[R-meta] Random-effects meta-analyses with inverse-variance weights cannot include studies with sample sizes of two?

Viechtbauer, Wolfgang (SP) wolfg@ng@viechtb@uer @ending from m@@@trichtuniver@ity@nl
Fri Nov 9 14:33:08 CET 2018

I see. This assumes independence between trials and is typically used when measuring a dichotomous outcome on N different subjects. But here it seems like a single subject went through N different trials. In that case, the independence assumption seems tenuous.

How does this relate to the two subjects you talked about earlier? Or is X based on the sum of two subjects?

Also, I don't see how you then derived the equation for the variance of z/sqrt(N). You wrote:

Variance = Sqr(1/(1-q)*(1-1/(1-q))/N

But let's go through this step-by-step:

Leaving aside the independence issue, we assume (for large enough N):

p ~ N(π, π(1-π)/N)

Then we compute:

z = (p - π₀) / sqrt(π₀(1-π₀)/N)

Under H0: π = π₀, it is easy to see that z ~ N(0,1), so z/sqrt(N) ~ N(0, 1/N).

But if H0 is not true, then some simple algebra shows:

Var(z) = π(1-π) / (π₀(1-π₀))

and hence Var(z/sqrt(N)) = π(1-π) / (N * π₀(1-π₀)), which we could estimate with:

Var(z/sqrt(N)) = p(1-p) / (N * π₀(1-π₀)).


-----Original Message-----
From: Patrizio Tressoldi [mailto:patrizio.tressoldi using unipd.it] 
Sent: Friday, 09 November, 2018 13:01
To: Viechtbauer, Wolfgang (SP); r-sig-meta-analysis using r-project.org
Subject: Re: [R-meta] Random-effects meta-analyses with inverse-variance weights cannot include studies with sample sizes of two?

Il 09/11/2018 12:05, Viechtbauer, Wolfgang (SP) ha scritto:

Maybe I am a bit dense here, but I still do not fully understand what you are computing. You wrote earlier that there are two participants that "contributed to a percentage of hits, e.g. .30 and .40". Ok, that sounds like these participants did a series of trials that could yield hits/successes. I assume N is the number of trials. So the first participant had .3*N hits and the second participant had .4*N hits. So far so good. But what is 'z = binomial z'? Where does that equation for the variance come from? It would also help if you could provide a fully reproducible example of the computations.

This is the z score obtained from a normal approximation of the binomial test binomial test: z= (X - µ)/σ ; where X = observed percentage;  µ = chance percentage; σ = Sqr(µ(1-µ)/N

ex. X= 31; N= 49; p = .5 = (31/46 - .5)/Sqr(.5*.5/49) = 2.4

Patrizio E. Tressoldi Ph.D.
Dipartimento di Psicologia Generale
Università di Padova
via Venezia 8
35131 Padova - ITALY

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