Skip to main content
6 events
when toggle format what by license comment
Nov 12, 2018 at 11:55 comment added mdewey @Michael That seems a different question from the original one you asked. The crude approach would be to convert your single value to a $z$-score using the mean and standard deviation of the values of $\tau$ you have in your comparison group.
Nov 11, 2018 at 9:06 comment added Wolfgang That's something entirely different (if you really mean the difference between means and not the difference between two $\tau^2$ values). For testing (differences of) means, Wald-type tests are usually fine (the Knapp & Hartung method is even better, but this is a refinement of the Wald-type tests, not an entirely different procedure).
Nov 10, 2018 at 18:38 comment added Mwatch Would the same apply in a 2 sample z test for checking difference in population means? Thanks.
Nov 10, 2018 at 15:15 comment added Wolfgang No, Wald-type test of variance components have terrible performance.
Nov 10, 2018 at 0:42 comment added Mwatch Thanks for the answer - I really appreciate it! I want to use the S.E (and Tau value or Tau-squared for that matter) in a z-test. Would this be appropriate? Thanks.
Nov 9, 2018 at 20:05 history answered Wolfgang CC BY-SA 4.0