comparison is made between two bus lines to determine if arrival times of their regular buses from Denver to Durango are off schedule by the same amount of time. For 51 randomly selected runs, bus line A was observed to be off schedule an average time of 53 minutes, with standard deviation 15 minutes. For 60 randomly selected runs, bus line B was observed to be off schedule an average of 60 minutes, with standard deviation 11 minutes. Do the data indicate a significant difference in average off-schedule times? Use a 5% level of significance. What are we testing in this problem? difference of proportionsdifference of means single meanpaired differencesingle proportion (a) What is the level of significance? State the null and alternate hypotheses. H0: ?1 = ?2; H1: ?1 < ?2H0: ?1 = ?2; H1: ?1 ≠ ?2 H0: ?1 > ?2; H1: ?1 = ?2H0: ?1 = ?2; H1: ?1 > ?2 (b) What sampling distribution will you use? What assumptions are you making? The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations.The Student's t. We assume that both population distributions are approximately normal with known standard deviations. The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations.The standard normal. We assume that both population distributions are approximately normal with known standard deviations. What is the value of the sample test statistic? (Test the difference ?1 − ?2. Round your answer to three decimal places.) (c) Find (or estimate) the P-value. P-value > 0.5000.250 < P-value < 0.500 0.100 < P-value < 0.2500.050 < P-value < 0.1000.010 < P-value < 0.050P-value < 0.010 Sketch the sampling distribution and show the area corresponding to the P-value. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level ?? At the ? = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.At the ? = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant. At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant.At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant. (e) Interpret your conclusion in the context of the application. There is sufficient evidence at the 0.05 to conclude that there is a difference in average off schedule times.There is insufficient evidence at the 0.05 to conclude that there is a difference in average off schedule times.

comparison is made between two bus lines to determine if arrival times of their regular buses from Denver to Durango are off schedule by the same amount of time. For 51 randomly selected runs, bus line A was observed to be off schedule an average time of 53 minutes, with standard deviation 15 minutes. For 60 randomly selected runs, bus line B was observed to be off schedule an average of 60 minutes, with standard deviation 11 minutes. Do the data indicate a significant difference in average off-schedule times? Use a 5% level of significance. What are we testing in this problem? difference of proportionsdifference of means single meanpaired differencesingle proportion (a) What is the level of significance? State the null and alternate hypotheses. H0: ?1 = ?2; H1: ?1 < ?2H0: ?1 = ?2; H1: ?1 ≠ ?2 H0: ?1 > ?2; H1: ?1 = ?2H0: ?1 = ?2; H1: ?1 > ?2 (b) What sampling distribution will you use? What assumptions are you making? The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations.The Student's t. We assume that both population distributions are approximately normal with known standard deviations. The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations.The standard normal. We assume that both population distributions are approximately normal with known standard deviations. What is the value of the sample test statistic? (Test the difference ?1 − ?2. Round your answer to three decimal places.) (c) Find (or estimate) the P-value. P-value > 0.5000.250 < P-value < 0.500 0.100 < P-value < 0.2500.050 < P-value < 0.1000.010 < P-value < 0.050P-value < 0.010 Sketch the sampling distribution and show the area corresponding to the P-value. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level ?? At the ? = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.At the ? = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant. At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant.At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant. (e) Interpret your conclusion in the context of the application. There is sufficient evidence at the 0.05 to conclude that there is a difference in average off schedule times.There is insufficient evidence at the 0.05 to conclude that there is a difference in average off schedule times.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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A comparison is made between two bus lines to determine if arrival times of their regular buses from Denver to Durango are off schedule by the same amount of time. For 51 randomly selected runs, bus line A was observed to be off schedule an average time of 53 minutes, with standard deviation 15 minutes. For 60 randomly selected runs, bus line B was observed to be off schedule an average of 60 minutes, with standard deviation 11 minutes. Do the data indicate a significant difference in average off-schedule times? Use a 5% level of significance.

What are we testing in this problem?

difference of proportionsdifference of means single meanpaired differencesingle proportion

(a) What is the level of significance?

State the null and alternate hypotheses.

H₀: ?₁ = ?₂; H₁: ?₁ < ?₂H₀: ?₁ = ?₂; H₁: ?₁ ≠ ?₂ H₀: ?₁ > ?₂; H₁: ?₁ = ?₂H₀: ?₁ = ?₂; H₁: ?₁ > ?₂

(b) What sampling distribution will you use? What assumptions are you making?

The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations.The Student's t. We assume that both population distributions are approximately normal with known standard deviations. The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations.The standard normal. We assume that both population distributions are approximately normal with known standard deviations.

What is the value of the sample test statistic? (Test the difference ?₁ − ?₂. Round your answer to three decimal places.)

(c) Find (or estimate) the P-value.

P-value > 0.5000.250 < P-value < 0.500 0.100 < P-value < 0.2500.050 < P-value < 0.1000.010 < P-value < 0.050P-value < 0.010

Sketch the sampling distribution and show the area corresponding to the P-value.

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level ??

At the ? = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.At the ? = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant. At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant.At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.

(e) Interpret your conclusion in the context of the application.

There is sufficient evidence at the 0.05 to conclude that there is a difference in average off schedule times.There is insufficient evidence at the 0.05 to conclude that there is a difference in average off schedule times.

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