1. 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)

2. 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)?

3. 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, and by yourself.)
7.1.2 Randomness and distributions describing randomness
“Model the system as simply (or as precisely) as possible (or necessary) for all practical purposes.”
The discussion of turning vehicles is very instructive. P.132 right column.
One new topic in 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.

4. Variance 2 = (x - µ)2P(x)
Connection between the typical computation and probability involving formulas for mean and variance
Mean µ = x*P(x)
Variance 2 = (x - µ)2P(x)
(Population)
(Sample)

5. 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
= 1.43
(Discuss the 3 procedures in p. 137 left column top)
From the sample normal distribution to the standard normal distribution.
from Table 7.3

6. Use of the standard normal distribution table, Tab 7-3
Z = 1.43
Most popular one is 95% within µ ± 1.96 
(Excel functions: NORMSDIST and NORMSINV)

7. 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).
True population mean - - X
Point estimate of X from a sample X
Two-sided interval estimate
X – tas/sqrt(n)
X + tas/sqrt(n)

8. 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 )  y  (µ )] = 0.95
Obviously we do not know µ and  . Hence we restate this in terms of the distribution of sample means:
P[( x E)  y  ( x E)] = 0.95
Where, E = s/SQRT(n), standard error of the mean _ _
When E is meant to mean tolerance, we use the symbol e.

9. 7.4 Sample size computations
For cases in which the distribution of means can be considered normal, the confidence range for 95% confidence is:
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:
and
By replacing 1.96 with z and 3.84 with z2, we can use this for any level of confidence.

10. 7.5 Addition of random variables
Summation of random variables:
Expected value (or mean) of the random variable Y:
Variance of the random variable Y:
These concepts are useful for statistical work. See the sample problems in page 140.

11. 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)
approaches x
X distribution
X ~ any (µ,  2)
distribution

12. 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
Has only two possible outcomes: Heads-tails, one-zero, yes-no
P(X = 1) = p
P(x + 0) = 1 - p
Probability mass function
Assumptions:
There is a single trial with only two possible outcomes.
The probability of an outcome is constant for each trial.
1 - p p 1
Event X

13. 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.
Outcome
Example: 3 trials of flipping a coin
No. of tails
Possible outcomes
Prob. of outcome
0 HHH (1/2)0(1/2)3
HHT HTH THH 3(1/2)1(1/2)2
TTH THT HTT 3(1/2)2(1/2)1
TTT (1/2)3(1/2)0
(See equation 7-14)
Read for a sample application of the Binomial distribution.
Mean: Np, Variance: Npq
Discuss

14. 7.7 The Poisson distribution (“counting distribution” or “Random arrival” discrete probability function)
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)

15. 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.)

16. Types of errors We want to minimize especially Type II error. Decision
Reality
Steps of the Hypothesis Testing
Reject H0
Accept H0
H0 is true
Type I error
Correct
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
Correct
Type II error
H1 is true
Fail to reject a false null hypothesis
Reject a correct null hypothesis
(see the binary case in p. 145/146. to get a feel of Type II error.)
P(type I error) =  (level of significance)
P(type II error ) = 

17. 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.
Here we are comparing means; hence divide σ by sqrt(n).
The t (or z) statistics is: t or z
7.8.1 Before-and-after tests with two distinct choices

18. 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.
0.025 each
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.
Two-sided
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.
0.05
One-sided upper

19. 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.)
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 change)
Alternative H1: 1 = 2
Null hypothesis H0: 1 = 2 (there is no increase)
Alternative H1: 1  2

20. Example The decision point (or typically zc: Existing
After improvement
Sample size 55
Mean
60 min
55 min
Standard Deviation
8 min
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.
At significance level  = 0.05 (See Table 7-3.)

21. 7.8.3 Other useful statistical tests
The t-test (for small samples, n<=30) – Table 7.6:
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.
(By definition the larger s is always on top.)
(See the samples in pages 149 and 151.

22. 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.
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.

23. 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.

24. 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 one-sided or one-tailed)
Not paired:
Paired:
|56.9 – 61.2| = 4.3 < 4.54 (=1.65*2.74)
Hence, H0 is NOT rejected.
4.3 increase > (=1.65*0.388)
Hence, H0 is clearly rejected.

25. 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.
The authors intentionally used the uniform distribution to make the computation simple. We will test a normal distribution I class using Excel.

26. 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:

27. 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.

28. What’s the Chi-square (2-) test testing?
You need to know how to pull out values from the assumed distribution to create the expected histogram.
Assumed distribution
Chi-square (2-) test
Expected distribution (or histogram)
Actual histogram