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 7.5 7.1 7.1 7.2 7.517.2 6.8 6.6 7.1 7.3 7.4 7.2 After 6.5 6.7 6.4 6.2 6.8 5.8 6.0 5.9 6.0 5.9 7.1 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, Hi-H2. 2) Ho: Mi - H2 = 0.825 1 3) H1 : Hi - H2 Find the test statistic. Round your answers to three decimal places (e.g. 98.765). ZO -1.429 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 ISH-425

Answer 1

ANSWER

solution We have to caloulate mean from given data n =12 ng=12 86 n X₂=EX n = 76.1 xi = 7.1666 X2 = 6.3416 2-01 The null and alternative hypothesis we Hol, -112=1 Hi: Mi-12-1 Test statistic z=(xi - X2 )-(-42) = (7.1666-6.3416) - | 0.3 Patta -0.825-1 -0.175 =-1.4297 0.1224 0.1224 2 = -1.429 pvalue? This is left tailed test. The p-value = 0.058674

2 since p-value 2010 so we reject null hypothesis Hoat o. l level of significance 1 Confidence Interval is 90% 2:1-0.90-01 critical value of 242 =1:645 90% GI for difference in the mean time delay is CI = (x^2) £24/2.0 + + = (7.1666-6.3416) +1.645X 0.3 N2 = 0.825 + 1.645 x 0.1224 =0.825 +0. 201348 . CI - (0.825, 0.201348)