An adventure company runs two obstacle courses, Fundash and Coolsprint. The designer of the courses suspects that the mean completion time of Fundash isnot equal to the mean completion time of Coolsprint. To test this, she selects 265 Fundash runners and 245 Coolsprint runners. (Consider these as randomsamples of the Fundash and Coolspring runners.) The 265 Fundash runners complete the course with a mean time of 63.6 minutes and a standard deviation of2.8 minutes. The 245 Coolsprint runners complete the course with a mean time of 63.0 minutes and a standard deviation of 2.9 minutes. Assume that thepopulation standard deviations of the completion times can be estimated to be the sample standard deviations, since the samples that are used to computethem are quite large. At the 0.01 level of significance, is there enough evidence to support the claim that the mean completion time, μ₁, of Fundash is not equalto the mean completion time, μ₂, of Coolsprint? Perform a two-tailed test. Then complete the parts below.Carry your intermediate computations to at least three decimal places. (If necessary, consult a list of formulas.)(a) State the null hypothesis H. and the alternative hypothesis H₁.HOH₁ :0(b) Determine the type of test statistic to use.Chi-SquareDegrees of freedom:(c) Find the value of the test statistic. (Round to three or more decimal places.)0(d) Find the two critical values at the 0.01 level of significance. (Round to three or more decimal places.)and(e) Can we support the claim that the mean completion time of Fundash is not equal to the meancompletion time of Coolsprint?Yes NoHX090#0OXS000=0 0≤0 0200<0Р

Followings are the Explanation of the question Use the two sample Z test Sample Mean X1=…

Answered: (a) State the null hypothesis H and the…

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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You are conducting a multinomial hypothesis test (α = 0.05) for the claim that all 5 categories are equally likely to be selected. Complete the table. Category ObservedFrequency ExpectedFrequency A 10 B 16 C 8 D 20 E 14 What is the chi-square test-statistic for this data? (Report answer accurate to three decimal places.)χ2=What are the degrees of freedom for this test? d.f.= What is the p-value for this sample? (Report answer accurate to four decimal places.) p-value = The p-value is... less than (or equal to) αα greater than αα This test statistic leads to a decision to... reject the null accept the null fail to reject the null accept the alternative As such, the final conclusion is that... There is sufficient evidence to warrant rejection of the claim that all 5 categories are equally likely to be selected. There is not sufficient evidence to warrant rejection of the claim that all 5 categories are equally likely to be selected. The sample data…
Can you please check my work
Question Help Previously, 10.4% of workers had a travel time to work of more than 60 minutes. An urban economist believes that the percentage has increased since then. She randomly selects 75 workers and finds that 17 of them have a travel time to work that is more than 60 minutes. Test the economist's belief at the α=0.1 level of significance. What are the null and alternative hypotheses? H0: ▼ pp muμ alphaα ▼ greater than> less than< not equals≠ equals= nothing versus H1: ▼ pp muμ alphaα ▼ not equals≠ less than< greater than> equals= nothing (Type integers or decimals. Do not round.) Because np01−p0=nothing ▼ equals= not equals≠ less than< greater than> 10, the normal model ▼ may not may be used to approximate the P-value. (Round to one decimal place as needed.) Find the P-value. P-value=nothing (Round to three decimal places as needed.) Is there sufficient evidence to support the…
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