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 Watch 55 or 11 p.m. News? 18 or less 19 to 35 36 to 54 Older Total Yes 49 59 61 84 253 No 201 191 189 166 747 Total 250 258 250 250 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 a = .05. (Round your answer to 3 decimal places.) so H0:Mdependence Suppose a manager implementing a smoke-free workplace policy is interested in whether smoking affects worker accidents. Since the company has complete reports of on-the-job accidents, she draws a sample of names of workers who were involved in accidents during the last year. A similar sample from among workers who had no reported accidents in the last year is drawn. She interviews members of both groups to determine if they are smokers or not. The sample results are given in the following table. OntheJob Accident Smoker Yes No Row Total. Heavy 13 10 23 Moderate 7 9 16 Nonsmoker 14 13 27 Column total 34 32 66 Expected counts are below observed counts Accident No Accident Total Heavy 13 10 23 11.85 11. 15 Mode rate 7 9 16 8. 24 7. 76 Nonsmoker 14 13 27 13.91 13. 09 Total 34 32 66 ChiSq = 0.62, DF = 2, PValue = 0.733 ' (a) For each row and column total in the above table. nd the corresponding row/column percentage. (Round your answers to 2 decimal places.) Column 2