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I am a complete novice and was hoping for some clarity around interpretation of my results. Note the response ratio I am interested in is the log coefficient of variation ratio (lnCVR). Please see a small sample of data - though it is unlikely to help provide meaningful results:

organism_detail veg_community   measure t_mean  t_sd    t_se    t_cv    t_quad_n    t_qsize_linearm2/time   t_qsize_m2  t_ssize_ha  r_mean  r_sd    r_se    r_cv    r_quad_n    r_qsize_linear  r_qsize_m2  r_ssize_ha  c_mean  c_sd    c_se    c_cv    c_quad_n    c_qsize_linear  c_qsize_m2  c_ssize_ha  age(rest)   site_size
Lepidoptera forest  Shan_div    21.4    14.9    2.483333333 0.696261682 32      NA  500 23.7    15.5    4.474464586 0.654008439 32      NA  500 NA  NA  NA  NA  NA      NA  NA  4   500
Lepidoptera forest  sp_rich 47.4    26  4.333333333 0.548523207 32      NA  500 62.8    38.2    11.02739014 0.608280255 32      NA  500 NA  NA  NA  NA  NA      NA  NA  4   500
grassland   grassland   even    0.777122531 0.071366376 0.02256803  0.091834136 10      1   1   0.8200173   0.045861093 0.014502551 0.055926982 10      1   1   0.782245149 0.051514245 0.016290235 0.065854349 10      1   1   3   1
grassland   grassland   even    0.761176052 0.040258669 0.012730909 0.052890089 10      1   1   0.8200173   0.045861093 0.014502551 0.055926982 10      1   1   0.782245149 0.051514245 0.016290235 0.065854349 10      1   1   3   1
grassland   grassland   even    0.727425674 0.047416787 0.014994505 0.065184374 10      1   1   0.8200173   0.045861093 0.014502551 0.055926982 10      1   1   0.782245149 0.051514245 0.016290235 0.065854349 10      1   1   3   1
grassland   grassland   even    0.848294027 0.080334923 0.025404133 0.094701743 10      1   1   0.8200173   0.045861093 0.014502551 0.055926982 10      1   1   0.782245149 0.051514245 0.016290235 0.065854349 10      1   1   3   1
grassland   grassland   Shan_div    4.190501692 0.396927242 0.125519415 0.094720697 10      1   1   4.722399781 0.312866417 0.098937048 0.066251574 10      1   1   3.438504231 0.304240111 0.096209171 0.088480365 10      1   1   3   1
grassland   grassland   Shan_div    4.079693163 0.30924688  0.09779245  0.075801505 10      1   1   4.722399781 0.312866417 0.098937048 0.066251574 10      1   1   3.438504231 0.304240111 0.096209171 0.088480365 10      1   1   3   1
grassland   grassland   Shan_div    4.004381201 0.37002005  0.117010614 0.092403803 10      1   1   4.722399781 0.312866417 0.098937048 0.066251574 10      1   1   3.438504231 0.304240111 0.096209171 0.088480365 10      1   1   3   1
grassland   grassland   Shan_div    4.519671132 0.592600506 0.187396734 0.131115846 10      1   1   4.722399781 0.312866417 0.098937048 0.066251574 10      1   1   3.438504231 0.304240111 0.096209171 0.088480365 10      1   1   3   1
grassland   grassland   sp_rich 36.5    4.352521619 1.376388188 0.119247168 10      1   1   42.8    3.938414797 1.245436113 0.092019037 10      1   1   34.49141414 6.397800753 2.02316224  0.185489662 10      1   1   1   1
grassland   grassland   sp_rich 37.6    5.796550698 1.833030278 0.154163582 10      1   1   42.8    3.938414797 1.245436113 0.092019037 10      1   1   34.49141414 6.397800753 2.02316224  0.185489662 10      1   1   1   1
grassland   grassland   sp_rich 39.6    6.801960502 2.150968774 0.171766679 10      1   1   42.8    3.938414797 1.245436113 0.092019037 10      1   1   34.49141414 6.397800753 2.02316224  0.185489662 10      1   1   1   1
grassland   grassland   sp_rich 32.33333333 4.415880433 1.396424004 0.136573622 10      1   1   42.8    3.938414797 1.245436113 0.092019037 10      1   1   34.49141414 6.397800753 2.02316224  0.185489662 10      1   1   1   1
grassland   grassland   sp_rich 39.5    4.600724581 1.454876856 0.11647404  10      1   1   50.22222222 4.381146476 1.385440163 0.087235217 10      1   1   34.49141414 6.397800753 2.02316224  0.185489662 10      1   1   2   1
grassland   grassland   sp_rich 39.9    6.172519745 1.95192213  0.154699743 10      1   1   50.22222222 4.381146476 1.385440163 0.087235217 10      1   1   34.49141414 6.397800753 2.02316224  0.185489662 10      1   1   2   1

I am trying to understand the cause of a change in the variance of the treatment group (t_mean, t_sd, etc.). There are two control groups, but for simplicities sake I am currently only interested in the "c" group (c_mean, etc.). Now, using the code I have pasted at the bottom of this question, I get the following output. I was after some advice on this.

Mixed-Effects Model (k = 344; tau^2 estimator: REML)

   logLik   deviance        AIC        BIC       AICc 
-359.4915   718.9830   728.9830   748.1277   729.1626   

tau^2 (estimated amount of residual heterogeneity):     0.1577 (SE = 0.0278)
tau (square root of estimated tau^2 value):             0.3971
I^2 (residual heterogeneity / unaccounted variability): 52.16%
H^2 (unaccounted variability / sampling variability):   2.09
R^2 (amount of heterogeneity accounted for):            38.54%

Test for Residual Heterogeneity:
QE(df = 340) = 857.1654, p-val < .0001

Test of Moderators (coefficients 2:4):
QM(df = 3) = 113.7937, p-val < .0001

Model Results:

            estimate      se     zval    pval    ci.lb    ci.ub 
intrcpt      -0.1758  0.0619  -2.8393  0.0045  -0.2972  -0.0545   ** 
r_cv          1.7251  0.2301   7.4979  <.0001   1.2742   2.1761  *** 
c_cv         -1.6457  0.1812  -9.0820  <.0001  -2.0009  -1.2906  *** 
t_qsize_m2   -0.0001  0.0000  -1.9625  0.0497  -0.0001  -0.0000    * 

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

My main questions are:

  1. Is it completely asinine to use the coefficient of variation (CV) of one of the groups as a moderator to understand the lnCVR? To elaborate, the question I am trying to ask by doing so is whether a high or low variance in the control group explains future high or low variance in the treatment group. Further to this, how do I understand the positive and negative intercepts (r_cv, and c_cv in the model)?
  2. Does the R^2 subtract this variance from the I^2? I.e. is the unnaccounted variability calculated taking into account the R^2? If so, why when I run without moderators is the I^2 not equal to the I^2+R^2 from the moderator model?
  3. The last thing I am confused about two aspects of the intercept. To understand the effect of t on c should I use the intercept from a random model without moderators, yes? I can't fully comprehend why this changes but I assume this is the intercept on data with moderators incorporated. I.e. the significant p-value indicates there is another variable moderating the variance to a significant degree that I am not accounting for?
  4. How do I interpret this figure? If I use exp(-0.2687) (this is the intercept when I run a random-effects model without moderators), I get 0.7643. Can I interpret this as the variance in the treatment group being 76.43% of the variance in the control group?

CVR<-escalc(measure = "CVR", n1i = dat$t_quad_n, n2i = dat$c_quad_n, 
           m1i = dat$t_mean, m2i = dat$c_mean, sd1i = dat$t_sd, sd2i = dat$c_sd)

dat <-bind_cols(dat, CVR)

dat[, "wi"] <- 1/dat$vi

v_rma <- rma(yi = yi, vi = vi, weights = wi,mods = ~r_cv+c_cv+t_qsize_m2, method = "REML", 
               data = dat)
summary(v_rma)
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  • $\begingroup$ For anyone else playing at home, I found a comment from Wolfgang linking to an explanation on these things in the metafor package documention here $\endgroup$ – joejonzz Apr 20 at 23:35
  • $\begingroup$ Glad to hear you found this useful. This aside, unless you have good reasons to do so, don't use custom weights (using the weights argument). The default behavior is to use inverse variance weights, but in a RE model, the weights are $w_i = 1 / (v_i + \hat{\tau}^2)$. $\endgroup$ – Wolfgang Apr 21 at 8:16
  • $\begingroup$ Oh, thanks for that. Noted. Did you have any comment on Q1? In the documentation I linked above, explain the moderators as effects of levels of factors on the intercept. I find this easier to conceptualize. How do you interpret a coefficient like above when it is numeric. Does c_cv in the model above say the average effect (lnCVR) is lower as c_cv increases? i.e. higher variance in control results in greater reduction of variation in treatment? Many thanks. $\endgroup$ – joejonzz Apr 21 at 22:11
  • $\begingroup$ Can't seem to edit my previous comment, so sorry for the spam. When I plot c_cv against yi, this seems to illustrate what I was assuming above. Greater reduction in variance (lnCVR) when initial (control) cv is higher. Pretty axiomatic but interesting to know that it explains a lot in the model. $\endgroup$ – joejonzz Apr 21 at 23:12
  • $\begingroup$ To examine 1. you should look into using a bivariate model. See: metafor-project.org/doku.php/analyses:vanhouwelingen2002 (different effect size / outcome measure, but the idea is the same). $\endgroup$ – Wolfgang Apr 25 at 9:02

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