Question

The time delay is measured for a city street. Variability in this value is estimated the rearrangement are shown below. so0.3 (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 Berore 7.27:1 777.37.1 6.9 6.8 7.2 7.2 7.47.1 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. meter of interect is the difforence in the mean time dalay before and after the rearrangcment at 2 2) Ho : H-2 = 3) H: -4 < three decimal places (e.q, 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). 90% confident that the difference in the mean time delay lies between

Answer 1

H₀: µ₁ - µ₂ = 1

H₁: µ₁ - µ₂ < 1

z_o = -1.698

P-value = 0.045

1.0000	hypothesized value
7.1417	mean Before
6.3167	mean After
0.8250	mean difference (Before - After)
0.3571	std. dev.
0.1031	std. error
12	n
-1.698	z
.045	p-value (one-tailed, lower)
0.655	confidence interval 90.% lower
0.995	confidence interval 90.% upper
0.1695	margin of error

0.655 <= µ₁ - µ₂ <= 0.995

Please give me a thumbs-up if this helps you out. Thank you!