A paper described a survey of 507 undergraduate students at a state university in the southwestern region of the United States. Each student in the sample was classified according to class standing (freshman, sophomore, junior, or senior) and body art category (body piercingsonly, tattoos only, both tattoos and body piercings, no body art).Use the data in the accompanying table to determine if there is evidence of an association between class standing and response to the body art question. Assume that it is reasonable to regard the sample of students as representative of the students at this university. Useα = 0.01.FreshmanSophomoreJuniorSeniorBodyPiercingsOnly64442021TattoosOnly711917Both BodyPiercings andTattoos1710726Calculate the test statistic. (Round your answer to two decimal places.)x² = |NoBodyArt86654657Use technology to calculate the P-value. (Round your answer to four decimal places.)P-value =

For the given data Perform chi square test

Answered: A paper described a survey of 507…

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 television station wishes to study the relationship between viewership of its 11 p.m. news program and viewer age (18 years or less, 19 to 35, 36 to 54, 55 or older). A sample of 250 television viewers in each age group is randomly selected, and the number who watch the station’s 11 p.m. news is found for each sample. The results are given in the table below. Age Group Watch11 p.m. News? 18 or less 19 to 35 36 to 54 55 or Older Total Yes 37 46 59 83 225 No 213 204 191 167 775 Total 250 250 250 250 1,000 (a) Let p1, p2, p3, and p4 be the proportions of all viewers in each age group who watch the station’s 11 p.m. news. If these proportions are equal, then whether a viewer watches the station’s 11 p.m. news is independent of the viewer’s age group. Therefore, we can test the null hypothesis H0 that p1, p2, p3, and p4 are equal by carrying out a chi-square test for independence. Perform this test by setting α = .05. (Round your answer to 3 decimal places.)…
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