# Sampling distributions: CLT, t-distribution

Here is my understanding:

1. If you have any distribution, mean unknown but variance known, the sampling distribution of the mean is normally distributed if $$n$$ is large, and this normal distribution has the mean of the population the samples came from, and variance $$\frac{\sigma}{\sqrt{n}}$$ (where $$\sigma$$ is the variance of the population).
1. If you have a normal distribution but you do not know the mean and variance, the sampling distribution of the mean will be a t-distribution with $$n-1$$ degrees of freedom

My question is first, are the above points correct?

Second, what happens in the case where you are sampling from any distribution, not necessarily normal, and you do not know the mean or variance? Would the sampling distribution not follow a normal distribution $$N\sim (\mu,\frac{s}{\sqrt{n}})$$, where $$s=\frac{1}{n-1}\sum_{i=1}^{n}(x_i-\bar{x})$$?

• On the right track, but more precise statements are needed. See my answer. Dec 12, 2021 at 17:34
• @fmtcs The CLT is about the distribution of standardized sample means (or sums) in the limit as n goes to infinity, under certain conditions. If you have joint normality then the distribution of sample means is normal Dec 12, 2021 at 22:48

In your #1, you need to say that the sampling distribution of $$\bar X$$ is approximately normal.

How large a sample is large enough for a good approximation depends on the shape of the population distribution. For uniform data, $$n=20$$ works well; for an exponential population distribution, $$n=200$$ may work OK, but $$n = 20$$ is not enough.

Illustrative simulations in R:

set.seed(1212)
a.unif = replicate(10^5, mean(runif(20)))
summary(a.unif);  sd(a.unif)
Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
0.2223  0.4558  0.4997  0.4998  0.5439  0.7524
[1] 0.06458699

a.ex20 = replicate(10^5, mean(resp(20)))
summary(a.ex20);  sd(a.ex20)
Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
0.3244  0.8408  0.9826  0.9988  1.1390  2.2612
0.2224916

a.ex200 = replicate(10^5, mean(rexp(200)))
summary(a.ex200);  sd(a.ex200)
Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
0.7353  0.9515  0.9984  1.0000  1.0467  1.3386
[1] 0.07083102


The figure below shows that $$n = 20$$ is not large enough to give a nearly normal distribution for the average of exponential data.

R code for above figure:

par(mfrow=c(1,3))
hdr1 = "n=20, UNIF(0.1): Sample Means; Normal Fit"
hist(a.unif,  prob=T, br=30, col="skyblue2", main=hdr1)
curve(dnorm(x, mean(a.unif), sd(a.unif)), add=T, lwd=2, col="red")
hdr2 = "n=20, EXP(1): Sample Means; Normal Fit"
hist(a.ex20,  prob=T, br=30, col="skyblue2", main=hdr2)
curve(dnorm(x, mean(a.ex20), sd(a.ex20)), add=T, lwd=2, col="red")
hdr3 = "n=200, UNIF(0.1): Sample Means; Normal Fit"
hist(a.ex200,  prob=T, br=30, col="skyblue2", main=hdr1)
curve(dnorm(x, mean(a.ex200), sd(a.ex200)), add=T, lwd=2, col="red")
par(mfrow=c(1,1))


It is the same story for your last paragraph, but then also $$S/\sqrt{n}$$ is only an approximate standard deviation based on your sample of size $$n.$$

• Thank you for the answer, but I'm still a bit confused about my last paragraph. If we can use $s/ \sqrt{n}$ for the standard error in the case where we don't know anything about the original distribution, why then do we get a t-distribution if we know the original distribution is normal but don't know the parameters? In that case couldn't we just say the sampling means are distributed with mean $\mu$ and standard deviation $s/ \sqrt{n}$? Dec 12, 2021 at 20:40
• If data are normal, then $T = \frac{\bar X-\mu}{S/\sqrt{n}} \sim \mathsf{T}(\nu = n-1).$ But for normal data $\bar X$ and $S$ are stochastically independent, so if data aren't normal it isn't quite enough for $\bar X$ to be essentially normal, in order for the T-statistic to truly have a t dist'n. // Maybe it would help if you can explain your interests/purposes in your last paragraph, which seems to be some combination of wrong and confusing. Dec 12, 2021 at 22:01