T-distribution versus normal distribution (sample means and linear inference) From course notes, I see that when working with a quantitative variable, we can standardize the sample mean to have a normal distribution (as per the central limit theorem) as long as the sample size is "large". As a result, the distribution of the sample means is normally distribution (whether we are working with $\sigma$ or s) as long as the sample size is "large":
$$\frac{\overline{X} - \mu}{\frac{\sigma}{\sqrt{n}}} \sim N(0,1),$$
$$\frac{\overline{X} - \mu}{\frac{s}{\sqrt{n}}} \sim N(0,1),$$
If the sample size is "small", then we will have a t-distribution:
$$\frac{\overline{X} - \mu}{\frac{s}{\sqrt{n}}} \sim t_{n-1}.$$
However, we recently started looking at inference for linear regression, and I see the following two equations:
$$\frac{\hat{\mu}_{y|x} - {\mu}_{y|x}} {\sigma{\sqrt{\frac{1}{n}+\frac{(x-\overline{x})^2}{\sum_i(x_i-\overline{x})^2  }}}} \sim N(0,1),$$
$$\frac{\hat{\mu}_{y|x} - {\mu}_{y|x}} {s{\sqrt{\frac{1}{n}+\frac{(x-\overline{x})^2}{\sum_i(x_i-\overline{x})^2  }}}} \sim t_{n-2}.$$
I am wondering if the second equation can be normally distributed if its sample size is "large". In other words, if we have a large sample size, then can we still use the central limit theorem and show that:
$$\frac{\hat{\mu}_{y|x} - {\mu}_{y|x}} {s{\sqrt{\frac{1}{n}+\frac{(x-\overline{x})^2}{\sum_i(x_i-\overline{x})^2  }}}} \sim N(0,1).$$
The course notes make it seem as though when working with $\hat{\mu}_{y|x}$ (a sample mean) we cannot use the central limit theorem like we can for $\overline{x}$ (a sample mean). In the case of linear regression, it seems that only $\sigma$ and s determine whether we have a normal distribution or t-distribution respectively. 
Is this correct, and if so, why can't we apply the central limit theorem in the linear regression case? 
 A: The difference is 'type of test' and not 'sample size'
The difference between the two formula, $\sigma$ vs $s$, is not in the difference of sample size.
The difference is whether $\sigma$ is known or estimated. The first formula uses a normalization with a "known" standard deviation, and the second formula uses a normalization with the sample estimate of the standard deviation. The first, $\sigma$, is a constant, the second, $s$, is a random variable (with chi-square distribution).
So the difference is:

*

*you use the t-distribution $\mathcal{N}(0,1)/\sqrt{\chi_{n-1}/(n-1)}$ to describe the distribution of the difference between a 'sample mean' and 'the population mean', if this difference is normalized based on the sample estimate of the standard deviation

*and you use the normal distribution $\mathcal{N}(0,1)$ to describe the distribution of the difference between a 'sample mean' and 'the population mean', if this difference is normalized based on the standard deviation of the population.


Note:
for large sample sizes you do get that the distribution of this chi-square denominator becomes closer to a peak around 1
$$\lim_{n \to \infty} \sigma_{\left(\frac{\chi_{n-1}}{n-1}\right)} = \sqrt{\frac{2}{n-1}}= 0 \qquad \mathrm{and} \qquad \mu_{\left(\frac{\chi_{n-1}}{n-1}\right)} = 1 $$
or in other words the sample estimate of the standard deviation is less variable
$$ \lim_{n \to \infty} s = \sigma$$
and the t-distribution becomes approximately a normal distribution
$$ \lim_{n \to \infty} t_n = \mathcal{N}(0,1)$$
So you could say that: for large sample sizes the formula that are used with sample estimated standard deviation approximate the formula that are used with known standard deviation.
This is a different thing than the central limit theorem in which a mean of sample of variables from a non-normal distribution becomes a normal distribution for large $n$.

Note:
The standard deviation is often not 'really' known. But it can be 'hypothetically' known. For instance in testing a hypothesis or in Bayesian inference you 'assume' a certain deviation.
(in the same way as $\mu$ is not known but you can still use it in the formula and use it hypothetically, for instance in determining confidence intervals)
