In: Statistics and Probability
A researcher is interested in the effects of “mindfulness training” on people’s sense of well being. She divides 15 people into three groups: one group is a waiting control group; one group does mindfulness training for a 45 day protocol; and the third group does the training for 90-days. Following the training she measures number of reports of “feeling well without significant anger, depression or anxiety for a full day” for 2-weeks. The results are below.
Number of reports of well being over 14-days
Control 45-days 90-days
5 8 6
4 1 7
3 1 7
0 1 9
1 3 8
A. State the Independent Variable and the Dependent Variable.
B. State the null hypothesis in words and symbols.
C. Compute the appropriate statistic.
D. What is your decision?
E. State the full conclusion in words, after computing, if necessary multiple comparisons.
A. The dependent varibale is say y denoting 'the number of reports ' of “feeling well without significant anger, depression or anxiety for a full day” and the independent variable is say x representing the three groups.
B. Null Hypothesis: there is no significant difference between the number of reports for each group.
H0 : m1 = m2 = m3 = 0
where mi is the mean number of reports for ith group, i = 1, 2, 3.
C. To compare the three groups perhaps you must be expecting to use ANOVA but thats an inappropriate tool in this case as the data is not continuous. The appropriate test is Kruskal-Wallis test. But I produce the result of both.
Using ANOVA the F statistic value is 9.9398. The ful ANOVA is given below.
Analysis of Variance Table
Response: num_of_reports
Df Sum Sq Mean Sq F value Pr(>F)
group 1 57.600 57.600 9.9398 0.007631
Residuals 13 75.333 5.795
Kruskal_Wallis Test
Kruskal-Wallis chi-squared = 7.1305, df = 2, p-value = 0.02829
The test Statistic value is H = 7.1305
D. Using the result of both the test we reach to the conclusion that at least one of the group differ significantly.
E. Following is the result of multiple comparison using t test assuming equal variances of each group. (I tested the equality of variances)
data: control and group45
t = -0.12172, df = 7.0687, p-value = 0.9065
data: control and group90
t = -4.5356, df = 6.2161, p-value = 0.003617
data: group45 and group90
t = -3.1743, df = 5.1083, p-value = 0.02397
from the above multiple comparison result we can say that the group which received 90 days training differ significantly from the other two with largest mean indicating that the 90 days of training results in feeling well most often.