Consider the information above and in question 1. It is conjectured that the mean number of people attending all 2017 Veterans Day celebrations is 145, and of interest is totest this conjecture versus the alternative that the mean number of people attending all 2017 Veterans Day celebrations is different from 145 peopleTo test the hypothesis, a simple random sample of 41 Veterans Day celebrations was selected and the number of people attending each was recorded. The mean number ofpeople attending for this sample of 41 celebrations was 157, with a standard deviation of 23 people. The distribution of the data was bimodal. If appropriate, use thisinformation to test the hypotheses stated in question 3 at the a= .05 level of significance.4Are the assumptions met?Yes, because there is a simple random sampleYes, because there is a simple random sample and the population is normally distributedYes, because there is a simple random sample and the sample size is large enough for the Central Limit TheoremYes, because the population is normally distributedNo, there is no simple random sample5Is the population standard deviation known?YesNo

Given that: The null hypothesis is that the mean number of people attending all 2017 Veterans Day…

Answered: Consider the information above and in…

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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An adventure company runs two obstacle courses, Fundash and Coolsprint. The designer of the courses suspects that the mean completion time of Fundash is not equal to the mean completion time of Coolsprint. To test this, she selects 250 Fundash runners and 285 Coolsprint runners. (Consider these as random samples of the Fundash and Coolspring runners.) The 250 Fundash runners complete the course with a mean time of 76.7 minutes and a standard deviation of 5.7 minutes. The 285 Coolsprint runners complete the course with a mean time of 77.5 minutes and a standard deviation of 5.6 minutes. Assume that the population standard deviations of the completion times can be estimated to be the sample standard deviations, since the samples that are used to compute them are quite large. At the 0.05 level of significance, is there enough evidence to support the claim that the mean completion time, u, of Fundash is not equal to the mean completion time, µa, of Coolsprint? Perform a two-tailed test.…
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USA TODAY (October 14, 2016) reported that Americans spend 4.1 hours per weekday checking work email. This was an estimate based on a survey of 1,004 white-collar workers in the United States.
An adventure company runs two obstacle courses, Fundash and Coolsprint. The designer of the courses suspects that the mean completion time of Fundash is not equal to the mean completion time of Coolsprint. To test this, she selects 240 Fundash runners and 200 Coolsprint runners. (Consider these as random samples of the Fundash and Coolspring runners.) The 240 Fundash runners complete the course with a mean time of 76.7 minutes and a standard deviation of 5.2 minutes. The 200 Coolsprint runners complete the course with a mean time of 77.5 minutes and a standard deviation of 5.4 minutes. Assume that the population standard deviations of the completion times can be estimated to be the sample standard deviations, since the samples that are used to compute them are quite large. At the 0.05 level of significance, is there enough evidence to support the claim that the mean completion time, μ1 , of Fundash is not equal to the mean completion time, μ2 , of Coolsprint?…
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