Page 29 - Statistics for Environmental Engineers

L1592_Frame_C02 Page 20 Tuesday, December 18, 2001 1:40 PM

Significance Tests
In Example 2.9 we knew that the nitrate population mean was truly 8.0 mg/L, and asked, “How likely
y
are we to get a sample mean as small as = 7.51 mg/L from the analysis of 27 specimens?” If this
result is highly unlikely, we might decide that the sample did not represent the population, probably
because the measurement process was biased to yield concentrations below the true value. Or, we might
decide that the result, although unlikely, should be accepted as occurring due to chance rather than due
to an assignable cause (like bias in the measurements).
Statistical inference involves making an assessment from experimental data about an unknown population
parameter (e.g., a mean or variance). Consider that the true mean is unknown (instead of being known as
in Example 2.9) and we ask, “If a sample mean of 7.51 mg/L is estimated from measurements on 27
specimens, what is the likelihood that the true population mean is 8.00 mg/L?” Two methods for making
such statistical inferences are to make a significance test and to examine the confidence interval of the
population parameter.
The significance test typically takes the form of a hypothesis test. The hypothesis to be tested is often
designated H o . In this case, H o is that the true value of the population mean is η = 8.0 mg/L. This is
sometimes more formally written as H o : η = 8.0. This is the null hypothesis. The alternate hypothesis
is H a : or η ≠ 8.0, which could be either η < 8.0 or η > 8.0. A significance level, α, is selected at which
the null hypothesis will be rejected. The significance level, α, represents the risk of falsely rejecting the
null hypothesis.
The relevant t statistic is:
(
statistic E statistic)
–
t = ---------------------------------------------------
(
V statistic)
where E(statistic) denotes the expected value of the statistic being estimated and V(statistic) denotes the
variance of this statistic.
A t statistic with ν degrees of freedom and significance level α is written as t ν,α .
Example 2.10

Use the nitrate data to test the hypothesis that η = 8.0 at α = 0.05. The appropriate hypotheses
are H o : η = 8.0 and H a : η < 8.0. This is a one-sided test because the alternate hypothesis involves
η on the lower side of 8.0.
The hypothesis test is made using:
(
statistic E statistic) y η– 7.51 8.0
–
–
t = --------------------------------------------------- = ------------ = ------------------------ = – 1.842
(
V statistic) s y 0.266
The null hypothesis will be rejected if the computed t is less than the value of the lower tail t
statistic having probability of α = 0.05. The value of t with α = 0.05 and ν = 26 degrees of
freedom obtained from tables is t ν=26,α=0.05 = – 1.706 . The computed value of t = –1.853 is
smaller than the table value of –1.706. The decision is to reject the null hypothesis in favor of
the alternate hypothesis.
Examples 2.9 and 2.10 are outwardly different, but mathematically and statistically equivalent. In Example
2.9, the experimenter assumes the population parameter to be known and asks whether the sample data
can be construed to represent the population. In Example 2.10, the experimenter assumes the sample data
are representative and asks whether the assumed population value is reasonably supported by the data. In
practice, the experimental context will usually suggest one approach as the more comfortable interpretation.
Example 2.10 illustrated a one-sided hypothesis test. It evaluated the hypothesis that the sample mean
was truly to one side of 8.0. This particular example was interested in the mean being below the true
value. A two-sided hypothesis test would consider the statistical plausibility of both the positive and
negative deviations from the mean.
© 2002 By CRC Press LLC