ANALYSIS OF WATER-QUALITY MEASUREMENTS - Water Quality Engineering in Natural Systems - page 364

EXAMPLE 10.21

=1 5 01. . The
sample standard deviation is calculated as S x = 2.067,
and application of Equation (10.107) yields

Based on these calculations b = ∑

k

b =

i i

Deviations of observed concentrations (in mg/l) from
those predicted by a numerical water-quality model are
as follows:

2

2

b
S N
x

5 012
2 067 20 1

.

=

W = = 0 309 .

− 1 . −

−0.393 1.807 0.971 2.558

0.467 1.476 2.598 0.924

At the 5% significance level, Table 10.6 gives the critical
values of the W statistic as 0.905. Since the calculated
statistic (= 0.309) is less than the critical statistic (= 0.905),
it can be asserted at the 5% significance level that the
data are not drawn from a normal distribution.

1.404 1.007 1.069 1.893

2.697 4.175 0.751 −0.434

3.276 2.048 2.642 1.164

Assess the normality of these random fluctuations at
the 5% significance level using the Shapiro-Wilk test.
For a sample size of 20, the Shapiro-Wilk constants are
(Shapiro and Wilk, 1965):

10.9.4.2 Shapiro-Francia Test. The Shapiro-Francia
test (Shapiro and Francia, 1972) is a slight modification
of the Shapiro-Wilk test that is recommended when the
sample size, N , is greater than 50. This test has the same
advantages as the Shapiro-Wilk test. The steps to be
followed in applying the Shapiro-Francia test are as
follows:

n a n n a n

20 0.4734 15 0.1334

19 0.3211 14 0.1013

18 0.2565 13 0.0711

Step 1. rank order the sample data.
Step 2. Compute the weighted sum of the observa-
tions using the relation

17 0.2085 12 0.0422

16 0.1686 11 0.0140

Solution

N

∑ m x

weighted sum =

(10.109)

i i

From the given data: N = 20 and k = 10. The first step is
to rank order the sample data, which yields:

i = 1

where the values of m i can be approximately com-
puted as

rank Data rank Data rank Data rank Data

i

1 −0.434 6 0.971 11 1.476 16 2.598

m

i =

Φ 1

(10.110)

2 −0.393 7 1.007 12 1.807 17 2.642

N + 1

3 0.467 8 1.069 13 1.893 18 2.697

where Φ −1 is the inverse of the standard normal
cumulative distribution function.
Step 3. Divide the square of the weighted sum by a
multiple of the sample standard deviation to
obtain the Shapiro-Francia statistic, W ′, defined as

4 0.751 9 1.164 14 2.048 19 3.276

5 0.924 10 1.404 15 2.558 20 4.175

Application of Equation (10.108) to calculate b i
from the ranked data are summarized in the following
table:

( )

2

∑

N

m x

i i

i N − i + 1 x N − i +1 − x i a N − i +1 b i

i = 1

(10.111)

W

′ =

∑

N

2 2

1 20 4.609 0.4734 2.182

( N − 1 ) S m

x i

i = 1

2 19 3.668 0.3211 1.178

3 18 2.230 0.2565 0.572

Values of W ′ for selected sample sizes at confi-
dence levels, α , of 0.01 and 0.05 are shown in Table
10.7.

4 17 1.890 0.2085 0.394

5 16 1.674 0.1686 0.282

6 15 1.587 0.1334 0.212

7 14 1.041 0.1013 0.106

8 13 0.824 0.0711 0.059

The hypothesis of normality is rejected at the α
significance level when the calculated Shapiro-
Francia statistic is less than the applicable value given
in Table 10.7.

9 12 0.642 0.0422 0.027

10 11 0.071 0.0140 0.001

Sum: 5.012