Question
(a) In a hypothetical study, 10 patients with kidney disease were randomly assigned to receive two dialysis methods (A and B). 5 patients received Dialysis-A followed by Dialysis-B, and the remaining 5 received Dialysis-B followed by Dialysis-A. The kidney function scores of these patients are given as:
| Group-I (AB Sequence) | |||||
|---|---|---|---|---|---|
| Patient No. | 1 | 2 | 3 | 4 | 5 |
| Dialysis-A | 69 | 73 | 50 | 69 | 76 |
| Dialysis-B | 68 | 79 | 68 | 72 | 85 |
| Group-II (BA Sequence) | |||||
|---|---|---|---|---|---|
| Patient No. | 6 | 7 | 8 | 9 | 10 |
| Dialysis-B | 57 | 70 | 75 | 70 | 69 |
| Dialysis-A | 51 | 71 | 53 | 62 | 64 |
Test whether the mean score for dialysis-A is different from the mean score of dialysis-B, assuming no carry-over effect.
(b) Describe descriptive and analytic epidemiology.
Answer :
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Skip in 10
Ten patients form a two-period crossover trial. For each patient the relevant observation, assuming no carry-over effect, is the within-subject difference between Dialysis-A and Dialysis-B. For Group-I (AB sequence), the differences A − B are: 1, −6, −18, −3 and −9. For Group-II (BA sequence), the differences A − B are: −6, 1, −22, −8 and −5. Combining all 10 patients, the differences are 1, −6, −18, −3, −9, −6, 1, −22, −8, −5. The sum of differences is −75, giving a mean difference of −7.5. Deviations from the mean (−7.5) are 8.5, 1.5, −10.5, 4.5, −1.5, 1.5, 8.5, −14.5, −0.5 and
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