What is the point of calculating a standard error?

Say I simulate from a normal distribution $$N(\mu, \sigma^2)$$ 10.000 times using a couple of different methods.

I can calculate the standard error as $$s/\sqrt{10000}$$ where $$s$$ is the sample standard deviation.

What I don't understand is .... $$s$$ is the standard deviation of the sample. The sample are generated by simulating a normal distribution. Hence, by that very fact, the sample standard deviation of all simulations will be equal to $$\sigma$$.

So, for all my simulation methods, the standard error will be the same. It will be roughly $$\sigma /\sqrt{10000}$$. So what is the point of calculating it? It is already known before-hand and is constant for all my simulations, as long as I am simulating the same distribution?

So what is the point of calculating this number? Let me give you an example.

Say you write in R:

sample_1 <- rnorm(10000, mu, sigma)

u <- runif(10000)
sample_2 <- qnorm(u)

Now, both sample_1 and sample_2 are 10.000 simulations from a $$N(\mu, \sigma^2)$$ distribution... In either case, the standard error will roughly be $$\sigma/\sqrt{10.000}$$. This number seems pointless to me. It tells me nothing about the relative performance of $$sample_1$$ vs $$sample_2$$ ....the standard error is the same?

• In many cases you have the sample but never know $\mu$ or $\sigma$. You then use $\bar x$ to estimate $\mu$, and use $s / \sqrt{n}$ to estimate the magnitude of the possible error in making that estimate of $\mu$ – Henry Sep 22 at 10:21
• @Henry , I think that's the answer. You might convert that comment to an answer. – Sal Mangiafico Sep 22 at 12:48

In many cases you have the sample but never know $$\mu$$ or $$\sigma$$.
You then use $${\bar x}$$ to estimate $$\mu$$, and use $$s / \sqrt{n}$$ to estimate the magnitude of the possible error in making that estimate of $$\mu$$
• But why use $s$ when you know $\sigma$? – Dave Sep 22 at 14:55
• @Dave If you know $\mu$ then this is all irrelevant. If you know $\sigma$ but not $\mu$ (how?) then use $\bar x$ and then $\sigma / \sqrt{n}$ for the standard error. If you know neither $\mu$ nor $\sigma$ then use $\bar x$ and then $s / \sqrt{n}$ for the estimate of the standard error – Henry Sep 22 at 16:25
• Z-test has known $\sigma$ and unknown $\mu$. – Dave Sep 22 at 16:30