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	43	60	60	76	239
No	207	190	190	174	761
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

We need to the hypothesis testing for the Independence of the attributes

"Age.group" and "Watch 11 PM news"

As p1, p2, p3, and p4 are the proportions of all viewers in each age group who watch the 11 p.m. news.

For each group we will be getting one proportion value.(Proportions of viewer)

We use R to solve this problem

a)

## Making data frame ready to use for chi_square test a=c(43,60,60,76,207,190,190,174) a m=data.frame("18_or_less"=c(43,207),"19-35"=c(60,190),"36-54"=c(60,190),"55_or_older"=c(76,174)) m rownames(m)=c("Yes","No") ## Chisquare test for independence cs=chisq.test(m) cs # Here we will conclude using p-value criteria # Our null hypothesis is # Ho : p1=p2=p3=p4 i.e in all age proportion is same for watching 11pm's News alpha=0.05 if (cs$p.value < alpha) {cat("Reject Ho at 5% l.o.s i.e not all proportions are same")} # i.e proportions are independent across all age groups

Output:-

> ## Making data frame ready to use for chi_square test > a=c(43,60,60,76,207,190,190,174) > a [1] 43 60 60 76 207 190 190 174 > m=data.frame("18_or_less"=c(43,207),"19-35"=c(60,190),"36-54"=c(60,190),"55_or_older"=c(76,174)) > m X18_or_less X19.35 X36.54 X55_or_older 1 43 60 60 76 2 207 190 190 174 > rownames(m)=c("Yes","No") > ## Chisquare test for independence > cs=chisq.test(m) > cs      Pearson's Chi-squared test data: m X-squared = 11.98, df = 3, p-value = 0.00745 > # Here we will conclude using p-value criteria > # Our null hypothesis is > # Ho : p1=p2=p3=p4 i.e in all age proportion is same for watching 11pm's News > alpha=0.05 > if (cs$p.value < alpha) + {cat("Reject Ho at 5% l.o.s i.e not all proportions are same")} Reject Ho at 5% l.o.s i.e not all proportions are same> # i.e proportions are independent across all age groups

b)

############# Confidence interval for (p1-p4)################ n=250 p1=43/250;p1 p4=76/250;p4 d=p1-p4;d lower.CL=round(d-1.96*sqrt((p1*(1-p1))/n+(p4*(1-p4))/n),3) lower.CL upper.CL=round(d+1.96*sqrt((p1*(1-p1))/n+(p4*(1-p4))/n),3) upper.CL Confidence.Interval=cat("95% CI:","[",lower.CL,"to",upper.CL,"]") # This will be the 95 % CI for the difference between p1-p4

Output:-

> ############# Confidence interval for (p1-p4)################ > n=250 > p1=43/250;p1 [1] 0.172 > p4=76/250;p4 [1] 0.304 > d=p1-p4;d [1] -0.132 > > lower.CL=round(d-1.96*sqrt((p1*(1-p1))/n+(p4*(1-p4))/n),3) > lower.CL [1] -0.206 > > upper.CL=round(d+1.96*sqrt((p1*(1-p1))/n+(p4*(1-p4))/n),3) > upper.CL [1] -0.058 > Confidence.Interval=cat("95% CI:","[",lower.CL,"to",upper.CL,"]") 95% CI: [ -0.206 to -0.058 ]> # This will be the 95 % CI for the difference between p1-p4