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 50 45 65 80 240 No 200 205 185 170 760 Total 250 250 250 250 1,000 (a) Let Pj. 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 Hg that p1. P2. P3- and Pa 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 = HO: independence so Reject Do not reject (b) Compute a 95 percent confidence interval for the difference between P1 and P4. (Round your answers to 3 decimal places. Negative amounts should be indicated by a minus sign.) 95% CI: [
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 50 45 65 80 240 No 200 205 185 170 760 Total 250 250 250 250 1,000 (a) Let Pj. 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 Hg that p1. P2. P3- and Pa 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 = HO: independence so Reject Do not reject (b) Compute a 95 percent confidence interval for the difference between P1 and P4. (Round your answers to 3 decimal places. Negative amounts should be indicated by a minus sign.) 95% CI: [
Chapter1: Making Economics Decisions
Section: Chapter Questions
Problem 1QTC
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7. 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.

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 ?
18 or less
19 to 35 36 to 54
Older
Total
Yes
50
45
65
80
240
No
200
205
185
170
760
Total
250
250
250
250
1,000
(a) Let P1. P2. P3, and Pa 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 Hg 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 =
so
H0: independence
Reject
Do not reject
(b) Compute a 95 percent confidence interval for the difference between p, and p4. (Round your answers to 3 decimal places.
Negative amounts should be indicated by a minus sign.)
95% CI: [
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