Question
I am completing a homework assignment. Week 5 of BUSI 820 for Liberty University. Note: Due to an issue with one of the data points
I am completing a homework assignment. Week 5 of BUSI 820 for Liberty University.
Note: Due to an issue with one of the data points not being available in the dataset we had to choose another data point, so there is not going to be a simple answer to pull from this question from previous students.
Below is the question I'm answering, the Chi-square IBM SPSS data I pulled, and then my interpretation below that.
I'm not sure how to interpret the significance of the following:
1) Under Chi-square tests: the Pearson Chi Square of 38.638, the likelihood ratio of 49.589, or the Linear-by-Linear Association of 25.432.
2) Under Symmetric measures: the Phi of .888, or Cramer's V of .888 with an approximate significance of <.001.
Can you help me understand how to interpret these, or understand their meaning in context of interpreting for this question? Thanks!
A.4.7.1.a. Run crosstabs and interpret the results of chi-square and phi (or Cramer's V), as discussed in Chapter 7 and in the interpretation of Output 7.1.
Figure 1. Have children and marital status
Case Processing Summary | ||||||
Cases | ||||||
Valid | Missing | Total | ||||
N | Percent | N | Percent | N | Percent | |
marital status * does subject have children | 49 | 98.0% | 1 | 2.0% | 50 | 100.0% |
marital status * does subject have children Crosstabulation | |||||
does subject have children | Total | ||||
no | yes | ||||
marital status | single | Count | 20 | 0 | 20 |
Expected Count | 9.4 | 10.6 | 20.0 | ||
% within does subject have children | 87.0% | 0.0% | 40.8% | ||
married | Count | 1 | 17 | 18 | |
Expected Count | 8.4 | 9.6 | 18.0 | ||
% within does subject have children | 4.3% | 65.4% | 36.7% | ||
divorced | Count | 2 | 9 | 11 | |
Expected Count | 5.2 | 5.8 | 11.0 | ||
% within does subject have children | 8.7% | 34.6% | 22.4% | ||
Total | Count | 23 | 26 | 49 | |
Expected Count | 23.0 | 26.0 | 49.0 | ||
% within does subject have children | 100.0% | 100.0% | 100.0% |
Chi-Square Tests | |||
Value | df | Asymptotic Significance (2-sided) | |
Pearson Chi-Square | 38.638a | 2 | <.001 |
Likelihood Ratio | 49.589 | 2 | <.001 |
Linear-by-Linear Association | 25.432 | 1 | <.001 |
N of Valid Cases | 49 | ||
a. 0 cells (0.0%) have expected count less than 5. The minimum expected count is 5.16. |
Symmetric Measures | |||
Value | Approximate Significance | ||
Nominal by Nominal | Phi | .888 | <.001 |
Cramer's V | .888 | <.001 | |
N of Valid Cases | 49 |
4.7.1.b Interpret the Results for Having Children and Marital Status
To investigate the relationship between participants with them being married and having or not having children, a chi-square statistic was conducted (Morgan et al., 2020). Figure 1 first shows the processing summary, a cross-tabulation, the chi-square tests, and symmetric measures (Morgan et al., 2020). The case processing summary shows that there were fifty participants, with one missing response, so 98% responding (Morgan et al., 2020). The cross tabulation showed that three categories: single, married and divorced; there were 20 single participants with 0 having children, and 20 not having children; there were 18 married participants with 17 having children, and 1 not having children; and there were 11 divorced participants with 9 having children and 2 not having children (Morgan et al., 2020). There was a large difference between the count and the expected count: for single all 20 participants had no children and the expected count was closer to an even split; the married participants, which had 17 of the 18 participants having children, and an expected count of a closer even split again; and divorced participants, which had 9 participants with children, and 2 without, but the expected count again was closer to an even split (Morgan et al., 2020). The study showed a definite relationship between being married or having been married and having children and being single and not having children (Morgan et al., 2020). The chi-square test found that 0 cells had expected counts less than 5, which is good as it shows that the conditions were met to use chi-square.
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