Question #152035
The following table represents a sample of 500 people and their opinion on Tax Reform. Do we have enough evidence to conclude that Party Affiliation and Opinion on Tax Reform are dependent on 5% level of significance.

Affiliation/Opinion on Tax | Reforms | Favor | Indifferent | Opposed
---------------------------------------------------------------------------------------------------------------
Democrat
---------------------------------------------------------------------------------------------------------------
Republican
1
Expert's answer
2020-12-25T12:44:03-0500


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:

χ2=nj=1mi=1(OijEij)2Eijχ2=nj=1mi=1Oij2EijNχ^2 = \sum_{n}^{j=1} \sum_{m}^{i=1} \frac{(O_{ij}-E_{ij})^2}{E_{ij}} \\ χ^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.

O​​​​​ij - observed frequency i​​​​​​throw and j​​​​​​th column

E​​​​​​ij - expected frequency i​​​​​​th row and j​​​​​​th column

Eij=(ith  row  total)×(jth  column  total)NE11=R1×C1N=285×202500=115.14E12=R1×C2N=285×150500=85.5E13=R1×C3N=285×148500=84.36E21=R2×C1N=215×202500=86.86E22=R2×C2N=215×150500=64.5E23=R2×C3N=215×148500=63.64χ2=1382115.14+83285.5+64284.36+64286.86+67264.5+84263.64500=22.0147E_{ij} = \frac{(i^{th} \; row \; total) \times (j^{th} \;column \;total)}{N} \\ E_{11} = \frac{R_1 \times C_1}{N} \\ = \frac{285 \times 202}{500} = 115.14 \\ E_{12} = \frac{R_1 \times C_2}{N} \\ = \frac{285 \times 150}{500} = 85.5 \\ E_{13} = \frac{R_1 \times C_3}{N} \\ = \frac{285 \times 148}{500} = 84.36 \\ E_{21} = \frac{R_2 \times C_1}{N} \\ = \frac{215 \times 202}{500} = 86.86 \\ E_{22} = \frac{R_2 \times C_2}{N} \\ = \frac{215 \times 150}{500} = 64.5 \\ E_{23} = \frac{R_2 \times C_3}{N} \\ = \frac{215 \times 148}{500} = 63.64 \\ χ^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 \\ = 22.0147

Critical value or table value at

(m1)(n1)=(21)(31)=1×2=2(m-1)(n-1)=(2-1)(3-1)= 1 \times 2=2 degrees of freedom and 5 % level of significance.

Tab  χ2=5.991Tab \; χ^2 = 5.991

Cal  χ2>Tab  χ2Cal \; χ^2 > Tab \; χ^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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