let's assume that I have 3 subjects with each of them producing 5 values. Now, I can calculate the mean, variance, standard deviation, and standard error for each subject. Each subject has its own mean value. However, how can I combine the VAR, SD, and SE, if I want to say what the average variance, average standard deviation and average standard error for an "average" subject is while neglecting that they have all different mean values?


Subject 1 produces the values {79,75,77,80,74} thus with the mean=77.0, VAR=5.2, SD=2.28, SE=1.02

Subject 2 produces the values {83,79,81,84,84} thus with the mean=82.2, VAR=3.8, SD=1.94, SE=0.87

Subject 3 produces the values {81,76,80,77,79} thus with the mean=78.6, VAR=3.4, SD=1.85, SE=0.83

Thank you for your help!

Edit: What I need is the standard deviation and standard error which can be expected within one random single subject. I certainly do not want to compare the subjects against each other but simply say how data within one subject will usually vary.

I have seen different approaches: I could calculate the SD and SE for every subject separately and then determine the mean values of the SDs and SEs.

In another thread, a different approach is used: The mean variance among all subjects is calculated with the average SD being deduced from the mean variance. Cf. How to 'sum' a standard deviation?

Both lead to similar but different results and the answer in the other thread is disputed. And it also does not answer how I deduce the average SE then.

Is there a >correct< approach?


If I understand correctly this can be done with a mixed model, where the subjects are the random effect. In R:

> dat=data.frame(subject=rep(1:3,each=5),
> library(nlme)
> summary(lme(value~1,random=~1|subject,data=dat))

Linear mixed-effects model fit by REML
 Data: dat 
       AIC      BIC    logLik
  75.28225 77.19942 -34.64112

Random effects:
 Formula: ~1 | subject
        (Intercept) Residual
StdDev:    2.461707  2.27303

Fixed effects: value ~ 1 
               Value Std.Error DF  t-value p-value
(Intercept) 79.26667  1.537675 12 51.54968       0

Standardized Within-Group Residuals:
       Min         Q1        Med         Q3        Max 
-1.4650929 -0.8858919  0.1332504  0.8572516  1.1745552 

Number of Observations: 15
Number of Groups: 3
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  • $\begingroup$ Thank you. However, I do not see where I get the average within-group SD and SE. $\endgroup$ – Toby Jan 30 '19 at 13:31
  • $\begingroup$ The SE in your output seems to be across all three subjects. I have standard errors of 1.02, 0.87, and 0.83 for subject 1, 2, and 3, respectively. So if I want to say what standard error is to be expected for a random subject i would expect it to be somewhere around this value, logically. $\endgroup$ – Toby Jan 30 '19 at 13:37
  • $\begingroup$ @Toby In a mixed model we separate the "within-group" variance or residual into multiple groups. In this case we get a separate SD (not standard error) for both "within-group" variances: subjects (Intercept) and the remainder (the unexplained variance). $\endgroup$ – user2974951 Jan 30 '19 at 13:38
  • $\begingroup$ Thank you but I still don't get it. It does not seem to solve what I am intending on doing. What I want is: I have x subjects and every subject created y values. Based on that I can determine an individual SD and SE for each subject based on the variance among his or her y values. I do not want to compare the values across the subjects but uniquely within subjects. Therefore I do not care how the mean values between the subjects differ from each other. One subject can create the values {15031,15029,15032} while another can create {31,29,32} and they will have the same SD anyway. $\endgroup$ – Toby Jan 30 '19 at 15:04
  • $\begingroup$ What I want to be able to say is what SD and what SE within one subject in average can be expected. But I do not think I may calculate the SD for each subject and determine the mean value out of it, may I? $\endgroup$ – Toby Jan 30 '19 at 15:05

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