Environmental Engineering Reference
In-Depth Information
Solution
N J
=
+
J
2
+
+
J
K
(10.104)
1
From the
g
iven data:
N
1
= 41,
Y
1
,
S
1
= 0.421,
= .
0 563
where the
J
k
need not be equal. The steps in applying
the KW test are as follows:
N
2
= 21,
Y
2
= .
,
S
2
= 0.403. Assuming that the
sample is drawn from a log-normal population, then the
t
statistic corresponding to the sample outcomes is given
by Equation (10.103) as
0 623
Step 1.
rank the
N
data points from the smallest to
the largest, assign rank 1 to the smallest datum. In
case of ties, assign the midrank. If nDs occur, treat
these as a group of tied values that are less than
the smallest numerical value in the data set.
Step 2.
Compute the sum of the ranks for each data
set. Denote this sum for the
k
th data set by
R
k
.
Step 3.
If there are no tied or nD values, compute
the KW statistic as follows
Y Y
−
1
2
t
=
1
2
2
2
1
1
(
N
−
1
)
S
+
(
N
−
1
)
S
1
2
+
N
N
(
N
−
1
)
+
(
N
−
1
)
1
2
1
2
0.563 0.623
−
=
1
2
2
)
2
1
41
1
21
(
40 0 421
)( .
)
+
(
20 0 403
)( .
+
(
40
)
+
(
20
)
K
−
12
R
J
2
∑
k
K
w
=
3
(
N
+
1
)
(10.105)
= −
0 539
.
N N
(
+
1
)
k
k
=
1
If the proposed hypothesis is true at the 5% significance
level, then
t
∈ [
t
0.975
,
t
0.025
], where the
t
-values are derived
for
N
1
+
N
2
− 2 = 60 degrees of freedom. using the criti-
cal
t
-values in Appendix C.2 gives
t
0.975
= −2.000 and
t
0.025
= 2.000. Hence, the proposed hypothesis is accepted
at the 5% significance level if
t
∈ [−2.000, 2.000]. Since
t
= −0.539 found in this particular case falls within the
required range, the hypothesis that the probability dis-
tributions before and after construction of the landfill
are the same is accepted at the 5% significance level.
Step 4.
If there are ties or nDs treated as ties,
compute a modified KW statistic by dividing
K
w
(Eq. 10.105) by a correction for ties, that is, compute
K
w
K
′
=
w
g
1
)
(10.106)
∑
1
−
t
(
t
2
−
1
k
k
N N
(
2
−
1
)
k
=
1
where
g
is the number of tied groups, and
t
k
is the
number of tied data in the
k
th group. Equation
(10.106) reduces to Equation (10.105) when there
are no ties.
Step 5.
For an
α
level test, reject
H
0
if
K
Equation (10.103) is derived using Equation (10.101)
based on the assumption that the means and variances
of the two data sets are the same. The significance of the
difference in the means could be evaluated separately
using Equation (10.94), and the significance of the dif-
ference in the variances could be evaluated using Equa-
tion (10.96). The
F
-test for variance ratio could also be
used to assess the difference in variance.
2
,
where
χ
α,
K
−
2
is the
α
significance level of the chi-
square distribution with
K
− 1 degrees of freedom.
w′
≥
χ
α,
K
−
1
EXAMPLE 10.20
Two sets of concentration data are collected from a
lake at during different periods of watershed develop-
ment. Each set of data has 25 samples, and the measured
concentrations in mg/l from each sample set are as
follows:
10.9.3.2 Kruskal-Wallis Test.
The Kruskal-Wallis
(KW) test is used to assess the differences between the
means of
K
independent data sets. These data sets need
not be drawn from underlying distributions that are
normal or even symmetric, but the
K
distributions are
assumed to be identical in shape (Gilbert, 1987). A mod-
erate number of tied and nondetect (nD) values can be
accommodated. The null hypothesis is given by
H
0
: The
populations from which the
K
data sets have been drawn
have the same mean. The alternative hypothesis is that
at least one population has a mean larger or smaller that
at least one other population. The data points used in
the KW test can be expressed as
x
k
,
j
, where
k
is the data
set,
k
∈ [1,
K
],
j
is the data point,
j
∈ [1,
J
k
], and
J
k
is the
number of data points within the
k
th data set. The total
number of data,
N
, can be expressed as
Set1 Set2 Set1 Set2 Set1 Set2 Set1 Set2 Set1 Set2
2.44 1.42 15.64 6.18 10.44 2.66 8.33 17.62 2.68 2.90
3.51 7.80
2.53 0.89 18.79 3.35 7.27
3.52 6.67 3.62
2.20 3.75
5.00 8.55
0.78 1.53 1.79
5.45 3.54 1.24
2.12 6.65
1.09 6.89
3.25 4.62 2.19
1.22 7.66 4.87
1.02 6.82
1.74 3.67
0.71 6.34 1.97
1.70 4.46 2.43
use the KW test to determine whether the popula-
tion means from which the data sets are drawn are
significantly different at the 5% level.
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