The highway department wants to estimate the proportion of vehicles on Interstate 25 between the hours of midnight and 5:00 A.M. that are 18-wheel tractor trailers. The estimate will be used to determine highway repair and construction considerations and in highway patrol planning. Suppose researchers for the highway department counted vehicles at different locations on the interstate for several nights during this time period. Of the 3,515 vehicles counted, 980 were 18-wheelers. a. Determine the point estimate for the proportion of vehicles traveling Interstate 25 during this time period that are 18-wheelers. b. Construct a 99% confidence interval for the proportion of vehicles on Interstate 25 during this time period that are 18-wheelers. Round your answers to 3 decimal places, the tolerance is +/-0.001. a. The point estimate is ________. b. ________ = p = __________.

Answer: a. 0.2788b)  0.260≤ p ≤ 0.298Step-by-step explanation:Given : Of the 3,515 vehicles counted, 980 were 18-wheelers.Here, Sample size : n= 3515Number of 18-wheelers = 980Then, the point estimate for the proportion of vehicles traveling Interstate 25 during this time period that are 18-wheelers : [tex]\hat{p}=\dfrac{980}{3515}=0.279[/tex]Critical z-value for 99% confidence interval ( Using z-value table) : [tex]z=2.576[/tex]Now, the 99% confidence interval for the proportion of vehicles on Interstate 25 during this time period that are 18-wheelers will be :[tex]\hat{p}\pm z\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]i.e. [tex]0.279\pm (2.576)(0.007565)[/tex]i.e. [tex]0.279\pm 0.019[/tex] [tex]\approx0.279\pm0.019=(0.279-0.019,\ 0.279+0.019)\\\\=(0.260,\ 0.298)[/tex]Hence, a 99% confidence interval for the proportion of vehicles on Interstate 25 during this time period that are 18-wheelers : 0.260≤ p ≤ 0.298