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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 55 or Older Watch 11 p.m. News? Yes 18 or less 19 to 35 36 to 54 59 191 49 201 61 189 84 166 Total 253 747 No Total 250 250 250 250 1,000

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 55 or Older Watch 11 p.m. News? Yes 18 or less 19 to 35 36 to 54 59 191 49 201 61 189 84 166 Total 253 747 No Total 250 250 250 250 1,000

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter10: Statistics
Section10.6: Summarizing Categorical Data
Problem 10CYU
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Therefore, we can test the null hypothesis Ho 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.) x^2 = 13.900 Reject HO: independence so (b) Compute a 95 percent confidence interval for the difference between Pi and p4. (Round your answers to 3 decimal places. Negative amounts should be indicated by a minus sign.) 95% Cl: [
Transcribed Image Text:Therefore, we can test the null hypothesis Ho 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.) x^2 = 13.900 Reject HO: independence so (b) Compute a 95 percent confidence interval for the difference between Pi and p4. (Round your answers to 3 decimal places. Negative amounts should be indicated by a minus sign.) 95% Cl: [
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? Yes No 18 or less 19 to 35 36 to 54 0lder 59 191 Total 61 49 201 84 166 253 747 189 Total 250 250 25Q 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 Ho 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.), X^2 = 13.900 Reject HO: independence so
Transcribed Image Text: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? Yes No 18 or less 19 to 35 36 to 54 0lder 59 191 Total 61 49 201 84 166 253 747 189 Total 250 250 25Q 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 Ho 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.), X^2 = 13.900 Reject HO: independence so
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