Homework9: Chapter 15
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11.Inthe general population, political party affiliation is 30%Republican, 55% Democratic, and 15% other. To determine whether thisdistribution is also found among the elderly, in a sample of 100senior citizens, we find 18 Republicans, 64 Democrats, and 18 other.
(a)What procedure should we perform?
Thebest procedure to perform in this question is to conduct a onewaychi square procedure since the data given has frequency of categorymembership along each variable.
(b)What are H0and Ha?
Itis claimed from the data to find out whether the affiliation to apolitical party is as follows
Republican– 30%
Democrats– 55%
Other– 15%
Withthis claim the null and alternative hypotheses will be:
Ho:The elderly affiliation to a political party is distribted with 30%Republican, 55% Democrats and 15% other
Ha:The elderly political party affiliation is not distributed in thisway
(c)What is fefor each group?
Weshall calculate fe for each of the three groups. To get fe wemultiply the percentage of the particular group with the total numberof senior citizens
Fevalue for the democrats therefore = 0.55 * 100 = 55
Fevale for the republicans = 0.3*100 = 30
Fefor others = 0.15*100 = 15
(d)Compute X2obt.
Wherefo is the observed frequency and fe represents the expected frequency
ThereforeX^{2}_{obt}=(1830)^{2}/30+(64 – 55)^{2}/55+(1815)^{2}/15
=144/30 +81/55 +9/15
= 4.8 + 1.47 + 0.6 = 6.87
(e)With α = .05, what do you conclude about party affiliation amongsenior citizens?
Beforewe can conclde, we have to compare the chi square obtained orcalculated with the value taken from the chi square tables
Sincewe are performing a oneway chi square procedure, the degrees offreedom is given by subtracting one from the number of categories. Inthis case
Df= 3 1 = 2
Fromthe tables, a chi sqare value with 0.05 significance level and 2degrees of freedom is 5.9915
Itis clear that the obtained chi square value (6.87) is bigger than thecritical value of 5.9915 at 0.05 level of significance. In this casewe therefore reject the null hypothesis thus concluding that theaffiliation of the elderly to the political party is not distributedwith 30% Republican, 55% Democrats and 15% other
14.Thefollowing data reflect the frequency with which people voted in thelast election and were satisfied with the officials elected:
Satisfied
YesNo
48 
35 
33 
52 
Yes
Voted
No

What procedure should we perform?
Inthis case, we have data with frequency of category membership alongtwo variable which suggests a twoway chisquare test

What are H0 and Ha?
Ho:People voted in the election and their satisfaction with the electedofficials is independent
Ha:People voted in the election and their satisfaction with electedofficials is dependent
thishypothesis will be tested at 0.05 level of significance

What is fe in each cell?
Fora two way chi square test, the expected frequency is got by dividingthe product of the total observed frequency of the row and that ofthe column and the total number
i.efe = (cell’s row total)(cell’s column total)/N
thereforewe have these total
forthose who voted yes row total = 48 + 35 = 83
forthose who voted No row total = 33+52 = 85
forcolmn total we have
SatisfiedYes = 48+33 = 81
SatisfiedNo = 35 + 52 = 87
TotalN = 83+85 = 81+87 = 168
Thereforefe for the first cell (voted yes Satisfied yes) = 83*81/168 = 40.018
Fefor the second cell (Voted Yes Satisfied No) = 83*87/168 = 42.985
Fefor the third cell (Voted No, satisfied Yes) = 85*81/168 = 40.982
Fefor the last cell (voted No, satisfied No) = 85*87/168 = 44.018this is summarized as follows

Yes
No
Row total
Yes
Fe = 40.018
Fe=42.982
83
No
Fe=40.982
Fe=44.018
85
Column total
81
87
N=168

Compute X 2obt.
=(4840.018)^{2}/40.018+ (3542.982)^{2}/42.982+(3340.982)^{2}/40.982+(52 44.018)^{2}/44.018

With α = .05, what do you conclude about the correlation between these variables?
Firstof all we calculate the degrees of freedom which is given as
Df= (nmber of rows – 1)(number of columns – 1) = (2 – 1) (2 –1) = 1
Readingthe critical value of the chi square with 1 degree of freedom and0.05 significant level we get 3.842
Itis therefore clear that the calculated value of the chi square =6.076 is greater than the tabulated value of the chi square at 1degree of freedom and 0.05 level of significance = 3.842. Thereforewe have the evidence to allege that the null hypothesis is not truethus we conclude that the two variables the people voted in theelection and their satisfaction of the elected officials aredependent.
(f)How consistent is this relationship?
Tocalculate the consistence of this relationship, we find the squareroot of obtained chi square divided by the total number (N)and theobtained chi square
Mathematically,C =
= = 0.2159
Thisindicates that the relationship is consistent
15.Astudy determines the frequency of the different political partyaffiliations for male and female senior citizens. The following dataare obtained:
RepublicanDemocrat Other
18 
43 
14 
39 
23 
18 
Male
Gender
Female

What procedure should we perform?
Atwoway chi square procedure is appropriate for this since themembership of the category is along two variables

What are H0 and Ha?
Ho:gender and political party affiliation are independent in thepopulation
Ha:gender and political party affiliation are not independent in thepopulation

What is fe in each cell?
Columntotal for republican = 57, for Democrat = 66, for other = 32
Rowtotal for male = 75, for female = 80
TotalN = 155
Fefor each cell is as follows
Forthe cell (Male, Republican) = 75*57/155 = 27.581
Cell(female, Republican) = 80*57/155 = 29.419
Cell(male, Democrat) = 75*66/155 = 31.935
Cell(female, Democrat) = 80*66/155 = 34.065
Cell(Male, Other) = 75*32/155 = 15.484
Cell(female, other) = 80*32/155 = 16.516
Ina table summary, we have the cell fe’s as

Republican
Democrat
Other
Row total
Male
27.581
31.935
15.484
75
Female
29.419
34.065
16.516
80
Colmn total
57
66
32
155

Compute X2obt.
X2obt= (18 – 27.581)^{2}/27.581+ (43 – 31.935)^{2}/31.936+ (14 – 15.484)^{2}/15.484+(39 – 29.419)^{2}/29.419+ ( 23 – 34.065)^{2}/34.065+ (18 – 16.516)^{2}/16.516
=3.328 + 3.834 + 0.142 +3.12 +3.594 +0.133
=14.152

With α = .05, what do you conclude about gender and party affiliation in the population of senior citizens?
Toconclude we must get the critical value of the chi square from thechi square tables with degrees of freedom = (2 – 1) ( 3 – 1) = 2and 0.05 level of significance
Fromthe tables, the critical value = 5.9915
Sincethe calculated value of the chi square (14.152) exceeds the criticalvalue at 0.05 significant level with 2 degrees of freedom (5.9915),we reject the null hypothesis and conclude that the gender andpolitical party affiliation are dependent in the population.

How consistent is this relationship?
Consistency= = 0.289 suggesting that the consistence of the relationship
17.Aftertesting 40 participants, a significant X2obt of 13.31 was obtained.With α = .05 and df=2, how would this result be reported in a publication?
Wehave the total number of participants N= 40
Thecalculated chi square value = X^{2}obt= 13.31
Levelof significance = 0.05 and degrees of freedom = 2
Wefirst of all state the null and alternative hypotheses which are
Ho:The variables presented are independent
Ha:The variables presented are not independent
Wetherefore go ahead and find the tabulated chi square value with 2degrees of freedom and 0.05 level of significance which is thepvalue
Fromthe table we get that X^{2}crit= 5.9915
Inconclusion as we can see, X^{2}obt= 13.31 is larger than X^{2}crit= 5.9915, a clear indication that the results are significant. Thuswe refuse to accept the null hypothesis and conclude that thevariables are not independent.
Ina report form this is reported as X^{2}(2, N =40) = 13.31, p<0.05