I'm trying to calculate the upper percentage points for the 0.99 percentile for samples drawn from a normal distribution, with a sample size of 500. How can I calculate the expected values for skewness, kurtosis, studentized range, which would be expected for a drawing of 500 values?
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asked Apr 7, 2012 at 18:03
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$\begingroup$ When you say "calculate" do you want mathematical formulae, or a way of getting a numerical approximation? (And is this homework?) $\endgroup$– guestApr 7, 2012 at 20:46
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$\begingroup$ en.wikipedia.org/wiki/Student_t $\endgroup$– GschneiderApr 7, 2012 at 20:48
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$\begingroup$ no this is no homework :P. yes a formula would be nice, I'm somehow unsure how to integrate the sample size here. The objective is to know the approximative numerical value which can be expected, when 500 values are drawn.. $\endgroup$– PeriodicProgrammerApr 7, 2012 at 21:05
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$\begingroup$ This question appears to contradict itself: are you drawing samples from a normal distribution or a Student t distribution? Which one? (PS: You're right: this site is a better choice for the question than the Mathematica site. Once you get your question straightened out here, if you need specific help with MMA in coding a solution you will know precisely what to ask over there as a follow-up.) $\endgroup$– whuber ♦Apr 7, 2012 at 21:08
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$\begingroup$ I could do an Confidence interval but i'm somehow missing the standard deviation which would be needed to derive theorethically? I assume it to be close to zero for 500 values for the skewness, respectively to 3 for the kurtosis..? $\endgroup$– PeriodicProgrammerApr 7, 2012 at 21:27
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1 Answer
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You can always simulate this kind of thing first, e.g. in R
set.seed(1)
samat <- matrix(rnorm(1000000), ncol=500)
means <- rowMeans(samat)
varis <- rowMeans((samat-means)^2)
skews <- rowMeans((samat-means)^3) / varis^(3/2)
kurts <- rowMeans((samat-means)^4) / varis^2 - 3
and find the quantiles
> quantile(means, probs=c(0.01, 0.5, 0.99) )
1% 50% 99%
-0.096977914 -0.001302285 0.109797211
>
> quantile(varis, probs=c(0.01, 0.5, 0.99) )
1% 50% 99%
0.8494985 0.9970259 1.1495887
>
> quantile(skews, probs=c(0.01, 0.5, 0.99) )
1% 50% 99%
-0.257665499 -0.002052328 0.260948808
>
> quantile(kurts, probs=c(0.01, 0.5, 0.99) )
1% 50% 99%
-0.4286435 -0.0320487 0.5399296
and while it is clear that you can probably only treat the first couple of decimal places as useful, it also shows that skewness of 0.60 would be extremely high. It also suggests that the "skewness 0.129" you quote is rather low for the 99th percentile; I make it nearer 0.35 or 0.36 for a sample of 250 and about 0.26 for a sample of 500.
