Environmental Engineering Reference
In-Depth Information
10.8 CONFIDENCE INTERVALS
EXAMPLE 10.14
Sampling theory is concerned with the probability dis-
tributions of sample statistics. Since measurements only
yield a single outcome of the sample statistic, calculated
statistics are regarded as only approximations to popu-
lation parameters. Consequently, in utilizing sample sta-
tistics to estimate population parameters, it is usually
desirable to also state the
confidence interval
of the
parameter estimate. Confidence intervals define the
range in which a stated percentage of possible outcomes
of a random function are expected to occur, and
confi-
dence limits
define the lower and upper bounds of the
confidence interval. Confidence intervals find their
primary application in identifying the accuracy of
sample statistics and in testing hypotheses of whether
observed outcomes are drawn from hypothetical popu-
lations. These applications are illustrated below.
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 mean.
Solution
From the given data:
Y
= 1 2.
,
S
y
= 0.827,
N
= 61, and
α
= 0.05. The required critical
t
-values for
N
− 1 = 60
degrees of freedom can be read directly from the table
in Appendix C.2 as
t
=
t
=
2 000
.
α/
2
0 025
.
t
=
t
= −
t
= −
2 000
.
1
−
α/
2
0 975
.
0 025
.
using these data in Equation (10.94) yields
S
N
S
N
y
y
10.8.1 Mean
Y t
−
≤
µ
≤
Y t
−
α
/
2
y
1
−
α
/
2
Theorem 10.1 states that if sample statistics
X
and
S
2
are obtained from
N
samples taken from a normal
distribution, then the variable
T
defined by the
equation
0 827
61
.
0 827
61
.
1 26
.
−
( .
2 000
)
≤
µ
y
≤
1 26
.
− −
(
2 000
.
)
1 05
.
≤
µ
y
≤
1 47
.
X
S
− µ
/
Therefore, the 95% confidence interval of the popula-
tion mean is the range 1.05-1.47.
x
T
=
(10.92)
N
x
has a Student's
t
distribution with
N
−
1 degrees of
freedom. Furthermore, if the confidence limits are
defined such that
α
is the probability of outcomes occur-
ring outside of the confidence interval, then for (1 −
α
)
of the outcomes, we expect that
10.8.2 Variance
Derivation of the confidence interval for the sample
variance follows the same procedure as for the sample
mean. From Theorem 10.2, it is known that the quantity
(
/σ
has a chi-square distribution with
N
− 1
degrees of freedom. Therefore, there is a 1 −
α
probabil-
ity that the outcomes are in the range
2
2
N
− 1
)
S
x
X
S
−
µ
x
x
t
≤
≤
t
(10.93)
1
−
α
/
2
α
/
2
/
N
x
where
t
α
is the value of
t
such that
P
(
T
≥
t
) =
α
.
Hence, based on any calculated value of
X
and
S
x
, it
is estimated that the 1 −
α
confidence interval for
the population mean is expressed by the following
inequality
(
N
−
1
)
S
x
2
2
2
χ
≤
≤
χ
(10.95)
1
−
α
/
2
α
/
2
σ
2
x
where
χ
2
is the value of
χ
2
with an exceedance probabil-
ity of
α
. rearranging Equation (10.95), the 1 −
α
confi-
dence interval of
σ
2
is given by
S
N
S
N
x
x
X t
−
≤
µ
≤
X t
−
(10.94)
α
/
2
x
1
−
α
/
2
S N
2
(
−
1
)
S N
2
(
−
1
)
x
x
2
≤
σ
≤
(10.96)
x
χ
2
χ
2
where, because of the symmetry in the
T
distribution,
t
α
/2
and
t
1−
α
/2
are of equal magnitude but opposite
sign. Values of
t
as a function of the exceedance
probability,
α
, and degrees of freedom,
ν
, are given in
Appendix C.2.
α
/
2
1
−
α
/
2
Values of
χ
2
as a function of the exceedance probabil-
ity,
α
, and degrees of freedom,
ν
, are given in Appendix
C.3.
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