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The article “Modeling Arterial Signal Optimization with

Statistics for Engineers and Scientists | 4th Edition | ISBN: 9780073401331 | Authors: William Navidi ISBN: 9780073401331 38

Solution for problem 4E Chapter 5.1

Statistics for Engineers and Scientists | 4th Edition

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Statistics for Engineers and Scientists | 4th Edition | ISBN: 9780073401331 | Authors: William Navidi

Statistics for Engineers and Scientists | 4th Edition

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Problem 4E

The article “Modeling Arterial Signal Optimization with Enhanced Cell Transmission Formulations” (Z. Li, Journal of Transportation Engineering 2011:445–454) presents a new method for timing traffic signals in heavily traveled intersections. The effectiveness of the new method was evaluated in a simulation study. In 50 simulations, the mean improvement in traffic flow in a particular intersection was 654.1 vehicles per hour, with a standard deviation of 311.7 vehicles per hour.

a. Find a 95% confidence interval for the improvement in traffic flow due to the new system.

b. Find a 98% confidence interval for the improvement in traffic flow due to the new system.

c. A traffic engineer states that the mean improvement is between 581.6 and 726.6 vehicles per hour.With what level of confidence can this statement be made?

d. Approximately what sample size is needed so that a 95% confidence interval will specify the mean to within ±50 vehicles per hour?

e. Approximately what sample size is needed so that a 98% confidence interval will specify the mean to within ±50 vehicles per hour?

Step-by-Step Solution:

Step 1 of 5</p>

Given mean

And Standard deviation s=311.7

a) We have to find 95% confidence interval

The 95% confidence interval is

                                                 =654.1

                                                 =

                                                 =(567.69, 740.51)

Hence the 95% confidence interval is (567.69, 740.51)

Step 2 of 5</p>

b) We have to find 98% confidence interval

The 98% confidence interval is

                                                 =654.1

                                                 =

                                                 =(551.38, 756.82)

Hence the 98% confidence interval is (551.38, 756.82)

Step 3 of 5</p>

c) Here we have to find the confidence level

The mean value is between 581.6 and 726.6

Then  

                 

Now the lower limit is =581.6

                654.1-(311.7/)=581.6

                                    (44.09)=654.1-581.6

                                               =1.64

      We will get this value at =1.64

                                      Then  =0.05

                                                     =0.1

                                                      =10%

Hence the the level of confidence is 1-=1-10

                                                                 =90%

Step 4 of 5

Chapter 5.1, Problem 4E is Solved
Step 5 of 5

Textbook: Statistics for Engineers and Scientists
Edition: 4
Author: William Navidi
ISBN: 9780073401331

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