Question

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

* State the alternative and null hypotheses (using the statistical notation for stating mathematical relationships) that represents the prediction that aggression will decrease from pre-intervention to post-intervention in these adolescent boys. * Calculate standard error of the mean difference and t_{obt.    *} Using an alpha of 0.05, what is t_crit? Would you reject or fail to reject the null hypothesis? Why? * What are the upper and lower boundaries of the range of mean differences that you can say with 95% confidence contains the mean difference represented by the above 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 1

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

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