Traffic and Highway Engineering - With Mindtap
Traffic and Highway Engineering - With Mindtap
5th Edition
ISBN: 9781305360990
Author: Garber
Publisher: CENGAGE L
Question
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Chapter 4, Problem 9P
To determine

Whether there was a statistically significant difference in the average speeds, the mean speed, standard deviation, 85th percentile speed and percentage of traffic exceeding the posted speed limit of 30 mi/h.

Expert Solution & Answer
Check Mark

Answer to Problem 9P

Significant reduction

  u¯1=34.9mi/h, u¯2=27.5mi/h, S1=±3.5mi/h, S2=±5.3mi/h, v851=36mi/h, v852=31.5mi/h, 76 %, 24 %

Explanation of Solution

Given:

Significance level of α=0.05

Traffic and Highway Engineering - With Mindtap, Chapter 4, Problem 9P , additional homework tip  1Traffic and Highway Engineering - With Mindtap, Chapter 4, Problem 9P , additional homework tip  2

Formula used:

  u¯=fiuifi

  u¯ is arithmetic mean

  fi is number of observations in each speed group

  ui is mid-value for the ith speed group

  S= f i ( u i u ¯ ) 2 N1

S is standard deviation

N is number of observations

  Sp2=(n11)S12+(n21)S22n1+n22.

Spis square root of the pooled variance

S1 and S2 are standard deviations of the populations

  n1 and n2 are sample means

  T=u¯1u¯2Sp1 n 1 +1 n 2

T is the test static

  u¯1 and u¯2 are sample means

Calculation:

Before an increase in speed enforcement activities:

The speed ranges from 28 to 40 mi/hgiving a speed range of 12. For five classes, the range per class is 2.4mi/h. A frequency distribution table can then be prepared, as shown below, in which the speed classes are listed in column 1 and the mid-values are in column 2. The number of observationsfor each class is listed in column 3, the cumulative percentages of all observations arelisted in column 6.

1234567
Speed class (mi/h)Class mid-value
  ui
Class frequency,
  fi
  fiuiPercentage of class frequencyCumulative percentage of class frequency  fi(uiu¯)2
28-302941161313139.24
31-33325160173042.05
34-36351242040700.12
37-39386228209057.66
40-4241312310100111.63
Total301047350.7

Determine the arithmetic mean speed:

  u¯= f i u i f i fi=30fiui=1047u¯1=104730=34.9mi/h

Determine the standard deviation:

  S= f i ( u i u ¯ ) 2 N1fi ( u i u ¯ )2=350.7N1=fi1=301=29S1= 350.7 29S1=±3.5mi/h

The 85th-percentile speed is obtained from the cumulative frequency distribution curve as 36 mi/h.

The percentage of traffic exceeding the posted speed limit of 30 mi/h is 76 %.

Below given figure shows the cumulative frequency distribution curve for the data given. In this case, the cumulative percentages in column 6 of the above Table are plotted against the upper limit of each corresponding speed class. This curve, therefore, gives the percentage of vehicles that are traveling at or below a given speed.

Traffic and Highway Engineering - With Mindtap, Chapter 4, Problem 9P , additional homework tip  3

After an increase in speed enforcement activities:

The speed ranges from 20 to 37 mi/h giving a speed range of 17. For six classes, the range per class is 2.83 mi/h. A frequency distribution table can then be prepared, as shown below in which the speed classes are listed in column 1 and the mid-values are in column 2. The number of observations for each class is listed in column 3, the cumulative percentages of all observations are listed in column 6.

1234567
Speed class (mi/h)Class mid-value
  ui
Class frequency,
  fi
  fiuiPercentage of class frequencyCumulative percentage of class frequency  fi(uiu¯)2
20-222161262020253.5
23-25248192274798
26-2827410813601
29-3130390107018.75
32-343351651787151.25
35-3736414413100289
Total30825811.5

Determine the arithmetic mean speed:

  u¯= f i u i f i fi=30fiui=825u¯2=82530=27.5mi/h

Determine the standard deviation:

  S= f i ( u i u ¯ ) 2 N1fi ( u i u ¯ )2=811.5N1=fi1=301=29S2= 811.5 29S2=±5.3mi/h

Below given figure shows the cumulative frequency distribution curve for the data given. In this case, the cumulative percentages in column 6 of the above Table are plotted against the upper limit of each corresponding speed class. This curve, therefore, gives the percentage of vehicles that are traveling at or below a given speed.

The 85th-percentile speed is obtained from the cumulative frequency distribution curve as 31.5 mi/h.

The percentage of traffic exceeding the posted speed limit of 30 mi/h is 24 %.

Traffic and Highway Engineering - With Mindtap, Chapter 4, Problem 9P , additional homework tip  4

Determine square root of the pooled variance:

  Sp2=( n 1 1)S12+( n 2 1)S22n1+n22n1=n2=N=30S1=3.5S2=5.3Sp2=( 301) 3.52+( 301) 5.3230+302Sp2=20.17Sp=4.49

Compute test static T:

  T= u ¯1 u ¯2Sp 1 n 1 + 1 n 2 u¯1=35.1mi/hu¯2=27.47mi/hT=35.127.474.49 1 30 + 1 30 =6.58

Determine whether u¯1 is significantly higher than u¯2 :

  Degree of freedom = 30 + 30 - 2 = 58

From Appendix A, theoretical tα= 1.672 for α=0.05

Since 6.58>1.672, then T > tα and the null hypothesis should be rejected.

Conclusion:

The increase in speed enforcement activities has resulted in a significant reduction in the mean speed on the street at a significance level of 0.05. The mean speeds before and after increase in speed enforcement activities are 35.1 and 27.47 mi/h respectively. The standard deviations are 3.5 and 5.3 mi/h respectively. The 85th percentile speeds are 36 mi/h and 31.5 mi/h and percentages of traffic exceeding the posted speed limit of 30 mi/h are 76 % and 24 %.

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