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A television station wishes to study the relationship between viewership of its 11 p.m. news program...

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Statistics

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 or less19 to 3536 to 5455 or OlderTotal
Yes45516784247
No205199183166753
Total2502502502501,000


(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 α = .05. (Round your answer to 3 decimalplaces.)


χ2χ2 =           

so (Click to select)Do not rejectReject 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: [  , ]

Answer & Explanation Solved by verified expert
4.0 Ratings (771 Votes)

a)

Applying chi square test of independence:
Expected Ei=row total*column total/grand total <18 19-35 36-54 >55 Total
yes 61.7500 61.7500 61.7500 61.7500 247.00
no 188.2500 188.2500 188.2500 188.2500 753.00
total 250.00 250.00 250.00 250.00 1000.00
chi square    χ2 =(Oi-Ei)2/Ei <18 19-35 36-54 >55 Total
yes 4.544 1.871 0.446 8.017 14.8785
no 1.490 0.614 0.146 2.630 4.8805
total 6.0339 2.4853 0.5928 10.6470 19.759
test statistic X2 = 19.759

Reject H0: independence

b)

estimated difference in proportion   =p̂1-p̂2   = -0.1560
std error Se =√(p̂1*(1-p̂1)/n1+p̂2*(1-p̂2)/n2) = 0.0385
for 95 % CI value of z= 1.960
margin of error E=z*std error = 0.0755
lower bound=(p̂1-p̂2)-E= -0.2315
Upper bound=(p̂1-p̂2)+E= -0.0805
from above 95% confidence interval for difference in population proportion =(-0.231,-0.081)
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