If is true that SQRT(Variance) = SD then one cannot have n-1 in the denominator and the other not but should be same?
SD formula:
Variance formula:
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3$\begingroup$ These are two different concepts. The first formula refers to the sample standard deviation $S$, and you are correct that the sample variance is simply $S^2$. The second quantity refers to the population variance, which is more generally written as $$\sigma^2 = E\left[(X-\mu)^2\right].$$ An important distinction is that the population mean $\mu$ is unknown in the first case and must be estimated with $\bar X$, while in the second case $\mu$ is known. $\endgroup$– knrumseyCommented Nov 21, 2019 at 21:01
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$\begingroup$ @knrumsey-ReinstateMonica aha that is very interesting. and why is it -1 in sample case afterall and not same like in case of population? so population is the case if it is finite and whole set of numbers and sample if it is only part of it. $\endgroup$– lukyCommented Nov 21, 2019 at 21:07
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2$\begingroup$ The estimated mean is chosen such that it minimizes the mse. therefore using 1/n will lead to a biased estimate (meaning it will tend to be smaller than the actual population variance) $\endgroup$– SebastianCommented Nov 21, 2019 at 21:10
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$\begingroup$ @luky, there are several excellent answers to that question (in your comment) in both of the linked questions above. $\endgroup$– knrumseyCommented Nov 21, 2019 at 21:12
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3$\begingroup$ I think you should just check out those other posts that contain great explanations $\endgroup$– SebastianCommented Nov 21, 2019 at 21:26
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1 Answer
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Using $n-1$ as denominator instead of $n$ is called 'Bessel's correction' and it is used to reduce bias in the estimation of variance and standard deviation of the population.
If the sample variance is computed with $n$ as denominator, the sample variance is a biased estimator of the population variance.
As you correctly said, it is coherent to use the same denominator in computing the sample variance and the sample standard error.
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$\begingroup$ You mean -1 in sample and and N in the population. But if i have eg 1,2,3 POPULATION then it make sense that SD is 2/2 = 1 because 2 is mean, so SD 1 is kind of correct SD, but if i compute it like population not sample then i do 2/3 = 0.66 SD which is not so much correct. $\endgroup$– lukyCommented Nov 21, 2019 at 21:11
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1$\begingroup$ Look at the answer to this question: stats.stackexchange.com/questions/3931/…. I think it explains well what you have unclear $\endgroup$ Commented Nov 21, 2019 at 21:18
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$\begingroup$ If 1,2,3 is a sample you actually do not know what is the mean value of the population. For this reason you prefer to overestimate the population variance and standard deviation using $n-1$. $\endgroup$ Commented Nov 21, 2019 at 21:22