Question 3: Is the default fixed effects model the right one to use?
To answer this you need to look at the heterogeneity statistics. Click "Next table" to see them.
Because these results form a very wide table, I have split them here. This is the first part of the table:
Model | Effect size and 95% interval | Test of null (2-Tail) | ||||
---|---|---|---|---|---|---|
Model | Number Studies | Point estimate | Lower limit | Upper limit | Z-value | P-value |
Fixed | 9 | 0.676 | 0.535 | 0.853 | –3.293 | 0.001 |
Random effects | 9 | 0.668 | 0.470 | 0.949 | –2.250 | 0.024 |
This is the second part of the table:
Heterogeneity | Tau-squared | ||||||
---|---|---|---|---|---|---|---|
Q-value | df (Q) | P-value | I-squared | Tau Squared | Standard Error | Variance | Tau |
12.763 | 8 | 0.120 | 37.317 | 0.089 | 0.129 | 0.017 | 0.298 |
Note that there are only nine studies, because Brooks et al. (2004) has been dropped.
The Q-value is the chi-squared test for heterogeneity test statistic. The heterogeneity is not significant, P = 0.12, even by the relaxed standard of P<0.10 sometimes used for this test. The I^{2} statistic is 37%, so there is some heterogeneity, but not as much as 50%, which would be moderate. I would suggest that a fixed effects model is acceptable here. Meier et al. (2010) preferred a random effects model. There is room for debate about this one and either approach would be accepted by most people, I think.
The point estimates for the two models are very similar, but the confidence interval is wider for the random effects model and the P value is larger, as we would expect.
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Last updated: 16 February, 2010.