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

Reality Decision

Reject H0 Accept H0

H0 is true

H1 is true

Type I error

Type II error

Correct

Correct

Reject a correct null hypothesis: Reject a true H0

Fail to reject a false null hypothesis: Accept a false H0

Steps of the Hypothesis Testing

State the hypothesisSelect the significance levelCompute sample statistics

and estimate parametersCompute the test statisticDetermine the acceptance

and critical region of the test statistics

Reject or do not reject H0

P(type I error) = (level of significance)

P(type II error ) =

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 zn

X

)(

Here we are comparing means; hence divide σ by sqrt(n).

(Read the handout given in class carefully to understand this. Notice we need to compute d value to find these.)

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.

One-sided upper

Two-sided

0.025 each

0.05

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 increase)

Alternative H1: 1 2

Null hypothesis H0: 1 = 2 (there is no change)

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

53.1558

558 22

2

22

1

21

nnY

96.12/ z 65.1zAt significance level = 0.05 (See Table 7-3.)

The decision point (or typically zc:

For two-sided: 1.96*1.53 = 2.998For one-sided: 1.65*1.53 =2.525

|µ1 - µ2| = |60-55| = 5 > zcBy either test, H0 is

rejected.

Y

XXz

21

You can compute z score and compare Z computed and Z critical values.

Use of the standard normal distribution table, Tab 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)

Table 7-3

7.8.3 Other useful statistical tests

21

21

11 nnsxxt

p

2

11

21

222

211

nn

snsns p

The t-test (for small samples, n<=30) – Table 7.6:

(See the samples in page 140)

tc = 2.101 for two-sided α = 0.05.

F-test - Table 7.7The 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.

22

21

ssF 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 one-sided or one-tailed)Not paired: Paired:

tcritical = 1.701 for df=28, a = 0.05. Computed 1.5807 < Critical t = 1.701

Hence, H0 is NOT rejected.

tcritical = 1.761 for df=14, a = 0.05. Computed 11.218 < Critical t = 1.761. Hence, H0 is clearly rejected.

Tab 7.6 for t values

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.

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

5.0-

5.2

5.2-

5.4

5.4-

5.6

5.6-

5.8

5.8-

6.0

6.0-

6.2

6.2-

6.4

6.4-

6.6

6.6-

6.8

6.8-

7.0

0

5

10

15

20

25

Observed Freq Theoretical Freq

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:

N

i i

ii

ffn

1

22 )(

15

N

i i

ii

ffn

1

22 )(

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 c

2 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)

Actual histogram

Chi-square (2-) test

You need to know how to pull out values from the assumed distribution to create the expected histogram.