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 655.6 vehicles per hour, with a standard deviation of 311.7 vehicles per hour. A traffic engineer states that the mean improvement is between 583.1 and 728.1 vehicles per hour. With what level of confidence can this statement be made? Express the answer as a percent and round to the nearest integer.

Answer 1

Answer:

With 89.9% we can say that the mean improvement is between 583.1 and 728.1 vehicles per hour.

Step-by-step explanation:

We are given that 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 655.6 vehicles per hour, with a standard deviation of 311.7 vehicles per hour.

A traffic engineer states that the mean improvement is between 583.1 and 728.1 vehicles per hour.

Let $\bar X$ = sample mean improvement

The z-score probability distribution for sample mean is given by;

            Z =  $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = mean improvement = 655.6 vehicles per hour

            $\sigma$ = standard deviation = 311.7 vehicles per hour

            n = sample of simulations = 50

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, Probability that the mean improvement is between 583.1 and 728.1 vehicles per hour is given by = P(583.1 < $\bar X$ < 728.1) = P($\bar X$ < 728.1) - P($\bar X$ $\leq$ 583.1)

  P($\bar X$ < 728.1) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{728.1-655.6}{\frac{311.7}{\sqrt{50} } }$ ) = P(Z < 1.64) = 0.9495

  P($\bar X$ $\leq$ 583.1) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{583.1-655.6}{\frac{311.7}{\sqrt{50} } }$ ) = P(Z $\leq$ -1.64) = 1 - P(Z < 1.64)

                                                            = 1 - 0.9495 = 0.0505                      

So, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 1.64 in the z table which has an area of 0.9495.

Therefore, P(583.1 < $\bar X$ < 728.1) = 0.9495 - 0.0505 = 0.899 or 89.9%

Hence, with 89.9% we can say that the mean improvement is between 583.1 and 728.1 vehicles per hour.