Question
Interpret the following chi-square analysis. What is the null hypothesis? Using an alpha level of .05, what do you conclude? If you reject the null
Interpret the following chi-square analysis.
- What is the null hypothesis?
- Using an alpha level of .05, what do you conclude?
- If you reject the null hypothesis, which cells appear to account for the decision?
Crosstab | ||||||
CAN PEOPLE BE TRUSTED | Total | |||||
CAN TRUST | CANNOT TRUST | DEPENDS | ||||
BOTH MEN AND WOMEN SHOULD CONTRIBUTE TO INCOME | Strongly Agree | Count | 28 | 97 | 4 | 129 |
Expected Count | 36.5 | 87.9 | 4.6 | 129.0 | ||
Std. Residual | -1.4 | 1.0 | -.3 | |||
Agree | Count | 65 | 208 | 11 | 284 | |
Expected Count | 80.3 | 193.5 | 10.2 | 284.0 | ||
Std. Residual | -1.7 | 1.0 | .2 | |||
Neither Agree nor Disagree | Count | 61 | 90 | 5 | 156 | |
Expected Count | 44.1 | 106.3 | 5.6 | 156.0 | ||
Std. Residual | 2.5 | -1.6 | -.3 | |||
Disgree | Count | 25 | 31 | 3 | 59 | |
Expected Count | 16.7 | 40.2 | 2.1 | 59.0 | ||
Std. Residual | 2.0 | -1.5 | .6 | |||
Strongly Disagree | Count | 2 | 10 | 0 | 12 | |
Expected Count | 3.4 | 8.2 | .4 | 12.0 | ||
Std. Residual | -.8 | .6 | -.7 | |||
Total | Count | 181 | 436 | 23 | 640 | |
Expected Count | 181.0 | 436.0 | 23.0 | 640.0 |
Value | df | Asymp. Sig. (2-sided) | |
Pearson Chi-Square | 24.118a | 8 | .002 |
Likelihood Ratio | 23.977 | 8 | .002 |
Linear-by-Linear Association | 9.333 | 1 | .002 |
N of Valid Cases | 640 | ||
a. 4 cells (26.7%) have expected count less than 5. The minimum expected count is .43. |
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