Question

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 11 p.m. News? 18 or less 19 to 35 36 to 54 55 or Older Total Yes 42 57 61 82 242 No 208 193 189 168 758 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.) χ2χ2 = so (Click to select)Do not rejectReject H0: independence (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: [ , ]

Answer 1

a)

Applying chi square test of independence:

Expected	Ei=row total*column total/grand total	p1	p2	p3	p4	Total
	Yes	60.5000	60.5000	60.5000	60.5000	242
	No	189.5000	189.5000	189.5000	189.5000	758
	total	250	250	250	250	1000
chi square    χ2	=(Oi-Ei)2/Ei	p1	p2	p3	p4	Total
	Yes	5.6570	0.2025	0.0041	7.6405	13.5041
	No	1.8061	0.0646	0.0013	2.4393	4.3113
	total	7.4631	0.2671	0.0055	10.0798	17.8155
test statistic X2 =		17.815

tReject H0: independence

b)

from above 95% confidence interval for population mean =(-0.234,-0.086)