Question

An article 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 653.5 vehicles per hour, with a standard deviation of 311.7 vehicles per hour.

1. Find a 95% confidence interval for the improvement in traffic flow due to the new system. Round the answers to three decimal places.

2. Find a 98% confidence interval for the improvement in traffic flow due to the new system. Round the answers to three decimal places.

3. Approximately what sample size is needed so that a 95% confidence interval will specify the mean to within ±55 vehicles per hour? Round the answer to the next integer.

4. Approximately what sample size is needed so that a 98% confidence interval will specify the mean to within ±55 vehicles per hour? Round the answer to the next integer.

Answer 1

1)

mean =653.5 , s = 311.7 , n = 50
The z value at 95% confidenc einterval is,

alpha = 1 - 0.95 = 0.05
alpha/2 = 0.05/2 = 0.025
Zalpha/2 = Z0.025 = 1.96

Margin of error = E =z *(s/sqrt(n))
= 1.96 *(311.7/sqrt(50))
= 86.399

The 95% confidence interval is

mean -E < mu < mean +E

653.5 - 86.399 < mu < 653.5 + 86.399

567.103 < mu < 739.897

2)

mean =653.5 , s = 311.7 , n = 50
The z value at 98% confidenc einterval is,

alpha = 1 - 0.98 = 0.02
alpha/2 = 0.02/2 = 0.01
Zalpha/2 = Z0.01= 2.326

Margin of error = E =z *(s/sqrt(n))
= 2.326*(311.7/sqrt(50))
= 102.548

The 98% confidence interval is

mean -E < mu < mean +E

653.5 - 102.548 < mu < 653.5 + 102.548

550.9522 < mu < 756.048

3)

ME = 55 , s = 311.7
z value at 95% = 1.96 which is calculated above

Using ME formula,

ME = z *(s/sqrt(n))
55 = 1.96 *(311.7/sqrt(n))
n = (1.96*311.7/55)^2
n = 123

4)

ME = 55 , s = 311.7
z value at 98% = 2.326 which is calculated above

Using ME formula,

ME = z *(s/sqrt(n))
55 = 2.326 *(311.7/sqrt(n))
n = (2.326*311.7/55)^2
n = 174