A friend of mine does program

A friend of mine does program evaluations for counseling centersthat deal with juvenile delinquents. As such, he does many simplestatistical analyses that compare some “pre-intervention”behavioral trait to that of the behavioral trait“post-intervention”. For example, say that some intervention isaimed at reducing aggression in adolescent boys. My friend mightthen quantify pre-intervention aggression (also known as “baseline”aggression) by administering something like The AggressionScale developed by Orpinas and Frankowski (2001), which yieldsa value between 0 and 66, with greater numbers indicating higheraggression. Then, after the adolescent boys complete theintervention program, my friend would then again administer TheAggression Scale to determine the effects of the interventionprogram on aggression (i.e., “post-intervention”).

* State the alternative and null hypotheses (using thestatistical notation for stating mathematical relationships) thatrepresents the prediction that aggression will decreasefrom pre-intervention to post-intervention in theseadolescent boys. * Calculate standard error of the mean differenceand tobt.    * Using an alpha of0.05, what is tcrit? Would you reject or failto reject the null hypothesis? Why? * What are the upper and lowerboundaries of the range of mean differences that you can say with95% confidence contains the mean difference represented by theabove sample?

Participant

Pre-Intervention Aggression

Post-Intervention Aggression

1

35

29

2

42

41

3

45

43

4

39

27

5

45

40

6

43

34

Answer:

Pre Post d = Post – Pre
35 29 -6
42 41 -1
45 43 -2
39 27 -12
45 40 -5
43 34 -9
d-bar = -5.833333333
s = 4.16733328

(a)

Let d = Post – Pre

Ho: d-bar ≥ 0 and Ha: d-bar < 0

(b)

Data:     

n = n1 = n2 = 6    

d-bar = -5.833333333    

s = 4.16733328    

Hypotheses:     

Ho: d-bar ≥ 0    

Ha: d-bar < 0    

Test Statistic:     

SE = s/√n = 4.16733328000853/√6 = 1.701306687   

t = d-bar/SE = -5.83333333333333/1.70130668735665 = -3.428737086  

(c)

α = 0.05    

Degrees of freedom = 6 – 1 = 5

Critical t- score = -2.015048372

Since -3.428737086 < -2.015 we reject Ho and accept Ha

(d)

n1 = n2 = n = 6

d-bar = -5.8333

s = 4.1673

% = 95

Degrees of freedom = n – 1 = 5

SE = s/√n = 1.701293101

t- score = 2.570581835

Width of the confidence interval = t * SE = 4.373313141

Lower limit of the confidence interval = d-bar – width =-10.20661314

Upper limit of the confidence interval = d-bar + width =-1.459986859

[Please give me a Thumbs Up if you are satisfied with myanswer. If you are not, please comment on it, so I can edit theanswer. Thanks.]


 
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