I have set of random values with the same distribution $y_1, \ldots, y_N$

$$T = \frac{1}{N}\sum_{j = 1}^{N} y_j$$

I want to find confidense interval for mathematical expectation $E(Y)$

I can use an approzimation $T∼N(μ,\frac{\sigma^2}{N})$ where $\sigma=\frac{\sum^N_{i=1}(y_i−T)^2}{N−1}$. This approximation is valid asymptotically. Then my confidence interval is $T±1.96 \frac{\sigma}{\sqrt{N}}$.

But how of course this method has a mistake, because I have only approximate Normal distribution and use sample variance.

How to get with this inaccuracy and make final confidence interval?

Thanks, Vasily

I have set of random values with the same distribution $y_1, \ldots, y_N$

$$T = \frac{1}{N}\sum_{j = 1}^{N} y_j$$

I want to find the confidence interval for the mathematical expectation $E(Y)$

I can use an approximation:

$T∼N(μ,\frac{\sigma^2}{N})$, where $\sigma=\frac{\sum^N_{i=1}(y_i−T)^2}{N−1}$. 

This approximation is valid asymptotically. Then my confidence interval is $T±1.96 \frac{\sigma}{\sqrt{N}}$.

Of course, this method has a mistake, because I only have an approximate Normal distribution and must use the sample variance.

How can I deal with this inaccuracy and make a final confidence interval?