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Assessing the Effectiveness of Violence Counseling Programs - Statistics Project Example

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Summary
The objective of the project "Assessing the Effectiveness of Domestic Violence Counseling Programs" is to solve three statistical tasks that feature testing of a given hypothesis and analysis of intermediate results. The presented tasks share the topic of physical force incidences…
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Assessing the Effectiveness of Violence Counseling Programs
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JUS520 Exam 2 You are interested in assessing the effectiveness of a domestic violence-counseling program designed to decrease the number of domestic abuse incidents among convicted abusers. The case files of ten offenders are randomly selected from the population of convicted abusers. The number of abuse incidents (verbal or physical) six reported six months before the implementation of the counseling program and six months following the completion of the program are reported below. Using a .01 level of significance, test the hypothesis that the number of abuse incidents was greater before participation in the counseling program. Number of Abuse Incidents Before and After Implementation of Domestic Violence Counseling Program Offender number Before After 1 4 2 2 3 1 3 1 1 4 4 0 5 1 2 6 3 1 7 4 3 8 4 2 9 4 2 10 2 0 Solution Step 1 The following is the set of hypothesis to be tested for the problem. H0: µBefore=µAfter, There is no significant difference between number of incidents before and after the program, (The program is not effective) H0: µBefore≠µAfter, There is a significant difference between number of incidents before and after the program, (The program is effective) Step 2 The hypothesis is tested at α= 0.01 Step 3 Paired sample t test is used to test the hypothesis because the data involves observation on same participants across the study and can therefore ne paired. Step 4 The following are the SPSS outputs for the paired sample t-test Paired Samples Statistics Mean N Std. Deviation Std. Error Mean Pair 1 before 3.0000 10 1.24722 .39441 after 1.4000 10 .96609 .30551 Paired Samples Test Paired Differences t df Sig. (2-tailed) Mean Std. Deviation Std. Error Mean 99% Confidence Interval of the Difference Lower Upper Pair 1 before - after 1.60000 1.34990 .42687 .21273 2.98727 3.748 9 .005 Step 5 From the analysis results, t(9)= 3.748, p= 0.005, p< 0.01 The null hypothesis is therefore rejected for the alternative hypothesis. This means that the program is effective because of existence of a significant difference in rates of reported cases before and after the program with after the program reporting lower rates. 2. You have been hired by the mayor of a mid-sized city who wants to see if senior citizens and younger citizens differ in their attitudes toward a neighborhood watch program. Fifteen citizens from each age group were randomly selected to give their opinions about the effectiveness of the program on a 7-point scale (the higher the number, the more favorable the attitude). Using the data below, test the hypothesis that there is no difference between the two groups of citizens. Younger Citizens Senior Citizens 5 3 5 4 7 5 7 7 7 4 6 4 7 3 6 4 5 4 7 3 5 5 7 4 6 2 7 3 6 2 Solution Step 1 The following is the set of hypothesis to be tested for the problem. H0: µYounger=µOlder, There is no significant difference attitudes toward a neighborhood watch program between the younger and the older population segments, (Age is not a significant factor) H0: µYounger=µOlder, There is a significant difference attitudes toward a neighborhood watch program between the younger and the older population segments, (Age is a significant factor) Step 2 The test is conducted at α= 0.05 Step 3 Independent ample t test is used for testing the hypothesis because the data sets cannot be paired as research participants are different in the two sets of data. Step 4 The following tables shows the SPSS results for the independent sample t test at α= 0.05. Group Statistics group N Mean Std. Deviation Std. Error Mean Attitude 1.00 14 6.2143 .89258 .23855 2.00 16 3.9375 1.34009 .33502 Independent Samples Test Levenes Test for Equality of Variances t-test for Equality of Means F Sig. t df Sig. (2-tailed) Mean Difference Std. Error Difference 95% Confidence Interval of the Difference Lower Upper Attitude Equal variances assumed .415 .525 5.391 28 .000 2.27679 .42236 1.41162 3.14195 Equal variances not assumed 5.536 26.273 .000 2.27679 .41127 1.43183 3.12175 Step 5 From the results and assuming equal variance, t(28)= 5.391, p= 0.000, p< 0.05 The null hypothesis is therefore rejected to the conclusion that there is a significant difference in attitude towards neighborhood watch program across the two groups. Younger citizens therefore value the program than older citizens do. 3. A researcher undertakes a study of physical force incidences among a sample of correctional officers. The research is specifically interested in determining whether the use of force differs depending on the height of the officer. The researcher selected a random sample of 45 officers in three different height categories and examines the number of times each of those officers used physical force within the past 12 months. Use the data below to answer the following questions: (a) Test the hypothesis that there is a significant difference, overall, between the three height groups and display the means in a bar graph. (30) (b) Using a post hoc analysis, write where the specific differences are between each of the possible pairwise comparisons of officers’ height. (10) (c) Calculate the strength of the effect for height on officers’ use of physical force (5) (d) Based on all of these results, would you make any recommendations to your state department of corrections regarding the hiring of correctional officers? Why or why not? (5) 5’0” – 5’10” (X1) 5’11” – 6’4” (X2) Over 6’4” (X3) 0 3 4 3 3 5 1 1 6 1 2 2 0 4 3 2 3 4 2 5 5 0 1 6 2 2 8 0 0 7 2 1 2 1 2 1 3 0 3 1 3 5 2 2 3 Solution a). Test of hypothesis Step 1 The following set of hypotheses is tested for the problem. Ho: µ1 = µ2 = µ3, There is no significant difference in number of incidences of used physical force across the groups (Height is not a significant factor in officers rates of physical force application) H1: µ1 ≠ µ2 ≠ µ3, There is a significant difference in number of incidences of used physical force across the groups (Height is a significant factor in officers’ rates of physical force application) Step 2 The test is conducted at α= 0.05 Step 3 ANOVA test is used for the analysis. Step 4 The following tables show results for the general test. Descriptives Force N Mean Std. Deviation Std. Error 95% Confidence Interval for Mean Minimum Maximum Lower Bound Upper Bound 1.00 15 1.3333 1.04654 .27021 .7538 1.9129 .00 3.00 2.00 15 2.1333 1.40746 .36341 1.3539 2.9128 .00 5.00 3.00 15 4.2667 1.98086 .51146 3.1697 5.3636 1.00 8.00 Total 45 2.5778 1.94806 .29040 1.9925 3.1630 .00 8.00 ANOVA Force Sum of Squares df Mean Square F Sig. Between Groups 68.978 2 34.489 14.781 .000 Within Groups 98.000 42 2.333 Total 166.978 44 Step 5 F(2, 42)= 14.781, P= 0.000 The null hypothesis is rejected to the effect that a significant difference exists between the groups. Height is therefore a significant determinant of the officers’ incidence of use of physical force. b). Source of the difference Following table for post hoc test allows for identification of sources of the observed difference across the groups. Multiple Comparisons Dependent Variable: Force Tukey HSD (I) Category (J) Category Mean Difference (I-J) Std. Error Sig. 95% Confidence Interval Lower Bound Upper Bound 1.00 2.00 -.80000 .55777 .333 -2.1551 .5551 3.00 -2.93333* .55777 .000 -4.2884 -1.5782 2.00 1.00 .80000 .55777 .333 -.5551 2.1551 3.00 -2.13333* .55777 .001 -3.4884 -.7782 3.00 1.00 2.93333* .55777 .000 1.5782 4.2884 2.00 2.13333* .55777 .001 .7782 3.4884 *. The mean difference is significant at the 0.05 level. The results shows that all the three groups contribute to the differences based on their difference from each other. With the criterion that identifies significance for p< 0.05, a significant difference exists between X1 and X3, and X2 and X3. c). Strength of effect The strength of association=sum of squares between groups/total sum of squares =68.978/166.978= 0.413 The strength of association is therefore moderate. d). Conclusion The results show that use of physical force increases with increase in officers’ height. The difference is significant across the groups with a moderate effect size. I therefore recommend that the state reduce the number of very tall recruits or develops policies for moderating use of physical force among the relatively taller officers. (Andrew, Pedersen and McEvoy, 2011) Reference Andrew, D., Pedersen, P. and McEvoy, C. (2011). Research methods and design in sport management. Champaign, IL: Human Kinetics. Read More
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