Find average and sample deviation of the set of samples Let we have a sample of $N$ idependent identical normally distributed random variables with expectation $a$ and variance $\sigma$. Now we take $k$ samples of initial $N$ to $m$ elements in each. I need to find the average and sample deviation of averages in 2 cases: (a) sampling with replacement (b) sampling without replacement.
I try to write an example to clarify the statement.
Let we have 1000 people and measure their height. We know that their height is disributed normally with average 175 sm and dispersion 4 sm. Now we 25 times choose 20 people out of 1000. And in each such group of 20 people compute the average height. So we will have 25 numbers that indicate the average height in each group. And I need to calculate the average and variance of the average heights of 25 groups in two following cases:
a) Every time we choose 20 random distinct people of 1000 (it can be two same samples for example)
b) We do not choose people who we chose before, ie, after the first sampling, we will only select from the remaining 980, then 960, etc.
I have an idea that in the case (a) every subsample will have the same average as the whole sample. So in the case (a) all averages will be the same and we have the answer $a$ and dipersion $0$. Am I right?
And I absolutely do not know what to do in case (b).
Great thanks for the help!
 A: In (a) and (b) each of the group sample means of $20$ individuals without replacement will have the same distribution with an expectation of $175$cm, and a standard error of $\frac{4}{\sqrt{20}}  \approx 0.894$cm
The difference between (a) and (b) is that the sample means in (a) will be independent of each other while the sample means in (b) will be negatively dependent given the big sample.  This will have the possibly counter-intuitive effects of making

*

*the dispersion of the $25$ sample means probably being wider in (b) than in (a)

*the average of the $25$ sample means in (b) being more likely to be closer to the population mean of $175$cm than the average of the $25$ sample means in (a)

As a check on these statements, here are $100000$ simulations in R
N <- 1000
popmean <- 175
popsd <- 4
smalln <- 20
numbersmall <- 25
sims <- 10^5
set.seed(2021)

# Check initial ≈175cm and ≈0.894cm statement for small samples

meansmallfrombig <- function(N, popmean, popsd, smalln){
  bigsample <- rnorm(N, popmean, popsd)  
  smallsample <- sample(bigsample, smalln)
  mean(smallsample)
  }
simmeansmallfrombig <- replicate(sims, 
                         meansmallfrombig(N, popmean, popsd, smalln)) 
mean(simmeansmallfrombig)  
# 175.0025
sd(simmeansmallfrombig)   
# 0.8947107

as expected
## Check (a) and (b) statements

groupsfrombig <- function(N, popmean, popsd, smalln, numbersmall){
  bigsample <- rnorm(N, popmean, popsd)  
  asample <- replicate(numbersmall, sample(bigsample, smalln))
  bsample <- matrix(sample(bigsample, smalln*numbersmall), nrow=smalln)
  ameans <- colMeans(asample)
  bmeans <- colMeans(bsample)
  c(meanofa=mean(ameans), meanofb=mean(bmeans), 
    sdofa=sd(ameans), sdofb=sd(bmeans))
  }
simgroups <- replicate(sims, 
               groupsfrombig(N, popmean, popsd, smalln, numbersmall))

## standard deviation of (a) tends to be less than of (b) 
mean(simgroups["sdofa",] < simgroups["sdofb",])
# 0.51959
c(mean(simgroups["sdofa",]), mean(simgroups["sdofb",]))
# 0.8764460 0.8846087
c(sqrt(mean(simgroups["sdofa",]^2)), sqrt(mean(simgroups["sdofb",]^2)))
# 0.8858420 0.8939059

confirming the dispersion of the $25$ sample means is slightly more likely to be wider in (b) than in (a), while
## dispersion of mean of (a) tends to be more than of (b) 
mean(abs(simgroups["meanofa",] - popmean) > 
     abs(simgroups["meanofb",]- popmean))
# 0.56806
c(sqrt(mean((simgroups["meanofa",] - popmean)^2)), 
  sqrt(mean((simgroups["meanofb",] - popmean)^2)))
# 0.2170935 0.1784308
c(sd(simgroups["meanofa",]), sd(simgroups["meanofb",]))
# 0.2170935 0.1784306

confirming the average of the $25$ sample means in (b) being slightly more likely to be closer to the population mean than the average of the $25$ sample means in (a)
