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Classwork 12.2 Test for Independence and Homogeneity of Proportion 1. The contingency table below shows the results of a rand
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Answer #1

Null hypothesis: Ho :Opinion and Party Affiliation are independent

Alternate hypothesis: Ha :Opinion and Party Affiliation are dependent

degree of freedom(df) =(rows-1)*(columns-1)= 4
for 4 df and 0.05 level of signifcance critical region       χ2= 9.488
Applying chi square test of independence:
Expected Ei=row total*column total/grand total Approve Disapprove No opinion Total
Republican 38.7600 22.8000 14.4400 76
Democrat 46.9200 27.6000 17.4800 92
Independent 16.3200 9.6000 6.0800 32
total 102 60 38 200
chi square    χ2 =(Oi-Ei)2/Ei Approve Disapprove No opinion Total
Republican 0.2708 0.3439 0.0134 0.6281
Democrat 0.2022 0.4696 0.0155 0.6872
Independent 2.4475 4.2667 0.0011 6.7152
total 2.9205 5.0801 0.0299 8.0305
test statistic X2 = 8.0305
since test statistic does not falls in rejection region we fail to reject null hypothesis
we do not have have sufficient evidence to conclude that :Opinion and Party Affiliation are dependent at 0.05 level of significance
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