed as though the neighborhoods are identical, although we under-
stand that this cannot be strictly true.
Sometimes a difference is statistically significant but small enough that, in practical terms, we do not
care. It is statistically significant, but unimportant. Suppose that the mercury concentrations in the city and
private waters had been 0.15 mg/L and 0.17 mg/L (not µg/L) and that the difference of 0.02 mg/L was
statistically significant. We would be concerned about the dangerously high mercury levels in both neigh-
borhoods. The difference of 0.02 mg/L and its statistical significance would be unimportant. This reminds
us that significance in the statistical sense and important in the practical sense are two different concepts.
In this chapter the test statistic used to compare two treatments was the difference of two averages
and the comparison was made using an independent t-test. Independent, in this context, means that all
sources of uncontrollable random variation will equally affect each treatment. For example, specimens
tested on different days will reflect variation due to any daily difference in materials or procedures in
addition to the random variations that always exist in the measurement process. In contrast, Chapter 17
explains how a paired t-test will block out some possible sources of variation. Randomization is also
effective for producing independent observations.

Exercises
18.1 Biosolids. Biosolids from an industrial wastewater treatment plant were applied to 10 plots
that were randomly selected from a total of 20 test plots of farmland. Corn was grown on
the treated (T) and untreated (UT) plots, with the following yields (bushels/acre).

UT 126 122 90 135 95 180 68 99 122 113
T 144 122 135 122 77 149 122 117 131 149
Calculate a 95% confidence limit for the difference in means.
© 2002 By CRC Press LLC