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 a = 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 7.1 7.1 7.2 7.4 7.3 6.6 6.7 7.3 7.2 7.4 7.2 After 6.3 6.7 6.3 6.1 6.7 6.0 5.7 6.0 6.1 5.9 7.2 6.9 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, #1 - 42. 2) H: MH2 = 3) H. : M- My < 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. 98.765). We are 90% confident that the difference in the mean time delay lies between SHi

Answer 1

n = sample size = 12 X = 7.15 Before a s 6.325 After and G=0,3, in the mean A. The parametes of interest is the difference in the delay before and after the rearrangement, MM2 2. Ho: M-M2 = 1 H,. M, - Men 3 Test statistic ne Before After 7.15-6,325 2 ん。 6 0.3 tt 12 -6,736 p-value =P P(Z < 6.736) = 1 3) 90% cf for the difference in the mean time delay is Bepre rathet a Z oth 0.10 :) =fa 7.17-6.325)+ 1; 282 * 0.3 ta tot 2 = 0 , 1.002) We are 90% confidence that the difference in the mean time delay his between <H4, 44, <!.002

0,=10, 52=6 n, Fll, n2 = 14 5 = 14.3 and =19.6 = 0,05 Hypothess : HoiM, FM2 against H,: M, CM2 a) Test statistic 5,- ła Zobs = dols 0 t 19 2 14.3-19.6 = -1,552 10² 6 T+ 14 p-value =P (24-1.552) = 0.0603 Mi-M2=4 and let (n = n) eritical - MD enit So, P. (za m Don't ) =0.05 Vio 69 14 M Dent 10 2 Tit 62 14 - m Derit = F 6.05) { "f" is the innere of CDP of aw1(0,1)] VES Type I error, B=P(G, ~ F2) 75.6006) = P(27-5.6006-4 + Plzy -2,8113) -1.64/102 MDonit =-5. 6006 14. 2 14 0.9975 Power I-BC1-0.99.75 = 0.0025

♡ ß P(2> -5.6006-4 ->0.05% 5=P(2> 5.6006-4 wo (136 = -0,05 -> p ( z< -5.6006+4 J136/ P(za -9.6006 -9.6006 < 1136 11367 -9.6006 -1.64 > -9.6006 V13/n = 1.64/36 1.64 2 > n = 164 - 9.6006 1136 => Chord og x 136 => n = 3.9.685 . Required sample size is 4. a) The nurl hypothesis is not rejected. The p-vahu is 0.0603. b) The poures is 0.0025 ♡ Ashan, ang=4