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	49	52	67	79	247
No	201	198	183	171	753
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)RejectDo not reject 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:

Expected	Ei=row total*column total/grand total	18 or less	19 to 35	36 to 54	55 or older	Total
	Yes	61.75	61.75	61.75	61.75	247
	No	188.25	188.25	188.25	188.25	753
	total	250	250	250	250	1000
chi square    χ2	=(Oi-Ei)2/Ei	18 or less	19 to 35	36 to 54	55 or older	Total
	Yes	2.6326	1.5395	0.4464	4.8188	9.437
	No	0.8635	0.5050	0.1464	1.5807	3.096
	total	3.496	2.044	0.593	6.400	12.5329

X² =12.533

reject H₀: independence

b)

95% CI: = -0.196 , -0.044