Question

Reserve Problems Chapter 10 Section 1 Problem 3 The time delay is measured for a city street. Variability in this value is estimated as o = 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 7.3 7.0 7.1 7.4 7.37.26.76.8 7.3 7.3 7.5 7.2 After 6.2 6.6 6.4 6.4 6.7 5.8 5.9 5.9 6.3 6.1 7.2 6.8 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. 1) The parameter of interest is the difference in the mean time delay before and after the rearrangement, H1 - H2- 2) Ho : Hi-H2 = 3) H1 : Hi - H2 Find the test statistic. Round your answers to three decimal places (e.g. 98.765). ZO 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. 98.765). We are 90% confident that the difference in the mean time delay lies between <H1 - H2 =

Answer 1

0 solution- Given that, 0.3 i = Exi 86.1 7.145 . 76.3 Exi "2 6.358 12 -0.1 ✓ The Parameter of interest is the difference in the mean time delay before and after the rearrangement, l, -₂ - ш- п, : 7.175 - 6.358 lli- ll₂ = 0.817 2 The hypothesie Holl- y = 1 H : - < The Test - statistic is z = (xyy)-(ll-lly) th han (7.145-6.358)-(1) 0.1830 1 0.1225 0.3 + + Z-1.494 0.0676 <0.1 P-value 0.067 It is clear that p-value < 0.1 Hence, we reject to

(b) for 90%. c.I 2:1-0.90 2:01 Ñ 1/2 = 0.05 The critical value of zis 142 20.05 1.645 . The 90% CD is (4-7) 742 + >> (7.175 -6.358) † 1.645 *0:3 12 12 0.8170 I0.2015 -) (0.8170 - 0.2015 0.8170+ 0.2015) 10.6155, 1.0185) :- The 90% confident that the difference in the mean time delay lies between 0.615 E lilly 51018