Question

1)

The time delay is measured for a city street. Variability in this value is estimated as = 0.3 (for time in minutes) by experience. Traffic signs and lights were recently altered to facilitate traffic and reduce delays. Delay times in minutes for a random car sample before and after the rearrangement are shown below. Before 7.5 7.0 7.1 7.3 7.7 7.0 6.8 6.6 7.0 7.3 7.5 7.2 After 6.5 6.6 5.3 6.4 7.0 5.7 6.0 5.9 5.8 6.1 7.2 5.8 Assume that the time delay variability is unaffected by the traffic signs change. If the difference in the time delay is 1 minute or less, we would like to detect it. (a) Formulate and test an appropriate hypothesis using a = 0.1. What is your conclusions? Find the P-value. i) The parameter of interest is the difference in the mean time delay before and after the rearrangement, ui - 12. 2) Ho : 1-1) = 3) HM-M2 Find the test statistic, Round your answers to three decimal places (e.g. 98.765). P-value = (b) Find a 90% confidence interval for the difference in the mean time delay. Round your answers to three decimal places (e.g. 99.765). We are 90% confident that the difference in the mean time delay lies between Ismi-MS

2)

Answer 1

Snecoer: Given Before 1705) 7.0.7-4170317-7|01681616 17-01-317-5/+2 After 16.5 6.6 6.3/6.470/5.7/6.0/5.9 5.8/6-1 + 2/ 6-8 03 0 1 sample mean ()EX, 37167 Standard deviation(s) = [<2-*)? = 0.314 N- Sample mean (A) = {12 6.358 n Standard deviation (Si) - - 0.481 N-1 Null hypothesis Ho: M-M₂ = 1 Alterative hypothesis Hy: M-M, 4 Test Statistic z =(x, -7) - (M-M) 1 6.314-0-481) -1 2-9152851 0-31 12 1Z0-9.529

time de is -0.3684 2 M-M, 50.0344 P-value = 0.000 P-value 20.1 So we reject the null typothesis at on level ay significance 90% CI for the difference in the means delay 33% = 68, - %) + 10** given 74/2 = 20% 26-314-6-421) [1545 (-38 = -0.167 I (0-4935) Z = (-0.3684, 0.0344) are 90% confident that the in the mean time delay lies between - 1.645 difference