The article “Modeling Arterial Signal Optimization with Enhanced Cell Transmission Formulations 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

The article “Modeling Arterial Signal Optimization with Enhanced Cell Transmission Formulations 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

Question
Modeling
asked 2021-03-07
The article “Modeling Arterial Signal Optimization with Enhanced Cell Transmission Formulations 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 \(\displaystyle{95}\%\) confidence interval for the improvement in traffic flow due to the new system.
b) Find a \(\displaystyle{98}\%\) 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 \(\displaystyle{95}\%\)
confidence interval will specify the mean to within \(\displaystyle\pm\ {50}\) vehicles per hour?
e) Approximately what sample size is needed so that a \(\displaystyle{98}\%\) confidence
interval will specify the mean to within \(\displaystyle\pm\ {50}\) vehicles per hour?

Answers (1)

2021-03-08

Step 1
Given:
\(\displaystyle{n}={S}{a}\mp\le\ {s}{i}{z}{e}={50}\)
\(\displaystyle\overline{{{x}}}={S}{a}\mp\le\ {m}{e}{a}{n}={654.1}\)
\(\displaystyle{s}={S}{a}\mp\le\ {s}{t}{a}{b}{d}{a}{r}{d}\ {d}{e}{v}{i}{a}{t}{i}{o}{n}={311.7}\)
a) \(\displaystyle{c}={C}{o}{n}{f}{i}{d}{e}{n}{c}{e}\ {c}{o}{e}{f}{f}{i}{c}{i}{e}{n}{t}={95}\%={0.95}\)
Since the sample size n is large (at least 30), it is appropriate to estimate the population standard deviation by the sample standard deviation.
\(\displaystyle\sigma\ \approx\ {s}\)
For confidence level \(\displaystyle{1}\ -\ \alpha\ {0.95},\) determine
\(\displaystyle{z}_{{\frac{\alpha}{{2}}}}={z}_{{{0.025}}}\) using the normal probability table in the appendix (look up 0.025 in the table, the z-score is then the found z-score with opposite sign):
\(\displaystyle{z}_{{\frac{\alpha}{{2}}}}={1.96}\)
The margine of error is then:
\(\displaystyle{E}={z}_{{\frac{\alpha}{{2}}}}\ \times\ {\frac{{\sigma}}{{\sqrt{{{n}}}}}}\ \approx\ {z}_{{\frac{\alpha}{{2}}}}\ \times\ {\frac{{{s}}}{{\sqrt{{{n}}}}}}={1.96}\ \times\ {\frac{{{311.7}}}{{\sqrt{{{30}}}}}}\ \approx\ {111.5404}\)
The boundaries of the confidence interval then become:
\(\displaystyle\overline{{{x}}}\ -\ {E}={654.1}\ -\ {111.5404}={542.5596}\)
\(\displaystyle\overline{{{x}}}\ +\ {E}={645.1}\ +\ {111.5404}={765.6404}\)
We are \(\displaystyle{95}\%\) confident that the improvement in traffic flow due to the new system is between 542.5596 vehicles per hour and 765.6404 vehicles per hour.
Step 2
b) \(\displaystyle{c}={C}{o}{n}{f}{i}{d}{e}{n}{c}{e}\ {c}{o}{e}{f}{f}{i}{c}{i}{e}{n}{t}={98}\%={0.98}\)
Since the sample size n is large (at leasr 30), it is appropriate to estimate the population standard deviation by the sample standard deviation.
\(\displaystyle\sigma\ \approx\ {s}\)
For confidence level \(\displaystyle{1}\ -\ \alpha={0.98},\ {\det{{e}}}{r}\min{e}\ {z}_{{\frac{\alpha}{{2}}}}={z}_{{{0.01}}}\) using the normal probability table in the appendix (look up 0.01 in the table, the z-score is then the found z-score with opposite sing):
\(\displaystyle{z}_{{\frac{\alpha}{{2}}}}={2.33}\)
The margin of error is then:
\(\displaystyle{E}={z}_{{\frac{\alpha}{{2}}}}\ \times\ {\frac{{\sigma}}{{\sqrt{{{n}}}}}}\ \approx\ {z}_{{\frac{\alpha}{{2}}}}\ \times\ {\frac{{{s}}}{{\sqrt{{{n}}}}}}={2.33}\ \times\ {\frac{{{311.7}}}{{\sqrt{{{30}}}}}}\ \approx\ {132.5965}\)
The boudaries of the confidenceinterval then become:
\(\displaystyle\overline{{{x}}}\ -\ {E}={654.1}\ -\ {132.5965}={521.5035}\)
\(\displaystyle\overline{{{x}}}\ +\ {E}={654.1}\ +\ {132.5965}={786.6965}\)
We are \(\displaystyle{98}\%\) confident that the improvement in traffic flow due to the new system is between 521.5035 vehicles per hour and 786.6965 vehicles per hour.
Step 3
c) Given confidence interval: \(\displaystyle{\left({581.6},\ {726.6}\right)}\)
The sampling distribution of the sample mean \(\displaystyle\overline{{{x}}}\) has mean
\(\displaystyle\mu\) and standard deviation
\(\frac{\sigma}{\sqrt{n}}.\)
The z-score is the value decreased by the mean, divided by the standard deviation
\(\displaystyle{z}={\frac{{\overline{{{x}}}\ -\ \mu_{{\overline{{{x}}}}}}}{{\sigma_{{\overline{{{x}}}}}}}}={\frac{{{581.6}\ -\ {654.1}}}{{\frac{{311.7}}{\sqrt{{{50}}}}}}}\ \approx\ -{1.64}\)
\(\displaystyle{z}={\frac{{\overline{{{x}}}\ -\ \mu_{{\overline{{{x}}}}}}}{{\sigma_{{\overline{{{x}}}}}}}}={\frac{{{726.6}\ -\ {654.1}}}{{\frac{{311.7}}{\sqrt{{{50}}}}}}}\ \approx\ {1.64}\)
The confidence level is the probability between the two z-scores. Determine the probability using the normal probability table:
\(\displaystyle{P}{\left(-{1.64}\ {<}\ {Z}\ {<}\ {1.64}\right)}={P}{\left({Z}\ {<}\ {1.64}\right)}\ -\ {P}{\left({Z}\ {<}\ -{1.64}\right)}\)
\(\displaystyle={0.9495}\ -\ {0.0505}\)
\(\displaystyle={0.8990}\)
Thus the confidence level is about \(\displaystyle{90}\%\)
Step 4
d) Given:
\(\displaystyle{c}={C}{o}{n}{f}{i}{d}{e}{n}{c}{e}\ lev{e}{l}={95}\%={0.95}\)
\(E=\text{Margin of error} =50\)
For confidence level \(1-\alpha=0.95,\ \text{determine}\ z_{\frac{\alpha}{2}}=1_{0.025}\) using the normal probability table in the appendix (look up 0.025 in the table, the z-score is then the found z-score with opposite sign):
\(\displaystyle{z}_{{\frac{\alpha}{{2}}}}={1.96}\)
Formula for the margin of error:
\(\displaystyle{E}={z}_{{\frac{\alpha}{{2}}}}\sigma\ \overline{{{x}}}={z}_{{\frac{\alpha}{{2}}}}\ {\frac{{{s}}}{{\sqrt{{{n}}}}}}\)
Multiply each side of the previous equation by \(\displaystyle\sqrt{{{n}}}\):
\(\displaystyle{E}\sqrt{{{n}}}={z}_{{\frac{\alpha}{{2}}}}{s}\)
Divide each side by ME:
\(\displaystyle\sqrt{{{n}}}={\frac{{{z}_{{\frac{\alpha}{{2}}}}{s}}}{{{E}}}}\)
Square each side:
\(\displaystyle{n}={\left({\frac{{{z}_{{\frac{\alpha}{{2}}}}\ \cdot\ {s}}}{{{E}}}}\right)}\)
Fill in known values and evaluate (round up to the nearest integer!):
\(\displaystyle{n}={\left({\frac{{{1.96}\ \cdot\ {311.7}}}{{{50}}}}\right)}^{{{2}}}\ \approx\ {150}\)
Step 5 e) Given:
\(\displaystyle{c}={C}{o}{n}{f}{i}{d}{e}{n}{c}{e}\ le{v}{e}{l}=\frac{{98}}{\%}={0.98}\)
\(E=\text{Margin of error}=50\)
For confidence level \(1-\alpha=0.98,\ \text{determine}\ z_{\frac{\alpha}{2}}=1_{0.01}\) using the normal probability table in the appendix (look up 0.01 in the table, the z-score is then the found z-score with opposite sign):
\(\displaystyle{Z}_{{\frac{\alpha}{{2}}}}={2.33}\)
Formula for the margin of error:
\(\displaystyle{E}={z}_{{\frac{\alpha}{{2}}}}\sigma\ \overline{{{x}}}={z}_{{\frac{\alpha}{{2}}}}\ {\frac{{{s}}}{{\sqrt{{{n}}}}}}\)
Multiply each side of the previous equation by \(\displaystyle\sqrt{{{n}}}\):
\(\displaystyle{E}\sqrt{{{n}}}={z}_{{\frac{\alpha}{{2}}}}{s}\)
Divide each side by ME:
\(\displaystyle\sqrt{{{n}}}={\frac{{{z}_{{\frac{\alpha}{{2}}}}{s}}}{{{E}}}}\)
Square each side:
\(\displaystyle{n}={\left({\frac{{{z}_{{\frac{\alpha}{{2}}}}\ \cdot\ {s}}}{{{E}}}}\right)}^{{{2}}}\)
Fill in the known values and evaluate (round up the nearest integer!):
\(\displaystyle{n}={\left({\frac{{{2.33}\ \cdot\ {311.7}}}{{{50}}}}\right)}^{{{2}}}\ \approx\ {211}\)

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