Traffic engineers compared rates of traffic collisions at intersections with raised medians and rates at intersections with two-way left-turn lanes. They found that out of 4,653 collisions at intersections With raised medians, 2,289 were rear-end collisions, and out of 4,606 collisions at two-way left-turn Lanes, 2,027 were rear-end collisions.
Assuming these to be random samples of collisions from the two types of intersections, construct a 95% Confidence interval for the difference between the proportions of collisions that are of the rear-end Type at the two types of intersection.
A) (0.032, 0.072)
B) (0.042, 0.061)
C) (0.472, 0.512)
D) (0.492, 0.440)

Question

Quizplus · Accepted Answer

To construct a 95% confidence interval for the difference between the proportions, we first calculate the sample proportions of rear-end collisions for each type of intersection. For raised medians, $p_1 = \frac{2289}{4653}$ and for two-way left-turn lanes, $p_2 = \frac{2027}{4606}$. The difference in sample proportions is $p_1 - p_2$. The standard error of the difference in proportions is calculated using the formula $\sqrt{\frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2}}$, where $n_1$ and $n_2$ are the sample sizes for the two types of intersections. Using the Z-score for a 95% confidence level (approximately 1.96), the confidence interval is given by $(p_1 - p_2) \pm 1.96 \times \text{standard error}$. The calculations yield a confidence interval that matches choice A, indicating that the true difference in proportions likely falls within this range.

Traffic Engineers Compared Rates of Traffic Collisions at Intersections with Raised