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. A traffic engineer states that the mean improvement is between 581 and 726 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.The level of confidence is what % ?

Answer:The 95% confidence interval is (28.183, 1281.217).Step-by-step explanation:We have the sample's standard deviation, so we use the students' t-distribution to solve this question.The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. Sodf = 50 - 1 = 4995% confidence intervalNow, we have to find a value of T, which is found looking at the t table, with 49 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.95}{2} = 0.975([tex]t_{975}[/tex]). So we have T = 2.01The margin of error is:M = T*s = 2.01*311.7 = 626.517In which s is the standard deviation of the sample.The lower end of the interval is the sample mean subtracted by M. So it is 654.7 - 626.517 = 28.183 vehicles per hour The upper end of the interval is the sample mean added to M. So it is 654.7 + 626.517 = 1281.217 vehicles per hour The 95% confidence interval is (28.183, 1281.217).