A $t$ - distribution's convergence to a Normal distribution

Question

I'm currently messing around with confidence intervals and I can't really understand how a $t$ distribution converges to a normal distribution for large $n$.

For example, suppose we want to construct a $95\%$ confidence interval when we have a sample mean $\bar{X} = 74.8$ and sample variance $S = 1.23$ with $n = 143$.

I would construct the confidence interval using,

$$(\bar{X} - z_{1 - \frac{\alpha}{2}} \frac{S} {\sqrt{n}},\bar{X} + z_{1 - \frac{\alpha}{2}} \frac{S} {\sqrt{n}})$$

since $n>30$. If $n<30$ I would have used a $t$-distribution.

My question is why does the $t$ - distribution approach a normal distribution for relatively large $n$?

Answer 1

The $t$ distribution arises because you estimate the population standard deviation $\sigma$ by the sample standard deviation $S$. For smaller $n$, there is some significant chance that $S$ is quite a bit smaller than $\sigma$; thus for fixed $c$ and $n$, there is significant probability that $\overline{X}$ is within $c \sigma$ of $\mu$ and not within $cS$ of $\mu$. (The reverse is possible too, but less likely.) But for large $n$, $S$ is essentially guaranteed to be very close to $\sigma$, because it is an asymptotically consistent estimator for $\sigma$. And of course, if $S$ is close to $\sigma$, $\overline{X}$ being within $c\sigma$ and being within $cS$ of $\mu$ are nearly equivalent.

Note that strictly speaking you should always use the $t$ distribution for confidence intervals from a normally distributed population with unknown standard deviation. It is just negligibly different from the same interval constructed with the normal distribution if $n$ is large enough. How large $n$ needs to be really depends on how small a difference can be treated as negligible.

Answer 2

Suppose $X_1,\ldots,X_n \sim \operatorname{i.i.d. N}(\mu,\sigma^2).$

Let $\overline X = \dfrac{X_1+\cdots+X_n} n.$

Let $S^2 = \dfrac 1 {n-1} \left( (X_1-\overline X)^2+\cdots+(X_n-\overline X)^2 \right). $

Then $ \dfrac{\bar X - \mu}{\sigma/\sqrt n} \sim N(0,1) $ and $\dfrac{\overline X - \mu}{S/\sqrt n} \sim t_{n-1}.$ The second one has $S$ where the first has $\sigma.$ If $n$ is large then the probability that $S$ is close to $\sigma$ is large, so these two random variables are nearly the same.