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 t_{obt. *} Using an alpha of0.05, what is t_{crit}? 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
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