Environmental Engineering Reference
In-Depth Information
EXAMPLE 10.8
M
∑
(
X
Np
Np
−
)
2
m
m
χ
2
=
It is proposed that the water-quality samples given in
Example 10.7 are drawn from a log-normal distribution
with a natural-log mean of 1.20 and a standard deviation
of 0.80. using the following categories of data, evaluate
whether the hypothesis of log-normality is supported at
the 5% significance level.
m
m
=
1
[
6 30 0 160
30 0 160
−
( .
)]
2
[
6 30 0 201
−
( .
)]
2
[
8 30 0 165
30 0 165
−
( .
)]
2
=
+
+
( .
)
30 0 201
( .
)
( .
)
2
2
[
5 30 0 170
30 0 170
−
( .
)]
[
5 30 0 304
30 0 304
−
( .
)]
+
+
( .
)
( .
)
=
4 042
.
Category
range (mg/l)
The confidence level corresponding to a chi-square
value of 4.042 with 4 degrees of freedom is 0.4004. Since
this confidence level is much greater than the 0.05 (i.e.,
5%), then the hypothesis that the data is drawn from a
log-normal distribution is accepted.
1
[0,1.5]
2
(1.5,2.5]
3
(2.5,3.5]
4
(3.5,5.0]
5
(5.0,∞]
10.5.2.2 Kolmogorov-Smirnov Test.
This test differs
from the chi-square test in that none of the parameters
from the theoretical probability distribution need to be
estimated from the observed data. In this sense, the KS
test is classified as a
nonparametric
test. The procedure
for implementing the KS test is as follows (Haan, 1977):
Solution
From the given data:
μ
y
= 1.2,
σ
y
= 0.8, and
N
= 30. The
analysis of the given data is summarized in the following
table:
Step 1.
let
P
X
(
x
) be the specified theoretical CDF
under the null hypothesis.
Step 2.
let
S
N
(
x
) be the sample CDF based on
N
observations. For any observed
x
,
S
N
(
x
) =
k
/
N
,
where
k
is the number of observations less than or
equal to
x
.
Step 3.
Determine the maximum deviation,
D
,
defined by
c
m
(mg/l)
X
(
m
)
z
F
(
z
)
p
m
1
[0,1.5]
6
[−∞,−0.993]
[0,0.160]
0.160
2
(1.5,2.5]
6
[−0.993,−0.355]
[0.160,0.361]
0.201
3
(2.5,3.5]
8
[−0.355,0.066]
[0.361,0.526]
0.165
4
(3.5,5.0]
5
[0.066,0.512]
[0.526,0.696]
0.170
5
5
[0.696,1.000]
0.304
(5.0,∞]
[0.512,−∞]
D
=
max
P x
( )
−
S x
( )
(10.55)
X
N
where
m
= category,
c
= measured concentrations,
X
(
m
) = number of observations in category
m
,
z
=
standard normal deviate corresponding to
c
,
F
(
z
) =
CDF of
z
, and
p
m
= probability of
c
occurring in cate-
gory
m
. The values of
z
and
p
m
are calculated using the
relations
Step 4.
If, for the chosen significance level, the
observed value of
D
is greater than or equal to the
critical value of the KS statistic tabulated in
Appendix C.5, the hypothesis is rejected.
An advantage of the KS test over the chi-square test
is that it is not necessary to divide the data into intervals;
thus, any error of judgment associated with the number
or size of the intervals is avoided (Haldar and Mahade-
van, 2000).
A minimum sample size of 50 is usually recom-
mended when using this test (McBean and rovers,
1998). The KS test is generally more efficient than the
chi-square test when the sample size is small (McCuen,
2002a). However, neither of these tests is very powerful
in the sense that the probability of accepting a false
hypothesis is quite high, especially for small samples
(Haan, 1977). Furthermore, probability distributions
that demonstrate a good fit with observed data might
ln
c
−
µ
ln
c
1 2
0 8
−
.
y
z
=
=
σ
.
y
m
=
(
)
−
(
)
p
F z
F z
U
L
where
z
u
and
z
l
are the upper and lower values
of
z
that bound the given category. Since there are
five categories and the parameters of the distribution
were not estimated from the data, then the number
of degrees of freedom is 5 − 1 = 4. The chi-square
statistic given by Equation (10.54) is calculated as
follows
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