Environmental Engineering Reference
In-Depth Information
EXAMPLE 10.15
The
F
distribution has two degrees of freedom, com-
monly denoted as
ν
1
and
ν
2
, and these degrees of
freedom must be specified separately in order to deter-
mine the
F
-value with a given exceedance probability,
α
. Values of
F
as a function of
ν
1
and
ν
2
for
α
= 0.05 are
given in Appendix C.4. In determining critical values of
F
, it is important to keep in mind that the order of the
degrees of freedom is important in that
F
α
(
ν
1
,
ν
2
) ≠
F
α
(
ν
2
,
ν
1
), and the first degree of freedom generally cor-
responds to the numerator of the ratio being evaluated,
and the second degree of freedom corresponds to the
denominator. The tabulated values of
F
α
(
ν
1
,
ν
2
) are
typically used to determine upper confidence limits, and
the following relationship is useful in determining the
corresponding lower confidence limits (Kottegoda and
rosso, 1997)
The natural logarithms of 61 concentration samples (in
mg/l) show a mean and standard deviation of 1.26 and
0.827, respectively. Determine the 95% confidence
interval of the population standard deviation.
Solution
From the given data:
Y
= 1 26
,
S
y
= 0.827,
N
= 61, and
α
= 0.05. The required critical
χ
2
values for
N
− 1 = 60
degrees of freedom can be read directly from the table
in Appendix C.3 as
.
χ
2
=
χ
2
=
83 298
.
α/
2
0 025
.
2
2
χ
=
χ
=
40 482
.
1
−
α
/
2
0 975
.
1
F
(
ν ν
,
)
=
)
(10.100)
using these data in Equation (10.96) yields
1
−
α
1
2
F
(
ν ν
,
α
2
1
S N
2
(
−
1
)
S N
2
(
−
1
)
This relationship can be used in determining the con-
fidence interval defined in Equation (10.99).
y
y
≤
σ
2
≤
y
2
2
χ
χ
α
/
2
1
−
α
/
2
( .
0 827 61 1
83 298
) (
2
−
)
( .
0 827 61 1
40 482
) (
2
−
)
2
≤
σ
y
≤
.
.
EXAMPLE 10.16
0 492
.
≤
σ
y
2
≤
1 014
.
Analysis of 61 water-quality samples taken from a lake
prior to the development of the surrounding watershed
show that the log-transformed concentrations have a
standard deviation of 0.683. Analysis of 41 postdevelop-
ment samples show a standard deviation of 0.752.
Assuming that the pre- and postdevelopment data
are both drawn from log-normal distributions, deter-
mine the 90% confidence interval of the ratio of prede-
velopment to postdevelopment standard deviation.
Has development of the watershed coincided with a
change in the natural water-quality fluctuations in the
lake?
0 701
.
≤
σ
y
≤
1 01
.
Therefore, the 95% confidence interval of the popula-
tion standard deviation is the range 0.701-1.01.
10.8.3 Variance Ratios
Theorem 10.3 states that if
N
and
M
samples are drawn
from two populations with variances
σ
2
and
σ
2
respec-
tively, and the sample variances are
S
2
and
S
2
, then the
quantity
F
, defined by the relation
S
S
2
/
/
σ
σ
2
1
1
F
=
(10.97)
Solution
2
2
From the given data:
N
= 61,
S
1
= 0.683,
M
= 41,
S
2
= 0.752, and
α
= 0.05. The degrees of freedom are
given by
has a
F
distribution with
N
− 1 and
M
− 1 degrees of
freedom. Defining the 1 −
α
confidence interval of
F
as
S
S
2
/
/
σ
σ
2
n N
1
=
− =
1
61 1
− =
60
1
1
(10.98)
F
≤
≤
F
1
−
α
/
2
α
/2
2
2
2
2
n M
2
=
− =
1
41 1
− =
40
leads to the following 1 −
α
confidence interval for
σ σ
referring to Appendix C.4,
F
0.05
(60,40) = 1.64,
F
0.05
(40,60) = 1.59, and hence Equation (10.100) gives
2
2
,
/
S
S
2
σ
σ
2
S
S
2
1
40 60
1
1 59
2
2
2
F
≤
≤
F
(10.99)
F
(
60 40
,
)
=
=
=
0 629
.
1
−
α
/
2
α
/
2
2
2
2
0 95
.
F
(
,
)
.
1
1
1
0 05
.
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