Download Example: Determining Speed Characteristics from a Set of Speed

Example: Determining Speed Characteristics from a Set of Speed Data
Table 4.2 shows the data collected along the South Luzon Expressway in Bulacan during
a speed study. Develop the frequency histogram and the frequency distribution of the
data and determine:
1.
2.
3.
4.
5.
6.
The
The
The
The
The
The
arithmetic mean speed
standard deviation
median speed
pace
mode or modal speed
85th-percentile speed
Table 4.2 Speed Data Obtained on a Rural Highway
Car No.
Speed
(kph)
Car No.
Speed
(kph)
Car No.
Speed
(kph)
Car No.
Speed
(kph)
1
35.1
23
46.1
45
47.8
67
56.0
2
44.0
24
54.2
46
47.1
68
49.1
3
45.8
25
52.3
47
34.8
69
49.2
4
44.3
26
57.3
48
52.4
70
56.4
5
36.3
27
46.8
49
49.1
71
48.5
6
54.0
28
57.8
50
37.1
72
45.4
7
42.1
29
36.8
51
65.0
73
48.6
8
50.1
30
55.8
52
49.5
74
52.0
9
51.8
31
43.3
53
52.2
75
49.8
10
50.8
32
55.3
54
48.4
76
63.4
11
38.3
33
39.0
55
42.8
77
60.1
12
44.6
34
53.7
56
49.5
78
48.8
13
45.2
35
40.8
57
48.6
79
52.1
14
41.1
36
54.5
58
41.2
80
48.7
15
55.1
50.2
54.3
45.4
55.2
45.7
54.1
54.0
37
51.6
51.7
50.3
59.8
40.3
55.1
45.0
48.3
59
48.0
58.0
49.0
41.8
48.3
45.9
44.7
49.5
81
61.8
56.6
48.2
62.1
53.3
53.4
16
17
18
19
20
21
22
38
39
40
41
42
43
44
Solution:
The arithmetic mean speed is computed as
60
61
62
63
64
65
66
82
83
84
85
86
The standard deviation is computed as
Figure 4.4 Histogram of Observed Vehicles’ Speeds



The median speed is obtained from the cumulative frequency distribution curve
(Figure 4.6) as 49 kph, the 50th – percentile speed.
The pace is obtained from the frequency distribution curve (Figure 4.5) as 45 to
55 kph.
The mode or modal speed is obtained from the frequency histogram as 49 kph
(Figure 4.4). It also may be obtained from the frequency distribution curve shown
in Figure 4.5, where the speed corresponding to the highest point on the curve is
taken as an estimate of the modal speed.
85th – percentile speed is obtained from the cumulative frequency distribution
curve as 54 kph (Figure 4.6).

Table 4.3 Frequency Distribution Table for Set of Speed Data
Frequency Distribution Table for Set of Speed Data
1
Speed Class
(kph)
34-35.9
36-37.9
38-39.9
40-41.9
42-43.9
44-45.9
46-47.9
48-49.9
50-51.9
52-53.9
54-55.9
56-57.9
58-59.9
60-61.9
62-63.9
64-65.9
Totals
2
Class
Midvalue,
ui
35
37
39
41
43
45
47
49
51
53
55
57
59
61
63
65
3
Class Frequency
(Number of
Observations in
Class), fi
2
3
2
5
3
11
4
18
7
8
11
5
2
2
2
1
86
4
5
fiui
70
111
78
205
129
495
188
882
357
424
605
285
118
122
126
65
4260
Percentage of
Observations in
Class
2.3
3.5
2.3
5.8
3.5
12.8
4.7
20.9
8.1
9.3
12.8
5.8
2.3
2.3
2.3
1.2
100
6
Cumulative
Percentage
of All
Observations
2.3
5.8
8.1
14.0
17.4
30.2
34.9
55.8
64.0
73.3
86.0
91.9
94.2
96.5
98.8
100.0
7
f(ui -umean)2
420.5
468.75
220.5
361.25
126.75
222.75
25
4.5
15.75
98
332.75
281.25
180.5
264.5
364.5
240.25
3627.5
Descriptive Statistics of the Data from EXCEL output
Speed (kph)
Mean
Standard Error
Median
Mode
Standard Deviation
Sample Variance
Kurtosis
Skewness
Range
Minimum
Maximum
Sum
Count
Largest(1)
Smallest(1)
Confidence Level(95.0%)
49.38953488
0.702590534
49.15
49.5
6.515556571
42.45247743
-0.088993003
-0.077281846
30.2
34.8
65
4247.5
86
65
34.8
1.396938183
Detailed solution of some of the descriptive statistics in the table
Standard error and confidence level
Considering large-sample ( 50 samples), in the confidence interval
the term
and
Hence, from the above example,
= confidence level
= 0.703.
In obtain the confidence level( at 95% level of confidence), consult first the Z
distribution for the value of
= 1.96, hence
= 1.96(.703) = 1.378.
Skewness – is a measure of symmetry, or more precisely, the lack of symmetry. A
distribution, or data set, is symmetric if it looks the same to the left and right of the
center point.
where: is the mean, S is the standard deviation, and N is the number of data points.
The skewness for a normal distribution is zero, and any symmetric data should have a
skewness near zero.
Negative values for the skewness indicate data that are skewed left meaning that the
left tail is long relative to the right tail.
Positive values for the skewness indicate data are skewed right meaning that the right
tail is long relative to the left tail.
Kurtosis is a measure of whether the data are peaked or flat relative to a normal
distribution. That is, data sets with high kurtosis tend to have a distinct peak near the
mean, decline rather rapidly, and have heavy tails. Data sets with low kurtosis tend to
have a flat top near the mean rather than a sharp peak. A uniform distribution would be
the extreme case.
where:
is the mean, S is the standard deviation, and N is the number of data points.
The kurtosis for a standard normal distribution is three. For this reason, some sources
use the following definition of kurtosis (often referred to as “excess kurtosis”).
Difference between time mean speed and space mean speed
The figure shoes vehicles traveling at constant speeds on a two-lane highway between
sections X and Y with their positions and speeds obtained at an instant of time by
photography. An observer located at point X observes the four vehicles passing point X
during a period T sec. The velocities of the vehicles are measured as 45, 40, 35, and 20
kph, respectively. Calculate the flow, the density the time mean speed, and the space
mean speed.
4
X
6
9
D
6
11
6
C
25
B
17
6
A
Direction
of flow
Y
90m
Solution:
Flow (q)
q = n x 3600 / T = 4 x 3600 / T = 14,400/T vph
With L equal to the distance between X and Y (m), density is obtained by
k = n/L = 4/90 x (1000) = 44.44 vpk
The time mean speed is found by
n
ut = (1/n) i=1 ui = (1/4)(20+35+40+45) = 35 kph
The space mean speed is found by
n
us = n / (i=1 1/ui) =
n
=
n L / (i=1 ti)
n
= 90n / (i=1 ti)
where ti is the time it takes the ith vehicle to travel from X to Y at speed u i, and L (m) is
the distance between X and Y.
t1 = 3.6 L / ui sec
tA = 3.6(90) / 45 = 7.2 sec
tB = 3.6(90) / 40 = 8.1 sec
tC = 3.6(90) / 35 = 9.26 sec
tD = 3.6(90) / 20 = 16.2 sec
ui = 4(90) / (7.2 + 8.1 + 9.26 + 16.2) = 8.83 m/sec
ui = 8.83 (3.6) = 31.80 kph
Comparison of Mean Speeds
Example: Significant Differences in Average Spot Speeds
Speed data were collected at a section of highway during and after utility maintenance
work. The speed characteristics are given as,
as shown below. Determine
whether there was any significant difference between the average speed at the 95%
confidence level.
Solution:
Use Eq. 4.6
Find the difference in means
38.7 – 35.5 = 3.2 kph
3.2 > (1.96)(0.65)
3.2 > 1.3 kph

It can be concluded that the difference in mean speeds is significant at the 95%
confidence level.
Review of Engineering Statistics
The problem above can be considered under the Tests on the Difference in Means,
Variances Known.
Solution:
Null Hypothesis:
Alternative Hypothesis:
From Table of Z-distribution, at 95% level of confidence,
Since
, reject H0, therefore it can be concluded that the difference in mean
speeds is significant at the 95% confidence level.