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Chi-Square Analysis of Conflict Resolution Styles and School Suspension Among Students Assignment #8

Solved assignment using chi-square tests to analyze how conflict resolution styles influence school suspensions.

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    SPSS Assignment #8 Chi - Square SPSS instructions: Chi - Square Test for Goodness of Fit: Open SPSS Remember that SPSS assumes that all the scores in a row are from the same participant. In the study presented in #1, there are 20 students, some of whom have been suspended for misbehavior. The primary conflict - resolution style used by each student is also entered. [Ignore the fi rst variable in this analysis.] When you have entered the data for all 20 students, move to the Variable View window and change the first variable name to “SUSPEND” and the second to “STYLE”. Set the number of decim als for both variables to zero. Click Analyze  Non - Parametric Tests  Chi - Square Click the variable “STYLE” and then the arrow next to the box labeled “Test Variable List” to indicate that the chi - square for goodness of fit should be conducted on the conflict - resolution style variable. Note that “All categories equal” is the default selection in the “Expected Values” box, which means that SPSS will conduct the goodness of fit test using equal expected frequencies for each of the four styles, in other words, SPSS will assume that the proportions of students each style are equa l. Click OK. Chi - Square Test for Independence: Open SPSS For #2, you need to add the variable “SUSPEND” to the analysis. Remember that in this problem, we are interested in whether there was an association between conflict - resolution style and having been suspended from school for misbehavior. Since the analysis will involve two nominal variables, the appropriate test is a chi - square test for independence. Click Analyze  Descriptive Statistics  Crosstabs Since “SUSPEND” is already selected, click the arrow next to the box labeled “Rows.” Click the variable “STYLE” and click the arrow next to the box labeled “Columns.” Click “Statistics” and click the box labeled “Chi - Square.” Click Continue. Click “Cells” and click the box labeled “Expected.” Click Continue. Click OK.

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    1. The following table includes the primary method of conflict resolution used by 20 students. Method Aggressive Manipulative Passive Assertive N of Students 8 2 2 8 a. Following the five steps of hypothesis testing, conduct the appropriate chi - square test to determine whether the observed frequencies are significantly different from the frequencies expected by change at the .05 level of significance. Clearly identify each of the five steps. b. Explain your response to some who has never had a course in statistics. a) Step 1 : Plan To investigate the fact that, in the primary method of conflict resolution used, whether , the observed frequencies are significantly different from the frequencies expected by change a study is carried out. Step 2 : Hypotheses Null hypothesis ) ( 0 H : The null hypothesis states that, the observed frequencies are significantly different from the frequencies expected by chance . Alternative hypothesis ) ( a H : The alternative hypothesis states that, the observed frequencies are significantly different from the frequencies expected are not by chance . Step 3 : Model I n this case 2  model is used to perform goodness - of - fit test . Step 4: Mechanics The test statistic can be defined as,          cells all E E O 2 2 ) (  where, the number of customers in each cell is O .

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    The expected number of each cell is E . To test the above hypotheses, SPSS software is used. Using the above proced ure of SPSS, the goodness of fit test is perfor med. The SPSS output is given below: Descriptive Statistics N Mean Std. Deviation Minimum Maximum Style 20 2.5000 1.39548 1.00 4.00 Style Observed N Expected N Residual Aggressive 8 5.0 3.0 Manipulative 2 5.0 - 3.0 Passive 2 5.0 - 3.0 Assertive 8 5.0 3.0 Total 20 Test Statistics Style Chi - Square 7.200 a df 3 Asymp. Sig. .066 a. 0 cells (.0%) have expected frequencies less than 5. The minimum expected cell frequency is 5.0. Step 5: Conclusion From the table, it is seen that, the value of the 2  statistic is 7.200. In this case, the p - value of 2  statistic is signified as Asymp. Sig. (2 - sided) . Here, the p - value of 2  statistic 7.200 with degrees of freedom 3 is 066 . 0 . Comparing the p - value with the significance level 05 . 0 , it is seen that the p - value is more than the significance level 05 . 0 . So, the null hypothesis cannot be rejected. Thus, it can be concluded that, there is a sufficient evidence to believe that, the observed frequencies are significantly different from the frequencies expected by chance .

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