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In order to receive credit for the homework, you must show all steps for doing a chi square test of independence. These steps
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The Expected value data are in the table below. Each Cell = Row total * Column Total/N. N = 500

Observed
Favor Indifferent Opposed Total
Democrat 138 83 64 285
Republican 64 67 84 215
Total 202 150 148 500
Expected
Favor Indifferent Opposed Total
Democrat 115.1400 85.5000 84.3600 285
Republican 86.8600 64.5000 63.6400 215
Total 202 150 148 500

The degrees of freedom (df): df = (r - 1) * (c - 1) = (2 - 1) * (3 - 1) = 1 * 2 = 2

_______________________________________________________________

The Hypothesis:

H0: There is no relation between opinion and voting reference.

Ha: There is a relation between opinion and voting reference.

_______________________________________________________________

The Critical Value:   The critical value at \alpha = 0.05, df = 2

\chi^2critical = 5.992

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The Test Statistic: The table below gives the calculation of \chi^2 .

Number Observed Expected (O-E) (O-E)2 (O-E)2/E
1 138 115.140 22.860 522.580 4.539
2 64 86.860 -22.860 522.580 6.016
3 83 85.500 -2.500 6.250 0.073
4 67 64.500 2.500 6.250 0.097
5 64 84.360 -20.360 414.530 4.914
6 84 63.640 20.360 414.530 6.514
Total 22.15247

\chi^2test = 22.15

__________________________________________________________

The p value: The p value at \chi^2 test = 22.15, df = 2 is;p value = 0.000

The Decision Rule: If \chi^2 test is > \chi^2 critical, then Reject H0.

Also, If p value is < \alpha then reject H0.

______________________________________________________________

The Decision:    Since If \chi^2 test (22.15) is > \chi^2 critical (5.992), We Reject H0.

Also since p value (0.0000) is < \alpha (0.05), We Reject H0.

_______________________________________________________________

The Conclusion: There is sufficient evidence at the 95% significance level to conclude that there is a relation between opinion and voting reference.

_________________________________________________________________

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