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
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The Hypothesis:
H0: There is no relation between opinion and voting reference.
Ha: There is a relation between opinion and voting reference.
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The Critical Value: The critical value at = 0.05, df = 2
critical = 5.992
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The Test Statistic: The table below gives the calculation of .
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 |
test = 22.15
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The p value: The p value at test = 22.15, df = 2 is;p value = 0.000
The Decision Rule: If test is > critical, then Reject H0.
Also, If p value is < then reject H0.
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The Decision: Since If test (22.15) is > critical (5.992), We Reject H0.
Also since p value (0.0000) is < (0.05), We Reject H0.
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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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In order to receive credit for the homework, you must show all steps for doing a chi square test of independence. T...
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