Chapter 7: Statistical Applications in
Traffic Engineering
Chapter objectives: By the end of these chapters the student will be
able to (We do skip simple descriptive stats because they were
covered in CE361 and in an undergrad statistics class.):
Lecture number
Lecture Objectives
(after these lectures you will be able to)
Section 7.1
through 7.7 are
reviews of basic
statistics topics
and not covered in
this class. Assume
you took Stat 201
or equivalent.
• Apply the basic principles of statistics contained in section 7.1 to traffic data analyses
• Explain the characteristics of the normal distribution and read the normal distribution
table correctly (section 7.2) and get necessary values from Excel.
• Explain the meaning of confidence bounds and determine the confidence interval of
the mean (section 7.3)
• Determine sample sizes of traffic data collection (section 7.4)
• Explain how random variables are added (section 7.5)
• Explain the implication of the central limit theorem (section 7.5.1)
• Explain the characteristics of various probabilistic distributions useful for traffic
engineering studies and choose a correct distribution for the study(section 7.6)
•Explain the special characteristics of the Poisson distribution and its usefulness to
traffic engineering studies (section 7-7)
We covers
these topics in
section 7-8
• Perform a hypothesis test correctly (two-sided, one-sided, paired
test, F-test)
• Conduct a Chi-square test to test hypotheses on an underlying
distribution f(x). This test will be covered again as part of the
lecture on speed data analysis in Chapter 10.
7.8 Hypothesis testing
Two distinct choices:
Null hypothesis, H0
Alternative hypothesis: H1
E.g. Inspect 100,000 vehicles, of which 10,000 vehicles are “unsafe.”
This is the fact given to us.
H0: The vehicle being tested is “safe.”
H1: The vehicle being tested is “unsafe.”
In this inspection,
15% of the unsafe vehicles are determined to be safe Type II error (bad error)
and 5% of the safe vehicles are determined to be unsafe  Type I error
(economically bad but safety-wise it is better than Type II error.)
Types of errors
Steps of the Hypothesis Testing
Decision
Reality
Reject H0
Accept H0
H0 is true
Type I error
Correct
H1 is true
Correct
Type II error
State the hypothesis
Select the significance level
Compute sample statistics
and estimate parameters
Compute the test statistic
Fail to reject a false null
hypothesis: Accept a false H0
Reject a correct null hypothesis:
Reject a true H0
P(type I error) =  (level of significance)
P(type II error ) = 
Determine the acceptance
and critical region of the test
statistics
Reject or do not reject H0
Dependence between , , and sample
size n
There is a distinct relationship between the two probability values 
and  and the sample size n for any hypothesis. The value of any one
is found by using the test statistic and set values of the other two.
 Given  and n, determine . Usually the  and n values are the most
crucial, so they are established and the  value is not controlled.
 Given  and , determine n. Set up the test statistic for  and  with H0
value and an H1 value of the parameter and two different n values.
(Read the handout given in class carefully to understand this. Notice we
need to compute d value to find these.)
The t (or z) statistics is: t
or z

(X  )

n
Here we are
comparing means;
hence divide σ by
sqrt(n).
7.8.2 Before-and-after tests with
generalized alternative hypothesis
 The significance of the hypothesis test is indicated by , the type I error
probability.  = 0.05 is most common: there is a 5% level of significance,
which means that on the average a type I error (reject a true H0) will occur 5 in
100 times that H0 and H1 are tested. In addition, there is a 95% confidence level
that the result is correct.
 If H1 involves a not-equal relation,
no direction is given, so the
significance area is equally divided
between the two tails of the testing
distribution.
 If it is known that the parameter can
go in only one direction, a one-sided
test is performed, so the significance
area is in one tail of the distribution.
0.025
each
Two-sided
0.05
One-sided upper
Two-sided or one-sided test
These tests are done to compare the effectiveness of an
improvement to a highway or street by using mean speeds.
 If you want to prove that the difference exists between the
two data samples, you conduct a two-way test. (There is no
change.)
Null hypothesis H0: 1 = 2 (there is no change)
Alternative H1: 1 ≠ 2
If you are sure that there is no decrease or increase, you
conduct a one-sided test. (There was no decrease)
Null hypothesis H0: 1 = 2 (there is no increase)
Alternative H1: 1  2
Example (p.137-138)
Existing
After
improvement
Sample size
55
55
Mean
60 min
55 min
Standard
Deviation
8 min
8 min
Y 
 12
 22
82 82



 1.53
n1 n2
55 55
z / 2  1.96
z  1.65
At significance level  = 0.05 (See Table 7-3.)
The decision point (or
typically zc:
For two-sided:
1.96*1.53 = 2.998
For one-sided:
1.65*1.53 =2.525
|µ1 - µ2| = |60-55| = 5 > zc
By either test, H0 is
rejected.
You can compute z
score and compare Z
computed and Z critical
values.
z
X1  X 2
Y
Use of the standard normal distribution
table, Tab 7-3
Table 7-3
Z = 1.43
Most popular one is a 95% confidence level
and both sided µ ± 1.96  . See section 7.2.2
for confidence interval.
Excel functions:
NORMSDIST(z)
NORMSINV(cum prob)
T.DIST.RT (Computed t, DF)
T.DIST.2T (Computed, DF)
7.8.3 Other useful statistical tests
The t-test (for small samples, n<=30) – Table 7.6:
t
sp
sp 
x1  x2
1 n1  1 n2
n1  1s12  n2  1s22
n1  n2  2
tc = 2.101 for twosided α = 0.05.
(See the samples in page 140)
F-test - Table 7.7
The F-Test to test if s1=s2
When the t-test and other similar means
tests are conducted, there is an implicit
assumption made that s1=s2. The F-test can
test this hypothesis.
s12
F 2
s2
The numerator variance > The
denominator variance when you
compute a F-value.
If Fcomputed ≥ Ftable (n1-1, n2-1, a), then s1≠s2 at
a asignificance level.
If Fcomputed < Ftable (n1-1, n2-1,a), then s1=s2 at a
asignificance level.
Discuss the problem in p.140.
Paired difference test
You perform a paired difference test only when you have a control
over the sequence of data collection.
e.g. Simulation  You control parameters. You have two different
signal timing schemes. Only the timing parameters are changed. Use
the same random number seeds. Then you can pair. If you cannot
control random number seeds in simulation, you are not able to do a
paired test.
Table 7-8 shows an example showing the benefits of paired testing
 The only thing changed is the method to collect speed data. The
same vehicle’s speed was measured by the two methods.
Paired or not-paired example (Table 7.8)
Method 1
Method 2
Difference
Estimated mean
56.87
61.20
4.33
Estimated SD
7.74
7.26
1.496
H0: No increase in test scores (means onesided or one-tailed)
Not paired:
tcritical = 1.701 for df=28, a = 0.05.
Computed 1.5807 < Critical t = 1.701
Hence, H0 is NOT rejected.
Tab 7.6
for t
values
Paired:
tcritical = 1.761 for df=14, a = 0.05.
Computed 11.218 < Critical t = 1.761.
Hence, H0 is clearly rejected.
Chi-square (2-) goodness-of-fit test
Example: Distribution of height data in Table 7-9.
H0:The underlying distribution is uniform.
H1: The underlying distribution is NOT uniform.
25
20
15
10
5
Observed Freq
Theoretical Freq
The authors intentionally used the uniform distribution to make the computation simple. We will
test a normal distribution in class using Excel.
6.8-7.0
6.6-6.8
6.4-6.6
6.2-6.4
6.0-6.2
5.8-6.0
5.6-5.8
5.4-5.6
5.2-5.4
5.0-5.2
0
Steps of Chi-square (2-) test
 Define categories or ranges (or bins) and
assign data to the categories and find ni = the
number of observations in each category i. (At
least 5 bins and each should have at least 5 observations.)
 Compute the expected number of samples for
each category (theoretical frequency), using the
assumed distribution. Define fi = the number of
samples for each category i.
 Compute the quantity:
2
(
n

f
)
2   i i
fi
i 1
N
(ni  f i )
 
fi
i 1
N
2
2
15
Steps of Chi-square (2-) test (cont)
 2 is chi-square distributed (see Table 7-11). If this
value is lower than the critical 2 value, our null
hypothesis is correct. Usually we use  = 0.05 (5%
significance level or 95% confidence level). When
you look up the table, the degree of freedom is f = N –
1 – g where g is the number of parameters we use in
the assumed distribution. For normal distribution g = 2
because we use µ and  to describe the shape of
normal distribution. (For the uniform distribution in the
example g = 0 (why?). Hence, f = 10 – 1 – 0 = 9.
 If the computed 2 value (43.20 for this example) is
smaller than the critical c2 value (16.92 for this
example), we accept H0.
Excel functions:
CHIDIST(Computed 2 , DF)  Probability
CHISQ.INV.RT (α-value, DF)  critical 2 value
What’s the Chi-square goodness of fit
(2-) test testing?
Assumed
distribution
Expected
distribution (or
histogram)
You need to know how to pull
out values from the assumed
distribution to create the
expected histogram.
Chi-square (2-)
test
Actual
histogram