Here are the data from a study in which adolescents were given counseling at the beginning of the school year to see if it had a positive impact on their tolerance for adolescents who were ethnically different from them. Assessments were made right before the treatment and then 6 months later. Did the program work? The outcome variable is scored on an attitude-toward-others test with possible scores ranging from 0 to 50; the higher the score, the more tolerance. Use SPSS or some other computer application to complete this analysis.
Before Treatment | After Treatment |
---|---|
45 | 46 |
46 | 44 |
32 | 47 |
34 | 42 |
33 | 45 |
21 | 32 |
23 | 36 |
41 | 43 |
27 | 24 |
38 | 41 |
41 | 38 |
47 | 31 |
41 | 22 |
32 | 36 |
22 | 36 |
34 | 27 |
36 | 41 |
19 | 44 |
23 | 32 |
22 | 32 |
Before Treatment | After Treatment | Difference |
45 | 46 | -1 |
46 | 44 | 2 |
32 | 47 | -15 |
34 | 42 | -8 |
33 | 45 | -12 |
21 | 32 | -11 |
23 | 36 | -13 |
41 | 43 | -2 |
27 | 24 | 3 |
38 | 41 | -3 |
41 | 38 | 3 |
47 | 31 | 16 |
41 | 22 | 19 |
32 | 36 | -4 |
22 | 36 | -14 |
34 | 27 | 7 |
36 | 41 | -5 |
19 | 44 | -25 |
23 | 32 | -9 |
22 | 32 | -10 |
Positive impact on their tolerance mean : before - after < 0
Sample mean of the difference using excel function AVERAGE(), x̅d = -4.1000
Sample standard deviation of the difference using excel function STDEV.S(), sd = 10.5925
Sample size, n = 20
Null and Alternative hypothesis:
Ho : µd ≥ 0
H1 : µd < 0
Test statistic:
t = (x̅d)/(sd/√n) = (-4.1)/(10.5925/√20) = -1.7310
df = n-1 = 19
p-value :
Left tailed p-value = T.DIST(-1.731, 19, 1) = 0.0498
Decision:
p-value < α, Reject the null hypothesis
Conclusion:
There is enough evidence to conclude that counseling had a positive impact on the tolerance for adolescents who were ethnically different from them at 0.05 significance level.
Here are the data from a study in which adolescents were given counseling at the beginning...
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