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 6 = 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 7.2 7.1 7.1 7.3 7.3 7.07 6.4 7.2 7.2 7.5 7.1 After 6.1 6.6 6.4 6.4 6.7 5.6 6.0 5.7 6.2 6.0 7.3 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. i) The parameter of interest is the difference in the mean time delay before and after the rearrangement, M1 - 12 2) Họ : H1 - 2 = 1 3) H: Hi-H2 1 Find the test statistic. Round your answers to three decimal places (e.g. 98.765). -1.494 Zo P-value = 0.068 (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 0.616 }<H1-12 1.019

Answer 1

First of all, calculate the differences of time delay from before to after to get the data as-

Before	7.2	7.1	7.1	7.3	7.3	7	7	6.4	7.2	7.2	7.5	7.1
After	6.1	6.6	6.4	6.4	6.7	5.6	6	5.7	6.2	6	7.3	6.9
Difference	1.1	0.5	0.7	0.9	0.6	1.4	1	0.7	1	1.2	0.2	0.2

Then calculate the mean of differences to get -

$\overline{d} = \frac{ \sum d_{i}}{n} = \frac{1.1+0.5+0.7+0.9+0.6+....+0.2}{12} = \frac{9.5}{12} = 0.791667$

As we want to test if the difference in the time delay is 1 minute or less, so the hypothesis to be tested is -

Null Hypothesis - $H_{0}: \mu_{1} - \mu_{2} = 1$

Alternate Hypothesis - $H_{1}: \mu_{1} - \mu_{2} < 1$

Then the test statistic is -

$Z_{0} = \frac{\overline{d} - (\mu_{1} - \mu_{2})}{\frac{\sigma}{\sqrt{n}}} = \frac{0.791667 - 1}{\frac{0.3}{\sqrt{12}}} = -2.40563$

So, test statistic is: Z₀ = -2.406

And as its a lower tailed test, so the p-value of test is -

P(Z < -2.406) = 0.008

As p-value is less than the significance level of 0.1, so we reject the null hypothesis and conclude that there is enough evidence in the data to support the claim that the mean population difference of delay time is less than 1 min.

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b)

Critical value of test-statistic = Z_0.05 = 1.645

Thus, required confidence interval is -

$\overline{d} \pm Z_{\frac{\alpha}{2}}(\frac{\sigma}{\sqrt{n}})$

Thus, required confidence interval is -

$\\ 0.791667 \pm 1.645(\frac{0.3}{\sqrt{12}})\\ \\ \Rightarrow 0.791667 \pm 0.142462$

Thus, required confidence interval is -

(0.649205, 0.934128)

Rounding up to 3 decimal places gives-

$0.649 < \mu_{1} - \mu_{2} < 0.934$

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