| Literature DB >> 11602018 |
S G Baker1, K S Lindeman, B S Kramer.
Abstract
BACKGROUND: Although a randomized trial represents the most rigorous method of evaluating a medical intervention, some interventions would be extremely difficult to evaluate using this study design. One alternative, an observational cohort study, can give biased results if it is not possible to adjust for all relevant risk factors.Entities:
Mesh:
Year: 2001 PMID: 11602018 PMCID: PMC57808 DOI: 10.1186/1471-2288-1-9
Source DB: PubMed Journal: BMC Med Res Methodol ISSN: 1471-2288 Impact factor: 4.615
Comparison of Several Methods of Evaluating a Medical Intervention
| Method | Strengths | Weaknesses | ||
| Randomized controlled trial | 1. | No temporal bias | 1. | Cost and effort |
| 2. | No selection bias | 2. | Recruitment | |
| Observational cohort study | 1. | No temporal bias | 1. | Cost of data collection |
| 2. | Selection bias if an | |||
| important risk factor is | ||||
| omitted or not | ||||
| adequately quantified | ||||
| Paired availability design | 1. | Lessens selection bias | 1. | Assumptions in Table |
Requirements for Paired Availablity Design
| Requirement | Specific Criteria | |
| Stable population | 1. | Hospital serves one geographic area or |
| is military medical center | ||
| 2. | No in- or out- migration | |
| 3. | Eligibility criteria constant over time | |
| 4. | No underlying change in prognosis | |
| over time | ||
| Stable treatment | 1. | Other patient management constant |
| over time | ||
| Stable evaluation | 1. | Evaluation criteria constant over time |
| Stable preference | 1. | No publicized credible reports |
| 2. | No direct-to-consumer advertising | |
| Effect of the intervention on outcome does | 1. | Effect of intervention does not depend |
| not change with a change in availability | on when the intervention was given | |
| (only applicable when some in "before" | during the course of disease | |
| group receive intervention) | 2. | No learning curve for the intervention |
Example of calculations from data in Baker and Lindeman [Reference 9]
| hospital | before" group data | after group data | estimate | std error | weight | ||||
| n1 | e1 | P1 | n2 | e2 | p2 | y | s | w | |
| 1 | 116 | .586 | .172 | 103 | .223 | .184 | -.033 | .143 | 44 |
| 2 | 180 | .290 | .080 | 180 | .440 | .090 | .067 | .196 | 24 |
| 3 | 373 | .131 | .110 | 421 | .587 | .100 | -.022 | .048 | 208 |
| 4 | 1000 | .100 | .040 | 1000 | .450 | .050 | .029 | .026 | 313 |
| 5 | 1298 | .000 | .074 | 1084 | .480 | .065 | -.019 | .022 | 333 |
| 6 | 1919 | .000 | .275 | 2073 | .316 | .229 | -.146 | .044 | 225 |
| 7 | 3195 | .010 | .030 | 3733 | .290 | .031 | .004 | .015 | 365 |
| 8 | 4778 | .008 | .194 | 4859 | .586 | .190 | -.006 | .014 | 369 |
| 9 | 4685 | .187 | .149 | 6170 | .551 | .125 | -.046 | .015 | 352 |
| 10 | 8108 | .467 | .248 | 9918 | .678 | .280 | .152 | .031 | 288 |
| 11 | 11159 | .328 | .209 | 11869 | .499 | .209 | .000 | .031 | 288 |
n1 (n2) = number of subjects in "before" ("after") group. el (e2) = fraction of subjects in "before" ("after") group that had epdiural analgesia, p1 (p2) = fraction of subjects in "before" ("after") group that had a Cesarean section, y= estimated effect of epidural analgesia on the probability of Cesarean section = (p2-p1)/(e2-e1), s= standard error of y= square root of (p2 (1-p2))/n2 + p1 (1-p1)/n1) /(e2-e1)2, w* = weight used in random effects meta-analysis. We computed the weights as follows. Let i index studies, so yi and si are the values of y and s for study i. It is convenient to define w1 = I/ si2. Following DerSimonian and Laird [Reference 19], to compute v, the variance of the true effect among the k studies, we set v equal to the larger of (Q-(k-1)) / (Σwi - Σwi2/Σwi) and 0, where Q = Σwi (yi - m)2, m = Σyi wi/Σwi. The random-effects weights are w*i= 1/(si2 + v), and the summary statistic is y* = Σyi w*i/Σw*i, with standard error s* = square root of 1/Σw*i. Following Proschan and Follman [reference 20], the 95% confidence interval is (y* - tk- s*, y*+ tk-1 s*), where tk-1 is the value of the 97 ½ percentile of a t-distribution with k-1 degrees of freedom. In this example, k = 11, Q = 50.1, v = .0025, m =-.007, s* = .019 y* = -.005, t10 = 2.23, y* = -.005 and the 95% confidence interval is (-.047, .037).