Homework SourseHealing heart patients with music imagery touch and prayer P
Healing heart patients with music, imagery, touch, and prayer.
Patients were randomly assigned to receive one of four types of treatment: 1.prayer, 2.MIT, 3.prayer and MIT, 4.standard care. Six months after therapy, the patients were evaluated for a major adverse cardiovasular event. The results of the study are summarized in the accompanying table and saved.
MIT
alpha = 0.10
a. Identify the two qualitative variables measured in the study.
b. State H0 and Ha for testing whether a major adverse cardiovascular event depends on type of therapy.
c. On the basis of this test, what can the resarchers infer about the effectiveness of music, imagey, and touch therapy and the effectiveness of healing prayer in heart patients.
| Therapy | Number of Patients with Major Cardiovascular events | Number of Patients with no events | Total |
| Prayer | 43 | 139 | 182 |
| Expected | 43.6 | 136.4 | 182 |
| MIT | 47 | 138 | 185 |
| Expected | 44.3 | 140.7 | 185 |
| Prayer and MIT | 39 | 150 | 189 |
| Expected | 45.2 | 143.8 | 189 |
| Standard | 50 | 142 | 192 |
| Expected | 45.9 | 146.1 | 192 |
Solution
ANSWER OF PART (a) prayer and MIT are two qualitative variables measured in the study.
ANSWER OF PART (b)null hypothesis H0=all the four treatmenst have same or equal effect
alternative hypothesis Ha=at least two treatments have different effect
ANSWER OF PART (c) we go here for goodness of fit test and calculate statistic =(O-E) 2/E will follow chi-square with n-1 degree of freedom , here n=4 (i.e. four group)
so (43-43.6)2/43.6 +(47-44.3)2/44.3+(39-45.2)2 /45.2 +(50-45.9)2 /45.9=1.3895
the tabulated value for chi-squre at alpha=0.1 is 6.251,
so we accept the null hypothesis as tabulated value is greater than calculated value.
