[R-meta] Inflated confidence intervals

Viechtbauer, Wolfgang (SP) wolfg@ng@viechtb@uer @ending from m@@@trichtuniver@ity@nl
Tue Sep 18 19:14:42 CEST 2018

Please always cc the mailing list.

I don't know what you mean by 'trivariate mean' or 'adjusted mean' (adjusted for what?). If you want the average of the three outcomes, then it would be:

intrcpt + 1/3 * MeasurementHDL + 1/3 * MeasurementTC

so in this case:

-0.2376 + 1/3 * 0.1982 + 1/3 * -0.0148 = -0.1764667

which you can also get with: predict(resMV, newmods = c(1/3, 1/3)).

I don't know what you mean by 'Univariate model'. A model that does not distinguish between the three outcomes? Why would this model give an estimate that is equal to the intercept of the model whose output you showed?


-----Original Message-----
From: Wasim Iqbal (UG) [mailto:W.Iqbal using newcastle.ac.uk] 
Sent: Sunday, 16 September, 2018 17:18
To: Viechtbauer, Wolfgang (SP)
Subject: Re: Inflated confidence intervals

Sorry the difference for both HDL and LDL from the intercept would be....-0.0541. -0.0394 would be just for HDL

From: Wasim Iqbal (UG)
Sent: 16 September 2018 16:13:16
To: Viechtbauer, Wolfgang (SP)
Subject: Re: Inflated confidence intervals 
Sorry I am getting easily confused. Essentially, if you were to work out the trivariate mean (or adjusted mean) for this model would this be the difference of both HDL and LDL from the intercept (i.e. -0.0394)? Therefore, if I was to show this in a table with the univariate model this would be: 

Univariate model: -0.2376
Trivariate model : -0.0394

I assumed that we must add the intercept to the slopes of HDL and LDL to get the overall adjusted mean? However, now having second thoughts based on what you said (and my textbooks) by adding the intercept ( i.e. -0.2376+0.1982+-0.0148) I am actually including the intercept again when it was already included? (silly me). 

Thank you for clarifying. Its the simple things that catch me!

Kind regards
From: Viechtbauer, Wolfgang (SP) <wolfgang.viechtbauer using maastrichtuniversity.nl>
Sent: 16 September 2018 14:30:01
To: Wasim Iqbal (UG)
Cc: r-sig-meta-analysis using r-project.org; Gavin Stewart; Chris Seal
Subject: RE: Inflated confidence intervals 
I apparently haven't had enough coffee today, so first a correction on my part:

predict(resMV, newmods = c(1,0)) gives the estimated (average) outcome for HDL (i.e., -0.2376 + 0.1982 = -0.0394). The coefficient for HDL (i.e., 0.1982) is already the difference between HDL and LDL.

But you seem to be after something different. Apparently, you want to add the intecept and the two coefficients together, so: -0.2376 + 0.1982 + -0.0148 =~ -0.0541, which indeed you would obtain with predict(resMV, newmods = c(1,1)). But what is the meaning of this?

If you want a marginal mean, that is, the average of the three outcomes, then you would want:

intercept + 1/3 * HDL + 1/3 * TC

(assuming the intercept corresponds to LDL, as in the output you showed), which you would get with predict(resMV, newmods = c(1/3,1/3)). But maybe I am still misunderstanding.


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