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I have set of random values with the same distribution $y_1, \ldots, y_N$ , $N = mN_1$. $ m \ge 4$,

$N_1$ is big enougth( $\approx 1000$ ).

I want to to estimeat $E(x)$.

How I do it:

I make $m$ groups of ($\{y_1,\ldots,y_{N_1}\},\{y_{N_1+1},\ldots,y_{2N_1}\},\ldots$).

$$x_i = \frac{1}{N_1} \sum_{j = iN_1+1}^{(i+1)N_1} y_j \approx N(\mu,\sigma^2).$$

Then I do the same thing as in Wikipedia

I'm really sorry for pasting hyperlink, but it is really difficult to print it all in TeX.

On wikipedia is said that $T$ has a Student's t-distribution. And as $$T = \frac{1}{mN_1}\sum_{j = 1}^{mN_1} y_j$$ $$ mN_1 = N$$ So $T$ can be an estimate for mathematical expectation. I can find confidence interval for my $\gamma = 0.95$ from Student's t-distribution.

But there is one thing. The article in Wiki is based on that $x_i \in N(\mu, \sigma^2)$ , but I of course have only $x_i \approx N(\mu, \sigma^2)$, and $x_i$ is closer to normal distribution with big $N_1$.

Now the main question. How can I make a correction to my confidence interval which occurs due to $x_i \approx N(\mu, \sigma^2)$.

This is a really important question for me and I'll be really happy if someone can advise me smth.

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    I think you are confused on a number of points. $T$ does not have a Students-t distribution because (1) you haven't normed it by an estimate of its standard deviation and (2) the underlying data aren't assumed to be normal. There is no need to divide your data into subsets; just use the approximation $T \sim N\left(\mu, \frac{\hat \sigma^2}{N}\right)$ where $\hat \sigma^2 = \frac{\sum (y_i - T)^2}{N - 1}$. This approximation is valid asymptotically. Then your confidence interval is $T \pm 1.96 \frac{\hat \sigma}{\sqrt N}$.2012-05-26
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    Yes. But it is only approximation. How can I calculate accuracy?2012-05-27
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    A straight-forward approximation how good the normal approximation is is given by the [Berry-Esseen theorem](http://en.wikipedia.org/wiki/Berry%E2%80%93Esseen_theorem). To apply this all you need is an estimate of the variance and skewness of the distribution. Essentially, the key is that the skewness is not too big relative to the variance after suitable scaling.2012-05-27
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    In previous comment, replace "skewness" with $E|X|^3$.2012-05-27

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