A television station wishes to study the relationship betweenviewership of its 11 p.m. news program and viewer age (18 years orless, 19 to 35, 36 to 54, 55 or older). A sample of 250 televisionviewers in each age group is randomly selected, and the number whowatch the station’s 11 p.m. news is found for each sample. Theresults are given in the table below.
| | Age Group | | |
Watch
11 p.m. News? | 18 orless | 19 to 35 | 36 to 54 | 55 orOlder | Total |
| Yes | 49 | 52 | 67 | 79 | 247 |
| No | 201 | 198 | 183 | 171 | 753 |
| Total | 250 | 250 | 250 | 250 | 1,000 |
|
(a) Let p1,p2, p3, andp4 be the proportions of all viewers in eachage group who watch the station’s 11 p.m. news. If theseproportions are equal, then whether a viewer watches the station’s11 p.m. news is independent of the viewer’s age group. Therefore,we can test the null hypothesis H0 thatp1, p2,p3, and p4 are equal bycarrying out a chi-square test for independence. Perform this testby setting ? = .05. (Round your answer to 3 decimalplaces.)
?2
=Â Â Â Â Â Â Â Â Â Â Â
so
H0: independence
(b) Compute a 95 percent confidence interval forthe difference between p1 andp4. (Round your answers to 3 decimalplaces. Negative amounts should be indicated by a minussign.)
95% CI: [ , ]
