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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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    In previous comment, replace "skewness" with $E|X|^3$.2012-05-27

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