Intuition Standard Error of the Mean Trying to understand the intuition behind the standard error of the mean.
Starting from this formula:
$Var(\bar{X}) = Var(\frac{\bar{X}_1+\bar{X}_2+...+\bar{X}_n}{n})$
The formula talks about a variance, specifically of observations divided by the number of observations (which is their mean).
The fact is that the mean of a single sample is a single value.
Shouldn't his variance  ($Var(\bar{X})$), be a single value, therefore equal to 0?
 A: This is my intuition behind it. The standard error of the mean is
$$
SEM = \frac{s}{\sqrt{n}}
$$
with $n$ being the sample size and $s$ the standard deviation of the sample.
The sample you have is one of many possible samples you could have collected from the entire population (which is typically unavailable or unknown). The mean of this sample will differ from the mean of the population due to (at least) the chance involved in collecting it. Therefore, you may want to know how much discrepancy you can expect between the mean of the sample and the true population mean and this is what the SEM is telling you. Intuitively, the larger the sample the more accurate it will be in estimating the population mean so $n$ is at the denominator. In contrast, if the variation in the sample ($s$) is large we should expect a more noisy estimate so $s$ is at the numerator.
Here's an example in R which should be understandable even if you don't know R:
set.seed(1234)
pop <- rnorm(n= 10000, mean= 0, sd= 1) # Entire population 
pop[1:10] # First 10 values
# -1.21  0.28  1.08 -2.35  0.43  0.51 -0.57 -0.55 -0.56 -0.89 

Collect 50 samples, each of size 50, and calculate mean and SEM for each:
N <- 50
means <- rep(NA, N)
stderr <- rep(NA, N)
for(i in 1:N) {
    set.seed(i)
    smp <- sample(pop, size= 50)
    means[i] <- mean(smp)
    stderr[i] <- sd(smp) / sqrt(length(smp))
}

Plot them together with true population mean (blue line, unknown in real life):
plot(x= 1:N, y= means, pch= 19, ylim= c(-0.5, 0.5))
segments(x0= 1:N, x1= 1:N, y0= means-stderr, y1= means+stderr)
abline(h= mean(pop), lty= 'dashed', col= 'blue')


As you would expect, samples are different from each other and from the true population mean
If the original population has little variation, you can get a more accurate estimate of the mean for the same sample size. This is the same plot but using pop <- rnorm(n= 10000, mean= 0, sd= 0.5) (sd reduced from 1 to 0.5):

