Environmental Engineering Reference
In-Depth Information
for
N
− 1 = 30 degrees of freedom. using the critical
t
-values in Appendix C.2 gives
t
0.975
= −2.042 and
t
0.025
= 2.042. Hence, the proposed hypothesis is accepted
at the 5% significance level if
t
∈ [−2.042, 2.042].
Since
t
= −0.467 found in this particular case falls within
the required range, the hypothesis that the population
mean is equal to 0.900 is accepted at the 5% significance
level.
46.979]. Since
χ
2
= 28.80 found in this particular case
falls within the required range, the hypothesis that the
standard deviation is equal to 0.500 is accepted at the
5% significance level.
10.9.3 Population Differences
In some cases, the objective of an analysis is to assess
whether two sets of data are drawn from the same popu-
lation or from populations with different statistics. These
tests are particularly applicable in cases where compari-
sons are to be made between water-quality data col-
lected at different times or places, or reported using
different measurement techniques. useful statistical
tests for testing the significance of population differ-
ences are given below.
10.9.2 Variance
Sampling theory has shown that if a sample of size
N
is
drawn from a normal population with variance
σ
2
, then
the random variable,
χ
2
, defined as
(
N
−
1
)
S
x
2
χ
2
=
(10.102)
2
σ
x
10.9.3.1 t-Test.
If samples of size
N
1
and
N
2
are drawn
from th
e
same
no
rmal population, and have the sample
means
X
1
and
X
2
and sample variances
S
2
and
S
2
, then
sampling theory states that the random variable
T
given
by
has a chi-square distribution with
N
− 1 degrees of
freedom. Consider the null hypothesis,
H
0
: The sample
is drawn from a normal population with variance
σ
2
.
If
χ
2
is calculated from a measured sample, and
χ
2
∈
−
[
χ
2
,
χ
2
]
/ /
, then
H
0
is accepted at the
α
signifi-
cance level, and if
χ
1
α
2
α
2
2
2
2
X X
−
∉
[
χ
,
χ
]
/ /
, then the alternate
hypothesis that the population is not normally distrib-
uted with variance
σ
2
is accepted.
1
2
1
−
α
2
α
2
T
=
1
2
(10.103)
1
1
(
N
−
1
)
S
2
+
(
N
−
1
)
S
2
1
1
2
2
+
N
N
(
N
−
1
)
+
(
N
−
1
)
1
2
1
2
EXAMPLE 10.18
has a Student's
t
distribution with
N
1
+
N
2
− 2 degrees
of freedom. Considering the null hypothesis,
H
0
:
The two samples are drawn from normal populations
having the same mean and variance, this hypothesis
would be accepted if the calculated outcome,
t
, were
such that
t
∈ [
t
1−
α
/2
,
t
α
/2
], and if
t
∉ [
t
1−
α
/2
,
t
α
/2
], the
null hypothesis would be rejected at the
α
significance
level.
Analysis of 31 log-concentration measurements show a
sample mean of 0.864 and a sample standard deviation
of 0.429. It is proposed that the population standard
deviation is equal to 0.500. Would you accept this
hypothesis at the 5% significance level?
Solution
From the given data:
N
= 31,
Y
= 0 86.
,
S
y
= 0.429, and
σ
y
= 0.500. Assuming that the sample is drawn from a
log-normal population, then the
χ
2
statistic correspond-
ing to the sample outcomes is given by Equation (10.102)
as
EXAMPLE 10.19
The concentration (in
μ
g/l) of a particular toxic sub-
stance in the groundwater downstream of a landfill are
shown to be drawn from a log-normal distribution.
Before construction of the landfill, 41 samples were
collected and the mean and standard deviation of the
log concentrations were found to be 0.563 and
0.421, respectively. After construction of the landfill,
21 samples were found have a mean and standard
deviation of the log-concentration equal to 0.623 and
0.403, respectively. Evaluate whether the postlandfill
distribution of concentrations is significantly different
from the prelandfill distribution at the 5% significance
level.
(
N
−
1
)
S
y
2
2
(
31 1 0 429
0 500
−
)( .
)
χ
2
=
=
=
28 8
.
σ
2
( .
)
2
y
If the proposed hypothesis is true at the 5% significance
level, then
χ
. .
, where the
χ
2
values are
derived for
N
− 1 = 30 degrees of freedom. using the
critical
χ
2
values in Appendix C.3 gives
χ
0 975
2
∈[
χ
2
,
χ
2
]
0 975
0 025
2
=
16 791
.
.
.=
. Hence, the proposed hypothesis is
accepted at the 5% significance level if χ
2
∈ [16.791,
and
χ
0 025
2
46 979
.
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