Homework SourseSevere anxiety often accompanies patients who must undergo c
Severe anxiety often accompanies patients who must undergo chronic hemodialysis. A study was undertaken to determine the effects of a set of progressive relaxation exercises on anxiety in hemodialysis patients. The treatment group consisted of 38 subjects who were shown a set of progressive relaxation videotapes. The control group was made up of 23 patients shown a set of neutral videotapes. A psychiatric questionnaire that measured anxiety revealed the posttest results reported in Table 12/14. Test the mean difference for significant.
Table 12.14 Posttest results
Group
n
Mean
Std. Dev.
Experimental
38
33.42
10.18
Control
23
39.71
9.16
| Table 12.14 Posttest results | |||
| Group | n | Mean | Std. Dev. |
| Experimental | 38 | 33.42 | 10.18 |
| Control | 23 | 39.71 | 9.16 |
Solution
Formulating the null and alternative hypotheses,
Ho: u1 - u2 = 0
Ha: u1 - u2 =/ 0
At level of significance = 0.05
As we can see, this is a two tailed test.
Calculating the means of each group,
X1 = 33.42
X2 = 39.71
Calculating the standard deviations of each group,
s1 = 10.18
s2 = 9.16
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):
n1 = sample size of group 1 = 38
n2 = sample size of group 2 = 23
Thus, df = n1 + n2 - 2 = 59
Also, sD = 2.524923362
Thus, the t statistic will be
t = [X1 - X2 - uD]/sD = -2.49116472
where uD = hypothesized difference = 0
Now, the critical value for t is
tcrit = +/- 2.000995378
Thus, as our t is not between the critical values, we decide to WE REJECT THE NULL HYPOTHESIS.
Thus, at 0.05 level, there is significant evidence that the difference betwen the means are significant.
