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If we compute the standard deviation of a data set composed of a single feature and then compute its confidence interval, then can we say that these computations have considered the stochastic variability of the data? What exactly is stochastic variability of the data?

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    $\begingroup$ Yes, indeed: exactly what do you mean by "stochastic variability"? This is not a standard term of statistics, so your definition or a clear reference with a definition would be helpful for understanding this question. $\endgroup$ – whuber Dec 24 '14 at 1:11
  • $\begingroup$ @whuber I believe (though not sure) that the randomness in the observations is stochastic variability. So putting this into context, the computation of standard deviation and its confidence interval will consider the stochastic variability of the data. Am I correct? What is the standard term in statistic for "stochastic variability"? $\endgroup$ – mani Dec 24 '14 at 1:20
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    $\begingroup$ Are you attempting to answer a question written by someone else? $\endgroup$ – Glen_b Dec 24 '14 at 6:42
  • $\begingroup$ @Glen_b Of course not Glen_b. I was explaining what I meant with stochastic variability. Since whuber said this is not a standard term of statistics so I asked him about the standard term for this concept. $\endgroup$ – mani Dec 25 '14 at 2:13
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    $\begingroup$ Okay, it's just that if the question is your own it would be odd to use a term you intend to mean something ("can we say that these computations have considered the stochastic variability") and then ask us to define the term you chose, as your question then does. That phrasing in the question made me wonder if it was an exercise. $\endgroup$ – Glen_b Dec 25 '14 at 2:16
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I presume the intent of 'stochastic variability' is simply 'random variation', and by "its confidence interval" you mean a confidence interval for $\sigma$ (since that's the only parameter you mention and you don't form a confidence interval for data).

If I understood the intent of the question correctly, then the standard deviation not only considers the variability, it's a measure of "how big" it is.

A suitable confidence interval for the standard deviation would be based in some direct sense on how variable that data is.

For example, if we assume normal random variables with constant mean and variance and mutual independence, then an interval for the standard deviation could be obtained by taking the square roots of the usual limits of the interval for the variance:

$\left[\frac{(n-1)s^2}{Q(1-\alpha/2)},\frac{(n-1)s^2}{Q(\alpha/2)}\right]$

where $Q(p)$ is the $p$th quantile of a $\chi^2_{(n-1)}$

As you see, the variance-interval itself depends on the sample variance.

Similarly a $1-\alpha$ interval for $\sigma$ would be (after taking out the common factor of $s$:

$\qquad s\left[\sqrt{\frac{(n-1)}{Q(1-\alpha/2)}},\sqrt{\frac{(n-1)}{Q(\alpha/2)}}\right]$

which clearly depends on $s$, which as mentioned before is a measure of the amount of random variability in the sample and an estimate of the amount of random variability in the population.

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  • $\begingroup$ hmm I see. good detailed answer. I have also seen the use of Chi-squared distribution for computation of confidence interval of standard deviation. What is its role exactly? Not sure, I need to put a separate question for that... $\endgroup$ – mani Dec 25 '14 at 9:24
  • $\begingroup$ There are several questions and answers here that relate to such a question; you might start best with a search using keywords in your comment there. $\endgroup$ – Glen_b Dec 25 '14 at 9:38
  • $\begingroup$ However, briefly -- it relates to the fact that a sum of squares of independent standard normal variates is chi-squared, and that in turn is why distributions of normal variances are related to the chi-squared. $\endgroup$ – Glen_b Dec 25 '14 at 9:43

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