Talk:Bessel's correction

message: The text is written in first person and unnecessarily wordy. Not written like an encyclopedia entry.  — Preceding unsigned comment added by 2A02:908:1087:60:E0EC:9DBF:195B:76D4 (talk) 16:34, 11 November 2019 (UTC)
thread_title:  The Source of the Bias (Intuitive Explanation) 

message: Why is the formula with Bessel's correction used as a default (ie in Matlab) to estimate standard deviation, instead of the unbiased version (such as Unbiased_estimation_of_standard_deviation) to estimate standard deviation? Frankmeulenaar (talk) 06:43, 12 July 2011 (UTC)
thread_title:  Bessel usage for standard deviation 

message: In the final formula:
:
\frac{1}{n-1}\left[\sum_{i=1}^n \sigma^2 - n(\sigma^2/n)\right]
Isn't:
:
n(\sigma^2/n) = \sigma^2 ?
And therefore the sum is zero?
Or is that factor outside the summation?
Even if it's outside, why not just put \sigma^2 ?  — Preceding unsigned comment added by 71.107.55.85 (talk) 08:12, 28 June 2012 (UTC)
replies: text: The first term reduces to n\sigma^2 while second is \sigma^2, so the result is not zero. The formula is presumably there because it relates to a formula above, with corresponding terms. Melcombe (talk) 12:46, 28 June 2012 (UTC)
thread_title:  Isn't n*(a/n) = a ? 

message: In the final section, when dicussing a particular random sampling, the article throws out the formula:
:\operatorname{Var}(\overline{x}) = \operatorname{E}((\overline{x}-\mu)^2) = \sigma^2/n. \, 
However, in the context of a single random sampling of the data, this formula doesn't make sense. It seems to want to measure the variance of the sample mean with respect to the real mean. However, in the case of a single random sampling, you just have one sample mean, and one number. Therefore the expected value is just that number.
Or is this formula stating the variance you'd get if you took a number of random samplings, took the sample mean of each of those samplings, and then calculated the variance of the the sample means of several random samplings to the true mean?
If the second is true (and it's the only explanation that makes sense to me), this formula has been thrown a bit of context, given the text above it, that dicusses a single random sampling, and then says "Also", then seems to throw the formula up on the wall, and see if it sticks...
Furthermore, it's unexplained nature makes it very difficult to make the leap to the final formula below it, which essentially justifies the results of the whole page.  — Preceding unsigned comment added by 71.107.55.85 (talk) 08:31, 28 June 2012 (UTC)
replies: text: This relates to the question above. The formula complained of is set down because it is the thing that is substituted in the earlier formula to get the final result. Melcombe (talk) 12:50, 28 June 2012 (UTC)
thread_title:  Formula doesn't make sense in context 

message: Do I understand that the "real" answer in the example would be 36/5 = 7.2, the observed answer is 16/5 = 3.2 and using Bessel's correction would give: 16/(5-1) = 4? Four being closer to the correct answer of 7.2 than 3.2 which is considered a low value per the article text? Thank you. Fotoguzzi (talk) 11:55, 1 February 2013 (UTC)
thread_title:  Putting the -1 in n-1. 

message: I had need to read this article. For me it read well overall. I found the article well set out in its flow of explanation for a person yet to understand this topic. Better than a number of other basic stats articles. Providing this feedback to those who have worked on this article. With my thanks. 
CitizenofEarth001 (talk) 10:15, 17 February 2014 (UTC)
thread_title:  Comment as a very occasional editor.  

message: The intuitive explanation in proof alternate #3 is very clear for even a lay person to understand, I think this makes it an ideal candidate for being in the main text. MATThematical (talk) 05:00, 26 May 2014 (UTC)
thread_title:  Proof Alternate #3 moved to #1 with intuition unhidden 

message: jump in the function? 1/n-1 is chosen arbitrarily to be different from 1/n (the whole/full set), it could be defined as 1/n+1, but it is always better to estimate the larger error rather than the smaller.  — Preceding unsigned comment added by 82.140.130.86 (talk) 10:58, 27 January 2019 (UTC)
thread_title:  The answer is yes except when the sample mean happens to be the same as the population mean. 

message: :s_n^2 = \frac {1}{n} \sum_{i=1}^n  \left(x_i - \overline{x} \right)^ 2 = \frac{\sum_{i=1}^n \left(x_i^2\right)}{n} - \frac{\left(\sum_{i=1}^n x_i\right)^2}{n^2}
That seems false to me :
:(a-b)^2 \neq a^2 - b^2
Perhaps would it be nice to include some more steps to better understand ?  — Preceding unsigned comment added by Neo 13 (talk • contribs) 18:31, 8 January 2020 (UTC)
thread_title:  Identity 

message: More concisely: Let {\rm Var\,} X = \sigma^2. Then, since the uncorrected variance is s_n^2=\overline{x^2}-(\overline x)^2 we have that 
{\rm E\,}s_n^2 = {\rm E\,}X^2 - \big({\rm Var\,\overline X}+({\rm E\,}\overline{X})^2\big).
But \rm E\,\overline X = \rm E\, X\,, \rm Var\,\overline X = \sigma^2/n\, and \,{\rm E\,}X^2 - ({\rm E\,}X)^2 = \sigma^2, so
{\rm E\,}s_n^2 = \,\frac{n-1}n\,\sigma^2\,.
Therefore we have to multiply s_n^2 by a factor \frac n{n-1} in order to get an unbiased estimation, which yields the usual variance s^2. José Luis Arregui (talk) 21:40, 2 May 2023 (UTC)
thread_title:  Proof of correctness 

message: I believe alternate proof #3 substitutes an incorrect expression for the population standard deviation σ. It sums over on the same number of elements as the sample but should sum over all the elements in the population (not the same as n) Antoine1619 (talk) 06:50, 12 December 2023 (UTC)
thread_title:  Alternate proof #3 is dubious 

message: The proof of correctness alternative #3 was better accessible for those with highschool backgrounds, and I hence request restoring it to before its deletion on 21 Jan 2024 please. CalebSohSweeKai (talk) 06:01, 8 February 2024 (UTC)
thread_title:  Reversion request: Restore alternative proof #3 present before 21 Jan 2024