[R-meta] Estimate variance from time series data
Dear Wolfgang, Exactly, and there's the problem, because the correlations are never reported. So what do do in this case? Best wishes, Arne
On 15-Aug-18 14:26, Viechtbauer, Wolfgang (SP) wrote:
Dear Arne, In this example, there are 15 observations in total. The code I provided shows how to obtain the mean and standard deviation of these 15 observations. However, these 15 observations are not independent and hence any sampling variance you compute for the combined mean (or some function thereof) would need to take the degree of correlation into consideration. Best, Wolfgang -----Original Message----- From: Arne Janssen [mailto:arne.janssen at uva.nl] Sent: Wednesday, 15 August, 2018 14:10 To: Viechtbauer, Wolfgang (SP); r-sig-meta-analysis at r-project.org Subject: Re: [R-meta] Estimate variance from time series data Dear Wolfgang, Thanks for your quick reply. The question really is what is the sample size. Suppose there are 3 time series of 5 data points through time each that I want to combine. Given are the average and s.d. of the 3 series per time (so 5 averages and s.d.). I would like to obtain an overall average and s.d. of these 3 time series. If we consider that N = 15, I can use the standard method to combine the 3 series. If we consider N to be 3, because there are only 3 time series, I would indeed need to know the correlation among the time series to estimate the s.d., but this correlation is unknown. Please advise. Thanks and best wishes, Arne On 14-Aug-18 22:50, Viechtbauer, Wolfgang (SP) wrote:
Hi Arne,
It is not entirely clear to me what you are trying to do. Do you want to know the mean and SD when throwing together the N1 measurements from timepoint 1 and the N1 measurements from timepoint 2 from the same group, such that there are 2*N1 measurements in total now for the group? (or 3*N1 if there were three timepoints and so on).
Reply: This is indeed what I want to do.
Then the same equation could be used as if there are independent subgroups.
For example:
### Suppose we have the mean, SD, and size of several subgroups, but we
### need the mean and SD of the total/combined groups. Code below shows
### what we need to compute to obtain this.
### simulate some data
n.total<- 100
grp<- sample(1:4, size=n.total, replace=TRUE)
y<- rnorm(n.total, mean=grp, sd=2)
### means and SDs of the subgroups
ni<- c(by(y, grp, length))
mi<- c(by(y, grp, mean))
sdi<- c(by(y, grp, sd))
### want to get mean and SD of the total group
mean(y)
sd(y)
### mean = weighted mean (weights = group sizes)
m.total<- sum(ni*mi)/sum(ni)
### SD = sqrt((within-group sum-of-squares plus between-group sum-of-squares) / (n.total - 1))
sd.total<- sqrt((sum((ni-1) * sdi^2) + sum(ni*(mi - m.total)^2)) / (sum(ni) - 1))
Here is my doubt: The sum(ni) is now larger than the number of replicates (4 time series, so 4 replicates, n should be 4), am I correct?
### check that we get the right values
m.total
sd.total
This would be the case for independent subgroups. Now let's simulate data for 50 individuals measured twice:
library(MASS)
Y<- mvrnorm(50, mu=c(0,0), Sigma=matrix(c(1, .8, .8, 1), nrow=2))
y<- c(t(Y))
grp<- c(1:50, 1:50)
### means and SDs of the subgroups
ni<- c(by(y, grp, length))
mi<- c(by(y, grp, mean))
sdi<- c(by(y, grp, sd))
### want to get mean and SD of the total group
mean(y)
sd(y)
### mean = weighted mean (weights = group sizes)
m.total<- sum(ni*mi)/sum(ni)
### SD = sqrt((within-group sum-of-squares plus between-group sum-of-squares) / (n.total - 1))
sd.total<- sqrt((sum((ni-1) * sdi^2) + sum(ni*(mi - m.total)^2)) / (sum(ni) - 1))
### check that we get the right values
m.total
sd.total
Still works. However, when it comes to computing the sampling variance for m.total (or some function thereof), one cannot treat these two cases as the same. In the first case, we really have sum(ni) independent measurements, so var(y) / sum(ni) would be the correct sampling variance of m.total, but not so for the second case. You would need to know the correlation between the measurements over time to compute an appropriate sampling variance of m.total in the second case.
Best,
Wolfgang
-----Original Message-----
From: R-sig-meta-analysis [mailto:r-sig-meta-analysis-bounces at r-project.org] On Behalf Of Arne Janssen
Sent: Monday, 13 August, 2018 19:23
To: r-sig-meta-analysis at r-project.org
Subject: [R-meta] Estimate variance from time series data
Dear list members,
I am doing a meta-analysis with data that are often presented as
repeated measures of population densities, but authors sometimes also
give overall averages and s.d. or s.e.. Because I want to combine these
data into one analysis, I am interested in the overall effect size of
the repeated measures, so would like to combine all data of the time
series into one average and s.d. The time series are repeated several
times, yielding data of the following form:
Time Treatment 1 Treatment 2
N Ave s.d. N Ave s.d.
1 N1 x1,1 sd1,1 N2 x2,1 sd2,1
2 N1 x1,2 sd1,2 n2 x2,2 sd2,2
...
...
...
What I want to obtain is one average and s.d. per treatment through time.
The average is straightforward, but I cannot come up with a calculation
for the s.d.
The formula normally used for calculating the combined variance of two
series of measurements:
Var = (s1^2(n1 -- 1) + s2^2(n2 -- 1) + n1(X-x1)^22 + n2(X-x2)^22)/( (n1
+ n2 -- 1)
does not seem to apply when combining the measurements through time,
because this increases the number of replicates, which in my opinion,
should be the number of time series and not the number of observations.
I hope I made myself clear, and would be very grateful if you could
advise me on this matter.
Thanks very much in advance.
Arne Janssen
.