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

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SPSSAssignment #8Chi-SquareSPSS instructions:Chi-Square Test for Goodness of Fit:Open SPSSRememberthat SPSS assumesthat all the scoresin a row are fromthe sameparticipant. In the study presented in #1, there are 20 students, some of whomhave been suspended for misbehavior. The primary conflict-resolution style used byeach student is also entered. [Ignore the first variable in this analysis.]When you have entered the data for all 20 students, move to the Variable Viewwindowandchangethefirstvariablenameto“SUSPEND”andthesecondto“STYLE”. Set the number of decimals for both variables to zero.Click AnalyzeNon-Parametric TestsChi-SquareClick the variable “STYLE” and then the arrow next to the box labeled “Test VariableList” to indicate that the chi-square for goodness of fit should be conducted on theconflict-resolution style variable.Note that “All categories equal” is the default selection in the “Expected Values”box,whichmeansthatSPSSwillconductthegoodnessoffittestusingequalexpected frequencies for each of the four styles, in other words, SPSS will assumethat the proportions of students each style are equal.Click OK.Chi-Square Test for Independence:Open SPSSFor #2, you need to add the variable “SUSPEND” to the analysis. Remember that inthisproblem,weareinterestedinwhethertherewasanassociationbetweenconflict-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CrosstabsSince“SUSPEND”isalreadyselected,clickthearrownexttotheboxlabeled“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 by20 students.MethodAggressiveManipulativePassiveAssertiveNof Students8228a.Following the five steps of hypothesis testing, conduct the appropriatechi-square test to determine whether the observed frequencies aresignificantly different from the frequencies expected by change at the .05level of significance. Clearlyidentifyeach of the five steps.b.Explain your response to some who has never had a course in statistics.a)Step 1 : PlanTo investigate the fact that,in the primary method of conflict resolution used, whether,theobserved frequencies are significantly different from the frequencies expected by changea studyis carried out.Step 2 : HypothesesNull hypothesis)(0H: The null hypothesis states that,the observed frequencies are significantlydifferent from the frequencies expected by chance.Alternative hypothesis)(aH: The alternative hypothesis states that,the observed frequencies aresignificantly different from the frequencies expectedare notby chance.Step 3:ModelIn this case2model is used to performgoodness-of-fit test.Step 4: MechanicsThe test statistic can be defined as,cellsallEEO22)(where, the number of customers in each cell isO.

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The expected number of each cell isE.To test the above hypotheses, SPSS software is used. Using the above procedure of SPSS, thegoodness of fit test is performed.The SPSS output is given below:Descriptive StatisticsNMeanStd. DeviationMinimumMaximumStyle202.50001.395481.004.00StyleObserved NExpected NResidualAggressive85.03.0Manipulative25.0-3.0Passive25.0-3.0Assertive85.03.0Total20Test StatisticsStyleChi-Square7.200adf3Asymp. Sig..066a.0cells(.0%)haveexpectedfrequencies less than 5. The minimumexpected cell frequency is 5.0.Step 5: ConclusionFrom the table, it is seen that, the value of the2statistic is 7.200.In this case, thep-value of2statisticis signified asAsymp. Sig. (2-sided). Here, thep-value of2statistic7.200with degrees of freedom 3 is066.0. Comparing thep-value with thesignificance level05.0, it is seen that thep-value is morethan the significance level05.0. So,the null hypothesis cannot be rejected.Thus, it can be concluded that, there is a sufficientevidence to believe that,the observed frequencies are significantly different from the frequenciesexpected by chance.

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