Chapter 7: Statistical Applications in
Traffic Engineering
Chapter objectives: By the end of these chapters the student will be
able to (We spend 3 lecture periods for this chapter. We do skip
simple descriptive stats because they were covered in CE361.):
Lecture
number
Lecture Objectives
(after these lectures you will be able to)
Lecture 3
(Chap 7a file)
• 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)
Lecture 4a
(Chap 7b file)
• Explain the special characteristics of the Poisson distribution and its usefulness to
traffic engineering studies (section 7-7)
• Conduct a hypothesis test correctly (two-sided, one-sided, paired test, F-test) (section
7-8)
Lecture 4b
(Chap 7 file)
• Conduct a Chi-square test to test hypotheses on an underlying distribution f(x)
(section 7-8)
Introduction
Traffic engineering studies: Infer the characteristics
in a population (typically infinite) by observing the
characteristics of a finite sample.
Statistical analysts is used to address the
following questions:
 How many samples are required?
 What confidence should I have in this estimate?
 What statistical distribution best describes the observed
data mathematically?
 Has a traffic engineering design resulted in a change in
characteristics of the population (hypothesis tests)?
7.1 An Overview of Probability Functions and Statistics
Most of the topics in this section are reviews of what we have learned
in CEEn 361. (Review 7.1.1, 7.1.3 and 7.1.4 by yourself.)
7.1.2 Randomness and distributions describing randomness
The discussion of
turning vehicles is very
instructive. P.132 right
column.
“Model the system as simply (or as
precisely) as possible (or necessary) for
all practical purposes.”
One new topic in 7.1.4 is a method to
estimate the standard deviation. This
is based on the normal distribution –
the probability of one standard
deviation from the mean is 68.3% in
the two-way analysis. 85%-15% =
70%, close enough.
sest
P85  P15

2
Connection between the typical computation and probability
involving formulas for mean and variance
(Population)
(Sample)
Mean µ = x*P(x) Variance 2 = (x - µ)2P(x)
1
x
N
N
x
i 1
s2 
i
P(x)
Data (Population)
Mean
Variance
3.50
4.25
2.70
2.70
3.65
5.50
3.72
0.93
0.17
0.17
0.17
0.17
0.17
0.17
Sum
2


x

x
 i
N 1
x*P(x)
(x-μ)^2*P(x)
0.58
0.71
0.45
0.45
0.61
0.92
3.72
0.01
0.05
0.17
0.17
0.00
0.53
0.93
7.2 The normal distribution and its applications
Mean = 55 mph, STD = 7 mph
What’s the probability the next
value will be 65 mph or less?
z = (x - µ)/ 
= (65 – 55)/7
(Discuss the 3
procedures in p.
137 left column
top)
= 1.43
From the sample
normal distribution
to the standard
normal distribution.
0.9236 from
Table 7.3
Use of the standard normal distribution
table, Tab 7-3
Table 7-3
Z = 1.43
Most popular one is 95% within µ ± 1.96 
(Excel functions:
NORMSDIST and
NORMSINV)
7.3 Confidence bounds (of the mean)
Point estimates: A point estimate is a single-values
estimate of a population parameter made from a sample.
Interval estimates: An interval estimate is a
probability statement that a population parameter is
between two computed values (bounds).
µ
-
-
X
True population mean
Point estimate of X from a
sample
X
X – tas/sqrt(n)
X + tas/sqrt(n)
Two-sided interval
estimate
7.3 (cont)
When n gets larger (n>=30), t can become z. The
probability of any random variable being within 1.96
standard deviations of the mean is 0.95, written as:
P[(µ - 1.96)  y  (µ + 1.96)] = 0.95
Obviously we do not know µ and  . Hence we restate
this in terms of the distribution of sample means:
_
_
P[( x - 1.96E)  y  ( x + 1.96E)] = 0.95
Where, E = s/SQRT(n), standard error of the mean
When E is meant to mean tolerance,
we use the symbol e.
7.4 Sample size computations
For cases in which the distribution of means can be
considered normal, the confidence range for 95%
confidence is:
s
 1.96
n
If this value is called the tolerance (or “precision”),
and given the symbol e, then the following equation
can be solved for n, the desired sample size:
s
e  1.96
n
and
s2
n  3.84 2
e
By replacing 1.96 with z and 3.84 with z2, we can use
this for any level of confidence.
7.5 Addition of random variables
Summation of random variables:
Y   ai X i
Expected value (or mean) of the random variable Y:
Y   ai  xi
Variance of the random variable Y:
  a 
2
Y
2
i
2
xi
These concepts are
useful for statistical
work. See the
sample problems in
page 140.
7.5.1 The central limit theorem
Definition: The population may have any unknown distribution with
a mean µ and a finite variance of  2. Take samples of size n from
the population. As the size of n increases, the distribution of sample
means will approach a normal distribution with mean µ and a
variance of  2/n.
F(x)
f (X )

X
approaches
µ
X distribution
X ~ any (µ,  2)
x
µ
X distribution
X ~ N ( ,  X 2 )
X
7.6 The Binomial Distribution Related to the
Bernoulli and Normal Distributions
7.6.1 Bernoulli and the Binomial distribution (discrete
probability functions))
 Discrete distribution
P(X = 1) = p
 Has only two possible outcomes:
Heads-tails, one-zero, yes-no
P(x + 0) = 1 - p
Assumptions:
Probability mass function
1-p
 There is a single trial with
only two possible outcomes.
 The probability of an
outcome is constant for each
trial.
p
1
0
Event X
Explanation of the Binomial distribution
Assumptions:
 n independent Bernoulli trials
 Only 2 possible outcomes on each trial
 Constant probability for each outcome on each trial
 The quantity of interest is the total number of X of positive
outcomes, a value between 0 and N.
0
1
2
Outcome
Example: 3 trials of flipping a coin
No. of tails
0
HHH
Possible outcomes
Prob. of outcome
(1/2)0(1/2)3
1
HHT
HTH
THH
3(1/2)1(1/2)2
2
TTH
THT
HTT
3(1/2)2(1/2)1
3
TTT
(See equation 7-14)
(1/2)3(1/2)0
Read 7.6.2 for a sample application of the Binomial distribution.
Mean: Np, Variance: Npq
3
Discuss 7.6.2.
7.7 The Poisson distribution (“counting
distribution” or “Random arrival” discrete
probability function)
x m
me
P( X  x) 
x!
With mean µ = m and
variance 2 = m.
If the above characteristic is
not met, the Poisson
theoretically does not apply.
 The binomial distribution tends to approach the Poisson distribution with
parameter m = np. Also, the binomial distribution approaches the normal
distribution when np/(1-p)>=9
 When time headways are exponentially distributed with mean  = 1/, the
number of arrivals in an interval T is Poisson distributed with mean = m =
T. Note that the unit  is veh/unit time (arrival rate).
(Read the sample problem in page 144, table 7.5)
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
We want to minimize
especially Type II
error.
Decision
Reality
Reject H0
Accept H0
H0 is true
Type I error
Correct
H1 is true
Correct
Type II error
Fail to reject a false
null hypothesis
Reject a correct null hypothesis
P(type I error) =  (level of
significance)
P(type II error ) = 
(see the binary case
in p. 145/146. to get a
feel of Type II error.)
Steps of the Hypothesis Testing
State the hypothesis
Select the significance level
Compute sample statistics
and estimate parameters
Compute the test statistic
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.
The t (or z) statistics is: t
or z

( X  )

n
7.8.1 Before-and-after tests with
two distinct choices
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
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.
7.8.3 Other useful statistical tests
The t-test (for small samples, n<=30) – Table 7.6:
t
sp
x1  x2
1 n1  1 n2
sp 
n1  1s12  n2  1s22
n1  n2  2
The F-test (for small samples) – Table 7.7:
In using the t-test we assume that the standard deviations of the
two samples are the same. To test this hypothesis we can use
the F-test.
s12
F 2
s2
(By definition the larger s is
always on top.)
(See the samples in pages 149 and 151.
7.8.3 Other useful statistical tests (cont)
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.151.
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 measure by the two methods.
Paired or not-paired example (table 7.8)
Method 1
Method 2
Difference
Estimated
mean
56.9
61.2
4.3
Estimated SD
7.74
7.26
1.5
H0: No increase in test scores (means onesided or one-tailed)
Not paired:
7.74 2 7.26 2
sY 

 2.74
15
15
Paired:
1.50
E
 0.388
15
|56.9 – 61.2| = 4.3 < 4.54
(=1.65*2.74)
4.3 increase > 0.642
(=1.65*0.388)
Hence, H0 is NOT rejected.
Hence, H0 is clearly rejected.
Chi-square (2-) test (So called
“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
The authors intentionally used the uniform distribution to make the computation simple. We will
test a normal distribution I class using Excel.
Theoretical Freq
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
Steps of Chi-square (2-) test (cont)
 2 is chi-square distributed (see Table 5-8). If this
value is low if our 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.
 If the computed 2 value is smaller than the
critical c2 value, we accept H0.
What’s the Chi-square (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