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I generated random tests to calculate standard deviations of populations and samples using both n-1 and n for the denominator of the variance formula with datasets like [5,20,25... numbers between 5-50ish and population lengths were like 200, 2000 and sample lengths were like 50,500 etc.

The standard deviations of using n-1 and n in the sample denominator appeared to be so indifferent/similar in these cases that I am confused about why using n-1 is useful. I read many explanations about this already and the logic makes sense although my tests doesn't make this correction seem significant.

When is this useful?

ima_prog
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    Try it for a sample size of $2$. – Dave Nov 26 '21 at 01:01
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    Why simulate when you can easily calculate the ratio between the two estimates exactly? – jbowman Nov 26 '21 at 02:52
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    I guess mainly habit and custom: Consider a random sample of size $n$ from a population distributed $\mathsf{Norm}(\mu,\sigma).$ (1) If $S^2 =\frac{1}{n-1}\sum_{i=1}^n (X_i - \bar X)^2,$ then $E(S^2) = \sigma^2,$ the population variance. (2) $\frac{(n-1)S^2}{\sigma^2}\sim\mathsf{Chisq}(\nu=n-1).$ // Some authors _have_ used denominator $n$ and that does lead to a few minor simplifications. Unfortunately, also to major inconvenient inconsistencies with other texts, printed tables, and software programs. – BruceET Nov 26 '21 at 03:29
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    @BruceET - that looks like a good *answer* to me! – jbowman Nov 26 '21 at 03:40
  • @jbowman. Thanks for your comment. Thought about making it an answer. But then I'd feel compelled to add a few more paragraphs to tell more of the story and digesting Thanksgiving dinner is a higher priority right now. Also, there's [this](https://stats.stackexchange.com/questions/35123/whats-the-difference-between-variance-and-standard-deviation?), linked in the margin. – BruceET Nov 26 '21 at 03:44
  • The duplicate thread thorougly discusses this issue. – whuber Nov 26 '21 at 16:05

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