Given independent and identically distributed (not necessarily normal) random variables $X_1, \dots, X_n$ with unknown mean $\mu$ and unknown variance $\sigma^2$, the  states that the sample mean (equiv. sum) approximately follows a normal distribution. In particular, we have that $$ \frac{\bar{X}-\mu}{\sigma/\sqrt{n}} \sim \mathcal{N}(0,1) $$ as the sample size $n$ gets large. Armed with this knowledge, we can create a level-$\alpha$ confidence interval for the mean: $$ \bar{X} \pm z_{1-\alpha/2}\frac{\sigma}{\sqrt{n}} $$ where $z_{1-\alpha/2}$ is the appropriate quantile of the standard normal distribution.  However, this interval depends on the population value $\sigma$ (equiv. $\sigma^2$); when we observe a sample, we don't know this value. What we do observe is the sample variance $S^2=\frac{1}{n-1}\sum_{i=1}^n(X_i-\bar{X})^2$, but my understanding is that $S^2$ doesn't follow a chi-square distribution unless the original $X_i$ came from a normal distribution (which is not part of our assumptions or else we don't need the CLT). So, we can't simply replace $\sigma$ with $s$ and $z_{1-\alpha/2}$ with $t_{1-\alpha/2;\ n-1}$ in our confidence interval.  My question is: how do we use the CLT to create a confidence interval if we don't know the population variance?  EDIT: In this question, one of the answers simply replaces $\sigma$ with $s$ and $z$ with $t$ like I mentioned above. Is this inaccurate?
Let $X$ be any distribution with defined mean, $\mu$, and standard deviation, $\sigma$.  The central limit theorem says that  $$ \sqrt{n}\frac{\bar{X} - \mu}{\sigma}  $$ converges in distribution to a standard normal distribution.  If we replace $\sigma$ by the sample standard deviation $S$, is there a theorem stating that  $$ \sqrt{n}\frac{\bar{X} - \mu}{S}  $$ converges in distribution to a t-distribution?  Since for large $n$ a t-distribution approaches a normal, the theorem, if it exists, may state that the limit is a standard normal distribution.  Hence, it would seem to me that t-distributions are not very useful - that they are useful only when $X$ is roughly a normal.  Is this the case?    If it is possible, would you indicate references that containing a proof of this CLT when $\sigma$ is replace by $S$?  Such a reference could preferably use measure theory concepts.  But anything would be great to me at this point.
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Solution of equation $\frac{x\cdot 2014^{\frac{1}{x}}+\frac{1}{x}\cdot 2014^x}{2} = 2014$ Solution of equation $\displaystyle \frac{x\cdot 2014^{\frac{1}{x}}+\frac{1}{x}\cdot 2014^x}{2} = 2014$   $\bf{My\; Try::}$ Clearly Here $x&gt;0$, Now Using $\bf{A.M\geq G.M}$  So Here $\displaystyle x\cdot 2014^{\frac{1}{x}}&gt;0$ and $\displaystyle \frac{1}{x}\cdot 2014^x&gt;0$ .  So $\displaystyle \left(\frac{x\cdot 2014^{\frac{1}{x}}+\frac{1}{x}\cdot 2014^x}{2}\right)\geq \left(x\cdot 2014^{\frac{1}{x}}\cdot \frac{1}{x}\cdot 2014^x\right)^{\frac{1}{2}} = \sqrt{2014^{\left(x+\frac{1}{x}\right)}} = 2014$   And equality Hold , When $\displaystyle x = \frac{1}{x}\Rightarrow x= 1$  Can we solve it  without Using $\bf{A.M\geq G.M}$  If Yes, Then please explain me.  Thanks