Mixed-Effects Model (k = 21; tau^2 estimator: SJ)

tau^2 (estimated amount of residual heterogeneity): 0.2301 (SE = 0.0752)
tau (square root of estimated tau^2 value): 0.4797
I^2 (residual heterogeneity / unaccounted variability): 93.30%
H^2 (unaccounted variability / sampling variability): 14.92
R^2 (amount of heterogeneity accounted for): 0.00%

Test for Residual Heterogeneity:
QE(df = 19) = 494.7195, p-val < .0001

Test of Moderators (coefficient 2):
F(df1 = 1, df2 = 19) = 0.4029, p-val = 0.5331

Model Results:

                      estimate      se     tval    pval    ci.lb   ci.ub 
intrcpt                 0.7497  0.1143   6.5572  <.0001   0.5104  0.9889  *** 
fidelity_assessmentY   -0.3183  0.5014  -0.6348  0.5331  -1.3676  0.7311      

Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Dear group,

Thank you for your replies regarding my previous post. I have one more question regarding the meta-regression results. I divided studies into "with fidelity assessment (Y)" and "without fidelity assessment (N)". The test of moderators shows that p-value is 0.5331. However p-value for the intrcpt is <.0001. Based on the previous post, I understand that this means that the group without fidelity assessment is significant. My question is that since the moderator test says that p-value is 0.5331, can I still draw the conclusion that the intercpt group is significant?


closed as unclear what you're asking by Michael Chernick, kjetil b halvorsen, Peter Flom Jul 26 at 12:33

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Based on the previous post, I understand that this means that the group without fidelity assessment is significant. Please clarify. $\endgroup$ – Subhash C. Davar Jul 21 at 12:35

In this case, the table is telling you that the estimate for the reference level "without fidelity assessment (N)" is significantly different from zero (p value <.0001). And the estimate of the other level "with fidelity assessment (Y)" is not significantly different from the reference level estimate (N) (p value 0.5331).

So yes, your reference level in the intercept is significant.

  • 1
    $\begingroup$ Thanks! But since the test of moderators shows that p-value is 0.5331, does this result make sense? $\endgroup$ – M. L Jul 12 at 8:34
  • $\begingroup$ If ML wants to test whether each level is different from zero then removing the intercept from the model is the way to go. $\endgroup$ – mdewey Jul 12 at 12:18
  • $\begingroup$ @mdewey I do not think that the exercise should help the OP. $\endgroup$ – Subhash C. Davar Jul 13 at 8:41
  • $\begingroup$ @user2974951 Your statement seems to sugget that <.0001 = 0. interesting and funny. Further, comparing to reference - what is being compared with - the p-value or estimate ? What will it represent? please avoid abrasive language that could mislead the reader like me who is not well-versed with Englsh. $\endgroup$ – Subhash C. Davar Jul 13 at 9:24
  • $\begingroup$ @M.L Can you run an ANOVA on your model? $\endgroup$ – user2974951 Jul 13 at 14:53

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