To test independence of two attributes we use chi-square test.
The given information is:
number of rows m=2
number of column n=3
Hypothesis:
H0 - The Party Affiliation and Opinion on Tax Reform are independent.
H1 - The Party Affiliation and Opinion on Tax Reform are dependent.
Test statistic:
"\u03c7^2 = \\sum_{n}^{j=1} \\sum_{m}^{i=1} \\frac{(O_{ij}-E_{ij})^2}{E_{ij}} \\\\ \n\n\u03c7^2 = \\sum_{n}^{j=1} \\sum_{m}^{i=1} \\frac{O_{ij}^2}{E_{ij}} - N"
Is follow χ2 distribution with (m-1)(n-1) degrees of freedom.
Oij - observed frequency ithrow and jth column
Eij - expected frequency ith row and jth column
"E_{ij} = \\frac{(i^{th} \\; row \\; total) \\times (j^{th} \\;column \\;total)}{N} \\\\\n\nE_{11} = \\frac{R_1 \\times C_1}{N} \\\\\n\n= \\frac{285 \\times 202}{500} = 115.14 \\\\\n\nE_{12} = \\frac{R_1 \\times C_2}{N} \\\\\n\n= \\frac{285 \\times 150}{500} = 85.5 \\\\\n\nE_{13} = \\frac{R_1 \\times C_3}{N} \\\\\n\n= \\frac{285 \\times 148}{500} = 84.36 \\\\\n\nE_{21} = \\frac{R_2 \\times C_1}{N} \\\\\n\n= \\frac{215 \\times 202}{500} = 86.86 \\\\\n\nE_{22} = \\frac{R_2 \\times C_2}{N} \\\\\n\n= \\frac{215 \\times 150}{500} = 64.5 \\\\\n\nE_{23} = \\frac{R_2 \\times C_3}{N} \\\\\n\n= \\frac{215 \\times 148}{500} = 63.64 \\\\\n\n\u03c7^2 = \\frac{138^2}{115.14} + \\frac{83^2}{85.5} + \\frac{64^2}{84.36} + \\frac{64^2}{86.86} + \\frac{67^2}{64.5} + \\frac{84^2}{63.64} - 500 \\\\\n\n= 22.0147"
Critical value or table value at
"(m-1)(n-1)=(2-1)(3-1)= 1 \\times 2=2" degrees of freedom and 5 % level of significance.
"Tab \\; \u03c7^2 = 5.991"
"Cal \\; \u03c7^2 > Tab \\; \u03c7^2"
Reject null hypothesis at 5 % level of significance.
There is enough evidence to conclude that the Party Affiliation and Opinion on Tax Reform are dependent at 5 % level of significance.
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