Question

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 7.1 7.5 7.4 6.96.4 7.2 7.2 7.6 7.1 After 6.5 6.6 6.3 6.2 6.8 6.2 6.0 5.7 6.2 5.8 7.4 6.7 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, H1-42. 2) Ho :ui-u2 = 3) H1 : Mi-42 < Find the test statistic. Round your answers to three decimal places (e.g. 98.765). 20 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 <M1 M2

Answer 1

Answer :- The time delay is measured for a city Street. Variability in this value is estimated as (σ = 0.3) (for time in minute) by experience. Traffic signs and lights were recently altered to facilitated traffic and reduce delays.

Given that :-

• σ = 0.3

• α = 0.1

• To find the sample mean (X̄1) :-

=> X̄1 = ∑X1 /n

= (7.5 + 7.1 + 7.1 + 7.1 + 7.5 + 7.4 + 6.9 + 6.4 + 7.2 + 7.2 + 7.6 + 7.1) /12

= 86.1 /12

= 7.175

[ X̄1 = 7.175 ]

• To find the sample mean (X̄2) :-

=> X̄2 = ∑X2 /n

= (6.5 + 6.6 + 6.3 + 6.2 + 6.8 + 6.2 + 6.0 + 5.7 + 6.2 + 5.8 + 7.4 + 6.7) /12

= 76.4 /12

= 6.367

[ X̄2 = 6.367 ]

a)

1) The parameter of interest is the difference in the mean time delay before and after the rearrangement :-

=> μ1 - μ2

= 7.175 - 6.367

= 0.808

[ μ1 - μ2 = 0.808 ]

2) Null Hypothesis,  Ho : μ1 - μ2 = 1

3) Alternative Hypothesis, H1 : μ1 - μ2 < 1

• The test statistic is calculated as follows :-

=> Zo = [(X̄1 - X̄2) - (μ1 - μ2)] /σ √(1/n1) + (1/n2)

= [(7.175 - 6.367) - 1] /0.3 √(1/12) + (1/12)

= (0.808 - 1) /0.3 √(0.0833 + 0.0833)

= -0.192 /0.3 √0.1666

= -0.192 /0.3(0.4082)

= -0.192 /0.1225

= -1.5673

[ Zo = -1.567 ]

•  P-value is calculated as follows :-

=> P-value = P(Z < Zo)

= P(Z < -1.567)

(from Z table)

= 0.059

[ P-value = 0.059 ]

b) 90℅ confidence interval (C.I.) for the difference in the mean time delay is calculated as follows :-

=> 90℅ Confidence Interval (C.I.)

• α = 0.1

• (α/2) = 0.05

• Z(α/2) :-

=> [ Z(α/2) = 1.645 ]

• Now, the 90℅ confidence interval for the difference in the mean time delay is :-

=> (X̄1 - X̄2) ± [Z(α/2) σ √(1/n1) + (1/n2)]

=> (7.175 - 6.367) ± [1.645 (0.3) √(1/12) + (1/12)]

=> 0.808 ± [1.645 (0.3) √(0.0833 + 0.0833)]

=> 0.808 ± [1.645 (0.3) √0.1666 ]

=> 0.808 ± [ 1.645 × 0.3 × 0.4082 ]

=> 0.808 ± 0.2014

=> 0.6066, 1.022