Page 274 - Statistics for Environmental Engineers

L1592_Frame_C30 Page 277 Tuesday, December 18, 2001 2:49 PM

Coefficients b 0 and b 1 were expected to be significant. There is nothing additional to note about them.
The interactions are not significant:

b 12 = 0.0192 ± 0.0224 b 13 = −0.0109 ± 0.0224
b 23 = 0.0004 ± 0.0224 b 123 = −0.0115 ± 0.0224

Method A gives results that are from 0.025 to 0.060 mg/L lower than Method B (on the log-transformed
scale). This is indicated by the coefficient b 2 = 0.0478 ± 0.0224. The difference between A and B on
1
the log scale is a percentage on the original measurement scale. Method A gives results that are 2.5 to
6% lower than Method B.
If a 5% difference between methods is not important in the context of a particular investigation, and
if Method B offers substantial advantages in terms of cost, speed, convenience, simplicity, etc., one
might decide to adopt Method B although it is not truly equivalent to Method A. This highlights the
difference between “statistically significant” and “practically important.” The statistical problem was
important to learn whether A and B were different and, if so, by how much and in which direction. The
practical problem was to decide whether a real difference could be tolerated in the application at hand.
Using PMA as a preservative caused no measurable effect or interference. This is indicated by the
confidence interval [−0.004, 0.041] for b 3 , which includes zero. This does not mean that wastewater
specimens could be held without preservation. It was already known that preservation was needed, but
it was not known how PMA would affect Method B. This important result meant that the analyst could
do nitrate measurements twice a week instead of daily and holding wastewater over the weekend was
possible. This led to economies of scale in processing.
This chapter began by saying that Method A, the widely accepted method, was considered to give
accurate measurements. It is often assumed that widely used methods are accurate, but that is not
necessarily true. For many analyses, no method is known a priori to be correct. In this case, finding that
Methods A and B are equivalent would not prove that either or both give correct results. Likewise,
finding them different would not mean necessarily that one is correct. Both might be wrong.
At the time of this study, all nitrate measurement methods were considered tentative (i.e., not yet
proven accurate). Therefore, Method A actually was not known to be correct. A 5% difference between
Methods A and B was of no practical importance in the application of interest. Method B was adopted
because it was sufficiently accurate and it was simpler, faster, and cheaper.

Comments
The arithmetic of fitting a regression model to a factorial design and estimating effects in the standard
way is virtually identical. The main effect indicates the change in response that results from moving
from the low level to the high level (i.e., from −1 to +1). The coefficients in the regression model indicate
the change in response associated with moving only one unit (e.g., from 0 to +1). Therefore, the regression
coefficients are exactly half as large as the effects.
Obviously, the decision of analyzing the data by regression or by calculating effects is largely a matter
of convenience or personal preference. Calculating the effects is more intuitive and, for many persons,
easier, but it is not really different or better.
There are several common situations where linear regression must be used to analyze data from a
factorial experiment. The factorial design may not have been executed precisely as planned. Perhaps
one run has failed so there is a missing data point. Or perhaps not all runs were replicated, or the number
of replicates is different for different runs. This makes the experiment unbalanced, and matrix multipli-
cation and inversion cannot be done by inspection as in the case of a balanced two-level design.

1
Suppose that Method B measures 3.0 mg/L, which is 1.0986 on the log scale, and Method A measures 0.0477 less on the
log scale, so it would give 1.0986 − 0.0477 = 1.0509. Transform this to the original metric by taking the antilog; exp(1.0509) =
2.86 mg/L. The difference 3.0 − 2.86 = 0.14, expressed as a percentage is 100(0.14/3.0) = 4.77%. This is the same as the effect
of method (0.0477) on the log-scale that was computed in the analysis.
© 2002 By CRC Press LLC